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

If $H$ is a Hopf algebra over a field $k$, the Hopf algebra cohomology $\mathrm{Ext}_H(k,k)$ is —as usual— an algebra, but the Hopfness of $H$ turns it into a Hopf algebra.

Is there a reference on this Hopf structure on $\mathrm{Ext}_H(k,k)$?

I am hoping someone wrote down all the basic details about this so as to avoid doing it myself...

share|cite|improve this question
Really? I think the usual thing structure on $\mathrm{Ext}_H(k,k)$ is the structure of homotopy-$E_2$ algebra. In particular, it is a Gerstenhaber algebra, and in characteristic $0$ by formality (hard!) there are no higher "Massey products". But it seems you "use up" the comultiplication on $H$ to define the $E_2$ structure on $\mathrm{Ext}_H(k,k)$. – Theo Johnson-Freyd Feb 7 '12 at 20:22
Theo, the coproduct is constructed much as one constructs the Pontryagin multiplication on the singular homology of a Lie group. (The Yoneda algebra is a commutative algebra under the Yoneda product, which is better than being am homotopy $E_2$-algebra iirc...) – Mariano Suárez-Alvarez Feb 7 '12 at 20:32
My intuition is that Theo is right. Does the construction you have in mind work for any coproduct, or does it have to be cocommutative? – James Griffin Feb 8 '12 at 15:01

Well, this might not be the answer you expected:

In general, there is no coproduct such that $Ext_H^\ast(k,k)$ (cup product) is a graded Hopf algeba.

For, let $k$ be a perfect field and suppose $A = Ext_H^\ast(k,k)$ is a graded Hopf algebra of finite type. By Borel's structure theorem on connected graded commutative Hopf algebras [A-M, VI.2.8], $A$ is (as $k$-algebra) isomorphic to the tensor product of algebras of the types $k[x]$ and $k[y]/(y^r)$. In particular, $A/rad(A)$ is a domain.

Now it's easy to find counterexamples. For instance let $\text{char}(k)=2$. Then the cohomology ring of the dihedral group $H^\ast(D_8;k) = k[x,y,z]/(xy),\;|x|=|y|=1, |z|=2$ has zero-divisors, but its radical is zero.

More generally: If $\text{char}(k) = p$ and $G$ is a $p$-group with at least two conjugacy classes of maximal elementary abelian subgroups, then there are non-nilpotent classes $x,y \in H^\ast(G;k)$ such that $xy=0$. Thus $Ext_{k[G]}^\ast(k,k) = H^\ast(G;k)$ can't be a Hopf algebra.

However, if $H$ is commutative, then the product induces a coproduct that makes $Ext_H^\ast(k,k)$ a commutative, cocommutative Hopf algebra. I guess searching for a reference will probably last much longer than the straightforward proof: Let $P \to k$ be a projective resolution over $H$. Then the operations on $H$ and the uniqueness of induced mappings (up to homotopy) induce a (kind of) DG-Hopf algebra $(P,\mu,\Delta)$ where the usual diagramms commute up to homotopy. Passage to cohomology now makes the diagramms commute and you have your Hopf algebra.

Note: $(P,\mu,\Delta)$ can be seen as an algebraic $H$-space analog.

One may wonder why the coproduct $\Delta: H \to H \otimes H$ always induces a product on $Ext_H^\ast(k,k)$ while the product $\mu: H \otimes H \to H$ induces a coproduct only if $H$ is commutative. The reason is that $\Delta$ is an algebra homomorphism while $\mu$ is an algebra homomorphism if and only if $H$ is commutative.

[A-M] Adem, Milgram: Cohomology of finite groups.

share|cite|improve this answer
Well, I would not call $\Delta$ a "Hopf algebra homomorphism" unless it is cocommutative. Certainly it is an algebra homomorphism, which is I think what you mean. – Theo Johnson-Freyd Feb 13 '12 at 2:41
Yes, thanks. In more detail: A hom. of $k$-algebras $A \to B$ induces a $k$-linear map $Ext_B \to Ext_A$. That is what is used in the last part. – Ralph Feb 13 '12 at 10:02

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.