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.

  • $\begingroup$ 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 $\endgroup$ Jun 7, 2013 at 23:19

1 Answer 1


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?

  • $\begingroup$ 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? $\endgroup$
    – user71815
    Jun 7, 2013 at 23:28
  • $\begingroup$ Sorry! This was stupid of me, of course $H$ is finite dimensional... $\endgroup$
    – user71815
    Jun 7, 2013 at 23:42
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    $\begingroup$ How can I see that the subspace generated by irreducible characters is the same as that generated by their conjugates? $\endgroup$
    – Wakaka
    Mar 1, 2016 at 9:08
  • $\begingroup$ @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. $\endgroup$
    – Michael
    Mar 25, 2021 at 15:15
  • $\begingroup$ ^ meant to say $\overline{X_1}$ is an irreducible character $\endgroup$
    – Michael
    Mar 25, 2021 at 15:29

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