TITLE: Computing multiplication in the Ext (for a simple example)
QUESTION [5 upvotes]: 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.

REPLY [15 votes]: 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 $P[y_1^*,\cdots,y_n^*]$.