Let $A$ be an associative algebra over a commutative ring $k$. I've read statements saying that Hochschild (co)homology is the "right" notion of (co)homology for associative algebras. When $A$ is projective over $k$, the Hochschild cohomology, say, can be written as $Ext^*_{A \otimes A^{op}}(A,A)$, where $A^{op}$ is the opposite algebra, i.e., $A$ with $xy$ redefined to be $yx$.

On the other hand, when $A$ is augmented, the ext-group $Ext^*_A(k,k)$ is also referred to as the cohomology of $A$. What is the difference between these two notions of cohomology, and why would I choose one over the other?

  • $\begingroup$ What is exactly $\textrm{Ext}_A^\ast(k,k)$? Don't you need to have a structure of $A$-module to be able to talk about the cohomology of $A$? $\endgroup$ – javier Nov 19 '09 at 22:31
  • $\begingroup$ Yes, I'm assuming $A$ comes with an augmentation in that case, sorry. $\endgroup$ – haakon Nov 19 '09 at 22:58

For good algebras, Hochschild cohomology computes ‘all’ other interesting cohomologies. For example, it is already in Cartan-Eilenberg that if $M$ and $N$ are left $A$-modules, then $\mathrm{Ext}_A^\bullet(M,N)=H^\bullet(A,\hom(M,N))$, where on the right $H^\bullet(A,\mathord-)$ is Hochschild cohomology with coefficients, and $\hom(M,N)$ is the space of homomorphisms over the base field turned into an $A$-bimodule using the left $A$-module structures of $M$ and $N$.

One general-nonsense explanation of the fact that Hochschild cohomology is somehow preferred is that an associative algebra is an algebra over the $\mathcal{A}ss$ operad, which is a Koszul operad, and a Koszul operad determines a canonical cohomology theory for its algebras: in the case of $\mathcal{A}ss$ you obtain in this way Hochschild cohomology. (Likewise, the operad $\mathcal{L}ie$ whose algebras are Lie algebras picks the usual Lie algebra cohomology, and $\mathcal{C}omm$, the operad of commutative algebras, picks the Harrison cohomology. This explanation breaks for, say, Hopf algebras—which are not the algebras of an operad—and then you do not have a clear winner among the cohomologies: then you have $\mathrm{Ext}_H^\bullet(k,k)$ and the Gerstenhaber-Schack cohomology as alternatives, both quite ‘preferred’...)


Regardless of my comment pointing out that since $k$ is not an $A$-module the groups $\text{Ext}_A^\ast(k,k)$ are not well defined, Hochschild cohomology is certainly the preferred options. Some reasons for picking it are

  • The Hochschild cohomology groups for a smooth commutative algebra coincide with the differential forms. This is the Hochschild-Kostant-Rosenberg isomorphism.
  • It measures separability (0-th group) of the algebra, formal smoothess (1st group), rigidity (2nd group) and obstructions to extend infinitesimal deformations (in the sense of Gerstenhaber) to complete formal deformations (3rd group).
  • It is related with cyclic cohomology via the SBI sequence, meaning that it can serve you as a good approximation to cyclic cohomology when that is too difficult to compute.

Don't be fooled by the $A\otimes A^{op}$ thing, $A$ seen as a module over $A\otimes A^{op}$ is just the same thing as seen as a bimodule over itself.

  • $\begingroup$ What coefficients are you using? For Gerstenhaber bracket stuff you surely want coefficients in the algebra; for the SBI sequence and links to cyclic cohomology you're going to want coefficients in the dual. $\endgroup$ – Yemon Choi Nov 20 '09 at 1:47
  • $\begingroup$ Sure, I was referring to the whole cohomology as in Hochschild cohomology theory, with coefficients in any bimodule; for formal smoothness you also need the vanishing of $H^2_A(M)$ for all bimodules $M$. $\endgroup$ – javier Nov 20 '09 at 5:13

In my view Hochschild cohomology is the most interesting cohomology on associative (and I dare say commutative algebras). So far all that has been said is about different methods of computation. But there are also many applications and ways of viewing it.

Skip the following paragraph if you want, it's just a side point.

The one that sticks in my mind is the application to deformation theory. The Hochschild cochain complex is actually the object of interest in deformation theory, its homology is just one invariant of it and captures the infinitesimal deformations. The cochain complex carries a specific algebraic structure; it's an algebra for the braces operad. And then there's the celebrated (and many times proved ;-)) Deligne conjecture which says that it may be viewed as a homotopy Gerstenhaber algebra. Finally there's Kontsevich's formality result which says that for smooth commutative algebras that looking at homology and its Gerstenhaber algebra structure actually does capture all information about the Hochschild cochains and hence the deformation theory of the algebra.

Anyway I didn't mean the write that, but just got overexcited, my point in writing this answer was to say that there are other homology theories.

For example there's the bar homology. This homology is little known which is a big pity because it's actually rather special! There's a very good reason why it's not studied though and that's because for a unital algebra its homology is always zero, but it is still interesting because it the chain complex a coalgebra and we're not interested in its homotopy type as a complex and so shouldn't be taking its homology at all! The coalgebra actually gives generators and relations for the algebra, it's the derived functor of $A \mapsto A/(A.A)$ from the category of associative algebras to vector spaces.

But you guys like taking homology, so I should give you a better reason for studying the bar homology. Suppose you have an augmented algebra, so we can split the identity off and write

$A = k\oplus A'$

Then the bar homology of $A'$ is not necessarily zero and gives interesting invariants of the algebra. In the char 0 commutative case this is well studied, you guys might know it as part of rational homotopy theory. The commutative bar homology of the cohomology ring of a nice space is the rational homotopy of the space.

  • $\begingroup$ Interesting stuff, I'll have to look into this :) $\endgroup$ – haakon Nov 21 '09 at 13:16
  • 4
    $\begingroup$ Just for the record: for an augmented algebra, the bar homology of A' is nothing but Tor_A(k,k) (see the normalized bar resolution of k) - so it's something not that much neglected as you are saying! $\endgroup$ – Vladimir Dotsenko Dec 7 '09 at 12:52

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