# Cohomology ring with non-commutative coefficient ring

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 $$A$$?

• 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. – LSpice Mar 23 '19 at 3:48
• Thanks for the comment. – Kiu Mar 23 '19 at 4:10
• 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? – user51223 Mar 24 '19 at 2:25

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