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

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

2012-11-04T06:32:34Z

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 $R(S^1)$?

Answer by Neil Strickland for Equivariant K-theory of $S^1$-action on $S^2$

2012-11-04T10:14:09Z

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.

Answer by Allen Knutson for Equivariant K-theory of $S^1$-action on $S^2$

2012-11-04T14:33:45Z

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 $p(1) = q(1)$.