when can we lift an action of Lie algebra? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T17:09:05Z http://mathoverflow.net/feeds/question/73564 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73564/when-can-we-lift-an-action-of-lie-algebra when can we lift an action of Lie algebra? unknown (google) 2011-08-24T13:36:25Z 2011-10-25T18:51:29Z <p>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 $G$?</p> <p>Here we may assume $G$ to be simply connected if needed. Thank you.</p> http://mathoverflow.net/questions/73564/when-can-we-lift-an-action-of-lie-algebra/73567#73567 Answer by Giuseppe for when can we lift an action of Lie algebra? Giuseppe 2011-08-24T14:46:27Z 2011-08-24T14:46:27Z <p>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.</p> <p>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 $m\mapsto\Phi(e^{-t.X},m)$</p> <p>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 $G\times M$. </p> <p>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 $M$.</p> <p>These results should be found in the work of Richard Palais on the Lie theory of transformation groups.</p> http://mathoverflow.net/questions/73564/when-can-we-lift-an-action-of-lie-algebra/73570#73570 Answer by Eric O. Korman for when can we lift an action of Lie algebra? Eric O. Korman 2011-08-24T14:58:04Z 2011-08-24T15:03:39Z <p>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.</p> <p>EDIT: I'm assuming $V$ is finite dimensional.</p> http://mathoverflow.net/questions/73564/when-can-we-lift-an-action-of-lie-algebra/73582#73582 Answer by Salvatore Siciliano for when can we lift an action of Lie algebra? Salvatore Siciliano 2011-08-24T16:14:02Z 2011-08-24T16:14:02Z <p>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.</p> http://mathoverflow.net/questions/73564/when-can-we-lift-an-action-of-lie-algebra/73631#73631 Answer by Alain Valette for when can we lift an action of Lie algebra? Alain Valette 2011-08-25T05:11:35Z 2011-08-25T05:11:35Z <p>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 $\mathbb{R}$.</p> http://mathoverflow.net/questions/73564/when-can-we-lift-an-action-of-lie-algebra/79104#79104 Answer by Joseph Wolf for when can we lift an action of Lie algebra? Joseph Wolf 2011-10-25T18:51:29Z 2011-10-25T18:51:29Z <p>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.</p>