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Unlike algebraic K-theory, equivariant K-theory of affine space (over a field $k$) can be quite nontrivial, depending on the action of the group in question. For example, if one takes the standard action of $\mathbb G_m$ on $\mathbb A^n$, then the $\mathbb G_m$-equivariant K-theory of $\mathbb A^n$ can be computed using a localization sequence corresponding to the inclusion of the origin inside $\mathbb A^n$. (In the localization sequence in equivariant K-theory, one uses that the quotient of $\mathbb A^n \backslash \{0\}$ is $\mathbb P^{n-1}$ and uses the projective bundle theorem, for example, to proceed.)

How does one compute $K_i^G(\mathbb A^n)$ in other cases? Is there a general procedure for doing so? For example, take $G = \mathbb G_m \times \cdots \times \mathbb G_m$ (product of $n$ copies) and consider the diagonal action on $\mathbb A^n$. How does one compute $K_i^G(\mathbb A^n)$ in this case? Is it easier to write down the description of $K_0^G(\mathbb A^n)$?

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In fact, just like in the case of non-equivariant algebraic $K$-theory, there is a homotopy-invariance for equivariant $K$-theory (of nonsingular varieties). Addressing your question specifically, for any action of an algebraic group $G$ on affine space, the answer is: $$ K^G_i({\Bbb A}^n) = R(G) \otimes K_i(k), $$ where $k$ is the ground field and $R(G) = K^G_0(k)$ is the representation ring. If $G$ is a (split) torus, the representation ring is just the group ring of the character group; choosing a basis $x_1,\ldots,x_n$ of characters identifies $R(G)$ with the Laurent polynomial ring ${\Bbb Z}[x_1^\pm,\ldots,x_n^\pm]$.

(The general statement is that when $f\colon X \to Y$ is a $G$-equivariant affine bundle, there is an isomorphism $$f^*\colon G^G_*(Y) \to G^G_*(X),$$ where $G^G_*$ denotes equivariant $K$-theory of coherent sheaves. For nonsingular varieties $X$ one has $G^G_*(X) = K^G_*(X)$. See Theorem 4.1 of Thomason's foundational paper "Algebraic K-theory of group scheme actions".)

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  • $\begingroup$ Is it standard to denote $K$-cohomology with the $*$ as subscript? Please say no. $\endgroup$
    – Allen Knutson
    Commented Feb 19, 2014 at 23:26
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    $\begingroup$ @Allen, very regrettably: yes. Worse, the standard convention of writing K-homology as "G" leads to the atrocity I committed above. For better or worse, after spending a little time with the literature I am starting to be convinced there are at least some good reasons for all this... $\endgroup$
    – Dave Anderson
    Commented Feb 20, 2014 at 1:30
  • $\begingroup$ @DaveAnderson: Thanks a lot for the answer. I didn't know about Thomason's this paper. $\endgroup$
    – Carl
    Commented Feb 20, 2014 at 7:34
  • $\begingroup$ @DaveAnderson: I am now having the following doubt: Consider the standard $\mathbb G_m$ action on $\mathbb A^n$ and consider the localization exact sequence as I had mentioned in the question. The part of the sequence I am worried about goes as follows: $ \cdots \to K_1^G(\mathbb A^n) \to K_1(\mathbb P^{n-1}) \to K_0^G(\{0\}) \to K_0^G(\mathbb A^n) \to K_0(\mathbb P^{n-1}) \to 0$. Now, the groups $K_0^G(\{0\})$ and $K_0^G(\mathbb A^n)$ are the same, however, the map between them in the localization sequence is not an isomorphism. $\endgroup$
    – Carl
    Commented Feb 20, 2014 at 14:19
  • $\begingroup$ So the map $K_1^G(\mathbb A^n) \to K_1(\mathbb P^{n-1})$ is not surjective. Is this right? $\endgroup$
    – Carl
    Commented Feb 20, 2014 at 14:23

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