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I am trying to nail down the following statement, which should be completeley standard, but right now I am not sure about some details:
For $G$ a compact Lie group acting smoothly and properly on a smooth manifold, there should be a $G$-representation $V_x$ associated with each point $x \in X^G$ in the fixed locus, obtained from a choice of linear structure on G-invariant tubular neighbourhoods.
This should follow for instance with Wasserman 69, Lemma 3.5. Maybe its made explicit in Bredon 72? In Kankaanrinta 07, p. 7 the statement seems to be taken for granted in the absence of boundaries, and the author seems to be concerned with establishing it also for boundaries, but the wording remains a little inconclusive.
Next, these $V_x$ should be independent of the choices made. Now as $x$ varies, do these representations arrange into a smooth $G$-vector bundle over $X^G$? Or what’s good sufficient conditions that they do?
The $G$-representation exists for instance by the local linearisation for proper Lie groupoids. Corollary 3.10 of https://arxiv.org/abs/1101.0180 might also be useful. $X^G$ is the smallest-dimensional stratum in the orbit-type stratification. $V_x$ will have underlying vector space $T_x X$, the normal bundle to each orbit=a point, so $T(X^G)$ should be the bundle you want.
Thanks! Okay, from the discussion on top of p. 5 in Crainic-Struchiner the essential uniqueness of the linear rep at a fixed point is pretty immediate.
It is clear that as vector spaces these glue to the normal bundle. But do we need an argument that the fiberwise $G$-action on that glues to a global $G$-action? Seems pretty clear, but maybe needs an argument.
Ah, no, of course that’s obvious, from their equation (1).
Okay, so we have canonical fiberwise linear $G$-action on the normal bundle $N(X^G)$.
[ edit: made that a Prop here ]
Next I’d want a $G$-equivariant diffeomorphism from this $N(X^G)$ onto a $G$-invariant tubular neighbourhood of $X^G$.
That maybe follows by using theorem 4.1 from arXiv:1101.0180 at all points in $X^G$ with one and the same metric on $X$?
I think so, you can get a tubular neighbourhood of the zero section of the normal bundle, and with the adapted metric (ie $G$-invariant) should be able to use the exponential map to get the tubular nhd. Since the Lie group is compact I don’t think the more general setting, where $G$-invariant tubular nhds only exist locally, is a worry.
found the required statement in Bredon, added a few more references (thanks to kind help from Markus Pflaum!) and then accordingly expanded the text here, and rearranged slightly.
Finally, I copied these statements over also to tubular neighbourhood into a new subsection Properties – Equivariant version
Another thought, maybe give me a sanity check:
I suppose the orthogonal G-representations which are the fibers of the normal bundle on $X^G$ never contain a direct summand of a trivial rep. Right? For if they did, that trivial rep summand would identify with a non-empty fixed locus in the tubular neighbourhood around and away from $X^G$, contradicting the definition of $X^G$.
That in turn should imply, by Schur’s lemma, that any equivariant map from $X$ to some representation sphere $S^V$, when linearised around $X^G$, should locally split as the Cartesian product of a map $X^G \to S^{(V^G)}$ and a linear map $N_x X^G \to V-V^G$.
Hm, might we have such a splitting even beyond the linear approximation?
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