# irreducible representation extending as projection onto matrix algebra

I want to ask a question about some statement in Peter Webb's "A course in finite group representation theory", p40 (book on his website, not the published version)

Let $\rho_1,\ldots,\rho_r$ be irreducible representations of finite group $G$ over $\mathbb{C}$ with degree $d_1,\ldots,d_r$ and character $\chi_1,\ldots,\chi_r$. It says that after extending by linearity $\rho_i:G\rightarrow M_{d_i}(\mathbb{C})$ to $\mathbb{C}$-algebra homomorphism $\rho_i: \mathbb{C}[G]\rightarrow M_{d_i}(\mathbb{C})$, we idetify $\mathbb{C}[G]$ with $\Pi_{j=1}^rM_{d_j}(\mathbb{C})$ by Artin-Wedderburn, then each $\rho_i:\Pi_{j=1}^rM_{d_j}(\mathbb{C})\rightarrow M_{d_i}(\mathbb{C})$ is a projection onto the $i$th matrix summand.

My question is why he can conclude that $\rho_i$ is a projection onto $M_{d_i}(\mathbb{C})$ or why $\rho_i$ is surjective. The book says it is because of the way we decompose $\mathbb{C}[G]$ as a sum of matrix algebra. But I have spent whole day on this problem and still can't figure it out.

• Doesn't $\rho_i:\Pi_{j=1}^r M_{d_j}(\mathbb{C})\rightarrow M_{d_i}(\mathbb{C})$ satisfy the definition of a projection? – bfhaha Mar 23 '18 at 23:19
• I don't know. Can we say that $\rho_i$ restricted on $M_{d_i}(\mathbb{C})$ is the identity map on it? – 廖信傑 Apr 3 '18 at 3:02

This argument is from Proposition 7.2, p.382, and Lemma 8.3, p387, in Grillet's Abstract Algebra 2/e.

• Recall that \begin{eqnarray*} \Bbb{C}G &\cong& M_{d_1}(\Bbb{C})\oplus M_{d_2}(\Bbb{C})\oplus \cdots\oplus M_{d_r}(\Bbb{C})\\ &\cong& \overbrace{S_1\oplus S_1\oplus \cdots\oplus S_1}^{d_1\text{-times}}\oplus \\ && \overbrace{S_2\oplus S_2\oplus \cdots\oplus S_2}^{d_2\text{-times}}\oplus \\ &&\cdots\oplus\\ && \overbrace{S_r\oplus S_r\oplus \cdots\oplus S_r}^{d_r\text{-times}} \\ &\cong& S_1^{d_1}\oplus S_2^{d_2}\oplus \cdots \oplus S_r^{d_r}. \end{eqnarray*} In fact, $$S_i\cong \left\{\left(0, ..., 0, \begin{pmatrix} 0 & \cdots & 0 & a_{1k} & 0 & \cdots & 0 \\ 0 & \cdots & 0 & a_{2k} & 0 & \cdots & 0 \\ \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & \cdots & 0 & a_{d_i k} & 0 & \cdots & 0 \\\end{pmatrix}, 0, ..., 0\right)\mid a_{1k}, a_{2k}, ..., a_{d_i k}\in \Bbb{C}\right\}$$ for some $k\in \{1, 2, ..., d_i\}$.

• Since $\rho_i$ is an irreducible representation of $G$ over $\Bbb{C}$, let $L_i$ be the corresponding simple $\Bbb{C}G$-module. That is, $\rho_i:G\to GL(L_i)$.

• $L_i$ is isomorphic to a simple $\Bbb{C}G$-submodule $S_i$ of $\Bbb{C}G$. (See Theorem C-2.33 in Rotman's Advanced Modern Algebra 3/e Part II.) So $\rho_i:G\to GL(S_i)$. In rigorous, we should write $L_i\cong 0\oplus \cdots \oplus 0\oplus S_i\oplus 0\oplus\cdots \oplus 0$.

• By the correspondence between the representations and modules, for any $\sum_{g\in G}z_g g\in \Bbb{C}G$ and $s\in S_i$, $$\tag{1} \left(\sum_{g\in G}z_g g\right)\cdot s =\sum_{g\in G}z_g (\rho_i(g)(s)).$$

• We define $\overline{\rho_i}:\Bbb{C}G\to \text{End}_{\Bbb{C}}{(S_i)}$ by $$\overline{\rho_i}\left(\sum_{g\in G}z_g g\right) =\sum_{g\in G}z_g \rho_i(g).$$

• For any \begin{eqnarray*} s &\in& S_i\\ &\cong& 0\oplus \cdots \oplus 0\oplus S_i\oplus 0\oplus\cdots \oplus 0\\ &\subseteq & 0\oplus \cdots \oplus 0\oplus M_{d_i}(\Bbb{C})\oplus 0\oplus\cdots \oplus 0\\ &\subseteq & \Bbb{C}G \end{eqnarray*} and \begin{eqnarray*} x=\sum_{g\in G}z_g g &\in& M_{d_j}(\Bbb{C}) \text{ (Notice the subscript!)}\\ &\cong& 0\oplus \cdots \oplus 0\oplus M_{d_j}(\Bbb{C})\oplus 0\oplus \cdots \oplus 0\\ &\subseteq & \Bbb{C}G, \end{eqnarray*} where $j\neq i$, we have $x\cdot s=0$ because they are in the different components $M_{d_i}(\Bbb{C})$ and $M_{d_j}(\Bbb{C})$. In this case, we view $x\cdot s$ as a prodcut of two elements $x$ and $s$ in $\Bbb{C}G$.
Then \begin{multline*} \overline{\rho_i}\left(x\right)(s)= \overline{\rho_i}\left(\sum_{g\in G}z_g g\right)(s) =\left(\sum_{g\in G}z_g \rho_i(g)\right)(s)\\ =\sum_{g\in G}z_g (\rho_i(g)(s)) \stackrel{(1)}{=}\left(\sum_{g\in G}z_g g\right)\cdot s =x\cdot s =0 \end{multline*} It follows that $\overline{\rho_i}\left(x\right)=0$ if $x\in M_{d_j}(\Bbb{C})$ for $j\neq i$.

• For any $s\in S_i$ and $x=\sum_{g\in G}z_g g\in M_{d_i}(\Bbb{C})$ (Notice the subscript!), \begin{multline*} \overline{\rho_i}\left(x\right)(s) =\overline{\rho_i}\left(\sum_{g\in G}z_g g\right)(s) =\left(\sum_{g\in G}z_g \rho_i(g)\right)(s)\\ =\sum_{g\in G}z_g (\rho_i(g)(s)) \stackrel{(1)}{=}\left(\sum_{g\in G}z_g g\right)\cdot s =x\cdot s. \end{multline*} In this case, we view the action of $x$ on $s$ as a linear transformation on $S_i$. It follows that $\overline{\rho_i}\left(x\right)=x$.

• In addition, $\text{End}_{\Bbb{C}}{(S_i)}\cong \text{End}_{\Bbb{C}}{(L_i)}\cong M_{d_i}(\Bbb{C})$. That is why the author says that $\overline{\rho_i}:\Bbb{C}G\to M_{d_i}(\Bbb{C})$ is a projection.