# Confusion regarding G-modules

• Jan 13th 2010, 07:53 PM
AtticusRyan
Confusion regarding G-modules
The following question illustrates the confusion I have (from The Symmetric Group by Bruce Sagan):

Let $X$ be a reducible matrix representation with block form given by $\left(\begin{array}{cc}A(g)&B(g)\\0&C(g)\end{array }\right)$ where $A,B,C$ are square matrices of the same size.

Let $V$ be a module for $X$ with submodule $W$ corresponding to $A$. Consider the quotient vector space $V/W=\{\mathbf{v}+W\mid \mathbf{v}\in W\}$.

Show that $V/W$ is a $G$-module with corresponding matrices $C(g)$. Furthermore, show that we have $V\cong W\oplus (V/W)$.

My confusion is in the definition of $W$ being a submodule (i.e. $gw\in W$ and the requirement that it is a module in its own right corresponding to matrices $A(g)$ i.e. $gw=A(g)w$.

How does one do this question? I have trouble getting started and confusion with all the definitions. Any help would be greatly appreciated!

Many thanks,

Atticus
• Jan 14th 2010, 12:30 AM
aliceinwonderland
Quote:

Originally Posted by AtticusRyan
The following question illustrates the confusion I have (from The Symmetric Group by Bruce Sagan):

Let $X$ be a reducible matrix representation with block form given by $\left(\begin{array}{cc}A(g)&B(g)\\0&C(g)\end{array }\right)$ where $A,B,C$ are square matrices of the same size.

Let $V$ be a module for $X$ with submodule $W$ corresponding to $A$. Consider the quotient vector space $V/W=\{\mathbf{v}+W\mid \mathbf{v}\in W\}$ (v in V?).

Show that $V/W$ is a $G$-module with corresponding matrices $C(g)$. Furthermore, show that we have $V\cong W\oplus (V/W)$.

My confusion is in the definition of $W$ being a submodule (i.e. $gw\in W$ and the requirement that it is a module in its own right corresponding to matrices $A(g)$ i.e. $gw=A(g)w$.

How does one do this question? I have trouble getting started and confusion with all the definitions. Any help would be greatly appreciated!

Many thanks,

Atticus

Let V be a G-module of dimension d with basis $B = \{w_1, w_2, \cdots , w_k, v_{k+1}, v_{k+2}, \cdots , v_d \}$, where a submodule W of dimension k has a basis $B' = \{w_1, w_2, \cdots , w_k\}$. Since the lower left corner of X(g) is zero and A corresponds to W, we see that $X(g)w_i \in W$ for all $1 \leq i \leq k$ and the last d-k coordinates of $X(g)w_i$ is zero. Since W is a G-submodule of V and $g(V/W)=\{gv + gW | v \in V\}=\{v' + W | v' \in V \}$, it follows that $V/W$ is a G-invariant subspace of V. Now V/W is an orthogonal complement of W, $V \cong W \oplus (V/W)$. If W corresponds to A, then the orthogonal complement of W, i.e. , V/W, should correspond to C.
• Jan 14th 2010, 12:56 AM
AtticusRyan
Why should the orthogonal complement correspond to C? And does the matrix B come into it at all? If not, why is it mentioned, i.e. how is this case different from a block-diagonal case?
• Jan 14th 2010, 05:41 PM
aliceinwonderland
Quote:

Originally Posted by AtticusRyan
Why should the orthogonal complement correspond to C? And does the matrix B come into it at all? If not, why is it mentioned, i.e. how is this case different from a block-diagonal case?

Let $V=\mathbb{C}^d$; let $W=\mathbb{C}\{e_1, e_2, \cdots, e_k\}$, where $e_i$ is the column vector with a 1 in the ith row and zeros elsewhere. Then the orthogonal complement of W, i.e., $V/W = \mathbb{C}\{e_{k+1}, e_{k+2}, \cdots, e_d \}$, where $X(g)e_i \in (V/W)$ for $k+1 \leq i \leq d$ and all $g \in G$. This accounts for the matrix C(g) in X(g) and zero elsewhere.

If X is a matrix representation with a block form $\left(\begin{array}{cc}A(g)&B(g)\\0&C(g)\end{array }\right)$, then X is a reducible representation. If X is a matrix representation with a block form $\left(\begin{array}{cc}A(g)&0\\0&C(g)\end{array}\r ight)$, then X is a decomposable representation. Note that there is an indecomposable representation but is reducible. For example, if $\mathbb{Re}$ is an additive group of real numbers, then $\rho: \mathbb{Re} \rightarrow GL_2(\mathbb{Re})$ defined by $a \mapsto \left(\begin{array}{cc}1&a\\0&1\end{array}\right)$.

Since V is a decomposable G-module, there is a decomposable matrix representation of X such that B(g) is a zero matrix by changing a basis of V.