most recent 30 from http://mathoverflow.net 2013-05-25T21:55:34Z http://mathoverflow.net/feeds/question/9611 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/9611/eigenvalue-characters S1 2009-12-23T11:18:53Z 2009-12-24T00:02:51Z

<p>This question is an addition to my <a href="http://mathoverflow.net/questions/9541/simultaneous-diagonalization" rel="nofollow">question</a> 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 $\chi_i:G_s \rightarrow \mathbb{G}_m$? </p> <p>If somebody knows what I'm talking about, then please change the title appropriately! :)</p>

http://mathoverflow.net/questions/9611/eigenvalue-characters/9612#9612 Pete L. Clark 2009-12-23T11:33:27Z 2009-12-23T11:40:48Z

<p>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) $\rho$. </p>