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.

  • $\begingroup$ Thank you. one question though, can I write the last equation using different notation than "$Hom$"? because I don't like it! $\endgroup$ – Math Apr 29 at 12:29
  • $\begingroup$ 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) $\endgroup$ – Adam Higgins Apr 29 at 12:33
  • $\begingroup$ 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 $\endgroup$ – Adam Higgins Apr 29 at 12:36
  • $\begingroup$ 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 $\mathsf{k}$. $\endgroup$ – Adam Higgins Apr 29 at 12:40
  • $\begingroup$ 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. $\endgroup$ – Adam Higgins Apr 29 at 12:42

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