Let $X$ be a $(G \times H)$-space. What is known about the connection between the groups $K_G(X)$, $K_H(X)$ and $K_{G \times H}(X)$? The $G$ and $H$ action on $X$ come from the canonical inclusions $G \to G \times H$ and $H \to G \times H$.

There is the Künneth theorem relating $K_G(X)$, $K_H(Y)$ and $K_{G \times H}(X \times Y)$, where $X$ is a $G$-space and $Y$ is a $H$-space. This only reduces to the above case when either $G$ or $H$ acts trivially (and the corresponding $X$ or $Y$ is a point).

I can't find anything in the literature, for K-theory or equivariant cohomology. Is there a specific reason why it is so hard to get information about $K_{G \times H}(X)$ from $K_G(X)$ and $K_H(X)$?

  • 1
    $\begingroup$ Let me mention that a $G\times H$-action carries potentially more structures than a $G$-action and an $H$-action. At least, there is a commutativity between two actions (and in higher algebra, such as the K-theory spectrum that you want to study, commutativity is an extra structure). $\endgroup$
    – Z. M
    Oct 18, 2023 at 13:33

1 Answer 1


It is exactly the content of a paper of Minami (At least for compact Lie groups)

A Künneth formula for equivariant $K$-theory. Haruo Minami. Osaka J. Math. 6(1): 143-146 (1969).

Following ideas of Atiyah's proof of the Kunneth theorem for K-theory (non-equivariant) he obtain exactly the result you want.

As a final comment note that if $G$ acts on spaces $X$ and $Y$ it is too much hard to obtain $K_G(X\times Y)$ in terms of $K_G(X)$ and $K_G(Y)$

  • 9
    $\begingroup$ OP is asking for the case when both $G$ and $H$ are acting on the same space $X$, and interested in the $G\times H$ equivariant $K$-theory of $X$ (and not that of $X \times X$). $\endgroup$
    – user43326
    Nov 13, 2022 at 19:49

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