Symmetric and Exterior products of sl(n,C)-module Let M be the $sl(n,C)$-representation of the inclusion $sl(n,C)\hookrightarrow gl(n,C)$.
Let q be a symbol.
$f(q)=1-M q + \wedge^2Mq^2-...+(-1)^n\wedge^nMq^n$
$g(q)=\sum_{i=0}^\infty Sym^iM \; q^i$
I want to prove that $f(q)g(q)=1$ which is equivalent to some isomorphism between many representations.
I am not sure if this choice of M is essential. Will any n-dim representation be enough? The worst method might be computing characters with character formulae...
btw product is tenor product. sum is direct sum.
 A: Let $SM$ be the symmetric algebra on $M$. There is an exact sequence
$$0\to SM\otimes\Lambda^nM\to SM\otimes\Lambda^{n-1}M\to\cdots SM\otimes\Lambda^2M\to SM\otimes M\to SM\to k$$ which is just the Koszul complex for $SM$. After appropriately shifting the degrees of the modules, this is an exact complex of $\mathbb Z$-graded $sl(n)$-modules. Taking the Euler characteristic of the Hilbert series with coefficients in the Grothendieck ring of finite dimensional $sl(n)$-modules should prove your identity.
A: It's easy to show for any $M$ using your "worst method," though in the definition of $f(q)$ we need to replace $n$ with $m=\dim M$. Let $A\in\mathrm{sl}(n,C)$ and suppose that $M\cdot A$ has eigenvalues $\theta_1,\dots,\theta_m$. Then the trace of $A$ acting on $f(q)$ is $(1-\theta_1 q)\cdots (1-\theta_m q)$, while on $g(q)$ is $1/(1-\theta_1 q)\cdots (1-\theta_m q)$. Since the character of a (finite-dimensional) representation of $\mathrm{sl}(n,C)$ determines the representation (up to equivalence), it follows that $f(q)g(q)=1$ as virtual representations. 
