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Aug 12, 2019 at 10:19 history edited YCor
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Nov 3, 2016 at 22:38 comment added David Roberts Regarding the map you wish to understand, I think that the KSp should be seen as coming from one specific KR. Btw web.archive.org/web/20230113145507/https://www.math.umd.edu/… should be rather useful, but I haven't been able to figure out the dimension correctly from there. Rosenberg uses a different grading to Atiyah.
Nov 3, 2016 at 21:55 comment added Willie Wong For your final question, the section titled "Ambient Space Construction" on this blog entry of mine is related. Basically it is a way to fix one conformal structure on $\mathbb{S}^6$.
Nov 3, 2016 at 19:59 history edited David C CC BY-SA 3.0
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Nov 1, 2016 at 21:28 comment added Mingcong Zeng @RyanBudney I suspect that the sentence "integers goes to 0" means integrable ACS goes to 0 in KR, while non-integrable ones goes to 1. It seems that he was using this map to separate integrable and non-integrable ACSs, it cannot be done if it is trivial. I feel that this might be the most crucial step, but do not have a clear idea what's going on...
Nov 1, 2016 at 8:19 comment added Tyrone Karoubi briefly treats KR-theory in "K-Theory: An Introduction" (page 177). The book is a good reference for K-Theory in general.
Nov 1, 2016 at 2:56 comment added Igor Belegradek I think basics of $KR$ theory is easy to learn from [1]. What is puzzling is the claim on p.5 of arxiv.org/abs/1610.09366 that "Because of the Atiyah-Singer theory, the linear algebra acquires a topological meaning, and that is embodied in $KR$ theory."
Nov 1, 2016 at 1:42 comment added David Roberts And the involution on the base space would be trivial.
Nov 1, 2016 at 1:41 comment added mme @DenisNardin You go from $E$ to $E \oplus \overline{E}$; your involution is the obvious one that swaps factors.
Nov 1, 2016 at 0:20 comment added Denis Nardin @RyanBudney Can you explain a little bit more how to treat a complex bundle as a real bundle? You need to throw in a C_2-action and I really don't see how to do that (this is a bit embarassing because I should know KR very well).
Oct 31, 2016 at 21:22 comment added Ryan Budney I think it's just a straight-forward computation, isn't it? I've only scanned the paper and this is my first time thinking about $KR$, but I suspect what's going on is that the generator for complex k-theory is the bundle with Euler class $1$. In $KR$ theory this is twice the generator, because the generator is one where the fixed-point set of the involution is a Moebius bundle. I haven't thought about this in detail but that's what I suspect is going on.
Oct 31, 2016 at 21:08 comment added David C Yes of course, but do you know why it is trivial?
Oct 31, 2016 at 21:07 comment added Ryan Budney The integers he's talking about is 6-dimensional complex k-theory of $S^6$, i.e. this group is infinite cyclic.
Oct 31, 2016 at 21:02 comment added David C This sentence is on p4 of Atiyah's preprint.
Oct 31, 2016 at 20:45 comment added Ryan Budney The forgetful map appears to be that you can treat a complex bundle as a real bundle, that gives your map from complex k-theory to this "KR" variant. Where is the sentence you are quoting?
Oct 31, 2016 at 20:43 history edited YCor
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Oct 31, 2016 at 20:22 comment added David C I am reading [1], it is beautifully written. However as I am a complete amateur, I would like to understand the relationships between topological complex K-theory and KR-theory. In particular this sentence: "There are natural forgetful maps from complex K-theory to KR-theory and in dimension 6 the integers go to 0".
Oct 31, 2016 at 20:05 comment added Ryan Budney It looks like you implicitly do not accept Atiyah's reference [1] in his paper. What isn't "nice" about that?
Oct 31, 2016 at 20:01 history asked David C CC BY-SA 3.0