User steven sivek - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T03:17:58Z http://mathoverflow.net/feeds/user/428 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/123472/are-there-non-compact-non-smoothable-manifolds/123476#123476 Answer by Steven Sivek for Are there non-compact, non-smoothable manifolds? Steven Sivek 2013-03-03T15:04:01Z 2013-03-03T15:04:01Z <p>The Cairns-Hirsch theorem says that a PL manifold $M$ is smoothable if and only if $M\times \mathbb{R}$ is smoothable, so you can take $M$ to be any one of the known compact, PL examples such as Kervaire's manifold and then $M\times\mathbb{R}^n$ is non-smoothable for $n \geq 1$.</p> http://mathoverflow.net/questions/112233/relation-of-sw-and-donaldson-invariant/112235#112235 Answer by Steven Sivek for Relation of SW and Donaldson Invariant Steven Sivek 2012-11-13T00:26:07Z 2012-11-13T00:26:07Z <p>This was conjectured by Witten in his paper in his paper <a href="http://arxiv.org/abs/hep-th/9411102" rel="nofollow">Monopoles and four-manifolds</a>. The conjecture says that if $X$ has Seiberg-Witten simple type (meaning that $SW_X(\mathfrak{s})$ is nonzero only when the moduli space associated to $X$ is 0-dimensional) and satisfies some mild homological conditions (including $b_1(X)=0$ and $b^+(X)>1$ odd), then $X$ has Donaldson simple type, its Donaldson and Seiberg-Witten basic classes coincide, and its Donaldson series has the form</p> <p>${\mathcal D}^w_X(h) = 2^{2+(7\chi(X)+11\sigma(X))/4}e^{Q(h)/2}\sum_{\mathfrak{s}} (-1)^{w^2+c_1(\mathfrak{s})\cdot w} SW_X(\mathfrak{s})e^{c_1(\mathfrak{s})\cdot h}$</p> <p>where $\mathfrak{s}$ ranges over Spin^c structures on $X$ and $Q(h)$ is the intersection form on $X$. Recall that the Donaldson series is defined as a formal power series by the sum $\sum_i D^w_X((1+\frac{x}{2})\frac{h^i}{i!})$, where $x$ is a point of $X$ and $h$ is an element of $H_2(X)$; this definition is originally due to Kronheimer and Mrowka, in <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.jdg/1214456482" rel="nofollow">Embedded surfaces and the structure of Donaldson's polynomial invariants</a>.</p> <p>There is a series of papers by Feehan and Leness proving many cases of the conjecture, although it has not been established in full. For an overview of this program, originally proposed by Pidstrigach and Tyurin, you might try their survey article <a href="http://arxiv.org/abs/dg-ga/9709022v2" rel="nofollow">PU(2) monopoles and relations between four-manifold invariants</a>. Notably, their most recent paper, <a href="http://arxiv.org/abs/math/0609530v2" rel="nofollow">Witten's conjecture for many four-manifolds of simple type</a>, proves it assuming that $c_1^2(X) \geq \chi_h(X)-3$, and before that, <a href="http://arxiv.org/abs/math/0203047" rel="nofollow">A general SO(3)-monopole cobordism formula relating Donaldson and Seiberg-Witten invariants</a> was used by Kronheimer-Mrowka to prove enough cases of the conjecture to establish the Property P conjecture, in <a href="http://arxiv.org/abs/math/0311489" rel="nofollow">Witten's conjecture and Property P</a>.</p> http://mathoverflow.net/questions/103049/knot-theory-question-bridge-number-vs-min-generators-of-fundamental-group-of-co/103051#103051 Answer by Steven Sivek for Knot theory question: bridge number vs. min generators of fundamental group of complement Steven Sivek 2012-07-24T22:04:33Z 2012-07-24T22:04:33Z <p>The (p,q) torus knot has a presentation with two generators, namely $\langle x,y \mid x^p = y^q\rangle$, but if $p,q>2$ then it's non-alternating and so it must have bridge index greater than 2.</p> http://mathoverflow.net/questions/88692/why-should-i-care-about-heegaard-floer-theory/88722#88722 Answer by Steven Sivek for Why should I care about Heegaard-Floer theory? Steven Sivek 2012-02-17T14:07:10Z 2012-02-17T14:07:10Z <p>It provides lots of computable invariants in contact geometry, in particular the <a href="http://arxiv.org/abs/math/0210127" rel="nofollow">contact invariant</a> defined by Ozsvåth and Szabó via open books and the Giroux correspondence. For example, on Seifert fibered spaces the question of whether tight contact structures exist was <a href="http://arxiv.org/abs/0709.0737" rel="nofollow">completely solved</a> by Lisca and Stipsicz, and the classification was completed in many of these cases in several other papers; and Ghiggini used it to exhibit contact structures which are <a href="http://arxiv.org/abs/math/0506380" rel="nofollow">strongly fillable but not Stein fillable</a>.</p> <p>Similarly, the <a href="http://arxiv.org/abs/0802.0628" rel="nofollow">LOSS invariant</a> (named after Lisca-Ozsváth-Stipsicz-Szabó) and the related, easily <a href="http://math.mit.edu/~petero/transverse.html" rel="nofollow">computable</a> Ozsváth-Szabó-Thurston <a href="http://arxiv.org/abs/math/0611841" rel="nofollow">grid diagram invariants</a> of Legendrian and transverse knots were used by Ng-Ozsváth-Thurston to successfully <a href="http://arxiv.org/abs/math/0703446" rel="nofollow">distinguish many pairs of knots</a> in the standard contact $S^3$, and been used to prove other properties such as the fact due to Etnyre and Vela-Vick that the complement of the binding of any open book of any contact structure has <a href="http://arxiv.org/abs/0909.3465" rel="nofollow">no Giroux torsion</a>. (According to recent work of Baldwin--Vela-Vick--Vértesi, <a href="http://arxiv.org/abs/1112.5970" rel="nofollow">these invariants are the same</a> for knots in the standard $S^3$.)</p> http://mathoverflow.net/questions/85976/why-tristram-levine-signature-jumps-at-the-zeros-of-alexander-polynomial/86014#86014 Answer by Steven Sivek for Why Tristram-Levine signature jumps at the zeros of alexander polynomial? Steven Sivek 2012-01-18T17:09:14Z 2012-01-18T17:09:14Z <p>If $A$ is a Seifert matrix for $K$ and $\omega \in \mathbb{C}$ has norm 1, then the Tristram-Levine signature $\sigma_\omega(K)$ is the signature of the matrix </p> <p>$(1-\omega)A + (1-\bar{\omega})A^T = (1-\bar{\omega})(A^T - \omega A),$</p> <p>which jumps when some eigenvalue of $A^T - \omega A$ crosses zero (i.e. changes sign). At these values of $\omega$ the product of the eigenvalues, which is $\det(A^T-\omega A) = \Delta_K(\omega)$, must therefore be zero.</p> http://mathoverflow.net/questions/60408/contact-manifolds-dimension-five/60410#60410 Answer by Steven Sivek for contact manifolds dimension five Steven Sivek 2011-04-03T05:40:13Z 2011-04-03T05:40:13Z <p>Let M<sup>5</sup> be a closed and oriented. A contact form &alpha; gives a 4-plane distribution with symplectic form d&alpha;, reducing the structure group of TM to U(2)&times;1; such a reduction is called an almost contact structure, and it exists iff the integral third Stiefel-Whitney class is zero (Gray, "Some global properties of contact structures", MR0112161). Equivalently, M is almost contact iff w<sub>2</sub>(M) lifts to an integral cohomology class.</p> <p>The existence problem for actual contact structures is much harder, though. Contact structures are known to exist in at least the following cases:</p> <ul> <li>$\pi_1(M)=0$ (Geiges, "Contact structures on 1-connected 5-manifolds", MR1147828)</li> <li>$\pi_1(M)=\mathbb{Z}/2$ and M spin (Geiges-Thomas, "Contact topology and the structure of 5-manifolds with $\pi_1=\mathbb{Z}_2$", MR1656012)</li> <li>$\pi_1(M)$ finite, of odd order with some other restrictions (Geiges-Thomas, "Contact structures, equivariant spin bordism, and periodic fundamental groups", MR1857135)</li> <li>M a product of lower-dimensional manifolds (Geiges-Stipsicz, "Contact structures on product five-manifolds and fibre sums along circles", arXiv:0906.5242)</li> <li>M a circle bundle over a symplectic (X<sup>4</sup>, &omega;) with Euler class [&omega;] (Boothby-Wang, "On contact manifolds", MR0112160)</li> </ul> <p>I'm not an expert, but I don't think there are any almost contact 5-manifolds which are known to not be contact.</p> http://mathoverflow.net/questions/4075/questions-about-analogy-between-spec-z-and-3-manifolds/4094#4094 Answer by Steven Sivek for Questions about analogy between Spec Z and 3-manifolds Steven Sivek 2009-11-04T14:47:55Z 2010-10-09T09:22:59Z <p>The analogy doesn't quite give a number theoretic version of the Poincare conjecture. See Sikora, "Analogies between group actions on 3-manifolds and number fields" (arXiv:0107210): the author states the Poincare conjecture as "S<sup>3</sup> is the only closed 3-manifold with no unbranched covers." The analogous statement in number theory is that Q is the only number field with no unramified extensions, and indeed he points out that there are a few known counterexamples, such as the imaginary quadratic fields with class number 1.</p> <p>The paper also has a nice but short summary of the so-called "MKR dictionary" relating 3-manifolds to number fields in section 2. Morishita's expository article on the subject, arXiv:0904.3399, has more to say about what knot complements, meridians and longitudes, knot groups, etc. are, but I don't think there's an explanation of what knot surgery would be and so I'm not sure how Kirby calculus fits into the picture.</p> <p>Edit: An <a href="http://arxiv.org/abs/math/0602064" rel="nofollow" title="arxiv">article</a> by B. Morin on Sikora's dictionary (and how it relates to Lichtenbaum's cohomology, p. 28): "he has given proofs of his results which are very different in the arithmetic and in the topological case. In this paper, we show how to provide a unified approach to the results in the two cases. For this we introduce an equivariant cohomology which satisfies a localization theorem. In particular, we obtain a satisfactory explanation for the coincidences between Sikora's formulas which leads us to clarify and to extend the dictionary of arithmetic topology." </p> http://mathoverflow.net/questions/28717/singmasters-conjecture/28718#28718 Answer by Steven Sivek for Singmaster's conjecture Steven Sivek 2010-06-19T03:23:56Z 2010-06-19T03:23:56Z <p>There is an upper bound of $O\left(\frac{(\log n)(\log \log \log n)}{(\log \log n)^3}\right)$ due to Daniel Kane: see "<a href="http://www.emis.de/journals/INTEGERS/papers/h53/h53.pdf" rel="nofollow">Improved bounds on the number of ways of expressing <em>t</em> as a binomial coefficient</a>," Integers 7 (2007), #A53 for details.</p> http://mathoverflow.net/questions/1489/cohomology-and-fundamental-classes/1495#1495 Answer by Steven Sivek for Cohomology and fundamental classes Steven Sivek 2009-10-20T21:13:54Z 2010-04-13T23:51:00Z <p>Rene Thom answered this in section II of "Quelques propriétés globales des variétés différentiables." Every class $x$ in $H_r(X; \mathbb Z)$ has some integral multiple $nx$ which is the fundamental class of a submanifold, so the homology is at least rationally generated by these fundamental classes.</p> <p>Section II.11 works out some specific cases: for example, every homology class of a manifold of dimension at most 8 is realizable this way, but this is not true for higher dimensional manifolds and the answer in general has to do with Steenrod operations.</p> http://mathoverflow.net/questions/20596/prime-decomposition-for-knots-in-manifolds/20625#20625 Answer by Steven Sivek for Prime decomposition for knots in manifolds Steven Sivek 2010-04-07T13:01:49Z 2010-04-07T13:01:49Z <p>This was addressed in Miyazaki, <a href="http://www.ams.org/journals/tran/1989-313-02/S0002-9947-1989-0997679-2/S0002-9947-1989-0997679-2.pdf" rel="nofollow">Conjugation and the prime decomposition of knots in closed, oriented 3-manifolds</a>, using the definition of connected sum suggested by Ryan Budney in the comments.</p> <p>The main results are that a prime decomposition of K exists iff a meridian of K is not null-homotopic in the complement of K, and if a prime decomposition of K does not contain a particular knot R in S<sup>1</sup>&times;S<sup>2</sup> then it is the unique decomposition of K, whereas knots with this summand R can admit several prime decompositions.</p> http://mathoverflow.net/questions/19745/counting-knots-with-fixed-number-of-crossings/19759#19759 Answer by Steven Sivek for Counting knots with fixed number of crossings Steven Sivek 2010-03-29T19:29:35Z 2010-03-29T19:29:35Z <p>There are some known exponential bounds on the number. For example, if k<sub>n</sub> is the number of prime knots with n crossings, then Welsh proved in "On the number of knots and links" (MR1218230) that </p> <blockquote> <p>2.68 &le; lim inf (k<sub>n</sub>)<sup>1/n</sup> &le; lim sup (k<sub>n</sub>)<sup>1/n</sup> &le; 13.5.</p> </blockquote> <p>The upper bound holds if you replace k<sub>n</sub> by the much larger number l<sub>n</sub> of prime n-crossing links.</p> <p>Sundberg and Thistlethwaite ("The rate of growth of the number of prime alternating links and tangles," MR1609591) also found asymptotic bounds on the number a<sub>n</sub> of prime alternating n-crossing links: lim (a<sub>n</sub>)<sup>1/n</sup> exists and is equal to (101+&radic;21001)/40.</p> http://mathoverflow.net/questions/17486/two-discs-with-no-parallel-tangent-planes/17497#17497 Answer by Steven Sivek for Two discs with no parallel tangent planes Steven Sivek 2010-03-08T17:31:51Z 2010-03-08T17:31:51Z <p>Suppose two such disks &Sigma;<sub>1</sub> and &Sigma;<sub>2</sub> exist, and pull back T&Sigma;<sub>2</sub> to &Sigma;<sub>1</sub> by some homeomorphism. Viewed as a subbundle of TR<sup>3</sup>|<sub>&Sigma;<sub>1</sub></sub>, this plane bundle intersects T&Sigma;<sub>1</sub> in a line bundle L over &Sigma;<sub>1</sub> since no two tangent planes are parallel. Furthermore, L|<sub>&part;&Sigma;<sub>1</sub></sub> is exactly the bundle of lines tangent to &part;&Sigma;<sub>1</sub>.</p> <p>Since a line bundle over a disk is trivial, we can take a nonzero section of L, and thus we get a nonvanishing vector field on the disk &Sigma;<sub>1</sub> which is tangent to the boundary at every point of &part;&Sigma;<sub>1</sub>. But then by doubling we can construct a nonvanishing vector field on S<sup>2</sup>, and this is impossible.</p> http://mathoverflow.net/questions/8890/flat-su2-bundles-over-hyperbolic-3-manifolds/8898#8898 Answer by Steven Sivek for Flat SU(2) bundles over hyperbolic 3-manifolds Steven Sivek 2009-12-14T19:08:52Z 2009-12-14T19:08:52Z <p>The figure eight knot is hyperbolic, so by Thurston all but finitely many 1/n-surgeries on it yield hyperbolic homology spheres. The Casson invariant of the 1/n surgery is (n/2)&Delta;''(1), where &Delta;(t) = -t + 3 - t<sup>-1</sup> is the Alexander polynomial of the figure eight, so it should be -n and therefore there are at least n distinct SU(2) representations.</p> <p>In practice you should be able to work these out explicitly from a presentation of the knot group: figure out the SU(2) representation variety of the knot group (see Klassen, "Representations of knot groups in SU(2)" for the figure eight), impose the relation &mu;&lambda;<sup>n</sup>=1, and compute what's left.</p> http://mathoverflow.net/questions/6845/a-random-walk-matrix-has-eigenvalue-1-with-multiplicty-1-why/6849#6849 Answer by Steven Sivek for A random walk matrix has eigenvalue 1 with multiplicty 1 - why? Steven Sivek 2009-11-25T22:03:01Z 2009-11-25T22:03:01Z <p>For large enough n, the matrix B = A + A<sup>2</sup> + ... + A<sup>n</sup> has positive entries since there's a path of length at most n connecting any two vertices. Thus by the Perron-Frobenius theorem B has a unique maximal eigenvalue and its multiplicity is 1.</p> <p>Any eigenvector of A with eigenvalue &lambda; is an eigenvector of B with eigenvalue f(&lambda;)=&lambda;+&lambda;<sup>2</sup>+...+&lambda;<sup>n</sup>, and f(&lambda;) is maximized on [-1,1] at &lambda;=1, so if there were two independent eigenvectors of A with &lambda;=1 then B would have two independent eigenvectors achieving the maximal eigenvalue f(1) and this is a contradiction.</p> http://mathoverflow.net/questions/6493/factorization-of-the-product-of-a-matrix-element-and-its-cofactor/6502#6502 Answer by Steven Sivek for factorization of the product of a matrix element and its cofactor Steven Sivek 2009-11-23T00:13:17Z 2009-11-23T00:36:40Z <p>For the main problem, we can use expansion by minors along the ith row to compute</p> <p><code>$\det(A_n) = (-1)^{i+1}(a(n)_{i1}\widetilde{a(n)_{i1}} - a(n)_{i2}\widetilde{a(n)_{i2}} + a(n)_{i3}\widetilde{a(n)_{i3}} - \dots)$</code></p> <p>or det(A<sub>n</sub>) = (-1)<sup>i+1</sup>c<sub>i</sub>(c<sub>1</sub> - c<sub>2</sub> + c<sub>3</sub> - ...).</p> <p>This is true for any i &lt;= n, so whenever c<sub>1</sub> - c<sub>2</sub> + ... - (-1)<sup>n</sup>c<sub>n</sub> != 0 we must have c<sub>i</sub> = (-1)<sup>i+1</sup>c<sub>1</sub> for all i &lt;= n. Therefore when c<sub>n</sub> != (-1)<sup>n+1</sup>c<sub>1</sub> for the first time, not only must we have c<sub>1</sub> - c<sub>2</sub> + ... - (-1)<sup>n</sup>c<sub>n</sub> = 0 but this alternating sum must vanish for all greater n, meaning that c<sub>j</sub> = 0 for all j > n.</p> <p>In conclusion: this is only possible if c<sub>k</sub> = (-1)<sup>k+1</sup>c<sub>1</sub> for all k &lt; K (K constant but possibly infinite) and then c<sub>k</sub> = 0 for all k > K.</p> http://mathoverflow.net/questions/6186/counting-solutions-to-xp1y4-in-a-finite-field/6196#6196 Answer by Steven Sivek for Counting solutions to x^{p+1}=y^4 in a finite field Steven Sivek 2009-11-19T23:36:55Z 2009-11-19T23:36:55Z <p>Let g be a generator of the multiplicative group of the field; assuming x and y are nonzero, we can write x=g<sup>a</sup> and y=g<sup>b</sup> with 0 &lt;= a,b &lt; p<sup>2</sup>-1, and then x<sup>p+1</sup>=y<sup>4</sup> becomes g<sup>a(p+1)</sup>=g<sup>4b</sup>, or equivalently a(p+1) = 4b (mod p<sup>2</sup>-1).</p> <p>From this we see that p+1 | 4b is necessary, and if 4b=k(p+1) then (a,b) gives a solution iff a=k (mod p-1). Since a can range from 0 to p<sup>2</sup>-2, then, there are either 0 solutions or p+1 solutions for any fixed b. The total number of nonzero solutions is therefore (p+1)* #{b | p+1 divides 4b}, and then (x,y)=(0,0) is the remaining solution.</p> <p>Now if p is 1 (mod 4) we have p+1 | 4b iff b is a multiple of (p+1)/2, and there are 2(p-1) such b up to p<sup>2</sup>-1, so there are 2(p-1)(p+1)+1 = 2p<sup>2</sup>-1 solutions.</p> <p>On the other hand, if p is 3 (mod 4) then p+1 | 4b iff b is a multiple of (p+1)/4, so we have 4(p-1) such b and there are 4p<sup>2</sup>-3 solutions.</p> http://mathoverflow.net/questions/5144/is-murasugis-conjecture-still-open/5155#5155 Answer by Steven Sivek for Is Murasugi's conjecture still open? Steven Sivek 2009-11-12T04:50:52Z 2009-11-12T04:50:52Z <p>It's actually a conjecture of Fox, sometimes known as the trapezoidal conjecture: the absolute values of the coefficients of $\Delta_K(t)$ are nonincreasing if K is an <em>alternating</em> knot. I think the original citation is Fox, "Some problems in knot theory."</p> <p>Murasugi apparently proved the conjecture for alternating algebraic knots -- see "On the Alexander polynomial of alternating algebraic knots", MR0802722, which doesn't seem to be online -- and Ozsváth and Szabó proved it for genus 2 alternating knots in "Heegaard Floer homology and alternating knots," arXiv:0209149, but it's still open in general.</p> http://mathoverflow.net/questions/3540/are-there-two-non-homotopy-equivalent-spaces-with-equal-homotopy-groups/3544#3544 Answer by Steven Sivek for Are there two non-homotopy equivalent spaces with equal homotopy groups? Steven Sivek 2009-10-31T13:53:51Z 2009-10-31T13:53:51Z <p>S<sup>3</sup> x RP<sup>2</sup> and RP<sup>3</sup> x S<sup>2</sup> are both smooth 5-manifolds with fundamental group Z/2 and universal cover S<sup>3</sup> x S<sup>2</sup>, so their homotopy groups are all the same. On the other hand, only the latter is orientable since RP<sup>3</sup> is orientable but RP<sup>2</sup> isn't, so they have different values on H<sub>5</sub> and therefore can't be homotopy equivalent. (I think this example is in Hatcher somewhere.)</p> http://mathoverflow.net/questions/2903/what-is-floer-homology-of-a-knot/2955#2955 Answer by Steven Sivek for What is Floer homology of a knot? Steven Sivek 2009-10-28T00:21:33Z 2009-10-28T00:21:33Z <p>I can say something about this for Heegaard Floer homology. Given a 3-manifold Y, you can take a Heegaard splitting, i.e. a decomposition of Y into two genus g handlebodies joined along their boundary. This can be represented by drawing g disjoint curves a<sub>1</sub>,...,a<sub>g</sub> and g disjoint curves b<sub>1</sub>,...,b<sub>g</sub> on a surface S of genus g; then you attach 1-handles along the a<sub>i</sub> and 2-handles along the b<sub>i</sub>, and fill in what's left of the boundary with 0-handles and 3-handles to get Y.</p> <p>The products T<sub>a</sub>=a<sub>1</sub>x...xa<sub>g</sub> and T<sub>b</sub>=b<sub>1</sub>x...xb<sub>g</sub> are Lagrangian tori in the symmetric product Sym<sup>g</sup>(S), which has a complex structure induced from S, and applying typical constructions from Lagrangian Floer homology gives you a chain complex CF(Y) whose generators are points in the intersection of these tori and whose differential counts certain holomorphic disks in Sym<sup>g</sup>(S). Miraculously, its homology HF(Y) turns out to be independent of every choice you made along the way. We can also pick a basepoint z in the surface S and identify a hypersurface {z}xSym<sup>g-1</sup>(S) in Sym<sup>g</sup>(S), and we can count the number n<sub>z</sub>(u) of times these disks cross that hypersurface: if we only count disks where n<sub>z</sub>(u)=0, for example, we get the hat version of HF, and otherwise we get more complicated versions.</p> <p>Given two points z and w on the surface S of any Heegaard splitting we can construct a knot in Y: draw one curve in S-{a<sub>i</sub>} and another in S-{b<sub>i</sub>} connecting z and w, and push these slightly into the corresponding handlebodies. In fact, for any knot K in Y there is a Heegaard splitting such that we can construct K in this fashion. But now this extra basepoint w gives a filtration on CF(Y); in the simplest form, if we only count holomorphic disks u with n<sub>z</sub>(u)=n<sub>w</sub>(u)=0 we get the invariant \hat{HFK}(Y,K), and otherwise we get other versions. The fact that this comes from a filtration also gives us a spectral sequence HFK(Y,K) => HF(Y).</p> <p>This was constructed independently by Ozsvath-Szabo and Rasmussen, and it satisfies several interesting properties. Just to name a few:</p> <ul> <li>for knots K in S<sup>3</sup> it has a bigrading (a,m), and the Euler characteristic \sum (-1)^m HFK<sub>m</sub>(S<sup>3</sup>,K,a) is the Alexander polynomial of K;</li> <li>there's a skein exact sequence relating HFK for K and various resolutions at a fixed crossing;</li> <li>the filtered chain homotopy type of CFK tells you about the Heegaard Floer homology of various surgeries on K;</li> <li>the highest a for which HFK<sub>*</sub>(S<sup>3</sup>,K,a) is nonzero is the Seifert genus of the knot;</li> <li>If Y-K is irreducible and K is nullhomologous, then HFK(Y,K,g(K)) = Z if and only if K is fibered (proved by Ghiggini for genus 1 and Ni in general, and later by Juhasz as well).</li> </ul> <p>For knots in S<sup>3</sup> it is also known how to compute HFK(K) combinatorially: see papers by Manolescu-Ozsvath-Sarkar and Manolescu-Ozsvath-Szabo-Thurston.</p> <p>The relation to other knot homology theories isn't all that well understood, but there are some results comparing it to Khovanov homology. For example, given a knot K in S<sup>3</sup>:</p> <ul> <li>Just as Lee's spectral sequence for Khovanov homology gave a concordance invariant s(K), the spectral sequence from HFK(K) to HF(S<sup>3</sup>) gives a concordance invariant tau(K), and both of these provide lower bounds on the slice genus of K. (Hedden and Ording showed that these invariants are not equal.)</li> <li>There's a spectral sequence from the Khovanov homology of the mirror of K to HF of the branched double cover of K.</li> <li>For quasi-alternating knots, both Khovanov homology and HFK are determined entirely by the Jones and Alexander polynomials, respectively, as well as the signature; this can be proven using skein exact sequences for both (Manolescu-Ozsvath).</li> </ul> <p>Anyway, that was long enough that I've probably made several mistakes above and still not been anywhere near rigorous. There's a nice overview that's now several years old (and thus probably missing some of the things I said above) on Zoltan Szabo's website, <a href="http://www.math.princeton.edu/~szabo/clay.pdf" rel="nofollow">http://www.math.princeton.edu/~szabo/clay.pdf</a>, if you want more details.</p> http://mathoverflow.net/questions/1781/lifting-bases-for-z-pzn-to-zn/1787#1787 Answer by Steven Sivek for Lifting bases for (Z/pZ)^n to Z^n Steven Sivek 2009-10-22T02:40:30Z 2009-10-22T05:21:58Z <p>Here's a unified argument based on my comments to Scott's post that doesn't use quadratic reciprocity in any form. Suppose n=2 and p >= 5, and lift each line of slope i in Y(2,p) to a point (a<sub>i</sub>+pb<sub>i</sub>, ia<sub>i</sub>+pc<sub>i</sub>).</p> <p>Since each pair of lifts should give a basis of Z<sup>2</sup> and thus a matrix with determinant \pm 1, taking each pair from among i=1,2,k+2 (with 1 &lt;= k &lt;= p-3) gives us conditions</p> <p>a<sub>1</sub>a<sub>2</sub> = \pm 1 (mod p)</p> <p>k*a<sub>2</sub>a<sub>k+2</sub> = \pm 1 (mod p)</p> <p>(k+1)*a<sub>1</sub>a<sub>k+2</sub> = \pm 1 (mod p).</p> <p>Combining the first two gives ka<sub>2</sub><sup>2</sup>*a<sub>1</sub>a<sub>k+2</sub> = \pm 1, or a<sub>2</sub><sup>2</sup> = \pm(1+1/k) (mod p).</p> <p>But for k=1 this gives us a<sub>2</sub><sup>2</sup> = \pm 2, and for k=2 we get a<sub>2</sub><sup>2</sup> = \pm (1 + (p+1)/2) = \pm (p+3)/2, so either (p+3)/2 = 2 (mod p) or (p+3)/2 = -2 (mod p). These imply p=1 and p=7, respectively, so already the only possible solution is p=7. But if p=7 then k=3 gives a<sub>2</sub><sup>2</sup> = \pm 6, which is not \pm 2 (mod 7), so that doesn't work either. Thus a lift with n=2 can only possibly exist if p is 2 or 3.</p> http://mathoverflow.net/questions/1237/is-s-6-the-automorphism-group-of-a-group/1245#1245 Answer by Steven Sivek for Is S_6 the automorphism group of a group? Steven Sivek 2009-10-19T17:57:48Z 2009-10-19T17:57:48Z <p>S_6 is not the automorphism group of a finite group. See H.K. Iyer, "On solving the equation Aut(X)=G", Rocky Mountain J. Math. 9 (1979), no. 4, 653--670, available online <a href="http://rmmc.eas.asu.edu/rmj/rmjVOLS/vol9/vol9-4/iyer.pdf" rel="nofollow">here</a>.</p> <p>This paper proves that for any finite group G, there are finitely many finite groups X with Aut(X)=G, and it explicitly solves the equation for some specific values of G. In particular, Theorem 4.4 gives the complete solution for G a symmetric group, and when n=6 there are no such X.</p> http://mathoverflow.net/questions/951/legendrian-homotopy-of-curves-in-a-contact-structure/954#954 Answer by Steven Sivek for Legendrian homotopy of curves in a contact structure? Steven Sivek 2009-10-17T23:24:14Z 2009-10-17T23:24:14Z <p>I don't have a general answer, but for the standard tight contact structure \xi on R^3, see "A contact geometric proof of the Whitney-Graustein Theorem" by Geiges (arXiv:0801.0046). Proposition 4 says that regular Legendrian curves in (R^3, \xi) are classified up to homotopy through Legendrian curves by their rotation number, so I guess the "Legendrian fundamental group" should be Z in this case. The proof is less than a page long and very straightforward.</p> http://mathoverflow.net/questions/849/conjugation-in-su2/942#942 Answer by Steven Sivek for Conjugation in SU(2) Steven Sivek 2009-10-17T21:10:54Z 2009-10-17T21:10:54Z <p>It's not hard to explicitly construct G using the quaternions, assuming P is not \pm Q, and I think this is worth working out in detail because I really like this picture of SU(2). Identify SU(2) with the unit quaternions by the isomorphism </p> <pre><code>[a b ] [-\bar{b} \bar{a}] --&gt; a+bj </code></pre> <p>and since the trace of such a matrix is 2*Re(a), the traceless matrices in SU(2) correspond exactly to the purely imaginary unit quaternions.</p> <p>Now if we consider R^3 as the space of all imaginary quaternions, we can describe the SU(2) action on it geometrically: Let v=cos(t)+sin(t)w, where w is purely imaginary and |w|=1. Then the conjugation map x -> vxv^{-1} is a rotation of the plane orthogonal to w -- you can easily check that it's in SO(3), and that it fixes w because vw=wv. It's also not hard to check that the angle of this rotation is 2t.</p> <p>Consider P and Q as imaginary quaternions, hence also as vectors in R^3. Then PQ = -(P.Q) + PxQ, so let G=Im(PQ)/|Im(PQ)|. Now G is a unit vector orthogonal to both P and Q, and conjugation by G is the same as rotating the plane through P and Q by \pi, so GPG^{-1}=-P and GQG^{-1}=-Q as desired.</p> <p>Finally, we just go back from the unit quaternions to SU(2): up to scale, G was supposed to be the imaginary part of the quaternion PQ, so the matrix G is some constant times the traceless part of the matrix PQ.</p> http://mathoverflow.net/questions/123141/vassilliev-invariants-of-knots-and-their-cables Comment by Steven Sivek Steven Sivek 2013-02-28T05:56:05Z 2013-02-28T05:56:05Z @Vivek: it's in section 9.2.2 of &quot;Introduction to Vassiliev Knot Invariants&quot; by Chmutov, Duzhin, and Mostovoy, available at <a href="http://www.pdmi.ras.ru/~duzhin/papers/cdbook/cdbook.pdf" rel="nofollow">pdmi.ras.ru/~duzhin/papers/cdbook/cdbook.pdf</a>. http://mathoverflow.net/questions/122359/computer-package-to-compute-homfly-polynomial Comment by Steven Sivek Steven Sivek 2013-02-21T18:14:44Z 2013-02-21T18:14:44Z @minimax: see <a href="http://katlas.org/wiki/Cabling" rel="nofollow">katlas.org/wiki/Cabling</a> for an example. You have to import the program CableComponent.m, and then I believe you want to use CableComponent[BR[TorusKnot[19,3]], Knot[3,1]]. (I haven't used CableComponent before, but this knot seems to at least have the right Alexander polynomial according to the formula $\Delta(t) = \Delta_{T_{3,19}}(t) * \Delta_{T_{2,3}}(t^3)$.) http://mathoverflow.net/questions/122359/computer-package-to-compute-homfly-polynomial Comment by Steven Sivek Steven Sivek 2013-02-20T00:21:59Z 2013-02-20T00:21:59Z The KnotTheory` Mathematica package, available with lots of documentation and examples at <a href="http://katlas.org/wiki/Setup" rel="nofollow">katlas.org/wiki/Setup</a>, can compute HOMFLY and many other knot polynomials and even produce cables for you automatically. http://mathoverflow.net/questions/113840/for-consecutive-primes-a-lt-b-lt-c-prove-that-ab-ge-c/113843#113843 Comment by Steven Sivek Steven Sivek 2012-11-20T00:42:34Z 2012-11-20T00:42:34Z You can get away with only using work of Chebyshev for large enough a: Let f(n) = \sum log(p) over all primes p up to n (usually denoted theta(n)). If a+b&lt;c then c&gt;2a and so there's at most one prime between a+1 and 2a, hence f(2a)-f(a) &lt; log(2a). He showed that f(a) &lt; a*log(4), and he proved a bound pi(N) &gt; 0.9N/log(N) for N large, so we should have f(a) &gt;= 0.7a for a large. Then for such a we have log(2a) &gt; f(2a)-f(a) &gt;= (1.4-log(4))a &gt; 0.0137a, which is impossible if a is large in the above sense and at least 505. http://mathoverflow.net/questions/99472/what-else-is-seiberg-witten-theory-equal-to/99479#99479 Comment by Steven Sivek Steven Sivek 2012-06-14T13:46:31Z 2012-06-14T13:46:31Z @Agol: Monopole Floer homology is another name for SW Floer homology, and in any case the two sutured Floer homologies are in fact isomorphic as abelian groups (i.e. maybe without a decomposition with respect to spin^c structures). This follows from work of Lekili (<a href="http://arxiv.org/abs/0903.1773" rel="nofollow">arxiv.org/abs/0903.1773</a>) together with the equivalence of Heegaard Floer and monopole Floer homologies proved by either Kutluhan-Lee-Taubes or Taubes + Colin-Ghiggini-Honda. http://mathoverflow.net/questions/81639/does-there-exist-infinitely-many-prime-knots Comment by Steven Sivek Steven Sivek 2011-11-22T20:07:19Z 2011-11-22T20:07:19Z Knots do decompose uniquely into a sum of primes, see e.g. chapter 2 of Lickorish's &quot;An introduction to knot theory.&quot; Also, since knot genus is additive under connected sum it follows that every genus 1 knot is prime, so take your favorite knot and consider all of its twisted Whitehead doubles; these have genus 1 and are distinguished by their Alexander polynomials. http://mathoverflow.net/questions/68063/are-there-any-very-hard-unlinks Comment by Steven Sivek Steven Sivek 2011-06-17T15:28:28Z 2011-06-17T15:28:28Z Dynnikov's paper &quot;Arc-presentations of links. Monotonic simplification&quot; (arXiv:0208153) was mentioned several times in answers to the unknot recognition question. The algorithm in that paper can also recognize split links and hence unlinks, and it does so without ever increasing the size of the diagram, but I don't think there are any good (e.g. subexponential) upper bounds known on the number of moves it requires. http://mathoverflow.net/questions/55769/are-infinite-dimensional-constructions-needed-to-prove-finite-dimensional-results/55782#55782 Comment by Steven Sivek Steven Sivek 2011-02-19T15:25:29Z 2011-02-19T15:25:29Z @Dan: Bizaca's paper &quot;An explicit family of exotic Casson handles&quot; shows that some Casson handles are exotic using the fact that iterated Whitehead doubles of the trefoil are not smoothly slice; at the time this required gauge theory, but it has since been done by Hedden (&quot;Knot Floer homology of Whitehead doubles&quot;) using the Ozsv&#225;th-Szab&#243; tau invariant. http://mathoverflow.net/questions/53399/spaces-with-same-homotopy-and-homology-groups-that-are-not-homotopy-equivalent Comment by Steven Sivek Steven Sivek 2011-01-27T13:18:49Z 2011-01-27T13:18:49Z It's the last comment on my answer here: <a href="http://mathoverflow.net/questions/3540/are-there-two-non-homotopy-equivalent-spaces-with-equal-homotopy-groups/3544#3544" rel="nofollow" title="are there two non homotopy equivalent spaces with equal homotopy groups">mathoverflow.net/questions/3540/&hellip;</a> http://mathoverflow.net/questions/53399/spaces-with-same-homotopy-and-homology-groups-that-are-not-homotopy-equivalent Comment by Steven Sivek Steven Sivek 2011-01-26T21:39:50Z 2011-01-26T21:39:50Z Since the homology groups of a closed 3-manifold are determined by the fundamental group, any pair of non-homotopy-equivalent 3-manifolds with the same homotopy groups should work, like the lens spaces L(5,1) and L(5,2) I mentioned in a comment at the above link. http://mathoverflow.net/questions/40887/polygons-arising-from-knot-diagrams/40891#40891 Comment by Steven Sivek Steven Sivek 2010-10-03T04:50:16Z 2010-10-03T04:50:16Z Most regions of a nice Heegaard diagram are bigons and squares, but the regions which contain basepoints can have arbitrarily many sides since the hat differential only counts disks which avoid those regions. http://mathoverflow.net/questions/33841/is-it-true-that-exotic-smooth-r4-cannot-be-diffeomorphic-to-rxn-where-n-is-a-3 Comment by Steven Sivek Steven Sivek 2010-07-29T20:18:30Z 2010-07-29T20:18:30Z See Andy Putman's first comment on <a href="http://mathoverflow.net/questions/24970/exotic-differentiable-structures-on-r4/25001#25001" rel="nofollow" title="exotic differentiable structures on r4">mathoverflow.net/questions/24970/&hellip;</a> -- he cites McMillan for the fact that it has the standard PL structure, and then Munkres to show that this implies its smooth structure is standard. http://mathoverflow.net/questions/26950/making-sure-that-you-have-comprehended-a-concept/26998#26998 Comment by Steven Sivek Steven Sivek 2010-06-04T17:28:49Z 2010-06-04T17:28:49Z @Mariano: see <a href="http://www.math.wayne.edu/~isaksen/Expository/carrying.pdf" rel="nofollow">math.wayne.edu/~isaksen/Expository/carrying.pdf</a> for &quot;a cohomological viewpoint on elementary school arithmetic.&quot; http://mathoverflow.net/questions/20331/how-not-to-write-an-nsf-proposal-poster-does-any-one-know-where-to-find-it-onl Comment by Steven Sivek Steven Sivek 2010-04-04T22:38:36Z 2010-04-04T22:38:36Z I don't know where to find the exact poster that was at MSRI, but there's a very similar one for astronomy: <a href="http://wia2009.gsfc.nasa.gov/contributed_posters/full_posters/poster27.pdf" rel="nofollow">wia2009.gsfc.nasa.gov/contributed_posters/&hellip;</a> http://mathoverflow.net/questions/20020/who-thought-that-the-alexander-polynomial-was-the-only-knot-invariant-of-its-kind/20022#20022 Comment by Steven Sivek Steven Sivek 2010-04-01T01:17:12Z 2010-04-01T01:17:12Z It's probably better to say Conway rediscovered the skein relation -- it's at the end of section 12 (&quot;Miscellaneous theorems&quot;) of Alexander's original paper, but apparently nobody noticed this for decades.