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S Jul 2, 2023 at 6:59 history suggested Jonas Linssen CC BY-SA 4.0
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S Jul 2, 2023 at 6:59
Nov 7, 2010 at 11:12 vote accept Ma Ming
Oct 19, 2010 at 18:45 comment added Johannes Ebert @Charles Rezk: Maybe that would open an expert's thread, but do you know of written account where the infinite-loop space viewpoint is explicitly combined with the Bott-periodicity? As far as I can see, this amounts to writing down all spaces of the Bott spectrum as $E_\infty$- or $\Gamma$-spaces, plus the Bott maps as maps preserving this structure.
Oct 19, 2010 at 18:40 comment added Johannes Ebert "though you'll certainly need periodicity to figure out what k ∗ and K ∗ look like" - that is what I mean
Oct 19, 2010 at 18:38 history edited Johannes Ebert CC BY-SA 2.5
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Oct 19, 2010 at 18:31 comment added Charles Rezk I'd disagree with the claim "without Bott periodicity, K-theory is not very interesting". There is a cohomology theory called "connective K-theory" $k^*(X)$, obtained merely using the fact that $Z\times BU$ is an $E_\infty$-space, as in Justin's answer. The groups $k^q(X)$ vanish when $X$ is a space and $q>0$. You can obtain $K^*(X)$ from $k^*(X)$ by inverting the "Bott element" $b\in k^{-2}(*)$, but $K^*$ does not determine $k^*$. You shouldn't need Bott periodicity to carry out these constructions (though you'll certainly need periodicity to figure out what $k^*$ and $K^*$ look like.)
Oct 19, 2010 at 13:07 comment added Ma Ming Aha! U(n) is homotopy-Abelian in U(2n) (See IOAN JAMES AND EMERY THOMAS "Which Lie Groups are Homotopy-Abelian"), therefore we may say U(\infty) is a homotopy-Abelian group (E_\infty space??). On the other hand, we know that Abelian groups are infinite loop space. Any chance to make this intuition be a concrete proof?
Oct 19, 2010 at 12:32 comment added Ma Ming Without Bott periodicity, any clue to BU can be delooped infinity times?
Oct 19, 2010 at 8:31 history answered Johannes Ebert CC BY-SA 2.5