Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.