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This question is an addition to my question on simultaneous diagonalization from yesterday and it is probably also obvious but I just don't know this: Let $G$ be a commutative affine algebraic group over an algebraically closed field $k$. Let $G_s$ be the semisimple part of $G$. Let $\rho:G \rightarrow GL_n(V)$ be an embedding. Then $\rho(G_S)$ is a set of commuting diagonalizable endomorphisms and I know from yesterday that I have unique morphisms of algebraic groups $\chi_i: \rho(G_s) \rightarrow \mathbb{G}_m$, $1 \leq i \leq r$, and a decomposition $V = \bigoplus _{i=1}^r E _{\chi_i}$, where $E_{\chi_i} = \lbrace v \in V \mid fv = \chi_i(f)v \ \forall f \in \rho(G_s) \rbrace$. Now, my question is: are the morphisms $\chi_i$ independent of $\rho$ so that I get well-defined morphisms $\chi_i:G_s \rightarrow \mathbb{G}_m$?

If somebody knows what I'm talking about, then please change the title appropriately! :)

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You should add the reference to the previous question. Thanks. –  user2330 Dec 23 '09 at 15:01
    
Done . –  user717 Dec 23 '09 at 15:05
    
I'm a bit confused by your statement. With the definitions I'm used to, an abelian group cannot be semisimple, so if $G$ is abelian, $G_s = 1$. Alternately, all abelians are completely reducible. In any case, the semisimple groups I know have no nontrivial one-dimensional representations. Better would be to take a nonabelian group, and look at a maximal abelian subgroup; then any representation of the big group picks out characters of the subgroup, although in general those characters do not extend to (one-dimensional reps of) the big one. –  Theo Johnson-Freyd Dec 23 '09 at 20:16
    
@TJF: You're right that a nontrivial commutative linear group is not semisimple. From the context of the question, it seems fairly clear that the OP means the subgroup consisting of semisimple elements, equivalently "the Levi subgroup" or the maximal torus. –  Pete L. Clark Dec 23 '09 at 20:44
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Unless I drastically misunderstand your question, of course the characters $\chi_i$ depend on the representation $\rho$. Try looking at the simplest nontrivial case: $G = \mathbb{G}_m$ acting on a one-dimensional vector space. In this case, there is exactly one $\chi_i$ and it is simply a character of $\mathbb{G}_m$, i.e., is of the form $x \mapsto x^n$ for a unique integer $n$. This integer $n$ is determined by (and determines) $\rho$.

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Oh, you are absolutely right! Thanks. –  user717 Dec 23 '09 at 11:42
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