MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
You should read what wikipedia has to say about the ring structure on Ext:…. It seems like whenever Yoneda products and products given by the "pick a resolution" method (see link) both exist, then they'll be the same (probably by an Eckman-Hilton argument?). In this case it should be graded-commutative. Right? – Dylan Wilson Jul 27 '10 at 6:39
This seems right; but it does not apply in several interesting cases (for example, the case of sheaves). – Angelo Jul 27 '10 at 6:44
Not at all. For $R$ a local commutative ring and $A$ its residue field, $\mathrm{Ext}(A,A)$ can be computed by a resolution but it is ver seldom graded commutative. It is always the enveloping algebra of a graded Lie algebra and the Lie algebra may contain for instance large free subalgebras. For instance for $R=k[x_1,...,x_n]/(x_1,...,x_n)^2$ the Lie algebra is free on $n$ generators (in degree $1$). – Torsten Ekedahl Jul 27 '10 at 8:53
As Torsten points out, graded commutativity of Ext(AA) will not hold for every object. However, it will hold when A is the unit object in an appropriate monoidal category, by an Eckman-Hilton argument. An important example of this is Hochschild cohomology of a ring or dga R, which is defined as Ext(RR) in the category of R-bimodules. To show this is graded commutative, you should think of R as the unit in the monoidal derived category of bimodules and apply Eckmann-Hilton. Here, you should think of the derived category as being enriched over graded vector spaces by summing up all Exts, or – Chris Brav Jul 27 '10 at 10:19
... even better, think of the derived category as enriched over cochain complexes (think of it as a dg category). Then you get that Hochschild cochains has two commuting products (Yoneda and tensor product), so is an $E_{2}$-algebra. Note that by Eckmann-Hilton, an $E_{2}$-algebra in graded vector spaces is just a graded commutative algebra. – Chris Brav Jul 27 '10 at 10:22

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 $2$.

share|cite|improve this answer

Another starting point is to think of ${\rm Ext}(A,A)$ as the derived endomorphism ring of the object $A$ and recall Schur's lemma. If $A$ is a finitely-generated simple module over a ring $R$, then ${\rm Hom}_R(A,A)$ is a division algebra. For example, if $R$ is a $k$-algebra over an algebraically closed field $k$, then ${\rm Hom}_R(A,A)$ is isomorphic to $k$ (so, in particular, it is commutative.) Via Freyd-Mitchell embedding, this should give some idea what to expect in degree $0$.

Going back the question, then, the examples one might have in mind are categories of modules over a group ring or enveloping algebra of a graded Lie algebra: in these cases, ${\rm Ext}(k,k)$ is group- or Lie algebra cohomology, respectively, and has a graded-commutative cup product, where $k$ is the trivial module.

Perhaps there is a suitable "semisimplicity" hypothesis one could impose on the category so that ${\rm Ext}(A,A)$ is graded-commutative for all simple objects $A$?

share|cite|improve this answer
A simple module is cyclic (1-generated). – Victor Protsak Jul 28 '10 at 19:56
You need some (finiteness...) hypothesis on $R$ for his to be true: take $k=\mathbb C$, $R=\mathbb C(t)$, and $A$ the $1$-dimensional $R$-vector space. – Mariano Suárez-Alvarez Jul 29 '10 at 21:10

Your Answer


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.