Homomorphism from noncompact semisimple Lie group to compact Lie group Is it true that there is no homomorphism from a semisimple Lie group without compact factor to a compact Lie group?
 A: See below a detailed version of the comment of @LSpice. (Edited taking into account a comment of @YCor.) This is an answer to the question on homomorpisms of real algebraic groups.

Proposition.
Let $\varphi\colon G\to K$ be a homomorphism of real algebraic groups, where $G$ is a connected semisimple real algebraic group without compact factors and $K$ is a compact real algebraic group.
Then $\varphi$ is trivial (identically 1).

Proof.  Such a homomorphism $\varphi$ induces an isomorphism
$$ G/{\rm ker}\,\varphi\,\overset\sim\longrightarrow\, {\rm im\,}\varphi.$$
The image   ${\rm im\,}\varphi$ is closed in $K$. Therefore, ${\rm im\,}\varphi$ is compact. On the other hand, from the theory of real semisimple algebraic groups we know that since $G$ is a connected semisimple real algebraic group without compact factors, it has no nontrivial compact quotients.  It follows that  $G/{\rm ker}\,\varphi=\{1\}$, and hence $\varphi$ is trivial.
Note that any compact real Lie group is algebraic. However,  a noncompact real semisimple Lie group might be non-algebraic. For example, any nontrivial cover of ${\rm SL}(2, {\Bbb R})$ is non-algebraic.
A: I think an elementary argument is that if such a homomorphism exists, one can pull back the Cartan-Killing form (or an extension of that, in case the compact group has positive dimensional center) to an  Ad-invariant definite inner product on  the Lie algebra of a quotient of the original group. This quotient is then compact.
