Given a connected Lie group, define its toral component as the maximal connected and compact subgroup of its center.

Is the toral component of a connected Lie group equal to the toral component of its solvable radical?

If not, is there a simple counter-example?

And if the connected Lie group is a linear algebraic one?


closed as off-topic by abx, Wolfgang, YCor, Chris Godsil, user1688 Jul 14 '16 at 8:02

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    $\begingroup$ Any normal connected solvable subgroup of a Lie group is contained in the solvable radical. $\endgroup$ – abx Jul 12 '16 at 17:14
  • $\begingroup$ @abx Is the answer obvious after that? I mean, the solvable radical may in principle have larger center... $\endgroup$ – მამუკა ჯიბლაძე Jul 12 '16 at 17:30
  • $\begingroup$ The direct product of SL(V) by the vector space V has trivial center and its solvable radical is V. But the toral component of V is also trial, so this is not a counter-example to my question. $\endgroup$ – Mauro Patrão Jul 12 '16 at 20:47
  • $\begingroup$ Such questions rather belong on Math StackExchange, where you should explain what you have tried. Here (and in other previous question) you ask between the equality of two subgroups, so you should study inclusions between these subgroups, say if you are able to prove one of the inclusions, etc. $\endgroup$ – YCor Jul 12 '16 at 23:22

Yes, the toral component of a connected Lie group is equal to the toral component of its solvable radical.

Let $G$ be a Lie group, $S$ its solvable radical, and $\mathrm{TC}(G)$ denote the toral component of $G$. As abx has noticed, $\mathrm{TC}(G)\subseteq S$, hence $\mathrm{TC}(G)\subseteq\mathrm{TC}(S)$. We show that $\mathrm{TC}(S)\subseteq\mathrm{TC}(G)$.

Since $S$ is a characteristic subgroup of $G$, we see that $\mathrm{TC}(S)$ is a characteristic subgroup of $G$, hence it is a normal subgroup of $G$, hence $G$ acts on $\mathrm{TC}(S)$ by conjugation. Since $T:=\mathrm{TC}(S)$ is a torus in the sense of Lie group theory, we have $T\simeq (\mathbb{R}/\mathbb{Z})^n$ for some natural $n$, hence $\mathrm{Aut}(T)\simeq\mathrm{GL}(n,\mathbb{Z})$, hence the connected component of the identity in $\mathrm{Aut}(T)$ is trivial. It follows that the connected Lie group $G$ acts trivially on $T$ when acting by conjugation. This means that $T=\mathrm{TC}(S)$ is contained in the center of $G$. Thus $\mathrm{TC}(S)\subseteq\mathrm{TC}(G)$, as required.

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    $\begingroup$ @MauroPatrão: If the answer is helpful, you can consider accepting it.... $\endgroup$ – Mikhail Borovoi Jul 13 '16 at 20:17

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