Proper compact connected subgroup of $Spin(n)$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T05:39:28Z http://mathoverflow.net/feeds/question/93771 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93771/proper-compact-connected-subgroup-of-spinn Proper compact connected subgroup of $Spin(n)$ berl13 2012-04-11T14:07:40Z 2012-04-11T18:10:06Z <p>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 $SO(n)$?</p> <p>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.</p> http://mathoverflow.net/questions/93771/proper-compact-connected-subgroup-of-spinn/93779#93779 Answer by Robert Bryant for Proper compact connected subgroup of $Spin(n)$ Robert Bryant 2012-04-11T15:53:28Z 2012-04-11T18:10:06Z <p>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.</p> <p><strong>Added after Mikhail's comment:</strong> 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 $n&lt;8$.) Thus, the above answer needs to be divided into parts when $n$ is even.</p> <p>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 <em>triality</em> as discovered by Cartan.</p> http://mathoverflow.net/questions/93771/proper-compact-connected-subgroup-of-spinn/93780#93780 Answer by Mikhail Borovoi for Proper compact connected subgroup of $Spin(n)$ Mikhail Borovoi 2012-04-11T16:16:26Z 2012-04-11T17:45:37Z <p>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.</p> <p>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 ${\rm dim}\ gl(5)=25$.</p>