I know that the steenrod algebra acts on the group cohomology. I would like to know whether this action extends to the Hochschild cohomology of the group algebra, as the latter contains group cohomology as a subalgebra.
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3 Answers
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Quite unlikely. The many solutions to Deligne's conjecture for Hochschild cohomology show that the Hochschild complex is an $E_2$-algebra, and this is the best one can get. It is known that the cohomology of an $E_\infty$-algebra carries an action of the Steenrod algebra. If you wish, this is why the standard group cohomology is an algebra over the Steenrod algebra, since it is the cohomology of the (singular) cochains on a space, which is an $E_\infty$-algebra. The homology of an $E_2$-algebra has a very low-dimensional $p$-power operation, where $p$ is the cgaracteristic of the ground field, and that's essentially all, see the references below.
PD I've edited this comment to correct some inaccuracies.
answered Apr 6, 2011 at 11:18
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$\begingroup$ Can you give some references of your statements above? Thanks a lot. $\endgroup$ Commented Apr 6, 2011 at 11:21
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$\begingroup$ MR0436146 (55 #9096) Cohen, Frederick R.; Lada, Thomas J.; May, J. Peter The homology of iterated loop spaces. Lecture Notes in Mathematics, Vol. 533. Springer-Verlag, Berlin-New York, 1976. vii+490 pp. (Reviewer: Peter J. Eccles), 55G25 (55D35) $\endgroup$ Commented Apr 6, 2011 at 12:39
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$\begingroup$ MR2240919 (2007d:55020) Tourtchine, Victor Dyer-Lashof-Cohen operations in Hochschild cohomology. Algebr. Geom. Topol. 6 (2006), 875–894 (electronic). (Reviewer: Vigleik Angeltveit), 55S12 (16E40 18D50 55P48) $\endgroup$ Commented Apr 6, 2011 at 12:44
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2$\begingroup$ I believe that Hochschild cohomology of the group algebra of $G$ can be described as the cartesian product, over conjugacy classes of $G$, of cohomology of the centralizer of a representative. This gives it Steenrod operations. There is some better way to say this, but I think the answer is yes. $\endgroup$ Commented Apr 6, 2011 at 12:46
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2$\begingroup$ @Tom: the isomorphism is not multiplicative if the group is non-abelian. $\endgroup$ Commented Apr 6, 2011 at 13:01
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In "An alegbraic approach to the Steenrod algebra" Peter May writes down pretty much all the situations where you could have the action of Steenrod "like" operations. The most relevant case here is that of the cohomology of a cocommutative Hopf algebra. In case you are unfamiliar with Hopf algebras, this mean that there is a product map $\mu :A \otimes A \to A$ and a coproduct map $\psi :A \to A \otimes A$. To be cocommutative $\psi=\tau \circ \psi$ where $\tau$ is the twist map. So it seems unlikely that there would be Steenrod type operations acting on the Hochschild cohomology of any algebra, but there are probably operations on the Hochschild cohomology of a cocommutative Hopf algebra. However, you would need to verify that if $A$ is a cocommutative Hopf algebra then $A \otimes A^{op}$ is as well. So that would insure that there are operations in $HH(A,k)$ but not in $HH(A)=HH(A,A)$. That I am not sure about.
answered Apr 6, 2011 at 12:34
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Yes, there is an action of the Steenrod algebra on HH^*(k[G]; k[G]), G a discrete group and k = Z/2 for the mod 2 Steenrod algebra. In my paper "A Comparison of Products in Hochschild cohomology," (on arXiv.org), I show how Steenrod's cup-i products act on the Hochschild cochain complex Hom_k(k[G]^*, k[G]) for an arbitrary coefficient ring k. The cup-i products can then be used to define an action of the Sq^i operators.
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$\begingroup$ I've taken a look at your very interesting paper and noticed that your cup-$i$ products seem to be non-compatible with the usual cup-product in Hochschild homology. $\endgroup$ Commented Aug 23, 2013 at 7:17
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$\begingroup$ Yes, there are two product structures on HH^*(k[G]; k[G]), one the Gerstenhaber product, the other the simplicial cup product. How these interact might lead to a new operad that acts on HH^* of a group ring. Of course, the cup-i products in the above paper lead to an E^{\infty} algebra for the simplicial cup product (not the Gerstenhaber product). $\endgroup$– J.LodderCommented Aug 27, 2013 at 15:06