Given a Lie group $\mathfrak{G}$ with finite centre and with Lie algebra $\mathfrak{g}$, I am looking at a simple proof that negative definite Killing form implies compactness. This proof is given here. A sketch of the proof is as follows:

1. The author claims that $\mathrm{Ad}:\mathfrak{G}\to GL(\mathrm{Lie}(\mathfrak{G})$ is a closed map;
2. The Killing form $K:\mathfrak{g}\times\mathfrak{g}\to\mathbb{R}$ is $\mathrm{Ad}$-invariant, so a negative definite $K$ allows us to define an inner product $-K$such that $\mathrm{Ad}(\gamma)$ is orthogonal wrt $-K$ i.e. $\mathrm{Ad}(\mathfrak{G})\subseteq SO(\dim \mathfrak{g}, -K)$.
3. $SO(\dim \mathfrak{g}, -K)$ being compact, and $\mathrm{Ad}(\mathfrak{G})$ a closed subgroup by 1., we conclude that $\mathrm{Ad}(\mathfrak{G})$ is itself compact;
4. $\mathfrak{G}$, being an $M$-fold cover of $\mathrm{Ad}(\mathfrak{G})$ (where $M$ is finite by dint of the finite centre), is thus also compact.

Crucial to this proof is the assertion that $\mathrm{Ad}(\mathfrak{G})$ is closed in $SO(\dim \mathfrak{g}, -K)$. I can prove this given nondegeneracy of the Killing form, (e.g. with Lemma 1 of G. Hochschild, "Complexification of Real Analytic Groups") but the author of the first document I linked seems to be saying that this is a much more general and well known property of $\mathrm{Ad}$. What am I missing here?: I think I'm making this harder than it should be through overlooking a simple fact. So, as in my title:

When is $\mathrm{Ad}:\mathfrak{G}\to GL(\mathrm{Lie}(\mathfrak{G}))$ a closed map and why?

Edit: It seems that this is not as trivial as I thought. Hence answers less than a full answer are helpful and acceptable to me. For example, interesting counterexamples (showing when $\mathrm{Ad}:\mathfrak{G}\to GL(\mathrm{Lie}(\mathfrak{G})$ is not closed) or sufficient conditions for it to be closed (such as semisimplicity of $\mathfrak{G}$).

• Here's an example in which it's not closed. Let $q$ be irrational, and consider the 5-dimensional Lie group $G=\mathbf{C}^2\rtimes\mathbf{R}$ where $t\in\mathbf{R}$ acts by $t\cdot (z,x)=(e^{it}z,e^{iqt}w)$. Then the adjoint map has a non-closed image. To see this, embed the latter as a dense normal Lie subgroup in $G'=(\mathbf{C}\rtimes U(1))^2$ and restrict the adjoint representation of $G'$ to a faithful rep of $G'$ on the Lie algebra of $G$. – YCor Sep 30 '14 at 18:31
• Two things to add to YCor's comment: (1) since that example has trivial center, it shows the author's claim #1 is false, (2) in practice, a very natural criterion which ensures closedness is that $\mathfrak{g}$ is semisimple, but the only proof I am aware of for closedness in that case is via non-trivial theorems from the theory of linear algebraic groups. – user27920 Sep 30 '14 at 23:56
• @user52824 So it seems that it's probably not just my overlooking something trivial. I had a gut feeling one had to assume semisimplicity (which of course follows by Cartan's criterion from the nondegeneracy of the Killing form) or something along those lines, but no idea how to prove it. – Selene Routley Oct 1 '14 at 0:38
• @WetSavannaAnimalakaRodVance: Yeah, I have no idea how to prove closedness directly, and in the context of differential geometry it seems rather delicate issue how to tell that a connected Lie subgroup is closed just from the data of the corresponding Lie subalgebra. The approach through algebraic geometry provides a genuinely extra technique which I have no idea how to replace from a purely $C^{\infty}$ point of view. The relation between Lie groups and matrix groups is quite essential to the theory beyond the compact case. – user27920 Oct 1 '14 at 1:41
• I guess that if a Lie group $G$ satisfies $\mathrm{Hom}(G,\mathbf{R})=0$, then every linear representation of $G$ has a closed image (this would be optimal since otherwise $G$ admits a 2-dimensional complex rep with non-closed image). – YCor Oct 1 '14 at 8:03

Let $G$ be a connected Lie group. Equivalences:

• (i) every linear representation of $G$ has a closed image
• (ii) $\mathrm{Hom}(G,\mathbf{R})=0$
• (iii) $G/\overline{[G,G]}$ is compact.

Here homomorphisms are meant continuous, and reps are in $\mathbf{GL}_n(\mathbf{R})$ for $n$ not fixed, or equivalently in $\mathrm{GL}_n(\mathbf{C})$ in view of the closed inclusions $\mathrm{GL}_n(\mathbf{R})\subset \mathrm{GL}_n(\mathbf{C})\subset \mathrm{GL}_{2n}(\mathbf{R})$. Note that the equivalence between (ii) and (iii) is obvious.

That (i) implies (ii) is clear since if $\mathrm{Hom}(G,\mathbf{R})\neq 0$, $G$ admits $\mathbf{R}$ as a quotient, and we can use a 2-dimensional complex representation of $\mathbf{R}$ whose image is dense in a 2-dimensional real torus.

The implication (ii)$\Rightarrow$(i) goes by reduction to the semisimple case.

Thus first assume that $G$ is semisimple. Fix a representation $\rho:G\to\mathrm{GL}_n(\mathbf{C})$. In the latter, the image of the representation is a (a priori not closed) Lie subgroup $H_1$, with Lie algebra $\mathfrak{h}$ (a real Lie aubalgebra of $\mathfrak{gl}_n(\mathbf{C})$.

Let $H_2$ be the closure of $H_1$ (in the real Zariski topology, or in the real topology, it does not matter). Then $H_2$ normalizes $\mathfrak{h}$ and $H_1$. Since up to finite index, the automorphisms of $H_1$ are inner, and since $H_2$ is connected, we deduce that $H_2=H_1C$ where $C$ is the centralizer of $H_1$ in $H_2$. Decompose $\mathbf{C}^n=\bigoplus V_i$, a sub of $H_1$-irreducible complex subspaces. Then since $C$ is in the closure of $H_1$, it preserve each $V_i$. Since the $H_1$-representation on the complex space $V_i$ is irreducible, its commutant is reduced to scalars. Moreover, since the restriction of $H_1$ on $V_i$ has determinant 1, it is also the case for $C$. We deduce that $C$ is a finite abelian group. Thus $H_1$ has finite index in $H_2$. Since $H_1$ is $\sigma$-compact, a Baire argument then shows that $H_1$ has non-empty interior in $H_2$, and hence $H_1$ is the unit component of $H_2$ in the real topology. Thus $H_1$ is closed.

(Another argument goes by using the fact that perfect Lie subalgebras of $\mathfrak{gl}_n(\mathbf{R})$ are always Lie algebras of some Zariski closed subgroup of $\mathrm{GL}_n(\mathbf{R})$.)

Now assume $G$ arbitrary satisfying (ii). Write $G=RS$ with $R$ the connected solvable radical and $S$ a semisimple Levi factor. A simple argument based on taking the complexification at the Lie algebra level shows that $[G,R]$ is normal and that each representation of $G$ maps $[G,R]$ to unipotents. Thus let $\rho$ be a representation of $G$, say in $\mathrm{GL}_n(\mathbf{R})$, and let $H$ be its image. Then $N=\rho\Big(\overline{[G,R]}\Big)$ is a connected, unipotent normal subgroup of $H$. Being a unipotent connected Lie subgroup, it is necessarily closed in $\mathrm{GL}_n(\mathbf{R})$, actually Zariski closed. If $M$ is the normalizer of $N$, then $H\subset M\subset \mathrm{GL}_n(\mathbf{R})$ and $M$ is Zariski closed. Moreover $M/N$ stands a Zariski closed subgroup of $\mathrm{GL}_m(\mathbf{R})$ for some $m$ (indeed, if $M=\mathbb{M}_\mathbf{R}$ and $N=\mathbb{N}_\mathbf{R}$ for some $\mathbf{R}$-subgroups of $\mathrm{GL}_n$, then $\mathbb{M}/\mathbb{N}$ is a linear algebraic group and hence is isomorphic to an $\mathbf{R}$-closed subgroup in some $\mathrm{GL}_m$, and $\mathbb{M}_\mathbf{R}/\mathbb{N}_\mathbf{R}\to (\mathbb{M}/\mathbb{N})_\mathbf{R}$ is a closed map in the real topology.)

Therefore, we are reduced to the case when $[G,R]=1$. In this case $G=RS$, with $S$ semisimple Levi factor, $R$ abelian, and $[R,S]=1$. Since every linear representation of $S$ factors through some finite index subgroup of its center, it is no restriction to assume that $S$ has a finite center, and in particular $S$ is closed. By (ii), we thus deduce that $R$ is compact. Hence since by the first case, $\rho(S)$ is closed, we deduce that $\rho(G)$ is closed as well.

• Very slick indeed. Most wonderful. Many thanks, Yves – Selene Routley Oct 5 '14 at 1:10
• Note that I didn't answer the question about closedness of the adjoint representation. This post only shows that it has closed image when $G/Z(G)$ has compact abelianization, and this is not optimal (e.g., when $G$ is algebraic it also has a closed image). – YCor Dec 4 '16 at 7:44