1
$\begingroup$

I am studying the section of Complete Reducibility of representations(Chapter 2) from Lie Algebras book by John Humphreys.

I got some remark which says: (Assuming $L$ is Lie algebra and $V$ is $L$-Module)

"Given a representation $\phi: L \rightarrow gl(V)$, the associative algebra(with 1) generated by $\phi(L)$ in End$V$ leaves invariant precisely the same subspaces as $L$. Therefore, all the usual results(e.g. Jordan-Holder Theorem) for modules over associative rings hold for $L$ as well."

I can't understand what "associative algebra with 1" they are referring to? What is 1?.

How all usual results for modules over associative rings hold for $L$?

If someone can explain, it will be a great help!

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Here, $1$ is the identity map $\operatorname{Id}\colon V\longrightarrow V$. Now, consider the set $\operatorname{End}V$. It has a natural structure of an associative and unitary algebra (the unit being $\operatorname{Id}$). So, consider the subalgebra of $\operatorname{End}V$ spanned by $\operatorname{Id}$ and by the set $\{\phi(X)\mid X\in L\}$. That's the “associative algebra with $1$” that Humphreys has in mind.

$\endgroup$
5
  • $\begingroup$ Thanks for clarifying. But what usual results for modules he is talking about? I see Jordan Holder Theorem which is not clear to me. Can you provide some another result, easy to understand which follows the last statement? $\endgroup$
    – Vats Y
    Jul 11, 2021 at 11:15
  • 1
    $\begingroup$ In this context, the Jordan-Hölder therem states that, for every finitely generated module $M$, there is an increasing sequence of sub-modules$$0=J_0\varsubsetneq J_1\varsubsetneq J_2\varsubsetneq \cdots\varsubsetneq J_n=M$$such that every quotient $J_k/J_{k-1}$ ($k\in\{1,2,\ldots,n\}$ is a simple module. Since this holds for modules over associative and unitary algebras, it also holds for Lie algebras, for the reason described by Humphreys. $\endgroup$
    – José Carlos Santos
    Jul 11, 2021 at 11:27
  • $\begingroup$ So, we are considering an arbitrary module over associative ring (in this case, set generated by $\phi(L)$) and then we are somehow replacing our $\phi(L)$ by $L$ since $\phi(L)$ leaves the same invariant subspaces as $L$. From this, we are getting all usual results of modules to be true if module is over a lie algebra. Am I right? $\endgroup$
    – Vats Y
    Jul 11, 2021 at 11:50
  • $\begingroup$ Correct, if you use “algebra” instead of “ring”. $\endgroup$
    – José Carlos Santos
    Jul 11, 2021 at 11:52
  • $\begingroup$ Thankyou for clarification! $\endgroup$
    – Vats Y
    Jul 11, 2021 at 11:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .