1
$\begingroup$

Now, I've been learning about character theory and I've been building up to showing that the character table is square; the number of irreducible characters is equal to the number of conjugacy classes.

During the proof of this, it was stated that the number of irreducible $G$-spaces is less than or equal to the number of conjugacy classes, with the following reasoning;

In a vector space of dimension $n$, you can't have more than $n$ non-zero vectors which are pairwise orthogonal.

Now, I know this must be referring to the irreducible characters being pairwise orthogonal, and I know the reasoning is clearly true, but I'm not entirely sure how it relates to the proof of the statement?

What vector space are they referring to here? How does it relate to conjugacy classes?

$\endgroup$
1
$\begingroup$

They are referring to the vector space of class functions. A class function is a function $f : G \to \mathbb C$ that is constant on each conjugacy class of $G$. (I'm telling the story as if you are working over $\mathbb C$ in order to avoid annoying technicalities...)

Together, the class functions form a vector space over $\mathbb C$, whose dimension is equal to the number of conjugacy classes of $G$. We can define an inner product on this vector space: $$\langle f\vert h\rangle = \frac{1}{|G|}\sum_{g \in G}f(g) h(g^{-1})$$

For any representation $\rho : G \to {\rm GL}(\mathbb C, n)$, the character $\chi(\rho) : G \to \mathbb C$, being constant on conjugacy classes, is a class function. The characters for irreducible representations are orthogonal with respect to the inner product defined above. So the number of irreducible representations must be less than or equal to the dimension of the vector space of class functions, which is equal to the number of conjugacy classes of $G$.

$\endgroup$
  • $\begingroup$ Ah, thank you very much! $\endgroup$ – the man Jan 1 at 19:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.