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Let $k$ be algebraically closed, let $T$ be a torus (diagonal group of some $\textrm{GL}_n$), and let $V$ be an affine variety on which $T$ acts morphically. Then we have an abstract representation $\tau: T \rightarrow \textrm{GL}(k[V])$ given by $\tau_t(f)(x) = f(t^{-1}x)$. Let $$k[V]_{\chi} = \{ f \in k[V] : \tau_t(f) = \chi(t)f, \textrm{ for all } t \in T \}$$ The claim is that $k[V]$ is the direct sum of the $k[V]_{\chi} : \chi$ is a character of $T$. I can see how the sum is direct, but I'm confused as to why the span of the $k[V]_{\chi}$ is $k[V]$. By comultiplication corresponding to the morphism $a: T \times V \rightarrow V, (t,x) \mapsto t^{-1}x$, you can show that for every $f \in k[V]$, there exist $g_{\chi} \in k[V] : \chi$ such that $$f(t^{-1}x) = \sum\limits_{\chi} \chi(t) g_{\chi}(x)$$ I'm using the fact that the characters of $T$ form a basis for $k[T]$. In particular, $f = \sum\limits_{\chi} \chi(t) g_{\chi}$. What I want to show now is that each $g_{\chi}$ lies in $k[V]_{\chi}$. Would anyone be able to give me a hint?

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    $\begingroup$ how about to consider the function $f(t^{-2}x)$, you can see it as $f(t^{-1}(t^{-1}x))$ or $f((t^2)^{-1}x)$. And then it is easy to see that $g(t^{-1}x)=\chi(t)g(t)$. $\endgroup$ – Chen Jiang Mar 15 '16 at 7:58
  • $\begingroup$ Sorry, but I still don't get it. How are you able to match up the individual terms? $\endgroup$ – D_S Mar 15 '16 at 16:08
  • $\begingroup$ Written into an answer :) $\endgroup$ – Chen Jiang Mar 16 '16 at 4:06
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Try to consider the equality

$$ \sum \chi(t)g_{\chi}(s^{-1}x)=f(t^{-1}s^{-1}x)=\sum \chi(ts)g_{\chi}(x)=\sum \chi(t)\chi(s)g_{\chi}(x). $$

Since as you said, $\chi(t)$ form a basis, it follows that

$$ g_{\chi}(s^{-1}x)=\chi(s)g_{\chi}(x). $$

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