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Suppose I have some smooth closed high-dimensional manifold $M$ acted on smoothly by a finite group $G$. By a metric averaging procedure, we can equip $M$ with a smooth Riemannian metric so that $G$ acts by isometries. I can't necessarily pick a $G$-invariant morse function $f:M\to\mathbb R$, but nevertheless, I can certainly pick a smooth function $f:M\to\mathbb R$ which, though perhaps not Morse, still has only isolated "nice" critical points in some precise sense. We therefore conclude:

There is a "handle" decomposition of $M$ (where I haven't said what I mean by "handle") which is preserved by $G$. Thus $G$ just permutes (and/or acts on individually) the handles.

I am interested in knowing to what extent this can be generalized to the case of an "action up to homotopy". More specifically, suppose we have $G\to\operatorname{Homeo}(M)/\text{homotopy}$. To what extent can we "decompose" M into simple pieces in a $G$-invariant way? If it helps, then it is OK to assume that the action of $G$ is "close to the identity" in a vague coarse sense.

(I am essentially just interested on the case of high-dimensional $M$, but of course the question makes sense in any dimension).

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    $\begingroup$ This does not answer your question, but provides a different approach to the Morse theory in the first part. A theorem of Illman says that if $M$ is a manifold and $G$ is a finite group acting on $M$, then $M$ has a $G$-invariant triangulation. You could get an equivariant handle decomposition from this, but it is in some sense more rigid. See MR1770606 (2001j:57032) Illman, Sören(FIN-HELS) Existence and uniqueness of equivariant triangulations of smooth proper G-manifolds with some applications to equivariant Whitehead torsion. J. Reine Angew. Math. 524 (2000), 129–183. $\endgroup$
    – Andy Putman
    Commented Aug 9, 2012 at 3:15
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    $\begingroup$ Can you be more precise about the meaning of "$G\to\operatorname{Homeo}(M)/\text{homotopy}.$" ? $\endgroup$
    – John Klein
    Commented Aug 9, 2012 at 13:40
  • $\begingroup$ Say that two elements of $\operatorname{Homeo}(M)$ are equivalent iff they are homotopic as maps $M\to M$. This is an equivalence relation, and the group operation on $\operatorname{Homeo}(M)$ descends to the quotient. $\endgroup$
    – John Pardon
    Commented Aug 9, 2012 at 15:45
  • $\begingroup$ @JohnPardon do you mean "two maps of $Homeo(M)$ are equivalent iff they are homotopic through homeomorphisms as maps $M \to M$? Because otherwise it's not clear that the inverses of equivalent maps are equivalent. $\endgroup$
    – Tim Campion
    Commented Apr 12, 2018 at 17:09
  • $\begingroup$ When $M=S^n$ is a sphere, then the space of orientation preserving homeomorphism is arc connected. In this case $G \to \textrm{Homeo}^+(M)/\textrm{homotopy} = pt$ seems to be no data at all? Is that really what you mean? $\endgroup$
    – Chris Schommer-Pries
    Commented Apr 12, 2018 at 17:58

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This foundational paper: Arthur G. Wasserman, Equivariant Differential Topology, Topology, 8(1967), 127-150, has section 4 dealing with equivariant Morse theory for manifolds with a smooth action of a compact Lie group, including equivariant handle attaching, equivariant Morse Lemma, etc. See MathSciNet Review. The last result of the paper concludes that a compact G-manifold is a union of "handle bundles over orbits".

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    $\begingroup$ This is a good reference for my boxed statement, but that's not what my question is about. I want to know about when the action is only defined "up to homotopy". Also, all my groups are finite. $\endgroup$
    – John Pardon
    Commented Aug 9, 2012 at 15:48
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    $\begingroup$ I'm upvoting because if somebody googles "equivariant handle decompositions" (which is, after all, the title of the question), they will be well-served to find this answer. Perhaps it's worth mentioning that an equivariant handle consists of an equivariant disc bundle over a homogenous space $G/H$. $\endgroup$
    – Tim Campion
    Commented Apr 12, 2018 at 17:02

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