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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeJan 31st 2012
    • (edited Jan 31st 2012)

    I have tried to start a table at maximal compact subgroup listing Lie groups and their max compact subgroups. But once again the table does not want to typeset properly.

    Have to run now, will try to fiddle with this later.

    • CommentRowNumber2.
    • CommentAuthorTodd_Trimble
    • CommentTimeJan 31st 2012

    I thought the existence and uniqueness up to conjugation of maximal compact subgroups was a theorem about finite-dimensional Lie groups, so I put that in.

    • CommentRowNumber3.
    • CommentAuthorUrs
    • CommentTimeJan 31st 2012

    Thanks, yes.

    (By “Lie group” I usually mean the finite dimensional case. Otherwise I’d say “Fréchet Lie group” or the like. )

    • CommentRowNumber4.
    • CommentAuthorTodd_Trimble
    • CommentTimeJan 31st 2012

    Added a comment about Fréchet Lie group to Lie group, and also added the crucial connectedness hypothesis to the statement of existence and uniqueness of maximal compact subgroups.

    Apropos of that, I also started Prüfer group, and added an example to counterexamples in algebra.

    • CommentRowNumber5.
    • CommentAuthorUrs
    • CommentTimeJan 31st 2012

    Thanks again!

    • CommentRowNumber6.
    • CommentAuthorUrs
    • CommentTimeJan 31st 2012
    • (edited Jan 31st 2012)

    Ah, stupid me. I finally realized why the tables wouldn’t do what I want. More often than not I would type

    a | b
    ------
    c | d 
    e | f 
    

    Instead of

    a | b
    --|---
    c | d 
    e | f 
    

    It’s only the second syntax that is recognized as a table. (This was clear to you all. I am just saying it for my own sake :-/ )

    • CommentRowNumber7.
    • CommentAuthorUrs
    • CommentTimeJan 31st 2012
    • (edited Jan 31st 2012)

    I now had some time to work on the entry.

    I have added a statement of the Malcev-Iwasawa theorem, and of a recent refinement by Antonyan.

    So “almost connectedness” is sufficient for the existence of a maximal compact subgroup.

    Todd, I have moved your counterexample of the Prüfer group to section “Examples”, subsection “Counterexamples”. Okay?

    • CommentRowNumber8.
    • CommentAuthorTodd_Trimble
    • CommentTimeJan 31st 2012

    More than okay, Urs – I really like what you’ve done! Nice work.

    I have half a mind to write up something on Hilbert’s fifth problem, which we don’t have an article on.

    • CommentRowNumber9.
    • CommentAuthorUrs
    • CommentTimeJan 31st 2012

    Thanks, Todd.

    I have further edited (corrected and expanded) the table of Lie groups and their max compact subgroups. I would like it to be more extensive, still. But I’l have to call it quits for tonight.

    • CommentRowNumber10.
    • CommentAuthorUrs
    • CommentTimeFeb 1st 2012

    added here a table with the maximal compact subgroups of the real forms of the exceptional EE-series of Lie groups.

    • CommentRowNumber11.
    • CommentAuthorTodd_Trimble
    • CommentTimeMar 30th 2019

    It looks like there’s a slight mistake in the row with SO(p,q)SO(p, q): it needs to be {T,U)O(p)×O(q)|det(T)=det(U)}\{T, U) \in O(p) \times O(q)|\;\det(T) = \det(U)\}. The group SO(p)×SO(q)SO(p) \times SO(q) is of index 22 in this.

    • CommentRowNumber12.
    • CommentAuthorUrs
    • CommentTimeMar 31st 2019

    Thanks for catching! Fixed now.

    • CommentRowNumber13.
    • CommentAuthorUrs
    • CommentTimeMar 31st 2019

    Thanks! Have added the reference now.

    diff, v20, current

    • CommentRowNumber14.
    • CommentAuthorUrs
    • CommentTimeMar 31st 2019
    • (edited Mar 31st 2019)

    Is the line after that correct, the one with

    MaxCompactSubgroup(Spin(q,p))=(Spin(q)×Spin(p))/ 2 MaxCompactSubgroup\big(\; Spin(q,p) \;\big) \;=\; \big( Spin(q) \times Spin(p) \big)/\mathbb{Z}_2

    ?

    • CommentRowNumber15.
    • CommentAuthorTodd_Trimble
    • CommentTimeMar 31st 2019

    Re #14: it looks like here too, you’ve written down the maximal compact connected subgroup. So the answer might be something like S(Pin(q)×Pin(p))/ 2S(Pin(q) \times Pin(p))/\mathbb{Z}_2.

    • CommentRowNumber16.
    • CommentAuthorUrs
    • CommentTimeMar 31st 2019

    Yeah, thanks. I have changed the table to speak about Pin(p,q)Pin(p,q) for the moment. Should try to sort of the Spin-case…

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