4
$\begingroup$

For a given commutative algebra $A$ over a field $\mathbb{K}$(with char=0) the algebra of differential operators on $A$ is the set of endomorphism $D$ of $A$ such for some $n$ we have that for any sequence $\left\lbrace a_i\right\rbrace_{1\leq i\leq n}$ for $a_i\in A$ one has $[\ldots[[D,a_0],a_1],\ldots,a_n]=0$.

By analogy, I think of Hochschild Cohomology as a sort of "algebraic differential forms"(maybe this isn't the right approach, but $C^{n}(A,A)=Hom(A^{\otimes n},A)$ and it gives a cohomology theory etc.), thus it seems like there should be a connection to differential operators on the algebra.

Sorry if the question is a little general, but I am open to accepting a variety of answers.

Thanks in advance!

  • 1
    The first sentence seems to contain some kind of structural error: I am not so knowledgeable in this field, but I doubt that the alebra of differential operators on $A$ is an endomorphism of $A$...2011-06-04
  • 0
    hehe, thanks @Pete L. Clark and @Mariano, the edit was what I intended. :)2011-06-04

3 Answers 3