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](https://www2.bc.edu/~reederma/Hansen.pdf). 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"](http://www.ams.org/journals/tran/1966-125-03/S0002-9947-1966-0206141-0/S0002-9947-1966-0206141-0.pdf)) but the author of the [first document I linked](https://www2.bc.edu/~reederma/Hansen.pdf) 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 $\mathfrak{G}$).



When is the Ad (Adjoint Representation) Morphism a Closed Map
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.


