Representations and Modules of Lie Algebra

Representations

Definition 1. Let $M_n(k)$ be the algebra of $n\times n$ matrices over $K$ and let
$$
\mathfrak{gl}_n(K):=[M_n(K)]
$$

be the corresponding Lie algebra, which is also called the general linear Lie algebra.

Definition 2. A representation of a Lie algebra $L$ over $K$ is a homomorphism of Lie algebra
$$
\rho:L\to \mathfrak{gl}_n(K).
$$

In this case, we say $\rho$ has degree $n$.

Definition 3. We say two representations $\rho,\rho’$ are equivalent if there exists a non-singular $n\times n$ matrix $T$ such that
$$
\rho’(x)=T^{-1}\rho(x)T \qquad \text{for all }x\in L.
$$

$L$-module

Definition 4. A left $L$-module is a vector space $V$ over $K$ with
$$
\begin{aligned}
L\times V&\to V\\\
(x,v)&\mapsto xv
\end{aligned}
$$

satisfying the axioms

(i) $(x,v)\mapsto xv$ is linear in $x$ and in $v$.

(ii)
$$
[x,y]v=x(yv)-y(xv)
\qquad \text{for all }x,y\in L,\ v\in V.
$$

Example 1. Let $V$ be a finite-dimensional $L$-module. Let $e_1,\dots,e_n$ be a basis for $V$. For $x\in L$, write
$$
xe_j=\sum_{i=1}^n \rho_{ij}(x)e_i
\qquad \text{for }\rho_{ij}(x)\in K.
$$

Now we claim

$$
\rho:x\mapsto \bigl(\rho_{ij}(x)\bigr)_{n\times n}
$$

is a representation of $L$.

Proof : We need to show:
$$
\rho([x,y])=[\rho(x),\rho(y)]
\qquad \text{for }x,y\in L.
$$

For $e_j$ in the chosen basis,
$$
[x,y]e_j=x(ye_j)-y(xe_j).
$$

Using
$$
ye_j=\sum_{k=1}^n \rho_{kj}(y)e_k,
\qquad
xe_j=\sum_{k=1}^n \rho_{kj}(x)e_k,
$$
we get
$$
[x,y]e_j
=x\left(\sum_{k=1}^n \rho_{kj}(y)e_k\right)-y\left(\sum_{k=1}^n \rho_{kj}(x)e_k\right).
$$

Since the action is linear,
$$
[x,y]e_j
=\sum_{k=1}^n \rho_{kj}(y)xe_k-\sum_{k=1}^n \rho_{kj}(x)ye_k.
$$

Again using
$$
xe_k=\sum_{i=1}^n \rho_{ik}(x)e_i,
\qquad
ye_k=\sum_{i=1}^n \rho_{ik}(y)e_i,
$$
we obtain
$$
[x,y]e_j
=\sum_{k=1}^n \rho_{kj}(y)\left(\sum_{i=1}^n \rho_{ik}(x)e_i\right)
-\sum_{k=1}^n \rho_{kj}(x)\left(\sum_{i=1}^n \rho_{ik}(y)e_i\right).
$$

Therefore
$$
[x,y]e_j
=\sum_{i=1}^n \left(\sum_{k=1}^n \rho_{ik}(x)\rho_{kj}(y)-\sum_{k=1}^n \rho_{ik}(y)\rho_{kj}(x)\right)e_i,
$$

where
$$
\sum_{k=1}^n \rho_{ik}(x)\rho_{kj}(y)
$$
is the $(i,j)$-entry of $\rho(x)\rho(y)$, and
$$
\sum_{k=1}^n \rho_{ik}(y)\rho_{kj}(x)
$$
is the $(i,j)$-entry of $\rho(y)\rho(x)$. Hence
$$
[x,y]e_j
=\sum_{i=1}^n \bigl(\rho(x)\rho(y)-\rho(y)\rho(x)\bigr)_{ij}e_i.
$$

Therefore,
$$
\rho([x,y])=\rho(x)\rho(y)-\rho(y)\rho(x)
=[\rho(x),\rho(y)].\quad \square
$$

Question: What if we choose another basis $f_1,\dots,f_n$ above? Let $\rho’$ denote the corresponding representation.

Proposition 1. $\rho’$ is equivalent to $\rho$.

Proof : Let $T$ be the transition matrix such that
$$
f_j=\sum_{i=1}^n T_{ij}e_i.
$$

Let $x\in L$. Then
$$
xf_j=\sum_{k=1}^n T_{kj}xe_k
$$
and hence
$$
xf_j
=\sum_{k=1}^n T_{kj}\left(\sum_{i=1}^n \rho_{ik}(x)e_i\right)
=\sum_{i=1}^n \left(\sum_{k=1}^n \rho_{ik}(x)T_{kj}\right)e_i.
$$

On the other hand,

$$
xf_j=\sum_{k=1}^n \rho’_{kj}(x)f_k
$$

and therefore

$$
xf_j
=\sum_{k=1}^n \rho_{kj}^\prime(x) \left(\sum_{i=1}^n T_{ik}e_i\right)
=\sum_{i=1}^n \left(\sum_{k=1}^n T_{ik}\rho_{kj}^\prime(x)\right)e_i.
$$

Comparing coefficients of $e_i$, we get

$$
\sum_{k=1}^n \rho_{ik}(x)T_{kj}
=
\sum_{k=1}^n T_{ik}\rho’_{kj}(x).
$$

Thus
$$
\rho(x)T=T\rho’(x),
$$
so
$$
\rho’(x)=T^{-1}\rho(x)T.
$$

Hence $\rho’$ is equivalent to $\rho$. $\square$

Proposition 2.: Define

$$
\begin{aligned}
L \times L &\to L,\\\
(x, y) &\mapsto [x, y]
\end{aligned}
$$

then $L$ is a left $L$-module.

Proof : For $x, y \in L$ and $v \in L$,
$$
[x, y], v] = [x, [y, v]] - [y, [x, v]]. \quad \square
$$

Remark: This is called the adjoint module of $L$. Define

$$
\begin{aligned}
ad \ x: L &\to L\\\
y &\mapsto [x, y]
\end{aligned}
$$

Then it is easy to show that

$$
ad\ [x, y] = (ad\ x)(ad\ y) - (ad\ y)(ad\ x).
$$

Submodule

Notation: Let $V$ be a left $L$-module, $U$ be a subspace of $V$, $H$ be a subspace of $L$. Let $HU$ be the subspace of $V$ spanned by elements $xu$ for $x \in H$, $u \in U$.

Definition 5. A submodule of $V$ is a subspace $U$ of $V$ such that
$$
LU \subseteq U.
$$

It is easy to show that $V$ and $\{0\}$ are submodules of $V$. A proper submodule of $V$ is a submodule of $V$ distinct from $V$ and $\{0\}$.

Definition 6. A $L$-module $V$ is irreducible if it has no proper submodules.

Definition 7. $V$ is completely reducible if $V$ is a direct sum of irreducible submodules.

Definition 8. $V$ is indecomposable if $V$ cannot be written as a direct sum of two proper submodules.

Proposition 3. If $V$ is irreducible then $V$ indecomposable.

Remark: The other direction is NOT true.

Counterexample: Lie algebra of upper triangular matrices. Let it act on the space of $n \times 1$ column vectors. It is indecomposable, but not irreducible.

Reference

This series of articles on Lie algebras is based on Carter R., Lie Algebras of Finite and Affine Type (Cambridge University Press, 2005), as well as the class notes for Introduction to Lie Algebras by Chenwei Ruan at BIMSA.

The cover image of this article was taken in Paris, France.