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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Dec 24, 2015 at 16:01 comment added Jiangnan Yu Your name is so interesting :D
Nov 19, 2015 at 14:26 answer added Robert Bruner timeline score: 4
Oct 12, 2015 at 17:04 history reopened Robert Bryant
Joonas Ilmavirta
Stefan Kohl
Andrey Rekalo
darij grinberg
Oct 12, 2015 at 3:02 review Reopen votes
Oct 12, 2015 at 17:04
Oct 6, 2015 at 12:50 comment added Gil Kalai Regardless of this particular question, in my opinion, "standard characteristic class stuff" like "standard applications of spectral sequences" like "standard facts about Floer homology" etc. are research-level mathematics as far as mathoverflow is concerned.
Oct 6, 2015 at 11:55 review Reopen votes
Oct 6, 2015 at 14:31
Oct 6, 2015 at 11:41 comment added Robert Bryant It it probably worth pointing out that there is no nontrivial normal subgroup $\mathrm{SSp}(\mathbb{H},n)\subset \mathrm{Sp}(n)$ (other than the center, which is discrete) because the group $\mathrm{Sp}(n)$ is simple. Thus, there is no interpretation of $p_1(\xi)=0$ as a condition for reducing the structure group of a quaternionic vector bundle $\xi$ to such a subgroup.
Oct 6, 2015 at 11:41 history edited Gil Kalai CC BY-SA 3.0
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Oct 6, 2015 at 10:41 vote accept QSR
Oct 6, 2015 at 9:20 history closed Allen Knutson
Qiaochu Yuan
David Roberts
Stefan Kohl
Marco Golla
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Oct 6, 2015 at 6:29 comment added QSR Sorry, Prof. Yes, this is what I want to express.
Oct 6, 2015 at 6:29 history edited QSR CC BY-SA 3.0
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Oct 6, 2015 at 6:27 comment added Sebastian Goette Could you please explain your definition $SU(\mathbb H,n)$? The notation is nonstandard. I did not find it on the wikipedia page in the link. And to relate to Sullivan's answer: quaternion vector bundle means vector bundle with an $\mathbb H$ action by scalars, doesn't it? So the group $Sp(n)\hat\times Sp(1)$ will not appear in this context.
Oct 6, 2015 at 5:52 history edited QSR CC BY-SA 3.0
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Oct 6, 2015 at 4:24 history edited QSR CC BY-SA 3.0
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Oct 6, 2015 at 4:18 comment added mme Crossposted from MSE.
Oct 6, 2015 at 4:01 answer added Dennis Sullivan timeline score: 9
Oct 6, 2015 at 2:19 review Close votes
Oct 6, 2015 at 9:20
Oct 6, 2015 at 2:06 comment added Allen Knutson "the structure group" isn't well-defined, unless you put a metric on it and ask about the holonomy group, in which case the answers are "false". You can ask whether the structure group can be reduced to $SU(n,\mathbb C$ or $\mathbb H)$, in which case the answers are "true". This is standard characteristic class stuff, as in Milnor's book or Husemoller's, not research-level.
Oct 6, 2015 at 1:00 history asked QSR CC BY-SA 3.0