19
$\begingroup$

First let's consider equivariant cohomology: if a compact Lie group $G$ acts on a compact manifold $M$. We have the equivariant cohomology $ H_G(M)$ defined as the cohomology of the cochain complex $((\mathbb{C}[\mathfrak{g^*}]\otimes \Omega^{\bullet}{M})^G, d_G)$ (for the definition of equivariant cohomology we can look at chapter 1 and 4 of Guillemin and Sternberg's book "Supersymmetry and Equivariant de Rham Theory"). $K< G$ is a closed subgroup, Let $M^K$ be the points of $M$ which has isotropy groups conjugated to $K$, obviously $M^K$ is a $G$-submanifold of $M$ and let $~i: M^K \rightarrow M$ denote the inclusion map. we have a version of localization theorem, see Guillemin and Sternberg's book "Supersymmetry and Equivariant de Rham Theory" chapter 11, especially Theorem 11.4.3 in page 178. In more details :

Consider the equivariant cohomology $ H_G(M)$ and $H_G(M^K)$ as $ S( \mathfrak{g^* })^G $ modules. Then the pullback map $$ i^*: H^ * _G(M)\rightarrow H^ *_G(M^K) $$ is an isomorphism after localizing at some certain prime ideals of $ S( \mathfrak{g^* })^G $.

On the other hand, we have the equivariant K-theory $K_G(M)$ and we also have the localization theorem in this side, see Segal "Equivariant K-theory" (1967) section 4, proposition 4.1, which also claims that Then the pullback map $$ i^*: K^ * _G(M)\rightarrow K^ *_G(M^K) $$ is an isomorphism after localizing at some certain prime ideals of $R(G)$, the representation ring of $G$.

We notice the similarity of the above two version of localization theorems. Nevertheless equivariant cohomology and equivariant K-theory are different. The first is the cohomology of a differential graded algebra and the latter is the Grothedieck group of modules of the cross product algebra $G \ltimes C(M)$.

My question is: is there any deep relation between them? Are they valid because of the same reason?

$\endgroup$
4
  • 2
    $\begingroup$ The localization theorem for equivariant K-theory is valid even when $G$ is a finite group. On the other hand, I think (correct me if I am wrong) that the localisation theorem for equivariant cohomology needs $G$ to have positive dimension to have a non-trivial content. This suggests that there is a conceptual difference between them. $\endgroup$ Sep 11, 2012 at 5:11
  • $\begingroup$ Thank you for your comments Damian! Yes, as far as I know the localization theorem is true essentially for compact abelian Lie groups. $\endgroup$ Sep 11, 2012 at 5:46
  • 2
    $\begingroup$ I'm still hoping someone who understands this better than me answers this, but... Yes - there is (at least in some cases) a "Chern map" taking (equivariant) vector bundles to their (equivariant) Chern classes, and this should be the relation you are looking for. $\endgroup$ Sep 11, 2012 at 17:55
  • $\begingroup$ @Alexander Thank you for your comment! Yes I think Chern character map may be the answer and there are a lot of interesting theory on it (for example the paper by J. Block and E. Getzler "Equivariant cyclic homology and equivariant differential forms"). I will think more carefully about this. $\endgroup$ Sep 17, 2012 at 15:18

2 Answers 2

8
$\begingroup$

This is very late and you've no doubt learned this in the last five years, but for completeness, the relation is indeed that they are linked by completion and the Chern character, as suggested in one of the comments. Given a compact $G$-space $X$, one can equip $EG \times X$ with the diagonal $G$-action, and then the projection $EG \times X \to X$ is equivariant. Pulling back an equivariant bundle $V \to X$ to $EG \times X$ gives an $G$-equivariant bundle, which descends to bundle $V_G \to X_G$ over the homotopy orbit space (Borel construction) $X_G = (EG \times X) / G$. Thus there is a natural induced ring map $$K^*_G(X) \longrightarrow K^*(X_G).$$ (The theorem of Atiyah and Segal is that that this can be identified with completion at the augmentation ideal $I(G)K^*_G(X)$.)

The Chern character $K^* \longrightarrow H^*(-;\mathbb Q)$ then gives a map to $H^*_G(X;\mathbb Q)$. The composition can be seen as an equivariant Chern character. Everything in sight is natural, so the inclusion $X^K \hookrightarrow X$ induces a commutative diagam

$$\require{AMScd}\begin{CD} K^*_G(X) @>>> H^*_G(X;\mathbb Q)\\ @V V V @VV V\\ K^*_G(X^K) @>>> H^*_G(X^K;\mathbb Q). \end{CD}$$

The claim is that the vertical maps become isomorphisms upon localization at certain ideals of $R(G)$ (resp. $H^*(BG;\mathbb Q)$). One checks now that this map, applied in the case $X = *$, sends the one ideal to the other.

Now, we've constructed natural transformations $$K^*_G \to K^*(-_G) \to H_G^*(-;\mathbb Q);$$ the final claim is that this process amounts to tensoring with $\mathbb Q$ and then completing at $I(G) K^*_G \otimes \mathbb Q$. One can extract this from the proof of the completion theorem, which proceeds by looking at compact approximations $X_{n,G}$ of $X_G$ and comparing the inverse limit to the ring completion; since these approximations are compact, the Chern character induces isomorphisms $K^*(X_{n,G}) \otimes \mathbb Q \to H^*(X_{n,G};\mathbb Q)$.

I have never checked this, but because $R(G)$ is Noetherian, I believe the localization theorem in equivariant cohomology at the level of $\mathbb Z/2$-graded rings then follows from the K-theoretic localization theorem by commutative algebra.

$\endgroup$
1
  • $\begingroup$ the target of the equivariant Chern character map is the completion of equivariant cohomology at the ideal generated by elements of positive degree. the issue is that $X_G$ will have unbounded cohomology, and the Chern character will live in every even degree $\endgroup$ May 27, 2022 at 12:55
4
$\begingroup$

The following is relatively standard. It is almost certainly not the "deep" reason you are asking for. I think it is useful point of view though.

Suppose I have some invariant $J$ which takes values on $G$-spaces. Assume it satisfies some kind of sum / long exact sequence / something more fancy for open-closed decompositions, and that its values on spaces without fixed points is "small". Then we expect $J(X)$ and $J(X^G)$ to differ by a "small" amount. Indeed, consider the open-closed decomposition involving $X$ and $X^G$, and apply $J$.

Example 1: $J$ is Euler characteristic for manifolds with $S^1$-action, where in this case "small" means "is zero". We deduce that $J(X) = J(X^{S^1})$.

Example 2: $J$ is Euler characteristic for manifolds with $\mathbb{Z}/p\mathbb{Z}$-action. Here "small" means divisible by $p$, and we deduce that $J(X) = J(X^{\mathbb{Z}/p\mathbb{Z}})$ modulo $p$ (so difference has small $p$-adic norm).

Example 3: $J$ is equivariant cohomology for spaces with $T$-action. Here "small" means "is torsion", and we recover the localization theorem in equviariant cohomology.

Example 4: $J$ is equivariant $K$-theory for spaces with $T$-action. Again "small" means "is torsion" and we recover the localization theorem.

For more detail, see Section 4 of this paper where I give a bit more detail and try to motivate Smith theory from this point of view: https://arxiv.org/abs/2001.04569.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.