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What is Borel-de Siebenthal theory?

Roman Fedorov
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    Whomever gave this a negative vote should state their objection (e.g., no context?). But I know this one. It's about studying the structure of identity components of centralizers of (possibly non-maximal, disconnected) mult.-type subgroups of a connected reductive group. It is described in terms of removing certain vertices/edges from the Dynkin diagram and inserting others. The original work of Borel and de Siebenthal was done for compact groups, but presumably nowadays the phrase refers to general (split?) connected reductive groups. Look at http://www.math.ens.fr/~gille/prenotes/bds.pdf – BCnrd Jun 20 '10 at 19:59
  • Unlike BCnrd, MO is not an encyclopedia. It is expected that users will do some research and provide context before asking their questions. See the FAQ for guidelines. (PS: I'm not the one who voted down.) – François G. Dorais Jun 20 '10 at 20:16
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    minus $n$ to whoever voted this down. "What is theory X" is a fine question. – T.. Jun 20 '10 at 20:16
  • Anyone can come onto MO and ask "what is X" about essentially any X. MO is not an encyclopedia; the first thing you should do when you have a question of the form "what is X" is to look it up in whatever sources you can. If you can't find an answer there, you should at least state the question in the form of asking for references, rather than asking for someone to write a Wikipedia article for you. – Qiaochu Yuan Jun 20 '10 at 20:17
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    I don't see anything in a quick web search for "Borel de Siebenthal theory" that is a sufficient self-help answer to the question in under 10 minutes (assuming reasonable background knowledge, as the questioner clearly possesses). – T.. Jun 20 '10 at 20:20
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    @T: Presumably the poster encountered the term somewhere before asking the question; this is often sufficient context. – François G. Dorais Jun 20 '10 at 20:24
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    The question is not optimally formulated, but Fedorov is a reputable young mathematician formerly a postdoc in my department. – Jim Humphreys Jun 20 '10 at 20:43
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    As one click of a mouse can confirm for those who doubt it, the questioner is an expert asking an expert-level question. His question is not equivalent to asking "what is real analysis" or "please write an encyclopedia article at my command", and it is absurd to treat it as such. Most of the questions on MO, even the excellent ones, would be eliminated or shrunken by a requirement to first perform extensive self-help, and even if such legwork were always feasible, it would degrade the discussions because questioners rarely know what they SHOULD be asking about, for lack of knowledge. – T.. Jun 20 '10 at 20:54
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    Also, +1 for BConrad's answer: amazing that nobody upvoted that (so far), but some of the catty meta-comments already got ++ed. Some users apparently are coming here for the non-math... – T.. Jun 20 '10 at 21:08
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    @T, Jim: I am not judging the questioner, nor even the content of the question. Neither am I suggesting that MO is a last resort. The only quibble is the form of the question and the usefulness to the community. The question I asked myself when making the above comment is: Why would someone other than the poster want to know the answer to this question? – François G. Dorais Jun 20 '10 at 21:13
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    @T, actually, this may be a 'fine' question in the sense of being acceptable, but it is certainly not in the sense of being a good question. There's a pretty well established consensus that a good question provides some context and motivation: see mathoverflow.net/howtoask. I'd encourage any further discussion of the suitability of this question to continue on meta. Please post a link if you start such a thread. – Scott Morrison Jun 20 '10 at 21:13
  • I'm giving this -1 as well. I don't think "what is X" questions are useful, for reasons given by Qiaochu. I'm nt criticizing the poster in any way- but this is not a good question in my opinion. – Daniel Moskovich Jun 20 '10 at 23:39
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    Actually, I prefer expert questions stated concisely, rather than rambling discussion of not always relevant motivations, $\textit{precisely}$ because MO is an answer resource not an encyclopaedia. – Victor Protsak Jun 20 '10 at 23:46
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    Thank you very much for all the mathematical responses, and let me apoologize to MO community for misusing the resource. I certainly did not ask for a wikipedia article (though Jim's answer is close to such an article). I did not ask for references because I just wanted to get a rough idea. I tried to google the answer before I asked. I did not think how to make the question valuable to other people though I assumed that those who are interested in Representation theory would read. – Roman Fedorov Jun 21 '10 at 02:24

2 Answers2

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In my head at least, part of it is this...

Let G be reductive. Consider the following algorithm:

  1. Extend one component of the Dynkin diagram to its affine diagram, by attaching the lowest root.

  2. From that component, remove one or more vertices. (If you want to stay semisimple, remove only one.)

  3. Repeat to taste. (If you want to stay semisimple, and you're at a disjoint union of $A_m$ diagrams, then you're definitely done.)

The result is the Dynkin diagram of a subgroup H of G of the same rank. Every such subgroup (up to finite factors) arises this way.

Allen Knutson
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  • I should have emphasized that the extended/affine diagrams and vertex removal is basic to the Borel-de Siebenthal classification of maximal subgroups, as well as some related developments: classification of maximal subgroups of semisimple algebraic groups or finite groups of Lie type in prime characteristic (Liebeck-Seitz, expanding on Dynkin's 1950s papers); centralizers of semisimple elements and pseudo-Levi subgroups or subalgebras (Carter-Deriziotis, Sommers); probably some of the geometry of affine Kazhdan-Lusztig cells, related in Lusztig's bijection to unipotent classes of dual groups.. – Jim Humphreys Jun 23 '10 at 15:51
  • P.S. I don't follow the parenthetic remarks in Allen's numbered steps, about staying semisimple. (And change "is" to "are" in my first line. Though I ran out of characters.) – Jim Humphreys Jun 23 '10 at 15:56
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    Steps 1+2 together replaces a root system by a sub root system. (Grr. I don't want to call it a sub-root system or a root subsystem. Anyway...) If you remove more than one vertex at step 2, the subsystem will be of lower rank, corresponding to a Levi subgroup of what you'd get from removing just one. E.g., if you start with G semisimple and remove all the vertices at step 2, the result is the *empty) Dynkin diagram of a maximal torus of G.

    As for the other remark, if you affinize and remove 1 vertex from an A_m diagram, you've done nothing. You're stuck; no more maximal rank s.s. subgroups.

     – Allen Knutson Jun 24 '10 at 01:48
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    Expanding a little on Allen's algorithm, if you remove a single node with prime label, or two nodes with label 1, from the extended Dynkin diagram $\tilde D$, then the resulting subgroup H is maximal. Furthermore (modulo the action of $Z_{sc}$ on the extended Dynkin diagram) this classifies the maximal, reductive subgroups of the same rank as G up to conjugacy.

    Then you get every reductive, equal rank subgroup by iterating. However I do not know any way to keep track of conjugacy classes of the resulting subgroups.

     – Jeffrey Adams Mar 23 '11 at 13:12
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I'm not sure the term "theory" is appropriate here, but the joint paper by Borel and de Siebenthal has had considerable influence in Lie theory over the years: MR0032659 (11,326d) Borel, A.; De Siebenthal, J., Les sous-groupes fermés de rang maximum des groupes de Lie clos. Comment. Math. Helv. 23, (1949). 200--221. (There was a short Comptes Rendus announcement in 1948.) This is found near the start of the Springer four-volume collected papers of Borel, with a couple of minor corrections appended.

To quote from the review by P.A. Smith: "Let $G$ be a compact Lie group, $G'$ a closed connected subgroup having the same rank as $G$. Let $Z'$ be the center of $G'$. The main object of this paper is to show that $G'$ is a connected component of the normalizator of $Z'$ in $G$." The proof involves
"a necessary and sufficient condition that a subsystem of root vectors of $G$ be the root vectors of a closed subgroup of $G$" and the subgroups of this type are found explicitly for all simple $G$. [Here $G$ is always assumed to be connected.]

The result on subsystems of root systems carries over in a natural way to the study of semisimple complex Lie (or algebraic) groups and their Lie algebras, for example the determination of subalgebras of maximal rank in the latter.

Jim Humphreys
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