In my earlier MO post, I proposed the double sum $\sum_{\mu\vdash n}\sum_{\lambda\vdash n}\chi_{\mu}^{\lambda}$ regarding characters of the symmetric group $\mathfrak{S}_n$. Soon after, I started considering the sum of squares $\sum_{\mu\vdash n}\sum_{\lambda\vdash n}(\chi_{\mu}^{\lambda})^2$ hoping to gain a better formula. A further look into older MO posts here and also here shows a Burnside-type Lemma $$\frac{1}{n!} \sum_{\alpha \in \frak{S}_n} \left( \sum_{\text{irreps}\ \chi} \chi(\alpha)^2 \right)^2.$$

After comparing the last two sums, I got prompted to ask:

QUESTION. The numerics suggest the below equality. Why is this true? $$\sum_{\mu\vdash n}\sum_{\lambda\vdash n}(\chi_{\mu}^{\lambda})^2 =\frac{1}{n!} \sum_{\alpha \in \frak{S}_n} \left( \sum_{\text{irreps}\ \chi} \chi(\alpha)^2 \right)^2.$$

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    $\begingroup$ It is clear that the right side is $p(n)n!$. $\endgroup$ Sep 15, 2021 at 17:41

2 Answers 2


The sum $\sum_\chi \chi(\alpha)^2$ is the size of the centralizer $z_\mu=\frac{n!}{|K_\mu|}=1^{m_1}m_1! 2^{m_2}m_2!\cdots n^{m_n} m_n!$ if $\alpha$ has cycle type $\mu=1^{m_1}2^{m_2}\ldots$, so $$\frac{1}{n!} \sum_{\alpha\in \mathfrak S_n} \left(\sum_\chi \chi(\alpha)^2\right)^2= \frac{1}{n!} \sum_{\mu\vdash n} |K_\mu| z_\mu^2=\sum_{\mu\vdash n} z_\mu =\sum_{\mu\vdash n}\sum_{\lambda\vdash n} (\chi^{\lambda}_\mu)^2,$$ as desired.

Although this identity is clear, there is something interesting here when you stand back. When you first come to representations, you'd maybe first think of $\mathfrak S_n$ acting on itself by left multiplication, which you quickly check affords the character $\rho$ which gives $|\mathfrak S_n|$ at the identity element and $0$ elsewhere. So, using the inner product, you immediately decompose this representation into irreducibles and find $$\rho=\sum_\chi \chi(1)\chi,$$ i.e. each irreducible shows up with multiplicity its dimension. Now you might try $\mathfrak S_n$ acting on itself by conjugation, which affords the character $\psi$ with value $z_\mu$ at any element of cycle type $\mu$. How does this one decompose? As we saw before, $$\langle \psi,\chi^\lambda\rangle=\sum_{\mu\vdash n}\chi^\lambda_\mu.$$ But using that $$\sum_{\lambda\vdash n} (\chi^\lambda_\mu)^2=z_\mu,$$ we also have the expression
$$\psi=\sum_{\chi} \chi^2,$$ since evaluating the RHS at an element with cycle type $\mu$ indeed gives $z_\mu$.

I should mention that the individual products $\chi^2$ are not generally well understood. For example, there is a conjecture that if $\lambda$ is a staircase shape $(n,n-1,n-2,\ldots,1)$, then the square of $\chi^\lambda$ contains at least one copy of each irreducible, i.e. $$\langle (\chi^\lambda)^2,\chi^\nu\rangle\geq 1$$ for all $\nu\vdash \binom{n+1}{2}$. This is a folklore conjecture due to Jan Saxl. (Of course, if one knew how to decompose arbitrary products $\chi\phi$, then one would know how to decompose the squares $\chi^2$, and if one knew how to decompose squares, then one would know how to decompose the conjugation representation.)

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    $\begingroup$ Wouldn't one also know how to decompose ordinary products if one knew how to decompose squares, since $2\chi\phi = (\chi + \phi)^2 - \chi^2 - \phi^2$? $\endgroup$
    – LSpice
    Sep 15, 2021 at 20:35

I think this is true. Let $X$ be the character table of $S_{n}$, viewed just as a $p(n) \times p(n)$ matrix. Then the left side is ${\rm trace}(XX^{T}) = {\rm trace}(X^{T}X).$ This is $\sum_{\alpha} |C_{S_{n}}(\alpha)|$, where $\alpha$ runs over a set conjugacy class representatives for $S_{n}.$ As mentioned in my comment, the right side is clearly $\frac{1}{n!} \left( \sum_{\alpha} |S_{n}| |C_{S_{n}}(\alpha)| \right),$ where $\alpha$ again runs through class representatives for $S_{n},$ so both sides are equal.

Later edit: For this result, the only special property of $S_{n}$ used is that its complex irreducible characters are all real valued. The following similar equality is true for a general finite group $G$ by a slight modification of the argument above (a straightforward consequence of the orthogonality relations).

$$\sum_{(\chi,x)}|\chi(x)|^{2} = \frac{1}{|G|}\sum_{y \in G} \left( \sum_{\chi} |\chi(y)|^{2} \right)^{2},$$ where $\chi$ runs through the distinct irreducible complex characters of $G$ and $x$ runs over a set of representatives for the conjugacy classes of $G$.


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