Heat kernels on Euclidean complexes

HEA T KERNELS ON EUCLIDEAN COMPLEXES A Dissertat ion Presen ted to the F aculty of the Gradua te Schoo l of Cor nell Universit y in Partial F ulﬁllm en t of the Requirem en ts for the Degr ee of Do cto r of Phi losophy b y Melanie Anne Piv a rski August 2006 c  2006 Melanie Ann e Piv arski ALL RIGHTS RESER VED HEA T KERNELS ON EUCLI DEAN COMPLEXES Melanie Anne Piv a rski, Ph.D. Cornell Univ ersity 2006 In this thesis w e describ e a t ype of metric space called an Euclidean p olyhedral complex. W e deﬁne a Diric hlet for m on it; this is us ed to giv e a corresp onding heat k ernel. W e provide a uniform small time P oincar´ e inequalit y for complexes with b ounded geometry and use this to determine unifo rm small time heat k ernel b ounds via a theorem of Sturm. W e then consider suc h complexes with an underlying ﬁnitely generated group structure. W e use tec hniques of Saloﬀ-Coste and Pittet to sho w a large time asymptotic equiv alence for the heat k ernel on the complex a nd the heat k ernel on the group. BIOGRAPH ICAL SKETCH Melanie Piv arski w as b orn on August 13, 1977 somewhere in the o utskirts of Pitts- burgh to Lynn and Thomas Piv arski. She grew up with her paren ts and t w o sisters, Kara and Janelle; her grandmother, Caroline Matovic k, lived a few blo c ks a w ay . They can all attest to the fact that y es, Melanie has alw a ys talk ed with her hands. She att ended Colfax Elemen tary Sc ho ol where she greatly enjo y ed M s. Kengor’s math classes; partly , this w as b ecause she could sp end time doing logic puzzles. She w as also in v olv ed in Girl Scouts with Chriss y . She then attended Springdale High Sc ho ol where she participated in many diﬀeren t activities , most notably art classes, Drama Club with Ms . F rauenholz, and Academic Games. F rom 1 995 through 1999, she attended Carnegie Mellon Unive rsit y , where she ma jored in mathematics and minored in computer science. While there, she to ok Prof. Mizel’s freshman analysis course out of Ap ostle. This class con vince d her that she needed to study mat hematics. In her analysis class she met Helen a and Ruth who b ecame her go o d friends and study partners. Computers ar e a part of the culture at CMU, and so s he found herself in a num b er of fun computer scie nce classes. Though she sp en t m uch of her time on math and computers, she f ound time to ta k e some ballet classes with her friend Rob ert and some intro ductory P o lish classes at t he Univ ersit y of Pittsburgh. Quite signiﬁcan tly , she met Jim McCann (now Jim McCann Piv arski) during freshman orientation. They b egan dating that fall and w ere married in June of 1999. Melanie a nd Jim mo ved to Ithaca in the summe r of 199 9, where they b ecame graduate studen ts in math and ph ysics r espective ly . Melanie has greatly enjo yed her time in the math departmen t; w hile there she w as in v olve d in the teach ing seminars, the outreach pro gram Expanding Y our Horizons, the w omen in math iii p otluc ks, the 120A Cafe, and the Gingerbread House con tests, sp ecializing in tow er constructions. She also studied some math. Outside the math departmen t, Melanie found many things to do. She’s tak en ballet classe s and sang in the c hoir at St. Catherine’s for muc h of her time here. She’s also b een in v olved in the St Catherine’s y oung adult group, the PreCana team, and sp ent a few y ears helping out at Loa v es and Fishes. Through all of this, she’s met a nu m b er of intere sting p eople and had a v ariet y of exp eriences. She considers hers elf to b e more mature than she’s ev er b een before, and she hop es to con t in ue gro wing and learning throughout her lif e. In the fall of 2006, Melanie and Jim will mov e to College Station where they will b e emplo ye d as p o stdo cs at T exas A & M Univ ersit y . iv T o F riendship! And most esp ecially to Jim! v A CKNO WLEDGEMENTS The mat h department at Cornell is full of w onderful p eople. The comm unit y here is splendiferous. I’m v ery grateful for the y ears spent here, the math learned, and esp ecially the friendships. Most imp ortantly , I ’d lik e to thank my a dvisor, Prof. Sa loﬀ-Coste, who g a v e me an in teresting problem to w ork on. He’s exp osed me to loads of co ol mathematics, and he has b een essen tial in m y learning some of it. He a lso has a nearly inﬁnite amoun t of patience , whic h comes in quite handy . I’d also lik e to thank m y committee m em b ers, Prof. Gross and Prof. Stric hartz, who also help ed me to dev elop mathematically through b oth courses and con v er- sations. Thanks go to Prof . F ulling as w ell, who found a mis tak e in an early draf t of this thesis . My mathematical siblings, Da vid Rev elle, Lee Gibson, Sharad Go el, Guan- Y u Chen, Evgueni Klebano v, P a ve l G yra, and Jessica Zuniga, are all great p eople. They’v e b een v ery encouraging and quite helpful to me in clarifying my though ts and deﬁnitions. I’v e lear ned a lot in our gro up mee tings! Thanks go to T o dd Kem p and T reve n W all, w ho helped me with v arious anal- ysis bits, Jim Belk, who gav e me a crash course in a lgebraic top o logy one summer, Kristin Camenga who help ed me with t he geometric deﬁnitions, a nd Mik e Koz- dron, who show ed me to v ar ious latex commands. I’d also lik e to thank Josh Bo wman, Jonathan Nee dleman, Rob yn Miller, Mia Minnes, and Brigitta V ermesi who, along with many of the other folk men tio ned ab ov e, help ed me to orga nize m y thoughts in to some kind of presen table form through v ario us con ve rsations. Maria Belk and Maria T errell should also b e thank ed for their e ncouragemen t and general go o d advice on how t o be a graduate studen t. vi T ABLE OF CONTE NTS 1 Setup for t he Complexes 1 1.1 In t ro duction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Geometry of the Comple xes . . . . . . . . . . . . . . . . . . . . . . 4 1.3 Analysis on the C omplexes . . . . . . . . . . . . . . . . . . . . . . . 8 1.3.1 The Diric hlet F orm . . . . . . . . . . . . . . . . . . . . . . . 8 1.3.2 The Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . 12 2 Lo cal P oincar ´ e Inequalities on X 14 2.1 W eak P oincar´ e Inequalities . . . . . . . . . . . . . . . . . . . . . . . 15 2.2 Whitney Co vers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 3 Small Time H eat Kernel Estimates for X 42 3.1 Small time Heat Kernel Asy mptotics . . . . . . . . . . . . . . . . . 42 3.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 4 Setup for Gr oups 64 4.1 Comparing distances in X and G . . . . . . . . . . . . . . . . . . . 67 4.2 Comparing functions on X with corresp onding ones on G . . . . . . 72 4.3 P o incar ´ e inequalities on X with unde rlying group structure . . . . . 76 4.4 Mapping functions on G to X . . . . . . . . . . . . . . . . . . . . . 79 4.5 P o incar ´ e inequalit y for v olume doubling ﬁnitely generated groups . 84 5 Comparing Heat Ker nels on X and G 87 5.1 Heat k ernels in the nonamenable case . . . . . . . . . . . . . . . . . 88 5.2 Heat Kernels in the Amenable Case . . . . . . . . . . . . . . . . . . 93 vii 5.2.1 Bounding those on G abov e b y those on X . . . . . . . . . . 94 5.2.2 Bounding those on X abov e b y those on G . . . . . . . . . . 97 Bibliograph y 108 viii LIST OF FIGURES 1.1 Example of a 2 dimensional Euclide an C omplex (left), its 1-sk eleton (cen ter), one possible triangula tion (righ t). . . . . . . . . . . . . . . 1 1.2 Examples of a complex whic h is not dimensionally ho mogeneous (left), one whic h is not 1-chainable (cen ter), and one whic h is ad- missible (righ t). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 2.1 Complex with shaded ball B ( left); the three w edges for B (righ t). . 19 2.2 Complex with shaded ball B (left); the t w o wed ges for B and a region whic h o v erlaps b oth of them(righ t). . . . . . . . . . . . . . . 20 3.1 Example of a s tar; here n=8. . . . . . . . . . . . . . . . . . . . . . 48 3.2 Example of t wo joined stars with la b els; here n=7 and m=3 . . . . . 54 3.3 A subset of the three dimensional grid. . . . . . . . . . . . . . . . . 6 1 3.4 Y (left); Y (dark er shading) and its four surrounding copies (ligh ter shading) (right) Eac h of the edges in these pictures should b e in- terpreted as ha ving length 1. . . . . . . . . . . . . . . . . . . . . . 6 3 4.1 X split into copies of Y(left);a cop y of Y co v ered b y 5 balls of ra dius .6 (ce n t er); a co v er for Y shifted by ﬁv e copies of G co ve r ev erything – the blac k lines represen t four copies that o v erlap exactly; the ﬁfth cop y is gra y(right). . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 ix LIST OF SYMBOLS • X the complex • X ( k ) the k -sk eleton of X • γ a path in X • L ( γ ) length of γ • d X ( · , · ) dis tance in X induced by the Euclidean metric • d X ( i ) ( · , · ), d i ( · , · ) dis tance in X ( i ) induced b y the E uclidean metric • µ measure on X • E ( · , · ) energy form constructed via Γ-limit • E ( · , · ) ene rgy form constructed via gradien ts • ∆, ∆ k Laplacian for X , X ( k ) • ∆ Ω Laplacian for Ω ⊂ X with Dir ic hlet boundary condition • ∇ f , F g radien t of f • dµ ( x ), dx equiv alent w ays of writing the diﬀere n t ial form • Lip( X ) the set of Lipsc hitz functions on X • W 1 ,p ( X ) the Sob olev space on X of functions in L p ( X ) with ﬁrst deriv ativ es in L p • C Lip 0 ( X ) the set of compactly supp orted Lipsc hitz functions on X • − R B a v erage in tegral ov er the set B • I Ω indicator function on Ω (1 if in Ω, 0 if not) • f E a v erage v alue of f on the set E • ∂ E the b oundary of the se t E • S ( k ) the k -sphere • W k , W j,k w edges of a ball in X • N ( j ) the list of indices of faces adjacen t to W j including j • M degree b ound on X • ℓ lo we r b ound on edge lengths of X x • α smallest in terior a ngle in X • κ constan t related to m ultiples of radii • R 0 b ound on radius deﬁned in terms of κ • C W eak constan t in w eak P oincar ´ e inequalit y • C P constan t f or the P oincar ´ e inequality • F collection of balls in the Whitney co v er • r B radius of the ball B • B z the cen tral ball in the Whitney co v er • F ( B ) a string of balls that tak es B z to B • h t ( x, y ), h k t ( x, y ) heat k ernel on X , X ( k ) • H t heat semigroup on X • τ exit time for Brow nian motion on a s et • X t random v ariable for lo cation of a Bro wnian motion in a subset of X • h Ω t ( x, y ) heat k ernel on Ω ⊂ X • H Ω t heat semigroup on Ω ⊂ X • p t ( x, y ), p A t ( x, y ) heat k ernel on G , A ⊂ G • K n , K n A n step transition op erator on G , A ⊂ G • λ Ω ( i ) ith eigen v alue for H Ω 1 • β A ( i ) ith eigen v alue for K 1 A • T r ( K n ) trace of K n • G group • S generating set for G • Y compact subse t of X with the prop erty X/ Y = G • | g | w ord length of g ∈ G • d G ( · , · ) dis tance in G with respect to word le ngth • d Y ( · , · ) distance in Y based o n Euc lidean paths in Y . xi • diam( Y ) diameter of Y with res p ect to d Y • Y ( i ) the i sk eleton of Y ; X ( i ) ∩ Y • | A | , # A n um b er of elemen ts in A • V ol, V ol X , V ol G v o lume (with respect to X , G ) • B r , B X ( r ), B G ( r ) ball of radius r (in X , G ) • B X ( x, r ), B G ( g , r ) ba ll in X ( G ) with radius r cente red at x ( g ) • E ( f , f ) energy form; for G this is 1 | S | P g ∈ G P s ∈ S | f ( g ) − f ( g s ) | 2 , fo r X this is −h∇ f , ∇ f i • |∇ f ( x ) | length of gradien t; on G this is q 1 | S | P s ∈ S | f ( x ) − f ( xs ) | 2 . • || f || p,A the L p norm restricted to a subset A ; for G this is ( P x ∈ A | f ( x ) | p ) 1 /p , for X this is ( R A | f ( x ) | p dx ) 1 /p • { γ i } N i =1 cen ters of the ba lls of radius δ cov ering Y • { g γ i } i =1 ..N ; g ∈ G cen ters of the ba lls of radius δ cov ering X • C N ,S maxim um n um b er of balls o v erlapping a p oint in X • group f a new function from ( G, N ) → R deﬁned to b e − R B X ( gγ i ,δ ) f ( x ) dx • C X G , C 0 constan ts use d to compare metrics in G and X • C H constan t for the Harnac k inequalit y • χ , χ g a smo oth function on X , χ translated b y g • C sup constan t used to bound the supp ort of χ • C O ve r constan t b ound on the maxim um num b er of χ g supp orted at any p oin t in X • C g b ound on the magnitude of the gradien t of χ • comp f ( x ) a new function from X → R deﬁned b y P g ∈ G f ( g ) χ g ( x ) • C g r ad constan t b ound comparing norms of gradien ts • ⌊ x ⌋ the largest in teger less than or equal to x • ⌈ x ⌉ the sm allest in teger greater than or equal to x • C 0 (Ω) con tinuous funcions whic h ar e compactly supported in Ω • U ( A ) subset of X dep ending on A xii Chapter 1 Setup for the C o mp l exes 1.1 In tro duction W e will study ho w local and global geometries aﬀect heat k ernels on a set of metric spaces called Euclidean p olyhedral complexes. Euclidean complexes are formed by taking a collection of n dimensional con v ex p olytop es and joining them along n − 1 dimensional faces. Within eac h p olytop e, w e will hav e the same metric structure as R n . When w e join them, w e will glue the faces of t w o p olytop es together so that p oints on one face are iden tiﬁed with p oin ts on the other face, and the metrics on those fa ces are preserv ed. W e will require that these structures ha v e a countable num b er of p olytop es, ar e lo cally ﬁnite, and ha v e lo w er b ound on the interior ang les a nd edge len gths. The complex formed b y lo oking at k dimensional faces is called the k -sk eleton. F or instance, the 0 -sk eleton is set of vertice s. A 1-sk eleton is a graph where the space includes b oth v ertices and p oints on t he edges; sometimes this is called a metric graph [24 ]. Note that w e can triangulate an y con ve x p olytop e to obta in a collection of simplices, and so this structure is ess en tially equiv alen t to looking at a simplicial complex. Figure 1.1: Example of a 2 dimensional Euclidean Complex (left), its 1-sk eleton (cen ter), one possible triangula tion (righ t). Let h k t ( x, y ) b e the heat kerne l on the k -sk eleton. This is t he fundamen tal 1 2 solution to the heat equation ∂ t u − ∆ u = 0 on the k -sk eleton. It can b e used to describe the probability that we tra v el from x to y in time t when our mo v emen t is restricted to the k -sk eleton. Theorem 1.1.1. F or X a uniform ly lo c al ly ﬁni te Euclide an c omplex of dimensio n n whose interior angles and e dge leng ths ar e b ounde d b elo w . Fix T ∈ (0 , ∞ ) . Ther e exist c , C ∈ (0 , ∞ ) such that for any x ∈ X a n d t < T we have: c t k / 2 ≤ h k t ( x, x ) ≤ C t k / 2 . Note that this claims that the heat k ernel on the k -sk eleton b ehav es, up to a constan t that is indep endent of where in X we are, lik e the heat k ernel on R k asymptotically when t → 0. The lo cal b eha vior reﬂects the lo cal geometry a nd structure of our space . Theorems in Sturm [30] can b e applied to Euclidean complexes to show that on an y compact subset of X ( k ) , the heat k ernel is lo cally lik e the one on R k , with constan ts that dep end on the choice of compact subset. The essen t ial diﬀerence in our theorem is that the constants are uniform throughout the en tire complex. An interes ting example of these comple xes comes from biology . In a pap er b y Billera, Holmes, and V o gtmann [3] they describ e a wa y of classifying distances b et w een ph ylogenetic trees, whic h are trees that describ e ev olutio n of sp ecies. O ne can fo rm an Euclidean complex, where each of the faces corresp onds to a diﬀeren t tree, and one mov es through the p oints in the face b y ch anging the edge lengths in the tree. One can then consider probabilit y distributions on this space to determine lik ely g enetic ancestry . Euclidean complexes a re also examples of f ractal “blow -ups”, whic h a re inﬁnite fractals that are lo cally nice but globally hav e a structure with rep etition. See 3 Kigami [22] for a description of these fractals. In this setting, our small time asymptotic estimates apply . Note that these examples need not b e compact. In [1], Barlow and K umagai studied the small time asymptotic of heat kerne ls for compact self-similar sets. Another collection of example s can b e found b y considering metric spaces , X , whic h are acted upon b y a ﬁnitely generated group, G of isometries. When w e tak e the space and mo d out by that g roup, w e obtain a compact set Y = X/G . When Y can b e expresse d as a ﬁnite Euclidean complex, then X is an Euclidean complex as w ell. Note that the k -sk eleton of Y will b e the ( k - sk eleton of X ) /G . A sim ple example of this is X = R 2 , Y = the unit square, and G = Z 2 . A more in teresting example o ccurs when G is the free gro up; t here the space is globally hy p erb olic, but lo cally Euclidean. With t his added g roup structure, w e can describ e the larg e time b eha vior of the heat ke rnel. W e write the heat k ernel on a gro up as p t ( · , · ). Prop osition 1.1.2. L et X b e a lo c al ly ﬁ n ite c ountable Euclide an c omplex of di- mension n and let G b e ﬁnitely gener a te d gr oup G . If X/G is a c omplex c om prise d of a ﬁnite numb er of p olytop es with Euclide an metric, we have: p t ( x, x ) ≃ h t ( x, x ) as t → ∞ . Our main result sa ys that, up to a constan t, the heat k ernel will b eha ve the same asymptotically as t → ∞ on b oth the gr oup and the complex. By transitivity , it will b eha v e the same asymptotically regardless o f which k -sk eleton we consider. This theorem relates to a pap er of Pittet and Salo ﬀ-Coste [26]. They show that a manifold M whic h has a ﬁnitely generated group of isometries G satisﬁes sup x h M t ( x, x ) ≃ h G t ( e, e ) for large t . In chapter one, w e desc rib e our set-up. W e pro vide deﬁnitions for the complex and sk eletons and then deﬁne an energy form and a Laplacian on them. In c hapter 4 t w o, w e will prov e the initial theorem b y ﬁrst sho wing that a series of P oincar´ e inequalities hold, starting with one for balls where the radius of the ball depends on the cen ter and generalizing un til the result is uniform in space. In chapter three, w e apply these inequalities to a r esult of Sturm [29] to yield a small time on dia gonal heat k ernel asymptotic with a uniform constant. W e also pro vide o ﬀ diagonal e stimates with constan ts that dep end on d ( x, y ), but not on where x a nd y are lo cated. W e giv e sev eral examples of heat ke rnels. In chapter four, w e consider complexes with underlying group structure. W e describ e ho w to compare metrics on the complex and those on the underlying group, as w ell as how t o switc h from a function o n a group to one on a complex and vic e ve rsa. W e then use the metric comparison as w ell as our small time P oincar´ e inequalit y to compare norms o f functions on complexes and their group coun terparts. In c hapter ﬁv e, we consider heat kerne ls on the g roup a nd the complex. W e split in to t w o cases ; nonamenable groups, whic h ha ve exp onentially fast heat k ernel deca y , and amenable groups. F or the amenable groups, w e look at h eat k ernels restricted to subsets of our space, and then tak e a Følner sequence t o limit to a b ound on the heat k ernels themselv es. In this w ay , w e prov e the second theorem. 1.2 Geometry of the Complexes W e will tak e our deﬁnitions of p olytop es and p olyhedral sets from G r ¨ un ba um’s Con vex P olytop es [16]. Deﬁnition 1.2.1. A p olyhedron K is a subset of R n formed by in tersecting a ﬁnite family of closed half spaces of R n . No te that this can be an unbounded set, but it will b e con v ex. 5 Deﬁnition 1.2.2. A set F is a face of K if F = ∅ , F = K , o r if F = H ∩ K where H is a supp orting h yp erplane of K . H is a supp orting hy p erplane of K if H ∩ K 6 = ∅ and H do es not cut K in to t w o pieces. Deﬁnition 1.2.3. A p oin t x ∈ K is an extreme p oint of a set K if the only y , z ∈ K whic h are solutions to x = λy + (1 − λ ) z fo r some λ ∈ (0 , 1) are x = y = z . That is, x cannot b e expressed as a c on v ex comb ination of p oints in K − { x } . Note that the extreme po in ts of K are faces for K . Deﬁnition 1.2.4. A p olytop e is a compact con v ex subset of R n whic h ha s a ﬁnite set of ex treme p o in t s. This is equiv alen t to s a ying it is a bounded p olyhedron. Deﬁnition 1.2.5. A p olyhedral complex X is the union of a collection, X , of con vex p olyhedra whic h are joined along low er dimensional faces. By t his w e mean that for an y tw o distinct p olyhedra P 1 , P 2 ∈ X , • P 1 ∩ P 2 is a polyhedron whose dimension satisﬁes dim( P 1 ∩ P 2 ) < max(dim( P 1 ) , dim( P 2 )) and • P 1 ∩ P 2 is a face of b oth P 1 and P 2 . W e allow this face to b e the empt y se t. W e do not ha v e a sp eciﬁc em b edding for the complex , X ; ho we v er, w e require eac h p olyhedra to ha v e a metric whic h is consisten t with that of its faces. Note that this deﬁnition implied P 1 ∩ P 2 is a connected set. This rules out expressing a c ircle as t wo edges whose ends are joined, but it allo ws us t o write it as a tria ngle of t hree edges. This is not v ery restrictiv e, as w e can triangulate the p olyhedra in order to form a complex whic h a v o ids the o v erlap. Simplicial complexes are an example of a p olyhedral complex; the diﬀerence here is that we allo w greater n um b ers o f sides. Note that we allow inﬁnite p olyhe- dra, not just ﬁnite polytop es. 6 Deﬁnition 1.2.6. Deﬁne a p -sk eleton, X ( p ) , for 0 ≤ p ≤ dim X to b e the union of all faces of dimension p or smaller. Note that this is a lso a p olyhedral complex. Deﬁnition 1.2.7. A maximal p o lyhedron is a p olyhedron that is not a prop er face of any o ther p olyhedron. The set of maximal p olyhedra of X is denoted X M AX . W e sa y X is dimensionally homogeneous if all of its maximal p olyhedra ha v e dimension n . Note that in com binat orics literature this is called pure. Deﬁnition 1.2.8. X is lo cally (n- 1)-c hainable if f or ev ery connected op en set U ⊂ X , U − X ( n − 2) is also connected. F or a dimensionally homogeneous complex X this is equiv alen t to the pro p ert y that any t w o n dimensional p olyhedra that share a low er dimensional face can b e joined b y a c hain of con t iguous ( n − 1) or n dimensional p olyhedra con taining that face. Deﬁnition 1.2.9. W e call X admissible if it is b o th dimensionally homogeneous and in some triangulation X is lo cally (n-1)- c haina ble. Figure 1.2: Examples of a complex whic h is not dimensionally homogeneous (left), one whic h is not 1-c hainable (cen ter), and one whic h is admissible (right). W e will be w o rking with connected admis sible complexes, and for our purposes, w e’d lik e to consider p olyhedra that ha v e an Euclidean metric. L et X b e an n- dimensional complex. When tw o p olyhedra share a face, w e require these metrics to coincide. F or p oin ts x and y in diﬀeren t p olyhedra, w e deﬁne the distance as follo ws. 7 Deﬁnition 1.2.10. Consider the s et of pa ths connecting x to y whic h consist of a ﬁnite n um b er of line segmen ts. W e can label eac h of the se by the p oin ts it crosses in t he ( n − 1 ) sk eleton. Set γ = { x = x 0 , x 1 , x 2 , .., x k = y } where x i ∈ ( n − 1)- sk eleton for i = 1 ..k − 1, and x i , x i +1 are b oth in the closure of the same maximal p olyhedron. Then set L ( γ ) = P k i =1 d ( x i − 1 , x i ). W e deﬁne the distance b et w een p oin ts in diﬀeren t p olyhedra to be d ( x, y ) = inf γ L ( γ ). Essen tially , w e are s plitting the path in to piece s, and letting the lengths of those pieces inside of the simplice s b e the standard lengths in R n . Since o ur complex is created using clos ed polyhedra, if the geometry of the p olyhedra is b ounded, the inf will b e realized. This w ill giv e us a length space; ie, one in whic h distances are realized b y geo desics in the space. Discussions of length spaces and other metric measure spaces can b e found in Heinonen [18] and Burago, Burago, and Iv ano v [6]. Deﬁnition 1.2.11. Let X = ∪ i P i , where the P i are the maximal p olyhedra. W e will set the measure of A , a Borel subset of X , to be µ ( A ) = P i µ i ( A ∩ P i ) where µ i is the Lebesgue measure on P i . Notice that the measure within the in terior of maximal p olyhedra is the same as Leb esgue measure on R n . This means that lo cally w e w ill hav e all of the s tructure of R n ; in particular, w e will ha v e v olume doubling for b alls contained in the in terior of the max imal p olyhedra. Since our complex is lo cally ﬁnite, v olume doubling will hold lo cally for all po in t s in the complex . Deﬁnition 1.2.12. An a dmissible p o lyhedral complex, X , equipp ed with distance, d ( · , · ) and measure µ is called an Euclidean p olyhedral complex. 8 F or brevit y , w e will often call this an Euclidean complex. A b o ok whic h de- scrib es t hese complexes is Harmonic Maps Betw een Riemannian P olyhedra [11]. In it, the authors deﬁne these structures with a Riemannian metric and provide analytic results on b oth the comple xes and functions whose domain and range are b oth complexes. 1.3 Analysis on the Complexes 1.3.1 The Diric hlet F orm No w that w e’v e deﬁned the space geometrically , w e will deﬁne a D iric hlet fo rm whose core consists of c ompactly supp orted Lipsc hitz functions. Deﬁnition 1.3.1. A function f on a metric space X is called L -Lipsc hitz (alter- nately , Lipsc hitz) if there e xists a constan t L ≥ 0 so that d ( f ( x ) , f ( y )) ≤ Ld X ( x, y ) for all x and y in X . The space of Lipsc hitz functions is denoted Lip( X ). The space of compactly suppor ted Lipsc hitz functions is denoted C Lip 0 ( X ). Note that Lipsc hitz functions are contin uous. By theorem 4 in section 5.8 of [12], for eac h B ǫ ( x ) ⊂ X − X ( n − 1) and f ∈ C Lip 0 ( X ), f restricted to B ǫ ( x ) is in the Sob olev space W 1 , ∞ ( B ǫ ( x )). This tells us that f has a gra dien t almost ev erywhere in X − X ( n − 1) . Since µ ( X ( n − 1) ) = 0, f has a gradien t for almost ev ery x in X . W e w ould lik e an energy form that acts like E ( u, v ) = R X h∇ u, ∇ v i dµ with domain F to deﬁne our op erato r ∆ with domain D om(∆). W e can deﬁne this in a very general manner whic h do es not depend on t he lo cal Euclidean structure b y follo wing a pap er of Sturm [31]. W e can also deﬁne it in a more straigh tforw ard manner whic h uses the geometry of X . W e do b oth, and then show that they coincide. 9 Sturm assumes that the space ( X , d ) is a lo cally compact separable metric space, µ is a Radon measure on X , and that µ ( U ) > 0 for ev ery nonempt y op en set U ⊂ X . These assumptions hold both in our space, X , and on the sk eletons, X ( k ) . W e begin by approximating E with a form E r deﬁned to b e: E r ( u, v ) := Z X Z B ( x,r ) −{ x } ( u ( x ) − u ( y ))( v ( x ) − v ( y )) d 2 ( x, y ) 2 N dµ ( y ) dµ ( x ) µ ( B r ( x )) + µ ( B r ( y )) for u , v ∈ Lip( X ) where N is the lo cal dimension. Note that whenev er x is in a region lo cally lik e R n , w e ha v e lim r → 0 N µ ( B r ( x )) Z B ( x,r ) −{ x } ( u ( x ) − u ( y )) 2 d 2 ( x, y ) dµ ( y ) = |∇ u ( x ) | 2 , and so this form looks v ery similar to E ( u, u ) = R X |∇ u | 2 dx . This form with domain C Lip 0 ( X ) is closable and symmetric on L 2 ( X ), and its closure ha s core C Lip 0 ( X ). See Lemma 3 .1 in [31]. One can tak e limits of these op erators in the follo wing w ay . The Γ-limit of the E r n is deﬁned to b e the limit that occurs when the follo wing lim sup and lim inf are equal for all u ∈ L 2 ( X , m ). See Dal Maso[9] for a thorough intro duction. Γ − lim sup n →∞ E r n ( u, u ) := lim α → 0 lim sup n →∞ inf v ∈ L 2 ( X ) || u − v ||≤ α E r n ( v , v ) Γ − lim inf n →∞ E r n ( u, u ) := lim α → 0 lim inf n →∞ inf v ∈ L 2 ( X ) || u − v ||≤ α E r n ( v , v ) . F or any seque nce { E r n } of these o p erators with r n → 0 , there is a subsequenc e { r n ′ } so tha t the Γ-limit of E r ′ n exists b y Lemma 4.4 in [31]. These lemmas are put together in to a theorem ( 5.5 in [31]) that tells us that this limit, E 0 , with domain C Lip 0 ( X ) is a closable and symmetric f orm, a nd its closure, ( E , F ), is a strongly lo cal regular Diric hlet form on L 2 ( X , m ) with core C Lip 0 ( X ). 10 Alternately , w e can deﬁne the energy form using the structure of the space. W e set E ( · , · ) t o the follo wing for f ∈ C Lip 0 ( X ): E ( f , f ) = X X M ∈M Z X M |∇ f | 2 dµ ( x ) . Lemma 1.3.2. E ( · , · ) is a closa b le form. That is, for any se quenc e { f n } ∞ n =1 ⊂ C Lip 0 ( X ) that c onver ges to 0 in L 2 ( X ) and is Cauch y in || · || 2 + E ( · , · ) we have lim n →∞ E ( f n , f n ) = 0 . Pr o of. T o show this, w e will ﬁrst loo k at what happ ens on one ﬁxe d polyhedron, and then lo ok at what happ ens on a complex. Let X M b e a maximal p olyhedron. Since { f n } ∞ n =1 is Cauc hy in the norm, we ha v e lim m,n →∞  Z X M ( f n − f m ) 2 dµ  1 2 +  Z X M ( ∇ f n − ∇ f m ) 2 dµ  1 2 = 0 . This give s us t w o functions, f and F which are the limits of f n and ∇ f n resp ectiv ely . W e hav e f = 0 by assumption. W e need to show that F = 0. F or almost ev ery x, y ∈ X M and line γ x ∼ y in X M w e ha v e Z γ x ∼ y ∇ f n dµ = f n ( y ) − f n ( x ) . Then w e can tak e the limit as n goes to inﬁnit y to get lim n →∞ Z γ x ∼ y ∇ f n dµ = lim n →∞ f n ( y ) − f n ( x ) = 0 . This giv es us lim n →∞ ∇ f n ( x ) = 0 for almost ev ery x ∈ X M . Since the c hoice of X M w as arbitrary , this show s lim n →∞ ∇ f n ( x ) = 0 for almost ev ery x ∈ X . Sho wing L 2 con vergence is a bit trickie r, as w e need to show that w e can in terc hange t he limit with the sum ov er the maximal p olyhedra. W e can do this 11 for |∇ f n − ∇ f m | b y F atou’s Lemm a. lim n →∞ X X M ∈M Z X M |∇ f n | 2 dµ = lim n →∞ X X M ∈M Z X M |∇ f n − lim m →∞ ∇ f m | 2 dµ = lim n →∞ X X M ∈M Z X M lim m →∞ |∇ f n − ∇ f m | 2 dµ ≤ lim n →∞ lim m →∞ X X M ∈M Z X M |∇ f n − ∇ f m | 2 dµ = 0 . This tells us that the form is closable. W e will show that the tw o energy forms are the same. T o do this, w e show that they are the same on the core C Lip 0 ( X ); this giv es equality on the domain. Lemma 1.3.3. Each function f ∈ C Lip 0 ( X ) satisﬁes E ( f , f ) = E ( f , f ) . Pr o of. W e can write X as ( X − X ( n − 1) ) ∪ X ( n − 1) ; this is a collection of maximal p olyhedra and a set of measure 0. The in terior of the maximal p olyhedra is a Riemannian ma nifold without b oundary . X is also a lo cally compact length space, and so it satisﬁes the conditions o f example 4G in [3 0]. This implies it has the strong measure con traction prop ert y with an exceptional s et. Corollary 5.7 in [30] sa ys that this then has E ( f , f ) = E ( f , f ) f or each f ∈ C Lip 0 ( X ). The equality is sho wn by appro ximating the forms using an increasing sequence o f op en subsets whic h limit to X − X ( n − 1) . As C Lip 0 ( X ) is a core for b oth E and E , the Dirichle t forms are the s ame. W e will explain more clearly where the doma in of this o p erator lies. The domain is the closure o f C Lip 0 ( X ) in the W 1 , 2 ( X ) norm. This domain is a subset of the set of functions whic h a re in W 1 , 2 of the in teriors of the maximal p olyhedra. 12 Lemma 1.3.4. F or E , C Lip 0 ( X ) ⊂ C ( X ) ∩  ∪ X M ∈X M AX M W 1 , 2 ( X o M )  wher e the closur e is taken with r esp e ct to the W 1 , 2 norm, || · || 2 + E ( · , · ) . X o M denotes the in terio r of X M . Pr o of. First note that C Lip 0 ( X ) ⊂ C ( X ). F or a n y f ∈ C Lip 0 ( X ), w e ha v e the compact subset Y = supp( f ). Then f restricted to X o M will b e in W 1 , 2 ( X o M ), since ||∇ f || 2 ,X o M ≤ || ∇ f || ∞ ,X M µ ( X o M ∩ supp( f )). This tells us f ∈ ∪ X M ∈X M AX M W 1 , 2 ( X o M ) . W e no w ha v e a con tainment without the closures: C Lip 0 ( X ) ⊂ C ( X ) ∩  ∪ X M ∈X M AX M W 1 , 2 ( X o M )  . As w e then close b ot h side s with respect to the same norm, w e ha v e: C Lip 0 ( X ) ⊂ C ( X ) ∩  ∪ X M ∈X M AX M W 1 , 2 ( X o M )  . 1.3.2 The Laplacian The Diric hlet form uniq uely determines a p ositiv e self-adjoin t op erator { ∆ , Dom(∆) } on L 2 where F = Dom(∆ 1 2 ) and E ( u, v ) = ( u, ∆ v ) fo r all u ∈ F and v ∈ Dom(∆). This is done b y deﬁning a collection of quadratic forms, E α ( u, v ) := E ( u, v ) + α ( u, v ) for α > 0. Then, b y the Reisz represen tation theorem, there will b e an op erator G α so that E α ( G α u, v ) = ( u, v ) for an y u,v in D om( E ). The set of these op erators forms a C 0 resolv ent. One can lo ok at inv erses, G − 1 α on 13 the image of G α . W e can then consider ∆ = G − 1 α − α on the space G α (Dom( E )). One can show that this deﬁnition is indep enden t o f α . The domain of the op erator is Dom(∆) = G α (Dom( E )). It’s diﬃcult to explic itly state exactly whic h functions are in Dom(∆), but the domain is dense in L 2 ( X ). See F ukushima et al [14] for the full ar gumen t; a ﬁne summary of this is done in T o dd Kemp’s lecture notes [20]. Note that this set-up will work on eac h of the sk eletons, and so w e can use it to deﬁne a diﬀerent Laplacian on eac h of them. When w e deﬁne the E r on a k-sk eleton, X ( k ) , w e’ll set N = k , integrate o v er X ( k ) , and let m b e a k-dimensional measure. This techniq ue will deﬁne ∆ k on a dense subs et of L 2 ( X ( k ) ). In the one dimensional case, this Laplacian giv es us a structure called a quan- tum graph. Here, the functions in the domain of the L aplacian should be con tin- uous and the in w ard p ointing deriv a tiv es should sum to zero at eac h v ertex. This is known as a Kirc hoﬀ condition. A nice in tro duction to these gra phs and their sp ectra as w ell as a wide v ariety of references to the litera ture on them can b e found in Kuc hmen t [24]. In the tw o dimensional case, this op erator is related to results in a pap er of Brin a nd K ifer [5 ] which constructs Brownian mot ion on t w o dime nsional Euclidean complexes. Bo uziane [4] constructs and prov es the exis tence of Brownian motion on a dmissible Reimannian complexes o f any dimension. It w ould be in teresting to determine whether these construc tions deﬁne the same op erator ; how ev er, that is not our fo cus. Chapter 2 Lo cal P oincar ´ e Inequalities on X In this c hapter we will sho w that a uniform lo cal P oincar ´ e inequalit y holds for a certain class of admissible complexes. Lo cal P o incar ´ e inequalities ha ve appeared in [33] and [11] for ﬁnite complex es or for compact subse ts of complexes. In White’s article [33], a global P oincar´ e inequalit y was sho wn for Lipsc hitz functions on an admissible complex made up of a ﬁnite n um b er of p olyhedra. The constan t in this pro of was linear in the num b er of p olyhedra in v olv ed, and so it do es not extend to an inﬁnite complex. A uniform w eak lo cal inequalit y for Lipsc hitz functions w as also sho wn on this ﬁnite comple x. This to o diﬀers from our ineq ualit y in its dep endence on a ﬁnite complex. In Eells and F uglede’s b o ok [11 ], they sho w that fo r an y relativ ely compact subset of an admissible complex, a lo cal Poincar ´ e inequalit y will hold for lo cally Lipsc hitz f unctions with a constan t that dep ends on the particular choice of com- pact subset. The larger complex itself can b e inﬁnite, but the constant in the inequalit y depends on our particular c hoice of compact subset. W e will show the follow ing f or f ∈ Lip( X ), under some assumptions on the geometry of X : k f − f B k p,B ≤ pP 0 r k∇ f k p,B where f B is the a v erage of f o v er B , B = B ( z , r ), r < R 0 . R 0 and P 0 are constan ts dep ending on the space, X . Our res ult sho ws that a uniform lo cal P oincar´ e ine qualit y will hold for Lipsc hitz con t in uous functions o n any ball of ra dius less than R 0 , where R 0 is ﬁxed and dep ends only on t he complex itself, not on the sp eciﬁc choice of ball. W e require 14 15 our comple x to b e admissible . Our complex can be inﬁnite, but w e b ound below the angles of the p olyhedra and the distance betw een t w o v ertices. W e a lso b ound ab ov e the n um b er of p olyhedra that join at a v ertex. In b oth White and Eells and F uglede these assumptions hold b ecause their sets are either ﬁnite or relativ ely compact. Connections betw een P oincar´ e inequalities and other analytic ineq ualities can b e found in Sobo lev met P oincar ´ e by Ha j lasz and Kosk ela [17]. 2.1 W eak P oincar´ e Inequalities W e would lik e to pro v e a lo cal Poincar ´ e inequalit y for an admissible Euclidean p olytopal complex. If w e lo o k at a conv ex subset of Euclidean space, this is a w ell kno wn statemen t. W e will show it ﬁrst in a conv ex space, and t hen w e will generalize it to our locally nonconv ex space. A no te on our notat ion: often w e will abbreviate dµ ( x ) by dx . Similarly , we will write the a verage in tegral of f ov er a set A b y − R A f dx . Lemma 2.1.1. L et Ω b e a c onne cte d c onvex set with Euclide an distanc e and struc- tur e and Ω 1 , Ω 2 b e c on vex subsets of Ω . F or f ∈ Lip(Ω) ∩ L 1 (Ω) , the fol lowin g h o lds: Z Ω 2 Z Ω 1 | f ( z ) − f ( y ) | dz dy ≤ 2 n − 1 diam(Ω) n ( µ (Ω 1 ) + µ (Ω 2 )) Z Ω |∇ f ( y ) | dy . Pr o of. The t yp e of argumen t used here can b e found in Aspects of Sob olev-T yp e Inequalities [28]. Let γ b e a path from z to y . The deﬁnition of a gradien t g iv es us: | f ( z ) − f ( y ) | ≤ Z γ |∇ f ( s ) | ds. Note that if w e are in a 1- dimensional space, a con v ex subset is a line. The desired inequalit y follows from expanding γ to Ω, and then noting that in tegrating o v er x 16 and y has the eﬀect of m ultiplying the righ t hand side b y µ (Ω 1 ) µ (Ω 2 ) ≤ diam(Ω)( µ (Ω 1 ) + µ (Ω 2 )). Because z and y are in the same conv ex region Ω with an Euclidean distance, w e can let the path γ b e a straigh t line: | f ( z ) − f ( y ) | ≤ Z | y − z | 0 |∇ f  z + ρ y − z | y − z |  | dρ. W e in tegr ate this ov er z ∈ Ω 1 , y ∈ Ω 2 . T o get a nice b o und, w e will use a tric k from Korev aar and Sc ho en [23]. W e split the path into t w o halv es. F or eac h half, w e switc h into and out of p o lar co ordinates in a w a y that av oids integrating 1 s near s = 0 . This a llo ws us t o ha ve a b o und whic h dep ends on the v olumes of Ω 1 and Ω 2 rather than Ω. First, w e consider t he half of the path whic h is closer to y ∈ Ω 2 . I Ω ( · ) is the indicator function for Ω. Z Ω 1 Z Ω 2 Z | y − z | | y − z | 2 |∇ f ( z + ρ y − z | y − z | ) | I Ω ( z + ρ y − z | y − z | ) dρdy dz . W e c hange of v ar iable so that y − z = sθ . That is, | y − z | = s a nd y − z | y − z | = θ . Note that diam(Ω) is an upper b ound on the distance b et w een y a nd z . ... = Z Ω 1 Z S n − 1 Z diam(Ω) 0 Z s s/ 2 |∇ f ( z + ρθ ) | I Ω ( z + ρθ ) s n − 1 dρdsdθ dz . W e switc h the order of in tegratio n. No w, ρ will be betw een 0 and diam(Ω) and s will b e b et wee n ρ and min(2 ρ, diam(Ω)). This allo ws us to in tegrate with respect to s . ... = Z Ω 1 Z S n − 1 Z diam(Ω) 0 Z min(2 ρ, diam(Ω)) ρ |∇ f ( z + ρθ ) | I Ω ( z + ρθ ) s n − 1 dsdρdθ dz = Z Ω 1 Z S n − 1 Z diam(Ω) 0 |∇ f ( z + ρθ ) | I Ω ( z + ρθ ) (min(2 ρ, diam(Ω))) n − ρ n n dρdθ dz . 17 No w w e rev erse the change of v ariables to set y = z + ρθ . Since our integral includes an indicator function at z + ρθ , w e ha ve y ∈ Ω. Z Ω 1 Z Ω |∇ f ( y ) | (min(2 | y − z | , diam(Ω))) n − | y − z | n n | y − z | n − 1 dy dz . Let’s consider the possible v alues of (min(2 | y − z | , diam(Ω))) n −| y − z | n n | y − z | n − 1 . If | y − z | < diam(Ω) 2 , then min(2 | y − z | , diam (Ω)) = 2 | y − z | . This giv es us: (min(2 | y − z | , diam(Ω))) n − | y − z | n n | y − z | n − 1 = 2 n | y − z | n − | y − z | n n | y − z | n − 1 = 2 n − 1 n | y − z | ≤ diam(Ω)(2 n − 1) 2 n . Otherwise, if | y − z | ≥ diam(Ω) 2 , then min(2 | y − z | , diam(Ω)) = diam(Ω). This gives us: (min(2 | y − z | , diam(Ω))) n − | y − z | n n | y − z | n − 1 = diam(Ω) n − | y − z | n n | y − z | n − 1 ≤ 2 n − 1 diam(Ω) n − | y − z | n n dia m(Ω) n − 1 ≤ 2 n − 1 diam(Ω) n n dia m(Ω) n − 1 = 2 n − 1 diam(Ω) n . Both cases are dominated b y 2 n − 1 diam(Ω) n . W e place this into the o riginal in tegral: Z Ω 1 Z Ω |∇ f ( y ) | (min(2 | y − z | , diam(Ω))) n − | y − z | n n | y − z | n − 1 dy dz ≤ Z Ω 1 Z Ω |∇ f ( y ) | 2 n − 1 diam(Ω) n dy dz = 2 n − 1 diam(Ω) n µ (Ω 1 ) Z Ω |∇ f ( y ) | dy . This is an upper b ound for Z Ω 1 Z Ω 2 Z | y − z | | y − z | 2 |∇ f ( z + ρ y − z | y − z | ) | I Ω ( z + ρ y − z | y − z | ) dρdy dz . 18 W e can apply t he same argumen t to t he half of the geo desic closest to z ∈ Ω 1 , after ﬁrst subs tituting ρ ′ = | y − z | − ρ : Z Ω 1 Z Ω 2 Z | y − z | 2 0 |∇ f ( z + ρ y − z | y − z | ) | I Ω ( z + ρ y − z | y − z | ) dρdy dz = Z Ω 2 Z Ω 1 Z | y − z | | y − z | 2 |∇ f ( y + ρ ′ z − y | z − y | ) | I Ω ( y + ρ ′ z − y | z − y | ) dρdz dy ≤ 2 n − 1 diam(Ω) n µ (Ω 2 ) Z Ω |∇ f ( y ) | dy . Com bining these with the or iginal inequalit y , w e ha v e Z Ω 2 Z Ω 1 | f ( z ) − f ( y ) | dz dy ≤ 2 n − 1 diam(Ω) n ( µ (Ω 1 ) + µ (Ω 2 )) Z Ω |∇ f ( y ) | dy . Notation 2.1.2. Let X b e an admissible Euclide an p olytopal complex of dimen- sion n . Deﬁnition 2.1.3. Let B b e a ball of radius r whos e cen ter is on a D -dimensional face with the prop ert y that B in tersects no other D -dimensional faces. W e deﬁne w edges W k of B to b e the closu res of eac h of the connected components of B − X ( n − 1) . Note that fo r an y p oin t z in X , a ball B ( z , r ) satisfying the ab ov e criteria exists: for eac h dimension D , w e can tak e any p oint z ∈ X D − X ( D − 1) and any r < d ( z , X ( D − 1) ) and create B = B ( z , r ) ⊂ X . Then B is a ball of radius r whos e cen ter is on a D -dimensional face, and B interse cts no other D -dimensional faces. In essence, the w edges, W k , ar e formed w hen the ( n − 1) sk eleton s lices the ball B in to pieces. This construction tells us that eac h W k has diameter at most 2 r , as eac h of the p oin ts in W k is within distance r of z , and z is included in W k . 19 Figure 2.1: Complex with shaded ball B (left); the three w edges for B (righ t). Example 2.1.4. In ﬁgure 2.1.4 w e hav e an example o f a 2 dimensional complex with a shaded ball cen tered a t a v ertex. This ball has three w edges; one for eac h of the t wo dimensional faces that share the v ertex. No te that eac h wedge is a fraction of a sph ere. Deﬁnition 2.1.5. W e sa y that X has degree b o unded b y M if M is the maximal n um b er of edges in X (1) that can share a v ertex in X (0) . Deﬁnition 2.1.6. W e sa y that X has edge lengths bo unded b elo w b y ℓ if 0 < ℓ ≤ inf v,w ∈ X (0) d ( v , w ) . Note that ha ving degree b o unded b y M implies that the maxim um n umber of k dimensional faces that can share a lo w er dimens ional face is also M . This tells us that suﬃcien tly small balls will be split into at most M wed ges. Note that if X has degree b o unded b y M , then X ( k ) will as w ell. When X ha s degree b ounded b y M , v olume doubling will o ccur lo cally with a uniform cons tan t. In particular, when the edge lengths a re b ounded b elo w b y ℓ the strong statemen t: µ ( B ( x, cr )) ≤ M c N µ ( B ( x, r )) will hold whenev er cr ≤ ℓ . F or balls in X , N will equal n , the dimension of X . If w e restric t to balls in X ( k ) , then this holds w ith N = k . T o sho w a lo cal P oincar ´ e inequalit y on X , we will split the ba lls, whic h are not necessarily con v ex, up in to smaller o v erlapping pieces whic h a re. W e will do this 20 using the w edges. W e can use a c haining argument in order to mov e thro ugh B from one of the W k to another. Note that this uses t he fact that our space X is admissible. W e will sa y W k and W j are adjacen t if they share an n − 1 dimensional face, and let N ( j ) b e the list of indices of faces adjacen t to W j including j . In order to create paths whic h w e can in tegrate o v er, w e need an ov erlapping region b et w een the adjacen t faces. F or k ∈ N ( j ), let W k ,j = W j,k b e the la rgest sub set of W k ∪ W j whic h has the property that W k ∪ W k ,j and W j ∪ W k ,j are b oth con v ex. Then, for each x in W k ,j there is a w ay of describing the rays b etw een x a nd W k in a distance prese rving manner as one would ha v e in R n . This will justify our use of the ρ in the calculation b elo w. Figure 2.2: Complex with shaded ball B (left); the t wo we dges for B and a region whic h o v erlaps b oth of them(righ t). Example 2.1.7. In ﬁgure 2.1.7 w e ha v e a complex and ball with t wo adja cen t w edges. The union of the wedges , W 1 and W 2 , is not con v ex, so w e form the region W 1 , 2 . In this example, b oth W 1 ∪ W 1 , 2 and W 2 ∪ W 1 , 2 are half c ircles. Theorem 2.1.8. L et X b e an admi s s ible Euclide an p olytop a l c omplex of dimension n with de g r e e b ounde d by M . F o r e a ch z ∈ X ther e exists r > 0 so that for B = B ( z , r ) an d i ts c orr esp ondi n g we dges, W i,j , the fol lowing hol d s for f ∈ Lip( X ) ∩ L 1 ( B ) : || f − f B || 1 ,B ≤ 2 M max k ,j ∈ N ( k )  µ ( B ) µ ( W k ) + 2  2 n r ( µ ( W k ) + µ ( W j,k )) nµ ( W j,k ) ||∇ f || 1 ,B . 21 Pr o of. F or a given z ∈ X let D b e the dimension suc h t hat z ∈ X ( D ) − X ( D − 1) . Pic k r < d ( z , X ( D − 1) ). Let B = B ( z , r ). F or x in B w e ha v e: | f ( x ) − f B | = | Z B 1 µ ( B ) f ( x ) dy − Z B 1 µ ( B ) f ( y ) dy | ≤ 1 µ ( B ) Z B | f ( x ) − f ( y ) | d y . W e w ould lik e to apply Lemm a 2.1.1 to this; ho we v er, B is not ne cessarily conv ex. W e will cons truct a path from x to y using a ﬁnite num b er of straigh t lines, where eac h o f the line segme n t s is contained in a conv ex region. F or simplicit y , w e will consider x ∈ W i and y ∈ W k . It is quite p ossible that these tw o wedges are not con t ained in a con v ex subset of B . W e need to use the fa ct that our spac e is lo cally ( n − 1)-c hainable b y lo o king at a c hain in B − { z } starting at W i and ending at W k . The pieces of the c hain will mo v e us fro m a p oin t in W j in to a connecting p oin t in W j,l , and then from that connecting p oin t in W j,l in to a point in W l . F ormulated more precis ely , there is a se quence of indices, σ (1) = i, ...σ ( l ) = k that corresp onds to this c hain of W ’s, so that for each j , W σ ( j ) and W σ ( j +1) are adjacen t, a nd none of the indices rep eat. W e can take p oin ts in these regions; z 1 ∈ W σ (1) , z 2 ∈ W σ (1) ,σ (2) , ... z 2 j − 1 ∈ W σ ( j ) and z 2 j ∈ W σ ( j ) ,σ ( j +1) . Note that eac h pair in this sequence is lo cated in a con v ex region– either W σ ( j ) ∪ W σ ( j ) ,σ ( j +1) or W σ ( j +1) ∪ W σ ( j ) ,σ ( j +1) . The line s egmen ts b et w een these p oints will deﬁne o ur path γ fro m x to y . | f ( x ) − f ( y ) | = | f ( x ) − f ( z 1 ) + f ( z 1 ) − ... + f ( z 2 l ) − f ( y ) | ≤ | f ( x ) − f ( z 1 ) | + | f ( z 1 ) − f ( z 2 ) | + ... + | f ( z 2 l ) − f ( y ) | = | f ( x ) − f ( z 1 ) | + l − 1 X j =1 ( | f ( z 2 j ) − f ( z 2 j − 1 ) | + | f ( z 2 j ) − f ( z 2 j + 1 ) | ) + | f ( z 2 l ) − f ( y ) | . 22 Since it didn’t matter whic h z ’s w e ch ose, as long as they w ere in the prop er sets, w e can av erage the pieces o v er a ll of the possible z ’s. | f ( x ) − f ( y ) | ≤ − Z W i,σ (1) | f ( x ) − f ( z 1 ) | dz 1 + l − 1 X j =1 − Z W σ ( j ) − Z W σ ( j ) ,σ ( j +1) | f ( z 2 j ) − f ( z 2 j − 1 ) | dz 2 j dz 2 j − 1 + − Z W σ ( j +1) − Z W σ ( j ) ,σ ( j +1) | f ( z 2 j ) − f ( z 2 j + 1 ) | dz 2 j dz 2 j + 1 ! + − Z W σ ( l ) ,k | f ( z 2 l ) − f ( y ) | dz 2 l . W e will not wan t to keep track of the exact path b etw een ev ery pair of regions, although in sp eciﬁc examples one may wan t to do that in order to achie v e a tigh ter b ound. Rather, it is useful sim ply in tegrate o ve r all pairs of neigh b oring w edges, as this will includ e ev erything in our path. | f ( x ) − f ( y ) | ≤ X l ∈ N ( i ) − Z W i,l | f ( x ) − f ( z ) | dz + X j X l 6 = i,l ∈ N ( j ) − Z W l − Z W j,l | f ( z ) − f ( w ) | dz dw + X j ∈ N ( k ) − Z W j,k | f ( z ) − f ( y ) | dz . This new inequ alit y will hold for x and y in an y pair of W i and W k with k 6 = i . If w e expand our notation so that W i,i = W i , then this will hold when x and y a re in the same set W k = W i . T o in tegrate ov er all y ∈ B , we can split the integral in to tw o parts; one where x and y are b oth in W i , a nd then add it to the second where y is in one of the W k 6 = W i . Similarly , w e can integrate o v er x in W i and 23 then sum o ver i . 1 µ ( B ) Z B Z B | f ( x ) − f ( y ) | d y dx ≤ 1 µ ( B )   X i,k X l ∈ N ( i ) Z W i Z W k − Z W i,l | f ( x ) − f ( z ) | dz d y dx + X i,k ,j X l ∈ N ( j ) Z W i Z W k − Z W j,l − Z W l | f ( z ) − f ( w ) | dw d z dy dx + X i,k X j ∈ N ( k ) Z W i Z W k − Z W j,k | f ( z ) − f ( y ) | dz dy dx   = X i X l ∈ N ( i ) Z W i − Z W i,l | f ( x ) − f ( z ) | dz dx + µ ( B ) X j X l ∈ N ( j ) − Z W j,l − Z W l | f ( z ) − f ( w ) | dw d z + X k X j ∈ N ( k ) Z W k − Z W j,k | f ( z ) − f ( y ) | dz dy . W e can com bine this in t o one double sum by setting x = w and y = w as w ell as reindexing so that i = j and l = k . ... ≤ X k X j ∈ N ( k )  µ ( B ) µ ( W k ) + 2  Z W k − Z W j,k | f ( z ) − f ( w ) | dz dw . Applying lemma 2.1.1 with Ω = W k ∪ W j,k , Ω 1 = W j,k , Ω 2 = W k , and diam(Ω) ≤ 2 r to eac h o f the piec es w e ﬁnd: ... ≤ X k X j ∈ N ( k )  µ ( B ) µ ( W k ) + 2  2 n r ( µ ( W k ) + µ ( W j,k )) nµ ( W j,k ) Z W k ∪ W j,k |∇ f ( y ) | dy . Note that points in the sets W k ∪ W j,k are coun ted at most 2 M times, since eac h of the W k has at most M neigh bo rs. This allows us to combine the s ums to ﬁnd: 1 µ ( B ) Z B Z B | f ( x ) − f ( y ) | d y dx ≤ 2 M max k ,j ∈ N ( k )  µ ( B ) µ ( W k ) + 2  2 n r ( µ ( W k ) + µ ( W j,k )) nµ ( W j,k ) Z B |∇ f ( y ) | dy . This is the desire d result. 24 No w that w e ha v e the ineq ualit y when p = 1, we can use a tric k to ex tend it to other v alues of p . Lemma 2.1.9. If fo r an y f ∈ Lip ( X ) we hav e : || f − f B || 1 ,B ≤ C r ||∇ f || 1 ,B for B = B ( z , r ) then inf c ∈ ( −∞ , ∞ ) || f − c || p,B ≤ pC r || ∇ f || p,B and || f − f B || p,B ≤ 2 pC r | |∇ f | | p,B . holds for 1 ≤ p < ∞ . Pr o of. Let g ( x ) = | f ( x ) − c f | p sign( f ( x ) − c f ). Note that g is in Lip( X ). Then if ∇ f is the gradien t of f , w e hav e that p | f ( x ) − c f | p − 1 |∇ f ( x ) | is the length of the gradien t of g . Pic k a v alue of c f so that g B = R B g ( x ) dx = 0. (One will exist; w e consider g B as a function of c f and apply the in termediate v alue theorem.) Applying our assumption to g , w e ha v e: Z B | g ( x ) − 0 | dx ≤ C Z B |∇ g ( x ) | dx = C Z B p | f ( x ) − c f | p − 1 |∇ f ( x ) | dx. No w w e use H¨ older to ﬁnd: Z B | f ( x ) − c f | p − 1 |∇ f ( x ) | dx ≤  Z B ( | f ( x ) − c f | p − 1 ) q dx  1 /q  Z B |∇ f ( x ) | p dx  1 /p . Since 1 p + 1 q = 1, w e ha v e ( p − 1) q = p . Com bining this with the ab o v e ineq ualit y giv es us:  Z B | f ( x ) − c f | p dx  1 /p ≤ pC r  Z B |∇ f ( x ) | p dx  1 /p . 25 When w e take an inﬁmum, w e ﬁnd that inf c  Z B | f ( x ) − c | p dx  1 /p ≤  Z B | f ( x ) − c f | p dx  1 /p . When w e combine thes e, w e ha ve inf c || f − c || p,B ≤ pC r ||∇ f || p,B . In the case where p = 2, it is easy to compute t he inﬁm um exactly . Consider h ( c ) = Z B | f ( x ) − c | 2 dx = Z B f ( x ) 2 dx − 2 c Z B f ( x ) dx + c 2 µ ( B ) . This is a pa rab ola whose minim um o ccurs at c = 1 µ ( B ) R B f ( x ) dx = f B . Its mini- m um is the same as that of p h ( c ), and so this giv es us || f − f B || 2 ,B ≤ 2 C r ||∇ f || 2 ,B . When p 6 = 2, w e can use Jensen’s inequalit y to get the av erage. W e do this b y noticing: − Z | f B − c | p dx = |− Z B ( f − c ) dx | p ≤ − Z B | f − c | p dx. This tells us that || f − f B || p,B ≤ inf c || f − c || p,B + || f B − c | | p,B ≤ 2 inf c || f − c || p,B ≤ 2 pC r ||∇ f || p,B . 26 Deﬁnition 2.1.10. W e sa y that X has solid angle bo und α if for eac h z ∈ X ( D ) − X ( D − 1) and r < d ( z , X ( D − 1) ) the w edges of the ball B ( z , r ) satisfy α ≤ µ ( W k ) µ ( r n S ( n − 1) ) ≤ 1 . Note t hat the rig h t ha nd side of the inequalit y reﬂects the fact that eac h of t he W k is a subset of an Euclidean ball. If w e hav e a uniform b ound on the solid angles formed, then the constan t in Theorem 2.1.8 will simplify . Corollary 2.1.11. Supp o s e X is an adm issible n-dimen s i o nal Euclide an p olytop al c omplex with soli d angle b ound α , an d f ∈ L ip( X ) . F or e ach z ∈ X ther e exists r > 0 s o that for B = B ( z , r ) we have || f − f B || p,B ≤ C X pr ||∇ f || p,B wher e the c onstant C X = 2 3 n +3 M 2 αn dep ends only on the sp ac e X . Pr o of. W e need t o b o und max k ,j ∈ N ( k )  µ ( B ) µ ( W k ) + 2  µ ( W k )+ µ ( W j,k ) µ ( W j,k ) from Theorem 2.1.8. Since w e will w an t to bound the µ ( W j,k ), we need a wa y to compare its size to the v o lume of the other W j . W e can sub divide the space initially b y cutting eac h piece in half in eac h of the n dimensions, so that there are at most M ′ = 2 n M pieces. When the W ′ k and W ′ j are a djacen t, this tells us that W ′ j,k has a v olume whic h is larger than min( µ ( W ′ j ) , µ ( W ′ k )). Th us µ ( W k )+ µ ( W j,k ) µ ( W j,k ) ≤ 2. T o bound µ ( B ) µ ( W k ) w e will need the solid angle b ound. Com bining the s olid a ngle b ound with the factor of 2 − n decrease in w edge s ize giv es us the mo diﬁed ine qualit y: µ ( W ′ k ) ≤ 2 − n µ ( r n S ( n − 1) ) ≤ µ ( W ′ k ) α . Summing the left hand side of t he inequalit y ov er k tells us that µ ( B ) ≤ M 2 n 2 − n µ ( r n S ( n − 1) ) . 27 If w e mu ltiply the righ t hand side of the ine qualit y b y M 2 n , w e hav e M µ ( r n S ( n − 1) ) ≤ M 2 n µ ( W ′ k ) α . Com bining these t w o inequalities, w e ﬁnd that: µ ( B ) µ ( W ′ k ) ≤ M 2 n α . W e can substitute these in to our constan t to get: 2 M ′ max k ,j ∈ N ( k )  µ ( B ) µ ( W ′ k ) + 2  2 n r ( µ ( W ′ k ) + µ ( W ′ j,k )) nµ ( W ′ j,k ) ≤ 2 M 2 n  M 2 n α + 2  2 n +1 r n . This com bined with theorem 2 .1.8 and lem ma 2.1.9 giv es us that: || f − f B || p,B ≤ 2 3 n +3 M 2 αn pr ||∇ f || p,B . W e w ould like to extend these theorems so tha t the radius is not dep enden t o n the cen ter of the ball. T o do so, w e will ﬁrst sho w a w eak er P oincar ´ e inequalit y , and then w e will extend it via a Whitney t yp e co ve ring to a stronger v ersion. Theorem 2.1.12. Supp ose X is an admissible n-dimensional Euclide an p olytop al c omplex with solid angle b ound α and e dge lengths b ounde d b elow by ℓ , and f ∈ Lip( X ) . When κ = 6  2 √ 2(1 − cos( α )) + 1  n , the fol lowi ng ine quality holds for e ach z ∈ X and e ach 0 ≤ r ≤ R 0 wher e R 0 := inf v,w ∈ X 0 d ( v,w ) 6 „ 2 √ 2(1 − cos( α )) +1 « n = ℓ κ . Z B ( z ,r ) | f ( x ) − f B ( z ,r ) | dx ≤ r C W eak Z B ( z ,κr ) |∇ f ( x ) | dx wher e C W eak = 2 3 n +3 M 3 κ N +1 αn . Pr o of. Let z ∈ X a nd r ≤ inf v,w ∈ X 0 d ( v,w ) 6( 2 √ 2(1 − cos( α )) +1) n b e giv en. If d ( z , X ( n − 1) ) > r , then the result follo ws as a w eak er ve rsion of Coro llary 2 .1.11. Otherwise, w e will need to 28 ﬁnd a p oin t v k whic h has the prop erty that it is on a k -sk eleton, and there are no other f aces in the k - sk eleton that are in tersected by B ( v , d ( v , z ) + r ) . W e will do this b y descending down the s k eletons. If there is a point within r of z with this prop ert y , w e will use it. If not, set r 0 = 3 r . Then there is a k suc h that the low est dimensional sk eleton that is in tersected b y B ( z , r 0 ) is X ( k ) , and X ( k ) is in tersected by B ( z , r 0 ) at at least t w o p oin ts v k and w k on tw o diﬀere n t f aces. If these faces did not in tersect, then they would b e at least distance inf v,w ∈ X (0) d ( v , w ) from one another. This would imply that inf v,w ∈ X (0) d ( v , w ) ≤ 2 r 0 = 6 r , a con tradiction. Th us those t w o faces in tersec t in a smaller j -dimensional face. Call v j the p oint on the j -dimensional face whic h minimize s min( d ( v j , v k ) , d ( v j , w k )). These three p oints form a triangle with angle v k v j w k ≥ α , where α is the smallest in terior angle in X . Note that this angle is b ounded by the assumption α ≤ µ ( W k ) µ ( r n S n − 1 ) . The triangle that w o uld maximize the minim um distance to this new p oint, min( d ( v j , v k ) , d ( v j , w k )) is an isosceles one with angle v k v j w k = α . The la w of cosines tells us that for the isosceles triangle, d ( v k , w k ) 2 = 2 d ( v k , v j ) 2 (1 − cos( α ) ), and so for a general tr iangle, d ( v k , v j ) ≤ d ( v k ,w k ) √ 2(1 − cos( α )) ≤ 2 r 0 √ 2(1 − cos( α )) . If this v j w or ks, w e’re done. Otherwise, we will hav e a t least tw o p oints, v j and w j within  2 √ 2(1 − cos( α )) + 1  r 0 of z . W e will rep eat the pro cess by taking new r ’s o f the form r i +1 =  2 √ 2(1 − cos( α )) + 1  r i un t il w e hav e a point whic h w orks. Note that eac h time we rep eat it, w e ﬁnd a p oint on a lo w er dime nsional sk eleton. The w o rst case scenario will ha v e us rep eat this n times until w e’re left with at least one p oint on X (0) . The largest radius that we could require is R =  2 √ 2(1 − cos( α )) + 1  n 3 r . Using this R , we can s ho w that B ( v , R ) do es not in tersect t wo v ertices. The con- 29 dition r ≤ inf v,w ∈ X 0 d ( v,w ) 6 „ 2 √ 2(1 − cos( α )) +1 « n tells us that R ≤ 1 2 inf v,w ∈ X 0 d ( v , w ). As t w o v ertices cannot b e closer than the closes t pair, B ( v , R ) con tains at most one v ertex. This construction giv es us a cen ter, v , on a k - dimensional face, and a radius, R ≤  2 √ 2(1 − cos( α )) + 1  n 3 r so that B ( v , R ) in tersects only the k -dimensional face that v is on. This allo ws us ﬁrst to rece n t er our ball around v and then to apply Corollary 2.1.1 1 to f on B ( v , R ) . Then, as κ = 6  2 √ 2(1 − cos( α )) + 1  n , w e ﬁnd B ( v , R ) ⊂ B ( z , κr ). Z B ( z ,r ) | f ( x ) − f B ( z ,r ) | dx = 1 µ ( B ( z , r )) Z B ( z ,r ) Z B ( z ,r ) | f ( x ) − f ( y ) | d xdy ≤ 1 µ ( B ( z , r )) Z B ( v, R ) Z B ( v, R ) | f ( x ) − f ( y ) | d xdy ≤ µ ( B ( v , R )) µ ( B ( z , r )) Z B ( v, R ) | f ( x ) − f B ( v, R ) | dx ≤ µ ( B ( v , R )) µ ( B ( z , r )) 2 3 n +3 M 2 αn R Z B ( v, R ) |∇ f ( x ) | dx ≤ µ ( B ( z , κr )) µ ( B ( z , r )) 2 3 n +3 M 2 αn κr Z B ( z ,κr ) |∇ f ( x ) | dx ≤ M κ N 2 3 n +3 M 2 αn κr Z B ( z ,κr ) |∇ f ( x ) | dx. 2.2 Whitney Co v ers W e w ould lik e to strengthen the w eak v ersion of the P oincar ´ e inequalit y . W e’ll do this by using a Whitney type co ve ring o f the ball, E = B ( z , r ). Once we hav e this co v er, we can us e a c haining arg umen t to allow us t o replace κ = 6  2 √ 2(1 − cos( α )) + 1  n with 1. Giv en our set w e will consider a collection F of balls suc h that (1) B ∈ F are disjoin t. 30 (2) If we expand the balls to ones with t wice the radius, w e cov er all of E . ∪ B ∈F 2 B = E . (3) F or any B ∈ F , its r adius is r B = 10 − 3 κ − 1 d ( B , ∂ E ). This implies 10 3 κB ⊂ E . Note that this also t ells us that the distance from the cen ter of B to the b oundary of E is (10 − 3 κ − 1 + 1) d ( B , ∂ E ) = (10 3 κ + 1) r B . (4) sup x ∈ E |{ B ∈ F | x ∈ 36 κB }| ≤ K . Note that the constan t κ dep ends on X but not on the sp eciﬁc c hoice of E . W e will sho w in the following lemm a that K is independent of E as w ell. Lemma 2.2.1. Pr op e rty 4 is sa tisﬁ e d for K ≤ C log 2 (8(10 3 κ +1)) vol . When E is a b al l in X which interse cts on l y one vertex, we have K ≤ M ( 8(1 + 10 3 κ )) N . Pr o of. Let a p oint x ∈ E b e giv en. Let B ( y , r y ) ∈ F b e a ball cen tered at y with the prop ert y that x ∈ 36 κB ( y , r y ). Then: d ( x, y ) ≤ 36 κr y = 36 κ 10 − 3 κ − 1 d ( y , ∂ E ) ≤ . 036( d ( x, y ) + d ( x, ∂ E )) . When w e solv e for d ( x, y ), w e ha ve : d ( x, y ) ≤ . 036 1 − . 0 36 d ( x, ∂ E ) . The triangle inequalit y tells us that d ( x, ∂ E ) − d ( x, y ) ≤ d ( y , ∂ E ) ≤ d ( y , x ) + d ( x, ∂ E ) . W e use this to bound d ( y , ∂ E ). 1 − . 0 72 1 − . 0 36 d ( x, ∂ E ) ≤ d ( y , ∂ E ) ≤ 1 1 − . 036 d ( x, ∂ E ) . 31 Because r y = (10 3 κ + 1) − 1 d ( y , ∂ E ), this tells us that the ra dius r y is b ounded b y (1 + 10 3 κ ) − 1 1 − . 072 1 − . 036 d ( x, ∂ E ) ≤ r y ≤ (1 + 10 3 κ ) − 1 1 − . 036 d ( x, ∂ E ) . These inequalities hold for a n y B ( y , r y ) ∈ F with x ∈ 36 κB ( y , r y ). Eac h of the balls, B y will b e contained in B x := B ( x, r 1 d ( x, ∂ E )) where r 1 = (1+10 3 κ ) − 1 + . 036 1 − . 036 . The B y ha v e radius at least r 2 d ( x, ∂ E ) where r 2 = (1 + 10 3 κ ) − 1 1 − . 072 1 − . 036 . W e also kno w that t he B y are disjoin t . This tells us that: |{ B ∈ F | x ∈ 36 κB }| min B y ∈F ∩ B x µ ( B y ) ≤ X B y ∈F ∩ B x µ ( B y ) ≤ µ ( B x ) . W e can use v olume doubling to compare the s ize o f min B y ∈F ∩ B x µ ( B y ) a nd µ ( B x ). Note that r 1 < 2 a nd 1 2(1+10 3 κ ) < r 2 . This tells us that B ( x, r 1 d ( x, ∂ E )) ⊂ B ( y , 8(1 + 10 3 κ ) r 2 d ( x, ∂ E )). µ ( B x ) ≤ C log 2 (8(1+10 3 κ )) vol µ ( B y ) . Com bining these inequ alities and taking the suprem um o ve r x ∈ E g iv es us: sup x ∈ E |{ B ∈ F | x ∈ 36 κB }| ≤ C log 2 (8(1+10 3 κ )) vol . Note that this only depends on κ and C vol . When E intersec ts only one v ertex w e ha v e: µ ( B x ) ≤ M 2 2 N (2(1 + 10 3 κ )) N µ ( B y ) . This giv es us a more reﬁned estimate. Here N is the dimension of X : sup x ∈ E |{ B ∈ F | x ∈ 36 κB }| ≤ M (8(1 + 1 0 3 κ )) N . 32 W e will ﬁrst describ e prop erties of this collection, and then we will use them to sho w a P oincar ´ e inequalit y . This is a mo diﬁed v ersion of the argument found in [2 8]. W e can a lso use this tec hnique to take a P oincar ´ e inequalit y on a small ball and extend it to one on a larger ball whenev er w e ha v e volume doubling. This increases the constant inv olv ed, so it cannot b e done indeﬁnitely , but given a ﬁxed radius w e will b e able to ha v e ineq ualities that hold up to balls of that size. W e b egin with a bit of notation. Let B z ∈ F b e a ball suc h that z ∈ 2 B z . Note that there ma y b e more than o ne; w e will pic k o ne arbitra rily . As z is the cen ter of E , w e will call B z the cen tral ball. F or a ball, B , call the cen ter x B , and ﬁx γ B , a distance minimizing curv e from z to x B . Lemma 2.2.2. F or any B ∈ F we have d ( γ B , ∂ E ) ≥ 1 2 d ( B , ∂ E ) = 1 2 κ 10 3 r B . If B ′ ∈ F has the pr op erty 2 B ′ ∩ γ B 6 = ∅ , then r B ′ ≥ 1 4 r B . Pr o of. This ﬁrst claim will follo w from m ultiple a pplications of the triangle in- equalit y . Let α b e the p oin t in γ B whic h is closest to the b oundary: d ( γ B , ∂ E ) = d ( α , ∂ E ). Then we can b ound r E : d ( z , α ) + d ( α, ∂ E ) ≥ d ( z , ∂ E ) = r E and w e can b ound the distance from B to ∂ E : d ( x B , α ) + d ( α, ∂ E ) ≥ d ( x B , ∂ E ) ≥ d ( B , ∂ E ) . Summing them, w e ﬁnd tha t: d ( z , α ) + d ( x B , α ) + 2 d ( α , ∂ E ) ≥ r E + d ( B , ∂ E ) . 33 As z is on γ B and α minimizes the distance to the boundary , w e ha v e d ( z , α ) + d ( α, x B ) = d ( z , x B ) ≤ r E . Putting this in to the inequalit y , w e ﬁnd: r E + 2 d ( α , ∂ E ) ≥ r E + d ( B , ∂ E ) d ( α, ∂ E ) ≥ 1 2 d ( B , ∂ E ) . The second part follo ws f rom the fact: 1 2 d ( B , ∂ E ) ≤ d ( γ B , ∂ E ) ≤ d ( γ B ∩ 2 B ′ , ∂ E ) . If α ′ is the p oin t in γ B ∩ 2 ¯ B ′ and β ′ is the point in ¯ B ′ that realize s the distance to the b oundary ∂ E , then we hav e d ( γ B ∩ 2 B ′ , ∂ E ) = d ( α ′ , ∂ E ) ≤ d ( α ′ , x B ′ ) + d ( x B ′ , β ′ ) + d ( B ′ , ∂ E ) ≤ 2 r B ′ + r B ′ + d ( B ′ , ∂ E ) . Com bining this with fa ct (3), w e hav e: 1 2 10 3 κr B ≤ 3 r B ′ + 10 3 κr B ′ 1 4 r B ≤ 10 3 κ 2(3 + 10 3 κ ) r B ≤ r B ′ . F or eac h ball B in F , we w ould like to deﬁne a string of balls, F ( B ), that takes B z to B . Set B 0 = B z . Then, for the ﬁrst p oint on γ B that is not contained in 2 B i , tak e a ball B i +1 in F suc h that that p oint is con tained in 2 B i +1 . As B i +1 is op en, this guaran tees that 2 B i ∩ 2 B i +1 6 = ∅ . W e con tinue in this manner until 2 B ℓ − 1 ∩ 2 B 6 = ∅ . Then w e set B ℓ = B . W e la b el F ( B ) = ∪ ℓ i =0 B i . Note that due to v o lume doubling, the chain will b e ﬁnite. This c hain will allow us to mov e from the cen tral ball to an y other ball in the cov er of E . It is useful b ecause neighboring balls are of comparable radii and volume. 34 Lemma 2.2.3. F or any B ∈ F and any B i , B i +1 ∈ F ( B ) we c an c omp ar e the r adii whe r e r j = r B j in the fol lowing m anner: (1 + 10 − 2 κ − 1 ) − 1 r i ≤ r i +1 ≤ (1 + 10 − 2 κ − 1 ) r i . We als o have B i +1 ⊂ 6 B i and B i ⊂ 6 B i +1 , a n d so µ (6 B i ∩ 6 B i +1 ) ≥ max { µ ( B i ) , µ ( B i +1 ) } . Pr o of. Let x i and x i +1 b e t he cen ters of B i and B i +1 resp ectiv ely . By our con- struction, 2 B i ∩ 2 B i +1 6 = ∅ . This tells us that d ( x i , x i +1 ) ≤ 2 r i + 2 r i +1 . d ( x i +1 , ∂ E ) ≤ d ( x i +1 , x i ) + d ( x i , ∂ E ) r i +1 + d ( B i +1 , ∂ E ) ≤ (2 r i + 2 r i +1 ) + ( r i + d ( B i , ∂ E )) r i +1 + 10 3 κr i +1 ≤ 2 r i +1 + 3 r i + 10 3 κr i r i +1 ≤ 3 + 10 3 κ 10 3 κ − 1 r i =  1 + 4 10 3 κ − 1  r i . This tells us that r i +1 ≤ (1 + 10 − 2 κ − 1 ) r i . By a symmetric argumen t , w e also get the lo wer b ound. T o sho w set inclus ions, we use the fact that d ( x i , x i +1 ) ≤ 2 r i + 2 r i +1 ≤ 2 r i + 2(1 + 10 − 2 κ − 1 ) r i . Since a n y p oint in B i +1 is within distance r i +1 of x i +1 , the triangle inequality tells us that it is within distance 2 r i + 2(1 + 10 − 2 κ − 1 ) r i + r i +1 ≤ 6 r i of x i . This giv es us the inclusion B i +1 ⊂ 6 B i . The rev erse holds b y a symme tric argumen t, and so µ (6 B i ∩ 6 B i +1 ) ≥ max { µ ( B i ) , µ ( B i +1 ) } follow s. Lemma 2.2.4. Given B ∈ F and A ∈ F ( B ) we have B ⊂ (10 3 κ + 9) A . 35 Pr o of. Let x B and x A b e the cen ters of B and A resp ectiv ely , and let α b e a point in 2 A ∩ γ B . By 2.2.2, w e k no w that 4 r A ≥ r B . Note that α o ccurs on the distance minimizing curv e γ B b et w een x B whic h is the center o f B and z , the cen ter of the large ball E . This tells us that d ( z , x B ) = d ( z , α ) + d ( α, x B ). Then d ( x A , x B ) ≤ d ( x A , α ) + d ( α , x B ) = d ( x A , α ) + ( d ( z , x B ) − d ( z , α ) ) ≤ 2 r A + d ( z , ∂ E ) − d ( z , α ) ≤ 2 r A + ( d ( z , α ) + d ( α, ∂ E )) − d ( z , α ) ≤ 2 r A + d ( α , x A ) + d ( x A , ∂ E ) ≤ 4 r A + ( r A + 10 3 κr A ) = (10 3 κ + 5) r A . Because all p oin ts in B are within r B ≤ 4 r A of x B , they will b e within (10 3 κ + 5) r A + 4 r A = (10 3 κ + 9) r A of x A . Th us, B ⊂ (10 3 κ + 9) A holds. W e no w ha v e a n um b er of lem mas that describe t he geometry of the cov ering. W e can use these to get an extension of our Poincar ´ e inequalit y . W e will do this b y using this c hain of ba lls to get a c hain of inequalities. Our ﬁr st step is to compar e the a verage of neigh b oring ba lls in the c hain. Lemma 2.2.5. F or B i and B i +1 neighb oring b al l s in a chain F ( B ) , we have | f 6 B i − f 6 B i +1 | ≤ C W eak 18 κ r i µ ( B i ) Z 36 κB i |∇ f ( x ) | dµ ( x ) whenever f satisﬁes || f − f 6 B i || 1 , 6 B i ≤ C W eak κ 6 r i ||∇ f || 1 ,κ 6 B i for al l B i ∈ F ( B ) . 36 Pr o of. W e can write: µ (6 B i ∩ 6 B i +1 ) | f 6 B i − f 6 B i +1 | = Z 6 B i ∩ 6 B i +1 | f 6 B i − f 6 B i +1 | dµ ( x ) ≤ Z 6 B i ∩ 6 B i +1 | f ( x ) − f 6 B i | + | f ( x ) − f 6 B i +1 | dµ ( x ) ≤ Z 6 B i | f ( x ) − f 6 B i | dµ ( x ) + Z 6 B i +1 | f ( x ) − f 6 B i +1 | dµ ( x ) ≤ C W eak 6 κ  r i Z 6 κB i |∇ f ( x ) | dµ ( x ) + r i +1 Z 6 κB i +1 |∇ f ( x ) | dµ ( x )  ≤ C W eak 6 κ  r i Z 6 κB i |∇ f ( x ) | dµ ( x ) + 2 r i Z 36 κB i |∇ f ( x ) | dµ ( x )  ≤ C W eak 18 κr i Z 36 κB i |∇ f ( x ) | dµ ( x ) . This string of inequalities holds b y the triangle inequalit y , set inclus ion, the we ak P o incar ´ e inequalit y (Theorem 2.1.12), and the comparisons in Lem ma 2.2.3. By Lemma 2.2.3 we know that µ ( B i ) ≤ µ (6 B i ∩ 6 B i +1 ), and so w e can rew rite this to get: µ ( B i ) | f 6 B i − f 6 B i +1 | ≤ 18 κr i C W eak Z 36 κB i |∇ f ( x ) | dµ ( x ) . Recall that w e ha v e shown C W eak = 2 3 n +3 M 3 κ N +1 αn , as in Theorem 2.1.12 for our small balls con taining only one vertex . W e are no w in a p o sition to pro ve our main theorem. Theorem 2.2.6. L et E b e set whose subsets satisfy volume doubling with c onstant C vol . Supp ose F is a Whitney typ e c over o f E and that f satisﬁes || f − f 6 B i || 1 , 6 B i ≤ C W eak κ 6 r i ||∇ f || 1 ,κ 6 B i 37 for al l B i ∈ F . T hen Z E | f ( x ) − f E | dµ ( x ) ≤ P 0 r Z E |∇ f ( x ) | dµ ( x ) holds wh e r e P 0 =  1 + 3 C 1+log 2 (10 3 κ +9) vol  K C W eak 12 κ 10 − 3 . Pr o of. W e w ant to b ound | f − f E | . In order to do this , w e will split this quan tit y in to t w o essen tially similar pieces, | f − f 6 B z | and | f E − f 6 B z | . A t the end of the pro of, we will sho w that | f E − f 6 B z | can b e b ounded b y | f − f 6 B z | . Because of this, w e only need consider | f − f 6 B z | . W e will tak e f min us its av erage on the cen tral ball a nd put it into a form where we can tak e adv an t age of the cov ering. This will in v olve splitting this further in to c ha ins of suﬃcien tly small balls, and then applying the w eak P oincar ´ e inequality to them. After a bit of w ork, this will giv e us the desire d inequalit y . First, w e will use the fact that ∪ B ∈F 2 B co v ers all of E to split the in tegral up in to pieces . Z E | f ( x ) − f 6 B z | dµ ( x ) ≤ X B ∈F Z 2 B | f ( x ) − f 6 B z | dµ ( x ) ≤ X B ∈F Z 2 B | f ( x ) − f 6 B | dµ ( x ) + Z 2 B | f 6 B − f 6 B z | dµ ( x ) . The ﬁrst piece can b e b ounded nicely using t he w eak P oincar ´ e inequality (The- orem 2.1.12). X B ∈ F Z 2 B | f ( x ) − f 6 B | dµ ( x ) ≤ X B ∈ F Z 6 B | f ( x ) − f 6 B | dµ ( x ) ≤ X B ∈ F C W eak 6 κr B Z 6 κB |∇ f ( x ) | dµ ( x ) ≤ K C W eak 6 κ 10 − 3 r E Z E |∇ f ( x ) | dµ ( x ) . The last part of the inequalit y fo llo ws from the fact that 6 κB ⊂ E (b y Lemma 2.2.2), and at most K balls in 6 κ F ov erlap an y giv en point in E . 38 The second piece can be rewritten as: X B ∈F Z 2 B | f 6 B − f 6 B z | dµ ( x ) = X B ∈F µ (2 B ) | f 6 B − f 6 B z | ≤ X B ∈F C vol µ ( B ) | f 6 B − f 6 B z | . No w let us consider what happ ens when w e ﬁx B . W e ha v e a chain, F ( B ), connecting B to the cen tra l ball; w e can use this and Lemma 2.2.5 to ﬁnd: | f 6 B − f 6 B z | ≤ ℓ − 1 X i =0 | f 6 B i − f 6 B i +1 | ≤ ℓ − 1 X i =0 C W eak 18 κ r i µ ( B i ) Z 36 κB i |∇ f ( x ) | dµ ( x ) = X A ∈F ( B ) C W eak 18 κ r A µ ( A ) Z 36 κA |∇ f ( x ) | dµ ( x ) . By lemma 2 .2.4 w e kno w t hat B ⊂ (10 3 κ + 9) A for an y A ∈ F ( B ), and so w e ha ve χ B = χ B χ (10 3 κ +9) A . Multiplying the previous inequ alit y by this, sum ming o v er the B , and then in tegrating o v er E giv es us: Z E X B ∈F | f 6 B − f 6 B z | χ B ( y ) dµ ( y ) ≤ Z E X B ∈F X A ∈F ( B ) C W eak 18 κ r A µ ( A ) Z 36 κA |∇ f ( x ) | dµ ( x ) χ B ( y ) χ (10 3 κ +9) A ( y ) dµ ( y ) . Since the B are disjoin t, w e hav e P B ∈F χ B ( y ) ≤ 1. This allo ws us to simp lify the righ t hand side. W e can then in tegra te. ... ≤ Z E X A ∈F C W eak 18 κ r A µ ( A ) Z 36 κA |∇ f ( x ) | dµ ( x ) χ (10 3 κ +9) A ( y ) dµ ( y ) = X A ∈F C W eak 18 κ r A µ ((10 3 κ + 9) A ) µ ( A ) Z 36 κA |∇ f ( x ) | dµ ( x ) . V olume doubling giv es us: ≤ X A ∈F C W eak 18 κr A C log 2 (10 3 κ +9) vol Z 36 κA |∇ f ( x ) | dµ ( x ) . 39 W e then use the bound from (4) to see: ≤ C W eak 18 κ 10 − 3 r E C log 2 (10 3 κ +9) vol K Z E |∇ f ( x ) | dµ ( x ) . Putting all of this together and fa ctoring, our original inequalit y becomes: Z E | f ( x ) − f 6 B z | dµ ( x ) ≤  1 + 3 C 1+log 2 (10 3 κ +9) vol  K C W eak 6 κ 10 − 3 r E Z E |∇ f ( x ) | dµ ( x ) . Let 1 2 P 0 =  1 + 3 C 1+log 2 (10 3 κ +9) vol  K C W eak 6 κ 10 − 3 . T hen w e can rewrite the inequalit y as: Z E | f ( x ) − f 6 B z | dµ ( x ) ≤ 1 2 P 0 r E Z E |∇ f ( x ) | dµ ( x ) . All that remains is to switc h fro m f 6 B z to the a ve rage on the en tire se t, f E . Z E | f E − f 6 B z | dµ ( x ) = µ ( E ) | f E − f 6 B z | = µ ( E ) | 1 µ ( E ) Z E f ( x ) − f 6 B z dµ ( x ) | ≤ Z E | f ( x ) − f 6 B z | dµ ( x ) ≤ 1 2 P 0 r E Z E |∇ f ( x ) | dµ ( x ) . Th us, the P oincar´ e ineq ualit y holds on the ball E = B ( z , r ) . Z E | f ( x ) − f E | dµ ( x ) ≤ Z E | f ( x ) − f 6 B z | dµ ( x ) + Z E | f 6 B z − f E | dµ ( x ) ≤ P 0 r E Z E |∇ f ( x ) | dµ ( x ) . Corollary 2.2.7. L et X b e an admissible n-di m ensional Euclide an c o mplex with de gr e e b ounde d ab ove by M , soli d angle b ounde d b y α , and e dge leng ths b ounde d b elow b y ℓ . L et E = B ( z , r ) wher e r < R 0 := ℓ κ . Then Z E | f ( x ) − f E | dµ ( x ) ≤ P 0 r Z E |∇ f ( x ) | dµ ( x ) 40 holds for f ∈ Lip( X ) ∩ L 1 ( E ) wher e κ = 6( 2 √ 2(1 − cos( α )) + 1) n and P 0 = (1 + 3 M 3 2 n (10 3 κ + 9) n ) M (8(1 + 10 3 κ )) n C W eak 6 κ . Pr o of. Apply Theorem 2.2.6 with K = M (8(1 + 10 3 κ )) n and C vol = M 2 n . Corollary 2.2.8. L et X b e an admissible n-di m ensional Euclide an c o mplex with de gr e e b ounde d ab ove by M , soli d angle b ounde d b y α , and e dge leng ths b ounde d b elow b y ℓ . F or f ∈ Lip( X ) ∩ L p ( E ) and r < R 0 we have inf c || f − c || p,E ≤ pP 0 r ||∇ f || p,E wher e E is a b al l of r adius r and 1 ≤ p < ∞ . Note that this implies: || f − f E || p,E ≤ 2 pP 0 r ||∇ f || p,E . Her e P 0 = (1 + 3 M 3 2 n (10 3 κ + 9) n ) M (8(1 + 10 3 κ )) n C W eak 6 κ , C W eak = 2 3 n +3 M 3 κ n +1 αn , and κ = 6( 2 √ 2(1 − cos( α )) + 1) n . Pr o of. Apply Lemma 2.1.9 to Corollary 2.2.7. Corollary 2.2.9. Assume p = 1 Poinc ar´ e ine quality || f − f B || 1 ,B ≤ P 0 r ||∇ f || 1 ,B holds for f ∈ Lip( X ) ∩ L p ( E ) on b al ls B = B ( x, r ) with r ≤ R . Assume volume doubling hol d s with c onstant C vol for b al ls with r adius less than C 0 R . The n || f − f E || p,E ≤ p 2 P 0  6(1 + 3 C 11 vol ) C 13 vol 10 − 3  ⌈ log 10 3 6 ( C 0 ) ⌉ r E ||∇ f || p,E also hold s for b al ls E with r adius less than C 0 R and 1 ≤ p < ∞ . Pr o of. Note that if E = B ( x, 1 6 10 3 r ), then the p = 1 P oincar ´ e inequalit y holds for all balls in the Whitney co v er dilated b y a factor of 6 with κ = 1. W e can a pply Theorem 2.2.6 whic h gives us || f − f E || 1 ,E ≤ 6(1 + 3 C 11 vol ) C 13 vol 10 − 3 P 0 r E ||∇ f || 1 ,E . 41 In particular, w e can repeat this to s ho w that the p = 1 P oincar ´ e inequalit y holds for balls up to radius C 0 R with constan t (6(1 + 3 C 11 vol ) C 13 vol 10 − 3 ) ⌈ log 10 3 6 ( C 0 ) ⌉ P 0 . T o get the p P oincar ´ e inequalit y , apply lemma 2.1.9. Note that b ounds on degree M , angles α and edge lengths ℓ giv e us uniform lo cal v olume doubling on our complex, X . F or an y ﬁxed R w e can then apply Lemma 2.1.9 to Corollary 2.2.9 to get the standard L 2 P o incar ´ e inequalit y for balls of radius up to R . In g eneral, this cannot b e extended to R = ∞ ; note that the constan t in the new P oincar´ e inequalit y go es to inﬁnity as C 0 go es to inﬁnit y . Chapter 3 Small Time Heat Kernel Estimates for X The heat k ernel, h t ( x, y ), is the fundamen tal solution to the heat equation ∂ t u = ∆ u. Note that our formulation do es not hav e factor s of − 1 or 1 2 , w hic h app ear in some of the literature. This t yp e of diﬀerential equation is parabo lic; one w a y of obtaining information a b out it is t hrough parab olic Harnack ineq ualities. Sturm [29] show s that lo cal v olume doubling and P oincar ´ e inequalities on a subset of a complete metric space imply a lo cal parab olic Harnac k inequalit y on that subset. He then uses this to ﬁnd G aussian estimates on the heat kernel. The equiv alence of the parab olic Harnack inequalit y with P oincar ´ e and v olume doubling had previously b een done in the Riemannian manifold case by G rigor’y an [1 5] and Saloﬀ- Coste [27]. 3.1 Small time Heat Kernel Asymptotics W e hav e show n a uniform lo cal P oincar ´ e ineq ualit y , and our complex is b oth com- plete and lo cally satisﬁes volume doubling. This tells us that w e’v e satisﬁed the h ypotheses of the following theorem of Sturm [29] whic h giv es a low er b ound on the diagonal. Theorem 3.1.1 (Sturm) . Assume Y is an op en subset of a c omplete sp a c e X that admits a Poinc ar ´ e ine quality with c o n stant C P and volume d o ubli n g with c onstant 2 N . Then ther e exists a c onstant C = C ( C P , N ) such that h t ( x, x ) ≥ 1 C µ ( B ( x, √ t )) 42 43 for al l x ∈ Y and al l t such that 0 < t < ρ 2 ( x, X − Y ) . Here, ρ refers to the in trinsic distance. In our complex, this will alw a ys satisfy ρ ( x, y ) ≥ d ( x, y ). Also note that since w e hav e a uniform local Poincar ´ e inequalit y and a uniform lo cal v olume doubling constant w e ha v e t he follo wing corollary: Corollary 3.1.2. L et X b e an admissibl e n -dime n sional Euclide an c omplex with de gr e e b ounde d ab ove by M , soli d angle b ounde d b y α , and e dge leng ths b ounde d b elow. F or any R 0 > 0 ther e is a c orr esp onding c onstant C = C ( X, R 0 ) s o that h t ( x, x ) ≥ 1 C M µ ( S ( n − 1) ) t n/ 2 for al l x ∈ X and a l l t such that 0 < t < R 2 0 . Pr o of. F or eac h x ∈ X apply 3.1.1 t o X w ith Y = B ( x, R 0 ). The distance compares easily: ρ ( x, X − Y ) ≥ d ( x, X − Y ) = R 0 . Because the constant C in 3.1 .1 dep ends only on C P and N , w e can use the fact that o ur constants C P and N do dep end only on the radius of our ball to obtain a univ ersal constan t, C . Sturm [29] also pro v es an upp er b ound f or the heat k ernel; this b ound is esp e- cially useful near the diagonal. Theorem 3.1.3 (Sturm) . Assume Y is an op en subset of a c omplete sp a c e X that admits a Poinc ar ´ e ine quality with c o n stant C P and volume d o ubli n g with c onstant 2 N . Then ther e exists a c onstant C = C ( C P , N ) such that for every x, y ∈ Y h t ( x, y ) ≤ C e − ρ 2 ( x,y ) 4 t q µ ( B ( x, √ T )) µ ( B ( y , √ T ))  1 + ρ 2 ( x, y ) t  N/ 2 e − λt (1 + λt ) 1+ N/ 2 wher e ρ is the intrinsic di s tanc e, R = inf ( ρ ( x, X − Y ) , ρ ( y , X − Y )) , T = min( t, R 2 ) and λ is the b ottom of the sp e ctrum of the self-adjoint op er ator − L on L 2 ( X , µ ) . 44 In general, w e can replace λ with 0, w hic h incre ases the v alue o f the right hand side. In our se tting, w e can sim plify this a bit more. Corollary 3.1.4. L et X b e an admissibl e n -dime n sional Euclide an c omplex with de gr e e b ounde d ab ove by M , soli d angle b ounde d b y α , and e dge leng ths b ounde d b elow. T hen for any R 0 we ther e exists a c onstant C = C ( X , R 0 ) so that for any x ∈ X and t > 0 we have: h t ( x, x ) ≤ C M µ ( S ( n − 1) )(min( t, R 2 0 )) n/ 2 . Pr o of. F or eac h x ∈ X apply 3.1.3 t o X w ith Y = B ( x, R 0 ). The distance compares easily: ρ ( x, X/ Y ) ≥ d ( x, X/ Y ) = R 0 . The d ( x, x ) terms drop out, as do the λ terms. Because the constan t C in 3 .1.3 dep ends only on C P and N , w e can use the fact that o ur constants C P and N do not dep end on our sp eciﬁc c hoice of ball to obtain a univ ersal constan t, C . Corollary 3.1.5. L et X b e an admissibl e n -dime n sional Euclide an c omplex with de gr e e b ounde d ab ove by M , soli d angle b ounde d b y α , and e dge leng ths b ounde d b elow. F o r any R 0 ther e exists a C = C ( X , R 0 ) so that for any x, y ∈ X and t > 0 we have: h t ( x, y ) ≤ C M µ ( S ( n − 1) )(min( t, R 2 0 )) n/ 2 e − d 2 ( x,y ) 4 t  1 + d 2 ( x, y ) t  N/ 2 . Pr o of. F or eac h x, y ∈ X apply 3.1.3 to X with Y = B ( x, R 0 ) ∪ B ( y , R 0 ). The distance compares easily: ρ ( x, X/ Y ) ≥ d ( x, X / Y ) = R 0 . Because the constant C in 3.1.3 dep ends only on C P and N , w e can use the fact that o ur constan ts C P and N do not dep end o n our sp eciﬁc c hoice of ball to obta in a univ ersal constan t, C . 45 Note that w e can rewrite this as a bound of the follo wing fo rm fo r some con- stan ts C, c : h t ( x, y ) ≤ C (min( t, R 2 0 )) n/ 2 e − c d 2 ( x,y ) t . Corollary 3.1.6. F or an admissi b le n -d i m ensional Euclide an c omp l e x X with de- gr e e b ounde d a b ove by M , soli d angle b o und e d by α , and e dge lengths b ounde d b elow , on X ( k ) we have 1 C k t k / 2 ≤ h k t ( x, x ) ≤ C k t k / 2 . This holds for al l t < R 2 0 and x ∈ X ( k ) , wher e C k dep ends on R 0 , α , M , k , and inf v,w ∈ X (0) d ( v , w ) . I n p articular, we c an take C = max k =1 ..n C k to have a uniform c onstant for e ach X ( k ) . Pr o of. Apply Corollaries 3 .1.2 and 3.1 .4 t o X ( k ) . This holds b ecause X ( k ) is also an admissible complex satisfying t he same b ounds as X . C k v aries sligh tly in eac h dimension due to the eﬀect of dimension on volume doubling, and hence P o incar ´ e. Oﬀ diagonal, the lo wer b ound is more complicated. Theorem 3.1.7 (Sturm) . Assume Y is an op en subset of a c omplete sp a c e X that admits a Poinc ar ´ e ine quality with c o n stant C P and volume d o ubli n g with c onstant 2 N . Then ther e ex i s ts a c onstant C = C ( C P , N ) such that for every x, y ∈ Y which ar e joine d by a curve γ of length ρ ( x, y ) h t ( x, y ) ≥ 1 C µ ( B ( x, √ T )) e − C ρ 2 ( x,y ) t e − C t R 2 wher e ρ is the intrinsic di s tanc e, R = inf 0 ≤ s ≤ 1 ( ρ ( γ ( s ) , X − Y )) , and T = min ( t, R 2 ) . In our setting, w e can ﬁnd a near-diagonal lo w er bound for an y complex. 46 Corollary 3.1.8. L et X b e an admissibl e n -dime n sional Euclide an c omplex with de gr e e b ounde d ab ove by M , soli d angle b ounde d b y α , and e dge leng ths b ounde d b elow. F or any R 0 > 0 ther e exists a C = C ( X , R 0 ) s o that for any x, y ∈ X with d ( x, y ) < R 0 and t > 0 we have: h t ( x, y ) ≥ 1 C µ ( B ( x, p min( t, R 2 0 ))) e − C d 2 ( x,y ) t e − C t R 2 0 . If X is volume do ubli n g and has a glob al Poinc ar ´ e ine quality, we c an set R 0 = ∞ to get: h t ( x, y ) ≥ 1 C µ ( B ( x, √ t )) e − C d 2 ( x,y ) t . Pr o of. F or eac h x, y ∈ X apply 3.1.7 to X with Y = B ( x, 2 R 0 ). Since w e hav e a length space, the distance compares easily: ρ ( γ ( s ) , X/ Y ) ≥ d ( γ ( s ) , X/ Y ) = R 0 . Because the constant C in 3.1.3 depends only on C P and N , w e can us e the fact that our constan ts C P and N do not dep end on our sp eciﬁc c hoice of ball to obta in a univ ersal constan t, C . 3.2 Examples Example 3.2.1. In R 1 the heat k ernel, h t ( x, y ) is the densit y fo r the t ransition probabilit y of Bro wnian motion. h t ( x, y ) = 1 √ 4 π t e − | x − y | 2 4 t . This is a normal densit y for y with exp ectation x and v ariance 2 t . In the probability literature, it is common for the he at eq uation to b e written with the time deriv ative m ultiplied b y an e xtra factor of 1 / 2 so that the v ariance is t . See for example F eller V olume 2 [13]. 47 W e can think of R 1 as an Euclidean complex. This mat c hes our on diagonal b ound exactly , but it is slightly nicer (by a factor of p 1 + d 2 ( x, y ) / t than our oﬀ diagonal b ound. Example 3.2.2. In R n , t he heat equation can b e solv ed using either a scaling argumen t or F ourier series. Alternately , it can be though t of as an n-dimens ional v ersion of Bro wnian motion. In the PDE literature, the heat kerne l is also called the Gauss k ernel or the fundamen tal solution to the heat eq uation. See Ev ans [1 2] for a deriv atio n. h t ( x, y ) = 1 (4 π t ) n/ 2 e − | x − y | 2 4 t . Note that R n is also an Euclidean complex, a nd that this k ernel is consisten t with our asymptotics. Example 3.2.3. W e can think of a circle of length 1 as a complex consisting of three edges o f length 1 / 3 joined in a triangle shap e. One can calculate the heat k ernel in terms of a sum using F ourier series; see Dym and McKean [10]. The heat k ernel here is h t ( x, y ) = 1 √ 4 π t ∞ X n = −∞ e − | x − y − n | 2 4 t . When x = y , this simpliﬁes to: h t ( x, x ) = 1 √ 4 π t + 2 √ 4 π t ∞ X n =1 e − n 2 4 t . F or s mall v alues of t , the dominant term is 1 √ 4 π t . This is the same b eha vior as our small time prediction. Note that once t = 1 / 4 , the ball of radius √ t will b e of size 1. By this p oin t in time, the asym ptotic will cease to b e useful. 48 Figure 3.1: Example of a star; here n=8. Example 3.2.4. W e will lo ok at the heat k ernel on a star shap ed gra ph, X , whic h has a cen tral ve rtex with n edges attac hed to it. When we compute functions on X f or x ∈ e ( a, b ), w e will let x represen t d ( x, a ). F or example, a w ould b e 0, the midp oint w ould b e 1 2 , and b w ould b e 1. All functions on X ar e of the form f ( x, j ) = P i f i ( x ) I j = i + f (0) I x =0 where x ∈ (0 , 1] and j = 1 , 2 , ...n . f i ( x ) represen ts the v a lue of the function alo ng the ith leg, and f (0) is the v a lue at the cen ter of the star. The meas ure on our star is dµ ( x, j ) = dx . When w e write the deriv ativ e d f dx w e will mean the usual deriv ative with respect to Leb esgue measure. d f dµ (0) = P i d f i dx (0) for functions f whic h are diﬀeren tiable for e ac h f i on (0 , 1]. W e’ll lo ok at a domain whe re our function has zero deriv ative at the boundar y p oin ts and has zero deriv ative in the ce n t er. Dom(∆) = { f ∈ C ( X ) : f i ∈ C 1 ((0 , 1]) , and d f dµ (0) = 0 , d f i dµ (1) = 0 } ; we will b e using an L 2 norm on this spac e. The symmetric set of eigenfunctions are cosine on ev ery leg: Φ k ( x ) = r 2 n cos( k π x ) for x ∈ e i , i = 1 ..n, 49 or, when k = 0, they are a constan t on every leg: Φ 0 ( x ) = 1 √ n for x ∈ e i , i = 1 ..n. These functions ha v e deriv ativ e zero a t eac h v ertex and are contin uous at t he cen tral vertex . The co eﬃcien t s are ch osen so that they ha v e an L 2 norm of 1. Note that if w e lo ok at the pro duct of these for p oints x and y on the star (regardless of whic h leg they o ccur on), w e ha ve Φ k ( x )Φ k ( y ) = 2 n cos( k π x ) cos ( k π y ) = 1 n (cos( k π ( x − y )) + cos( k π ( x + y ))) . If w e sum e − λ 2 t Φ( x )Φ( y ) we ﬁnd: 1 n + 1 n ∞ X k =1 e − ( kπ ) 2 t (cos( k π ( x − y )) + cos( k π ( x + y ))) = 1 2 n ∞ X k = −∞ e − ( kπ ) 2 t (cos( k π ( x − y )) + cos( k π ( x + y ))) . In or der to simplify this, we will use Jacobi’s ide n t it y (see D ym for deriv ation): ∞ X k = −∞ e − ( x − k ) 2 2 s = √ 2 π s ∞ X k = −∞ e − 2 π 2 k 2 s e 2 π ik x . Because this sums to a r eal n um b er, we can rewrite it as: ∞ X k = −∞ e − ( x − k ) 2 2 s = √ 2 π s ∞ X k = −∞ e − 2 π 2 k 2 s cos(2 π k x ) . This giv es us: 1 2 n ∞ X k = −∞ e − ( kπ ) 2 t (cos( k π ( x − y )) + cos( k π ( x + y ))) = 1 2 n √ π t ∞ X k = −∞  e − ( x − y − 2 k ) 2 4 t + e − ( x + y − 2 k ) 2 4 t  . W e will also ha v e ones that form an n − 1 dimensional ba sis on the legs. These are the “o dd” eigenfunctions. These will b e either sine or 0 along the legs. Since 50 sin(0) = 0, they will b e con tin uous at the cen ter. W e require the deriv ativ es at the v ertices to b e 0, and so the p ossible sine functions are sin  (2 k +1) π 2 x  . Note that they m ust b e normalized acc ording to an L 2 norm; this means that P n i =1 1 2 b 2 i = 1, where the b i are the co eﬃcien ts. Since they will need to ha v e deriv ativ e zero at the cen tral v ertex, we will nee d P n i =1 b i = 0. The combination of these t wo restrictions, along with the f act that these eigenfunctions m ust b e orthogonal to one a nother, determine the co eﬃcien ts. F or i = 1 ... ⌊ n 2 ⌋ w e hav e eigenfunctions of the form: ˜ Φ k ,i ( x ) =            sin  (2 k +1) π 2 x  for x ∈ e 2 i − 1 − sin  (2 k +1) π 2 x  for x ∈ e 2 i 0 otherwise. These func tions are trivially orthogonal to one another. The factors of ± 1 giv e us deriv ativ e 0 at the cen ter. F or i = 1 ... ⌊ n 2 ⌋ − 1 w e hav e: ˜ Φ k , ⌊ n 2 ⌋ + i ( x ) =              1 √ i ( i +1) sin  (2 k +1) π 2 x  for x ∈ e j , j = 1 .. 2 i − i √ i ( i +1) sin  (2 k +1) π 2 x  for x ∈ e j , j = 2 i + 1 , 2 i + 2 0 otherwise. In the case where there is an o dd num b er of legs w e ha v e: ˜ Φ k ,n − 1 ( x ) =        √ 2 √ n ( n − 1) sin  (2 k +1) π 2 x  for x ∈ e j , j = 1 ..n − 1 − ( n − 1) √ 2 √ n ( n − 1) sin  (2 k +1) π 2 x  for x ∈ e n Note that edges 2 i and 2 i − 1 ha ve constan ts with the same sign, w hic h forces the functions to b e orthogonal to the ﬁrst set. The pattern of + and - allo w t hem to be orthogonal to one a nother. The other factors guaran tee that the deriv ativ e at the cen ter is zero. 51 If w e lo ok at the product of the ˜ Φ k for p oin ts x and y on the star, w e hav e ˜ Φ k ( x ) ˜ Φ k ( y ) = c sin  (2 k + 1) π 2 x  sin  (2 k + 1) π 2 y  = c 2  cos  (2 k + 1) π x − y 2  − cos  (2 k + 1) π x + y 2  . If w e sum these “o dd” eigenfunctions, m ultiplied by e − (2 k +1) 2 π 2 4 t , w e hav e: ∞ X k =0 e − (2 k +1) 2 π 2 4 t c 2  cos  (2 k + 1) π x − y 2  − cos  (2 k + 1) π x + y 2  . W e can rewrite t he x − y terms a s follows; a similar calculation will work for the x + y terms. Fir st w e add in terms with 2 k . W e note that cosine is an ev en function, and so we can extend this to negat iv e k . W e a lso hav e the zero term, since it will cancel b et w een t he t wo su ms. ∞ X k =0 e − ((2 k +1) π ) 2 t 4 c 2 cos  (2 k + 1) π x − y 2  + ∞ X k =1 e − (2 kπ ) 2 t 4 c 2 cos  2 k π x − y 2  − ∞ X k =1 e − (2 kπ ) 2 t 4 c 2 cos  2 k π x − y 2  = ∞ X k =1 e − ( kπ ) 2 t 4 c 2 cos  k π x − y 2  − ∞ X k =1 e − ( kπ ) 2 t c 2 cos ( k π ( x − y )) = ∞ X k = −∞ e − ( kπ ) 2 t 4 c 4 cos  k π x − y 2  − ∞ X k = −∞ e − ( kπ ) 2 t c 4 cos ( k π ( x − y )) . W e will apply Jacobi’s iden tity to eac h of the sums. The ﬁrst sum yields: c 4 ∞ X k = −∞ e − ( kπ ) 2 t 4 cos  2 k π x − y 4  = c 4 q 2 π t 8 ∞ X k = −∞ e − ( x − y 4 − k ) 2 2 t 8 = c 2 √ π t ∞ X k = −∞ e − ( x − y − 4 k ) 2 4 t . The second sum giv es us: − c 4 ∞ X k = −∞ e − ( kπ ) 2 t cos  2 k π x − y 2  = c 4 √ π t ∞ X k = −∞ − e − ( x − y − 2 k ) 2 4 t . 52 Similarly , for the x + y terms w e hav e: − ∞ X k =1 e − (2 k +1) 2 π 2 4 t c 2 cos  (2 k + 1) π x + y 2  = − c 2 √ π t ∞ X k = −∞ e − ( x + y − 4 k ) 2 4 t + c 4 √ π t ∞ X k = −∞ e − ( x + y − 2 k ) 2 4 t . T o get the heat k ernel, w e sum the e − λ 2 t Φ( x )Φ( y ): h t ( x, y ) = 1 2 √ π tn ∞ X k = −∞  e − ( x − y − 2 k ) 2 4 t + e − ( x + y − 2 k ) 2 4 t  + c 2 √ π t ∞ X k = −∞  e − ( x − y − 4 k ) 2 4 t − e − ( x + y − 4 k ) 2 4 t  + c 4 √ π t ∞ X k = −∞  e − ( x + y − 2 k ) 2 4 t − e − ( x − y − 2 k ) 2 4 t  . Note that the v alue of c dep ends on whic h edges x and y are on. If they are on the same edge, c = 2  1 − 1 n  . If x and y are on diﬀerent edges, then c = − 2 n . W e a re no w in a position to see what happ ens on the star near t = 0. On the diagonal, the heat k ernel will limit to inﬁnit y as t go es to zero with the following asymptotics: h t (0 , 0) ≈ 1 n √ π t , h t (1 , 1) ≈ 1 √ π t , and h t ( x, x ) ≈ 1 2 √ π t for x 6 = 0 , 1 . T o determine what happ ens for t near zero when we hav e t w o diﬀeren t po in ts, w e need to consider the relativ e p ositions of x and y . W e kno w tha t the heat k ernel will limit to zero as t go es to zero. Without lo ss of generalit y , we will lo ok a t when d ( x, 0) < d ( y , 0). When x = 0 a nd y 6 = 1, the k = 0 terms will dominate. They giv e us: h t (0 , y ) ≈ 1 n √ π t e − y 2 4 t . 53 When x = 0 and y = 1, w e will only hav e terms in volving e − 1 4 t con t ributing. After cancellation, this giv es us : h t (0 , 1) ≈ 2 n √ π t e − 1 4 t . When x 6 = 0, t he dominant terms will inv olv e e − ( x − y ) 2 4 t . If the co eﬃcien t for those terms is zero, w e then ha v e e − ( x + y ) 2 4 t instead. The relev an t terms are: h t ( x, 1 ) ≈  2 n − c  1 2 √ π t e − ( x +1) 2 4 t +  2 n + c  1 2 √ π t e − ( x − 1) 2 4 t , and h t ( x, y ) ≈  2 n − c  1 4 √ π t e − ( x + y ) 2 4 t +  2 n + c  1 4 √ π t e − ( x − y ) 2 4 t for x 6 = y . When x and y a re o n diﬀerent edges, c = − 2 n , and so a ll of the term s with co eﬃcien t 2 n + c disapp ear. Note that the notation giv es us d ( x, y ) = x + y : h t ( x, 1 ) ≈ 2 n √ π t e − ( x +1) 2 4 t for x 6 = 1, and h t ( x, y ) ≈ 1 n √ π t e − ( x + y ) 2 4 t for x 6 = y , y 6 = 1 . When x and y are on the s ame edge, 2 n + c = 2 n + 2  1 − 1 n  = 2. This g iv es us the same asym totic as in R . h t ( x, 1 ) ≈ 1 √ π t e − ( x − 1) 2 4 t , and h t ( x, y ) ≈ 1 2 √ π t e − ( x − y ) 2 4 t for x 6 = y . Example 3.2.5. Consider the graph consisting of t w o cen tra l v ertices which a re joined by one edge with edges coming oﬀ of them. W e will calculate t he b ehaviour of the heat k ernel b etw een the t w o central v ertices for small times. Consider tw o star- lik e graphs joined together b y a cen tral edge, e ( v 1 , v 2 ). The star cen tered at v 1 connects to n edges in addition to the cen tral edge. These are lab eled e ( v 1 , w i ) i = 1 ..n . F or the star cen tered at v 2 , there are m suc h edges 54 Figure 3.2: Example of tw o joined stars with labels; here n=7 and m=3. whic h a re lab eled e ( v 2 , w ′ i ) i = 1 ..m . This graph ha s three diﬀeren t t ypes of eigenfunctions. When w e compute these functions for x ∈ e ( a, b ), w e will let x represen t d ( x, a ). So a w ould b e 0, the midp oin t w ould b e 1 2 , and b w ould b e 1. The ﬁrst kind corresp ond to eigen v alues of the for m (2 k +1) 2 π 2 4 . They form an m − 1 + n − 1 dimensional space. The corresp onding eigenfunctions are of the fo rm c ( e ( v , w )) sin( (2 k +1) π 2 x ) on the edges e ( v 1 , w i ) and e ( v 2 , w ′ i ) and are 0 on the cen tral edge, e ( v 1 , v 2 ). Since eac h eigenfunction is zero o n the cen tral edge, they will not con t ribute when w e compute h t ( x, y ) for x ∈ e ( v 1 , v 2 ). The second kind corresp ond to eigen v alues o f the form ( k π ) 2 . When k 6 = 0, these are ± √ 2 √ m + n +1 cos( k π x ) along eac h edge in the graph, with sign chosen to preserv e con tin uit y . W e write them: Φ k ( x ) =      √ 2 √ m + n +1 cos( k π x ) for x ∈ e ( v 1 , v 2 ) or e ( v 1 , w i ) i = 1 ..n sgn(cos( k π )) √ 2 √ m + n +1 cos( k π x ) for x ∈ e ( v 2 , w i ) i = 1 ..m. The pro duct of this function with itself a t the v ertices v 1 and v 2 is Φ k ( v 1 )Φ k ( v 2 ) = 2 m + n + 1 cos( k π ) . When k = 0, w e hav e Φ 0 ( v 1 )Φ 0 ( v 2 ) = 1 m + n +1 . 55 If w e lo ok at the su m of Φ k ( v 1 )Φ k ( v 2 ) e − k 2 π 2 t , w e can write it as follo ws: ∞ X k =0 Φ k ( v 1 )Φ k ( v 2 ) e − k 2 π 2 t = 1 m + n + 1 + ∞ X k =1 2 m + n + 1 cos( k π ) e − k 2 π 2 t = 1 m + n + 1 ∞ X k = −∞ cos( k π ) e − k 2 π 2 t . The third kind are more complicated. They are of the form f ( x ) = c 1 ( e )  sin  √ λ (1 − x )  + c 2 ( e ) sin  √ λx  along each edge, where c 1 ( e ) and c 2 ( e ) will depend on the edge e ; see [24]. F or these t o b e in the domain, they mus t b e con t in uo us and ha v e zero deriv ativ e at eac h v ertex. W e hav e 2 ( m + n + 1) constan ts c i ( e ) to determine, as well as the p ossible v alues of λ . T o guara n tee contin uit y , w e need the function to ha v e the same v alue at v 1 regardless of whic h edge w e’re considering. T o get this, w e set c 1 ( e ( v 1 , w i )) = c 1 ( e ( v 1 , v 2 )) . Similarly , for con tin uity at v 2 , w e need c 1 ( e ( v 2 , w ′ i )) = c 1 ( e ( v 1 , v 2 )) c 2 ( e ( v 1 , v 2 )) . This brings us do wn to m + n + 2 diﬀeren t c i ( e ) that w e need to determine. No w w e need zero deriv ativ e at the v ertices. Note that for an edge e f ′ ( x ) = √ λc 1 ( e )  − cos  √ λ (1 − x )  + c 2 ( e ) cos  √ λx  . F or e dges of the f orm e ( v , w ), we ﬁnd that setting c 2 ( e ( v , w )) = 1 cos ( √ λ ) will al- lo w f to satisfy f ′ ( w ) = 0. F or f ′ ( v 1 ) = 0, w e need c 2 ( v 1 , v 2 ) = ( n +1) cos 2 ( √ λ ) − n cos ( √ λ ) . W e no w hav e only tw o things that can b e used to determine our function; c 1 ( e ( v 1 , v 2 )) 56 and λ . V arying the v a lue of c 1 ( e ( v 1 , v 2 )) will m ultiply the en tire function by a constan t. W e’ll need to determine λ in order to hav e f ′ ( v 2 ) = 0. F or this, w e need to solv e √ λc 1 ( e ( v 1 , v 2 ))   m ( n + 1) cos 2  √ λ  − n cos  √ λ    − cos  √ λ  + 1 cos  √ λ    + ( − 1)   − 1 + ( n + 1) cos 2  √ λ  − n cos  √ λ  cos  √ λ      = 0 . The ( − 1) comes from the fact that w e w an t the sum of the inw ard p o in t ing deriv a- tiv es a long the edges to sum to zero. W e can simplify this to: m   − ( n + 1) cos 2  √ λ  + 2 n + 1 − n cos 2  √ λ    +  n + 1 − ( n + 1) cos 2  √ λ  = 0 . This can be written as a quadratic equation in cos 2  √ λ  ; when solv ed, the o nly p ossibilit y for cos 2  √ λ  in [0 , 1] is cos 2  √ λ  = mn ( m +1)( n +1) . Finally , w e set c 1 ( e ( v 1 , v 2 )) so that | | f || 2 = 1. This holds when c 1 ( e ( v 1 , v 2 )) = √ m ( m +1) m + n +1 . Putting all of this information together, w e ﬁnd that the third ty p e of eigen- function has the form: ˜ Φ k ( x ) =                                      √ m ( m +1) m + n +1  sin  √ λ k (1 − x )  + ( n +1) cos 2 ( √ λ k ) − n cos ( √ λ k ) sin  √ λ k x   for x ∈ e ( v 1 , v 2 ) √ m ( m +1) m + n +1  sin  √ λ k (1 − x )  + 1 cos ( √ λ k ) sin  √ λ k x   for x ∈ e ( v 1 , w i ) , i = 1 ..n √ m ( m +1)(( n +1) cos 2 ( √ λ k ) − n ) cos ( √ λ k ) ( m + n +1)  sin  √ λ k (1 − x )  + 1 cos ( √ λ k ) sin  √ λ k x   for x ∈ e ( v 2 , w ′ i ) , i = 1 ..m. 57 These eigenfunctions corresp ond to eigen v alues λ k , where cos 2  √ λ k  = mn ( m +1)( n +1) . Let √ λ 0 b e the square ro ot of the eigenv alue in (0 , π 2 ). All other √ λ are of the form √ λ 0 + k π and − √ λ 0 + k π , where k is a p o sitiv e in teger. When k is ev en, the cosine is p ositiv e; for k odd, cosine is negativ e. When w e lo ok at the product of ˜ Φ k ( v 1 ) and ˜ Φ k ( v 2 ), this expression simpliﬁes greatly . ˜ Φ k ( v 1 ) ˜ Φ k ( v 2 ) = − sgn  cos  p λ k  √ mn p ( m + 1)( n + 1)( m + n + 1) . W e then know that the sum of ˜ Φ k ( v 1 ) ˜ Φ k ( v 2 ) e − λ 2 k t is √ mn √ ( m +1)( n +1)( m + n +1) m ulti- plied b y the following: ∞ X k =0 − sgn  cos  p λ k  e − ( λ k ) 2 t = ∞ X k =0 e − ((2 k +1) π − √ λ 0 ) 2 t − e − ((2 k +2) π − √ λ 0 ) 2 t − e − (2 kπ + √ λ 0 ) 2 t + e − ((2 k +1) π + √ λ 0 ) 2 t = ∞ X k =0 e − ( √ λ 0 − (2 k +1) π ) 2 t − e − ( √ λ 0 − (2 k +2) π ) 2 t − e − ( √ λ 0 +2 k π ) 2 t + e − ( √ λ 0 +(2 k +1) π ) 2 t = ∞ X k = −∞ e − ( √ λ 0 − (2 k +1) π ) 2 t − e − ( √ λ 0 − 2 k π ) 2 t . W e can sum Φ( v 1 )Φ( v 2 ) e − λ 2 t o v er all of t he eigenfunctions to ﬁnd an explicit ex- pression for the heat k ernel at ( v 1 , v 2 ). h t ( v 1 , v 2 ) = 1 m + n + 1 ∞ X k = −∞ e − k 2 π 2 t cos( k π ) + r mn ( m + 1)( n + 1) ∞ X k = −∞ e − ( √ λ 0 − (2 k +1) π ) 2 t − ∞ X k = −∞ e − ( √ λ 0 − 2 k π ) 2 t !! . W e can rewrite this using Jacobi’s identit y (see D ym [10]). The iden tit y is: ∞ X k = −∞ e − ( x − k ) 2 2 s = √ 2 π s ∞ X k = −∞ e − 2 π 2 k 2 s e 2 π ik x . 58 Because the left hand side is a real n um b er, w e can r ewrite the e 2 π ik x on the righ t hand side to obta in: ∞ X k = −∞ e − ( x − k ) 2 2 s = √ 2 π s ∞ X k = −∞ e − 2 π 2 k 2 s cos(2 π k x ) . W e can apply this iden tity to the three series . F or the ﬁrs t, w e use the reve rse of the equalit y with x = 1 2 and s = t 2 in the righ t hand side. This giv es us: ∞ X k = −∞ e − k 2 π 2 t cos( k π ) = 1 √ π t ∞ X k = −∞ e − (1 − 2 k ) 2 4 t . In the second, let x = √ λ 0 + π 2 π and s = 1 8 π 2 t . This giv es us ( x − k ) 2 2 s = ( √ λ 0 + π − 2 π k ) 2 (2 π ) 2 2 1 8 π 2 t = ( √ λ 0 + (1 − 2 k ) π ) 2 t . When we put that into the iden tit y , w e ha v e: ∞ X k = −∞ e − ( √ λ 0 +(1 − 2 k ) π ) 2 t = 1 2 √ π t ∞ X k = −∞ e − k 2 4 t cos  k  p λ 0 + π  . Note that P ∞ k = −∞ e − ( √ λ 0 − (2 k +1) π ) 2 t = P ∞ k = −∞ e − ( √ λ 0 +(1 − 2 k ) π ) 2 t b y reindexing, and so this is a wa y of rewriting our original sum. F or the third, w e use x = √ λ 0 2 π and s = 1 8 π 2 t whic h, b y a similar computatio n, giv es us: ∞ X k = −∞ e − ( √ λ 0 − 2 k π ) 2 t = 1 2 √ π t ∞ X k = −∞ e − k 2 4 t cos  k p λ 0  . W e can com bine these three to rewrite h t ( v 1 , v 2 ) as follo ws: h t ( v 1 , v 2 ) = 1 m + n + 1 1 √ π t ∞ X k = −∞ e − (1 − 2 k ) 2 4 t + r mn ( m + 1)( n + 1) 1 2 ∞ X k = −∞ e − k 2 4 t  cos  k p λ 0 + k π  − cos  k p λ 0  ! . W e can use the a ngle sum form ula for cosines to see that cos  k p λ 0 + k π  = sin  k p λ 0  sin ( k π ) + cos  k p λ 0  cos ( k π ) = cos  k p λ 0  cos ( k π ) . 59 As cos ( k π ) = 1 for k ev en and − 1 f or k o dd, t he term cos  k √ λ 0 + k π  − cos  k √ λ 0  reduces to 0 for ev en k and − 2 cos  k √ λ 0  for o dd k . This allo ws us to write: h t ( v 1 , v 2 ) = 1 m + n + 1 1 √ π t ∞ X k = −∞ e − (2 k +1) 2 4 t + r mn ( m + 1)( n + 1) 1 2 ∞ X k = −∞ e − (2 k +1) 2 4 t ( − 2) cos  (2 k + 1) p λ 0  ! = 1 m + n + 1 1 √ π t ∞ X k = −∞ e − (2 k +1) 2 4 t 1 − √ mn cos  (2 k + 1) √ λ 0  p ( m + 1)( n + 1) ! . When w e consider the limit as t approac hes 0, the k = 0 and k = − 1 terms will dominate. By symmetry , thes e terms are equal. This giv es us the appro ximation: h t ( v 1 , v 2 ) ≈ 1 m + n + 1 2 √ π t e − 1 4 t 1 − √ mn cos  √ λ 0  p ( m + 1)( n + 1) ! = 1 m + n + 1 2 √ π t e − 1 4 t   1 − √ mn q mn ( m +1)( n +1) p ( m + 1)( n + 1)   = 1 ( m + 1)( n + 1) 2 √ π t e − 1 4 t . Note that when m = n = 0 w e hav e a line; this is the correct asymptotic there. Similarly , when m = n = 1, w e hav e a line of length 3. This w o rks out there to o. W e can also ﬁnd the on- diagonal asymptotic at v 1 , one of the cen tr al v ertices. Here w e hav e Φ 2 k ( v 1 ) = 2 m + n + 1 and ˜ Φ 2 k ( v 1 ) = p m ( m + 1) √ m + n + 1 ! sin 2  p λ k  = m ( m + n + 1)( n + 1) . 60 Using the same st yle of manipulations as b efore w e ﬁnd that h t ( v 1 , v 1 ) is h t ( v 1 , v 1 ) = 1 m + n + 1 ∞ X k = −∞ e − k 2 π 2 t + m ( m + n + 1)( n + 1) ∞ X k = −∞ e − ( λ 0 − k π ) 2 t . W e c an use a Jacobi transform here with x = 0 , t = 2 s in the ﬁrst sum and x = √ λ 0 π and t = 1 2 π 2 s in the second to obtain: h t ( v 1 , v 1 ) = 1 ( m + n + 1) √ π t ∞ X k = −∞  1 + m n + 1 cos(2 k p λ 0 )  e − k 2 t . As t approache s zero, the k 6 = 0 terms also approach zero. The dominant k = 0 term is 1 ( m + n + 1) √ π t  1 + m n + 1  = 1 ( n + 1) √ π t . This is the same asymptotic as w e hav e o n a single star whose center connects to n + 1 edges. This is not surprising, since small v alues of t correspo nd to the lo cal geometry of a space. Example 3.2.6. The heat k ernel on an inte rv al is messier than that on a line. Dym [10] calculates it for ∂ ∂ t u − 1 2 ∆ u = 0 on the interv a l [0 , 1] to be: p t ( x, y ) = 1 + 2 ∞ X n =1 e − n 2 π 2 t/ 2 cos( nπ x ) cos( nπ y ) = 1 √ 2 π t ∞ X n = −∞ e − ( x − y − 2 n ) 2 2 t + e − ( x + y − 2 n ) 2 2 t . W e would like to hav e the heat k ernel for ∂ ∂ t ˜ u − ∆ ˜ u = 0 on [0 , L ]. W e can ﬁnd it b y noting that if u ( t, x ) is a solution to ∂ ∂ t u − 1 2 ∆ u = 0 on the in terv al [0 , 1], then ˜ u ( t, x ) = cu ( at, bx ) has deriv ativ es  ∂ ∂ t ˜ u  ( t, x ) =  ca ∂ ∂ t u  ( at, bx ) and (∆ ˜ u )( t, x ) = ( cb 2 ∆ u )( at, bx ) =  2 cb 2 ∂ ∂ t u  ( at, bx ) . 61 Then ( ∂ ∂ t ˜ u )( t, x ) − (∆ ˜ u )( t, x ) = 0 when a = 2 b 2 . W e change the interv a l length b y setting b = 1 L . This give s us ˜ p t ( x, y ) = cp 2 t L 2 ( x L , y L ). The constant c is use d to normalize so that R L 0 ˜ p t ( x, y ) d x = 1 holds for eac h t > 0. Since this integral is R L 0 cp 2 t L 2 ( x L , y L ) dx = R 1 0 cp 2 t L 2 ( z , y L ) Ldz = cL , w e hav e c = 1 L . W e can write this as ˜ p t ( x, y ) = 1 √ 4 π t ∞ X n = −∞ e − ( x − y − 2 nL ) 2 4 t + e − ( x + y − 2 nL ) 2 4 t . This description allo ws one to lo ok a t terms with n near 0 to ﬁnd the small t ime asymptotic. Note that for L = 3, x = 1 and y = 2 we hav e: ˜ p t (1 , 2) = 1 √ 4 π t ∞ X n = −∞ e − ( − 1 − 6 n ) 2 4 t + e − (3 − 6 n ) 2 4 t . When t is s mall, this b eha v es lik e the ﬁrst n = 0 term: ˜ p t (1 , 2) ≈ 1 2 √ π t e − 1 4 t . This is consisten t with the asymptotic for the t wo star case when m = n = 1. Note also that when L = 1, x = 0, y = 1 t hat this giv es 2 √ π t e − 1 4 t whic h is consisten t with the star with one e dge. Figure 3.3: A subset of the three dimensional grid. Example 3.2.7. Complex es c an hav e an underlying group structure. F or example, Z 3 , the group consisting of triplets of in tegers, can b e used to create a 3 dimensional 62 complex b y connecting a cub e, [0 , 1] 3 , to itself w here eac h pair o f triples whic h diﬀer b y one b y a line segmen t. This grid is the 1- sk eleton, and the p oin ts in the group form a 0-sk eleton. W e can use the 1- sk eleton to create a space that lo oks lik e a bunc h of empty b oxe s b y ﬁlling in the faces formed b y lo ops of four edges. This is the 2-sk eleton. If w e then ﬁll in the b o xes in the 2-sk eleton, w e’ll ha v e a 3-sk eleton, whic h is R 3 . Lo cally , w e’v e show n that if h t is the heat k ernel on the k -sk eleton, 1 C t k / 2 ≤ h t ( x, x ) ≤ C (min( t, R 2 0 )) k / 2 for all x ∈ X and all t suc h that 0 < t < R 2 0 . W e prov e in c hapter 5 that the heat k ernel on each of these k-sk eletons globa lly b eha v es lik e the heat k ernel on R 3 . Example 3.2.8. There a re also complexes with underlying gro up structure whose geometry is no t globally Euclide an. Let G b e the free gro up on t w o eleme n t s; this is a gro up of w ords formed by letters a a nd b a nd their inv erses a − 1 and b − 1 where the only cancellations are aa − 1 = a − 1 a = 1 and bb − 1 = b − 1 b = 1, a nd a and b don’t comm ute. Let Y b e the complex f ormed by three squares joined in to an L shap e (see F ig. 3 .4). W e ha ve a larger structure X whic h has copies of Y connected to eac h ot her via the group G . That is, eac h cop y of Y will b e connected to four other copies of Y ; the top of the L connects to the b otto m edge o f the lo w er left square of the L, and the right of t he L connects to the left edge of the b ottom square. Note that w e can not isometrically em b ed X into R 2 . The stretc hing of the edges in Fig. 3.4 is to allo w y ou to see distinct edges and v ertices. This structure globally it acts lik e a hy p erb olic space, but lo cally it is Euclidean. F or a small ball with R < . 5, we hav e a t w o dimensional circle (possibly missing 63 Figure 3.4: Y (left); Y (dark er shading) and it s fo ur surrounding copies (lighter shading) (righ t) Eac h of the edges in these picture s should be in terpreted as having length 1. a w edge) whose v olume is ≈ π R 2 . F o r t < . 5 , corollary 3.1.6 tells us: h t ( x, x ) ≈ C (4 π t ) . F or a g iv en copy of Y , there are four neighbors, eac h of whic h ha s three additional neighbors. F or a ball o f radius R > 1, there will b e appro ximately 1 + 4 + 4(3) + · · · + 4(3 R ) ≈ 2(3 R +1 ) copies of Y . This tells us that for large R , w e hav e exp onen tial v olume growth. In particular, this group is nonamenable. W e deﬁne t his in c hapter 5 and sho w that the large time b ehavior o f the heat k ernel is sup x ∈ X h t ( x, x ) ≈ C 0 e − t/C 1 . Chapter 4 Setup for Groups A ﬁnite pro duct of elemen ts from a set S is called a word. If a w or d is written s 1 s 2 ...s k , w e say it has length k . A ﬁnitely generated gro up is a gr oup with a generating set, S , where eve ry elemen t in the group can b e written as a ﬁnite w ord using elemen ts of S . Although for a giv en g ∈ G it is computationa lly diﬃcult to determine whic h w ord is the smallest one represen ting g , suc h a w ord (or w ords) ex ists. If this word has len gth k , then w e write | g | = k . W e deﬁne the v olume of a subset o f G to b e the n umber of elemen ts of G con t ained in that subset. W e write | B r | to denote t he v olume of a ball o f radius r , B r := { g ∈ G : | g | ≤ r } . F or groups, v olume is translation in v arian t, a nd so w e do not lose an y generality b y ha ving it cen tered at the iden tit y . F or a function f whic h maps elemen ts of a group to the reals, we deﬁne the Diric hlet form on ℓ 2 ( G ) to be E ( f , f ) = 1 | S | P g ∈ G P s ∈ S | f ( g ) − f ( g s ) | 2 . Although we’d need to sp ecify directions if we w ere to deﬁne a gra dien t, w e can deﬁne an ob ject whic h b eha ves lik e the length o f the gradient of f on G . W e write this as |∇ f ( x ) | = q 1 | S | P s ∈ S | f ( x ) − f ( xs ) | 2 . Notationally , this means tha t E ( f , f ) = P g ∈ G |∇ f ( g ) | 2 . Discrete L p norms restricted to a subs et A ⊂ G are written as || f || p,A =  P x ∈ A | f ( x ) | p  1 /p . When A = G , w e will write || f || p,G . One can show a P oincar´ e type inequality on a volume doubling ﬁnitely gener- ated group. The arguments used in this can b e found in [7]. Lemma 4.0.9. L et G b e a ﬁnitely gener ate d gr oup with gener ating set S . F or an y 64 65 f : G → R , the fol low ing ine quality holds on b al ls B r : k f − f B r k 1 ,B r ≤ | B 2 r | | B r | 2 r p | S |k∇ f k 1 ,B 3 r . If the gr oup is volume doubling, this is a we ak Poinc ar´ e ine quality on b al ls for p = 1 : k f − f B r k 1 ,B r ≤ 2 r C D oubling p | S |k∇ f k 1 ,B 3 r . Pr o of. Let G be a ﬁnitely g enerated group with a symmetric set of g enerators, S . Let B r b e a ball of ra dius r ; for brevit y , w e will not explicitly w rite the cen ter. W e can write the norm of f min us its a verage as follo ws. k f − f B r k 1 ,B r = X x ∈ B r | f ( x ) − 1 | B r | X y ∈ B r f ( y ) | ≤ 1 | B r | X x ∈ B r X y ∈ B r | f ( x ) − f ( y ) | . F or eac h y ∈ B r , there exis ts a g ∈ G with | g | ≤ 2 r suc h that y = xg . W e make this substitution and sum o v er all g ∈ G with | g | ≤ 2 r . 1 | B r | X x ∈ B r X y ∈ B r | f ( x ) − f ( y ) | ≤ 1 | B r | X x ∈ B r X g : | g |≤ 2 r | f ( x ) − f ( xg ) | = 1 | B r | X g : | g |≤ 2 r X x ∈ B r | f ( x ) − f ( xg ) | . W e will b egin with the innermost quantit y , and then simplify t he sums. W e can write g = s 1 ..s k as a reduced word with k ≤ 2 r . W e rewrite the diﬀerence o f f at x and xg b y splitting the path b et w een t hem in to pieces. | f ( x ) − f ( xg ) | ≤ | g | X i =1 | f ( xs 1 ...s i − 1 ) − f ( xs 1 ...s i ) | . W e ﬁx g and sum ov er all x ∈ B r . X x ∈ B r | f ( x ) − f ( xg ) | ≤ X x ∈ B r | g | X i =1 | f ( xs 1 ...s i − 1 ) − f ( xs 1 ...s i ) | = | g | X i =1 X x ∈ B r | f ( xs 1 ...s i − 1 ) − f ( xs 1 ...s i ) | . 66 W e can change v aria bles by letting z = xs 1 ...s i − 1 . Not e that | xs 1 ..s i − 1 | < 3 r as i ≤ 2 r . Then z ∈ B 3 r . | g | X i =1 X x ∈ B r | f ( xs 1 ...s i − 1 ) − f ( xs 1 ...s i ) | ≤ | g | X i =1 X z ∈ B 3 r | f ( z ) − f ( z s i ) | . Since s i ∈ S , w e can sum o v er all s ∈ S instead of the s i in g . T o do this, w e m ust accoun t for the multiplicit y of the s i . W e could hav e at most | g | copies of an y generator; | g | ≤ 2 r , and so w e will multiply b y 2 r . | g | X i =1 X z ∈ B 3 r | f ( z ) − f ( z s i ) | ≤ 2 r X s ∈ S X z ∈ B 3 r | f ( z ) − f ( z s ) | . Jensen’s inequalit y allo ws us to rewrite this in terms of the gradient. 2 r X s ∈ S X z ∈ B 3 r | f ( z ) − f ( z s ) | ≤ 2 r X z ∈ B 3 r s | S | 1 | S | X s ∈ S | f ( z ) − f ( z s ) | 2 = 2 r X z ∈ B 3 r p | S ||∇ f ( z ) | . W e’ll use this calculation to get the desired inequality . W e no w ha ve X x ∈ B r | f ( x ) − f ( xg ) | ≤ 2 r X z ∈ B 3 r p | S ||∇ f ( z ) | . Dividing b y | B r | and summing o ver the g giv es us 1 | B r | X g : | g |≤ 2 r X x ∈ B r | f ( x ) − f ( xg ) | ≤ 1 | B r | X g : | g |≤ 2 r X z ∈ B 3 r 2 r p | S ||∇ f ( z ) | = | B 2 r | | B r | 2 r p | S | X z ∈ B 3 r |∇ f ( z ) | . This reduces to k f − f B r k 1 ,B r ≤ | B 2 r | | B r | 2 r p | S |k∇ f k 1 ,B 3 r . Note that in g eneral, | B 2 r | | B r | will dep end on the radius, r . If the group is volume doubling, this giv es us a w eak P o incar ´ e inequalit y . k f − f B r k 1 ,B r ≤ C D oubling 2 r p | S |k∇ f k 1 ,B 3 r . 67 4.1 Comparing distances in X and G Let X b e a complex, and G b e a ﬁnitely generated g roup of isomorphisms on the complex suc h that X/G = Y is an admiss ible complex c onsisting of a ﬁnite n umber of p olytop es. One example of this type of complex is a Cay ley g raph; this is the graph where eac h v ertex corres p onds to a group eleme n t, a nd tw o v ertices are connecte d b y an edge if they diﬀer b y an elemen t of the generating set. In this case, Y is the unit in terv al. W e w ould lik e to b e able to compare functions deﬁned on the group, G , with functions deﬁned on the complex, X . T o do this, w e will lo ok at wa ys to transfer a function deﬁned on G to a function deﬁne d on X that roughly preserv es t he norm of both the function and its energy form. W e seek to do the rev erse a s w ell. W e will use a tec hnique that originated with Kanai [19] and additionally w as used b y Coulhon and Saloﬀ-Coste [8]. W e also w an t a w a y of c hanging from r eal v alued functions whic h take v alues in X to ones that tak e v alues in G . W e will do this b y taking a cop y of Y and splitting it into man y smaller pieces. G iv en δ ≤ diam( Y ), w e can ﬁnd a ﬁnite co v ering o f Y by balls of radius δ suc h tha t balls of radius δ / 2 are disjoin t in Y . As Y is a ﬁnite p olytopal comple x, v olume doubling on Y implies that at most a ﬁnite n um b er of balls of radius δ will o v erlap. X can b e written b y taking a copy of Y for eac h elemen t of G , and so this cov er can b e expanded to a cov er o f X . Note that once w e hav e a copy of the co v er of Y for eac h elemen t of G , balls of radius δ / 2 in this larger cov er ma y o v erlap. F or X , the o v erlap is also ﬁnite; call the n umber o f o ve rlapping balls C N ,S . Call the cen ters of the ba lls co v ering Y { γ i } N i =1 and the ba lls co v ering X 68 { g γ i } i =1 ..N ; g ∈ G . Note that eac h x ∈ X is within δ of at least one of the g γ i . As w e are frequen tly switc hing b et w een X and G , w e will use B X for balls in X and B G for balls in G . Figure 4.1: X split into copies of Y(left);a copy o f Y cov ered b y 5 balls of radius .6 (ce n ter); a co ver for Y shifted b y ﬁv e copies of G co v er ev erything – the blac k lines represen t fo ur copies that o verlap exactly; the ﬁfth copy is g ra y(righ t ). Example 4.1.1. Let X = R 2 , G = Z 2 and Y = [0 , 1] 2 . F o r δ = . 6 , balls of radius δ / 2 = . 3 ce n t ered at the corners of Y are disjoint, but not all points in the plane are cov ered. W e can in tro duce another copy of G that’s shifted b y ( . 5 , . 5). All p oin ts in Y are co v ered b y some ball of radius delta, but the balls of radius . 3 will not o v erlap. W e can c heck this b y comparing the distance along the diagonal from (0 , 0) to (1 , 1) with the length cov ered b y the radii along the same diagonal. W e hav e d ((0 , 0) , (1 , 1)) = √ 2 ≈ 1 . 4. F or the balls, there’s the ra dius of the one cen tered at (0 , 0), the one cente red at (1 , 1) a nd the diameter of the one cen tered at ( . 5 , . 5 ). This sums to 4 δ = 1 . 2. The n umber of ov erlapping balls in X is C N ,S = 10. This happ ens at (0 , . 5) where there are four balls cen tered at (0 , 0), one centere d at ( . 5 , . 5), one centere d at ( − . 5 , . 5 ), and four cen tered at (0 , 1). In this example, γ 1 = ( 0 , 0), γ 2 = (0 , 1), γ 3 = (1 , 0), γ 4 = (1 , 1), and γ 5 = ( . 5 , . 5). 69 W e deﬁne group f ( g , i ) = 1 µ ( B X ( g γ i , δ )) Z B X ( gγ i ,δ ) f ( x ) dx = − Z B X ( gγ i ,δ ) f ( x ) dx. W e can view group f as a collection of N functions deﬁned on G . F or an y ﬁxed i , w e can treat gr oup f ( · , i ) as a function on the group, and so t he norm of gro up f can b e found b y summing these o v er i . It’s impo rtan t to no te that the sets { g : g ∈ B X ( r ) ∩ G } and B G ( r ) are p oten- tially diﬀeren t. W e can describe this by comparing distances with some explicit constan ts. Lemma 4.1.2. We c an c omp ar e distanc es in G and X in the fol low ing manner. Ther e ex i s t c on stants C X G and C 0 so that for any g , h ∈ G we have: 1 C 0 d X ( g , h ) ≤ d G ( g , h ) ≤ C X G d X ( g , h ) . This tel ls us that b al ls c enter e d at p oints in G c omp ar e as: G ∩ B X ( r C X G ) ⊂ B G ( r ) ⊂ G ∩ B X ( C 0 r ) . Her e, C 0 = max v 1 ,v 2 ∈ Y (0) d Y (1) ( v 1 , v 2 ) a nd C X G = 1 min v 1 ,v 2 ∈ Y (0) d X (1) ( v 1 ,v 2 ) Q n i =1 max  q 2 1 − cos( α ) , max y 1 ,y 2 ∈ Y ( i − 1) d X ( i − 1) ( y 1 ,y 2 ) min v 1 ,v 2 ∈ Y (0) d X (1) ( v 1 ,v 2 )  wher e α is the smal le s t interior angle in X . Pr o of. W e’ll start with the easy direction. An y path in X (1) is also a pat h in X , and so d X ( g , h ) ≤ d X (1) ( g , h ). T o compare this with distances in G , whic h coun t n um b ers o f p oin ts in paths, w e use lengths of edges betw een them. d X (1) ( g , h ) ≤ max v 1 ,v 2 ∈ Y (0) d Y (1) ( v 1 , v 2 ) d G ( g , h ) . Set C 0 = max v 1 ,v 2 ∈ Y (0) d Y (1) ( v 1 , v 2 ). Since whenev er C 0 d G ( g , h ) ≤ r w e also ha v e d X ( g , h ) ≤ r , w e know that an y p oint in B G ( r C 0 ) is also in B X ( r ). This tells us that B G ( r ) ⊂ G ∩ B X ( C 0 r ). 70 W e can use the fact that X can b e sub divided into copies of Y to relate distances in the other direction as w ell. W e will compare distances in the diﬀeren t sk eletons of X . In order to simplify notation, d k ( x, y ) will refer to the distance b et wee n x and y whe n w e restrict to paths in X ( k ) . Let g , h ∈ G b e give n. Then there is a shortest path in X b et w een them. If there are multiple suc h paths, pic k one. Lab el it { x 0 = g , x 1 , x 2 , .., x k = h } where x i ∈ X ( n − 1) for i = 1 ..k − 1, and x i , x i +1 are b o th in the b oundary of the same maximal p olytop e, although they are on diﬀeren t faces. In particular, b oth are in X ( n − 1) . Then w e kno w that the length of this path is P k i =1 d n ( x i − 1 , x i ). W e will compare d n ( x i − 1 , x i ) with d n − 1 ( x i − 1 , x i ), and use this to relate the distances betw een sk eletons who diﬀer b y 1 dime nsion. This will allow us to w ork our w ay from n dimensions down to X (0) . Then w e can compare X (0) with G . Lo ok a t x i and x i +1 whic h are in ( n − 1) dimens ional f aces F i and F i +1 . Either F i ∩ F i +1 is nonempt y and so they share a lo w er dimensional p o in t , or else it is empt y and they do not. If they do not s hare a p oin t, then: min v 1 ,v 2 ∈ Y (0) d 1 ( v 1 , v 2 ) ≤ d n ( x i , x i +1 ) . W e get t his b ound b ecause the diameter of an y subp o lyhedra of Y must b e b o unded b elo w b y the length of the smallest edge of that polyhedra. W e can also b ound the ( n − 1) distance: d n − 1 ( x i , x i +1 ) ≤ max y 1 ,y 2 ∈ Y ( n − 1) d n − 1 ( y 1 , y 2 ) . Putting this together, w e get: d n − 1 ( x i , x i +1 ) ≤ max y 1 ,y 2 ∈ Y ( n − 1) d n − 1 ( y 1 , y 2 ) min v 1 ,v 2 ∈ Y (0) d 1 ( v 1 , v 2 ) d n ( x i , x i +1 ) . 71 Otherwise, if F i and F i +1 in tersec t in a low er dimensional fa ce, we will call v the p oint on the in tersection whic h minimizes d n − 1 ( v , x i ) + d n − 1 ( v , x i +1 ). These three p oin ts form a triangle with angle x i v x i +1 = θ ≥ α , where α is t he smalles t in terior angle in Y as w ell as in X . Note that this a ngle is b ounded b ecause Y is made up of a ﬁn ite n um b er of polytop es. W e w ould lik e to determine a relationship b et w een d n ( x i , x i +1 ) and inf v ∈ F i ∩ F i +1 d n − 1 ( v , x i ) + d n − 1 ( v , x i +1 ). T o ﬁnd this, w e will use a simple deriv atio n. F or p ositiv e n um b ers a a nd b w e ha v e ( a − b ) 2 ≥ − cos( α )( a − b ) 2 a 2 + b 2 − 2 ab cos( α ) ≥ 2 ab − a 2 cos( α ) − b 2 cos( α ) 2 a 2 + 2 b 2 − 4 ab cos( α ) ≥ a 2 + b 2 + 2 ab − 2 ab cos( α ) − a 2 cos( α ) − b 2 cos( α ) a 2 + b 2 − 2 ab cos( α ) ≥ ( a + b ) 2  1 − cos( α ) 2  . This is helpful b ecause when w e apply the law of cosines to the triangle w e ha v e: d 2 n ( x i , x i +1 ) = d 2 n − 1 ( x i , v ) + d 2 n − 1 ( v , x i +1 ) − 2 d n − 1 ( x i , v ) d n − 1 ( v , x i +1 ) cos( θ ) . W e can form an inequalit y by replacing cos ( θ ) with the larger cos( α ): d 2 n ( x i , x i +1 ) ≥ d 2 n − 1 ( x i , v ) + d 2 n − 1 ( v , x i +1 ) − 2 d n − 1 ( x i , v ) d X ( n − 1) ( v , x i +1 ) cos( α ) . Then w e can apply our fact with a = d n − 1 ( x i , v ) and b = d n − 1 ( v , x i +1 ). d 2 n ( x i , x i +1 ) ≥ ( d n − 1 ( x i , v ) + d n − 1 ( v , x i +1 )) 2  1 − cos ( α ) 2  . This leads us to the conc lusion that d n ( x i , x i +1 ) ≥ ( d n − 1 ( x i , v ) + d n − 1 ( v , x i +1 )) r 1 − cos( α ) 2 ≥ d n − 1 ( x i , x i +1 ) r 1 − cos ( α ) 2 . 72 When w e combine the cases where faces in tersect with the cas e where they do not, w e get the following ineq ualit y: d n − 1 ( x i , x i +1 ) ≤ max s 2 1 − cos ( α ) , max y 1 ,y 2 ∈ Y ( n − 1) d n − 1 ( y 1 , y 2 ) min v 1 ,v 2 ∈ Y (0) d 1 ( v 1 , v 2 ) ! d n ( x i , x i +1 ) . W e can sum and use the fact that w e had a distance minimizing path in X ( n ) to get d n − 1 ( g , h ) ≤ max s 2 1 − cos ( α ) , max y 1 ,y 2 ∈ Y ( n − 1) d n − 1 ( y 1 , y 2 ) min v 1 ,v 2 ∈ Y (0) d 1 ( v 1 , v 2 ) ! d n ( g , h ) . W e can rep eat this argument for the lo wer dimensions (do wn to dimension 1) to get: d 1 ( g , h ) ≤ n Y i =1 max s 2 1 − cos ( α ) , max y 1 ,y 2 ∈ Y ( i − 1) d i − 1 ( y 1 , y 2 ) min v 1 ,v 2 ∈ Y (0) d 1 ( v 1 , v 2 ) ! d X ( n ) ( g , h ) . T o compare with the distance in G , w e see that min v 1 ,v 2 ∈ Y (0) d 1 ( v 1 , v 2 ) d G ( g , h ) ≤ d 1 ( g , h ) . W e will deﬁne C X G to b e C X G = 1 min v 1 ,v 2 ∈ Y (0) d 1 ( v 1 , v 2 ) n Y i =1 max s 2 1 − cos( α ) , max y 1 ,y 2 ∈ Y ( i − 1) d i − 1 ( y 1 , y 2 ) min v 1 ,v 2 ∈ Y (0) d 1 ( v 1 , v 2 ) ! . This giv es us the inequalit y: d G ( g , h ) ≤ C X G d X ( g , h ) . The inequalit y implies the con tainmen t G ∩ B X ( r C X G ) ⊂ B G ( r ) by the argumen t from the start of this pro of. 4.2 Comparing functions on X with corresp on d ing ones on G W e can compare the norm of f with the norm of group f , a s w ell as the norm of ∇ f with that o f it s analogue. Note that giv en a radius, R , Corollary 2.2.9 tells us 73 that w e hav e a uniform P oincar ´ e inequalit y for f on balls of radius at most R . This will b e helpful for our comparison. In particular, w e will use t his where C P is the constan t asso ciated to t he Poinc ar ´ e inequalit y for balls of radius up t o 3 diam( Y ). Note that if w e to ok δ = diam( Y ), w e could co v er Y with exactly one ball. Lemma 4.2.1. L et B X ( r ) := B X ( g ′ , r ) b e a b al l in X c enter e d at g ′ ∈ G . F or any c ∈ R , we c an c omp ar e f : X → R with group f : G N → R in the fol lowing manner: || f − c || p p,B X ( r ) ≤ C  δ p ||∇ f || p p,B X ( r +2 δ ) + || gro up f − c || p p,B G ( C X G ( r + δ +diam ( Y )))  . When r = ∞ , this says that: || f − c || p p,X ≤ C  δ p ||∇ f || p p,X + || gro up f − c || p p,G  . The c on stant C de p ends on X , p , and δ . Pr o of. W e b egin b y rewriting the norm using the f act that balls of r adius r + δ cen tered a t g γ i form a co ve r. || f − c || p p,B X ( r ) = Z B X ( r ) | f ( x ) − c | p dx ≤ X i X g γ i ∈ B X ( r + δ ) Z B X ( gγ i ,δ ) | f ( x ) − c | p dx. W e’ll use the fa ct that | f ( x ) − c | p ≤ 2 p | f ( x ) − g roup f ( g , i ) | p + 2 p | group f ( g , i ) − c | p to split t his in to t w o pieces. In the ﬁrst piece, w e can simplify using the lo cal P o incar ´ e inequalit y in X . X i X g γ i ∈ B X ( r + δ ) 2 p Z B X ( gγ i ,δ ) | f ( x ) − group f ( g , i ) | p dx ≤ X i X g γ i ∈ B X ( r + δ ) 2 p δ p C P Z B X ( gγ i ,δ ) |∇ f ( x ) | p dx ≤ 2 p δ p C P C N ,S Z B X ( r +2 δ ) |∇ f ( x ) | p dx. 74 In the second, w e ﬁrst not e that there is no x dep endence in the in tegrand. W e in tegrate to get the v o lume of the ball. This will b e dominated by the largest suc h v o lume. X i X g γ i ∈ B X ( r + δ ) 2 p Z B X ( gγ i ,δ ) | group f ( g , i ) − c | p dx = X i X g γ i ∈ B X ( r + δ ) 2 p µ ( B X ( g γ i , δ )) | group f ( g , i ) − c | p ≤ 2 p  max g γ i ∈ B X ( r + δ ) µ ( B X ( g γ i , δ ))  X i X g γ i ∈ B X ( r + δ ) | group f ( g , i ) − c | p This giv es us the deﬁnition o f the p norm in X . W e switc h to the norm in G using the distance comparisons from Lemma 4.1.2. ... = 2 p max g γ i ∈ B X ( r + δ ) µ ( B X ( g γ i , δ )) || gro up f − c || p p,B X ( r + δ ) ≤ 2 p max g γ i ∈ B X ( r + δ ) µ ( B X ( g γ i , δ )) || gro up f − c || p p,B G ( C X G ( r + δ +diam( Y ))) . When w e put these together w e ha ve for s ome constan t C : || f − c || p p,B X ( r ) ≤ C  δ p ||∇ f || p p,B X ( r +2 δ ) + || gro up f − c || p p,B G ( C X G ( r + δ +diam ( Y )))  . Note that the uniformity of X tells us that µ ( B X ( g γ i , δ )) can b e b ounded b y a constan t. In particular, w e use the f act that µ ( B X ( g γ i , δ )) = µ ( B X ( hγ i , δ )) for any g , h ∈ G . W e can also b ound the gradien t s of f and group f in their resp ectiv e norms. Lemma 4.2.2. L et f ∈ Lip( X ) and B G ( r ) b e given. F or 1 ≤ p < ∞ , w e have ||∇ group f || p p,B G ( r ) ≤ C ( δ ) || ∇ f | | p p,B X ( C 0 r +2 diam( Y )) . When r = ∞ , this is: ||∇ group f || p p,G ≤ C ( δ ) || ∇ f | | p p,X . 75 C ( δ ) = max γ i µ ( B X ( gγ i ,R )) µ ( B X ( gγ i ,δ )) 2 N max x ∈ X # { B X ( g , R ) | x ∈ B X ( g , R ) } C P R p . Note that this c ons tant dep ends on X , p , N , and δ . Pr o of. Here, R is a large enough radius so that for an y g and γ i b oth B X ( g γ i , δ ) and B X ( g sγ i , δ ) are co v ered b y B X ( g γ i , R ). Note that R = diam( Y ) + δ will w ork, but to remo v e the dep endence on δ , w e can tak e R = 2 diam( Y ). W e start b y explicitly writing out the gradien t a nd then mo ving the p/ 2 in to the in tegra l via Jensen. ||∇ group f || p p,B G ( r ) = X i X g ∈ B G ( r ) 1 | S | X s ∈ S | group f ( g , i ) − gro up f ( g s , i ) | 2 ! p/ 2 ≤ X i X g ∈ B G ( r ) 1 | S | X s ∈ S | group f ( g , i ) − group f ( g s, i ) | p = X i X g ∈ B G ( r ) 1 | S | X s ∈ S |− Z B X ( gγ i ,δ ) f ( x ) dx − − Z B X ( gs γ i ,δ ) f ( y ) dy | p . W e apply Jensen again; this time to the absolute v alue. ... ≤ X i X g ∈ B G ( r ) 1 | S | X s ∈ S − Z B X ( gγ i ,δ ) − Z B X ( gs γ i ,δ ) | f ( x ) − f ( y ) | p dxdy . The regularit y of the space a nd the cov er tell us µ ( B X ( g γ i , δ )) = µ ( B X ( g sγ i , δ )). ... = X i X g ∈ B G ( r ) 1 | S | X s ∈ S Z B X ( gγ i ,δ ) Z B X ( gs γ i ,δ ) | f ( x ) − f ( y ) | p dxdy µ ( B X ( g γ i , δ )) 2 . W e expand the sets we are integrating ov er to B X ( g γ i , R ). This larger set contains b oth B X ( g γ i , δ ) and B X ( g sγ i , δ ) b y construction. W e then rewrite the sum ov er S , and c ha nge one in tegral to an av erage integral. ... ≤ 1 | S | X i X g ∈ B G ( r ) X s ∈ S 1 µ ( B X ( g γ i , δ )) 2 Z B X ( gγ i ,R ) Z B X ( gγ i ,R ) | f ( x ) − f ( y ) | p dxdy = 1 | S | X i X g ∈ B G ( r ) | S | µ ( B X ( g γ i , δ )) µ ( B X ( g γ i , δ )) 2 Z B X ( gγ i ,R ) − Z B X ( gγ i ,R ) | f ( x ) − f ( y ) | p dxdy . 76 W e now apply a lo cal p Poinc ar ´ e inequalit y on X to f . The constan t for this is C P . ... ≤ X i X g ∈ B G ( r ) C P R p µ ( B X ( g γ i , R )) µ ( B X ( g γ i , δ )) 2 Z B X ( gγ i ,R ) |∇ f ( x ) | p dx. W e combine the sums and in tegral in to a single in tegral. All of the B X ( g γ i , R ) for g ∈ B G ( r ) are con tained in B X ( C 0 r + R ) by our distance comparison b et ween G and X . W e m ultiply this in tegral b y the num b er of o v erlapping balls in our sum. C M is N t imes the maxim um n um b er of balls B X ( g γ i , R ) whic h ov erlap at a point in X . ... ≤  max γ i µ ( B X ( g γ i , R )) µ ( B X ( g γ i , δ )) 2  C M C P R p Z B X ( C 0 r + R ) |∇ f ( x ) | p dx = C ( δ ) ||∇ f || p p,B X ( C 0 r + R ) . 4.3 P oincar´ e inequalities on X with un d erlying group struc- ture The b ounds in the previous section can b e used to tra nsfer inequalities b et w een X and G . W e can com bine them with the w eak P oincar ´ e ineq ualit y on G to get a n inequalit y on X . Theorem 4.3.1. L et X , a volume do ubli n g Euclide an c omplex and G , a ﬁnitely gener ate d gr oup with X/G = Y , a ﬁnite adm issible p olytop al c ompl e x b e giv en. X admits a Poinc ar´ e in e quality with uniform c onstant at al l sc a l e s. L et f ∈ L ip( X ) . F o r 1 ≤ p < ∞ , we have: inf c || f − c || p,B X ( r ) ≤ C r ||∇ f || p,B X ( r ) . 77 Note that this implie s : || f − f B X ( r ) || p,B X ( r ) ≤ 2 C r ||∇ f || p,B X ( r ) . Her e the b al ls c an b e c enter e d at any p oint in X . Pr o of. Note that w e ch ose C P so that the P oincar ´ e inequalit y holds f or balls of radius up to 3 diam( Y ). W e need to sho w that it also holds for balls of radius greater than 3 diam( Y ). Let r ≥ 3 diam( Y ) b e g iv en. T o start, w e will assume that the ce n t er of B X ( r ) is in G . Pic k δ = diam( Y ); this will force N = 1. This will allo w us to sp lit things up in suc h a w a y that we can use the w eak P oincar ´ e inequalit y on G . If w e had m ultiple copie s of G , w e w ouldn’t necessarily ha v e the same a v erage on eac h of them. By c ho osing a v alue o f c , w e obtain some thing at least as large as t he inﬁm um: inf c || f − c || 1 ,B X ( r ) ≤ || f − group f B G (2 C X G r ) || 1 ,B X ( r ) . Then w e can use our ﬁrst b ound to get || f − group f B G (2 C X G r ) || 1 ,B X ( r ) ≤ C  δ || ∇ f | | 1 ,B X ( r + δ ) + || gro up f − group f B G (2 C X G r ) || 1 ,B G ( C X G ( r + δ +diam( Y )))  ≤ C  δ || ∇ f | | 1 ,B X (1 . 5 r ) + || gro up f − gro up f B G (2 C X G r ) || 1 ,B G (2 C X G r )  . Happily , w e can apply the w eak P o incar ´ e inequalit y on groups (Lemma 4.0.9) to the second term: || group f − group f B G (2 C X G r ) || 1 ,B G (2 C X G r ) ≤ 3 r p | S |||∇ gro up f || 1 ,B G (6 C X G r ) . Then, we can use the b ound we hav e on the g radien ts to get an ineq ualit y o n ∇ f . Setting R = 2 diam( Y ) < r t ells us that B X ((6 r + R ) C 0 C X G ) ⊂ B X (7 C 0 C X G r ). ||∇ group f || 1 ,B G (6 C X G r ) ≤ C ( δ ) || ∇ f | | 1 ,B X (7 C 0 C X G r ) . 78 Com bining these, w e ha v e: inf c || f − c || 1 ,B X ( r ) ≤ C  δ || ∇ f | | 1 ,B X (1 . 5 r ) + 3 r p | S | C ( δ ) | |∇ f | | 1 ,B X (7 C 0 C X G r )  ≤ C 0 r ||∇ f || 1 ,B X (7 C 0 C X G r ) . As in lemma 2.1.9, we ha ve : || f − f B X ( r ) || 1 ,B X ( r ) ≤ 2 inf c || f − c || 1 ,B X ( r ) . If the cen ter, x , w ere not in G , there is some g ′ ∈ G suc h that the cen ter is within diam( Y ) of g ′ . That is, d X ( x, g ′ ) ≤ diam( Y ). By inclusions of balls, w e know that: inf c || f − c || 1 ,B X ( x,r ) ≤ inf c || f − c || 1 ,B X ( g ′ ,r +diam( Y )) . As r + diam( Y ) ≤ 1 . 5 r and B X ( g ′ , 7 C 0 C X G 1 . 5 r ) ⊂ B X ( x, 1 2 C 0 C X G r ), w e can switc h centers b y increasing the radius: ||∇ f || 1 ,B X ( g ′ , 7 C 0 C X G 1 . 5 r ) ≤ ||∇ f | | 1 ,B X ( x, 12 C 0 C X G r ) . This tells us that any complex X with the underlying group structure a dmits a w eak p = 1 P oincar´ e inequalit y . X is v olume doubling, and so this we ak inequalit y can b e turned into a strong p inequalit y via repeated application of a Whitney co v er, using Corollary 2.2.9. In [32] V aro p olous sho w ed that groups with polynomial growth of degree d hav e on diagonal b eha vior p 2 n ( e, e ) ≈ n − d/ 2 . W e show that a similar result holds for v o lume doubling complexes with underlying group structure. Theorem 4.3.2. Assume X is a volume doubling Euclide an c omplex and G is a ﬁnitely ge n er ate d gr oup w i th X/ G = Y , wher e Y is a ﬁnite adm i ssible p olytop al c omplex. 79 Then X satisﬁes the on diagonal he at kernel estimates: 1 C µ ( B ( x, √ t )) ≤ h t ( x, x ) ≤ C µ ( B ( x, √ t )) X also satisﬁes the oﬀ diagonal h e at k e rnel lower b ound: 1 C µ ( B ( x, √ t )) exp  − C d 2 X ( x, y ) t  ≤ h t ( x, y ) , as wel l as the upp er b ound: h t ( x, y ) ≤ C q µ ( B ( x, √ t )) µ ( B ( y , √ t )) exp  − d 2 X ( x, y ) 4 t   1 + d 2 X ( x, y ) t  n/ 2 . Pr o of. T o get the heat k ernel b ounds, apply Sturm [29] theorems 3.1.1 and 3.1.3, noting that we ’v e satisﬁed bot h v olume doubling and a P oincar ´ e inequ alit y a t all scales uniformly . 4.4 Mapping functions on G to X No w w e will lo ok at ho w to t ak e functions on G to smoot h ve rsions on X . Let f b e a function mapping G to the reals. W e’ll lo ok a t a partitio n o f unity on t he complex, X , whic h is created b y translating a smo oth function χ by g ∈ G . Then P g ∈ G χ g ( x ) = 1 . W e require the follow ing: • χ g ( x ) = 1 if d X ( x, g ) ≤ 1 4 • χ g ( x ) = 0 if d X ( x, g ) ≥ C sup • |∇ χ g ( x ) | ≤ C g W e kno w that |{ g ∈ G : d X ( g , e ) ≤ C sup }| is ﬁnite; when Y is nice and C sup = 1 this will b e | S | . W e also kno w that f or an y x ∈ X , |{ g ∈ G : χ g ( x ) 6 = 0 }| is ﬁnite. In particular, there is a unifor m b ound, C O ve r . 80 This allo ws us to deﬁne a nic e smoo th f unction, comp f ( x ), mapping X to the reals: comp f ( x ) = X g ∈ G f ( g ) χ g ( x ) . Its L p norm is comparable to that of f . Theorem 4.4.1. L et f : G → R b e given . If we limit ourselve s to a b a l l, B G ( r ) , with r adius at le ast 1, we c an c omp ar e L p norms in the fo l lowing way: C 1 || f − c || p,B G ( r − . 25 C 0 ) ≤ || comp f − c || p,B X ( r ) ≤ C 2 || f − c || p,B G ( C X G ( r + C sup )) This hold s f o r any c ∈ R . Note that when r = ∞ and c = 0 we hav e a nic e b ound on the norms: C 1 || f || p,G ≤ || comp f || p,X ≤ C 2 || f || p,G . F o r b oth of these ine qualities, C 1 = µ ( B X ( e, 1 4 )) 1 p and C 2 = C ( p − 1) /p O ve r || χ e || p,X . Pr o of. If w e limit ourselv es to a ball, B G ( r ), with radius a t least 1, w e hav e a comparison. W e ﬁrst write out the deﬁnition of the norm, and then we use the fact that for ev ery x , P g ∈ G χ g ( x ) = 1 . || comp f − c || p,B X ( r ) = Z B X ( r ) | X g ∈ G f ( g ) χ g ( x ) − c | p dx ! 1 p = Z B X ( r ) | X g ∈ G ( f ( g ) − c ) χ g ( x ) | p dx ! 1 p . F or eac h x , a t most C O ve r of the χ g ( x ) are nonzero. This allo ws us to apply a discrete v ersion of Jensen to mo v e the e xp onen t in to the sum. ... ≤ Z B X ( r ) C p − 1 O ve r X g ∈ G | f ( g ) − c | p χ g ( x ) p dx ! 1 p . 81 The only g with a nonzero χ g ( x ) will b e those within X distance C sup of a p oint in B X ( r ). W e can inte grate o v er g ∈ G ∩ B X ( r + C sup ), a nd switc h the ﬁnite in tegral and sum. ... ≤ C ( p − 1) /p O ve r   X g ∈ G ∩ B X ( r + C sup ) | f ( g ) − c | p Z B X ( r ) χ g ( x ) p dx   1 p . The quan tity R B X ( r ) χ g ( x ) p will b e b ounded ab o v e b y R X χ g ( x ) p = R X χ e ( x ) p . ... ≤ C ( p − 1) /p O ve r || χ e || p,X   X g ∈ G ∩ B X ( r + C sup ) | f ( g ) − c | p   1 p . W e then use the dis tance comparisons from lemma 4.1.2 to get a norm with respect to distance in G . ... ≤ C ( p − 1) /p O ve r || χ e || p,X || f − c || p,B G ( C X G ( r + C sup )) . No w we will sho w the other inequality . By deﬁn ition, w e can write the norm in G as: || f − c || p,B G ( r ) =   X g ∈ B G ( r ) | f ( g ) − c | p   1 p . W e intro duce χ g b y noting χ g ( x ) = 1 for x in B X ( g , 1 4 ), a nd integrating ov er t his set. Due to the regularit y of X , µ ( B X ( g , 1 4 )) do es not dep end on g , and so w e write it as µ ( B X ( e, 1 4 )). ... =   X g ∈ B G ( r ) 1 µ ( B X ( e, 1 4 )) Z B X ( g, 1 4 ) | f ( g ) − c | p χ g ( x ) dx !   1 p . W e no w will switc h the in tegral and the sum. W e are in tegrating only ov er x in ba lls cen tered at p oints in B G ( r ) of radius 1 / 4. This set can b e written ∪ h ∈ B G ( r ) B X ( h, 1 4 ). ... ≤ 1 µ ( B X ( e, 1 4 )) Z ∪ h ∈ B G ( r ) B X ( h, 1 4 ) X g ∈ G | f ( g ) − c | p χ g ( x ) dx ! 1 p . 82 F or x in ∪ h ∈ B G ( r ) B X ( h, 1 4 ), χ g ( x ) = 1 for exactly one g ∈ G , and it is z ero o therwise. This tells us P g ∈ G | f ( g ) − c | p χ g ( x ) = | P g ∈ G ( f ( g ) − c ) χ g ( x ) | p . W e then can write the ab o v e as ... = 1 µ ( B X ( e, 1 4 )) Z ∪ h ∈ B G ( r ) B X ( h, 1 4 ) | X g ∈ G ( f ( g ) − c ) χ g ( x ) | p dx ! 1 p . W e us e the distance comparisons from L emma 4.1.2 to see that ∪ h ∈ B G ( r ) B X ( h, 1 4 ) ⊂ B X ( C 0 r + . 25). ... ≤ 1 µ ( B X ( e, 1 4 )) Z B X ( C 0 r + . 25) | X g ∈ G ( f ( g ) − c ) χ g ( x ) | p dx ! 1 p . No w w e rewrite this using the fact that P g ∈ G χ g ( x ) = 1. ... = 1 µ ( B X ( e, 1 4 )) Z B X ( C 0 r + . 25) | X g ∈ G f ( g ) χ g ( x ) − c | p dx ! 1 p =  1 µ ( B X ( e, 1 4 ))  1 p || comp f − c | | p,B X ( C 0 r + . 25) . W e’d also lik e to compare the norms of the gradien ts. T o do this, w e w a n t to write the gradient in suc h a wa y that w e can compare it with the one on G . W e ﬁrst note that: ∇ X g ∈ G χ g ( x ) ! = ∇ 1 = 0 . This allo ws us to write the gradient o f comp f ( x ) as ∇ comp f ( x ) = X g ∈ G f ( g ) ∇ χ g ( x ) = X g ∈ G ( f ( g ) − f ( h )) ∇ χ g ( x ) . Lemma 4.4.2. L et f : G → R b e given. Then for any r we have: ||∇ comp f ( x ) || p p,B X ( r ) ≤ C | |∇ f || p p,B G ( C X G ( r +3 C sup )) 83 wher e C = C p g µ ( B X ( e, C sup )) V ol G ( G ∩ B X ( e, 2 C sup )) p | S | p . If we have B ( e, 2 C sup ) = S , the gener ating set, then this is: ||∇ comp f ( x ) || p p,B X ( r ) ≤ C | |∇ f | | p p,B G ( C X G ( r + C sup )) wher e C = C p g µ ( B X ( e, C sup )) | S | p . When r = ∞ , this is: ||∇ comp f ( x ) || p p,X ≤ C | |∇ f || p p,G . Pr o of. W e can co v er X with ba lls of radius C sup . This lets us rewrite the norm a s follo ws: ||∇ comp f ( x ) || p p,B X ( r ) = Z B X ( r ) |∇ comp f ( x ) | p dx ≤ X h ∈ G ∩ B X ( r + C sup ) Z B X ( h,C sup ) |∇ comp f ( x ) | p dx = X h ∈ G ∩ B X ( r + C sup ) Z B X ( h,C sup ) | X g ∈ G ( f ( g ) − f ( h )) ∇ χ g ( x ) | p dx F rom its deﬁnition, w e know that ∇ χ g ( x ) will b e nonzero only when d X ( x, g ) < C sup . As w e’re in tegrating o ve r x with d X ( x, h ) ≤ C sup , w e can restrict our po ssible g to those w ith d X ( g , h ) < 2 C sup . Then w e use the fact that | ∇ χ g ( x ) | ≤ C g . X h ∈ G ∩ B X ( r + C sup ) Z B X ( h,C sup ) | X g ∈ B X ( h, 2 C sup ) ( f ( g ) − f ( h )) ∇ χ g ( x ) | p dx ≤ C p g X h ∈ G ∩ B X ( r + C sup ) | X g ∈ G ∩ B X ( h, 2 C sup ) ( f ( g ) − f ( h )) | p µ ( B X ( h, C sup )) . Note that b y in v a riance, µ ( B X ( h, C sup )) = µ ( B X ( e, C sup )). A t this po in t , if w e had G ∩ B X ( e, 2 C sup ) = S , the g enerating set, w e could pro ceed as follo ws. O therwise, 84 w e’ll nee d to expand things a little bit more. C p g µ ( B X ( e, C sup )) | S | p X h ∈ G ∩ B X ( r + C sup ) | X s ∈ S 1 | S | ( f ( hs ) − f ( h )) | p ≤ C p g µ ( B X ( e, C sup )) | S | p X h ∈ G ∩ B X ( r + C sup ) | X s ∈ S 1 | S | ( f ( hs ) − f ( h )) 2 | ! p/ 2 = C p g µ ( B X ( e, C sup )) | S | p ||∇ f || p p,G ∩ B X ( r + C sup ) ≤ C p g µ ( B X ( e, C sup )) | S | p ||∇ f || p p,B G ( C X G ( r + C sup )) . If G ∩ B ( e, 2 C sup ) 6 = S , w e could mo dify this b y noting that: X h ∈ G ∩ B X ( r + C sup ) | X g ∈ G ∩ B X ( h, 2 C sup ) ( f ( g ) − f ( h )) | p = X h ∈ G ∩ B X ( r + C sup ) | X g ∈ G ∩ B X ( h, 2 C sup ) k − 1 X i =0: s 0 ..s k = h − 1 g ( f ( hs 0 ..s i ) − f ( hs 0 ..s i +1 )) | p ≤ X h ∈ G ∩ B X ( r +3 C sup ) V ol G ( G ∩ B X ( e, 2 C sup )) p | X s ∈ S ( f ( hs ) − f ( h )) | p . This will yield the inequalit y: ||∇ comp f ( x ) || p p,B X ( r ) ≤ C | |∇ f || p p,B G ( C X G ( r +3 C sup )) for C = C p g µ ( B X ( e, C sup )) V ol G ( G ∩ B X ( e, 2 C sup )) p | S | p . Note that in these , C g is the bo und on the g radien t of χ g . 4.5 P oincar´ e inequalit y for v olume doubling ﬁnitely gen- erated groups W e c an us e these estimates along w ith our know ledge of complexes in order to sho w that v olume do ubling ﬁnitely generated groups admit a strong P oincar ´ e Inequalit y . This is not a new fact, but it is a cute pro of. 85 Theorem 4.5.1. L et G b e a ﬁnitely g ener ate d volume doubling gr o up. L et f : G → R and B G ( r ) ⊂ G b e given. Th e n || f − f B G ( r ) || 1 ,B G ( r ) ≤ C r ||∇ f || 1 ,B G ( r ) . Her e C = 4 C P C g | S | C sup wher e C P is the c onstant in the glob al Poinc ar´ e ine quality for X . Pr o of. T ak e an y suc h gro up, and let X b e its Ca yley gra ph. Theorem 4.3 .2 sho w ed that strong P oincar ´ e inequalities hold on X . W e happily note that b ot h C 0 and C X G are 1 on a Ca yley gra ph, and so w e can omit them f rom our calculation. W e form a c hain o f inequalities as follows . F rom Theorem 4 .4.1, we can set c = (comp f ) B X ( r + . 25) to get: || f − (comp f ) B X ( r + . 25) || 1 ,B G ( r ) ≤ 1 µ ( B X ( g , 1 4 )) || comp f − (comp f ) B X ( r + . 25) || 1 ,B X ( r + . 25) . Note tha t for ev ery g ∈ G , µ ( B X ( g , 1 4 )) = 1 / 4 | S | . F ro m Theorem 4.3.2, w e know that: || comp f − (comp f ) B X ( r + . 25) || 1 ,B X ( r + . 25) ≤ C P r ||∇ comp f || 1 ,B X ( r + . 25) . Then w e transfer back , using the fact that G ∩ B X ( e, 2 C sup ) = S . ||∇ comp f ( x ) || 1 ,B X ( r + . 25) = C g µ ( B X ( C sup )) | S | ||∇ f || 1 ,B G ( r + . 25+ C sup ) . W e can ev aluate this as X is a Cay ley graph: µ ( B X ( e, C sup )) = | S | C sup . Since X is a graph whose edges hav e unit length, C sup < 1. In particular, w e can pic k C sup = . 74. Since our o riginal ball, B G ( r ) is on the group, without loss of generalit y we know that r is an integer. Then B G ( r + C sup + . 25) = B G ( r + . 99) = B G ( r ) on the group. Com bining this, w e ha v e: || f − comp f B X ( r + . 25) || 1 ,B G ( r ) ≤ 1 . 25 | S | C P r C g | S | C sup | S | | |∇ f || 1 ,B G ( r ) . 86 W e can get the desired left hand side fro m || f − f B G ( r ) || 1 ,B G ( r ) ≤ || f − comp f B X ( r + . 25) || 1 ,B G ( r ) . W e can use the g raph structure to reduce this to: || f − f B G ( r ) || 1 ,B G ( r ) ≤ 4 C P C g | S | C sup r ||∇ f || 1 ,B G ( r ) . Chapter 5 Comparing Heat Kernels on X and G The main goal of this chapter is to sho w that for la rge times, the heat k ernel on the gro up is comparable to the heat k ernel on the complex. The comparison w a s sho wn for g roups and manifolds b y Saloﬀ-Coste and Pittet [26]. Notation 5.0.2. T o simplify notation, w e us e p t for the heat k ernel on the group, and h t when it is on the complex. On a ﬁnitely generated group, the heat k ernel can b e used to describe a sym- metric random walk . This is a w alk whe re fro m a p oint g ∈ G , the pro babilit y of mo ving to g s in o ne step is 1 | S | for each generator s ∈ S . The v alue o f the heat k ernel on the diagonal, p 2 n ( e, e ) giv es us the probability of returning to the same p oin t a fter 2 n steps. W e are in terested in this for ev en n um b ers of steps because this a voids parit y issues. The set-up for these w alks can b e found in [21]. Deﬁnition 5.0.3. W e sa y f ( t ) ≈ g ( t ) if there exist p ositiv e ﬁnite constan ts C 1 , C 2 , C 3 , and C 4 so that C 1 f ( C 2 t ) ≤ g ( t ) ≤ C 3 f ( C 4 t ) . W e will sho w that the follo wing holds when t ≥ 1: p 2 ⌊ t ⌋ ( e, e ) ≈ sup x ∈ X h t ( x, x ) . Note that it do esn’t mak e sense to compare them for small times, since p t is only deﬁned for in teger v alues of t . An imp ortan t notion in this proo f is that of a menabilit y . 87 88 Deﬁnition 5.0.4. A Følner sequ ence is a sequence of ﬁnite subse ts, F ( i ), with the follo wing prop erties: (1) F or an y g ∈ G there exists i suc h that g ∈ F ( i ), (2) F ( i ) ⊂ F ( i + 1), and (3) F or an y ﬁnite subset Q ⊂ G , lim i →∞ #( QF ( i )) # F ( i ) = 1. Here, QF ( i ) refers to the set { g : g = q f with q ∈ Q, f ∈ F ( i ) } . Deﬁnition 5.0.5. G is amenabl e if and only if G admits a F ølner sequenc e. Example 5.0.6. The group of integers, Z , is amenable. Here, the sets [ − i, i ] form a Følner seque nce. In o rder to sho w this, w e will split it into t w o cases. In the ﬁrst, w e lo ok at when G is nonamenable. Here, p 2 ⌊ t ⌋ ( e, e ) ≈ e − t . Then we will lo o k at when G is amenable. W e will ﬁrst show p t is approx imately less than o r equal to h t , a nd then w e will show the rev erse. 5.1 Heat k ern els in the nonamenable case W e no w lo o k at t he b ehav ior of the heat kernel on X a nd G when G is nonamenable. W e call H t is the semigroup form fo r the heat k ernel on X . It is related to h t ( x, y ) b y H t f ( x ) = R X f ( y ) h t ( x, y ) d y . It is also written as H t = e − t ∆ . Alterna- tiv ely , h t is called the transition function for H t . Estimates on norms of functions and their deriv ativ es can give us estimates on || H t || 2 → 2 . Lemma 5.1.1. || f || 2 ≤ C | |∇ f | | 2 89 wil l b e true for al l f ∈ Dom(∆) if and onl y if for al l t > 0 , || H t || 2 → 2 ≤ e − t/C . Pr o of. W e will sk etc h the pro of . W e can show the forw ard implication by using in tegration b y parts: ||∇ f || 2 2 = Z |∇ f | 2 = Z f ∆ f = Z √ ∆ f √ ∆ f = || √ ∆ f || 2 2 . This tells us that for an y non-zero f ∈ Dom(∆), w e ha v e: || √ ∆ f || 2 2 || f || 2 2 ≥ 1 C . W e can tak e a square ro ot and then an inﬁm um to g et: inf f 6 =0 || √ ∆ f || 2 || f || 2 ≥ 1 √ C . This tells us that 1 √ C is a low er bo und on eigenv alues of √ ∆. Sp ectral t heory tells us that 1 C is a low er b ound on eigen v alues o f ∆, a nd e − t C is an upp er b ound on eigen v alues of H t = e − t ∆ . This yields || H t || 2 → 2 ≤ e − t/C . F or the rev erse, consid er the fact that E ( f , f ) = lim t → 0 h ( H t − I ) f , f i t = −h∇ f , ∇ f i . W e can use our bound to get: lim t → 0 h ( H t − I ) f , f i t ≤ lim t → 0 h ( e − t/C − 1) f , f i t = h ( − 1 /C ) f , f i . Then since E ( f , f ) = −h∇ f , ∇ f i , 90 w e ha v e −h∇ f , ∇ f i ≤ h ( − 1 /C ) f , f i whic h giv es us || f || 2 ≤ C | |∇ f | | 2 . W e can transfer b etw een estimates on | | H t || 2 → 2 and h t ( x, y ). Since the b ound on t he norm of f will hold for nonamenable groups, w e will com bine lemmas 5.1.1 and 5.1.2 to get our heat kerne l estimates. Lemma 5.1.2. If | | H t || 2 2 → 2 ≤ e − 2 t/C , then for al l z ∈ X and t ≥ t ′ : h t ( z , z ) ≤ h t ′ ( z , z ) e − t/C . Pr o of. Apply H t to f ( y ) = h s ( y, z ) || h s ( · ,z ) || 2 . This giv es y ou H t h s ( x, z ) || h s ( · , z ) || 2 = Z X h s ( y , z ) || h s ( · , z ) || 2 h t ( x, y ) d y = h t + s ( x, z ) || h s ( · , z ) || 2 . Our estimate then tells us: || H t || 2 2 → 2 ≥ Z X h t + s ( x, z ) 2 || h s ( · , z ) || 2 2 dx = Z X h t + s ( x, z ) h t + s ( z , x ) || h s ( · , z ) || 2 2 dx = h 2 t +2 s ( z , z ) || h s ( · , z ) || 2 2 . Note that || h s ( · , z ) || 2 2 = Z X h s ( y , z ) 2 dy = h 2 s ( z , z ) . When w e combine this with the ine qualit y for H t , w e hav e: h 2 t +2 s ( z , z ) h 2 s ( z , z ) ≤ e − 2 t/C . 91 Fix z and let u ( t ) = h t ( z , z ). This can b e written as: u ( t + s ) ≤ u ( s ) e − t/C . This is equiv alent to: u ( t + s ) − u ( s ) t ≤ u ( s ) e − t/C − 1 t . T aking the limit as t → 0 + giv es us: u ′ ( s ) ≤ ( − 1 /C ) u ( s ) . This giv es us the estimate that u ( t ) ≤ u ( t 0 ) e − t/C for any t ≥ t 0 . Rewriting this, w e ha v e the long time dec a y for all z ∈ X and t ≥ t ′ : h t ( z , z ) ≤ h t ′ ( z , z ) e − t/C . Note that t he con v erse is essen tially true as well. If h t ( z , z ) ≤ h t ′ ( z , z ) e − t/C for t ≥ t ′ , then w e can construct an upp er b ound for || H t || 2 2 → 2 whenev er t ≥ t ′ . || H t || 2 2 → 2 = sup || f || 2 =1 Z X  Z X f ( y ) h t ( x, y ) d y  2 dx. Note that R X f ( y ) p t ( x, y ) ≤ || f || 2 || p t ( x, · ) || 2 holds b y H¨ older. ... ≤ sup || f || 2 =1 Z X || f || 2 2 || p t ( x, · ) || 2 2 dx = Z X || p t ( x, · ) || 2 2 dx = Z X Z X h t ( x, y ) h t ( y , x ) dy d x = Z X h 2 t ( x, x ) d x ≤  Z X h t ′ ( x, x ) d x  e − 2 t/C . 92 This giv es us || H t || 2 → 2 ≤ C ′ e − t/C for t ≥ t ′ where C ′ = R X h t ′ ( x, x ) d x dep ends only on X and t ′ . In the case where G is not amenable, it is we ll known that the heat kerne l deca ys expo nen tially . This result w as s ho wn by K esten [21]. In particular, for any f ∈ Dom( E ), we know that: || f || 2 ,G ≤ C G ||∇ f || 2 ,G . W e can use a v eraging to sho w that this will hold o n X as w ell. Lemma 5.1.3. If G is not amenable and X/G = Y , then for any t ′ > 0 ther e exist c onstants C 0 = sup y ∈ Y h t ′ ( y , y ) a n d C 1 = p C ( δ 2 + C 2 G C ( δ )) so that for al l x, y ∈ X h t ( x, y ) ≤ C 0 e − t/C 1 holds for al l t ≥ t ′ . Note that C , C ( δ ) ar e as in L emma s 4.2.1 and 4.2.2. Pr o of. Applying Lemma 4.2.1 with p = 2 , r = ∞ giv es us: || f || 2 2 ,X ≤ C  δ 2 ||∇ f || 2 2 ,X + || gro up f || 2 2 ,G  . The inequalit y for g roups then tells us this is less than || f || 2 2 ,X ≤ C  δ 2 ||∇ f || 2 2 ,X + C 2 G ||∇ (group f ) || 2 2 ,G  . W e can then b ound the gradien t in G b y the gradient in X using Lemma 4.2.2 with p = 2, r = ∞ : || f || 2 2 ,X ≤ C  δ 2 ||∇ f || 2 2 ,X + C 2 G C ( δ ) ||∇ f || 2 2 ,X  . Putting this together, w e hav e: || f || 2 ,X ≤ q C ( δ 2 + C 2 G C ( δ )) ||∇ f || 2 ,X . 93 W e can apply this with C 1 = p C ( δ 2 + C 2 G C ( δ )) to the ﬁrst argument to get || H t || 2 → 2 ≤ e − t/C 1 on our complex, X . Then, apply Lemma 5.1.2 to g et t he on- diagonal heat k ernel b ound for an y ﬁxed z . h t ( z , z ) ≤ h t ′ ( z , z ) e − t/C 1 . Because X/G = Y , w e can shift z by elemen ts of G , and it won’t aﬀect our heat k ernel. Sp eciﬁcally , t his means h t ( z , z ) = h t ( z + g , z + g ) for any g ∈ G . This allo ws us to consider only v alues o f h t ( y , y ) fo r p o in t s y ∈ Y . This tells us that the suprem um in Y dominates: sup y ∈ Y h t ( y , y ) ≥ h t ( z , z ) . Set C 0 = sup y ∈ Y h t ′ ( y , y ). Because Y is compact a nd h t ′ ( y , y ) is con tinuous in y , for ﬁxed t ′ > 0 w e will hav e C 0 < ∞ . As su p x,y h t ( x, y ) = sup y h t ( y , y ), this will giv e us our o v erall bound. Corollary 5.1.4. If G is not amenable and X/G = Y , then for t ≥ 1 sup x ∈ X h t ( x, x ) ≈ p 2 ⌈ t ⌉ ( e, e ) . Pr o of. Kesten [21] sho w ed that nonamenable groups hav e heat k ernel b eha vior p 2 ⌈ t ⌉ ( e, e ) ≈ e − t/C . By lemma 5.1.3, w e ha v e h t ( x, x ) ≤ c e − t/C . Since sup x ∈ X h t ( x, x ) ≥ c ′ e − t/C ′ , w e ha v e the eq uiv alence. 5.2 Heat Kernels in the Amenable Case This is a mo diﬁed version of the argumen t in LSC-Pittet pap er [26] which sho ws that the o n diagonal heat k ernel on a group is b ounded a b ov e (in some sense) b y the one on a manifold. The basic argument inv olv es comparing eigen v alues a nd traces of the heat equation restricte d to a ﬁnite set. W e iterate through these s ets using Følner sequenc es, and then w e compare the heat k ernels themselv es. 94 5.2.1 Bounding those on G ab o v e b y those on X Theorem 5.2.1. L et G b e an am e n able gr oup and X the as so ciate d c omplex. F or times t > 1 , we have c onstants C , C 0 so that p ⌈ C t ⌉ ( e, e ) ≤ C 0 sup x ∈ X h t ( x, x ) . Her e C = 2 C 1 C 2 wher e C 1 and C 2 ar e the c onstants i n 4 .4.1 and4 . 4 .2 and C 0 = | S | C X G R 0 / min g 6 = h d G ( g, h ) . Pr o of. Let A be a ﬁnite subset of G , and let A 0 b e the set of p oin ts in X which surround it. That is, A 0 := { x ∈ X | d ( x, A ) < R 0 } . Because R 0 ≥ C sup , w e will ha ve functions f : G → R whic h are suppo rted in A map t o functions comp f : X → R whic h are supp orted in A 0 . Using lemmas 4.4.1 and 4.4.2, w e know that || f || 2 2 ≤ C 1 || comp f || 2 2 and ||∇ comp f || 2 2 ≤ C 2 E ( f , f ). Com bining these, w e get: ||∇ comp f || 2 2 || comp f || 2 2 ≤ C 1 C 2 E ( f , f ) || f || 2 2 = C 1 C 2 h I − K A f , f i || f || 2 2 = C 1 C 2 1 − || K 1 / 2 A f || 2 2 || f || 2 2 ! . Here, w e use d the fact that K A is self-adjoin t. W e can apply the min-max principle in order to compare eigen v alues. Let λ A 0 ( i ) b e the ith e igen v alue for H t on A 0 ⊂ X (denoted H A 0 t ) and β A ( i ) ith eigenv alue for K on A ⊂ G (denoted K A ). F or eigen v alues 1 .. | A | , w e ha v e: λ A 0 ( i ) ≤ C 1 C 2 (1 − β A ( i )) W e can rewrite this as: β A ( i ) ≤ 1 − 1 C 1 C 2 λ A 0 ( i ) . 95 As λ A 0 ( i ) will b e bounded b elo w b y 0, w e can use 1 − x ≤ e − x to get: β A ( i ) ≤ e − 1 C 1 C 2 λ A 0 ( i ) . W e can use this to compare the traces. Recall T r( H A 0 t ) = X i e − tλ A 0 ( i ) T r( K n A ) = X i β n A ( i ) . When β A ( i ) ≥ 0, w e hav e: β 2 n A ( i ) ≤ e − 1 C 1 C 2 λ A 0 ( i )2 n . W e will compare the negative β A ( i ) terms with the p ositive ones. W e kno w that 0 ≤ T r( K 2 n +1 A ). This means w e c an split the sum in to tw o pie ces and subtract the part with negativ e eigen v alues from b oth sides: X β A ( i ) < 0 | β 2 n +1 A ( i ) | ≤ X β A ( i ) > 0 β 2 n +1 A ( i ) . Since all of the eigenv alues are bet w een -1 and 1, we ha ve : X β A ( i ) < 0 | β 2 n +2 A ( i ) | ≤ X β A ( i ) < 0 | β 2 n +1 A ( i ) | ≤ X β A ( i ) > 0 β 2 n +1 A ( i ) ≤ X β A ( i ) > 0 β 2 n A ( i ) . This tells us that T r( K 2 n +2 A ) ≤ X β A ( i ) | β 2 n +2 A ( i ) | ≤ 2 X β A ( i ) > 0 β 2 n A ( i ) ≤ 2 X i β 2 n A ( i ) . W e can compare the ﬁrst | A | terms in the t w o sums, and the extra t erms in T r( H A 0 t ) will only help us : T r( K 2 n +2 A ) ≤ 2 T r( H A 0 2 n C 1 C 2 ) . W e are now in a go o d sp ot. W e will compare the heat ke rnels with the resp ectiv e traces. Fix n . Let F ( i ) b e a F ølner sequenc e in G , and recall S n is the set of w or ds 96 in G of length at mo st n . F or eac h i w e will ha v e a set A = S n F ( i ) = { g : g = f u, f ∈ F ( i ) , u ∈ S n } . In Lsc-Pittet [26], they show ed that for an amenable group G w e ha v e the comparison: p 2 n +2 ( e, e ) ≤ 1 | F ( i ) | T r( K 2 n +2 A ) . By the deﬁnition of the trace, w e kno w tha t on the comple x w e ha ve: T r( H A 0 t ) = X i e − tλ A 0 ( i ) = Z A 0 h A 0 t ( x, x ) d x ≤ µ ( A 0 ) sup x ∈ A 0 h A 0 t ( x, x ) ≤ µ ( A 0 ) sup x ∈ X h t ( x, x ) . When w e combine thes e, w e ﬁnd that: p C 1 C 2 (2 n +2) ( e, e ) ≤ µ ( A 0 ) | F ( i ) | sup x ∈ X h 2 n ( x, x ) . W e can compare µ ( A 0 ) with V ol G ( A ). Since A 0 := { x ∈ X | d X ( x, A ) < R 0 } , each elemen t in A can expand to at most | S | C X G R 0 / min g 6 = h d G ( g, h ) new elemen ts in A 0 . This tells us: p C 1 C 2 (2 n +2) ( e, e ) ≤ | S | C X G R 0 / min g 6 = h d G ( g, h ) | A | | F ( i ) | sup x ∈ X h 2 n ( x, x ) . W e can no w let i go to inﬁnit y; since w e ha v e a Følner sequen ce, | A | | F ( i ) | = | S n F ( i ) | | F ( i ) | will b ecome 1. This leav es us with: p C 1 C 2 (2 n +2) ( e, e ) ≤ | S | C X G R 0 / min g 6 = h d G ( g, h ) sup x ∈ X h 2 n ( x, x ) . 97 5.2.2 Bounding those on X ab o v e b y those on G W e’d lik e to sh o w the rev erse inequalit y . W e will do this using a chain of compar- isons. First, w e will compare h t ( x, x ) with h W t ( x, x ), where h W t is the diﬀusions in an o p en subset W ⊂ X . Then w e will compare eigen v a lues of h W t ( x, x ) and p W ′ t ( e, e ), where p W ′ t ( e, e ) represen ts probabilit y of a random w alk restricted to a set W ′ ⊂ G returning t o the iden tit y , using our b ounds on nor ms and minimax inequalities. Lastly , w e use a comparison for p W ′ t ( e, e ) and p t ( e, e ). A t this po in t , w e will remo ve some of the dep endence on W , and limit a w ay other factors to get the ﬁnal result. W e would lik e to lo ok at what happ ens t o diﬀusions in an op en subset W ⊂ X . Let τ b e the exit time fo r this set: τ = inf { t : t ≥ 0 , X t ∈ W } . Then b y the strong Mark ov prop ert y w e hav e a restricted heat k ernel: h W t ( x, y ) = h t ( x, y ) − E x ( h t − τ ( X τ , y )1 τ ≤ t ) . Here, X t is a random v ariable whic h at time t = τ will b e the p oin t on ∂ W where X t exits W . The term h t − τ ( X τ , y ) repres en ts going from the p oint on the b oundary to y in the time t − τ whic h is left after exiting W . W e tak e the exp ected v alue of this where X 0 = x . W e can b ound the expected v alue abov e b y the maxim um v alue. Since E x ( h t − τ ( X τ , y )1 τ ≤ t ) ≤ sup 0 0 ther e exists a > 0 so that for al l op en subse ts W ⊂ X and for al l t ≥ 6 r 2 1 we know that h W t ( x, x ) ≥ C − 1 H sup y ∈ X h t − 3 r 2 1 ( y , y ) − ε 1 . 98 for al l x ∈ { x ∈ W : d ( x, ∂ W ) > at 1 / 2 } . Her e, r 1 = diam( Y ) . Pr o of. By Corollary 3.1.5 w e know that there are constan ts C 1 and C 2 so that h t ( x, y ) ≤ C 1 min( t, 1 ) d/ 2 e − C 2 d 2 ( x,y ) t . This estimate allo ws us to b ound sup 0 C 1 /d 1 ( d 12 r 2 1 C 2 e ) 1 / 2 ε − 1 /d 1 . If t > s > 1, then the maxim um o ccurs when s = t : h s ( x, y ) ≤ C 1 e − C 2 a 2 t s ≤ C 1 e − C 2 a 2 . W e kno w that C 1 e − C 2 a 2 < ε 1 whenev er a > q 1 C 2 ln( C 1 ε 1 ). Th us, whenev er a > max  C 1 /d 1 ( d 12 r 2 1 C 2 e ) 1 / 2 ε − 1 /d 1 , q 1 C 2 ln( C 1 ε 1 )  w e ha v e sup 0 6 r 2 1 , w e can set T = t + r 2 , α = 1, β = 2, γ = 4, and δ = 5 . Then Q + = ( T − r 2 , T + r 2 ), Q − = ( T − 4 r 2 , T − 2 r 2 ), and Q = ( T − 5 r 2 , T + r 2 ). Applying Sturm here give s us sup y ∈ B r ( x ) h t − 3 r 2 ( y , y ) ≤ C H inf y ∈ B r ( x ) h t ( y , y ) ≤ C H h t ( x, x ) . Due to the symm etry of the space X , h s ( y , y ) is the same as h s when y is translated b y an ele men t of G . F or r = diam( Y ), w e ha ve a copy of Y ⊂ B r ( x ) ⊂ B 2 r ( x ) ⊂ Y 1 for ev ery x ∈ Y . This tells us that sup y ∈ B r ( x ) h s ( y , y ) = sup y ∈ X h s ( y , y ) . W e can b ound the in tegra l of h W t ( x, x ) ab o v e b y an analogue of Lemma 5.3 in LSC-Pittet [26]. 100 Lemma 5.2.3. F or subsets W ⊂ X , B > 0 , and t ≥ 1 , Z W h W t ( x, x ) d x ≤ X λ W ( i ) ≤ 1 /B e − tλ W ( i ) + C 1 2 d/ 2 µ ( W ) e − t/ (2 B ) . Her e, C 1 and d ar e deﬁne d as in C or ol lary 3.1.5, and λ W ar e the eigenva lues of h W t . Pr o of. When a, b ≥ 1 we ha v e the inequalit y ab ≥ a/ 2 + b/ 2. Let a = t and b = B /λ W ( i ). Then for t ≥ 1 and λ W ( i ) ≥ 1 /B w e ha v e tB /λ W ( i ) ≥ t/ 2 + B / (2 λ W ( i )) . If w e mu ltiply through b y − 1 /B and exp onen tiate w e ﬁnd e − tλ W ( i ) ≤ e − t/ (2 B ) − λ W ( i ) / 2 . This allo ws us to b ound the sum o v er the larger eigen v alues: X λ W ( i ) ≥ 1 /B e − tλ W ( i ) ≤ X λ W ( i ) ≥ 1 /B e − t/ (2 B ) − λ W ( i ) / 2 ≤ e − t/ (2 B ) X λ W ( i ) e − λ W ( i ) / 2 = e − t/ (2 B ) Z W h W 1 / 2 ( x, x ) d x ≤ e − t/ (2 B ) C 1 2 d/ 2 µ ( W ) . In the last step, w e used the b ound in 3.1.5 whic h tells us h W 1 / 2 ( x, x ) ≤ C 1 2 d/ 2 . Using the eigen v alue expansion, we can compare the in tegral of the heat k ernel at times greater than one with the sum ov er small eigen v alues plus our b ound on the sum o v er larger eigen v alues: Z W h W t ( x, x ) d x = X λ W ( i ) e − tλ W ( i ) ≤ X λ W ( i ) ≤ 1 /B e − tλ W ( i ) + C 1 2 d/ 2 µ ( W ) e − t/ (2 B ) . 101 Let’s consider what it means to hav e a Laplacian, ∆ Ω , deﬁned for functions restricted to a set, Ω with a p olygonal b oundary . Let the domain of ∆ Ω b e the closure of the interse ction of Dom(∆) and the contin uous functions whic h are compactly supp orted on Ω; t hat is, D om(∆ Ω ) = Dom(∆) ∩ C C 0 (Ω). Note that since Dom(∆) ∩ C 0 (Ω) ⊂ Dom(∆) and Dom(∆) is closed, we kno w that Dom(∆ Ω ) ⊂ Dom(∆). F or functions f ∈ Dom(∆ Ω ), we set ∆ Ω f = ∆ f . ∆ Ω inherits man y properties from ∆. It is self-adjoint with a discrete sp ectrum, and as we will see in the follo wing lemma, for t he Ω t hat w e are intere sted in there will b e only ﬁnitely man y eigen v alues whic h are close to 0. W e can show this b y comparing op erators r estricted to subsets of X to op erators restricted to subsets of G . Let A ⊂ G b e give n. Let Ω = U ( A ) b e a subset of X with p o lygonal b oundary so that an y function f whose supp ort is in U ( A ) has an asso ciated function group f whose support is in ( A, { 1 ..N } ). In pa rticular, w e w ould lik e U ( A ) to b e close in size to A . Since group f ( g , i ) = − R B X ( gγ i ,δ ) f ( x ) dx a v erages ov er neigh b orho o ds of points in X , w e can guaran tee a set with v olume estimate: min y ∈ Y µ ( B ( y , δ )) N # A ≤ µ ( U ( A )) ≤ µ ( Y ) N # A. The follo wing lemm a will giv e us a comparison f or small eigen v alues on h U ( A ) t . Lemma 5.2.4. L et A ⊂ G and U ( A ) ⊂ X b e given as ab ove. Eigenvalues of h U ( A ) t ( x, x ) and p n ( e, e ) ar e c omp ar abl e i n the fol low ing manner: X i :0 ≤ λ U ( A ) ( i ) ≤ 1 /B e − 2 nλ U ( A ) ( i ) ≤ # AN p 2 ⌊ n/ ( B (2 − √ 2)) ⌋ ( e, e ) . Her e B = 4 C ( q 1 2 C gr a d ) / (2 − √ 2) wher e C 1 is the c onstant in L emma 4.2.1 and C ( · ) i s the c onstant fr om L emma 4.2.2. N d e p ends on δ = q 1 2 C gr a d . 102 Pr o of. Supp ose u is a solution to ∆ Ω u = λu on a set Ω ⊂ X with p olygonal b oundary , a nd u = 0 on ∂ Ω. Set u = 0 outside of Ω. F or u ∈ Dom(∆ Ω ) and λ 6 = 0, a formal argumen t using integration b y parts tells us: h u, u i = 1 λ h λu, u i = 1 λ h ∆ Ω u, u i = 1 λ h∇ u, ∇ u i + h∇ u , u i| ∂ Ω = 1 λ h∇ u, ∇ u i . This give s us || u | | 2 = | 1 λ |||∇ u | | 2 . W e kno w that suc h eigenfunctions exist b e- cause ∆ Ω is self-adjoin t. W e will com bine this with t he inequalit y in Lemma 4.2.1 f or eigenfunctions f on the se t U ( A ): || f || 2 2 ,X ≤ C g r ad ( δ 2 ||∇ f || 2 2 ,X + || gro up f || 2 2 ,G ) = C g r ad ( δ 2 | λ ||| f || 2 2 ,X + || gro up f || 2 2 ,G ) . This tells us: (1 − C g r ad δ 2 | λ | ) || f || 2 2 ,X ≤ || gr oup f | | 2 2 ,G . If δ is less than q 1 λC gr a d , w e hav e a nice b ound fo r that λ . In particular, δ = q 1 2 C gr a d giv es us a simple bound for all λ ≤ 1 b ecause 1 − (1 / 2) | λ | > 1 / 2. || f || 2 2 ,X ≤ 2 | | group f || 2 2 ,G . Lemma 4.2.2 tells us ||∇ group f || 2 2 ,G ≤ C ( δ ) || ∇ f | | 2 2 ,X . 103 W e ha v e that for C ′ = 2 C ( q 1 2 C gr a d ): ||∇ group f || 2 2 ,G || group f || 2 2 ,G ≤ C ′ ||∇ f || 2 2 ,X || f || 2 2 ,X . W e can rewrite ∇ group f in terms of K 1 / 2 A,N . 1 − || K 1 / 2 A,N group f || 2 2 ,G || group f || 2 2 ,G ≤ C ′ ||∇ f || 2 2 ,X || f || 2 2 ,X . This will allow us to compare the ﬁrst k eigen v alues of h U ( A ) t with the absolute v alues of those for K A,N , where k = min(# N A, # { λ U ( A ) ( i ) ∈ [0 , 1] } ). The min- max deﬁnition will give us these eigen v alue comparisons. F or simplicit y , w e will use λ U ( A ) ( i ) to refer to t he ith s mallest eigen v alue of h U ( A ) t , and | β A ( i ) | to refer to the ith largest absolute v alue of the eigen v alue of K A,N . W e hav e 1 − | β A ( i ) | ≤ C ′ λ U ( A ) ( i ) whic h can b e written as 1 − C ′ λ U ( A ) ( i ) ≤ | β A ( i ) | . When 1 / 2 ≤ x ≤ 1, w e kno w x ≥ e − 2(1 − x ) . Applying that to x = 1 − C ′ λ U ( A ) ( i ), w e ha v e e − 2 C ′ λ U ( A ) ( i ) ≤ | β A ( i ) | for i ≤ k with 0 ≤ λ U ( A ) ( i ) ≤ 1 / (2 C ′ ). W e can exp onen tiate to get: e − 2 nλ U ( A ) ( i ) ≤ | β A ( i ) | n/C ′ . W e will hav e this b ound for all of the λ U ( A ) ∈ [0 , (2 − √ 2) / (2 C ′ )] provide d we can sho w that we hav e an i with C ′ λ U ( A ) ( i ) > (2 − √ 2) / 2. If we knew that (2 − √ 2) / 2 ≤ 1 − | β A ( i ) | for some i , then this would b e shown. This means w e w ant to ha v e | β A ( i ) | 2 ≤ 1 / 2 f or some i . W e kno w that K A,N is an # AN by # AN matrix whose en tries are either 1 / | S | or 0 and that there are | S | nonzero en tries 104 p er ro w. When w e lo ok at its square, w e hav e another # AN b y # AN matrix whose en tries are at most | S | / | S | 2 = 1 / | S | and at least 0. K 2 A,N has eigen v alues | β A ( i ) | 2 . This means that the largest T r( K 2 A,N ) could p ossibly b e is # AN / | S | , and so P # AN i =1 | β A ( i ) | 2 ≤ # AN/ | S | . The av erage v alue of an eigen v alue | β A | 2 is 1 / | S | . Since | β A ( i ) | 2 ∈ [0 , 1], we mus t hav e at least one | β A | 2 whic h is smaller tha n 1 / | S | in order to ha v e that as the av erage. This tells us that t here is some i with | β A ( i ) | ≤ 1 / p | S | ≤ 1 / √ 2. In this w a y , w e ha v e g uaran teed the b ound f or all λ U ( A ) ∈ [0 , ( 2 − √ 2) / (2 C ′ )]. Note that this also sho ws that there are at most # AN suc h eigenv alues. Summing o v er λ U ( A ) ( i ) ∈ [0 , ( 2 − √ 2) / (2 C ′ )] giv es us: X i :0 ≤ λ U ( A ) ( i ) ≤ (2 − √ 2) / (2 C ′ ) e − 2 nλ U ( A ) ( i ) ≤ X i :0 ≤ λ U ( A ) ( i ) ≤ (2 − √ 2) / (2 C ′ ) ( β A ( i )) n/C ′ . Note that the β A ( i ) in this sum ar e p ositiv e. W e can compare these to p ositiv e eigen v alues in the trace by us ing K 2 ⌊ n/ (2 C ′ ) ⌋ A,N . X i :0 ≤ λ U ( A ) ( i ) ≤ (2 − √ 2) / (2 C ′ ) ( β A ( i )) n/C ′ ≤ T r( K 2 ⌊ n/ (2 C ′ ) ⌋ A,N ) . Com bining these yields : X i :0 ≤ λ U ( A ) ( i ) ≤ (2 − √ 2) / (2 C ′ ) e − 2 nλ U ( A ) ( i ) ≤ T r( K 2 ⌊ n/ (2 C ′ ) ⌋ A,N ) . W e kno w that by its deﬁnition T r( K 2 ⌊ n/ (2 C ′ ) ⌋ A,N ) = X g ∈ A,j =1 ..n p 2 ⌊ n/ (2 C ′ ) ⌋ ( g , g ) ≤ # AN p 2 ⌊ n/ (2 C ′ ) ⌋ ( e, e ) . This giv es us the result: X i :0 ≤ λ U ( A ) ( i ) ≤ (2 − √ 2) / (2 C ′ ) e − 2 nλ U ( A ) ( i ) ≤ # AN p 2 ⌊ n/ (2 C ′ ) ⌋ ( e, e ) . 105 If w e w an t to simplify the notation on the left, w e ma y set B = 2 C ′ / (2 − √ 2). This means n/ (2 C ′ ) = n/ ( B (2 − √ 2)). Hence: X i :0 ≤ λ U ( A ) ( i ) ≤ 1 /B e − 2 nλ U ( A ) ( i ) ≤ # AN p 2 ⌊ n/ ( B (2 − √ 2)) ⌋ ( e, e ) . Theorem 5.2.5. F or t > 6 r 2 1 we get: sup y ∈ X h t − 3 r 2 1 ( y , y ) ≤ C p 2 ⌊ t B log | S | ⌋ ( e, e ) wher e C = C H  1 min y ∈ Y µ ( B ( y ,δ )) + µ ( Y ) µ ( B ( y ,δ )) C 1 2 d/ 2  . Pr o of. W e’ll use these lemmas and Følner se quences to build this inequalit y . Recall Lemma 5.2.2 told us: h W t ( x, x ) ≥ C − 1 H sup y ∈ X h t − 3 r 2 1 ( y , y ) − ε 1 . for all x ∈ { x ∈ W : d ( x, ∂ W ) > at 1 / 2 } when t > 6 r 2 1 . Set T = { g ∈ G : d X ( e, g ) ≤ √ ta + 10 R 0 } . Then AT = { g ∈ G : g = th for t ∈ T , h ∈ A } . W e’ll apply this to W = U ( AT ). Note that U ( A ) ⊂ { x ∈ U ( AT ) : d ( x, ∂ U ( AT )) > at 1 / 2 } . When w e take the a v erage o v er U ( A ) we ha v e: sup y ∈ X h t − 3 r 2 1 ( y , y ) ≤ C H  − Z U ( A ) h U ( AT ) t ( x, x ) d x + ε 1  F rom Lemma 5.2.3 w e kno w how to b ound the in tegral in terms of λ W ≤ 1 /B : Z U ( A ) h U ( AT ) t ( x, x ) d x ≤ Z U ( AT ) h U ( AT ) t ( x, x ) d x ≤ X λ U ( AT ) ( i ) ≤ 1 /B e − tλ U ( AT ) ( i ) + C 1 2 d/ 2 µ ( U ( AT )) e − t/ (2 B ) . 106 Putting them together giv es us: sup y ∈ X h t − 3 r 2 1 ( y , y ) ≤ C H   1 µ ( U ( A )) X λ U ( AT ) ( i ) ≤ 1 /B e − tλ U ( AT ) ( i ) + µ ( U ( AT )) µ ( U ( A )) C 1 2 d/ 2 e − t/ (2 B ) + ε 1   . By Lemma 5.2.4 w e hav e: X i :0 ≤ λ U ( AT ) ( i ) ≤ 1 /B e − 2 nλ U ( AT ) ( i ) ≤ #( AT ) N p 2 ⌊ n B (2 − √ 2) ⌋ ( e, e ) . When w e set n = t/ 2, this giv es us: sup y ∈ X h t − 3 r 2 1 ( y , y ) ≤ C H  #( AT ) N µ ( U ( A )) p 2 ⌊ t 2 B (2 − √ 2) ⌋ ( e, e ) + µ ( U ( AT )) µ ( U ( A )) C 1 2 d/ 2 e − t/ (2 B ) + ε 1  . On G , w e can b ound b elow the probabilit y of returning to the start b y noting that b ecause S = S − 1 , after mo ving n steps , w e hav e a 1 | S | n c hance of exactly retracing our path. p 2 n ( e, e ) ≥ 1 | S | n = e − n log | S | . A more con ve nien t time giv es us p 2 ⌊ n 2 B log | S | ⌋ ( e, e ) ≥ e − n/ (2 B ) . When w e place this into the inequalit y , we ha v e: sup y ∈ X h t − 3 r 2 1 ( y , y ) ≤ C H  #( AT ) N µ ( U ( A )) p 2 ⌊ t 2 B (2 − √ 2) ⌋ ( e, e ) + µ ( U ( AT )) µ ( U ( A )) C 1 2 d/ 2 p 2 ⌊ t 2 B log | S | ⌋ ( e, e ) + ε 1  . W e can use the fact tha t p t ( e, e ) ≤ p s ( e, e ) whenev er t > s noting tha t b oth 2 ⌊ t 2 B (2 − √ 2) ⌋ and 2 ⌊ t 2 B log | S | ⌋ are larger tha n 2 ⌊ t B log | S | ⌋ . ... ≤ C H  #( AT ) N µ ( U ( A )) + µ ( U ( AT )) µ ( U ( A )) C 1 2 d/ 2  p 2 ⌊ t B log | S | ⌋ ( e, e )) + ε 1  . 107 W e tak e a Følner sequence for G , and set A = F ( i ). W e can use our v olume estimates to ﬁnd: #( AT ) N µ ( U ( A )) ≤ #( AT ) N min y ∈ Y µ ( B ( y , δ )) N # A = #( AT ) # A min y ∈ Y µ ( B ( y , δ )) and µ ( U ( AT )) µ ( U ( A )) ≤ µ ( Y ) N #( AT ) min y ∈ Y µ ( B ( y , δ )) N # A = #( AT ) µ ( Y ) # A min y ∈ Y µ ( B ( y , δ )) . When w e take the limit of #( AT ) # A = #( F ( i ) T ) # F ( i ) as i → ∞ , w e ﬁnd it is 1. This giv es us: sup y ∈ X h t − 3 r 2 1 ( y , y ) ≤ C H 1 + µ ( Y ) min y ∈ Y µ ( B ( y , δ )) C 1 2 d/ 2 p 2 ⌊ t B log | S | ⌋ ( e, e ) + C H ε 1 . No w let ε 1 go to zero. This yields the comparison. W e can com bine these three results in to a single theorem. Theorem 5.2.6. L et G b e a ﬁnitely gener ate d gr oup and X the asso ciate d c omplex. F o r times t > 1 , we have the c omp arison p 2 ⌈ t ⌉ ( e, e ) ≈ sup x ∈ X h t ( x, x ) . Note that by tr an s itivity, this holds for the he at kernels on the skeletons a s wel l. Pr o of. If G is amenable, apply theorems 5.2.1 and 5.2 .5. If G is nonamenable, apply corollary 5.1.4. This theorem gives a comparison of heat k ernel b eha vior at large times. It do es not; how ev er, tell y ou what tha t b eha vior is for a giv en group. Ev en though the pro of tells you the asymptotic f or nonamenable groups, it is not easy to determine amenabilit y . F or example, it is unk no wn whether T hompson’s group F is ame nable or not. (See Belk [2].) BIBLIOGRAP HY [1] Martin T. Barlo w and T ak ashi Kumagai. T ransition dens it y asymptotics for some diﬀusion pro cesses with m ulti- fractal structures. Ele ctr on. J. Pr ob ab. , 6:no. 9, 23 pp. (electronic), 20 01. [2] James Belk. Thompson’s group F. Ph.D. D issertation, Cornel l University , 2004. [3] Louis J. Billera, Susan P . Holmes, and Kar en V ogtmann. Geometry of the space of ph ylogenetic trees. A dv. in Appl. Math. , 27 (4):733–76 7, 2 001. [4] T aouﬁk Bouziane. Brow nian motio n in Riemannian admissible complexes. Il linois J. Math. , 49(2):559–5 80 (electronic), 200 5. [5] Mic hael Brin and Y uri Kifer. Brownian motion, harmonic functions and hy- p erb olicity fo r Euclide an complexes. Math. Z. , 237(3):421–468 , 2001. [6] Dmitri Burago, Y uri Burago, and Sergei Iv anov. A c ourse in m etric ge om e- try , v olume 33 of Gr aduate Studies in Mathematics . American Mathematical So ciet y , Pro vidence, RI, 2001. [7] Thierry Coulhon and Lauren t Saloﬀ-Coste. Isop´ erim ´ etrie p our les group es et les v ari ´ et ´ es. R ev. Mat. Ib er o americ ana , 9(2):293–314, 1993. [8] Thierry Coulhon and Lauren t Saloﬀ-Coste. V ari ´ et ´ es riemanniennes isom ´ etriques ` a l’inﬁni. R ev. Mat. Ib er o a m eric ana , 1 1(3):687–7 26, 1995. [9] Gianni Da l Maso. A n intr o duction to Γ -c onv er genc e . Progress in Nonlin- ear Diﬀeren tial Equations and their Applications, 8. Birkh¨ auser Boston Inc., Boston, MA, 1993. [10] H. Dym and H. P . McKean. F o urier Series and Inte gr als . Academis Press, New Y ork, NY, ﬁrst edition, 1972. [11] James Eells and Bent F uglede. Harmonic Maps Betwe en Riemannian Poly- he dr a . Cam bridge Univ ersit y Pres s, Cam bridge, UK, 2001. [12] La wrence C. Ev ans. Partial diﬀer ential e quations , volume 19 of Gr aduate Studies in Mathematics . American Mathematical So ciet y , Prov idence, RI, 1998. [13] William F eller. A n intr o duction to pr ob ab i l i ty t he ory and its applic ations. Vol. II. Second edition. John Wiley & Sons Inc., New Y ork, 1 971. [14] Masatoshi F ukushima, Y¯ oichi ¯ Oshima, and Masa yoshi T aked a. Dirich let form s and symmetric Markov p r o c esses , v olume 19 of de Gruyter Studies in Mathe- matics . W alter de Gruyter & Co., Berlin, 1994. 108 109 [15] A. A. Grigor ′ y a n. The heat equation on noncompact Riemannian manifolds. Mat. Sb. , 182(1 ):55–87, 1991. [16] Brank o Gr ¨ un baum. Convex p olytop es , v olume 221 of Gr aduate T exts in Math- ematics . Springer-V erlag, New Y ork, second edition, 2003. Prepared and with a preface b y V olker K aib el, Victor Klee and G ¨ un ter M. Ziegler. [17] Piotr Ha j lasz and P ekk a Kosk ela. Sob o lev met Poincar ´ e. Mem. Amer. Math. So c. , 145(688):x+101, 2000. [18] Juha He inonen. L e ctur es on analysis on m e tric sp ac es . Univers itext. Springer- V erlag, New Y ork, 2001. [19] Masahik o Kanai. Rough isometries, and com binatorial appro ximations of geometries of no ncompact R iemannian manifo lds. J. Math. So c. Jap an , 37(3):391– 413, 19 85. [20] T o dd (Singing Sensation) Kemp. Lecture notes on the theorem of Beurling and Den y . 2005 . [21] Harry Kesten. Symmetric r andom w a lks on groups. T r ans. A mer. Math. So c. , 92:336–35 4, 1959 . [22] Jun Kigami. Analysis on fr actals , v olume 143 of Cambridge T r acts in Math- ematics . Cam bridge Univ ersity Press, Cam bridg e, 2001. [23] Nic holas J. Kor ev aar and R ic har d M. Sc ho en. Glo bal existence theorems for harmonic ma ps to non-lo cally compact s paces. Co mm. A nal. Ge om. , 5(2 ):333– 387, 1997. [24] P eter Kuc hmen t. Quantum graphs. I. Some basic s tructures. Waves R andom Me dia , 14(1):S107–S12 8, 2 004. Sp ecial section on quan tum graphs. [25] J. Moser. On a point wise estimate for parab olic diﬀeren tial equations. Comm. Pur e Appl. Math. , 24:727–740, 1971. [26] Christophe Pittet and Lauren t Saloﬀ-Coste. On the stabilit y of the b eha vior of random w alks on groups. J. Ge om. A nal. , 10(4):713–73 7, 2000. [27] Lauren t Saloﬀ-Coste. A note on Poincar´ e, Sob olev, and Harnack inequalities. Internat. Math. R es. Notic es , (2):27–38, 1992. [28] Lauren t Salo ﬀ-Coste. Asp e cts of Sob olev-typ e ine qualities , v olume 28 9 of L on - don Mathematic al So ciety L e ctur e Note Seri e s . Cambridge Univ ersit y Press, Cam bridg e, 2002. [29] K. T. Sturm. Analysis on local Diric hlet spaces . I I I. The parabo lic Harnac k inequalit y . J. Math. Pur es Appl. (9) , 75(3):273–2 97, 199 6. 110 [30] K. T. Sturm. Diﬀusion pr o cesses and heat k ernels on metric spaces. Ann. Pr ob ab. , 26(1):1–55 , 1 998. [31] K. T. Sturm. How to construct diﬀusion pro cesses on metric spaces. Potential A nalysis , no. 8:149–161, 1998. [32] Nic holas Th. V arop oulos. Th ´ eorie du p otentie l sur des group es et des v ari ´ et ´ es. C. R. A c ad. S ci. Paris S ´ er. I Math. , 302( 6):203–20 5, 1986. [33] Brian White. Inﬁma of energy functionals in homotop y classes of mappings. J. Diﬀer e n tial Ge om. , 23(2):12 7–142, 1986.

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