This is a follow-up to [Computing Ext in Exterior algebra (related to Koszul duality)](https://mathoverflow.net/questions/134595/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.



Computing multiplication in the Ext (for a simple example)
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 $P[y\_1^\*,\cdots,y\_n^\*]$.


