User satoshi nawata - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T11:10:21Z http://mathoverflow.net/feeds/user/17644 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/129106/links-which-homfly-homology-distinguish-but-the-homfly-polynomial-does-not/129120#129120 Answer by Satoshi Nawata for Links which HOMFLY homology distinguish but the HOMFLY polynomial does not. Satoshi Nawata 2013-04-29T17:04:08Z 2013-04-29T17:04:08Z <p>Although $5_1$ and $10_{132}$ cannot be distinguished by Jones, Alexander and (uncolored) HOMFLY-PT polynomials, their HOMFLY homologies do tell them. (See the <a href="http://arxiv.org/abs/1211.6075" rel="nofollow">review</a> by Gukov-Saberi.)</p> <p>In addition, some mutant pairs can be distinguished by Khovanov homology. (See the <a href="http://xxx.lanl.gov/pdf/math/0301312v1.pdf" rel="nofollow">paper</a> by Wehrli.)</p> http://mathoverflow.net/questions/126289/categorification-of-wrt-invariants-of-integral-homology-spheres Categorification of WRT invariants of integral homology spheres Satoshi Nawata 2013-04-02T16:19:03Z 2013-04-02T16:29:46Z <p>First, I would like to know how many definitions are there for categorification of WRT invariants. In addition, I wonder if the categorified version of WRT invariants have been explicitly computed for some integral homology spheres. </p> <p>In the case of ordinary WRT invariants, sl(2) invariants can be expressed by the summation of colored Jones polynomials with the modular S-matrices over all the colors. Could the categorified invariants for integral homology spheres be computable if the colored sl(2)-homology, say, of torus knots or twist knots, are known? If so, what replaces the modular S-matrices?</p> http://mathoverflow.net/questions/124838/khovanov-rozansky-homology-and-spectral-sequences/126283#126283 Answer by Satoshi Nawata for Khovanov-Rozansky homology and spectral sequences Satoshi Nawata 2013-04-02T15:57:22Z 2013-04-02T15:57:22Z <p>In <a href="http://arxiv.org/abs/1112.0030" rel="nofollow">the paper by Gukov and Stosic</a>, they formulate the axioms which colored HOMFLY homology (triply-graded homology) is supposed to satisfy, assuming there exists such homology. If you apply the $d_M$ differential on colored HOMFLY homology, then you will obtain colored $sl(M)$-homology. The action of $d_M$ differential is trivial for thin knot while they acts non-trivially on thick knot homology. So far, it has been proven that the axioms work consistently in the classes of the (2,2p+1)-torus knots and the twist knots.</p> http://mathoverflow.net/questions/108965/what-is-knot-contact-homology What is knot contact homology? Satoshi Nawata 2012-10-05T22:43:20Z 2013-02-20T02:15:17Z <p>Recently, it was conjectured by the <a href="http://arxiv.org/abs/1204.4709" rel="nofollow">paper</a> of Aganagic and Vafa that the $Q$-deformed $A$-polynomials can be identified with the augmentation polynomials of the knot contact homology. The $Q$-deformed $A$-polynomial of a knot $K$ can be obtained by finding the difference equation of minimal order for the colored HOMFLY polynomials of the knot $K$. This conjecture seems to hold true for torus knots and twist knots. However, I do not understand what the knot contact homology is. </p> <p>First of all, the knot contact homology describes knot invariants as invariants of the Legendrian submanifolds in the contact manifold. A knot is realized by an intersection of the cosphere bundle $ST^∗M$ of a 3-manifold $M$ with the unit conormal bundle $\Lambda_K$ where $ST^∗M$ admits a contact structure.</p> <blockquote> <p>1) Is there any way to visualize an intersection of $ST^∗M$ with $\Lambda_K$?</p> </blockquote> <p>The knot contact homology is constructed by the Legendrian differential graded algebra (DGA)</p> <blockquote> <p>2) Why do you need DGA to obtain homology theory invariant under Legendrian isotopy?</p> </blockquote> <p>From the combinatorial definition (Appendix B of the <a href="http://arxiv.org/abs/1205.1515" rel="nofollow">paper</a>), I cannot see the reason why this is isomorphic to Legendrian DGA. Although the differentials are determined by the braiding data of a knot, it seems to me that there is no contact structure involved.</p> <blockquote> <p>3) Could the isomorphism between the two DGA be explained in layman's terms?</p> </blockquote> <p>I do not understand what the augmentation polynomials of the knot contact homology are. </p> <blockquote> <p>4) Is there any relation between augmentation polynomials and Porincare-Chekanov polynomials?</p> </blockquote> <p>In addition, </p> <blockquote> <p>5) I would like to know if there is an explicit connection of knot contact homology to other knot homologies such as Khovanov-Rozansky and HOMFLY homology.</p> </blockquote> http://mathoverflow.net/questions/119110/proof-of-generalized-cauchy-formula Proof of generalized Cauchy formula Satoshi Nawata 2013-01-16T20:11:36Z 2013-02-16T01:41:44Z <p>I would like to know if there is a proof for the identity used in the superconformal index of 4d ${\cal N}=2$ gauge theory. In the <a href="http://arxiv.org/pdf/1104.3850v1.pdf" rel="nofollow">paper</a> by Rastelli el al, it was discovered that Eq. (10) is equal to the right hand side in the previous equation. $$\exp\left[\sum_{n>0}\frac{1}{n}\frac{q^{n/2}}{1-q^n}\chi_{[1]}(a_1^n)\chi_{[1]}(a_2^n)\chi_{[1]}(a_3^n) \right]=\frac{(q)_\infty}{1-q}\prod^3_i \eta_2^{-1/2}(a_i) \sum_{R}\frac{\chi_R(a_1)\chi_R(a_2)\chi_R(a_3)}{\dim_q R}$$ wheret $$\eta_2(x)=\exp\left[-2\sum_{n>0} \frac{1}{n}\frac{q^n}{1-q}(\chi_{[1]}^2(x)-1)\right]$$ Note that $\chi_R$ is a character of the irreducible representation of ${\mathfrak sl}_2$ with highest weight $R$. If there are only two characters involved, it could be seen as the Cauchy formula in representation theory. I wonder if somebody know a proof for this identity.</p> http://mathoverflow.net/questions/106735/how-much-do-homological-knot-invariants-improve-the-classification-problem-of-kno How much do homological knot invariants improve the classification problem of knots? Satoshi Nawata 2012-09-09T15:32:21Z 2012-09-10T01:09:19Z <p>The mutation operation in knots appears to be detected by the Floer homological invariants. See the papers by <a href="http://xxx.lanl.gov/abs/math/0303225" rel="nofollow">Ozsvath, Szabo</a> and by <a href="http://xxx.lanl.gov/abs/math/0610167" rel="nofollow">Baldwin, Gillam</a>. In addition, the Khovanov homology turns out to be able to detect the unknot. See the papers by <a href="http://arxiv.org/abs/0807.1432" rel="nofollow">Elisenda Grigsby, Wehrli</a> and by <a href="http://arxiv.org/abs/1005.4346" rel="nofollow">Kronheimer, Mrowka</a>.</p> <p>My question is the following.</p> <blockquote> <p>How much do homological knot invariants improve the classification problem of knots? Is there something homological invariants cannot distinguish?</p> </blockquote> http://mathoverflow.net/questions/105056/6j-symbols-for-u-q-mathfraksl-n-and-colored-homfly-polynomials $6j$-symbols for $U_q({\mathfrak{sl}}_n)$ and colored HOMFLY polynomials Satoshi Nawata 2012-08-19T21:17:52Z 2012-08-23T10:08:45Z <p>Explicit expression of quantum $6j$-symbolos for $U_q({\mathfrak{sl}_2})$ have been known due to the work of Kirillov and Reshitikhin. </p> <p>My Question:</p> <blockquote> <p>How much are known about quantum $6j$-symbolos for $U_q({\mathfrak{sl}_n})$? Are there partial results, say when the highest weights are given by small rank symmetric representation (a few horizontal boxes of Young Tableaux)?</p> </blockquote> <p>I do not know how one can calculate colored HOMFLY-PT polynomials for non-torus knots without information about quantum $6j$-symbols for $U_q({\mathfrak{sl}_n})$. Physicists guessed the form of HOMFLY-PT polynomials colored by symmetric representation for the figure-eight (See the <a href="http://arxiv.org/abs/1203.5978" rel="nofollow">paper</a> by the ITEP group and the recent <a href="http://arxiv.org/abs/1205.1515" rel="nofollow">paper</a>). However, I want to know more examples for colored HOMFLY-PT polynomials.</p> <blockquote> <p>Are there explicit formulae of colored HOMFLY-PT polynomials of non-torus knots? </p> </blockquote> <p>In addition, </p> <blockquote> <p>is there any way to calculate colored HOMFLY-PT polynomials of non-torus knots without using quantum $6j$-symbolos for $U_q({\mathfrak{sl}_n})$?</p> </blockquote> <p>A similar question can be found <a href="http://mathoverflow.net/questions/15800/calculating-6j-symbols-aka-racah-wigner-coefficients-for-quantum-groups" rel="nofollow">here</a></p> http://mathoverflow.net/questions/100882/s-matrix-for-the-homfly-hecke-category/105054#105054 Answer by Satoshi Nawata for S-matrix for the HOMFLY/Hecke category Satoshi Nawata 2012-08-19T20:45:39Z 2012-08-19T23:11:42Z <p>The $S$-matrix is given by $$\frac{S_{ij}}{S_{00}}=S_{R_i}(q^{\rho})S_{R_j}(q^{\rho+R_i})$$ where $S_{R}(x_1,\cdots,x_N)$ is the Schur polynomial with highest weight $R$, $S_{R}(q^{\rho})=S_{R}(q^{\rho_{1}},...,q^{\rho_{N}})$ and $\rho$ is the Weyl vector. Furthermore, the <a href="http://arxiv.org/abs/1105.5117" rel="nofollow">paper</a> by Aganagic and Shakirov propoesed the refinement (categorification) of the $S$-matrix $$\frac{S_{ij}}{S_{00}}=M_{R_i}(t^{\rho})M_{R_j}(t^{\rho}q^{R_i})$$ where $M_{R}(x_1,\cdots,x_N;q,t)$ is the Macdonald polynomial with highest weight $R$ and $M_{R}(t^{\rho}q^{R})=M_{R}(t^{\rho_{1}}q^{R_{1}},...,t^{\rho_{N}}q^{R_{n}};q,t)$. It reduces to the above equation for $q=t$. By using the refined topological vertex, Iqbal and Kozcaz showed that the Khovanov-Rozansky polynomial of the Hopf link is actually proportional to the refined $S$-matrix $$KhR_{ij}({\rm Hopf},q,t)\propto M_{R_i}(t^{\rho})M_{R_j}(t^{\rho}q^{R_i})$$ See Eq.(4.10) and appendix B in the <a href="http://arxiv.org/abs/1111.0525v1" rel="nofollow">paper</a>.</p> http://mathoverflow.net/questions/103561/on-finding-a-polynomials On finding A-polynomials Satoshi Nawata 2012-07-30T23:38:32Z 2012-08-02T13:36:40Z <p>I have two questions to obtain the explicit forms of A-polynomials.</p> <p><a href="http://arxiv.org/abs/math/0401068" rel="nofollow">Takata</a> used the mathematica pacage qMultisum.m to obtain the recursion relation of the colored Jones polynomials for twist knots. As Stavros Garoufalidis and Xinyu Sun pointed out in <a href="http://de.arxiv.org/abs/0802.4074" rel="nofollow">this paper</a>, the simple use of the mathematica pacage qZeil.m, qMultisum.m does not give the recursion relation of minimal order. They made use of the method, so-called creative telescoping, to get the recursion relation of minimal order by using the certificat function.</p> <blockquote> <blockquote> <ol> <li>How do you implement this method in Mathematica, say, to get the recursion relation for $5_2$ and $6_1$ knots as in p.4 of <a href="http://de.arxiv.org/abs/0802.4074" rel="nofollow">the paper</a>?</li> </ol> </blockquote> </blockquote> <p>Recently, <a href="http://arxiv.org/abs/1205.1515v2" rel="nofollow">Gukov, Sulkowski and Fuji</a> conjecture that, in the limit, $$q = e^{\hbar} \to 1 \,, \qquad a = \text{fixed} \,, \qquad t = \text{fixed} \,, \qquad x = q^n = \text{fixed}$$ the $n$-colored superpolynomials $P_n (K;a,q,t)$ exhibit the following large color'' behavior: $$P_n (K;a,q,t) \;\overset{{n \to \infty \atop \hbar \to 0}}{\sim}\; \exp\left( \frac{1}{\hbar} \int \log y \frac{dx}{x} \,+\, \ldots \right)$$ where ellipsis stand for regular terms (as $\hbar \to 0$) and the leading term is given by the integral on the zero locus of the super-$A$-polynomial: $$A^{\text{super}} (x,y;a,t) \; = \; 0 \ .$$</p> <p>For example, the critical points of the leading term of colored superpolynomials of torus knots $T^{2,2p+1}$ are give by \begin{eqnarray} 1 \; &amp;=&amp; \; -\frac{t^{-2-2p}(x-z_0)z_0^{-1-2p}(-1+t^2z_0)(1+ at^3 xz_0)}{(-1+z_0)(atx+z_0)(-1 + t^2 x z_0)} \cr y(x,t,a)&amp;=&amp; \frac{a^p t^{2 + 2 p} (-1 + x) x^{1 + 2 p} (atx + z_0) (1 + a t^3 x z_0)}{(1 + a t^3 x) (x - z_0) (-1 + t^2 x z_0)} , \end{eqnarray} which is written in Eq.(2.35) and (2.36). By eliminating $z_0$, you will obtain the super-$A$-polynomials for torus knots $T^{2,2p+1}$. Off course, it should be doable in principle, but</p> <blockquote> <blockquote> <p>$2$. how can it be implemented explicitly to obtain the super-$A$-polynomials as in Table 5 of <a href="http://arxiv.org/abs/1205.1515v2" rel="nofollow">this paper</a>? In other words, how do you explicitly eliminate $z_0$ in such a way that you will obtain the super-$A$-polynomials?</p> </blockquote> </blockquote> <p>I have the same problem to obtain the $Q$-deformed $A$-polynomials from Eq.(A.21) in <a href="http://arxiv.org/abs/1203.2182v1" rel="nofollow">this paper</a>.</p> http://mathoverflow.net/questions/102280/on-expressions-of-colored-jones-polynomials On expressions of colored Jones polynomials Satoshi Nawata 2012-07-15T11:01:48Z 2012-07-15T22:35:08Z <p>In the paper by <a href="http://arxiv.org/pdf/math/0211044.pdf" rel="nofollow">Masbaum</a>, it was shown that the colored Jones polynomials for a twist knot $K_p$ can be written as</p> <p>\begin{eqnarray} J_{n}(K_p;q)&amp;=&amp;\sum_{k=0}^{\infty} {\cal C}_{K_p}(k) \frac { \lbrace n-k\rbrace\lbrace n-k+1\rbrace \cdots \lbrace n+k\rbrace} {\lbrace n\rbrace} \cr &amp;=&amp;\sum_{k=0}^{n-1} {\cal C}_{K_p}(k) q^{n k} (q^{-n-1};q^{-1})_k (q^{-n+1};q)_k \ , \end{eqnarray}</p> <p>where $\def\usc{_}$</p> <p>\begin{eqnarray} {\cal C}_{K_p}(k) &amp;=&amp;(-1)^{k+1} {q_I}^{k(k+3)/2}\sum_{l=0}^k (-1)^l q^{l(l+1)p} \lbrace 2l+1\rbrace \frac {\lbrace k \rbrace! }{\lbrace k+l+1 \rbrace! \lbrace k-l\rbrace!}\cr &amp;=&amp; (-1)^{k+1} {q}^{k(k+3)/2}\sum_{l=0}^k (-1)^l q^{l(l+1)p+l(l-1)/2}(q^{2l+1}-1) \frac{(q;q)\usc{k}}{(q;q)\usc{k+l+1} (q;q)_{k-l}} \ . \end{eqnarray}</p> <p>For the trefoil ${\bf 3_1}$, the twist number $p$ is equal to 1. Then <a href="http://arxiv.org/pdf/math/0211044.pdf" rel="nofollow">the paper</a> says ${\cal C}_{K_1}(k)=(-1)^{k} {q}^{k(k+3)/2}$. My first question: how can it be shown? Namely, does anybody have idea how to prove this? $$\sum_{l=0}^k (-1)^l q^{l(l+1)+l(l-1)/2}(q^{2l+1}-1) \frac{(q;q)\usc{k}}{(q;q)\usc{k+l+1} (q;q)_{k-l}}=-1$$</p> <p>In addition, <a href="http://arxiv.org/abs/1205.1515" rel="nofollow">the recent paper</a> (footnote in p.13) showed a simpler form of the colored Jones polynomial for the trefoil $$J_n({\bf 3_1};q)=\sum_{k=0}^{n-1} q^{n(k+1)-1} (q^{n-1};q^{-1})_k \ .$$</p> <p>My second question: how can one prove that these two expressions are the same? $$\sum_{k=0}^{n-1} q^{n(k+1)-1} (q^{n-1};q^{-1})\usc{k} =\sum_{k=0}^{n-1} (-1)^{k} q^{k(k+3)/2} q^{n k} (q^{-n-1};q^{-1})_k (q^{-n+1};q)_k$$</p> <p>Although I can check both the identities for $k=2,3$ or $n=2,3$, I have had a hard time to prove them.</p> <p>Some notations are fixed. \begin{eqnarray} &amp; &amp; \lbrace n\rbrace={q_I}^{n}-{q_I}^{-n}, \; {q_I}^2=q, \ [n]=\frac {\lbrace n\rbrace}{\lbrace 1\rbrace}, \cr &amp; &amp; \lbrace n\rbrace!=\lbrace n\rbrace\lbrace n-1\rbrace\cdots \lbrace 1\rbrace, \; (x;q)_n=(1-x)(1-x q)\cdots (1-x q^{n-1}) \end{eqnarray}</p> http://mathoverflow.net/questions/99931/aj-conjecture-for-links AJ conjecture for links Satoshi Nawata 2012-06-18T18:51:48Z 2012-06-22T11:18:01Z <p><a href="http://arxiv.org/abs/math/0306229" rel="nofollow">Garoufalidis</a> proposed a conjecture on $q$-difference equations for the colored Jones polynomials of knots. $$\hat{A}_K(\hat{l},\hat{m};q)J_n(K;q)=0$$ where the actions of the operators $\hat l$, $\hat m$ are defined by $$\hat{l}J_n(K;q)=J_{n+1}(K;q) , \quad \hat{m}J_n(K;q)=q^{n/2}J_n(K;q) \ .$$</p> <p>Are there analogous $\hat{A}$-polynomials which define $q$-difference equations for the colored Jones polynomials of links? If there are, they should be polynomials with $2n$ variables $\hat{l}_i, \hat{m}_i$ ($i=1,\cdots,n$) for a link with $n$ components. Are their actions to the colored Jones polynomials of links known?</p> <p>More simply, is there a paper which expresses the classical $A$-polynomial $A(l,m)$ or the character variety for the Hopf link in $S^3$?</p> http://mathoverflow.net/questions/97091/what-is-the-stringy-kahler-moduli-space/97164#97164 Answer by Satoshi Nawata for what is the stringy Kähler moduli space? Satoshi Nawata 2012-05-16T23:06:37Z 2012-05-16T23:06:37Z <p>I have written an very brief introduction to <a href="http://physics.stackexchange.com/questions/28197/cy-moduli-fields/28272#28272" rel="nofollow">Calabi-Yau moduli</a>. This is covered by standard textbooks of string theory. (See the chapter 9 of <a href="http://www.amazon.com/String-Theory-M-Theory-Modern-Introduction/dp/0521860695" rel="nofollow">BBS</a>.) In addition, there are many good reviews such as <a href="http://arxiv.org/abs/hep-th/9702155v1" rel="nofollow">this</a> on this topic.</p> <p>The metric deformations $\delta g$ of a Calabi-Yau 3-fold $M$ are classified by the one of type $(1,1)$ and type $(2,1)$. The Ricci flat condition $$R_{mn}(g+\delta g)=0 \ ,$$ requires the infinitesimal $(1,1)$-form $\delta g$ to be harmonic. Therefore the K\"ahler moduli space is locally described by the vector space of $H_{1,1}$. The K\"ahler metric defines the K\"ahler form $J = ig_{ij}dz^i\wedge dz^j$, and positivity of the metric is equivalent to $$\int_M J\wedge J\wedge J >0$$ The K\"ahler moduli space has a cone structure since if $J$ satises the positivity condition, so does $sJ$ for any positive number $s$.</p> <p>In fact, in the study of string theory, the K\"ahler form of a Calabi-Yau 3-fold arises as one of the moduli fields of the associated 2d conformal field theory (Landau-Ginzburg model). Investigation of 2d CFT moduli space reveals that the corresponding geometrical description necessarily involves configurations in which the (supposed) K\"ahler form lies outside of the K\"ahler cone of the particular Calabi-Yau being studied. This may be thought of as residing in a new K\"ahler cone which shares a common wall with the original one. It was first shown by Witten in the <a href="http://arxiv.org/abs/hep-th/9301042" rel="nofollow">seminal paper</a> that this is a flop transition as you move from the one cone to the next cone, which is a famous example of topology change in string theory.</p> <p>In string theory, there is NS-NS two form $B$ which is an element of $H^{1,1}$. It is natural to combine the $B$-field with the K\"ahler form $J$, providing the complexied K\"ahler form $B+iJ$.</p> <p>As for the stability condition in the context of string theory, the <a href="http://arxiv.org/abs/hep-th/0403166v1" rel="nofollow">paper</a> by Aspinwall is a good point to start.</p> http://mathoverflow.net/questions/96793/the-equivariant-index-of-dirac-operator The equivariant index of Dirac operator Satoshi Nawata 2012-05-12T19:12:07Z 2012-05-14T19:20:48Z <p>Let us consider the Dirac complex $$D_{\rm Dirac}:S^+\to S^-$$ where $S^{\pm}$ are the chiral-spinor bundles on $\mathbb{R}^4$. Using the fact that the bundle $S^+$ is given by $\Omega^{0,0} \oplus \Omega^{0,2}$ twisted by $K^{1/2}$ while $S^-$ is given by $\Omega^{0,1}$ twisted by $K^{1/2}$ where $K$ is the canonical bundle, the equivariant index of the Dirac complex with respect to $T=U(1)_1\times U(1)_2$ action $(z_1,z_2)\mapsto (t_1 z_1,t_2,z_2)$ can be computed by<br> \begin{eqnarray} {\rm ind} D_{\rm Dirac}&amp;=&amp; \frac{ t_1^{1/2} t_2^{1/2} + t_1^{-1/2} t_2 ^{-1/2} - ( t_1^{1/2} t_2^{-1/2} + t_1^{-1/2} t_2^{1/2})} { (1-t_1)(1 -t_1^{-1})(1-t_2)(1-t_2^{-1})} \cr &amp;=&amp; \frac { t_1^{1/2} t_2^{1/2}}{ (1 -t_1)(1-t_2)} \end{eqnarray} I would like to know the reason why the spinor bundles are equivalent to the Dolbeault complex twisted by the square root $K^{1/2}$ of the canonical bundle. Why is the index of the Dirac operator equal to the one of the twisted Dolbeault operator?</p> <p>This question comes from the computation of one-loop determinant done by Pestun. (see p.35-36 in the <a href="http://arxiv.org/abs/0712.2824" rel="nofollow">paper</a> and p.34 in the <a href="http://arxiv.org/abs/1105.2568" rel="nofollow">paper</a>) It was shown in those papers that, to compute one-loop determinant $$\frac{\det_{{\rm Coker} D} T}{\det_{{\rm Ker} D} T} \ ,$$ one can use the Atiyah-Singer index theorem for transversally elliptic operators.</p> http://mathoverflow.net/questions/79666/deriving-the-hilbert-spaces-for-chern-simons-tqfts-with-complex-gauge-group/79703#79703 Answer by Satoshi Nawata for Deriving the Hilbert spaces for Chern-Simons TQFTs with complex gauge group Satoshi Nawata 2011-11-01T11:51:57Z 2011-11-01T11:51:57Z <p>The quantization procedure is proposed by Gukov by using A-polynomials</p> <p><a href="http://arxiv.org/abs/hep-th/0306165" rel="nofollow">http://arxiv.org/abs/hep-th/0306165</a></p> <p>This quantization is shown to be true for $SL(2,\mathbb{C})$ character variety of hyperbolic knots in $S^3$.</p> <p><a href="http://arxiv.org/pdf/math/0604057v2" rel="nofollow">http://arxiv.org/pdf/math/0604057v2</a></p> http://mathoverflow.net/questions/126289/categorification-of-wrt-invariants-of-integral-homology-spheres Comment by Satoshi Nawata Satoshi Nawata 2013-04-02T17:28:16Z 2013-04-02T17:28:16Z Thanks for your comments, Ben. Are the expressions for categorified invariants expected to have all positive integer coefficients? In the paper by Aganagic and Shakirov (arxiv.org/abs/1105.5117), they propose the refined modular S-matrix which is written $M_{R_i}(t^{\rho})M_{R_j}(t^{\rho}q^{R_i})$ where $M_{R}(x_1,\cdots,x_N;q,t)$ is the Macdonald polynomial with highest weight $R$. Naively thinking, this is a nice candidate for what replaces the modular S-matrix. http://mathoverflow.net/questions/108965/what-is-knot-contact-homology/122369#122369 Comment by Satoshi Nawata Satoshi Nawata 2013-03-03T18:45:41Z 2013-03-03T18:45:41Z Thank you very much for your elaborate answers! http://mathoverflow.net/questions/119110/proof-of-generalized-cauchy-formula/121940#121940 Comment by Satoshi Nawata Satoshi Nawata 2013-02-16T01:37:15Z 2013-02-16T01:37:15Z Thank you very much! Although I knew the papers, I have not read the parts you mentioned carefully. It's good to hear from one of the original authors. http://mathoverflow.net/questions/100882/s-matrix-for-the-homfly-hecke-category/105054#105054 Comment by Satoshi Nawata Satoshi Nawata 2012-08-19T23:35:12Z 2012-08-19T23:35:12Z Oh, sorry. I did not understand your question correctly. But, that's all I know. http://mathoverflow.net/questions/102280/on-expressions-of-colored-jones-polynomials/102315#102315 Comment by Satoshi Nawata Satoshi Nawata 2012-07-16T02:16:23Z 2012-07-16T02:16:23Z Thank you very much for your answer. This is very helpful. http://mathoverflow.net/questions/14714/what-do-heat-kernels-have-to-do-with-the-riemann-roch-theorem-and-the-gauss-bonne/14723#14723 Comment by Satoshi Nawata Satoshi Nawata 2012-05-15T05:14:58Z 2012-05-15T05:14:58Z The statement about supersymmetry breaking in the last is actually wrong. If the Witten index is not zero, which means there is a supersymmetric ground state with zero energy, it does NOT occur that supersymmetry is spontaneously broken. http://mathoverflow.net/questions/96793/the-equivariant-index-of-dirac-operator/96939#96939 Comment by Satoshi Nawata Satoshi Nawata 2012-05-15T04:54:18Z 2012-05-15T04:54:18Z Thank you very much for explaining me and telling me the reference. It is very much appreciated. I also found the paper by Atiyah very beautiful.