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I am wondering if there is a nice presentation of the Hochschild cohomology of $A_n$ the Weyl algebra. It is known that $H^m(A_n,A_n)=0$ for $m>0$, and thus it is rigid. A proof can be found in Sridharan, but this proof seems to be doing a lot more and is fairly complicated.

I was wondering if there was a simpler way to see this fact specifically. Essentially, I am being a bit lazy.

Thanks!

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2 Answers 2

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(I write $HH^\bullet(\Lambda)$ what you write $H^\bullet(\Lambda,\Lambda)$. When I become Emperor of Notation, everyone will!)

Since $A_n\cong A_1\otimes\cdots\otimes A_1$, the Künneth formula for Hochschild cohomology (which is proved in Cartan-Eilenberg, Theorem XI.3.1, for example) tells you that $HH^\bullet(A_n)\cong HH^\bullet(A_1)^{\otimes n}$. It is enough, then, to compute $HH^\bullet(A_1)$.

The computation of the whole of $HH^\bullet(A_1)$ is not difficult to carry out directly.

You can climb on the shoulders of others and do the following, too. First, it is easy to check that the center of $A_1$ is $k$, so that $HH^0(A_1)=k$. Second, Jacques Dixmier shows, in his Algèbres enveloppentes, that every derivation of $A_1$ is inner, so that $HH^1(A_1)=0$. Finally, the algebra $A_1$ is a Calabi-Yau algebra of global dimension $2$, so in particular it satisfied van den Bergh duality and $HH^2(A_1)=HH_0(A_1)$. The latter vector space is $A_1/[A_1,A_1]$, and a pleasurable computation shows this is zero.

Alternatively, in view of Calabi-Yau-ness of $A_n$, now of global dimension $2n$, we have that $HH^\bullet(A_n)=HH_{2n-\bullet}(A_n)$, and a theorem of Wodzicki tells us that $HH_{2n-\bullet}(A_n)$ is isomorphic to the algebraic de Rham cohomology of $n$-dimensional affine space $\mathbb A^n$, which is just $k$. This way we get $HH^\bullet(A_n)=k$ in one big swoop.

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    $\begingroup$ You're a cool dude. And to be honest, I also prefer HH. Thanks. :) $\endgroup$
    – B. Bischof
    Commented Jun 28, 2011 at 23:01
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    $\begingroup$ @Mariano, in your paper on cohomology of GWAs, does your result hold for higher "dimensional" GWAs(I mean multiple central elements and automorphisms)? I may be being dumb. $\endgroup$
    – B. Bischof
    Commented Jul 25, 2011 at 17:56
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    $\begingroup$ Additionally, after finding this paper: I reiterate my previous comment. $\endgroup$
    – B. Bischof
    Commented Jul 25, 2011 at 18:02
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    $\begingroup$ @Bischof: regarding GWAs: the method should work, but I don't know what the end result will be. In particular, you surely can get a resolution by the same means. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jul 25, 2011 at 18:19
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    $\begingroup$ It is not true that HH^1(A)=0 implies A is a projective bimodule. It is very easyy to construct counterexamples for that claim. (It is true that if H^1(A,M)=0 for all A-bimodules M, then A is projective, but that is rather irrelevant here) In any case, the Weyl álgebras are definitely not projective as bimodules over themselves. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Apr 13, 2014 at 21:25
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Since the Weyl algebra is a deformation of the polynomial ring in two variables, there is a short transparent proof using deformation theory, cf. M. Gerstenhaber and A. Giaquinto, On the cohomology of the Weyl algebra, the quantum plane, and the q-Weyl algebra, arXiv:1208.0346.

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  • $\begingroup$ Thanks Murray. After asking this question many moons ago, I chatted with Anthony about this. I also chatted with you about this once, long ago. I appreciate the reference, however. $\endgroup$
    – B. Bischof
    Commented Mar 11, 2016 at 15:31
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    $\begingroup$ Welcome to MO, Professor Gerstenhaber. $\endgroup$
    – Todd Trimble
    Commented Mar 11, 2016 at 15:31
  • $\begingroup$ Although I'm not certain I've looked at this paper carefully, so I am again thankful for the reference. $\endgroup$
    – B. Bischof
    Commented Mar 11, 2016 at 15:33

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