Let $R_G$ be the representation ring of $G$ a connected, simply connected Lie group, $I_G$ the augmentation ideal and $\mathbb{Z}=R_G/I_G$. $R_G$ acts on $\mathbb{Z}$ via $V \cdot n = (\dim V ) n$. I want to (somewhat) compute $K_G^*(G * G)$ (where the star denotes the topological join). From Milnor's old paper on classifying spaces this can be reduced to a computing $\text{Tor}_*^{R_G} (\mathbb{Z}, \mathbb{Z})$. I don't know a resolution of $\mathbb{Z}$ over $R_G$, so I can't really get much beyond $\text{Tor}_1^{R_G}(\mathbb{Z}, \mathbb{Z})=I/I^2$.
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$\begingroup$ What does $G \ast G$ denote? To me that symbol means the free product. $\endgroup$– Qiaochu YuanCommented Nov 17, 2014 at 4:28
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$\begingroup$ In any case, if $G$ is in addition assumed to be connected then $R_G$ is very close to a polynomial algebra and so you should be able to use some kind of Koszul resolution. $\endgroup$– Qiaochu YuanCommented Nov 17, 2014 at 4:31
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$\begingroup$ Yea, it's close to a polynomial algebra. It's finitely generated but the exact resolution will probably depend on $G$. $\endgroup$– Sven CattellCommented Nov 17, 2014 at 4:43
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1$\begingroup$ Can we say that $R_G$ being very close to polynomial, $Tor ^{R_G}(Z,Z)$ is very close to an exterior algebra? $\endgroup$– user43326Commented Nov 17, 2014 at 10:21
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$\begingroup$ Yea, in the case of $SU(n)$, $R_G$ is a polynomial algebra on a suitable basis of symmetric polynomials on $n$-generators, so in this case it reduces to that. $\endgroup$– Sven CattellCommented Nov 17, 2014 at 17:05
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