# Total sum of squares of characters of the symmetric group $\mathfrak{S}_n$

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? $$\sum_{\mu\vdash n}\sum_{\lambda\vdash n}(\chi_{\mu}^{\lambda})^2 =\frac{1}{n!} \sum_{\alpha \in \frak{S}_n} \left( \sum_{\text{irreps}\ \chi} \chi(\alpha)^2 \right)^2.$$

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

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.)

• Wouldn't one also know how to decompose ordinary products if one knew how to decompose squares, since $2\chi\phi = (\chi + \phi)^2 - \chi^2 - \phi^2$? 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 $$G$$.