Suppose I have a Lie group $G$. How can one determine whether there is more than one conjugacy class in $G$ of subgroups isomorphic to a given Lie subgroup $H$?

Put another way:

Fix a Lie subgroup $H \leq G$. How can one determine whether there is a subgroup of $G$ isomorphic to $H$ but not conjugate to $H$?

I'm particularly interested in the case (in the first formulation of the problem) $$G := SO(3, 4), \qquad H := SL(3, \Bbb R) .$$ One way to get a group isomorphic to $H$ is as follows: Fix a copy of $G_2 < SO(3, 4)$; then, the stabilizer in $G_2$ of a timelike vector under the usual action on $\Bbb R^{3, 4}$ is $SL(3, \Bbb R)$. (This claim follows from translating Corollary 2.4(2) in I. Kath, $\smash{G_{2(2)}^*}$-Structures on pseudo-Riemannian manifolds, from spinor language.)

Prima facie this seems harder than the corresponding question for Lie algebras, as one can sometimes (as in this example) produce subgroups of $G$ that have the same Lie algebra as $H$ but, for reasons of global topology, are not isomorphic to $H$.

  • 1
    $\begingroup$ The one case when this question is analyzed in great detail for the last 100 years is when G is GL(n,C); one does it by looking at characters of group representations. But your situation is the opposite. $\endgroup$ – Moishe Kohan May 26 '16 at 21:44

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