# The symmetric algebra as a g-module

I'm quite sure that this question is not difficult, but I think that my understanding of the definitions is just not deep enough yet:

Given a lie algebra $g$, and a $g$-module $V$ (or equivalently a representation $\rho:g\rightarrow\mbox{End}\left(V\right)^{-}$), I need to show that $S\left(V\right)$ (the symmetric algebra) is a graded $g$-module. i.e. that it is graded, and that the action of $g$ respects the grading.

I can easily understand why $S\left(V\right)$ is a graded module. If I'm not mistaking, it's quite clear from the definition of it (and the fact that we can easily grade $T\left(V\right)$). But I don't understand how to define the action of $g$ on $S\left(V\right)$.

Like I said, I'm very new to this, and would greatly appreciate answers without too big steps, as I'm pretty sure that it's not a difficult trick or idea that is needed here, but just a better understanding of all the definitions.

Thanks alot!

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Here are some hints that should get you going in the right direction. If you really want a formalized proof, I actually have written one some time ago for no real reason (pbwlong.pdf Proposition 5.4), but I am sure that understanding it will waste more of your time than doing the following steps on your own:

Let $n\in\mathbb N$. We must show that the $n$-th symmetric power $S^n\left(V\right)$ of the $g$-module $V$ is a $g$-module.

1. I assume you know that the $n$-th tensor power $V^{\otimes n}$ becomes a $g$-module by

$X\rightharpoonup\left(u_1\otimes u_2\otimes ...\otimes u_n\right)$ $= \sum\limits_{k=1}^n u_1\otimes u_2\otimes ...\otimes u_{k-1}\otimes \left(X\rightharpoonup u_k\right)\otimes u_{k+1}\otimes u_{k+2}\otimes ...\otimes u_n$

for all $X\in g$ and $u_1,u_2,...,u_n\in V$, where $\rightharpoonup$ denotes the action of $g$ on $V$ (for various reasons I consider denoting this action by $\cdot$ harmful). The well-definedness of this $g$-module is easily checked (e. g., by induction over $n$).

2. The symmetric power $S^n\left(V\right)$ can be defined as the quotient of the tensor power $V^{\otimes n}$ modulo the subspace

$K_n\left(V\right) := \left< v_1\otimes v_2\otimes ...\otimes v_n - v_{\sigma\left(1\right)}\otimes v_{\sigma\left(2\right)}\otimes ...\otimes v_{\sigma\left(n\right)} \mid v_1,v_2,...,v_n\in V,\ \sigma\in S_n\right>$.

Thus, in order to make $S^n\left(V\right)$ into a $g$-module, it is enough to prove that $K_n\left(V\right)$ is a $g$-submodule of $V^{\otimes n}$. This is a rather straightforward task: if you expand

$X\rightharpoonup\left(v_1\otimes v_2\otimes ...\otimes v_n - v_{\sigma\left(1\right)}\otimes v_{\sigma\left(2\right)}\otimes ...\otimes v_{\sigma\left(n\right)} \right)$,

you will get a difference of two sums. Examining these two sums more closely, you will see that we can match the terms of the first sum to terms of the second sum in such a way that corresponding terms are permutations of each other. (I call two terms "permutations of each other" if they are tensor products, and the factors of one term are a permutation of the factors of the other. I hope it is clear how this looks like.) A difference between two such terms always lies in $K_n\left(V\right)$ (by the very definition of $K_n\left(V\right)$). So we get a sum of $n$ elements of $K_n\left(V\right)$, thus an element of $K_n\left(V\right)$. This shows that $K_n\left(V\right)$ is a $g$-submodule of $V^{\otimes n}$, and we are done.

Now I have made a simplifying assumption in the above: namely, I assumed that you define $S^n\left(V\right)$ as the quotient of the tensor power $V^{\otimes n}$ modulo the subspace $K_n\left(V\right)$. This is only one possible definition of $S^n\left(V\right)$. Another (apparently more popular) definition proceeds by defining $S^n\left(V\right)$ as the $n$-th degree component of the quotient of the graded tensor algebra $T\left(V\right)$ by the ideal generated by the elements $v\otimes w-w\otimes v$ for $v,w\in V$. With this definition, you can still proceed as in the above, but this time you have to show that this ideal is a $g$-submodule of $T\left(V\right)$ instead of proving that $K_n\left(V\right)$ is a $g$-submodule of $V^{\otimes n}$. This turns out to be even simpler if you first prove that $T\left(V\right)$ is a $g$-algebra (because then, in order to prove that an ideal is a $g$-submodule, it is enough to check this on generators of the ideal). What I mean by "$g$-algebra" here is a $g$-module with an algebra structure satisfying the "Leibniz equality"

$X\rightharpoonup\left(ab\right)=\left(X\rightharpoonup a\right)b+a\left(X\rightharpoonup b\right)$ for all $X\in g$ and $a,b$ in the algebra.

You'd have to prove that $T\left(V\right)$ is a $g$-algebra, but that is very easy.

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