# Commutant Algebra of Matrix Representation

I am currently working on Bruce Sagan's The Symmetric Group.

In the following example they show that for a representation that contains 2 different subrepresentations the commutative algebra Com$X$ has the form

\begin{equation} \text{Com}X = \left\{ c_1I_{d_1} \oplus c_2I_{d_2}: c_1, c_2 \in \mathbb{C} \right\}. \end{equation}

Suppose that $X$ is a matrix representation such that

\begin{equation} X = \begin{pmatrix}X^{(1)}&0\\0&X^{(2)}\end{pmatrix} = X^{(1)} \oplus X^{(2)}, \end{equation}

where $X^{(1)}, X^{(2)}$ are inequivalent and irreducible of degrees $d_1, d_2,$ respectively. What does Com$X$ look like?

Suppose that

\begin{equation} T = \begin{pmatrix}T_{1,1}&T_{1,2}\\T_{2,1}&T_{2,2}\end{pmatrix} \end{equation}

is a matrix partitioned in the same way as $X$. If $TX = XT$, then we can multiply out each side to obtain

\begin{equation} \begin{pmatrix}T_{1,1}X^{(1)}&T_{1,2}X^{(2)}\\T_{2,1}X^{(1)}&T_{2,2}X^{(2)}\end{pmatrix} = \begin{pmatrix}X^{(1)}T_{1,1}&X^{(1)}T_{1,2}\\X^{(2)}T_{2,1}&X^{(2)}T_{2,2}\end{pmatrix}. \end{equation}

Equating corresponding blocks we get

\begin{equation} T_{1,1}X^{(1)} = X^{(1)}T_{1,1}, \end{equation} \begin{equation} T_{1,2}X^{(2)} = X^{(1)}T_{1,2}, \end{equation} \begin{equation} T_{2,1}X^{(1)} = X^{(2)}T_{2,1}, \end{equation} \begin{equation} T_{2,2}X^{(2)} = X^{(2)}T_{2,2}. \end{equation}

Using Corollaries 1.6.6 and 1.6.8 along with the fact that $X^{(1)}$ and $X^{(2)}$ are inequivalent, these equations can be solved to yield

\begin{equation} T_{1,1} = c_1I_{d_1}, T_{1,2} = T_{2,1} = 0, T_{2,2} = c_2I_{d_2}, \end{equation}

where $c_1, c_2 \in \mathbb{C}$ and $I_{d_1}, I_{d_2}$ are identity matrices of degrees $d_1, d_2$. Thus

\begin{equation} T = \begin{pmatrix} c_1I_{d_1}&0\\0&c_2I_{d_2} \end{pmatrix}. \end{equation}

We have shown that when $X = X^{(1)} \oplus X^{(2)}$ with $X^{(1)} \neq X^{(2)}$ and irreducible, then

\begin{equation} \text{Com}X = \left\{ c_1I_{d_1} \oplus c_2I_{d_2}: c_1, c_2 \in \mathbb{C} \right\}, \end{equation}

where $d_1 = \text{deg}X^{(1)}, d_2 = \text{deg}X^{(2)}$.

Corollary 1.6.6 is the matrix version of Schur's Lemma and Corollary 1.6.8 says: "Let $X$ be an irreducible matrix representation of $G$ over the complex numbers. Then the only matrices $T$ that commute with $X(g)$ for all $g \in G$ are those of the form $T = cI$ - i.e., scalar multiples of the identity matrix."

There are three things that I do not understand in this example:

$1$. How can we assume that $T$ has the block form

\begin{equation} T = \begin{pmatrix}T_{1,1}&T_{1,2}\\T_{2,1}&T_{2,2}\end{pmatrix}? \end{equation}

Would it not be possible that $T$ does not have such a block form?

$2$. How do we know that $T_{1,2} = T_{2,1} = 0$? Schur's Lemma says that either $T_{1,2}, T_{2,1}$ are invertible or they are zero matrices. How do we know that they are not invertible?

Thank you very much for your help!

Write $V=V_1\oplus V_2$, a direct sum of two irreducible representations of $G$ with dimensions $d_1,d_2$, respectively. We are showing the intertwiners in ${\rm End}_G(V)$ look like $c_1I_{V_1}\oplus c_2 I_{V_2}$.
(1) Every $(n+m)\times(n+m)$ matrix can be written as a block matrix $(\begin{smallmatrix} A & B \\ C & D \end{smallmatrix})$ with the appropriate matrices $A\in M_{n\times n}$, $B\in M_{n\times m}$, $C\in M_{m\times n}$, $D\in M_{m\times m}$. This is actually trivial. What do you think the matrices $A,B,C,D$ are? They're already right there in the matrix already! We just need to put parentheses (or brackets if that's your thing) around them. For instance:
$$\quad \begin{pmatrix} \color{Blue}{1} & \color{Blue}{2} & \color{Blue}{3} & \color{Purple}{4} & \color{Purple}{5} \\ \color{Blue}{6} & \color{Blue}{7} & \color{Blue}{8} & \color{Purple}{9} & \color{Purple}{10} \\ \color{Blue}{11} & \color{Blue}{12} & \color{Blue}{13} & \color{Purple}{14} & \color{Purple}{15} \\ \color{Magenta}{16} & \color{Magenta}{17} & \color{Magenta}{18} & \color{Red}{19} & \color{Red}{20} \\ \color{Magenta}{21} & \color{Magenta}{22} & \color{Magenta}{23} & \color{Red}{24} & \color{Red}{25}\end{pmatrix}\longrightarrow \begin{pmatrix} \color{Blue}{\begin{pmatrix} 1 & 2 & 3 \\ 6 & 7 & 8 \\ 11 & 12 & 13\end{pmatrix}} & \color{Purple}{\begin{pmatrix} 4 & 5 \\ 9 & 10 \\ 14 & 15\end{pmatrix}} \\ \color{Magenta}{\begin{pmatrix} 16 & 17 & 18 \\ 21 & 22 & 23 \end{pmatrix}} & \color{Red}{\begin{pmatrix}19 & 20 \\ 24 & 25\end{pmatrix}} \end{pmatrix}.$$
(2) We're not using Schur's lemma to prove $T_{1,2}$ and $T_{2,1}$ are zero. Instead, as the text explicitly says, we're using that fact that $V_1$ and $V_2$ are inequivalent. Look at the first equation, which reads as $T_{1,2}X^{(2)}=X^{(1)}T_{1,2}$. This tells us $T_{1,2}$ is a morphism $V_2\to V_1$ of representations. Can there be any nonzero morphisms of representations between inequivalent irreducible representations?