In my [earlier MO post](https://mathoverflow.net/questions/403957/total-sum-of-characters-of-the-symmetric-group-fraks-n), 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](https://mathoverflow.net/questions/162428/the-simultaneous-conjugacy-problem-in-the-symmetric-group-s-n#comment414650_162428) and also [here](https://mathoverflow.net/questions/41337/a-general-formula-for-the-number-of-conjugacy-classes-of-mathbbs-n-times-m?rq=1) 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?
> $$\sum\_{\mu\vdash n}\sum\_{\lambda\vdash n}(\chi\_{\mu}^{\lambda})^2
> =\frac{1}{n!} \sum\_{\alpha \in \frak{S}\_n} \left( \sum\_{\text{irreps}\ \chi} \chi(\alpha)^2 \right)^2.$$
> 
> 
> 



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


