# Irreducible characters form orthonormal basis of set of class functions

I am reading Serre's book (Linear Representations of Finite Groups). Theorem 6 in chapter 2 says that the irreducible characters $\chi_1,\dotsc,\chi_h$ of a finite group $G$ form an orthonormal basis of $H$, the set of class functions on $G$. It says in the proof (given the $\chi_i$ form are orthonormal) that 'it is enough to show that every element of $H$ orthogonal to the $\chi_i^\ast$ is zero', where $\ast$ is the complex conjugate. Why is this so? Will appreciate any hints etc.

• The author is trying to show that characters form a complete system, that is, that their orthogonal complement is null. This proves that they span the whole space Jun 7, 2013 at 23:19

If $V$ is a finite dimensional inner product space, and $W$ is a subspace, then:

$$V=W\oplus W^{\perp}$$

Now, let $V$ be the space of class functions, and let $W$ be the subspace generated by the irreducible characters. Then the statement you've written means to show that $W^{\perp}=0$ from which it follows that $V=W$.

Perhaps we should also note that the subspace generated by irreducible characters is the same as that generated by their conjugates. Can you see why this is true?

• But I thought the proof was to demonstrate that $H$ is finite dimensional and that the characters form an orthonormal basis? How do we know a priori that $H$ is finite dimensional?
– user71815
Jun 7, 2013 at 23:28
• Sorry! This was stupid of me, of course $H$ is finite dimensional...
– user71815
Jun 7, 2013 at 23:42
• How can I see that the subspace generated by irreducible characters is the same as that generated by their conjugates? Mar 1, 2016 at 9:08
• @Wakaka If $X$ is the character of a representation $p$, it's not too hard to show that $\overline{X}$ is also the character of some irreducible representation (if $g \in G$, then consider taking the inverse transpose of $p(g)$). So, if $X_1, \ldots, X_k$ are all the irreducible characters of $G$, then b/c $\overline{X_1}$ is also an irreducible representation, it must be among the original $X_i$. From this, it isn't too hard to deduce that the set of characters $X_i$ and the set of their conjugates are the same. Mar 25, 2021 at 15:15
• ^ meant to say $\overline{X_1}$ is an irreducible character Mar 25, 2021 at 15:29