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kiran
  • Member for 4 years, 4 months
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Can a phantom map have finite cofiber?
I was thinking: filter X as a colimit of finite spectra X_n such that the first stage is F. Then all other stages have their identity map factoring through F so theyre summands of F, so X is also.
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Can a phantom map have finite cofiber?
You could also argue via the fiber, which would be a finite F-->X through which all maps finite-->X factor. I think that forces X to be finite already?
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If homotopy groups of spaces are identical, then stable ones are also identical?
Also, your example pair is a counter-example, since those two spaces don't have the same rational stable homotopy groups (=rational homology)
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Why did Ravenel define a ring spectrum to be flat if its smash-square splits into copies of itself?
Why does (3) require highly-structured associativity? That definition appears in Adams blue book from the 70s which, it was my understanding, doesn't deal with any high structure
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Why did Ravenel define a ring spectrum to be flat if its smash-square splits into copies of itself?
Is this equivalent to definition (3)? One direction seems to follow from pi_* commuting with filtered colims. For the other direction if D is the filtered diagram of free E_-modules whose colimit is E_*E, then E_*( )(x)_{E_}D is a diagram of homology theories which I think lifts to a diagram of spectra (the nodes are free E-modules) whose colimit has a map to E(x)E by universality, and by another pi_*-commuting with filtered colims that map is an equivalence I think.
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Does every complex orientable $E_\infty$-ring admit an $E_\infty$ complex orientation?
@JeremyHahn I hope not, because it's exactly such an E_infty map I seem to have constructed!
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Does every complex orientable $E_\infty$-ring admit an $E_\infty$ complex orientation?
@FernandoMuro 10 bucks? Deal! On a more serious note, can you expand? (I think I have a proof of the positive answer, very much need to check the details)
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