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Let A be a non-commutative algebra and let X be some geometric space (such as a topological space or an algebraic variety or scheme). Is there a notion of cohomology ring of X with coefficients in A? What is the correct set up to consider cohomologies with non-commutative coefficients?

If a topological group $G$ acts on a space $X$ one can construct its equivariant cohomology ring $H^*_G(X)$ (say with coefficients in $\mathbb{R}$). Is there a notion of equivariant cohomology for $(X, G)$ with coefficients in a non-commutative algebra $A$?

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    $\begingroup$ Although it's not what you're asking for, the Galois cohomology sets $\operatorname H^1(E/F, \mathbb G(E))$ with coefficients in the $E$-rational points of an algebraic group $\mathbb G$ make sense even if $\mathbb G$ is non-Abelian. $\endgroup$ – LSpice Mar 23 at 3:48
  • $\begingroup$ Thanks for the comment. $\endgroup$ – Kiu Mar 23 at 4:10
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    $\begingroup$ I think a related question to ask is that are there examples of cohomology theories such that the non-commutativity of the coefficient ring helps to detect something more that the case when the coefficient ring is commutative? $\endgroup$ – user51223 Mar 24 at 2:25
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Is there a notion of cohomology ring of X with coefficients in A?

Yes, and nothing new is needed. The underlying additive group of $A$ is abelian so you take cohomology with coefficients in that abelian group; then the multiplication on $A$ is a bilinear map $A \times A \to A$ which induces a map

$$H^n(X, A) \times H^m(X, A) \to H^{n+m}(X, A)$$

in the usual way. That is, the construction is exactly the same as for cohomology with coefficients in a commutative ring; commutativity of the multiplication is not actually used in the construction. So this isn't actually "nonabelian cohomology" in the sense that term is usually meant.

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