Let $A$ be a finite semisimple algebra over $\mathbb{k}$, a **perfect** field. Is true that the second Hochschild cohomology group vanishes, i.e. $$HH^2(A) = 0?$$


In order to make this question a little bit more complete, it would also be interesting to discuss whether $$HH^{\*>0}(A) = 0?$$



Hochschild cohomology of finite semisimple algebras
By corollary 18 of <https://projecteuclid.org/journals/nagoya-mathematical-journal/volume-12/issue-none/On-the-dimension-of-modules-and-algebras-VIII-Dimension-of/nmj/1118799929.full>


we have (using the perfect condition which ensures the seperability condition needed for corollary 18 to hold) that the enveloping algebra $A^e=A \otimes\_k A^{op}$ is again semisimple (which is equivalent to having global dimension 0).
Thus $Ext\_{A^e}^i(M,N)=0$ for all $A^e$-modules $M,N$ and $i>0$.
Especially: All Hochschild cohomology spaces $Ext\_{A^e}^i(A,A)$ vanish in positive degree.


