I've just begin a course on character theory. Juste to repeat we say :

Let $G$ be a finite group. Then a character $\chi$ is a morphism from $G \to \mathbb{C}^*$. We then have some property on dual group $G^{\wedge}$ :

For example with the inner product : $\langle \chi_1, \chi_2 \rangle = \frac{1}{\mid G \mid } \sum_{g \in G} \chi_1(g) \overline{\chi_2(g)}$ all the element of the dual group are orthogonal which means $\langle \chi_1, \chi_2 \rangle = 0, \chi_1 \ne \chi_2$

My problem is that I don't understand all these definitions which are a bit cumbersome for me.

Why are we defining the dual of group as all the morphism from $G$ to $\mathbb{C}^*$ and not from $G$ to an other group ?

In linear algebra I have a really good sens of what an inner product is, and what it represent (it's a projection between two vectors). Here I don't understand what geometrically this inner product represent. The factor $\frac{1}{\mid G \mid}$ in the formula maybe is there to say : we are kind of looking at the barycenter of the element of the dual ?

With some intuition is it then possible that : the orhtogonality of the element of the dual is an obvious fact ?

Thank you !

  • $\begingroup$ I'd say a morphism (character) $\chi : \mathbb{Z}/n_1\mathbb{Z} \times \ldots \times \mathbb{Z}/n_l \mathbb{Z} \to \mathbb{C}^*$ is of the form $\chi(a) = \prod_{j=1}^l e^{2i \pi a_j k_j / n_j}$. A character $G \to \mathbb{C}^*$ is a morphism $G \to G^{ab} \to \mathbb{C}^*$ where $G^{ab} = G / [G,G]$ (abelianization) is the group where any morphism to an abelian group factors through. $\endgroup$ – reuns Nov 22 '18 at 23:27

The fantastic thing about the dual group $G^\vee = \operatorname{Hom}(G, \mathbb{C}^\times)$ is that it is in fact a group, and so a lot of questions can be reduced to simply asking whether something is the identity or not. Here are some facts about linear characters:

  • The identity character is $\chi_{\mathrm{id}}(g) = 1$ for all $g \in G$.
  • If $\chi$ is a character, so is $\overline{\chi}$, and since characters are valued on the unit circle, $\overline{\chi(g)} = \chi(g)^{-1} = (\chi^{-1})(g)$.

After this we can prove the relation $$\sum_{g \in G} \chi(g) = \begin{cases} |G| & \text{if } \chi = \chi_{\mathrm{id}} \\ 0 & \text{otherwise} \end{cases}$$ the first case is clear, so let $\chi \neq \chi_{\mathrm{id}}$. Then there is some $g_0 \in G$ such that $\chi(g_0) \neq 1$, and we have $$\chi(g_0) \sum_{g \in G} \chi(g) = \sum_{g \in G} \chi(g_0 g) = \sum_{g \in G} \chi(g)$$ and hence $\sum_{g \in G} \chi(g) = 0$.

After this, the orthgonality relations are clear, since $\langle \chi_1, \chi_2 \rangle$ is just plugging in $\chi_1 \overline{\chi_2} = \chi_1 \chi_2^{-1}$ into that sum up above.

If you read further into the representation theory of finite groups, it will turn out that for any irreducible representation $V$ of $G$ (not just one-dimensional representations), there is a character $\chi_V$, and we have $\langle \chi_V, \chi_W \rangle = 1$ when $V$ and $W$ are isomorphic representations, and 0 when they are different. This is a good motivating example for defining the inner product.


To me, characters are interesting insofar as they are $1$-dimensional versions of representation-theoretic characters.

Starting from a field $k$ and a group $G$ you may define finite-dimensional $k$-representations of $G$ as morphisms $\rho: G\to GL(V)$ for some finite dimensional $k$-vector space $V$. Then the character of $\rho$ is $\mathrm{tr}\circ \rho : G\to k$. In dimension $\geq 2$, this is not multiplicative in general, but in dimension $1$, $\mathrm{tr}\circ\rho$ is naturally identified with $\rho$, and it becomes a group morphism $G\to k^\times$.

So we are tempted to define more general duals as $\hom (G, k^\times)$ for some fields $k$. However $k=\mathbb{C}$ is interesting for at least two reasons : it is algebraically closed which allows to use reduction techniques on $\rho$; but more importantly here, character theory works great on $\mathbb{C}$ because you have a nice hermitian product which has great properties, in particular you may define $\langle \chi, \psi\rangle$ for characters (or more generally central functions) and there are many representation-theoretic properties of $\rho$ that are translated to hermitian properties of $\mathrm{tr}\circ\rho$ : you can find the dimension, or determine the irreducibility of a representation knowing only its character !

If you want an example, you can prove quite nicely that for representations $\rho_1,\rho_2$ with characters $\chi_1,\chi_2$ , $\langle \chi_1,\chi_2\rangle = \dim \hom_G(\rho_1,\rho_2)$. For more general fields you cannot have such properties (e.g. in positive characteristic, even if you manage to define a nice inner product, you lose information because you can at most recover dimensions and other integer invariants modulo the characteristic)

Another direction for generalization (that also shows that it's interesting to consider $k=\mathbb{C}$) is to consider continuous representations $G\to GL(V)$ for some topological group $G$ with a Borel measure $\mu$; e.g. compact Hausdorff abelian groups. For these you also have an interesting dual group, and you have a duality called Pontryagin duality

  • $\begingroup$ I think most people would not call composition with trace the character unless the field has characteristic 0. $\endgroup$ – Tobias Kildetoft Nov 23 '18 at 17:12
  • $\begingroup$ @TobiasKildetoft : Interesting, that's how I learned it though $\endgroup$ – Max Nov 23 '18 at 19:07
  • $\begingroup$ I see. Do you have some reference to a book that defines it this way? It just seems highly unusual since you would lose so much information. $\endgroup$ – Tobias Kildetoft Nov 23 '18 at 19:51
  • $\begingroup$ @TobiasKildetoft : No I actually only learned it in class. But if you find it so unusual, it might simply be that my teachers defined the characters when thinking about characteristic $0$ before fully moving on to that (they usually started in arbitrary characteristic before endikg up in $\mathbb{C}$) $\endgroup$ – Max Nov 23 '18 at 20:13

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.