Is there a non-injective trace-preserving endomorphism of a matrix group?

Question

Abstractly, I'm wondering if you can have a Lie group $G$ and a non-injective homomorphism $\phi: G \to G$, and a faithful representation $\rho_G$ of $G$ which is such that $\chi_{\rho_G} = \chi_{\rho_G \circ \phi}$. The idea is that since the trace only refers to diagonal elements, I feel like you should be able to fully characterize it via it's values on a subgroup. Specifically: we know that each maximal torus $T$ of a (compact, connected) Lie group $G$ contains elements of the conjugation classes of every element of $G$, and since trace is only defined up to conjugation class, we should be able to "project" elements to representatives of their conjugation class in $T$ without losing trace information. I'm wondering if we can do so homomorphically, rather than using an algorithm like "$\chi_G (g) = {\chi_{G}}_{\upharpoonright H} (a)$ where $a \in H \subset G$ is conjugate to $g$".

But, e.g., projecting onto the diagonal, or a map $A \mapsto tr(A)I$ aren't homomorphisms, and the only type of trace-preserving homomorphism I know of is a unitary transformation, which is an isomorphism.

Can anyone give an example of such a map, or does anyone know why it's not possible if it's not? I feel like I'm missing something obvious here.

Answer 1

For the revised question: Take $G$ to be the group of strictly upper triangular matrices in $SL(n, {\mathbb R})$ (strictly here means only $1$'s on the diagonal). Now, let $\phi: G\to G$ be any endomorphism, say, the trivial representation; let $\rho_G$ be the identity representation $G\to G< GL(n, {\mathbb R})$. You can find other examples by looking at the Borel subgroup $B< SL(n, {\mathbb R})$ (the subgroup of upper triangular matrices and taking $\rho_G=Id$, $\phi$ any endomorphism equal to the identity on the maximal torus in $G$.

Answer 2

These things are much clearer if you make the choice of representation explicit. You have two Lie groups $G, H$ and two representations $\rho_G : G \to GL(V)$ and $\rho_H : H \to GL(W)$ of those Lie groups. You want to know if we can have a morphism $\phi : G \to H$ such that $\chi_H(\phi(g)) = \chi_G(g)$ which is not an isomorphism, or maybe not injective.

And the answer is, of course we can: given any choice of $G, H, \rho_H, W$, and $\phi$, we can take $\rho_G$ to be $\rho_G(g) = \rho_H(\phi(g))$ (so that $V = W$).