# subgroups with the same number of roots that the group.

When thinking in terms of Dynkin diagrams, I am naively used to see that the diagram for a subgroup can be extracted from the diagram of the group by removing some roots. Now, I noticed that for SO(10) and its subgroup SU(4)xSU(2)xSU(2), no such removal happens, the Dynkin diagrams have the same number of nodes.

How usual is this? For which cases we can find subgroups with the same number of roots (the same number of nodes in the Dynkin diagram, I mean) that the main group?

For the aforementionated case, if one tries to remove a further root then the inclusion relationship does not hold anymore: from SO(10) we got to SO(8), but from 4-2-2 we go either to SU(3)xSU(2)xSU(2) or to SU(4)xSU(2), and neither of them are subgroups of SO(8).

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Removing nodes from the Dynkin-Diagram gives you the parabolic subgroups but nothing more. –  Johannes Hahn Sep 30 '11 at 10:46
What you are seeing here is the following fact: one can obtain maximal-rank subgroups by taking the extended Dynkin diagram and then removing a node. This gives subgroups of the same rank. The case of SU(4)xSU(2)xSU(2) arises in this way, for example. –  Chuck Hague Sep 30 '11 at 15:26
@Chuck Nice! I had not read of this technique; the fact that SU(4)xSU(2)xSU(2) seems very much as a removal of edges from SO(10) is then a red herring, isn't it? –  arivero Sep 30 '11 at 22:31
In connection with Chuck Hague's remark: this technique was formalized in the paper of Borel and de Siebenthal in the Commentarii Mathematici Helvetici, 1949. –  Joseph Wolf Oct 16 '11 at 19:39

Removing a random edge and keeping the nodes do not give you a subgroup. The subgroups you mention are obtained by adding one node (and a few edges) and then by removing an inner node. Perhaps you should see Borel-Siebenthal's paper on `maximal subgroups of maximal rank in compact Lie groups' for more on this process.

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I'm not sure which modern treatment is most helpful, but anyway the original 1949 paper seems to be available online: retro.seals.ch/digbib/view?rid=comahe-002:1949:23::16. There is further discussion in a previous MO question: mathoverflow.net/questions/28878 –  Jim Humphreys Oct 1 '11 at 15:32
As noted above, it's worth mentioning that the adding of nodes & edges comes from the extended Dynkin diagram. –  Chuck Hague Oct 6 '11 at 17:17