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Suppose $G$ is a Lie group, $\mathfrak{g}$ its Lie algebra, if we have a smooth representation $(\pi,V)$, then it induces an action of $\mathfrak{g}$ on $V$. Now conversely, if we have a nice (with properties you may assume) action of $\mathfrak{g}$ on $V$, can we say such action arises from some unique smooth action of $G$?

Here we may assume $G$ to be simply connected if needed. Thank you.

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Have you looked at the classical paper by Edward Nelson? Nelson, Edward Analytic vectors. Ann. of Math. (2) 70 1959 572–615. – Alain Valette Aug 24 '11 at 14:14
What is $V$ here? – José Figueroa-O'Farrill Aug 24 '11 at 14:16

Let $V$ be the space of smooth, compactly supported functions on $\mathbb{R}$, which vanish at $0$ together with all their derivatives. Define $X:V\rightarrow V$ by $Xf=f'$. Then $X$ defines an action of the Lie algebra of $\mathbb{R}$ on $V$, which does not integrate to an action of $\mathbb{R}$.

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Let $\pi$ represent a finite dimensional real Lie algebra $\mathfrak g$ on a Hilbert space $\mathcal H$ by skew-adjoint operators. Then $\pi$ integrates to the connected simply connected Lie group $G$ with Lie algebra $\mathfrak g$ if, and only if, the elements of $\pi(\mathfrak g)$ have a common invariant dense domain. This is an old result of Moshe Flato, Daniel Sternheimer and others. My apologies to the mathematical physicists whose names I have omitted.

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For a reference about the well-known fact that finite-dimensional representations of a connected and simply connected Lie group are in one-to-one correnspondence with finite-dimensional representations of its Lie algebra, the OP is referred e.g. to "Fulton-Harris: Representation Theory", Section 8.1.

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Let be $G$ a simply connected Lie group, $\mathfrak{g}$ its Lie algebra and $M$ an arbitrary smooth manifold. Let be $\zeta$ a smooth action of $\mathfrak{g}$ on a $M$, i.e. $\zeta:X\in\mathfrak{g}\to\mathfrak{X}(M)$ is a Lie algebra homomorphism.

Then there exists a local left action $\Phi$ of $G$ on $M$ such that, for any $X\in\mathfrak{g}$, the t-time local flow of $\zeta(X)$ is given by $m\mapsto\Phi(e^{-t.X},m)$

In general the action of $\mathfrak{g}$ on $M$ can only be lifted to a local left action $\Phi$ of $G$ on $M$, i.e. defined only on a neighborhood of $\{e\}\\times M$ in $G\times M$.

But, if $\zeta(X)$ is a complete vector on $M$ for any $X\in\mathfrak{g}$, then $\zeta$ can be lifted to a global left action of $G$ on $M$.

These results should be found in the work of Richard Palais on the Lie theory of transformation groups.

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Yes if $G$ is connected and simply connected, since in that case there is a one to one correspondence between Lie group homomorphisms $G\to H$ and Lie algebra homomorphisms $\mathfrak g \to \mathfrak h$. Since a representation of $\mathfrak g$ is just a Lie algebra homomorphism $\mathfrak g \to \mathfrak{gl}(V)$, your desired result follows.

EDIT: I'm assuming $V$ is finite dimensional.

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Presumably (this applies to Giuseppe's answer also, I believe) the OP is interested in infinite dimensional $V$. For instance, what is the representation of $SL_2$ that corresponds to the Verma module $M(0)$ for $sl_2$? I don't think there is one. – Stephen Aug 24 '11 at 15:04

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