An n!xn! determinant. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T15:51:40Z http://mathoverflow.net/feeds/question/88399 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88399/an-nxn-determinant An n!xn! determinant. Igor Makhlin 2012-02-14T03:49:08Z 2012-03-11T19:36:50Z <p>Let us consider the matrix $A$ with its rows and columns enumerated by the elements of $S_n$ with $A_{\sigma\tau}=x^{c(\sigma\tau^{-1})}$ where $c()$ is the number of cycles in a permutation's decomposition. I'm interested in $|A|$. More specifically I aim to prove that all of its roots as of a polynomial in $x$ are integers between $-n+1$ and $n-1$ but the roots' multiplicities would also be nice to know.</p> http://mathoverflow.net/questions/88399/an-nxn-determinant/88412#88412 Answer by Gjergji Zaimi for An n!xn! determinant. Gjergji Zaimi 2012-02-14T09:31:45Z 2012-02-14T16:01:18Z <p>I might as well write an answer with the proof I referenced to above (found as theorem 110 <a href="http://qspace.library.queensu.ca/bitstream/1974/5235/1/Novak_Jonathan_I_200909_PhD.pdf" rel="nofollow">here</a>). Hopefully Darij will write a more detailed answer tomorrow.</p> <p>The first thing to observe is that your matrix $A$ is the image of the element $$\omega=\sum_{\sigma\in S_n} x^{|\pi|}\pi$$ in the regular representation of the group algebra $\mathbb C[S(n)]$. Next notice that this element factors as $$\omega=(x+J_1)(x+J_2)\cdots(x+J_n)$$ where $J_k$ are the Jucys-Murphy elements defined as $J_k=\sum_{ s &lt; k} (s,k)$.</p> <p>Let's denote $\Xi_n=\lbrace J_1,J_2,\dots,J_n,0,0,\dots\rbrace$. It is a theorem that for any symmetric function $f\in \Lambda$, the mapping $f\to f(\Xi_n)$ sends symmetric polynomials onto elements of the class algebra $\mathcal Z(n)$. </p> <p>Now since $f(\Xi_n)\in \mathcal Z(n)$, by Schur's lemma it acts as a scalar on any irreducible representation $V^{\lambda}$ of $\mathbb C[S(n)]$. Jucys theorem says that the central character of $f(\Xi_n)$ acting on $V^{\lambda}$ can be obtained by simply substituting the alphabet $\Xi_n$ with the content alphabet $$A_{\lambda}=\lbrace c(\square): \square\in \lambda\rbrace.$$ These two facts are proved in Jucys' article, <a href="http://www.sciencedirect.com/science/article/pii/0034487774900196" rel="nofollow">"Symmetric polynomials and the center of the symmetric group ring"</a>.</p> <p>So, in particular, the central character of $\omega$ is $$\prod_{\square\in \lambda}(x+c(\square)),$$ and, putting things together, from the decomposition $\mathbb C[S(n)]=\bigoplus_{\lambda \vdash n} (\dim \lambda)V^{\lambda}$, we obtain $$\det(A)=\prod_{k=1}^{n-1}(x^2-k^2)^{r_k},$$ where $$r_k=\sum_{\lambda\vdash n, k\in A_{\lambda}} \dim \lambda.$$</p> http://mathoverflow.net/questions/88399/an-nxn-determinant/88421#88421 Answer by David Speyer for An n!xn! determinant. David Speyer 2012-02-14T11:23:58Z 2012-02-14T11:23:58Z <p>This determinant came up, and was evaluated, in the <a href="http://sbseminar.wordpress.com/2009/02/25/a-warmup-gl_t-for-t-not-an-integer/#comment-4870" rel="nofollow">comments of the Secret Blogging Seminar</a>. The motivation there was that it vanishes if and only if $V^{\otimes n}$ has neglible endomorphisms in Deligne's category of "$GL_x$ representations for noninteger $x$". Here $V$ is the "$x$-dimensional representation of $GL_x$". See that post for more.</p> http://mathoverflow.net/questions/88399/an-nxn-determinant/88451#88451 Answer by Jonathan Novak for An n!xn! determinant. Jonathan Novak 2012-02-14T19:10:38Z 2012-02-14T19:19:28Z <p>Hello, I really think it helps to think about the entries of this matrix does $q^{d(\sigma,\tau)}$ where d denotes the distance between two permutations on the Cayley graph of the symmetric group generated by the set of all transpositions. In this light you can compare the determinant you're interested in with the one computed by Don Zagier in his article "realizability of a model in infinite statistics." In that paper, Zagier computes the analogous determinant corresponding to the Cayley graph of the symmetric group generated by adjacent transpositions (i.e. the Coxeter generators). This is a very different geometry on permutations but Zagier finds a similar factorization showing that the determinant has roots on the unit circle. Zagier's result is based on a clever factorization in the group algebra, just like the factorization into JM elements discussed above. </p> <p>Another point of view on Zagier's result, relating it to hyperplane arrangements, is explained in Stanley and Hanlon's article "a q-deformation of a trivial symmetric group action." Also, if you want to see similar results in the context of the Brauer algebra, you should look up Paul Zinn-Justin's paper <a href="http://arxiv.org/abs/0907.2719" rel="nofollow">"Jucys-Murphy elements and Weingarten matrices."</a> I saw that a user above posted a link to my Phd thesis - I was young and foolish then and more polished versions of those results have since appeared in a paper written jointly with Sho Matsumo called <a href="http://arxiv.org/abs/0905.1992" rel="nofollow">"Jucys-Murphy elements and unitary matrix integrals."</a> </p> http://mathoverflow.net/questions/88399/an-nxn-determinant/88893#88893 Answer by darij grinberg for An n!xn! determinant. darij grinberg 2012-02-19T05:04:52Z 2012-03-11T19:36:50Z <p><a href="http://mit.edu/~darij/www/algebra/MO88399.pdf" rel="nofollow">Here is the proof</a> that I hinted at in the comments section in more details. Repeated mistakes absorbed most of the time I spent writing it, which is why it took four days; let me apologize for this.</p>