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APPEARED IN BULLETIN OF THE

arXiv:math/9307233v2 [math.GT] 24 Mar 2000RESEARCH ANNOUNCEMENTAPPEARED IN BULLETIN OF THEAMERICAN MATHEMATICAL SOCIETYVolume 29, Number 1, July 1993, Pages 51-59QUASIPOSITIVITY AS AN OBSTRUCTION TO SLICENESSLee RudolphAbstract. For an oriented link L ⊂S3 = ∂D4, let χs(L) be the greatest Euler char-acteristic χ(F ) of an oriented 2-manifold F (without closed components) smoothlyembedded in D4 with boundary L. A knot K is slice if χs(K) = 1.

Realize D4 inC2 as {(z, w) : |z|2 + |w|2 ≤1}. It has been conjectured that, if V is a nonsingularcomplex plane curve transverse to S3, then χs(V ∩S3) = χ(V ∩D4).

Kronheimerand Mrowka have proved this conjecture in the case that V ∩D4 is the Milnor fiberof a singularity. I explain how this seemingly special case implies both the generalcase and the “slice-Bennequin inequality” for braids.

As applications, I show thatvarious knots are not slice (e.g., pretzel knots like P(−3, 5, 7); all knots obtained froma positive trefoil O{2, 3} by iterated untwisted positive doubling). As a sidelight, Igive an optimal counterexample to the “topologically locally-flat Thom conjecture”.1.

A brief history of slicenessA link is a compact 1-manifold without boundary L (i.e., finite union of simpleclosed curves) smoothly embedded in the 3-sphere S3; a knot is a link with onecomponent. If S3 is realized in R4 as, say, the unit sphere, then a natural way toconstruct links is to intersect suitable two-dimensional subsets X ⊂R4 with S3;one may then ask how constraints on X are reflected in constraints on the linkX ∩S3.For instance, Fox and Milnor (c. 1960) considered, in effect, the case that Xis a smooth 2-sphere intersecting S3 transversally; at Moise’s suggestion, Fox [5]adopted the adjective slice to describe the knots and links X ∩S3 so constructed.Fox and Milnor [6] gave a criterion for a knot K to be slice: its Alexander polynomial∆K(t) ∈Z[t, t−1] must have the form F(t)F(t−1).

This shows that, for instance,the two trefoil knots O{2, ±3} are not slice (since ∆O{2,±3} = t−1−1+t is not of theform F(t)F(t−1)), but it says nothing about the two granny knots O{2, 3}O{2, 3},O{2, −3}O{2, −3} (indeed, both granny knots share the Alexander polynomial1991 Mathematics Subject Classification. Primary 57M25; Secondary 32S55, 14H99.Key words and phrases.

Doubled knot, quasipositivity, slice knot.Received by the editors June 10, 1992 and, in revised form, November 1, 1992c⃝1993 American Mathematical Society0273-0979/93 $1.00 + $.25 per page1

2LEE RUDOLPH(t−1−1+t)2 with the square knot O{2, 3}O{2, −3}, which is slice), and Fox couldonly aver [5] that “it is highly improbable that the granny knot is a slice knot.”By the end of the 1960s, several mathematicians [30, 14, 31] had found invariantswhich could be applied to show that, for instance, the granny knots are not slice.For any knot K, all these invariants (signatures of various families of hermitianforms), as well as ∆K(t), can be calculated from the Seifert pairing θF : H1(F, Z)×H1(F, Z) →Z determined by any Seifert surface F for K (i.e., a smooth, oriented,2-submanifold-with-boundary F ⊂S3 without closed components, with K = ∂F).In particular, if K is slice, then (for any F) there is a subgroup N ⊂H1(F, Z)with rank(N) = 12 rank(H1(F, Z)) on which θF vanishes identically. Any knot forwhich such a subgroup exists is called algebraically slice (briefly, A-slice).

Levineshowed [13] that in higher odd dimensions, A-slice knots are slice. Whether thiswere true for knots in S3 was unknown until 1975, when Casson and Gordon [1, 2]developed “second-order” obstructions to sliceness (again using signatures, but ofmore subtly constructed forms that are not determined just by a Seifert pairing)and used them to show that many A-slice knots are not slice.

Their methods werepowerless, however, to prove nonsliceness of any knot K with ∆K(t) = 1 (such aknot is A-slice, as observed by L. Taylor, cf. [9 (1978), problem 1.36]).The subject took surprising turns in the 1980s after Donaldson and Freedmanrevolutionized the theory of 4-manifolds and, not so incidentally, the theory of knotsand links in S3.

In fact, let X ⊂R4 now be a 2-sphere, still transverse to S3, whichis, however, assumed no longer smooth but merely topologically locally-flat (i.e., inlocal C0 charts it looks like R2 ⊂R4); then the link X ∩S3 is topologically locally-flatly slice (briefly, T -slice). T -slice implies A-slice.

Freedman [8] proved that anyknot K with ∆K(t) = 1 (e.g., any untwisted double) is T -slice. Nonsliceness resultsflowed from Donaldson’s restrictions on intersection forms of smooth, as distinctfrom topological, 4-manifolds: Casson proved the existence of a nonslice knot Kwith ∆K(t) = 1 (cf.

[9 (1984), problem 1.36]); Akbulut gave an explicit example ofsuch a knot, the untwisted positive double D(O{2, 3}, 0, +) (cf. [3]); Cochran andGompf [3] found large classes of knots K such that D(K, 0, +) is not slice; and Yu[32], building on work of Fintushel and Stern, found many A-slice Montesinos knotswhich are not slice.In §4 I give many examples of nonslice knots: for example—recovering some ofFintushel and Stern’s results—all pretzel knots P(p, q, r) ̸= O with Alexander poly-nomial 1, and—considerably generalizing [3, Corollary 3.2]—all iterated untwistedpositive doubles of any knot K ̸= O which is a closed positive braid.

The methodin each case is to show that the knot under consideration is strongly quasipositive,then to use the fact that a strongly quasipositive knot K ̸= O is not slice, whichfollows from a corollary to a recent result of Kronheimer and Mrowka [10]. In §3I state their result and establish that corollary, as well as a superficially stronger(actually equivalent) corollary, the “slice-Bennequin inequality” for braids.

Section2 is preliminary material on quasipositivity, etc. Section 5 is a sidelight, using anexample from §2 to produce a topologically locally-flat surface in CP2, of algebraicand geometric degree 5, with genus 5 = 12(5 −1) × (5 −2) −1: this is an optimalcounterexample to the “topologically locally-flat Thom conjecture”.Remarks.

(1) Note that it is not Kronheimer and Mrowka’s machinery, but “only”their (spectacular) result which is used.In particular, one can understand thepresent note while staying totally disengaged from gauge theory.

QUASIPOSITIVITY AS AN OBSTRUCTION TO SLICENESS3(2) Although Kronheimer and Mrowka in [10] do not discuss the slice- Bennequininequality, they do draw explicit attention to a (strictly weaker) corollary of theirmain result, namely, the affirmative answer to the “question of Milnor” [15] on theunknotting number of a link of a singularity. The nonsliceness results of the presentpaper have nothing to do with unknotting number.

(3) W. M. Menasco has recently announced a proof of the unknotting resultwhich, in marked contrast to that in [10], uses purely three-dimensional techniques(somewhat in the style of [0]); should such techniques someday be used successfullyto establish the slice-Bennequin inequality, then the present nonsliceness resultswill have a purely three-dimensional proof as well.2. QuasipositivityTransverse C-links and quasipositive Seifert surfaces.

When constructinglinks as intersections X ∩S3, instead of restricting the topological type of X as in§1, one might restrict the nature of the embedding X ֒→R4. In particular, if R4is identified with C2 ⊃S3 := {(z, w) : |z|2 + |w|2 = 1} and X is required to be acomplex plane curve, then one can obtain many interesting links.Definitions.

A complex plane curve is any set Vf := f −1(0) ⊂C2, where f(z, w) ∈C[z, w] is nonconstant; Vf is a smooth, oriented 2-submanifold of C2 except at afinite set S(Vf) ⊂Vf of singularities. If Vf is transverse to S3, then the orientedlink Kf := Vf ∩S3 is a transverse C-link [22, 29].Examples.

Replacing S3 by a round sphere of sufficiently small radius centeredat a point of S(Vf ), one sees that any link of a singularity of a complex plane curveis a transverse C-link; replacing S3 by a round sphere of sufficiently large radius,one sees that any link at infinity of a complex plane curve is a transverse C-link.Links of singularities and links at infinity, though very interesting (cf. [15, 11, 4,23, 17], etc.

), are highly atypical transverse C-links (for instance, while the unknotO is the only slice knot which is a link of a singularity [11] or a link at infinity [23],many nontrivial slice knots are transverse C-links [19]). A much broader class oftransverse C-links is easily defined using braid theory.Definitions.

In the n-string braid groupBn := gpσi, 1 ≤i ≤n −1[σi, σj] = σ−1j σi,[σi, σj] = 1,|i −j| = 1|i −j| ̸= 1,a positive band is any conjugate wσiw−1 (w ∈Bn, 1 ≤i ≤n−1); a positive embeddedband is one of the positive bands σi,j := (σi · · · σj−2)σj−1(σi · · · σj−2)−1, 1 ≤i

Up to ambient isotopy, every quasipositive linkis a transverse C-link [19].Question. Is every transverse C-link quasipositive?Remarks.

(1) There are non-quasipositive knots, for example, the figure-8. Thisfollows, for instance, from a result of Morton [16] and Franks and Williams [7]

4LEE RUDOLPHabout the oriented link polynomial of a closed braid (cf. [26]).

(Note, however, thatevery Alexander polynomial, and indeed every Seifert pairing, can be realized by aquasipositive knot or link [21]. )(2) There are knots which are not transverse C-links; the figure-8 is again anexample.

Biding an affirmative answer to the above question, I know of no way toshow this without using the methods of the present paper.Any specific expression of a quasipositive braid as a product of positive bands,β = w1σi1w−11 w2σi2w−12· · · wkσikw−1k∈Bn, gives a recipe for constructing a quasi-positive braided Seifert ribbon S(w1σi1w−11 , . .

. , wkσikw−1k ) ⊂D4, that is, a smoothsurface (actually “ribbon-embedded”, a refinement we can ignore) bounded by theclosed braid bβ.

The isotopy carrying bβ onto a transverse C-link Kf can be chosento carry S(w1σi1w−11 , . .

. , wkσikw−1k ) onto the (nonsingular) piece of complex planecurve Vf ∩D4.

The Euler characteristic of S(w1σi1w−11 , . .

. , wkσikw−1k ) is n −k.

Ifβ = σi1,j1σi2,j2 · · · σik,jk is strongly quasipositive, then S(σi1,j1, σi2,j2, . .

. , σik,jk) ⊂D4 is the “push-in” of a quasipositive braided Seifert surface, abusively indicatedby the same notation; Figs.

1 and 2 give a sufficient idea of the construction.A Seifert surface is quasipositive if it is ambient isotopic to a quasipositive braidedSeifert surface. (See [20–22] for more on braided surfaces and quasipositivity.) Asubset of a surface is full if no component of its complement is contractible.Theorem [18].

A full subsurface of a quasipositive Seifert surface is quasipositive.Plumbing; quasipositive doubles. For K a knot, τ ∈Z, let A(K, τ) ⊂S3 be anannulus of type K with τ twists; that is, K ⊂∂A(K, τ) and θA(K,τ) has matrix (τ).Let A(K, τ)∗A(O, ±1) be a Seifert surface formed by plumbing A(O, ±1) to A(K, τ);that is, there is a 3-cell B ⊂S3 such that A(K, τ) ⊂B, A(O, ±1) ⊂S3 \ Int B,and A(K, τ) ∩A(O, ±1) = A(K, τ) ∩∂B = A(O, ±1) ∩∂B is a quadrilateral 2-cell whose sides are, in order, contained in alternate components of ∂A(K, τ) and∂A(O, ±1).

The knot D(K, τ, ±) := ∂(A(K, τ) ∗A(O, ∓1)) is the τ-twisted positive(resp. negative) double of K. A matrix for θD(K,τ,±) isτ10 ∓1, so ∆D(K,τ,±)(t) =1 ∓τ(t −2 + t−1), and D(K, 0, ±) is A-slice for any K.Lemma 1.

If K ̸= O is strongly quasipositive, then A(K, 0) is quasipositive.Proof. This follows from the last theorem; for a collar of the boundary of a quasi-positive Seifert surface F ̸= D2 bounded by K is an annulus A(K, 0), and full.□Example.

O{2, 3} = ∂S(σ1, σ1, σ1); A(O{2, 3}, 0) is isotopic to the quasipositivebraided surface S(σ3,6, σ1,4, σ3,5, σ4,6, σ2,5, σ1) pictured in Fig. 1.Lemma 2.

If the knot K ̸= O is strongly quasipositive, then D(K, 0, +) is stronglyquasipositive, being the boundary of a quasipositive braided Seifert surface of Eulercharacteristic −1.Proof. This follows from Lemma 1 and a theorem in [25]: for any Seifert surface S,annulus A, and proper arc α ⊂S, the plumbed surface S∗αA is quasipositive if bothS and A are quasipositive.

A proof in the present case, where S is itself an annulus,α is a transverse arc of S, and A = A(O, −1), was given in [22]; the reader canreadily recreate it after comparing the following example to the preceding one.□Example. D(O{2, 3}, 0, +) = ∂S(σ6, σ3,6, σ6, σ1,4, σ3,5, σ4,6, σ2,5, σ1).

QUASIPOSITIVITY AS AN OBSTRUCTION TO SLICENESS5Fig. 1 S(σ3,6, σ1,4, σ3,5, σ4,6, σ2,5, σ1).Fig.

2 F(−3, 5, 7) on the Seifert surface of O{5, 5}.Quasipositive pretzels. Let p, q, r ∈Z.

A diagram for the pretzel link P(p, q, r)is obtained from a braid diagram for βp,q,r := σ−p1 σ−q3 σ−r5∈B6 by forming the platof βp,q,r (using the pairing (16)(23)(45) at top and bottom). If p, q, r are all odd,then P(p, q, r) is a knot, and (once it is oriented) the obvious surface F(p, q, r) thatit bounds (two 0-handles attached by three 1-handles) is a Seifert surface.Example.

P(1, 1, 1) = O{2, 3}; F(1, 1, 1) = S(σ1, σ1, σ1) (up to ambient isotopy).Lemma 3. For p, q, r all odd, F(p, q, r) is quasipositive iff(*)min{p + q, p + r, q + r} > 0.Proof.

For −τ ∈{p + q, p + r, q + r}, F(p, q, r) contains A(0, τ) as a full subsurface(omit each 1-handle in turn). It is proved in [29] that A(O, τ) is quasipositive iffτ < 0; therefore, by the theorem of [18] quoted above, if F(p, q, r) is quasipositive,then (∗) is true.

Conversely, if (∗) is true, then either min{p, q, r} > 0, or exactly oneof p, q, r is negative and it is of strictly smaller absolute value than the other two. Inthe first case, F(p, q, r) is obtained (up to ambient isotopy) from the quasipositiveSeifert surface S(σ1, σ1, σ1) by applying nonpositive twists to the three 1-handles,so, according to [21] (or [22]), F(p, q, r) is quasipositive; a similar, only slightly lessstraightforward, twisting argument applies in the second case.□Example.

F(−3, 5, 7) is ambient isotopic toS(σ1, σ2, σ2,4, σ3,6, σ1,4, σ5, σ2,5).

6LEE RUDOLPH3. Kronheimer-Mrowka Theorem; “slice-Bennequin inequality”If L ⊂S3 is an oriented link, let χs(L) be the greatest Euler characteristic χ(F)of an oriented 2-manifold F (without closed components) smoothly embedded inD4 with boundary L; so, for a knot K, χs(K) = 1 iffK is slice.If Kf ⊂S3ǫ is the link of the singularity (0, 0) ∈S(Vf), then its Milnor fiber[15] is the nonsingular piece of complex plane curve Vf−δ ∩D4ǫ (for any sufficientlysmall δ > 0); of course, Kf is isotopic to Kf−δ = ∂Vf−δ ∩D4ǫ.

The following is arestatement of [10, Corollary 1.3] in the present terminology.Kronheimer-Mrowka Theorem. If Kf is the link of a singularity, then χs(Kf)is the Euler characteristic of its Milnor fiber.This is a special case of the next proposition, which, however, it implies!Proposition.

If Kf ⊂S3 is a transverse C-link and S(Vf )∩D4 = ∅, then χs(Kf)= χ(Vf ∩D4).Proof. Without loss of generality (after perturbing f slightly) we may assume thatthe projective completion Γ ⊂CP2 of Vf in CP2 ⊃C2 is nonsingular and transverseto the line at infinity.

Then the link at infinity of Vf is isotopic to O{d, d}, d = deg Γ.Assuming χs(Kf) > χ(Vf ∩D4), we would then also have χs(O{d, d}) > χ(Vf).Yet O{d, d} is also a link of a singularity (namely, zd + wd at the origin), and theinterior of its Milnor fiber is diffeomorphic to Vf, so our assumption is inconsistentwith the Kronheimer-Mrowka Theorem.□Corollary. If β = w1σi1w−11· · · wkσikw−1k∈Bn is quasipositive, then χs(bβ) =n −k.□This corollary—in fact, its special case that a strongly quasipositive knot K ̸= Ois not slice—suffices for the applications in §4.

It is easy, however, to go further.Let e : Bn →Z be abelianization (exponent sum with respect to the standardgenerators σi).Slice-Bennequin Inequality. For every n, for every β ∈Bn, χs(bβ) ≤n −e(β).Proof.

The preceding corollary asserts the slice-Bennequin inequality (with equal-ity) for β quasipositive. Now apply the following lemma.□Lemma 4 [28].

If the slice-Bennequin inequality holds for all quasipositive β, thenit holds for all β.Proof. Since [28] is somewhat obscure, I resuscitate the proof.

Letβ = σǫ1i1 · · · σǫkik ∈Bn,ǫj ∈{1, −1},have p (resp. ν = k −p) indices j with ǫj = 1 (resp.

ǫj = −1); so e(β) = p −ν.If 1 ≤j1 < j2 < · · · < jp ≤k are the positive indices, let γ = σij1· · · σijp ; so γ isquasipositive (in fact, positive), and χs(bγ) = n −p. There is a smoothly embeddedsurface Q ⊂S3×[0, 1] of Euler characteristic −ν (Q is a union of annuli with ν extra1-handles attached somehow) such that ∂Q ∩S3 × {0} = bγ and ∂Q ∩S3 × {1} = bβ;so |χs(bβ) −χs(bγ)| ≤ν, and χs(bβ) ≤n −p + ν = n −e(β).□Remark.

Bennequin [0] proved that χ(S) ≤n −e(β) for all β ∈Bn and all Seifertsurfaces S bounded by bβ, and conjectured the slice-Bennequin inequality.

QUASIPOSITIVITY AS AN OBSTRUCTION TO SLICENESS74. Nonsliceness resultsProposition.

If the knot K ̸= O is strongly quasipositive, then none of the knotsD1(K) := D(K, 0, +), Di(K) := D(Di−1(K), 0, +), i ≥2, is slice.Proof. If K ̸= O is strongly quasipositive, then, by Lemma 2 and the corollaryto the Kronheimer-Mrowka Theorem, D(K, 0, +) is strongly quasipositive and notslice (because χs(D(K, 0, +)) = −1); the proof is completed by induction.□Remark.

Cochran and Gompf [3, Corollary 3.2] show that if the knot K ̸= O is theclosure of a positive braid, then Di(K) is not slice for 1 ≤i ≤6. The present resultis infinitely stronger.

It would be interesting to understand the relation betweenbeing (strongly) quasipositive and “being greater than or equal to T” in the senseof [3].Proposition. If p, q, r are all odd, {1, −1} ̸⊂{p, q, r}, and(∗∗)qr + rp + pq = −1,then P(p, q, r) is not slice.Remarks.

(1) For p, q, r odd, {1, −1} ⊂{p, q, r} iffP(p, q, r) is an unknot, and (∗∗)iff∆P(p,q,r)(t) = 1. (2) This corollary, which answers problem 1.37 in [9], is a special case of resultsin [32].Proof.

Not all of p, q, r have the same sign; we may assume p < 0 < q ≤r.By Lemma 3, if (∗) is true, then P(p, q, r) bounds a quasipositive Seifert surfaceof Euler characteristic −1, so by §3 it is not slice.Suppose (∗) is false; thenp+q = −a, r−q = b with a, b ≥0, so by (∗∗), −1 = qr+rp+pq = −(q2 +2aq+ab),whence q = 1, a = 0, p = −1, and {1, −1} ⊂{p, q, r}, contrary to hypothesis.□5. The “topologically locally-flat Thom conjecture”The “Thom conjecture” says that (|da(S)| −1)(|da(S)| −2)/2 ≤g(S) for anyclosed, oriented surface S smoothly embedded in CP2 of (algebraic) degree da(S)and genus g(S).

This conjecture is not known to be true, but it certainly becomesfalse if it is strengthened by replacing “smoothly embedded” with “topologicallylocally-flatly embedded” (briefly, T -embedded). Let the geometric degree dg(S) ofa T -embedded surface S ⊂CP2 be the minimum number of points of intersectionof a surface S′ isotopic to S that intersects CP1∞transversally.Claim.

There is a T -embedded surface S ⊂CP2 with g(S) = da(S) = dg(S) = 5.Remark. Lee and Wilczy´nski [12] show the existence, for every d > 0, of a T -embedded surface Wd ⊂CP2 with da(Wd) = d and g(Wd) = gt(d), where gt(d)is the lower bound for g(S) provided by classical estimates (Hsiang and Szczarba,Rohlin, etc.) if S ⊂CP2 is T -embedded and da(S) = d; gt(d) = (d−1)(d−2)/2 for1 ≤d ≤4, and gt(5) = 5, so the claim is a sharp counterexample.

The techniquesof [12] appear to give no control over dg(Wd). It would be interesting to know ifWd can always be taken to have geometric degree d.Proof (sketch).

Follow [27]; instead of replacing a copy of A(O{2, 3}, 0) ∗A(O, −1)embedded on the quasipositive Seifert surface of O{6, 6} by a T -embedded diskwith the same boundary, do the same with the copy of F(−3, 5, 7) embedded onthe quasipositive Seifert surface of O{5, 5} illustrated in Fig. 2.

(By an oversight,in [27] the embedding actually given was of A(O{2, 3}, 1) ∗A(O, −1). )□

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