On the number of extreme measures with fixed marginals

On the n um b er of extreme measures with fixed marginal s M.G.Nadk arni and K.Gow ri Na v ada In t ro duction: In his pap er [3], K . R. Parthasarath y giv es a b ound for the n um- b er of extreme p oin ts of the con vex set of all G − inv arian t pro ba - bilit y measures on X × Y with giv en marginals of full supp ort. The purp ose of this pap er is to improv e this b ound. Section 1: Let X and Y b e finite sets with | X | = m and | Y | = n. Let G b e a group acting on X a nd Y . Let G act on X × Y b y g ( x, y ) = ( g ( x ) , g ( y )) for all g ∈ G and ( x, y ) ∈ X × Y . Let X/G b e t he set of G orbits of X . W rite | X / G | = m 1 , | Y /G | = n 1 and | ( X × Y ) /G | = m 12 . Let π 1 and π 2 denote the pro jection maps from X × Y to X and Y resp ectiv ely . The sets G ( x ) , G ( y ) and G ( x, y ) resp ectiv ely denote the G − orbits of x ∈ X , y ∈ Y and ( x, y ) ∈ X × Y . Let µ 1 and µ 2 b e G − in v ariant probabilit y measures with full supp ort on X and Y resp ectiv ely . Then K ( µ 1 , µ 2 ) denotes the con vex set of all G -in v arian t probability measures µ on X × Y with marginals µ 1 and µ 2 . Note that for any measure µ ∈ K ( µ 1 , µ 2 ) , the supp ort S ( µ ) of µ is G − inv ariant. Let E ( µ 1 , µ 2 ) denote the set of extreme p oints of K ( µ 1 , µ 2 ). In [3], K.R.P arthasarathy giv es an estimate for the n umber of p o in ts in E ( µ 1 , µ 2 ) : | E ( µ 1 , µ 2 ) | ≤ X max( m 1 ,n 1 ) ≤ r ≤ m 1 + n 1  m 12 r  . In this note we prov e that | E ( µ 1 , µ 2 ) | ≤  m 12 m 1 + n 1 − 1  (1) whic h considerably impro ves the ab o v e b ound. Indeed  m 12 m 1 + n 1 − 1  is one of the terms in t he ab o ve sum. Mor eov er, if G acts trivially or if n um b er of G orbits in G ( x ) × G ( y ) is indep enden t of x and y , then 1 | E ( µ 1 , µ 2 ) | /  m 12 m 1 + n 1 − 1  − → 0 as m 1 , n 1 − → ∞ . In [3], K.R. P arthasara th y has prov ed the following theorem: Theorem ([3], Theorem 3.5): A pr ob a bility me a sur e w ∈ K ( µ 1 , µ 2 ) is extr eme if and only if ther e is no nonzer o r e al value d function ζ on S ( w ) such that (i) ζ ( g ( x ) , g ( y )) = ζ ( x, y ) for al l ( x, y ) ∈ S ( w ) , g ∈ G ; (ii) P y ζ ( x, y ) w ( x, y ) = 0 for al l x ; (iii) P x ζ ( x, y ) w ( x, y ) = 0 for al l y . Definition : A G − in v a rian t subset S ⊂ X × Y is said to b e G − go o d if any G − in v ariant real (or complex) v alued function f defined on S can b e written as f ( x, y ) = u ( x ) + v ( y ) for all ( x, y ) ∈ S for some G − in v arian t functions u and v on X and Y respective ly . Prop osition 1: The supp ort S = S ( µ ) of a me asur e µ ∈ K ( µ 1 , µ 2 ) is G − go o d i f and only if µ ∈ E ( µ 1 , µ 2 ) . Pro of: Let µ ∈ E ( µ 1 , µ 2 ) and assume that S is not G − go o d. Then there exists a G − inv arian t function f o n S whic h cannot b e written as f = u + v where u and v are G − inv arian t. Let L 2 G ( S, µ ) denote the Hilb ert space of all G − inv arian t functions defined on S. Let Λ ⊂ L 2 G ( S, µ ) denote the set of all G − in v ariant functions f whic h hav e represen tation f = u + v with u, v G − in v arian t functions on X and Y resp ectiv ely . Then Λ is a prop er subspace of L 2 G ( S, µ ). Hence there exists a nonzero ζ ∈ L 2 G ( S, µ ) whic h is orthogonal to Λ . Then X ζ ( x, y ) u ( x ) µ ( x, y ) = 0 and X ζ ( x, y ) v ( y ) µ ( x, y ) = 0 for all G − in v arian t u and v defined on X and Y respective ly . In particular X X u ( x ) X Y ζ ( x, y ) µ ( x, y ) = 0 for all G − in v ariant u on X . F or an y x 0 ∈ X , taking u ( x ) = 1 on G ( x 0 ) and u ( x ) = 0 on all other orbits in X, w e get X Y ζ ( x 0 , y ) µ ( x 0 , y ) = 0 . 2 Hence X Y ζ ( x, y ) µ ( x, y ) = 0 for all x and similarly , X X ζ ( x, y ) µ ( x, y ) = 0 for a ll y whic h con tradicts the theorem a b o v e. Con v ersely , supp ose S is G − go o d. Then Λ = L 2 G ( S, µ ) . Let ζ ∈ L 2 G ( S, µ ) satisfy the conditions of the ab ov e theorem with resp ect to the measure µ. Let f ∈ L 2 G ( S, µ ) b e any f unction. Then f can b e written as f = u + v , for some G − inv arian t functions u and v on X and Y resp ectiv ely . By condition (ii) of the ab o v e theorem, X X × Y u ( x ) ζ ( x, y ) µ ( x, y ) = X x u ( x ) X y ζ ( x, y ) µ ( x, y ) = 0 . Similarly b y (iii), X ζ ( x, y ) v ( y ) µ ( x, y ) = 0 . Both these equations together imply X ζ ( x, y ) f ( x, y ) µ ( x, y ) = 0 . Since f is arbitrary in L 2 G ( S, µ ) , ζ = 0 . By the ab ov e theorem µ ∈ E ( µ 1 , µ 2 ) , whic h prov es the prop osition. Remark 1: F or a ny x ∈ X and y ∈ Y , the G − in v a r ia n t set G ( x ) × G ( y ) can b e written as union of G − orbits on X × Y whose first pro jection is G ( x ) and second pro jection is G ( y ) . G ( x ) × G ( y ) = ∪{ G ( z , w ) | π 1 ( G ( z , w )) = G ( x ) and π 2 ( G ( z , w )) = G ( y ) } . This is b ecause the or bit G ( z , w ) of ( z , w ) ∈ X × Y has π 1 ( G ( z , w )) = G ( x ) and π 2 ( G ( z , w )) = G ( y ) if a nd only if G ( z , w ) ⊂ G ( x ) × G ( y ) . Remark 2: If S is a G − g o o d set then ( G ( x ) × G ( y )) ∩ S contains atmost one G − orbit. This is b ecause S canno t contain tw o distinct orbits with the same pro jections: for , if G ( z , w ) and G ( a, b ) are tw o suc h orbits with π 1 ( G ( z , w )) = π 1 ( G ( a, b )) = G ( x ) and π 2 ( G ( z , w )) = π 2 ( G ( a, b )) = G ( y ) then for an y G − in v arian t f = u + v defined 3 on S with f ( z , w ) 6 = f ( a, b ) w e will hav e f ( z , w ) = u ( z ) + v ( w ) = u ( x ) + v ( y ) and similarly , f ( a, b ) = u ( x ) + v ( y ) , a con tra diction. Section 2: Let X 1 and Y 1 b e tw o finite sets with | X 1 | = m 1 and | Y 1 | = n 1 . A subset S ⊂ X 1 × Y 1 is called go o d (r ef. [1]) if ev ery real (or complex) v a lued function f on S can b e expressed in the form f ( x ) = u ( x ) + v ( x ) fo r all ( x, y ) ∈ S Let X 1 = X/G and Y 1 = Y /G. Then X 1 × Y 1 can b e iden ti- fied with a set whose p oints a re G ( x ) × G ( y ) , x ∈ X, y ∈ Y . The G − in v ariant functions on X and on Y are in one-to-one corresp on- dence with the functions on X 1 and Y 1 resp ectiv ely . Let e S ⊂ ( X × Y ) /G denote the set of all G − orbits in a G − in v ar ia n t subset S of X × Y . Define φ : e S − → X 1 × Y 1 b y φ ( G ( x, y )) = G ( x ) × G ( y ). One can sho w that subsets of g o o d sets are go o d and ev ery go o d set S ⊂ X 1 × Y 1 is con tained in a maximal go o d subset of X 1 × Y 1 . F urt her ev ery maximal go o d set of X 1 × Y 1 con ta ins m 1 + n 1 − 1 elemen ts. ( r ef [1]) Prop osition 2: A G − invariant subset S ⊂ X × Y is G − go o d if and only if φ is one-to-one on e S an d φ ( e S ) is go o d in X 1 × Y 1 . F urther, S is maxima l G − go o d set if and only if φ is on e-to-one o n S and φ ( e S ) is maximal go o d set in X 1 × Y 1 . Pro of: Assume S is G − go o d. By r emark 2, if S is G − go o d, then φ is one-to -one on e S . L et f b e an y real (or complex) v alued function defined on φ ( e S ) . D efine g on e S b y g = f ◦ φ. This map g giv es rise to a G − in v ariant map on S, again denoted b y g . W r it ing g = u + v , where u and v are G − in v ariant functions o n X and Y resp ectiv ely , and noting that u and v are constant on eac h orbit , w e can define e u and e v on X 1 and Y 1 b y e u ( G ( x )) = u ( x ) and e v ( G ( y )) = v ( y ) . It is easy t o see t hat f = e u + e v . So φ ( e S ) is go o d. Con ve rsely , let S ⊂ X × Y b e suc h that φ is one-to-one on e S and φ ( e S ) is go o d. Since φ is one-to-one, an y G ( x ) × G ( y ) in tersects S in atmost one orbit. Giv en a function g o n S w e can define f on φ ( e S ) as f = g ◦ φ − 1 . Since f is defined on the go o d set φ ( e S ) w e can write f a s f = e u + e v where e u, e v are defined on π 1 ( φ ( e S )) and π 2 ( φ ( e S )) respectiv ely . Defining u ( x ) = e u ( G ( x )) a nd 4 v ( y ) = e v ( G ( y )) w e get G − in v aria n t functions u and v with g = u + v . No w supp ose S is a maximal G − go o d set. W e kno w from t he first part of the theorem that φ is one-to -one on e S . I f φ ( e S ) is not a maximal go o d set, there exists a p oint, sa y G ( a ) × G ( b ) / ∈ φ ( e S ) , suc h that φ ( e S ) ∪{ G ( a ) × G ( b ) } is go o d. Then, since { G ( a ) × G ( b ) } ∩ S = ∅ , the map φ is one-to-one on e T where T = G ( a, b ) ∪ S. Using t he first part of the theorem, T is G − go o d con tradicting the maximality of S. The con v erse can b e prov ed in a similar manner. This completes the pro of of the prop osition. By corollary 3.6 of [3], differen t extreme p oints of K ( µ 1 , µ 2 ) hav e distinct supp orts. As p oin ted out by the referee, this fa ct is also a consequenc e of prop osition 1: Assume that µ, ν ∈ E ( µ 1 , µ 2 ) , with µ 6 = ν , hav ing the same supp ort S. By prop osition 1 S is G − go o d. But this is a con tra diction since S is a lso the supp ort of ( µ + ν ) / 2 , whic h is not extreme. F urther, for µ and ν ∈ E ( µ 1 , µ 2 ) the measure ( µ + ν ) / 2 ∈ K ( µ 1 , µ 2 ) is not extreme, and so b y Prop osition 1 its supp ort S ( µ ) ∪ S ( ν ) is not a G − go o d set. F urther, for µ and ν ∈ E ( µ 1 , µ 2 ) the measure ( µ + ν ) / 2 ∈ K ( µ 1 , µ 2 ) is not extreme, and so b y Prop osition 1 its suppo r t S ( µ ) ∪ S ( ν ) is not a G − go o d set. This show s that support s o f differen t measures in E ( µ 1 , µ 2 ) are con ta ined in different maximal G − go o d sets of X × Y : Because, if µ 6 = ν ∈ E ( µ 1 , µ 2 ) suc h that S ( µ ) ⊂ S and S ( ν ) ⊂ S for some maximal G − go o d set S then t he measure ( µ + ν ) / 2 ∈ K ( µ 1 , µ 2 ) has its suppo rt S ( µ ) ∪ S ( ν ) con tained in S. Since S is G − go o d, S ( µ ) ∪ S ( ν ) is a lso G − go o d a con tra diction to prop osition 1 a s ( µ + ν ) / 2 is not extreme . Therefore, | E ( µ 1 , µ 2 ) | is b o unded b y the n um b er of maximal G − go o d sets of X × Y . Let S b e a maximal G − go o d set in X × Y . By Prop o sition 2, φ ( e S ) is a maximal go o d set in X 1 × Y 1 . Since φ is one-to- one on e S , e S con tains m 1 + n 1 − 1 orbits of G . Since the nu m b er of orbits in X × Y is m 12 , and an y maximal G − go o d set in X × Y is of the form φ ( e S ) , the total n umber of maximal G − go o d sets in X × Y is less than or equal to  m 12 m 1 + n 1 − 1  . This pro ve s (1) . W e give an example to show that the a b o v e b ound is sharp. Let G b e the gro up S n , the p erm uta tion g roup on n elemen ts. Let X = { 1 , 2 , ..., n } and Y b e the set S n . Here | X | = n a nd | Y | = n ! . Then G a cts on X in the ob vious manner and on Y b y g ( h ) = g ◦ h. The only G − inv arian t subset of X is X itself and t he only G − inv ariant subset of Y is Y itself. Then G also acts on X × Y diago nally . That is, g ( x, y ) = ( g ( x ) , g ( y )) . F or any ( x, y ) ∈ X × Y , the set G ( x, y ) = { ( g ( x ) , g ( y )) | g ∈ G } is a G − in v ariant subset o f X × Y with n ! n umber of elemen ts and 5 G ( x ) × G ( y ) is the whole set X × Y . In this case, | X/ G | = m 1 = 1 and | Y /G | = n 1 = 1 and | ( X × Y ) / G | = m 12 = n . There fore,  m 12 m 1 + n 1 − 1  = n. The only G − inv ariant pro babilit y measures on X and Y are uni- form measures. That is, µ 1 ( x ) = 1 n for all x ∈ X and µ 2 ( y ) = 1 n ! for all y ∈ Y . So the only G − in v ariant f unctions on X and Y are constan t functions. If µ ∈ E ( µ 1 , µ 2 ) , t hen the supp ort S of µ should b e G − go o d. An y G − inv arian t function f defined on S, can b e written a s f = u + v where u, v are G − inv ariant functions on X a nd Y resp ectiv ely . This sho ws that f must be constant, whic h means S consists of a single orbit, say S = G ( x, y ) . Then µ (( g ( x ) , g ( y )) = 1 n ! for all g ∈ G. Observ e that the collection { g ( y ) | g ∈ G } has a ll n ! differen t eleme n ts whereas in t he collection { g ( x ) | g ∈ G } eve ry v alue of g ( x ) is repeated ( n − 1)! times. This sho ws that ev ery suc h uniform measure µ supp orted on an y single orbit G ( x, y ) has marginals µ 1 and µ 2 . Since there are n orbits in X × Y , we get | E ( µ 1 , µ 2 ) | = n. No w w e state some results ab out go o d subsets of X 1 × Y 1 not necessarily G -go o d sets ( ref. [1], [2] ). Consider any tw o p oin ts ( x, y ) , ( z, w ) ∈ S ⊂ X 1 × Y 1 where S is an y (not necessarily go o d) subset of X 1 × Y 1 . W e say that ( x, y ) , ( z , w ) are linke d if there exists a seque nce of p oin t s ( x 1 , y 1 ) = ( x, y ) , ( x 2 , y 2 ) ... ( x n , y n ) = ( z , w ) o f po in ts of S suc h that ( i ) for an y 1 ≤ i ≤ n − 1 exactly one of the following equalities hold: x i = x i +1 or y i = y i +1 ; ( ii ) if x i = x i +1 then y i +1 = y i +2 , and if y i = y i +1 then x i +1 = x i +2 , 1 ≤ i ≤ n − 2 . W e also call this a link joining ( x, y ) t o ( z , w ) . A non trivial link joining ( x, y ) to itself is called a lo op . Theorem (ref. [1], cor. 4.11): A subset S ⊂ X 1 × Y 1 is go o d if and only if S contains no lo ops. Remark 3: Let the orbits in e S b e G ( x 1 , y 1 ) , G ( x 2 , y 2 ) , ..., G ( x m 1 + n 1 − 1 , y m 1 + n 1 − 1 ) . Then S ∩ ( G ( x i ) × G ( y i )) = G ( x i , y i ) for 1 ≤ i ≤ m 1 + n 1 − 1 . Let G ( z , w ) b e an y other orbit in G ( x i ) × G ( y i ) and let S ′ = ( S \ G ( x i , y i )) ∪ G ( z , w ) . It is clear that S ′ is maximal G − go o d set with φ ( e S ) = φ ( e S ′ ). If α i denote the num b er of orbits in G ( x i ) × G ( y i ) , then there are α 1 α 2 ... α m 1 + n 1 − 1 man y maximal G − go o d sets in X × Y with image φ ( e S ) under φ. 6 It seems lik ely that | E ( µ 1 , µ 2 ) | /  m 12 m 1 + n 1 − 1  − → 0 as m 1 , n 1 − → ∞ . W e show this in the case G = ( e ) and more generally when n umber of G orbits in G ( x ) × G ( y ) is indep enden t of x and y . F or that w e first prov e the following theorem. Theorem: L et X = { x 1 , x 2 , ...x m } and Y = { y 1 , y 2 , ...y n } b e two finite sets. Then ( i ) the numb er of maximal go o d sets c ontaine d in X × Y is m n − 1 n m − 1 . ( ii ) the numb er of maximal go o d sets among them with exac tly k fixe d p oi n ts, (s a y ( x i , y j 1 ) , ... ( x i , y j k ) ) hav i ng a fixe d first c o or din ate say x i is : k n m − 2 ( m − 1) n − k , 1 ≤ k ≤ n. ( iii ) the numb e r of maxim al go o d sets wi th exactly k fixe d p oints having a fixe d se c ond c o or dinate s ay y j is: k m n − 2 ( n − 1) m − k , 1 ≤ k ≤ m. Pro of: W e use induction on m + n. The result is true f o r m = 1 and n = 1 . Assume the result for a ll v alues of | X | ≤ m and | Y | ≤ n. W e pr ov e the result for | X | = m a nd | Y | = n + 1 . Let X = { x 1 , x 2 , ...x m } and Y = { y 1 , y 2 , ...y n , y n +1 } . Consider a m × ( n + 1) grid of m ( n + 1) cells with m ro ws corresp onding to { x 1 , x 2 , ..., x m } and n + 1 columns corresp ondiong to { y 1 , y 2 , ..., y n +1 } Asso ciate ( i, j )th cell with the p oint ( x i , y j ) ∈ X × Y . W e sa y that ( x i , y j ) ∈ ( i, j )th cell. T o prov e ( iii ) let S b e a maximal g o o d set in X × Y . Then | S | = m + n. Supp ose S con tains exactly k po in ts with fixed second co ordinate, say y n +1 . Without loss of generalit y w e assume them to b e ( x 1 , y n +1 ) , ( x 2 , y n +1 ) , ... ( x k , y n +1 ) . Denote K = { ( x 1 , y n +1 ) , ( x 2 , y n +1 ) , ... ( x k , y n +1 ) } . (i) A t least one of these first k r ows contain atleast t w o p oints of S, i.e., there exist a p o in t ( x i , y j ) of S with 1 ≤ i ≤ k and 1 ≤ j ≤ n. Pr o of: Ot herwise leav ing these k row s and the la st column, the remaining p oints of S will b e a go o d set with m + n − k p o in ts using m + n − k co ordinates whic h is not p ossible. (ii) If ( x i , y j ) ∈ S with 1 ≤ i ≤ k and 1 ≤ j ≤ n. Then the j the column (whic h contains the p o in t ( x i , y j )) ha s no other p oin t ( x l , y j ) o f S with 1 ≤ l 6 = i ≤ k b ecause the fo ur p oin ts { ( x i , y j ) , ( x i , y n +1 ) , ( x l , y n +1 ) , ( x l , y j ) } form a lo op. (iii) Supp ose ( x i , y j ) ∈ S f o r some 1 ≤ i ≤ k and 1 ≤ j ≤ n. Then the set got b y dropping the p oin t ( x i , y j ) and adding ( x l , y j ) , 1 ≤ l 6 = i ≤ k to S clearly contain no lo op and so is maximal go o d. 7 Let S ′ b e the maximal go o d set obtained in this wa y b y replacing all the p oints ( x i , y j ) , 1 ≤ i ≤ k and 1 ≤ j ≤ n of S b y ( x 1 , y j ) , 1 ≤ j ≤ n . Then eac h of the rows corresp onding to x 2 , ..., x k con ta ins exactly one p oint of S ′ . The set S ′′ got f r om S ′ b y dropping t hese ro ws and the last column will b e a maximal go o d set in { x 1 , x k +1 , ..., x m } × { y 1 , y 2 , ...y n } and con tains m + n − k elemen ts. By induction h yp othesis, the n um b er of maximal go o d sets in { x 1 , x k +1 , ..., x m } × { y 1 , y 2 , ...y n } ha ving exactly r p o in ts in r fixed p ositions in the first row, is: r n m − k − 1 ( m − k ) n − r , for 1 ≤ r ≤ n. Con- sider any such maximal go o d set, sa y A. F urther add the dropp ed ro ws and the last column. Enlarge A by adding the first k p oin ts of the ( n + 1)th column, call this set B . It is a maximal go o d set in { x 1 , x 2 , ..., x m } × { y 1 , y 2 , ...y n +1 } . An y p oin t ( x 1 , y j ) , 1 ≤ j ≤ n in B can b e replaced b y ( x l , y j ) , for an y 1 ≤ l ≤ k and the result- ing set will con tin ue to remain ma ximal go o d in { x 1 , x 2 , ..., x m } × { y 1 , y 2 , ...y n +1 } . In this wa y each one of r n m − k − 1 ( m − k ) n − r max- imal go o d set A giv es rise to k r maximal go o d sets in the original m × ( n + 1) matrix. . F urther, we can c ho ose the r p oin ts in the first ro w in  n r  w ays . Adding ov er r , the total n umber of maximal go o d sets with exactly k cells in k fixed p ositions of the la st column is: n X r =1  n r  k r r n m − k − 1 ( m − k ) n − r = k n m − k − 1 n n − 1 X r =0  n − 1 r  k r − 1 ( m − k ) n − r = k n m − k ( m − k + k ) n − 1 = k n m − k m n − 1 whic h is ( iii ) for m × ( n + 1) matrix. T o pro ve ( i ) , since we can c ho ose the k p oints in the last column in  m k  w ays , the total n umber of maximal go o d sets with exactly k p oin ts from the last column is:  m k  k n m − k m n − 1 . The tota l nu m b er of maximal go o d sets in X × Y is got b y adding these n umbers as k v aries fro m 1 to m : m X k =1  m k  k n m − k m n − 1 = m − 1 X k =0  m − 1 k  n m − k − 1 m n = m n ( n +1) m − 1 . ( ii ) can b e pro ved in a similar w ay as ( iii ). This completes the pro of of the theorem. 8 Next, we pro v e that a s m, n → ∞ the rat io m n − 1 n m − 1 /  mn m + n − 1  → 0 as m, n → ∞ . lim m,n →∞ m n − 1 n m − 1 /  mn m + n − 1  = lim m,n →∞ m n − 1 n m − 1 ( m + n − 1)!(( mn − m − n + 1)! / ( mn )! By Sterling’s for mula, we kno w that n ! ∼ √ 2 π n n + 1 2 e n for larg e n. Using this expres sion o ne can sho w that m n − 1 n m − 1 /  mn m + n − 1  ≤ C (1 − 1 m ) mn (1 + n m ) m (1 − 1 n ) mn (1 + m n ) n (1 − 1 m ) n (1 − 1 n ) m ( m + n ) 1 2 (2) for some constant C . If m n ≥ 1, since (1 − 1 m ) m increases to e − 1 , the righ t hand side of (2) tends to 0 as m, n − → ∞ . The case where m n ≤ 1 is similar b ecause the the express ion on the righ t hand side of (2) is symmetric with resp ect to m and n. If G = ( e ) , the maximal G - g o o d sets in X × Y are just the maximal go o d sets and the n umber of maximal g o o d sets, b y the previous theorem is, m n − 1 n m − 1 . In this case m 12 = m 1 n 1 . Therefore | E ( µ 1 , µ 2 ) | /  m 12 m 1 + n 1 − 1  ≤ m n − 1 n m − 1 /  m 1 n 1 m 1 + n 1 − 1  − → 0 as m and n − → ∞ . No w suppose that the n umber of G -o rbits in G ( x ) × G ( y ) is a constan t, say a, for all x and y . Then b y remark 3, the n um b er of maximal G - go o d sets in X × Y is a m 1 + n 1 − 1 m n 1 − 1 1 n m 1 − 1 1 and m 12 = am 1 n 1 . Therefore, | E ( µ 1 , µ 2 ) | /  m 12 m 1 + n 1 − 1  ≤  a m 1 + n 1 − 1 m n 1 − 1 1 n m 1 − 1 1  /  am 1 n 1 m 1 + n 1 − 1  ≤  a m 1 + n 1 − 1 m n 1 − 1 1 n m 1 − 1 1  /a m 1 + n 1 − 1  m 1 n 1 m 1 + n 1 − 1  9 =  m n 1 − 1 1 n m 1 − 1 1  /  m 1 n 1 m 1 + n 1 − 1  − → 0 as m and n − → ∞ . Note: The maximal go o d sets in X × Y can b e asso ciated in a one-to-one manner with the spanning trees of a complete bipartite graph. Consider the complete bipartit e gr a ph K m,n where | X | = m and | Y | = n. A subset S ⊂ X × Y is maximal go o d if and only if | S | = m + n − 1 and in the grid correspo nding to X × Y , S con ta ins no lo o ps. Construct an m × n matrix corresp onding to an y spanning tr ee T in K m,n as follo ws: Iden tifying the elemen ts of X a nd Y with the ve ritces of K m,n , let V = ( X , Y ) denote the v ertices of K m,n . Whene v er the edge ( x i , y j ) ∈ T , put ( i, j )th en try in the mat r ix equal t o one; otherwise ( i, j ) th entry is zero. Since T is a spanning t r ee, there are exactly m + n − 1 nonzero en tries in the matrix. As T con tains no cycles, the nonzero entries in the matrix donot fo rm a lo o p. Th erefore the nonzero entries of the matrix corresp ond to a maximal go o d set in the g r id corresp onding to X × Y . This corresp ondence is one-to-one. In [5], it is pro v ed that the n um b er o f spanning trees of K m,n is m n − 1 n m − 1 . But the pro of mak es use of t he determinan t of the matrix and is differen t from the one giv en here. Ac kno wledgemen t : The sec ond author thanks CSIR for fund- ing the pro ject of whic h this pap er forms a part. Sincere thanks to IMSc, Chennai and ISI, Bangalore for pro viding short visiting app oin tmen ts. References: [1] Co wsik R C, Klop otowsk i A and Nadk arni M G, When is f ( x, y ) = u ( x ) + v ( y )?, Pro c. Indian Acad. Sci. (Math. Sci.) 109 (1999) 57-64. [2] Nadk a rni M G, Kolmogorov’s sup erp o sition theorem a nd sums of algebras, The Jo urnal of Analysis v ol. 12 (2004) 21- 6 7. [3] K. R P arthasarathy , Extreme p oin ts of the con ve x set of join t probability distributions with fixed marginals, Pro c. Indian Acad. Sci. (Math. Sci.) 117 (2007) 505-5 16. [4] D. B. W est, In tro duction to Gr a ph Theory , Second Edition, P earson Education. [5] Lo v asz L., Com binatorial problems and exercis es, Second Edition, Elsevier Science Ltd. Address: 10 1. Prof . M.G. Na dk arni Departmen t of Mathematics Univ ersit y of Mum bai Kalina Campus, San tacruz East Mum bai-4 0 0098 email: mgnadk arni@ g mail.com 2. K . Go wri Nav ada Departmen t of Mathematics P eriy ar Unive rsit y Salem-636011 T amil Nadu email: gna v ada@yahoo .com 11