One-parameter subgroup of SL(n) is diagonalizable Given a group homomorphism $\rho\colon \mathbb{C}^\times \rightarrow SL(n,\mathbb{C})$. Why is it true that the subgroup $\rho(\mathbb{C}^\times)$ is diagonalizable, i.e. that there is a basis of $\mathbb{C}^n$ such that $\rho(\mathbb{C}^\times)$ is a subgroup of the group of diagonal matrices in this basis?
I found this in a proof and suppose that it is a special case of a more general fact in representation theory, but unfortunately I do not have enough background in this area. It probably can be derived somehow from the Jordan Chevalley decomposition.
Thanks!
 A: You need some assumptions on $\rho$ for this to be true.  E.g.
$$z \mapsto \begin{pmatrix} 1 & \log |z| \\ 0 & 1 \end{pmatrix}$$
is a continuous group homomorphism from $\mathbb C^{\times}$ to $SL_2(\mathbb C)$
with non-diagonalizable image.
The standard hypothesis made to ensure that the image be diagonalizable is that $\rho$ be a map of algebraic groups, i.e. be given by expressions that are polynomial in $z^{\pm 1}$.  (Polynomials in $z^{\pm 1}$ and $\overline{z}^{\pm 1}$ would also be okay; this corresponds to a map of algebraic groups over $\mathbb R$.)
The claim follows from a classification of the possible algbraic representations of $\mathbb C^{\times}$: if you let
$A = \mathbb C[z,z^{-1}]$ (the ring of Laurent polynomials on $\mathbb C^{\times}$) with $\mathbb C^{\times}$ acting via the regular representations, you an easily check that $A$ is semi-simple, i.e. is a direct sum of one-dimensional representations.  (More precisely, it is the direct sum of the representations $z \mapsto z^n$, for $n \in \mathbb Z$; this essentially corresponds to interpreting the elements of $A$ as Fourier polynomials.)
Similarly, $B = \mathbb C[z,\overline{z}, z^{-1},\overline{z}^{-1}]$ is semi-simple, being a direct sum of the representations $z \mapsto z^p\overline{z}^q$, for $p,q \in \mathbb{Z}$.
Since $A$ and $B$ are the "regular" algebraic representations of $\mathbb C^{\times}$ (thought of either as an algebraic group over $\mathbb C$ or $\mathbb R$) and they are semi-simple, a standard argument shows that any algebraic representation of $\mathbb C^{\times}$ is semi-simple.
In particular, the algebraic $n$-dimensional representation of $\mathbb C^{\times}$ that arises from an algebraic homomorphism $\mathbb C^{\times} \to SL_n(\mathbb C)$ must be semi-simple, which is to say that the image  must consists of simultaneously diagonalizable matrices.  This is the claim that you asked about.
