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Victor TC's user avatar
Victor TC's user avatar
Victor TC
  • Member for 7 years, 5 months
  • Last seen more than 2 years ago
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Stable homotopy type of $BG^{\wedge}_p$ in algebraic terms
I think I managed to find a shorter proof under a modern approach (fusion systems, biset functors), details coming soon.
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Stable homotopy type of $BG^{\wedge}_p$ in algebraic terms
@Kuhn I am not aware of the result from representation theory you refer to, could you please let me know?
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Stable homotopy type of $BG^{\wedge}_p$ in algebraic terms
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Stable homotopy type of $BG^{\wedge}_p$ in algebraic terms
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Stable homotopy type of $BG^{\wedge}_p$ in algebraic terms
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Stable homotopy type of $BG^{\wedge}_p$ in algebraic terms
I added a likely, but uncomplete proof
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Stable homotopy type of $BG^{\wedge}_p$ in algebraic terms
@GregoryArone you are referring to the unstable homotopy characterization, I guess. This is the stable one, there is no mistake in their proof so far (Ragnarsson found one, then it was corrected).
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Stable homotopy type of $BG^{\wedge}_p$ in algebraic terms
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Stable homotopy type of $BG^{\wedge}_p$ in algebraic terms
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 Teacher
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$(B\mathbb Z/p^{\infty})^{\wedge}_p\rightarrow (BS^1)^{\wedge}_p$ induced by inclusion is an equivalence
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$p$-completeness of the function spectrum $F(\Sigma^{\infty} BS, \Sigma^{\infty} BK)$
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$p$-completeness of the function spectrum $F(\Sigma^{\infty} BS, \Sigma^{\infty} BK)$
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$p$-completeness of the function spectrum $F(\Sigma^{\infty} BS, \Sigma^{\infty} BK)$
@Drew Yes, I did but found no mention of that result so far. I might have to take a closer look at that article.
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 Yearling
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$p$-completeness of the function spectrum $F(\Sigma^{\infty} BS, \Sigma^{\infty} BK)$
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$(B\mathbb Z/p^{\infty})^{\wedge}_p\rightarrow (BS^1)^{\wedge}_p$ induced by inclusion is an equivalence
Fine!, I was wrong, sorry. Thank you again.
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$(B\mathbb Z/p^{\infty})^{\wedge}_p\rightarrow (BS^1)^{\wedge}_p$ induced by inclusion is an equivalence
but $B\mathbb{R}$ is contractible, and $B\mathbb{Q}$ is $p$-good. So far, their $p$-completions cannot be homotopy equivalent.
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$(B\mathbb Z/p^{\infty})^{\wedge}_p\rightarrow (BS^1)^{\wedge}_p$ induced by inclusion is an equivalence
Thank you!, do you mean $B\mathbb{Q}^{\wedge}_p\rightarrow B\mathbb{R}^{\wedge}_p$ is a homotopy equivalence?
asked
$(B\mathbb Z/p^{\infty})^{\wedge}_p\rightarrow (BS^1)^{\wedge}_p$ induced by inclusion is an equivalence
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