## Return to Question

5 edited tags
4 added 2 characters in body

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 $\chi_i:G \chi_i:G_s \rightarrow \mathbb{G}_m$?

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

3 added 82 characters in body

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 $\chi_i:G \rightarrow \mathbb{G}_m$?

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

2 removed the word "stupid"
1