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.

  • $\begingroup$ 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? $\endgroup$ – bfhaha Mar 23 '18 at 23:19
  • $\begingroup$ I don't know. Can we say that $\rho_i$ restricted on $M_{d_i}(\mathbb{C})$ is the identity map on it? $\endgroup$ – 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 $s\in S_i$, \begin{equation} \tag{1} \left(\sum_{g\in G}z_g g\right)\cdot s =\sum_{g\in G}z_g (\rho_i(g)(s)). \end{equation}

  • 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.


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.