5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Have you looked at the classical paper by Edward Nelson? Nelson, Edward Analytic vectors. Ann. of Math. (2) 70 1959 572–615. $\endgroup$ Aug 24, 2011 at 14:14
  • 1
    $\begingroup$ What is $V$ here? $\endgroup$ Aug 24, 2011 at 14:16

5 Answers 5

7
$\begingroup$

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 $\mathbb{R}$.

$\endgroup$
7
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ 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. $\endgroup$
    – Stephen
    Aug 24, 2011 at 15:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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