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I am trying to find the embedding and the branching rules for some group decompositions. For example, I consider $E_7$ and its maximally compact subgroup $SU(8)$ and I want to "see" how the Dynkin diagram of $E_7$ is modified to get $A_7$. I tried to take a linear combination of roots to produce $A_7$. And also following other analog questions, with this I have (almost) no issues.

The problem is for an embedding like $SU(8)⊃SU(6)×U(1)$. How do I get this? As all of you can imagine, my doubts arise when the embedding involves abelian subgroups, that I am not able to see from Dynkin diagrams.

All these issues comes from the calculus of centraliser groups. For instance, I want the centraliser group $\mathcal{C}_{E_7}(SU(3))$ of $SU(3)_{\mathrm{diag}} \subset SU(3) \times SU(3) \subset SU(6) \subset SU(8) \subset E_7$. I know it is $$ \mathcal{C}_{E_7}(SU(3)) = SU(3) \times SU(2) $$ but I do not have a procedure to get something involving $U(1)$ factors.

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    $\begingroup$ What you are looking for originates in the older work of Borel and deSiebenthal: gdz.sub.uni-goettingen.de/dms/load/img/… (which figures in some previous questions here that you can find by searching for "Siebenthal"). The main point is to look for subdiagrams of the extended Dynkin diagram, as Andre indicates in a relevant example. Here you get "pseudo-Levi subgroups", which are not necessarily actual Levi factors in parabolic subgroups of the given group. $\endgroup$
    – Jim Humphreys
    Commented Jun 17, 2015 at 13:24
  • $\begingroup$ P.S. Maybe I should add that the pseudo-Levi subgroups of maximal rank (say in a compact Lie group) provide the needed centralizers of semisimple elements (which applies to all elements of a compact Lie group). Strictly speaking, in some situations these centralizers may fail to be connected, but for example not in the simply connected groups. $\endgroup$
    – Jim Humphreys
    Commented Jun 17, 2015 at 14:07
  • $\begingroup$ Concerning the proposed embedding in your second paragraph, why do you think this should exist? $\endgroup$
    – Jim Humphreys
    Commented Jun 17, 2015 at 15:55
  • $\begingroup$ Do you mean $SU(6) \subset SU(8)$ or $SU(3) \times SU(3)\subset SU(6)$? $\endgroup$
    – Oscar
    Commented Jun 17, 2015 at 19:29
  • $\begingroup$ Yes, this is where it all gets confusing for me. I'm still uncertain where the embedding comes from in your second paragraph. $\endgroup$
    – Jim Humphreys
    Commented Jun 17, 2015 at 22:37

1 Answer 1

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"I want to see how the Dynkin diagram of E7 is modified to get A7":

Take $E_7$:

$\bullet -\bullet -\bullet -\stackrel{\stackrel{\textstyle\bullet}{\textstyle |}}{\bullet} -\bullet -\bullet$

Add the lowest negative root (these roots are no longer linearly independent):

$\bullet -\bullet -\bullet -\stackrel{\stackrel{\textstyle\bullet}{\textstyle |}}{\bullet} -\bullet -\bullet -\bullet$

Remove the top root to get another basis of $\mathfrak t^*$:

$\bullet -\bullet -\bullet -\bullet -\bullet -\bullet -\bullet$

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  • $\begingroup$ Thanks a lot. I connected the top root to one of the extremal, and then I "cutted" the line between the top root and the 3rd one. As in mathoverflow.net/questions/25371/chopping-up-dynkin-diagrams Thank you very much, however, this does not quite answer my question. I am interested in the $U(1)$ sugroups, or in general in a method to compute centraliser groups. $\endgroup$
    – Oscar
    Commented Jun 17, 2015 at 13:12
  • $\begingroup$ The "remove the top root" step amounts to "take the centralizer in $E_7$ of a certain element of adjoint order $2$". Look around MO for Borel-de Siebenthal theory, which is about subgroups of the same rank as the big group. $\endgroup$
    – Allen Knutson
    Commented Jun 17, 2015 at 13:37

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