A simple proof of the Weyl algebra's rigidity. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T15:22:49Z http://mathoverflow.net/feeds/question/69059 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69059/a-simple-proof-of-the-weyl-algebras-rigidity A simple proof of the Weyl algebra's rigidity. B. Bischof 2011-06-28T22:03:26Z 2011-06-29T01:11:42Z <p>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 <a href="http://www.ams.org/journals/tran/1961-100-03/S0002-9947-1961-0130900-1/S0002-9947-1961-0130900-1.pdf" rel="nofollow">in Sridharan</a>, but this proof seems to be doing a lot more and is fairly complicated. </p> <p>I was wondering if there was a simpler way to see this fact specifically. Essentially, I am being a bit lazy. </p> <p>Thanks!</p> http://mathoverflow.net/questions/69059/a-simple-proof-of-the-weyl-algebras-rigidity/69065#69065 Answer by Mariano Suárez-Alvarez for A simple proof of the Weyl algebra's rigidity. Mariano Suárez-Alvarez 2011-06-28T22:55:09Z 2011-06-28T23:01:22Z <p>(I write $HH^\bullet(\Lambda)$ what you write $H^\bullet(\Lambda,\Lambda)$. When I become Emperor of Notation, everyone will!)</p> <p>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 $HH^\bullet(A_1)$.</p> <p>The computation of the whole of $HH^\bullet(A_1)$ is not difficult to carry out directly.</p> <p>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 <em>Algèbres enveloppentes</em>, 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.</p> <p>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.</p>