Question about the definition of hamiltonian group action. - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T04:41:15Z http://mathoverflow.net/feeds/question/76071 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/76071/question-about-the-definition-of-hamiltonian-group-action Question about the definition of hamiltonian group action. Gigou 2011-09-21T18:46:57Z 2012-01-27T07:04:49Z <p>So I'm reading the part in Ana Cannas da Silva's book "Lectures on Symplectic Geometry" available (on her website) about hamiltonian group actions on a symplectic manifold. She starts by defining $\mathbb{R}$-actions and $S¹$ actions by saying that the vector field on M that they generate must be hamiltonian. Then she defines hamiltonian $T^n$-actions (p.154) by the requirement that the restriction of the action to each circle ${1} \times\ldots\times S¹\times\ldots\times{1}$ be hamiltonian (plus the requirement that each of the n corresponding hamiltonian functions be invariant une the action of the rest of $T^n$)</p> <p>And then, finally, she defines a hamiltonian action of a general Lie group G as one having a "moment map". This is a natural generalisation because the existence of a moment map is equivalent (I believe!) to the fact that for each $X\in Lie(G)$, the vector field $X^*$ on M induced by $X$ is hamiltonian (i.e. $X^*_p=\frac{d}{dt}(\exp(tX)\cdot p)(0)$).</p> <p>For indeed, if that it that case, then on can just define the moment map $\mu : M \rightarrow Lie(G)^*$ by setting $&lt;\mu(p),X>:=\mu^X(p)$, where $\mu^X$ is a hamiltonian function for $X^*$ chosen so that $\mu$ is G-equivariant with respect to the coadjoint action on $Lie(G)^*$. (Note that in the case where $G$ is commutative such as $G=T^n$, this last condition boils down to $\mu$ being $G$-invariant.)</p> <p><strong>The question:</strong> I am trying to prove that the ad-hoc definition implies the general definition in the $T^n$ case. The problem I am having is that we only know that n vector fields $X_1^*,\ldots,X_n^*$ (one for each subcircle $1\times\ldots\times S¹\times\ldots\times 1 \subset T^n$) are hamiltonian. Knowing that the $X_i$'s form a basis of $Lie(T^n)$, does this imply that every $X\in Lie(T^n)$ induces a hamiltonian $X^*$. Does something like '$(X+Y)^* =X^* +Y^*$' hold? </p> <p>More generally, given a Lie group $G$ acting on a symplectic manifold $M$, is it necessary to check that each $X\in Lie(G)$ induce a hamiltonian vector field on $M$, or is it sufficient to check this for a basis of $Lie(G)$ ? </p> <p>Thanks.</p> http://mathoverflow.net/questions/76071/question-about-the-definition-of-hamiltonian-group-action/76072#76072 Answer by Igor Rivin for Question about the definition of hamiltonian group action. Igor Rivin 2011-09-21T19:09:16Z 2011-09-21T19:09:16Z <p>Check out:</p> <p><a href="http://www.math.nyu.edu/~kessler/teaching/group/Talk4.pdf" rel="nofollow">http://www.math.nyu.edu/~kessler/teaching/group/Talk4.pdf</a></p> http://mathoverflow.net/questions/76071/question-about-the-definition-of-hamiltonian-group-action/76078#76078 Answer by JosÃ© Figueroa-O'Farrill for Question about the definition of hamiltonian group action. JosÃ© Figueroa-O'Farrill 2011-09-21T19:56:35Z 2011-09-21T19:56:35Z <p>Your belief is only partially correct. The existence of a momentum map requires <em>two</em> conditions:</p> <ol> <li><p>First, as you point out, if $X^*$ is the fundamental vector field corresponding to $X \in \mathfrak{g}$, then $i_{X^*} \omega = d\mu^X$ should be exact. This defines $\mu^X$ up to a locally constant function.</p></li> <li><p>But also you need equivariance, which boils down to $\lbrace \mu^X,\mu^Y \rbrace = \mu^{[X,Y]}$.</p></li> </ol> <p>This second condition is not automatic, and indeed it is possible that the first condition holds, but not the second. The reason is that from the first condition it follows that $$c(X,Y) := \lbrace \mu^X,\mu^Y \rbrace - \mu^{[X,Y]}$$ is a locally constant function. It follows that $c$ so defined is a Lie algebra cocycle, whence it defines a class in the Lie algebra cohomology $H^2(\mathfrak{g};H^0(M))$, where $H^0(M)$ is the trivial $\mathfrak{g}$-module of locally constant functions.</p> <p>Concerning your actual question, the map $X \mapsto X^*$ is a Lie algebra homomorphism $\mathfrak{g} \to \mathfrak{X}(M)$ to the vector fields on $M$ and hence, in particular, it is linear. The map $c$ is clearly then bilinear, so again it is enough to check on a basis.</p> http://mathoverflow.net/questions/76071/question-about-the-definition-of-hamiltonian-group-action/76092#76092 Answer by Paul Skerritt for Question about the definition of hamiltonian group action. Paul Skerritt 2011-09-21T23:11:10Z 2012-01-27T07:04:49Z <p>Just so you're aware, not every author insists that a momentum map be infinitesimally equivariant (Prof. Figueroa-O'Farrill's condition 2), although it is part of da Silva's definition (edit: actually, on checking, da Silva requires the slightly stronger condition of equivariance, i.e. $\mu(g\cdot p)=\mu(p)\circ\mathrm{Ad}_{g^{-1}}$). The literature isn't uniform - for example, Marsden just requires the first condition. I'm biased towards this definition since Jerry Marsden, who unfortunately passed away a year ago today, was my advisor.</p> <p>To answer your main question, yes the map $X\in\mathfrak{g}\mapsto X^* \in\mathfrak{X}(M)$ is always linear, regardless of whether the action is Hamiltonian. To see this explicitly, let $\Phi^p:G\rightarrow M$ be the map $g\mapsto g\cdot p$. By definition $X^* _p$ is precisely $T_e\Phi^p(X)$ (you can see this agrees with your definition by writing $\exp(tX)\cdot p$ as $\Phi^p(\exp(tX))$, and using the chain rule to calculate $\frac{d}{dt}$), and derivative maps $T_xf$ are always linear. So yes, it's enough to check condition 1 on a basis.</p>