# How to prove column orthogonality of character table

This is a standard result in representation theory of finite groups. A poof of the column orthogonality of the character table can be found here. By orthogonality of columns I mean that $$\sum_{V}\overline{\chi}_V(g)\chi_V(h)=\begin{cases} \vert G \vert/\vert c_g\vert& \text{if }g\sim h\\ 0& \text{otherwise. }\end{cases}$$ Here $$G$$ is a finite group, and $$c_g$$ is the conjugacy class of $$g\in G$$. Also, $$V$$ is supposed to be running in the irreducible representations of $$G$$.The prove mentioned before, although clever, it seems a bit artificial. In the sense that treating a character table as a matrix is not obvious at all. I think there must be a more pedestrian way to deduce this result from the row orthogonality $$\sum_g\overline\chi_V(g)\chi_W(g)=\begin{cases}\vert G\vert&\text{if }V\simeq W\\ 0& \text{otherwise.} \end{cases}$$ There must be an algebraic manipulation of equation that lead to this result. Is there one?

• What is wrong with treating it as a matrix? Anyway, what is a rigorous definition for the character table itself? Answer: you define it as a matrix. If $\chi_1,...,\chi_r$ are the irreducible characters and $g_1,...,g_r$ are representatives for the conjugacy classes of $G$ then you simply define a matrix by $(A)_{ij}=\chi_i(g_j)$. This is absolutely well defined. So the character table is a matrix by definition. (if you want to define it formally)
– Mark
Sep 1, 2020 at 21:45
• @Mark Yes. It is well defined, but I mean that it is not obvious, right away, that we should construct a table, let alone a matrix. If all we knew was that characters of different irreducible representation are orthonormal; how could we come up with the other result? Sep 1, 2020 at 22:02
• I don't know other proofs. I mean, multiplying matrices is exactly the "algebraic manipulation" that you ask about. When you multiply matrices, the corresponding equations give you the result. I think it is a pretty natural idea to construct a matrix when we work with finite orthogonal sequences.
– Mark
Sep 1, 2020 at 22:10

Lemma: The characters of irreducible representations form a basis for the space $$\mathbb C_{class}(G)$$ of class functions $$f:G\to \mathbb C$$.
Proof: We prove that whenever a class function $$f$$ is orthogonal to every character of irreducible representations, then it is the zero map. Let $$\rho:G\to\text{GL}(V)$$ be an irreducible representation of $$G$$. If $$f\in \mathbb C_{class}(G)$$, then the map $$\phi=\sum_{g\in G}f(g)\rho(g^{-1})$$ is a representation homomorphism. By Schur's Lemma $$\phi=\lambda I$$ for some $$\lambda\in \mathbb{C}$$. The trace of $$\phi$$ vanishes by assumption. Thus $$\phi$$ is the zero map in any irreducible representation and therefore in any representation of $$G$$. Setting the representation $$\rho$$ as the regular representation we have $$\sum_{g\in G}f(g)\rho(g^{-1})=0$$, and all group actions in the regular representation are linearly independent. Then $$f=0$$. $$\blacksquare$$
1. Fix $$g\in G$$, let $$f(h)=1$$ if $$h\sim g$$ and $$f(h)=0$$ otherwise.
2. Write $$f=\sum_{i}a_i\chi_i$$, we consider the irreducible representations $$V_i$$ with characters $$\chi_i$$.
3. Compute $$a_i$$ by means of the Hermitian product (use orthonormality).