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Let $G$ be a Lie group and $\mathfrak g$ its Lie algebra. Suppose that $\rho_\mathfrak{g}$ is a representation of $g$ on a vector space $V$. Is it true that the mapping $\rho$ from the identity component of $G$ to linear operators on $V$ defined by $\rho(e^X) = e^{\rho_\mathfrak{g}(X)}$ is a representation of the identity component of $G$ on $V$? My hunch is yes, but I'm having trouble proving it.

EDIT As pointed out to me below in the comments, the answer to my original question is no. As a follow-up question: under what assumptions on $G$ will the mapping I attempt to construct be a representation on the identity component of $G$?

Thank you!

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Since in general the exponential map from the Lie algebra to the group is not surjective (take the group connected, for simplicity), your candidate representation of the latter is not even defined on all of the group! – Mariano Suárez-Alvarez Jan 28 '13 at 8:24
As I wrote, the map is not surjective even if the group is connected. Just take $G=SL_2(\mathbb R)$, for example. – Mariano Suárez-Alvarez Jan 28 '13 at 8:27
Yea my apologies; I deleted the comment almost as soon as I wrote it. Thanks @MarianoSuárez-Alvarez – joshphysics Jan 28 '13 at 8:27
If the group is compact, the exponential is surjective, but now you have a problem with definition. Just take the simplest possible example where $G=S^1$, the $1$-dimesional torus. – Mariano Suárez-Alvarez Jan 28 '13 at 8:30
@MarianoSuárez-Alvarez Please excuse my being slow, but could you be more explicit/specific with what you mean by a problem with definition? Also, do you know what assumptions are strong enough to ensure the result I'm looking for? – joshphysics Jan 28 '13 at 8:39

In the book by J.F.Cornwell, Group Theory in Physics, Chapter 11, sec. 7, theorem 1) it is stated that for any representation $\mathcal{\Gamma_L}$ of the Lie algebra $\mathcal{L}$ there is a representation $\Gamma_\mathcal{G}$ of a simply connected Lie group with the same Lie algebra $\mathcal{G}$ (also proven to exists) so that

$\Gamma_\mathcal{L}(a) = \ \frac{d}{dt} \ \Gamma_\mathcal{G} \exp \{t \ a\}|_{t=0} $.

Note that this fails to be true for the Lie groups $SO(3)$ and $U(1)$, both not simply connected.

$U(1)$ has been discussed in the comments, to see why this happens for $SO(3)$ consider the representation of its algebra given by the identification $so(3) = su(2)$ and note that by taking the exp of $diag\{I t ,-I t\}$ you can obtain both the matrices $1$ and $-\mathbb{1}$ whilst $SO(3)$ possesses a single element commuting with all the others.

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