# Explanation of a theorem

Theorem. Let $$V$$ be a linear representation of $$G$$, with character $$\phi$$ and suppose $$V$$ decomposes into a direct sum of irreducible representations: $$V= W_1 \oplus \cdots \oplus W_k$$ Then if $$W$$ is an irreducible representation with character $$\chi$$, the number of $$W_i$$ isomorphic to $$W$$ is equals $$(\phi|\chi)$$.

Can someone give me a concrete but simple example of this theorem?

Consider the cyclic group $$G = \langle s \mid s^{2} \rangle$$, and consider the three representations $$\rho_1 : G \to \operatorname{GL}(\mathbb{C}) = \mathbb{C}^{\times} \ : \ s \mapsto 1,$$ $$\rho_2 : G \to \operatorname{GL}(\mathbb{C}) = \mathbb{C}^{\times} \ : \ s \mapsto -1,$$ $$\rho_3 : G \to \operatorname{GL}(\mathbb{C}^{3}) \ : \ s \mapsto \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$$

Then the characters $$\chi_i$$ of $$\rho_i$$ are given by

$$\begin{array}{c | c c c} G & 1 & s \\ \hline \chi_1 & 1 & 1 \\ \chi_2 & 1 & -1 \\ \chi_3 & 3 & -1 \\ \end{array}$$

We calculate that

\begin{align*} & \langle \chi_1, \chi_1 \rangle = \frac{1 + 1}{2} = 1, & \langle \chi_2, \chi_2 \rangle = \frac{1 + (-1)^{2}}{2} = 1, & & \langle \chi_3, \chi_3 \rangle = \frac{3^{2} + (-1)^{2}}{2} = 5, \\ & \langle \chi_1, \chi_2 \rangle = \frac{1 - 1}{2} = 0, & \langle \chi_1, \chi_3 \rangle = \frac{3 - 1}{2} = 1, & & \langle \chi_2, \chi_3 \rangle = \frac{3 + 1}{2} = 2. \\ \end{align*}

Now notice that $$\rho_1, \rho_2$$ are irreducible (dimension one), and $$\rho_3$$ is reducible since $$\langle \chi_3, \chi_3 \rangle = 5$$. We have two possibilities for $$\rho_3$$, either $$\rho_3$$ is the direct sum of a single two dimensional representation and a single one dimensional, or is the direct sum of one one-dimensional representation, and two copies of the a different one dimensional representation. Moreover, since $$\langle \chi_1, \chi_3 \rangle, \langle \chi_2, \chi_3 \rangle \neq 0$$, with $$\rho_1, \rho_2$$ irreducible, we must have that $$\rho_1, \rho_2$$ are subrepresentations of $$\rho_3$$. It follows that $$\rho_3$$ is equivalent to either $$\rho_1 \oplus \rho_1 \oplus \rho_2$$, or $$\rho_1 \oplus \rho_2 \oplus \rho_2$$. Since $$\langle \chi_2, \chi_3 \rangle = 2$$, it follows that $$\rho_3$$ is equivalent $$\rho_1 \oplus \rho_2 \oplus \rho_2$$. Now in this contrived example it is easy to see this directly. But the key take home is that if $$\rho$$ is a representation of a finite group $$G$$ over an algebraically closed field $$\mathsf{k}$$ satisfying $$\rho = \bigoplus_i \rho_i^{\oplus n_i}$$ for $$\rho_i, \rho_j$$ pairwise non-isomorphic irreducible representations, then

\begin{align*} \langle \chi, \chi_k \rangle = \operatorname{dim}_{\mathsf{k}}\operatorname{Hom}_{\mathsf{k}\left[G\right]}(\rho, \rho_k) & = \operatorname{dim}_{\mathsf{k}}\left( \bigoplus_{i} n_i \operatorname{Hom}_{\mathsf{k}\left[G\right]}(\rho_i, \rho_k) \right) \\ & = \sum_i n_i \operatorname{dim}_{\mathsf{k}}\left(\operatorname{Hom}_{\mathsf{k}\left[G\right]}(\rho_i, \rho_k) \right) \\ & = \sum_i n_i \langle \chi_i, \chi_k \rangle \\ & = \sum_i n_i \delta_{i,k} = n_k, \end{align*} where the penultimate line is due to Schur's lemma.

• Thank you. one question though, can I write the last equation using different notation than "$Hom$"? because I don't like it! – Math Apr 29 at 12:29
• You're welcome, may I ask what it is you don't like about it? I'll explain what is going on, and why the notation is as it is. A representation of a group $G$ is a pair $(\rho, V)$, where $V$ is some vector space (lets say over a field $\mathsf{k}$), and $\rho$ is a group homomorphism $\rho : G \to \operatorname{GL}(V)$, where $\operatorname{GL}(V)$ denotes the group of $\mathsf{k}$-linear automorphisms of $V$ (if $V$ is finite dimensional, these are invertible matrices after picking some basis) – Adam Higgins Apr 29 at 12:33
• What this definition is encoding is that you want $G$ to acts on the vector space $V$ in a way that commutes with the vector space structure of $V$, and the $k$-linear maps are exactly the maps that commute with the vector space structure. Now lets add a degree of abstraction, have seen the group algebra $\mathsf{k}\left[ G \right]$ of a finite group $G$ over a field $\mathsf{k}$? Another way of describing a representation is simply as a module over the group algebra, and this is where the notation $\operatorname{Hom}$ comes from – Adam Higgins Apr 29 at 12:36
• If $R$ is a ring (or perhaps an algebra), and $M,N$ are $R$-modules, then $\operatorname{Hom}_R\left(M,N\right)$ is the set of $R$-linear module homomorphisms from $M,N$. The representations $\rho_i$ above I'm actually using as shorthand for $V_{\rho_i}$, that is the representation space for $\rho_i$, and then $V_{\rho_i}$ is a $\mathsf{k}\left[ G \right]$-module. And the definition of $\langle \chi_\rho, \chi_{\tau} \rangle_G$ $\textbf{is}$ $\dim_{\mathsf{k}} \operatorname{Hom}_{\mathsf{k}\left[G\right]}\left(V_{\rho}, V_{\tau} \right)$, where $\tau, \rho$ are reps of $G$ over $\mathsf{k}$. – Adam Higgins Apr 29 at 12:40
• It is a theorem that $\langle \chi_\rho, \chi_\tau \rangle_G = \frac{1}{\left| G \right|}\sum_{g \in G} \chi_\rho(g^{-1})\chi_\tau(g)$, this is not the definition. – Adam Higgins Apr 29 at 12:42