sum of the character of the symmetric group - MathOverflow most recent 30 from http://mathoverflow.net2013-05-22T18:12:01Zhttp://mathoverflow.net/feeds/question/56795http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/56795/sum-of-the-character-of-the-symmetric-groupsum of the character of the symmetric groupHanxiong Zhang2011-02-27T05:56:13Z2011-02-27T13:18:55Z
<p>Suppose $\mu$ is a fixed partition of $n$ of length $l(\mu)$, and I was encountered with the following sum, namely
$\sum_{\nu} \chi_{\nu}(\mu)$.</p>
<p>I did some calculation using the character table that I can find (mainly Fulton & Harris's book, they have the character table up to $S_5$), and found that the sum does not vanish only if $\mu$ has an even number of even parts(someone call such $\mu$ an orthogonal partition). </p>
<p>This is actually very simple to prove, only use the fact that
$\chi_{\nu^t}(\mu)=(-1)^{n-l(\mu)} \chi_{\nu}(\mu)$, if $\mu$ has an odd number of even parts, then $n-l(\mu)$ is odd.</p>
<p>But my calculation indicates more: the sum is nonzero only if every even part of $\mu$ occurs even times.(someone also call such partition an orthogonal partition, and I donot know which is the correct definition...can anyone help?)</p>
<p>I checked this for $n \leq 6$ and also for $n=11$ (I found the charater table of $S_{11}$ in some paper...)</p>
<p>I donot know whether this is just an coincidence or this is always true. </p>
<p>Since my knowledge of symmetric group is very limited, I donot hesitate to ask for help on MO. Hopefully, someone will give me an answer. Thank you all!</p>
<p>p.s. (1) My second question, which is quite related to the above one. We know that
$\sum_{\nu} s_{\nu}(x) s_{\nu}(y)=\prod_{i,j} \frac{1}{1-x_i y_j}$, where $s_{\nu}$ is the Schur function.
Is there a similar expression for $\sum_{\nu} s_{\nu}$?</p>
<p>(2) My third question: Is there a similar expression for $\sum_{\nu} (\frac{|\nu|!}{dim R_{\nu}})^k s_{\nu}$? Here $R_{\nu}$ is the irreducible representation indexed by $\nu$, and $k$ is a positive integer.</p>
http://mathoverflow.net/questions/56795/sum-of-the-character-of-the-symmetric-group/56797#56797Answer by Philippe Nadeau for sum of the character of the symmetric groupPhilippe Nadeau2011-02-27T06:50:46Z2011-02-27T13:18:55Z<p>Your conjecture is correct; as a matter of fact, it is possible to compute these sums $C(\mu):=\sum_\nu\chi_\nu(\mu)$ exactly for any $\mu$:</p>
<p>Suppose $\mu$ has $m_i$ parts of length $i$, for $i=1,2,\ldots$. Then $C(\mu)=\prod_{i>0} c_{i,m_i}$, where $c_{i,m_i}$ is the coefficient of $t^{m_i}/({m_i}!)$ in $\exp(t+\frac{1}{2}it^2)$ if $i$ is odd, or in $\exp(\frac{1}{2}it^2)$ if $i$ is even. </p>
<p>A first proof can be found in Macdonald's <em>Symmetric functions and Hall polynomials</em>, ex.11 p.122; it relies on symmetric function techniques. A second one is given in Stanley's <em>Enumerative Combinatorics Vol. 2</em>, ex. 7.69, and is based on the fact that, from a general character theory result, $C(\mu)$ is the number of square roots of a given permutation of cycle-type $\mu$.</p>