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I am trying to get used to Hochschild cohomology of algebras by proving its properties. I am currently trying to show that the cup product is graded-commutative (because I heard this somewhere); however, my trouble is that I have no idea what the exact conditions are for this to hold.

As always in Hochschild cohomology, we start with an algebra A and an A-bimodule M. Is A supposed to be unital? In the original article by Hochschild, it is not, and the proofs would even become harder if we suppose it to be. On the other hand, I have some troubles working with non-unital algebras - they seem just so uncommon to me. Is it still standard to use non-unital algebras 60 years after Hochschild's articles?

Anyway, this is not the main problem. The main problem is that I have no idea what we require from A and M. Of course, M should be an A-algebra, not just a module, for the cup product to make sense. Now:

  • Must A be commutative?

  • Must M be commutative?

  • Must M be a symmetric A-bimodule? (That is, am = ma for all a in A and m in M.)

I have some doubts that if we require this all, then we still get something useful (in fact, the most common particular case of Hochschild cohomology is group cohomology, and it is as far from the "symmetric A-bimodule" case as it can be), but I may be completely mistaken.

As there seem to be different definitions of Hochschild cohomology in literature, let me record mine:

For a k-algebra A and an A-bimodule M, we define the n-th chain group $C^n \left(A, M\right)$ as $\mathrm{Hom}\left(A^{\otimes n}, M\right)$, with differential map

$\delta : C^n \left(A, M\right) \to C^{n+1} \left(A, M\right)$,

$\left(\delta f\right) \left(a_1 \otimes ... \otimes a_{n+1}\right) = a_1 f\left(a_2 \otimes ... \otimes a_{n+1}\right)$ $ + \sum\limits_{i=1}^{n} f\left(a_1 \otimes ... \otimes a_{i-1} \otimes a_{i}a_{i+1} \otimes a_{i+2} \otimes ... \otimes a_{n+1}\right) + \left(-1\right)^{n+1} f\left(a_1 \otimes ... \otimes a_n\right) a_{n+1}$.

The cohomology is then the homology of the resulting complex, and if M is an A-algebra, the cup product is given by

$\left(f\cup g\right)\left(a_1 \otimes ... \otimes a_{n+m}\right) = f\left(a_1 \otimes ... \otimes a_n\right) g\left(a_{n+1} \otimes ... \otimes a_{n+m}\right)$.

I am aware of this article by Arne B. Sletsjøe, but it defines Hochschild cohomology differently (by using $\left(-1\right)^{n+1} a_{n+1} f\left(a_1 \otimes ... \otimes a_n\right)$ in lieu of $\left(-1\right)^{n+1} f\left(a_1 \otimes ... \otimes a_n\right) a_{n+1}$); this definition is only equivalent to mine if M is a symmetric A-module, so it won't help me find out whether this is necessary to assume.

Thanks for any help, and sorry if the counterexamples are so obvious that I am an idiot not to find them on my own...

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1 Answer 1

up vote 8 down vote accepted

The cup product in Hochschild cohomology$H^\bullet(A,A)$ is graded commutative for all unitary algebras. If $M$ is an $A$-bimodule, then the cohomology $H^\bullet(A,M)$ with values in $M$ is a symmetric graded bimodule over $H^\bullet(A,A)$.

(If $M$ itself is also an algebra such that its multiplication map $M\otimes M\to M$ is a map of $A$-bimodules, then in general $H^\bullet(A,M)$ is not commutative (for example, take $A=k$ to be the ground field, and $M$ to be an arbitrary non-commutative algebra! I do not know of a criterion for commutatitivity in this case)

These results originally appeared in [M. Gerstenhaber, The cohomology structure of an associative ring, Ann. of Math. (2) 78 (1963), 267–288.], and they are discussed at length in [Gerstenhaber, Murray; Schack, Samuel D. Algebraic cohomology and deformation theory. Deformation theory of algebras and structures and applications (Il Ciocco, 1986), 11--264, NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., 247, Kluwer Acad. Publ., Dordrecht, 1988.] Both references give proofs of a rather computational nature.

You can find an element-free proof of the graded commutativity in this paper, which moreover applies to the cup-products of many other cohomologies.

As for what happens with non-unital algebras, I do not know. But they are very much used today as they were before. One particular context in which they show up constantly is in the intersection of K-theory and functional analysis, where people study `algebras' which are really just ideals in rings of operators of functional spaces---one egregius example is the algebra of compact operators in a Hilbert space.

By the way, you say that group cohomology is a special case of Hochschild cohomology: it is only in a sense... There is a close relationship between the group cohomology $H^\bullet(G,\mathord-)$ of a group $G$, and the Hochschild chomology $H^\bullet(kG,\mathord-)$ of the group algebra, but they are not the same. You can use the second to compute the first (because more generally if $M$ and $N$ are $G$-modules, then $\mathord{Ext}_{kG}^\bullet(M,N)=H^\bullet(kG, hom(M,N))$, where on the right we have Hochschild cohomology of the group algebra $kG$ with values in the $kG$-bimodule $\hom(M,N)$), but the «principal» group cohomology $H^\bullet(G,k)$ is only a little part of the «principal» Hochschild cohomology $H^\bullet(kG,kG)$.

Finally, in the paper by Sletsjøe the definition for the boundary is given as you say because he only considers commutative algebras and only principal coefficients, that is $H^\bullet(A,A)$.

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Sorry, I should have been more precise when talking about group cohomology. But if we allow $k$ to be an arbitrary commutative ring (rather than a field), then the group cohomology is the Hochschild cohomology over $kG$ for $k=\mathbb Z$, right? –  darij grinberg Dec 12 '09 at 15:09
    
What exactly do you mean by their being the "same"? If you make a concrete statement, it would be easier to know hwo to say no :P –  Mariano Suárez-Alvarez Dec 12 '09 at 15:34
    
Okay, I was imprecise as always. What I meant is: In group cohomology (written the inhomogenous way), a n-cochain is an element of $\mathrm{Hom}_{\mathrm{Set}}\left(G^n,\mathbb{Z}\right)$. (If it's already here that I am wrong, sorry.) By the universal property of the free $\mathbb{Z}$-module, $\mathrm{Hom}_{\mathrm{Set}}\left(G^n,\mathbb{Z}\right)\cong\mathrm{Hom}_{\mathb‌​b{Z}}\left(\mathbb{Z}G^n,\mathbb{Z}\right)$ canonically. Finally, $\mathbb{Z}G^n\cong\left(\mathbb{Z}G\right)^{\otimes n}$. –  darij grinberg Dec 12 '09 at 15:37
    
And, if I am not mistaken, these isomorphisms commute with $\delta$. –  darij grinberg Dec 12 '09 at 15:38
    
Sorry, the hom doesn't go to $\mathbb{Z}$, but to the group module $M$ (which is turned into a $\mathbb{Z}G$-bimodule as follows: The left action of $\mathbb{Z}G$ on $M$ is the linearization of the group action that we started with; the right action of $\mathbb{Z}G$ on $M$ is the linearization of $mg=m$ for every $m\in M$ and $g\in G$). –  darij grinberg Dec 12 '09 at 15:46

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