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Let $G$ be a linear algebraic group over $\mathbb C$. Let $\psi$ be a finite dimensional regular representation of $G$ into $GL(V)$.

Suppose $G$ is connected. I would like to show for $v$ in $V$ the following are equivalent:

1) $\psi(g)(v)=v$ for all $g$ in $G$.

2) $d\psi(X)(v)=0$ for all $X$ in the Lie algebra of $G$.

For 1) implies 2), can I just say that the representation is "constant" for all $g$ and so the derivative is 0? Is that the right intuition? Is there a more formal way to show it? Where does connectedness come in?

For the other way I dont see how to use any of the theorems I know to take the Lie algebra information and bring it to the group. My intuition is that, because we are connected, and "locally constant", we are constant everywhere.

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up vote 1 down vote accepted

1) implies 2) is true without much further ado: just differentiate the equation with respect to $g$.

2) implies 1) is true for $G$ connected, since 2) tells you the derivative of $\psi(g)(v)$ with respect to $g$ is $0$, so the expression is locally constant with respect to $g$, and you know that $\psi(e)(v)=v$. As you suspected, one of the ways to define "connected" is that locally constant functions are globally constant.

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