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For a (closed and oriented) manifold $M$, the first Lie algebra cohomology $H^1(\mathrm{Vect}(M),C^\infty(M))$ of the space of vector fields with coefficients in smooth functions is isomorphic to $H^1(M,\mathbb{R})\oplus\mathbb{R}\operatorname{div}$, where $\operatorname{div}$ is the divergence associated with an arbitrary volume form. (For the reference, see Theorem 2.4.11 of "Cohomology of Infinite-dimensional Lie Algebras" by D. B. Fuks.)

Question: Are there any analogues in non-commutative geometry? For example, is $H^1(\operatorname{Der}(A),A/[A,A])$ isomorphic to $\mathrm{HDR}^1(A)\oplus K$ for a unital associative $K$-algebra $A$ with $\mathrm{HDR}^0(A)=K$?

Here $\mathrm{HDR}^1$ is the Karoubi-de Rham cohomology of $A$. I suppose some kind of smoothness or finite-dimensionality would be required.

Thank you.

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    $\begingroup$ It's not true for $A=\{0\}$, since the left-hand term is zero and the right-hand term is not (because of $\oplus K$). You might also compare with the case of $M$ with $n$ connected components, which corresponds to $A$ being a product of $n$ indecomposable commutative algebras. $\endgroup$
    – YCor
    Commented Dec 11, 2023 at 10:27
  • $\begingroup$ @YCor Thanks for the comment. Some assumptions are added. $\endgroup$
    – Qwert Otto
    Commented Dec 11, 2023 at 10:53
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    $\begingroup$ I am a bit surprised that you take $A/[A,A]$ as coefficients - at least if you want to imitate the divergence statement, since the conceptually meaningful divergence of a derivation takes values in the commutator quotient of the universal enveloping algebra, not in the commutator quotient of the algebra itself. $\endgroup$
    – Vladimir Dotsenko
    Commented May 19 at 8:51
  • $\begingroup$ @VladimirDotsenko I thought that substituting $B^e = A$ and pulling back by $\mathrm{Der}(B)\to \mathrm{Der}(A)$ recovers enough information for my computation, but you're right. I should've taken $A^e/[A^e,A^e]$ for better formulation. Thanks. $\endgroup$
    – Qwert Otto
    Commented May 19 at 9:20

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