Sometimes, given an object A in an Abelian category, the Yoneda product on Ext(A, A) is graded-commutative, for example in cases where it coincides with the cup-product in singular cohomology. Are there any nice theorems about when the Yoneda product is graded-commutative in general? Thanks in advance.



When is the Yoneda product graded commutative?
I move this to a more proper answer to discuss some subtle points of the
question. The Eckman-Hilton argument (or more concrete calculations) shows, as
Chris points out, that $\mathrm{Ext}(A,A)$ is commutative when $A$ is the unit
for a monoidal category. The subtleties appear when we consider for instance the
ring $R=k[x]/(x^2)$ for $k$ a field and $A=k$. Then $A$ has a uniform resolution
$\dots\xrightarrow{x}R\xrightarrow{x}R\xrightarrow{x}R\to k\to 0$ giving
$\mathrm{Ext}^i(A,A)=k$ for all $i$. Using the definition of the Yoneda product
in terms of maps of resolutions we get that $\mathrm{Ext}(A,A)$ is the
polynomial ring on $\mathrm{Ext}^1(A,A)$. This is graded commutative only when
the characteristic is $2$ (and then it is not graded commutative in the strict
sense of the square of odd degree elements being zero). However, it is exactly
in characteristic $2$ that $R$ is the affine algebra of a finite group scheme
(with $x\mapsto x\otimes1+1\otimes x$ as coproduct) with $k$ the unit for the
associated monoidal structure on the category of $R$-modules. Hence we have a
monoidal reason for the $\mathrm{Ext}$-algebra being graded commutative in
characteristic. On the other hand we have a uniform description of the
$\mathrm{Ext}$-algebra in all characteristics which just happens to fulfil the
definition of being graded commutative in characteristic $2$.


