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What are the proper compact connected subgroups of $Spin(n)$ of maximal rank where $Spin(n)$ is the spin group, that is, the universal cover of the special orthogonal group $SO(n)$?

In fact, I am only interested in the highest dimension of a compact connected subgroup of $Spin(n)$ of maximal rank. I am not sure if this is an easier question.

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What's wrong with a maximal torus? That satisfies all of your conditions and is certainly the one of minimum dimension. – Robert Bryant Apr 11 '12 at 15:15
Yes, it is an easier question. I assume that by $Spin(n)$ you mean the compact group $G=Spin(n)$ over $\mathbf{R}$. Then a connected subgroup of maximal rank (i.e. containing a maximal torus) of minimal dimension is a maximal torus. Its dimension is $rk(G)$, equal to $n/2$ or $(n-1)/2$ depending on whether $n$ is even or odd. The "next lowest" dimension of a connected subgroup is $rk(G)+2$. – Mikhail Borovoi Apr 11 '12 at 15:16
Sorry, for the confusion. I made a mistake which I edited. In fact, I am looking for the subgroup of highest dimension which satisifies these properties. – berl13 Apr 11 '12 at 15:20
up vote 7 down vote accepted

I think that the answer here is just the double cover of the obvious answer for $SO(n)$, which is $U(n/2)$ when $n$ is even and $SO(n{-}1)$ when $n$ is odd. You can double-check this by consulting the Dynkin tables of maximal subgroups.

Added after Mikhail's comment: Mikhail actually went to the tables and checked (which I had not) and observed that, when $n$ is even, the maximal subgroup $SO(n{-}2)\times SO(2)$ of maximal rank has larger dimension than $U(n/2)$ when $n>8$. (They have equal dimension when $n=8$ and the former has smaller dimension when $n<8$.) Thus, the above answer needs to be divided into parts when $n$ is even.

By the way, the double covers of the subgroups $SO(6)\times SO(2)$ and $U(4)$ in $Spin(8)$ are actually conjugate by an outer automorphism of $Spin(8)$, so they are essentially the same. This is a consequence of triality as discovered by Cartan.

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Ok, thank you for your help. – berl13 Apr 11 '12 at 15:59
I double-checked this. When $n\ge 8$ is even, the answer is (the cover of) $SO(n−2)\times SO(2)$, which differs from $U(n/2)$ when $n>8$. – Mikhail Borovoi Apr 11 '12 at 17:55
@Mikhail: Yes, I agree that you are right. $U(n/2)$ is a maximal subgroup of maximal rank, but it's not the largest dimension, even among the ones of rank $n/2$ (except when $n$ is small). – Robert Bryant Apr 11 '12 at 18:03

A subgroup of maximal rank of maximal dimension is certainly a maximal subgroup of maximal rank. Maximal connected subgroups of maximal rank in $Spin(n)$ correspond to maximal reductive Lie subalgebras of maximal rank in $so(n)_{\mathbf{C}}$. Such subalgebras in semisimple Lie algebras were classified by Dynkin in 1952, see Onishchik and Vinberg (Eds.), Lie Groups and Lie Algebras III, Encyclopaedia of Mathematical Sciences, vol. 41, Tables 5 and 6. For $so(n)$ all such subalgebras are $so(2k)\oplus so(n-2k)$, and also $gl(n/2)$ for $n$ even. The subalgebras of highest dimension are probably $so(n-1)$ for $n$ odd and $gl(n/2)$ for $n$ even.

EDIT: For $n=2l\ge 10$, the subalgebra of highest dimension and of maximal rank in $so(n)$ is $so(n-2)\oplus so(2)$ of dimension $2l^2-5l+4=l^2+l(l-5)+4$, and NOT $gl(n/2)$ of dimension $l^2$. For example, for $n=10$ we have ${\rm dim} (so(8)\oplus so(2))=29$, while ${\rm dim}\ gl(5)=25$.

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