Equivariant Homotopy


by Kreizhn
Tags: equivariant, homotopy
Kreizhn
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#1
Feb12-13, 02:22 PM
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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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mathwonk
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#2
Feb12-13, 03:03 PM
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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.
Kreizhn
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#3
Feb12-13, 03:08 PM
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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.

mathwonk
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#4
Feb12-13, 08:38 PM
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Equivariant Homotopy


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.
lavinia
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#5
Feb13-13, 03:41 PM
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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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