Suppose $\mu$ is a fixed partition of $n$ of length $l(\mu)$, and I was encountered with the following sum, namely $\sum_{\nu} \chi_{\nu}(\mu)$.

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).

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.

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?)

I checked this for $n \leq 6$ and also for $n=11$ (I found the charater table of $S_{11}$ in some paper...)

I donot know whether this is just an coincidence or this is always true.

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.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 $\sum_{\nu} s_{\nu}$?

(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.

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    $\begingroup$ (1) is answered in www-igm.univ-mlv.fr/~berstel/Lothaire/AlgCWContents.html "The Plactic Monoid" as well. $\endgroup$ – darij grinberg Feb 27 '11 at 12:44
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    $\begingroup$ Actually I have requested a categorification of (1) on MO some time ago: mathoverflow.net/questions/45065/… . $\endgroup$ – darij grinberg Feb 27 '11 at 12:47
  • $\begingroup$ For (0), check Theorem 2.1 in math.ipm.ac.ir/pournaki/Main%20Files/1-Papers/Accepted-3/… . I am pretty sure it does NOT require $g\in G^{\prime}$. It yields a formula for the sum of all irreducible characters applied to some group element. It is zero if the group element is not a commutator. The thing should now boil down to proving that a permutation is a commutator if and only if its cycle partition has an even number of even parts. Am I right? $\endgroup$ – darij grinberg Feb 27 '11 at 12:54
  • $\begingroup$ Oh wait. This is not an answer to (0). Still it solves a similar sum (namely, $\sum_{\nu} \frac{\chi_{\nu}}{\chi_{\nu}\left(1\right)}$. $\endgroup$ – darij grinberg Feb 27 '11 at 13:03
  • $\begingroup$ Thanks for the link and reference, Darij. But I do not think that Theorem 2.1 actually simplifies anything, it replace the sum with something not explicit and still intractable... $\endgroup$ – Hanxiong Zhang Feb 28 '11 at 2:59

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 $\mu$:

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.

A first proof can be found in Macdonald's Symmetric functions and Hall polynomials, ex.11 p.122; it relies on symmetric function techniques. A second one is given in Stanley's Enumerative Combinatorics Vol. 2, 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 $\mu$.

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    $\begingroup$ Thanks for the reference, Philippe! And I think you really mean Macdonald's Symmetric functions and Hall polynomials :) My second question is also solved, which is clearly written in Macdonald's book, page 76. $\endgroup$ – Hanxiong Zhang Feb 27 '11 at 7:44
  • $\begingroup$ Indeed that's what I meant. $\endgroup$ – Philippe Nadeau Feb 27 '11 at 13:19

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