Every element $g \in GL(n,\mathbb{C})$ has a unique Jordan decomposition $$ g = g_u g_s $$ where $g_u$ is unipotent, $g_s$ is semisimple (i.e. diagonalizable over $\mathbb{C})$ and $g_ug_s=g_sg_u$. It is well known that if $G \subseteq GL(n,\mathbb{C})$ is an algebraic group, then it contains the unipotent and semisimple part of all of its elements. Moreover, if $\varphi \colon G \to H$ is a map given by polynomials, then $\varphi(g)_u = \varphi(g_u)$ and $\varphi(g)_s = \varphi(g_s)$ for all $g \in G$. (a reference for this is chapter 15.3 in Humphrey's book on Lineare Algebraic Groups)

Now a semisimple element can be further decomposed into an elliptic and a hyperbolic part, where elliptic means all eigenvalues have norm $1$ and hyperbolic means all eigenvalues are real and positive. This can be done by writing every eigenvalue $\lambda \in \mathbb{C}$ of the semisimple element as $\lambda = ae^{i\phi}$ where $a > 0$ and $|\phi| = 1$. So we have a unique decomposition $$g_s = g_eg_h$$ where $g_e$ is elliptic, $g_h$ is hyperbolic and $g_ug_h=g_hg_u$.

My question is:

If $\varphi \colon G \to H$ is a homomorphism between real algebraic groups given by real polynomials, do we have $\varphi(g_e) = \varphi(g)_e$ and $\varphi(g_h) = \varphi(g)_h$ for all $g \in G$ and if so, why? I could not find a proof of this statement anywhere.


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