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In equivariant homotopy theory, it seems like one tends to consider "genuine" $G$-spaces or $G$-spectra, rather than spaces (spectra) with a $G$-action.

My (rather soft) question is : why is that a reasonable thing to consider ?

For instance, a baby example of a difference between both is the (unique) $\mathbb Z$-equivariant map $\mathbb R\to *$. This is a weak equivalence in the model structure that produces "spaces with a $G$-action", but not for "genuine $G$-spaces", e.g. because $\mathbb R^\mathbb Z = \emptyset \not\simeq * = *^\mathbb Z$ (where $X^\mathbb Z$ denotes the set of fixed points for a $\mathbb Z$-space $X$).

What I'd like to understand is for instance why we might want that map not to be an equivalence.

Is it simply because we want an equivariant Whitehead theorem ? Or is there something deeper to understand here.

From my (very naïve) perspective, it seems like the naïve weak equivalences are very natural to think of, and preserve the most important aspects of $G$-spaces : the homotopy fixed points and the homotopy orbits.

My guess is that there are other types of "equivariant" structures that aren't well-behaved with respect to these equivalences, so it makes sense to require more - and I simply don't know enough equivariant homotopy theory (yet) to know about them. Is that the case ? What would be examples of such structures ?

Perhaps a related (broad) question, which might shed some light is : what are we trying to capture with genuine $G$-spaces (spectra) ?

Maxime Ramzi
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    I really like this question because the people using equivariant homotopy theory in application rarely use genuine $G$-spacess/spectra. Even looking at this years Handbook of Homotopy Theory, many of the cutting edge theories are being developed in the context of naive spaces/spectra. Although genuine equivariant homotopy theory seems like the 'correct' thing to do, I also don't understand what 'correct' is actually meant to mean. – Tyrone Jun 13 '20 at 12:33
  • This doesn't directly address your question, but one reason to care about genuine things is for dualities. Non equivariantly the Spanier-Whitehead dual of a space is obtained by embedding it in a sphere and taking the normal bundle. To do this equivariantly you need to embed in a representation sphere, so then in the stable category you invert representation spheres and not just spheres with a trivial action. – J Cameron Jun 16 '20 at 03:49
  • Another lower brow reason to not want $\mathbb{R} \to $ to be a $\mathbb{Z}$-equivariant homotopy equivalence is that there are no $\mathbb{Z}$-equivariant maps $ \to \mathbb{R}$, so a homotopy inverse won't be equivariant. It's natural to want to set up equivariant homotopy theory just like regular homotopy theory but where everything in sight is equivariant, so you would want an equivariant homotopy equivalence to have an equivariant homotopy inverse. – J Cameron Jun 16 '20 at 03:56
  • @JamesCameron : thanks for your comments - I don't really know a lot about Spanier-Whitehead duality so that's something I'll try to read about; for the second comment it amounts essentially to what I said about the equivariant Whitehead theorem - that's the part of the motivation I understand but it seemed a bit "thin" to me so I was wondering if there were other reasons – Maxime Ramzi Jun 16 '20 at 17:23
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    @MaximeRamzi One nice property of these duals that is nice is that if $X$ is dualizable and $DX$ is the dual of $X$ then there is an equivalence $DX \wedge Y \to Y^X$. This lets you reduce questions about a general map to questions about homotopy groups. Being dualizable is like being finite dimensional (e.g. the dualizable vector spaces are the finite dimensional ones) and to make the suspension spectrum of a finite $G$-CW complex dualizable you need to invert representation spheres. There are similar (but more complicated) issues with Poincare duality. – J Cameron Jun 16 '20 at 19:21
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    One thing that is perhaps good to point out is that the distinction between spaces with a G-action and 'genuine' G-spaces is not the same as the distinction between naive G-spectra and genuine G-spectra (which is what the formulation of the question seems to suggest a little bit.) For spectra, you effectively have three layers: 'doubly naive spectra' (i.e. spectra with a G-action, not sure if anyone calls it like this), 'naive G-spectra' (i.e. the stabilization of G-spaces) and 'genuine G-spectra' (i.e. the stabilization of G-spaces in which representation spheres are inverted.) – Bastiaan Cnossen Feb 20 '21 at 22:58

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