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?

  • Would you be able to give references for how these maps detect equivariant K-theory? – A. S. May 17 at 2:57
  • 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? – jdc May 17 at 3:45
  • 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 – A. S. May 17 at 3:57
up vote 9 down vote accepted

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).

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.