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There are at least two fixed point functors that characterize stable equivalences of orthogonal G-spectra: the geometric fixed points and the naive fixed points of a fibrant replacement. Is this true also for fibrations? Precisely, is it true that a map of orthogonal G-spectra is a fibration if and only if its (geometric) fixed points is a fibration of orthogonal spectra for every subgroup of G?

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Nice question. I think the answer is no, but I don't have time to think seriously. The term ``naive'' for the categorical or genuine fixed points (they often go under another name) is not a good one. –  Peter May Jun 21 '13 at 11:40
    
If it was true, it might give a chance to define the model structure on orthogonal G-spectra as a transport construction from the model structure on orthogonal spectra. I think this could be quite convenient. –  Emanuele Dotto Jun 21 '13 at 13:22
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Are you aware of Marc Stephan's master's thesis, titled Elmendorf's Theorem in General Model Categories? It gives a method of transporting a model structure which might do what you're looking for without need to look deeply at the fibrations. Of course, there's still work to be done. You have to prove that orthogonal spectra have cellular fixed point functors (defined in the thesis) but that seems likely to be true, at least to me. No idea about the fibrations, though. –  David White Jun 21 '13 at 14:32
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Yeah, I know his work! I was thinking precisely of a variation of his argument. The fixed point he uses are defined as a limit, and this gives the wrong notion for G-spectra (they are just the levelwise fixed points). But the geometric fixed points functors give a family of adjunctions that might define a model structure. For it to be the right one I need the question I asked to be true :) –  Emanuele Dotto Jun 21 '13 at 14:38
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