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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...

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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
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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
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1 Answer

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.

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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
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