Timeline for A reductive group is the complexification of a compact subgroup even if not connected?
Current License: CC BY-SA 4.0
28 events
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| Sep 9 at 20:35 | comment | added | brunoh | @WillSawin You are absolutely right. The key part of your and YCor answer was a consequence of Mostow theorem. I was just talking about a small step in your development. | |
| Sep 9 at 20:31 | comment | added | brunoh | @WillSawin so my comment was that to check in your answer the fact that every smooth representation of $K$ extend to an algebraic one of $G$ is done in this book. But I like your direct and to the point approach (even if the algebraicity of the extension was not obvious for me). In any case I accepted your answer and gave you the bounty. | |
| Sep 9 at 20:25 | comment | added | Will Sawin | @brunoh So then I guess it doesn't handle the main part of YCor's and my answer, which is the existence of a compact group whose complexification (in the sense of Tannaka-Krein duality) is a given reductive group. | |
| Sep 9 at 20:21 | comment | added | brunoh | @WillSawin Actually, the result of III.8 of this book is the building of the complexification of a compact group, mainly using Tannaka-Krein duality, and proving that it is a reductive complex algebraic group, with every smooth representation of the compact one extending to an algebraic one of its complexifciation, without apparently any connectedness hypothesis on the compact group. | |
| Sep 9 at 20:12 | history | edited | brunoh | CC BY-SA 4.0 |
Typo in reference
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| S Sep 9 at 20:06 | history | bounty ended | brunoh | ||
| S Sep 9 at 20:06 | history | notice removed | brunoh | ||
| Sep 9 at 6:22 | vote | accept | brunoh | ||
| Sep 9 at 0:36 | answer | added | Will Sawin | timeline score: 4 | |
| Sep 9 at 0:23 | comment | added | Will Sawin | @brunoh I didn't check the reference to that book, but you said it answers your third question, which I interpreted as a statement of the form "For each compact real Lie group there exists a complex reductive group such that ..." and not "For each reductive linear algebraic group there exists a compact group such that...." which YCor's reference was about. What's the result in the book say exactly? | |
| Sep 8 at 19:42 | comment | added | YCor | @WillSawin I'm happy if you post an answer. | |
| Sep 8 at 18:44 | comment | added | brunoh | @WillSawin simple and clear, but I thought the result of chapter III.6 of the book of Tammi Tom Dieck was sufficient to demonstrate existence. Am I mistaken? In any case you or YCor should give an answer to my question. If you don’t a catastrophic and frustrating event will happen: the hard earned bounty points will disappear into nothingness. | |
| Sep 8 at 17:53 | comment | added | Will Sawin | @YCor To extend a representation of a disconnected reductive group $G$ from the maximal compact $K$, first extend from the identity component of $K$ the identity component of $G$, then induce to $G$. Restricting to $K$ you get an induction of restriction which has the original representation as a summand, and Hom-spaces over $K$ and $G$ are equal so you can pick out the original representation as a summand over $G$. | |
| Sep 7 at 17:41 | comment | added | brunoh | @YCor A complex reductive linear algebraic group has a finite number of irreducible components. The result in the book of Tammo tom Dieck mentioned in my question show that the existence of extension of reps is clear. So is it sufficient to answer my question? If you think so, would you mind transforming your comment into an answer so that I can give you the bounty? | |
| Sep 7 at 12:18 | comment | added | YCor | If $G$ is any (real) Lie group with finitely many components, then it is still true (Mostow 1955) that compact maximal subgroups meet all connected components and are all conjugate. In case $G$ is complex reductive and $K$ is maximal compact, then it might hold that $G$ is the complexification of $K$ (in the sense that complex reps of $K$ uniquely extend to $G$); uniqueness of extensions of reps is clear from Mostow's result plus the connected case, but I'm not sure about existence. | |
| Sep 7 at 11:54 | history | edited | brunoh | CC BY-SA 4.0 |
Better formatting
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| S Sep 2 at 16:42 | history | bounty started | brunoh | ||
| S Sep 2 at 16:42 | history | notice added | brunoh | Draw attention | |
| Aug 30 at 21:07 | history | edited | brunoh | CC BY-SA 4.0 |
Partial answer
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| Aug 30 at 20:37 | history | edited | brunoh | CC BY-SA 4.0 |
Some precisions and change of title
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| Aug 28 at 13:18 | history | edited | brunoh | CC BY-SA 4.0 |
added 130 characters in body
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| Aug 28 at 13:15 | comment | added | brunoh | @WillSawin Once again you are right! sorry for my mistake. Will edit the two questions with the link. | |
| Aug 28 at 13:11 | comment | added | Will Sawin | Please wait a few days between reposting between MathSE and MO to give people time to answer on the first site, and please also link between the two questions so people can easily check if one is answered before answering the other. | |
| Aug 28 at 12:39 | history | edited | brunoh | CC BY-SA 4.0 |
Correcting a mistake in the question
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| Aug 28 at 12:38 | comment | added | brunoh | @WillSawin Oopss ... you are right I edited my question! | |
| Aug 28 at 12:30 | comment | added | Will Sawin | There is no equivalence of categories between compact real Lie groups and complex linear algebraic reductive groups. Just consider automorphism groups on each side. The equivalence of categories is only between the categories of representations. | |
| Aug 28 at 11:43 | history | edited | brunoh | CC BY-SA 4.0 |
Link to MathSE
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| Aug 28 at 11:01 | history | asked | brunoh | CC BY-SA 4.0 |