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.



Symmetric and Exterior products of sl(n,C)-module
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. 


