I've run across something that surprises me, so I'm wondering (1) Is it true? and (2) Is it well known? (And if the answers are affirmative, why didn't I know this already?)

Let $G$ be a compact Lie group and let $H$ and $K$ be closed subgroups. So that the question isn't trivial, you can assume that $H$ is subconjugate to $K$. Then we can consider the $WH = NH/H$ space $(G/K)^H$. The claim is that this is a disjoint union of orbits of $WH$.

The first proof I found depends on the Mongtomery-Zippin theorem via a consequence that appears as Lemma 1.1 in Peter May's "Equivariant Orientations and Thom Isomorphisms": If $j\colon \alpha\to\beta$ is a homotopy between $G$-maps $G/H\to G/K$, then $j$ factors as the composite of $\alpha$ and a homotopy $c\colon G/H\times I \to G/H$, where $c(eH,t) = c_tH$ for a path $c_t$ in the centralizer $C_GH$ of $H$, starting at the identity. Reinterpreting in terms of $(G/K)^H$, this implies that, if $\alpha$ and $\beta$ are two points in the same path component of $(G/K)^H$, then $\beta = c\alpha$ for some $c$ in the identity component of $C_GH$. In particular, they are in the same $WH$-orbit. Hence, each $WH$-orbit consists of a union of path components of $(G/K)^H$.

I think another proof could be found by looking at the tangent plane at an $H$-fixed point in $G/K$, as an $H$-representation, and observing that the only $H$-trivial directions are those in the direction of the $NH$-action. (But I haven't worked this out completely.)

So, is this true? Is it already well known? If it is true, it has some interesting implications for equivariant ordinary homology.


To answer my own question: I happened to run across the answer just now. Yes, it's true and known. tom Dieck quotes it as II.5.7 in Bredon's Introduction to compact transformation groups (which I don't happen to have handy).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.