For a given topological group $G$ there are natural transformations $$K^* \leftarrow K^*_G \overset a\to H^{**}(EG \times_G -;\mathbb Q)$$ from equivariant K-theory, the first forgetting the $G$-structure of a bundle and $a$ inducing from an equivariant bundle over $X$ a nonequivariant one over $EG \times_G X$ and taking the Chern character. These maps do respectably at detecting equivariant K-theory classes in nice cases, so I'm interested in how they can fail.

Let us simplify matters by rationalizing and taking $X$ and $G$ compact. Then the forgetful map factors as $$K^*_G(X;\mathbb Q) \overset a\to H^{**}(EG \times_G X;\mathbb Q) \to H^{**}(X;\mathbb Q) \overset\cong\to K^*(X;\mathbb Q),$$ so we really only want the kernel of $a$. As $a$ can be viewed as completion of the $R(G) \otimes \mathbb Q$-module $K_G^*(X;\mathbb{Q})$ at the augmentation ideal $I(G;\mathbb Q)$, its kernel consists of classes annihilated by some element of $1 + I(G;\mathbb Q)$. Such classes exist when $G$ is discrete (already in $R(G) \otimes \mathbb Q$), but I don't know any where $G$ is connected.

Does anyone have an explicit example of this happening when $G$ is connected?

  • $\begingroup$ Would you be able to give references for how these maps detect equivariant K-theory? $\endgroup$ – A. S. May 17 '18 at 2:57
  • $\begingroup$ I'm not sure if I'm using "detect" wrong: all I mean is that often one can determine equivariant bundles over a $G$-space $X$ exist by knowing something about the image of the forgetful map and about $K^*(X)$, or else by knowing the Borel cohomology groups $H^*(X_G;\mathbb Q)$. Does this make sense? $\endgroup$ – jdc May 17 '18 at 3:45
  • $\begingroup$ I guess a better question would be something like "is it in fact known for some general classes of (X, G) that the image via these two maps recovers equivariant K-theory?" References for where statements like this are proved is what I was asking for I guess $\endgroup$ – A. S. May 17 '18 at 3:57

The simplest example of what you are looking for occurs when $G = S^1$ and $X=S^1/C_6$, where $C_6$ is the group of 6th roots of unity. Then the map $$ K_G(X) \rightarrow K(EG\times_G X)$$ identifies with $$ R(C_6) \rightarrow K(BC_6),$$ which has free abelian kernel of rank 2 (corresponding to the two conjugacy classes in $C_6$ of order not a power of a prime).


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