On finding A-polynomials - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T18:57:07Z http://mathoverflow.net/feeds/question/103561 http://www.creativecommons.org/licenses/by-nc/2.5/rdf 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>