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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.

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Does this really have anything to do with geometric Langlands? –  darij grinberg Nov 17 '12 at 1:34
    
BTW I'm interested in characteristic-free proofs (or refutations...) of this $f\left(q\right)g\left(q\right)=1$ equality (viewed as an equality in the ring of formal power series in $q$ over the Grothendieck ring). –  darij grinberg Nov 17 '12 at 1:35
    
This should be Koszul duality for the symmetric algebra on $M$ in the category of $sl(n)$-modules, or something. –  Mariano Suárez-Alvarez Nov 17 '12 at 1:44
    
Mariano, could you explain this to me? I happen to need some nice examples to understand Koszul duality... –  darij grinberg Nov 17 '12 at 1:50
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2 Answers

up vote 2 down vote accepted

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.

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yes, good, this is elementary enough. –  Jeep Wrangler Nov 17 '12 at 22:08
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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.

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Very nice! So this gives a simple proof over any field or ring and for modules over any cocommutative Hopf algebra instead of $\mathfrak{sl}\left(n\right)$-modules. Only one question: where is Koszul duality here? I just see a Koszul complex. –  darij grinberg Nov 17 '12 at 2:16
    
Well, Koszul duality is the relationship between a Koszul quadratic algebra $A$ and its Koszul dual $A^!$, which can be described in many ways, one of which is that $A\otimes A^!$ can be made, by grading appropiately, into a resolution of the trivial module for $A$ (or for $A^!$, for that matter) When $A$ is the symmtric algebra on a vector space, then $A^!$ is the exterior algebra, and the corresponding compex is the Koszul complex (which is an $SM$-projective resolution of $k$; with a different grading, it becomes a $\Lambda M$-projective resolution of $k) –  Mariano Suárez-Alvarez Nov 17 '12 at 2:23
    
I didn't understand this complex yet. Can you give any reference about the maps in this complex? –  Jeep Wrangler Nov 17 '12 at 22:08
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