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Let $V = \mathbb{C}^n$, $A = \Lambda^{\bullet}(\mathbb{C}^n)$ is a graded algebra (with $A_0 = \mathbb{C}, A_1 = V$, etc).

Consider $A_0$ as a left $A$-module, how do we compute the graded ring $\text{Ext}^{\bullet}_A(A_0, A_0)$? (Doing the $n=3$ example should be enough; then it would be easy to generalize.)

(I was trying to understand Koszul duality for symmetric/exterior algebras from [BGS]; and this is the first step.)

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Should this be migrated to MSE? –  Sean Tilson Jun 25 '13 at 20:29
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up vote 6 down vote accepted

Consider the Koszul complex $$ \dots \to S^3V\otimes A(-3) \to S^2V\otimes A(-2) \to V\otimes A(-1) \to A \to A_0 \to 0, $$ where $(-i)$ is the shift of grading. This is a free resolution of $A_0$. Using this to compute $Ext$ you obtain $Ext^\bullet(A_0,A_0) = S^\bullet(V^*)$.

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So what are the maps $S^3 V $ –  Vinoth Jun 25 '13 at 15:30
    
The maps are the compositions of the partial polarization $S^nV \to S^{n-1}V\otimes V$ with the multiplication in $A$ (wedge multiplication) $V\otimes A = A_1\otimes A \to A(1)$. –  Sasha Jun 25 '13 at 20:21
    
1. What is the partial polarization $S^n V \rightarrow S^{n-1}V \otimes V$? For instance, when $n=3$, where does $v_1 v_2 v_3$ go (where $v_1, v_2, v_3$ are basis vectors)? 2. How do you use this free resolution to compute the $\text{Ext}^{\bullet}(A_0, A_0)$? (Sorry for asking trivial questions. Thanks!) –  Vinoth Jun 26 '13 at 11:45
    
1. $v_2v_2v_3 \mapsto v_1v_2\otimes v_3 + v_1v_3\otimes v_2 + v_2v_3\otimes v_1$. In general a polinomial $f(v_1,\dots,v_n)$ goes to $\sum \partial (f/\partial v_i) \otimes v_i$. 2. You apply the functor $Hom(-,A_0)$ to the resolution termwise. As a result you get a complex. Its cohomology are the the $Ext$-spaces. –  Sasha Jun 26 '13 at 16:45
    
Thanks - then it's obvious that the composition of the two adjacent maps is 0. Can you give me a hint about how to compute the cohomology of Ext spaces? I understand that by $\text{Hom}$ you mean as left $A$-modules (not as vector spaces).. I tried calculating but it got out of hand. –  Vinoth Jul 1 '13 at 14:08
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