When $G$ is a linear algebraic connected solvable Lie group and has a normal maximal torus $T$ (in fact, $T$ is the unique maximal torus of $G$), it follows that $T$ is a subgroup of the center of $G$.

In fact, since $G$ is a linear algebraic connected solvable Lie group, it follows that $G = TU$, where $U$ is a normal simply-connected subgroup and $T \cap U = 1$. Since $T$ is also normal, $G = TU$ is a direct product and, since $T$ is abelian, it follows that $T$ is a subgroup of the center of $G$.

And when $G$ is a connected solvable Lie group which has a normal maximal torus $T$, is it true that $T$ is a subgroup of the center of $G$?

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    $\begingroup$ In the Lie-group setting is a "torus" compact? If so, then a normal torus $T$ in any connected Lie group $G$ (solvable or not) is central since the automorphism group of a (compact) torus is discrete. More generally, the only continuous action $\alpha:G \times T\rightarrow T$ of a connected topological group $G$ on a (compact) torus $T$ (e.g., conjugation on a subgroup) is trivial. Indeed, if two automorphisms of the compact $T$ are "close" then they're equal, so a neighborhood $U$ of 1 in $G$ acts trivially on $T$ by continuity of $\alpha$, and $U$ generates $G$ by connectedness. $\endgroup$ – nfdc23 Aug 20 '16 at 19:06
  • $\begingroup$ I cannot resist mentioning in this connection the related fact, to which @nfdc23 alludes, that also a torus in the Zariski sense admits no non-trivial, Zariski connected group of automorphisms. That this is in fact a highly general phenomenon is evidenced by the fact that, when I went Googling for a reference to this (the so called rigidity of tori), I found that it also holds for quantum tori: arxiv.org/abs/1204.3218 . $\endgroup$ – LSpice Aug 20 '16 at 20:26

This is an elaboration of nfdc23's comment. Claim: A normal (compact) torus $T$ in a connected Lie group (solvable or not) is central.

Proof: Let $\exp:\mathfrak t=\text{Lie}\ T\to T$ be the exponentail map. Its kernel $\Gamma$ is a lattice in $\frak t$. Since $T$ is normal, the group $G$ acts on $T$ and therefore $\mathfrak t$ by conjugation. Moreover, it preserves $\Gamma$. For any $\gamma\in\Gamma$, the set $\{g\gamma\mid g\in G\}$ is connected (since $G$ is) and contained in the discrete set $\Gamma$. So it equals $\{\gamma\}$ which means that $G$ acts trivially on $\Gamma$. But then $G$ acts also trivially on $\mathfrak t$ and therefore on $T$, i.e., $T$ is central.

  • $\begingroup$ I like also to observe that one can come at $T$ from "the other side": that is, via homomorphisms to circles, not just (implicitly) from circles (because the exponential map factors through $\mathfrak t/2\pi i\mathfrak t$). That is, a map into $T$ is completely determined by its post-composition with all possible maps $T \to \mathrm S^1$; and then, as you observe for co-characters, the character group also is discrete. $\endgroup$ – LSpice Aug 20 '16 at 20:29
  • $\begingroup$ @Friedrich Knop: Thank you! A related question: If an element $g$ of a connected solvable Lie group $G$ normalizes a compact maximal torus $T$, then $g$ centralizes $T$? $\endgroup$ – Mauro Patrão Aug 20 '16 at 20:32

This is a long comment rather than an answer, since I don't understand precisely what the question really means.

I'm still confused about the formulation here, though Knop and others have contributed what they can to sorting it out. For example, the initial wording "linear algebraic connected solvable Lie group" doesn't immediately make sense to me, since Lie groups are originally real manifolds whereas linear algebraic groups usually require an algebraically closed ground field. I guess you might be looking at the real points of a connected solvable linear algebraic group, but if so then your further comments suggest that you intend the group to be $\mathbb{R}$-split. Otherwise the rational points of a maximal $\mathbb{R}$-torus (in the algebraic group sense) might be of any size, even trivial.

What do you mean here by "maximal torus" of a connected solvable Lie group? The study of solvable Lie groups (for instance by Hochschild and Mostow) is fairly delicate, and the role of topological tori is not obvious outside the context of reductive or semisimple Lie groups. The use of the term "torus" in both algebraic and topological settings is perhaps unfortunate, but you need to be clear about the distinction. The Jordan decomposition for linear algebraic groups doesn't work well for all Lie groups, since the notions of "semisimple" and "unipotent" for a Lie group (or "nilpotent" in its Lie algebra) are not intrinsic.

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    $\begingroup$ Thak you very much for your comments. The expression "linear algebraic connected solvable Lie group" should be replaced by "linear connected solvable Lie group". The "maximal torus" here is a maximal compact connected abelian subgroup. $\endgroup$ – Mauro Patrão Aug 22 '16 at 13:53

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