I'm reading Atiyah and Segal's article "Equivariant K-theory and Completion" and need a little help understanding the notation they use. At various points in the paper they talk about objects of the form $K_G(X)$, $K^q_G(X)$ (where $q$ presumably signifies an integer), and $K^*_G(X)$.

This third notation confuses me: I've read papers that define it as the direct sum of the first two $K$-rings, but in the first paragraph they quote an isomorphism between $R(G)^{\wedge}$ (I'm assuming this signifies the completion of $R(G)$ at the augmentation ideal $I_G$) and a ring they call $K^*(B_G)$, where $B_G$ is a classifying space for the group $G$. As far as I can see, taking $X$ to be a point would then give $K^*(X)$ as the direct sum of two copies of this ring since the suspension of a point is an interval, and in particular is contractible.

I've also seen sources that define $K^*_G(X)$ simply as the system of rings $\{K^n_G(X): n\in\mathbb{Z}\}$, which in the complex case consists of only two rings. There is also a comment on page 7 following a lemma about $K_G(X)$ saying can be replaced by $K^*_G(X)$ by considering the product $X\times S^1$ instead of X.

Any help with this would be greatly appreciated.


Although not given in the Atiyah-Segal paper, I suspect the definition is as follows:

The functor $K_G$ is defined on pairs of $G$-spaces $(X,A)$, since it is a cohomology theory (I guess it's the Grothendieck group of $G$-vector bundles over $X$ that are trivialized on $A$). Set $$ K^{-q}_G(X) := K_G(X\times D^q,X\times S^{q-1}) \, , $$ where $X\times D^q$ is given the diagonal action in which $G$-acts trivially on the disk $D^q$. (Alternatively, $K_G(X\times D^q,X\times S^{q-1}) = \tilde K_G(\Sigma^q (X_+))$, where the latter is the reduced equivariant $K$-theory of the $q$-fold suspension of the based $G$-space $X \amalg +$.)

As is, this gives a cohomology theory graded over negative natural numbers. To extend to a grading over the integers, one needs the equivariant Bott periodicity result (due to Atiyah and Segal) $$ K^q_G(X) \cong K^{q-2}_G(X)\, . $$ In particular this gives an inductive definition of $K^q$ when $q > 0$.

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  • $\begingroup$ Thank you, I suspect you're right. In this case what does $K^*_G(X)$ signify? $\endgroup$ – theinvariant Mar 21 '18 at 17:54
  • 3
    $\begingroup$ It stands for (depending on context) "the collection of groups $K_G^q(X)$" or "the same collection, assembled in some way into a single group" (e.g., as $\bigoplus_q K_G^q(X)$, the associated graded group). $\endgroup$ – LSpice Mar 21 '18 at 17:56
  • $\begingroup$ I see. This generates the following question for me: I understand the six-term exact sequence that Bott periodicity generates, but when they state on page 6 "Consider the exact sequence for the pair $(X\times D^n,X\times S^{n+1})$" and then use the Thom isomorphism, which as I understand it is defined in terms of the $0$th $K$-ring of an equivariant bundle over X, could someone tell me what comes after $K^*(X\times S^{n+1})$ in the sequence they point to? The paper I'm talking about can be found here projecteuclid.org/download/pdf_1/euclid.jdg/1214428815 $\endgroup$ – theinvariant Mar 21 '18 at 18:19
  • $\begingroup$ The long exact sequence of the pair has a connecting homomorphism $K^\ast_G(X\times S^{n-1}) \to K^{\ast+1}_G(X\times D^n,X\times S^{n-1})$. The Thom isomorphism identifies the target of this with $K^{\ast-n+1}_G(X)$. In their case $n$ is even so the latter group is identified with $K^{\ast+1}_G(X)$ by periodicity. The image of the connecting map with respect to these identifications is what they are denoting by ${}_\xi K$. $\endgroup$ – John Klein Mar 21 '18 at 20:25
  • $\begingroup$ That makes sense to me. My only question is then on the notation they choose: when they write $_{\xi}K:=\{x\in K:\xi\cdot x=0\}$ they're talking about a submodule of the $R(\mathbb{T})$-module $K^{\ast+1}_G(X)$, but if I understand correctly, when they write $K/\xi\cdot K$ at the start of this exact sequence, by $K$ they mean $K^{\ast}_G(X)$. Is this a mistake on their part or is there something I'm not getting? $\endgroup$ – theinvariant Mar 22 '18 at 22:51

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