Equivariant Homotopy: M & N Manifolds, G-Action & Cohomology Rings

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Discussion Overview

The discussion centers on the relationship between homotopy equivalence and G-equivariant homotopy equivalence for two manifolds M and N, both endowed with a G-action. Participants explore whether homotopy equivalence implies G-equivariant homotopy equivalence and the implications for equivariant cohomology rings.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant questions if homotopy equivalent manifolds M and N must also be G-equivariantly homotopy equivalent, particularly regarding their equivariant cohomology rings.
  • Another participant expresses skepticism, arguing that the homotopy equivalence does not inherently involve the G-action, suggesting that this relationship is not guaranteed.
  • A different participant agrees with the skepticism, proposing that a compatibility condition between the G-actions might be necessary for G-equivariant homotopy equivalence to hold.
  • One suggestion is made to explore simple examples of equivariant cohomology to investigate potential differences, specifically mentioning the construction of non-trivial actions on R^n.
  • Another example is provided involving the circle and the Z2 action, illustrating that different actions can lead to different homotopy equivalences despite the spaces being homotopy equivalent.

Areas of Agreement / Disagreement

Participants generally disagree on whether homotopy equivalence implies G-equivariant homotopy equivalence, with multiple competing views expressed regarding the necessity of compatibility between actions.

Contextual Notes

Participants note the need for definitions and examples to clarify the concepts discussed, indicating that the discussion may depend on specific interpretations of G-actions and equivariant cohomology.

Who May Find This Useful

This discussion may be of interest to those studying algebraic topology, particularly in the context of equivariant cohomology and manifold theory.

Kreizhn
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This might be a really silly question, but suppose that you have two (possibly Frechet) manifolds M and N both endowed with a G-action. If M and N are homotopy equivalent, is it necessary that they will be G-equivariantly homotopy equivalent?

Edit: That is, should I expect them to have the same equivariant cohomology rings?
 
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I don't know what any of those words mean but I will say no, because why on Earth would this be true when the hypothesis (homotopy equivalent) does not contain the key word "G". I mean if this were true why would G actions be interesting? Maybe if you give us some of the relevant definitions, or at least a few examples, we can see more what is going on.
 
I am leaning towards agreeing with your assessment: it seems unreasonable that one should imply the other, especially since I could put really weird actions on each manifold.

Perhaps there needs to be some sort of added "compatibility" requirement between the actions. Upon thinking about it, that compatibility is probably exactly the fact that a homotopy equivalence be equivariant.

Anyway, thanks for the response. I ask because I'm reading a paper which makes some assertions on equivariant cohomology, but only utilizes the fact that two spaces are homotopy equivalent. I was wondering if perhaps I was missing something obvious.
 
work out some very simple examples of equivariant cohomology, and see if they can differ. e.g. try to cook up a non trivial action that gives a non trivial equivariant cohomology on R^n, which of course is homotopic to a point.
 
The circle is homotopy equivalent to itself. Let Z2 act on it in two ways: first trivially by the identity map and second by reflection around the y axis. These maps are not homotopically equivalent.
 

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