What are the places where character table values of $S_n$ occurs naturally? one such an example is when we write power sum symmetric function of order n in terms of Schur function of order n the coefficient of each monomial will be the character value of the respective conjugacy class and respective irreducible representation in the character table of $S_n$.

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    $\begingroup$ One is tempted to say: In the representation theory of the symmetric groups! ;-) $\endgroup$ – Johannes Hahn Apr 12 '13 at 17:54
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    $\begingroup$ It is very weird that you accepted an answer so soon: you asked for what is essentially a list... $\endgroup$ – Mariano Suárez-Álvarez Apr 12 '13 at 18:21
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    $\begingroup$ Should be CW it would seem. $\endgroup$ – Benjamin Steinberg Apr 12 '13 at 18:32
  • $\begingroup$ I am new to mathoverflow. cant I accept more than one answer? I found that answer interesting. $\endgroup$ – GA316 Apr 12 '13 at 18:35

My favourite place where the character table of $S_n$ occurs naturally is in the enumeration of branched covers of Riemann surfaces (Hurwitz numbers).

In more detail, let $\Sigma$ be a closed Riemann surface of $g$ with $k$ marked points $x_1, \ldots, x_k$, and fix an integer $n\geq 1$. For each marked point $x_i$ pick a partition $\mu^i$ of $n$.

We define $\mathcal H_n(g,\mu_1 , \ldots ,\mu_k)$ to be the sum of $1/|Aut(f)|$ for each isomorphism class of branched covers $f: \widetilde{\Sigma} \to \Sigma$ which are ramified precisely at $x_i$ with ramification type $\mu^i$.

Then there is a formula due to Hurwitz (or perhaps Frobenius, or Burnside, depending on which version...?):

$\mathcal H_n(g,k) = \sum_{\nu \in P(n)} \left(\prod_{i=1} ^k|ccl(\mu^i)| \frac{\chi_\nu(\mu^i)}{\chi_\nu(1)}\right) \left( \frac{\chi_\nu(1)}{n!} \right)^{2-2g}$.

Here $\nu \in P(n)$ indexes the irreducible representations $\chi_\nu$ of $S_n$, and the $\mu^i$ are identifed with conjugacy classes in $S_n$.

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  • $\begingroup$ Of course, this isn't really specific to the symmetric group... we have the same formula for every finite group $G$ (enumerating $G$-Galois covers). But the symmetric group case has historical significance. $\endgroup$ – Sam Gunningham Apr 12 '13 at 18:09
  • $\begingroup$ what is ccl in the expression for $Hn$? $\endgroup$ – GA316 Apr 12 '13 at 18:14
  • $\begingroup$ That means the size of the conjugacy class in $S_n$ corresponding to the partition $\mu ^i$. There is a formula for this, I just didn't want to write it down... $\endgroup$ – Sam Gunningham Apr 12 '13 at 18:20
  • $\begingroup$ @SamGunningham I don't think your first comment is quite correct. You count degree $n$ covers unbranched away from $p_1,...,p_k$ by counting homomorphisms $\pi_1(\Sigma\setminus\{p_1,...,p_k\})\to S_n$, and those with ramification data $\mu_1,...,\mu_k$ by forcing the images of loops around $p_i$ to lie in the right conjugacy class. For $\Sigma=\mathbf{P}^1$ a result of Frobenius then gives your formula. Maps $\pi_1(\Sigma\setminus\{p_1,...,p_k\})\to G$ do not correspond to $G$-Galois covers (otherwise all degree $n$ covers would be $S_n$ Galois), but those with $\text{Aut}(f)\subseteq G$. $\endgroup$ – Meow Mar 20 '19 at 14:56
  • $\begingroup$ @Meow Apologies, you are correct. To make sense of what I say, you should replace the phrase "$G$-Galois cover" with "principle $G$-bundle" (I am not assuming the total space is connected). Note that there is an equivalence of categories between principal $S_n$-bundles and $n$-fold covers (taking the associated bundle with respect to the standard $S_n$-action on the set $\{1,2, \ldots, n\}$). I apparently have been using the terminology wrong... $\endgroup$ – Sam Gunningham Mar 20 '19 at 15:27

As a somewhat trivial example, the sign of a permutation is a character of $S_n$, and the sign function appears in many contexts.

For instance, the sign function is used when calculating the determinant of a matrix by summing over permutations in $S_n$. Replacing the sign character in this formula with another character of $S_n$ gives the definition of the immanant of a matrix.

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  • $\begingroup$ thanks. but I want little more natural answer. $\endgroup$ – GA316 Apr 12 '13 at 18:51

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