# Equivariant K-theory of $S^1$-action on $S^2$

Is there any references for the structure of the equivariant K-theory $K_{S^1}(S^2)$ where the action of $S^1$ on $S^2$ is defined to be rotation about the $z$-axis? What is the ring structore of $K_{S^1}(S^2)$ and the module structure over the representation ring $R(S^1)$?

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What reference are you using for the definition of equivariant k-theory? –  Ryan Budney Nov 4 '12 at 6:39
@Ryan: You can find the definition in Segal's 1968 paper "Equivariant K-theory". –  Zhaoting Wei Nov 4 '12 at 7:31
Your response confuses me -- you supplied the appropriate reference that answers your question on your own. Did you want to ask a different question? –  Ryan Budney Nov 4 '12 at 10:25

Let $L$ denote $\mathbb{C}$ with $S^1$ acting by multiplication, and let $\mathbb{C}$ denote $\mathbb{C}$ with trivial $S^1$-action. Then the projective space $P(L\oplus\mathbb{C})$ is homeomorphic to $S^2$, and the natural $S^1$-action is the one that you mentioned. Thus, your problem is a special case of calculating $K_G(PV)$, where $V$ is a complex representation of a compact Lie group $G$. There is an evident map from $R(G)=K_G(\text{point})$ to $K_G(PV)$, and the tautological bundle $T$ also gives an element of $K_G(PV)$, so the polynomial ring $R(G)[T]$ maps to $K_G(PV)$. Put $f(t)=\sum_{k=0}^{\text{dim}(V)}(-1)^k\Lambda^k(V^*)t^k$. The constant bundle with fibre $V$ splits as $T\oplus T^\perp$, and using this one can check that $f(T)=0$ in $K_G(PV)$. With more work it can be shown that $K_G(PV)=R(G)[T]/f(T)$. This is stated as Proposition 3.9 in Segal's "Equivariant K-Theory"; the proof relies on a result that Segal states as Proposition 3.8, but does not prove; for that, see Proposition 4.9 of Atiyah's "Bott periodicity and the index of elliptic operators". A more direct argument is possible for the case that you mention, but the result above gives the general context.
For Hamiltonian actions (e.g. on smooth complex projective varieties), one can use equivariant localization, as in http://front.math.ucdavis.edu/0503.5609 . Let $R(S^1) = Z[t^\pm]$, so the restriction map $K_T(S^2) \to K_T($fixed points$) = Z[t_1^\pm] \oplus Z[t_2^\pm]$ hits those pairs $(p(t_1),q(t_2))$ such that $p(1) = q(1)$.