Timeline for Is Hochschild cohomology finitely generated?
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| Aug 16, 2013 at 15:54 | comment | added | user36931 | @ Marci You might be thinking about an example of a commutative dg-algebra whose Hochschild cohomology is not finitely generated \emph{as a ring}. This is certainly possible, but not the question. | |
| Aug 15, 2013 at 16:52 | comment | added | Marci | Yes, it should be $\wedge^i T_A$. Btw, if I remember well, there is an dg-alg. A (finitely generated over k) s.t. its Hochschild cohomology is not finitely generated over A (which does not answer the question). | |
| Aug 15, 2013 at 15:18 | comment | added | Mariano Suárez-Álvarez | If $M$ and $N$ are symmetric $A$-bimodules, then $\hom_{A^e}(M,N)$ is also a symmetric $A$-bimodule. IOW, the action of $A^e$ on this $\hom$ space factors through the multiplication map $A\otimes A\to A$, and the answer to your question follows at once, no? | |
| Aug 15, 2013 at 14:10 | review | Suggested edits | |||
| Aug 15, 2013 at 14:20 | |||||
| Aug 15, 2013 at 13:49 | comment | added | user36931 | But let me ask an even more naive question, I had always assumed that that the above $A\otimes_k A$ module structure should factor through the diagonal A. If you know that that automatically implies what you want, so perhaps your saying that is false? P.S. A nice reference for this Quillen fact is the following paper arxiv.org/pdf/math/0606730.pdf. Also in Marci's comment the result should be $\wedge^i T_A$ for cohomology. | |
| Aug 15, 2013 at 12:54 | comment | added | user36931 | Seems like maybe you can just use the adjunction and $Ext_{A\otimes A}(A,M)=Ext_A(\Delta^*\Delta_{*}(A),M).$ So it seems like you further have the fact that $\Delta^*\Delta_{*}(A)= \oplus Sym^p(L_{A/k})$, due to Quillen (L_{A/k} is the cotangent complex). So the question is does this $\oplus Sym^p(L_{A/k})$ have coherent cohomology in each degree. It seems the answer is yes, since the cotangent complex should also have coherent cohomology(getting confused with gradings...), but either way the answer should be pretty easy to see... | |
| Aug 15, 2013 at 12:35 | comment | added | Marci | I do not think it is true. However, in the case of smooth A, we have the Hochschild-Konstant-Rosenberg isomorphism, which tells that there is an isomorphism between $HH^i(A)$ and $\Omega^i_A$. In particular, $HH^i(A)$ is finitely generated. | |
| Aug 15, 2013 at 11:45 | review | First posts | |||
| Aug 15, 2013 at 12:09 | |||||
| Aug 15, 2013 at 11:27 | history | asked | Hom | CC BY-SA 3.0 |