On expressions of colored Jones polynomials - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T17:20:34Z http://mathoverflow.net/feeds/question/102280 http://www.creativecommons.org/licenses/by-nc/2.5/rdf 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/102280/on-expressions-of-colored-jones-polynomials/102315#102315 Answer by Ian Agol for On expressions of colored Jones polynomials Ian Agol 2012-07-15T22:35:08Z 2012-07-15T22:35:08Z <p>If you look at the paper by Garoufalidis and Le "<a href="http://front.math.ucdavis.edu/0309.5214" rel="nofollow">The colored Jones function is q-holonomic</a>", they prove that the colored Jones function is $q$-holonomic. Using techniques of Zeilberger, one can then verify such identities algorithmically. So you need only check the first few terms are equal, and check that both sides of the equation satisfy the same recursion relation, which is given by the non-commutative $A$-polynomial which is <a href="http://front.math.ucdavis.edu/0802.4074" rel="nofollow">computed for twist knots</a>.</p>