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](http://www.ams.org/journals/tran/1961-100-03/S0002-9947-1961-0130900-1/S0002-9947-1961-0130900-1.pdf), 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!



A simple proof of the Weyl algebra's rigidity.
(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.


