# Abstract Nonsense

## The Square Root Function and its Relation to Irreducible Characters

Point of post: In this post we describe what can best be verbally described as “the number of square roots” function for a group and a way which it relates to the irreducible characters of the group.

Motivation

Recall that in our last post that we found an interesting property involving the characters: namely, we characterized real, complex, and quaternionic irreps in terms of their character. In this characterization a sum come up whose summand had the form $\chi\left(g^2\right)$. That said, since the major theorems we have thus far developed involve summands of the form $\chi(g)$ it would, of course, be preferable to change the summand in our characterization of real, complex, and quaternionic irreps into a summand involving $\chi(g)$. The way we can do this is clear, namely for each $h\in G$ we define the ‘square root’ of $h$, denoted $\sqrt{h}$, to be equal to $\#\left\{g\in G:g^2=h\right\}$. Then, with this it’s clear that our characterization can be rewritten as a sum with summand $\sqrt{h}\chi(h)$. It turns out though that the interplay goes much farther than this, to the point where we can actually express $\sqrt{h}$ entirely in terms of irreducible characters…and thus make it possible to compute $\sqrt{h}$ from a groups character table.

The Square Root Function and its Relation to Irreducible Characters

Let $G$ be a finite group and define the map $\sqrt{\text{ }}:G\to\mathbb{N}\cup\{0\}$ by $\sqrt{g}=\#\left\{h\in G:h^2=g\right\}$. We call this the square root function on $G$. Then, our previous characterization of real,complex, and quaternionic irreps can be phrased in terms of $\sqrt{\text{ }}$. Indeed, let $\rho^{(\alpha)}:G\to\mathcal{U}\left(\mathscr{V}\right)$ be an irrep of the finite group $G$. Then,

$\displaystyle \frac{1}{|G|}\sum_{g\in G}\sqrt{g}\chi^{(\alpha)}(g)=c_\alpha\quad\quad\mathbf{(1)}$

Where $c_\alpha$ is $1,0,$ and $-1$ if $\rho^{(\alpha)}$ is real, complex, and quaternionic respectively. Note that the notation $c^{(\alpha)}$ is appropriate (in the sense that it only depends on $\alpha$), since being a real, complex, or quaternionic irrep is invariant under equivalence.

To derive our main result we need to know very little about $\sqrt{\text{ }}$ itself. In fact, we only need the following basic theorem:

Theorem: Let $\sqrt{\text{ }}:G\to\mathbb{N}\cup\{0\}$ be defined as above. Then, $\sqrt{\text{ }}$ is a class function.

Proof: Let $g,h\in G$ be arbitrary. Define then $f:\left\{k\in G:k^2=g\right\}\to\left\{k\in G:k^2=hgh^{-1}\right\}$ by $k\mapsto hkh^{-1}$. This mapping is well defined (in the sense that the image of the map really lies in the codomain) since if $k^2=g$ then $\left(hkh^{-1}\right)=hk^2h^{-1}=hgh^{-1}$. Evidently then $f$ is injective (being the restriction of the inner automorphism $i_h$) and thus it suffices to prove that $f$ is surjective. To see this suppose that $k$ is an element of the codomain, then $k^2=hgh^{-1}$ and so $h^{-1}k^2h=g$ but, of course $h^{-1}k^2h=\left(h^{-1}kh\right)^2$ and so $h^{-1}kh$ is an element of the codomain. But, a quick check shows then that $f\left(h^{-1}kh\right)=k$. Since $k\in\text{codom }f$ was arbitrary the conclusion follows. Thus, since $f$ is a bijection we may conclude that

$\sqrt{g}=\#\left\{k\in G:k^2=g\right\}=\#\left\{k\in G:k^2=hgh^{-1}\right\}=\sqrt{hgh^{-1}}$

Since $g,h\in G$ were arbitrary we may conclude that $\sqrt{\text{ }}\in\text{Cl}(G)$ as required. $\blacksquare$

With this theorem in mind we are now well-equipped to prove the main result of this post. Namely:

Theorem: Let $G$ be a finite group and $h\in G$. Then,

$\displaystyle \sqrt{h}=\sum_{\alpha\in\widehat{G}}c_\alpha\chi^{(\alpha)}(h)$

Proof: Citing once again the formula $\mathbf{(1)}$ we may conjugate both sides to obtain

$\displaystyle \frac{1}{|G|}\sum_{g\in G}\sqrt{g}\overline{\chi^{(\alpha)}(g)}=c_\alpha$

Then, multiplying both sides by $\chi^{(\alpha)}(h)$ gives

$\displaystyle \frac{1}{|G|}\sum_{g\in G}\sqrt{g}\chi^{(\alpha)}(h)\overline{\chi^{(\alpha)}(g)}=c_\alpha \chi^{(\alpha)}(h)$

Then, summing over $\widehat{G}$ and performing minimal manipulations to the left-hand gives

$\displaystyle \frac{1}{|G|}\sum_{g\in G}\sqrt{g}\sum_{\alpha\in\widehat{G}}\chi^{(\alpha)}(h)\overline{\chi^{(\alpha)}(g)}=\sum_{\alpha\in\widehat{G}}c_\alpha\chi^{(\alpha)}(h)$

But, using the second orthogonality relation and denoting the conjugacy class of $h$ by $\mathcal{C}_h$ we may rewrite this as

$\displaystyle \frac{1}{|G|}\sum_{g\in G}\sqrt{g}c(g,h)\frac{|G|}{\#\left(\mathcal{C}_h\right)}=\sum_{\alpha\in\widehat{G}}c_\alpha\chi^{(\alpha)}(h)$

Which, of course may be rewritten as

$\displaystyle \frac{1}{\#\left(\mathcal{C}_h\right)}\sum_{g\in\mathcal{C}_h}\sqrt{g}=\sum_{\alpha\in\widehat{G}}c_\alpha \chi^{(\alpha)}(h)$

But, by our previous theorem we know that $\sqrt{\text{ }}$ is a class function and so the left-hand side reduces to

$\displaystyle \frac{1}{\#\left(\mathcal{C}_h\right)}\#\left(\mathcal{C}_h\right)\sqrt{h}=\sqrt{h}$

The conclusion follows. $\blacksquare$

References:

1. Isaacs, I. Martin. Character Theory of Finite Groups. New York: Academic, 1976. Print.

2. Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.