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Let $G$ be a linear algebraic group, $T$ a subtorus of $G$, and $\mathfrak g$ the Lie algebra of $G$. Then there exist characters $\chi_1, ... , \chi_t$ of $T$, and subspaces $V_i = V_{\chi_i}$ of $\mathfrak g$, such that $\mathfrak g$ is the direct sum of the $V_i$, and $\textrm{Ad }x(v) = \chi_i(x)v$ for any $x \in T, v \in V_i$. The characters $\chi_i$, excluding the trivial character if it occurs, are called the weights of $T$.

Fix a weight $\chi$, and let $g \in N_G(T)$. Then $g$ induces a character $\chi'$ of $T$ given by $\chi'(x) = \chi(gxg^{-1})$. I am trying to understand why $\chi'$ is also a weight of $T$. In other words, given that there exists a $0 \neq v \in \mathfrak g$ such that $\textrm{Ad }x(v) = \chi(x)v$ for all $x \in T$, why is it the case that there exists a $0 \neq w \in \mathfrak g$ such that $\textrm{Ad }x(w) = \chi(gxg^{-1})w$ for all $x \in T$?

D_S
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  • If $v \in V_\chi$ then try to show that $gv \in V_{\chi'}$. – Matthew Towers May 04 '16 at 11:54
  • Do you mean $\textrm{Ad } g(v)$? I had tried this, but it doesn't appear to work. You are saying that if $x \cdot v = \chi(x)v$ for all $x \in T$, then $x \cdot (g \cdot v) = \chi(gxg^{-1})v$ for all $x \in T$? – D_S May 04 '16 at 15:09
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    It does work. $x\cdot (g \cdot v) = g \cdot (g^{-1}xg) \cdot v = g \cdot \chi(g^{-1}xg) v = \chi(g^{-1}xg)g\cdot v$, valid for $v$ in the $\chi$-weight space of any $G$-module. – Matthew Towers May 04 '16 at 15:24
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    Ah thanks. I was so tired last night when I asked this question – D_S May 04 '16 at 22:56

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