6
$\begingroup$

Let $k$ be a field. Given a commutative $k$-algebra $Z$ and an associative algebra $A$ that is Azumaya over $Z$, do we have an isomorphism of Hochschild cohomologies: $HH^*(A) \cong HH^*(Z)$?

This is true in characteristic zero by Weibel and Cortiñas, but their proof doesn’t generalise to prime characteristic. Is there a characteristic-independent proof, or is there a counter example in positive characteristic?

$\endgroup$
1
  • 2
    $\begingroup$ The paper Noncommutative motives of Azumaya algberas by Tabuada and Van den Bergh proves this when the rank of $A$ is invertible in $k$. I do not know a counterexample in the case when the characteristic divides the rank. $\endgroup$ Commented Jun 17, 2020 at 15:08

1 Answer 1

4
$\begingroup$

It is true that $HH^*(A) \cong HH^*(Z)$ when $Z$ is a commutative $k$-algebra (i.e. in the setting of affine varieties). The following argument is due to Vadim Vologodsky, although any errors are mine: $\DeclareMathOperator{\Spec}{Spec}\DeclareMathOperator{\Hom}{Hom}$

Work in schemes over $k$. Let $X = \Spec Z$, and let $\Delta \subseteq X \times X$ be the diagonal. The Hochschild cohomology of $Z$ is $R\Hom_{X \times X}(\mathcal O_\Delta, \mathcal O_\Delta),$ but in fact this agrees with $R\Hom_{\widehat{X \times X}}(\mathcal O_\Delta, \mathcal O_\Delta)$ where $\widehat{X \times X}$ is the formal neighborhood of $\Delta \subseteq X \times X$. Similarly, $HH^*(A) = R\Hom_{\widehat{A \boxtimes A^{op}}}(A,A)$, where $\widehat{A \boxtimes A^{op}}$ means completion of $A \boxtimes A^{op}$ on $X \times X$ along the diagonal. So, it suffices to show $\widehat{A \boxtimes A^{op}}$ is a split Azumaya.

Note that $\widehat{A \boxtimes A^{op}}$ is split on $\Delta \subseteq \widehat{X \times X}$, with splitting bundle $A$. Viewing $A$ as defining a $\mathbb G_m$-gerbe, the obstruction to extending a splitting from $\Delta$ to $\widehat{X \times X}$ is in $$ H^2(\widehat{X \times X}, ker(\mathbb G_{m, \widehat{X \times X}} \to \mathbb G_{m,\Delta}))$$ But by filtering according to powers of the ideal of $\Delta$, we see $ker(\mathbb G_{m,\widehat{X \times X}} \to \mathbb G_{m,\Delta})$ has a complete filtration where each piece of the associated graded is coherent. As $X$ is affine, this cohomology vanishes, and the splitting on $\Delta$ extends to $\widehat{X \times X}$.

Remark: if $X$ is not affine, the result is false, see e.g. Counterexamples to Hochschild-Kostant-Rosenberg in characteristic p by Antieau, Bhatt, and Mathew.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .