Given a representation $\phi: L \rightarrow \mathfrak {gl}(V)$ $\phi(L)$ in End $V$ leaves invariant precisely the same subspaces as $L$.

Question

I'm studying Humphrey's book: Lie Algebras and R. Theory on my own, and I'm trying to understand a claim and a definition (both stated imprecisely in page 26) about $L$-modules given in the book. I've not taken any course on groups yet, so modules are very new to me. I'll quote what is written and my interpretation of it. I hope someone can clarify them.

"Of course, the terminology "irreducible" and "completely reducible" applies equally well to representations of $L$."

This is said right after defined what is an irreducible and complety reducible $L$-module. Trying to follow these definitions, given a representation $\phi: L \rightarrow \mathfrak{gl}(V)$ I would say that a irreducible representation is a vector space $V$ which has exactly two $\phi(L)$-submodules: $0$ and itself; and completly reducible if $V$ is a direct sum of irreducible $\phi(L)$-submodules.

"Given a representation $\phi:L \rightarrow \mathfrak{gl}(V),$ the associative algebra (with $1$) generated by $\phi(L)$ in End $V$ leaves invariant precisely the same subspaces as $L$."

Since the author mentioned invariant subspaces leaved by a Lie Algebra $L$, I suppose he is talking about $L$-submodules. So I guess the claim is that the algebra generated by $\phi(L)$ has the same $\phi(L)$-submodules as $L$. I've tried to prove this result, but I could not get far: let $W$ be a $L$-submodule. Then, given $x\in L,w\in W$, we have $x\cdot w\in W.$ Now, take any $\phi(x)\in \phi(L)$, I can't see why $\phi(x)\cdot w$ would lie in $W$.

Can someone clarify those quotes for me? Thanks in advance.

Answer 1

The notion of an $L$-module is just the notion of a Lie algebra representation $\rho\colon L\rightarrow {\rm End}(V)$ via the convention that $x\cdot v=\rho(x)v$ for all $x\in L$ and $v\in V$. So you do not need "any course on groups". Irreducible for representations corresponds to "simple" for modules, i.e., no proper non-trivial submodules. You have just confused "reducible" with "irreducible" in what you wrote. And yes, an invariant space $U$ means $\rho(L)(U)\subseteq U$, or $L\cdot U\subseteq U$, i.e., an $L$-submodule.