# Finite-dimensional unitary representations of $SL_n(\mathbb{R})$

In Proposition 2.6.4 of his book Automorphic Forms and Representations, Bump is trying to prove that $SL_n(\mathbb{R})$ has no non-trivial finite-dimensional unitary representations. His argument is as follows:

If $\rho$ is such a representation on a finite-dimensional complex vector space $V$, then (identifying $V$ with $\mathbb{C}^m$) then image of $\rho$ is contained in the compact group $U(m)$, hence is compact. Therefore $SL_n(\mathbb{R})/\ker(\rho)$ is compact, and since the only normal subgroups of $SL_n(\mathbb{R})$ are $\{I\}$, $SL_n(\mathbb{R})$, and $\{\pm I\}$ if $n$ is even, this implies that $\rho$ is trivial.

Now I may be missing something, but this argument seems incomplete. Why does the fact that $U(m)$ is compact imply that the image of $\rho$ is? It is certainly not true in general that the image of a continuous homomorphism from an arbitrary topological group $G_1$ to a compact group $G_2$ is compact. For instance, the map $x\mapsto (e^{2\pi ix},e^{2\pi i\alpha x})$ is a continuous embedding of the real line into the torus if $\alpha$ is irrational.

$\mathfrak{sl}_n(\mathbb{R}) \to \mathfrak{su}_m$,
for some $m$.
This can't exist, for example by consideration of the adjoint representation. There are vectors of $\mathfrak{sl}_n(\mathbb{R})$ with nonzero real eigenvalues, but $\mathfrak{su}_m$ is a compact Lie algebra, so the only nonzero eigenvalues that occur in the adjoint representation are purely imaginary.