Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This is a follow-up to Computing Ext in Exterior algebra (related to Koszul duality) .

Let $V = \mathbb{C} x$, $A = \Lambda^{\bullet}(V) = A_0 \oplus A_1$ is graded (with $A_0 = \mathbb{C}, A_1 = V$). Consider $A_0$ as a left $A$-module, how do we compute the multiplication in the algebra $\text{Ext}^{\bullet}_A(A_0, A_0)$? I've heard the word "Yoneda product" mentioned, but I don't know how it works.

We can compute $\text{Ext}^{k}_A(A_0, A_0)$ as follows (see the answer in my above post for more details). We have a Koszul complex $\cdots \rightarrow S^2 V \otimes \Lambda^{\bullet}V[-2] \rightarrow V \otimes \Lambda^{\bullet}V[-1] \rightarrow \Lambda^{\bullet}V \rightarrow A_0$; and we have to compute the cohomology of the complex $\text{Hom}_A(A_0, A_0) \rightarrow \text{Hom}_A(A, A_0) \rightarrow \text{Hom}_A(V \otimes A[-1], A_0) \rightarrow \cdots$. It's easy to check that $\text{Hom}_A(S^k V \otimes A[-k], A_0)=(S^k V)^*[-k]$, and that all maps in this chain complex are $0$ (except for the first); this means $\text{Ext}^k_A(A_0, A_0)=(S^k V)^*[-k]$.

I'm also interested in multiplication for the general case (where $\text{dim } V>1$), but I'm guessing it follows from using the exact same method.

share|improve this question
4  
There's a delightful (and short!) method, published by Tate in his Homology of Noetherian rings and local rings, for computing the product structures on Ext and Tor in these sorts of situations. The basic idea is to build the projective resolution in such a way that you can further equip it with the structure of a DGA, which then induces the product structure you want. The word "Koszul" does not appear, but nonetheless it's fun and easy and you can learn it in an afternoon. projecteuclid.org/euclid.ijm/1255378502 –  Eric Peterson Aug 7 '13 at 16:13
    
Thanks! I had a look - but I couldn't find the specific part which talks about the product structures on Ext. Do you know which section it's in? –  Vinoth Aug 8 '13 at 8:13
add comment

1 Answer

up vote 13 down vote accepted
+100

This is a standard computation, if I understand the question correctly. You are interested in computing $Ext_A(k,k)$ as an algebra, starting with an exterior algebra $A$ on $n$ generators $x_1,\cdots, x_n$ over a field $k$ (no reason to restrict to $\mathbb C$); $A$ is graded with $k$ in degree $0$ and the $x_i$ in degree $1$. It is a Hopf algebra, with the $x_i$ primitive. Let $\Gamma$ denote the divided polynomial Hopf algebra on generators $y_i$, $1\leq i\leq n$; it is bigraded with the $y_i$ in homological degree $1$ and also internal degree $1$. We are mainly interested in the coproduct on $\Gamma$, and that is defined so that its vector space dual is the polynomial algebra on $n$ generators $y_i^*$. The basis elements dual to $(y_i^*)^r$ are denoted $\gamma_r(y_i)$. Write $\otimes = \otimes_k$ and form $K = \Gamma \otimes A$. It is a differential graded $A$-algebra with differential defined on the $\gamma_r(y_i)\otimes 1$ by $$ d(\gamma_r(y_i)\otimes 1) = \gamma_{r-1}(y_i)\otimes x_i. $$ With the natural augmentation to $k$, $K$ is a free $A$-resolution of $k$. Therefore $$ Hom_A(K,k)\cong Hom_k(\Gamma,k) \cong P[y_1^*,\cdots,y_n^*] $$ is a cochain complex suitable for computing $Ext_A(k,k)$, and its differential is zero. We are entitled to conclude that $$ Ext_A(k,k) = P[y_1^*,\cdots,y_n^*] $$ as an algebra (indeed as a Hopf algebra, with the $y_i^*$ primitive). Indeed, via the coproduct on $A$, $K\otimes K$ is a chain complex of $A$-modules, and it is clearly a free $A$-resolution of $k\cong k\otimes k$. The coproduct $K\longrightarrow K\otimes K$ is a map of chain complexes over the identification $k\longrightarrow k\otimes k$. We have $$Ext_A(k,k) \otimes Ext_A(k,k) \cong Ext_{A\otimes A}(k\otimes k,k) $$ The product on $A$ induces a product $$ Ext_{A\otimes A}(k\otimes k,k)\longrightarrow Ext_{A}(k,k).$$ The composite is the product in question, and it is computed using $$ Hom_A(K,k) \otimes Hom_A(K,k) \cong Hom_{A\otimes A}(K\otimes K,k) \longrightarrow Hom_A(K,k). $$ This is nothing but the multiplication on the polynomial algebra $P[y_1^*,\cdots,y_n^*]$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.