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Consider a finite-dimensional Lie group $G$ and two conjugacy classes $H$ and $I$ of isomorphic subgroups of $G$.

Question. Is there some finite-dimensional Lie overgroup of $G$ which fuses $H$ and $I$ into a single conjugacy class?

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No. Take $G={\rm SL}(2,{\Bbb R})$, $\ H=\{\,h(\lambda)={\rm diag}(\lambda, \lambda^{-1})\ |\ \lambda\in {\Bbb R}, \lambda>0\,\}$, $$U=\bigg\{ u(a)= \begin{pmatrix} 1 &a\\ 0&1 \end{pmatrix}\ \ \bigg|\ \ a\in {\Bbb R}\ \bigg\}.$$ The Lie groups $U$ and $H$ are isomorphic via the isomorphism $u(a)\mapsto h(e^a)$.

In any Lie overgroup $G'$ containing $G$, the subgroups $H$ and $U$ of $G$ act on $L' ={\rm Lie\,}G'$ via a smooth finite dimensional representation $${\rm Ad}\colon\, G\hookrightarrow G'\to {\rm GL}(L').$$ of the Lie group $G$. Since any such representation of $G$ is polynomial (I don't know a reference!), we see that each element of ${\rm Ad}(U)$ is unipotent (has all eigenvalues 1), whereas each element of ${\rm Ad}(H)$ is semisimple. We conclude that ${\rm Ad}(U)$ and ${\rm Ad}(H)$ are not conjugate in ${\rm GL}(L')$, and therefore $U$ and $H$ are not conjugate in $G'$.

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  • $\begingroup$ This is a great example! Maybe I'm underthinking or overcomplicating it, but doesn't polynomial-ity follow from the fact that the finite-dimensional representations of $\operatorname{SL}(2, \mathbb R)$ are explicitly described, and can be seen to be polynomial? $\endgroup$
    – LSpice
    Commented Nov 15, 2022 at 17:25
  • $\begingroup$ @LSpice: Yes, it does, but I don't know a reference. $\endgroup$
    – Mikhail Borovoi
    Commented Nov 15, 2022 at 17:40

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