Functional equation of irreducible characters

Question

I am preparing to an exam in representations of finite groups. I am trying to tackle a problem regarding a characterization of irreducible characters: Let $f$ be a complex-valued function on a finite group $G$. Then: $$\forall x,y\in G: f(x)f(y) = \frac{f(1)}{|G|}\sum_{z\in G} f(yzxz^{-1}) \leftrightarrow f \text{ is proportional to an irreducible character}$$

I have made some progress in $\rightarrow$: if $f$ is identically 0, then RHS is implied. Otherwise, it is easy to see that $f$ is a class function (choose $y$ such that $f$ doesn't vanish on it, and notice that $\{yzxz^{-1} | z \in G \}$ is invariant under conjugation of $x$. I thought about proving $\leftarrow$ and then writing $f$ as a combination of irreducible characters.

I also noticed the following: $f(1)$ is the degree of the representation when $f$ is a character (and it appears often in orthogonality relations). Also, the projection on an isotypical component (corresponding to an irreducible representation $\pi$) of a representation $\rho$ is $\frac{\chi(1)}{|G|}\sum_{g} \overline{\chi(g)}\rho(g)$ (where $\chi$ is $\pi$'s irreducible character).

To sum it all: I have some directions, but haven't used orthogonality relations explicitly, for example. Can you help?

EDIT: Some more advances: I've noticed that if $\phi$ is a class function, the equation is equivalent to $$\sum_{z\in G} \phi(x) \phi(z^{-1} y z) \sim \sum_{z \in G} \phi(x (z^{-1}yz))$$ which looks nicer ($\sim$ implies that LHS is proportional to RHS). By putting $y=1$ it is easy to deduce that $\sum_{z\in G} \phi(x) \phi(z^{-1} y z) = \phi(1) \sum_{z \in G} \phi(x (z^{-1}yz))$ (i.e., $\phi(1)$ is the ratio). Some directions in $\leftarrow$: if $\rho$ is an irreducible representation with a simple character $\chi$, then we can let $\rho ' = \sum_{z \in G} \rho_{z^{-1}yz}$. The equation becomes $Tr(\rho_x)Tr(\rho ') = \chi(1) Tr(\rho_x \rho ')$. The product of characters might suggest use of tensor product.

Answer 1

user8268 already pointed out how to finish your proof for $\leftarrow$. For $\rightarrow$, too, your ideas lead in the right direction.

As you showed, if $f$ satisfies the equation, it's a class function; and as you realized, $f$ can then be written as a linear combination of irreducible characters. So let

$$f(x)=\sum_ic_i\chi_i(x)\;,$$

where the sum runs over all irreducible characters and the $c_i$ are complex constants; and let $\rho_i (x)$ be irreducible representations with characters $\chi_i(x)$. Then

$$ \begin{eqnarray} \sum_{z\in G}f(yzxz^{-1}) &=& \sum_{z\in G}\sum_ic_i\chi_i(yzxz^{-1}) \\ &=& \sum_{z\in G}\sum_ic_i\mathrm{Tr}\rho_i(yzxz^{-1}) \\ &=& \sum_{z\in G}\sum_ic_i\mathrm{Tr}\left(\rho_i(y)\rho_i(zxz^{-1})\right) \\ &=& \sum_ic_i\mathrm{Tr}\left(\rho_i(y)\sum_{z\in G}\rho_i(zxz^{-1})\right)\;. \end{eqnarray} $$

Now, again, by Schur's lemma $\sum_{z\in G}\rho_i(zxz^{-1})$ is a multiple of the identity, since it commutes with all $\rho_i(g)$:

$$\sum_{z\in G}\rho_i(zxz^{-1})\rho_i(g)=\sum_{z\in G}\rho_i(zxz^{-1}g)=\sum_{z\in G}\rho_i(gzxz^{-1})=\rho_i(g)\sum_{z\in G}\rho_i(zxz^{-1})\;.$$

Thus $\sum_{z\in G}\rho_i(zxz^{-1})=\kappa_i(x)\mathbb I$, with a complex constant $\kappa_i(x)$ that we can evaluate by taking the trace,

$$\kappa_i(x)=\frac{\mathrm{Tr}\sum_{z\in G}\rho_i(zxz^{-1})}{\dim\rho_i}=\frac{\chi_i(x)}{\chi_i(1)}|G|\;,$$

so $\kappa_i(x)$ as a function of $x$ is proportional to the irreducible character $\chi_i(x)$. Thus we have

$$\sum_{z\in G}f(yzxz^{-1})=\sum_ic_i\kappa_i(x)\mathrm{Tr}\rho_i(y)=\sum_ic_i\kappa_i(x)\chi_i(y)\;,$$

and the equation becomes

$$f(x)\sum_ic_i\chi_i(y)=\frac{f(1)}{|G|}\sum_ic_i\kappa_i(x)\chi_i(y)\;.$$

This is an equation between two linear combinations of the irreducible characters $\chi_i(y)$. Since these are linearly independent, we can equate the coefficients individually:

$$f(x)c_i=\frac{f(1)}{|G|}c_i\kappa_i(x)\;.$$

For each $i$ with $c_i\neq0$, this yields

$$f(x)=\frac{f(1)}{|G|}\kappa_i(x)\;.$$

But $\kappa_i(x)$ is proportional to the irreducible character $\chi_i$, and thus so is $f(x)$. (It follows that all the $c_i$ except at most one are zero.)

Answer 2

The lecturer provided another proof - slightly different from joriki proof. It is less intuitive and it uses convolution. As it happens in representation theory, many problems might eventually be equivalent and just phrased differently, but it is still cool, and avoids a direct use of Schur's lemma:

If $f(1)=0$, it follows that $f \equiv 0$, and $0$ is the zero multiple of the trivial character.

From now on we'll assume $f(1) \neq 0$. As was observed in my first comment, by putting $y=1$ we get that $f$ is a class function, so we can write $f = \sum_{i=1}^{n} a_i \chi_i$ ($\chi_i$ - the simple characters that span the class functions, $a_i$ - complex coefficients). Using $\chi_i * \chi_j = \frac{|G|}{\chi_j(1)} \chi _j \delta_{ij}$, we get: $$f * \chi_j = \frac{a_j |G|}{\chi_j(1)}\chi_j$$

• Lemma: if $a_j \neq 0$, $f * \chi_j$ is proportional to $f$.

Since $f \neq 0$, $a_j \neq 0$ for some $j$. By the lemma (which we'll soon prove) : $\frac{a_j |G|}{\chi_j(1)}\chi_j$ is proportional to $f$. It means that $f$ is proportional to $\chi_j$ (and the ratio is $a_j$), QED.

• Proof of lemma: we extend $f$ linearly to $\mathbb{C}G$, i.e.: $f(\sum_{g\in G} b_g g) = \sum_{g \in G} b_g f(g)$. We assume $f(1) = 1$ WLOG. The equation implies: $$f(y)f(x) = f(yx) \forall y \in \mathbb{C}G, x \in Z(\mathbb{C}G)$$ since the center of group algebra is spanned by the indicators of conjugacy classes.

Put $x = \sum_{g \in G} \chi_j(g^{-1}) g \in Z(\mathbb{C}G)$ (as a linear combination of class functions). $$\forall y \in G: (f * \chi_j)(y) = \sum_{g \in G} f(yg)\chi_j(g^{-1}) = f(y \sum_{g\in G} \chi_j(g^{-1})g) = f(yx) $$ $$=f(y)f(x)=f(y)\sum_{g\in G} \chi_j(g^{-1})f(g) = |G| \langle f, \chi_j \rangle f(y) = |G| a_j f(y)$$