# Are maximal connected semisimple subgroups automatically closed?

(Yet another question in a series demonstrating my rather embarrassing ignorance of standard Lie theory... I hope this is not too basic for MO!)

To be a little more precise: let $G$ be a real connected Lie group, not necessarily simply connected, and let $R$ be its solvable radical. Then (see the EOM entry, for instance) $G$ can be written as $R\cdot S$ for some maximal semisimple, connected Lie subgroup $S$. Having looked online, I get the impression that $S$ need not be closed. Is this impression correct?

(As mentioned in the link above, if $G$ is simply connected then a Levi--Malcev decomposition of its Lie algebra exponentiates to give $G=R\rtimes S$ and $S$ is closed.)

The motivation is in some joint work, where we have a property of locally compact groups that is known to pass to closed subgroups, and which we have shown is not satisfied by any connected semisimple Lie group. It would be nice if we could deduce from this that a connected Lie group with this property has to be solvable...

• If by a Lie subgroup you mean that $S$ has the induced topology, then every Lie subgroup is closed, since the complement of $S$ in its closure is a union of orbits, and all orbits are submanifolds of the same dimension. – Angelo Jan 10 '13 at 8:52
• @Angelo: perhaps my terminology is incorrect or ambiguous. I meant that $S$ is an immersed submanifold of $G$ (as in Alain Valette's answer). – Yemon Choi Jan 10 '13 at 22:56

Your impression is correct. Let $H$ be the universal cover of $SL_2(\mathbb{R})$, it has infinite cyclic center, let $z$ be a generator of the center. Consider the product $H\times U(1)$ (where $U(1)$ is the group of complex numbers of modulus 1), and mod out by the (discrete, central) subgroup generated by $(z,e^{2\pi i\theta})$, where $\theta$ is irrational. You get a reductive group with radical $S^1$, in which the Levi factor $H$ is dense.