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 $\mathbb S^1$ actions by saying that the vector field on $M$ that they generate must be hamiltonian. Then she defines hamiltonian $\mathbb T^n$-actions (p.154) by the requirement that the restriction of the action to each circle $\{1\} \times\ldots\times\mathbb S^1\times\ldots\times\{1\}$ be hamiltonian (plus the requirement that each of the $n$ corresponding hamiltonian functions be invariant under the action of the rest of $\mathbb T^n$)

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 $X\in\operatorname{Lie}(G)$, the vector field $X^* $ on $M$ induced by $X$ is hamiltonian (i.e. $ X^*_p=\frac{d}{dt}(\exp(tX)\cdot p)(0)$).

For indeed, if that it that case, then on can just define the moment map $\mu : M \rightarrow\operatorname{Lie}(G)^* $ by setting $\langle\mu(p),X\rangle:=\mu^X(p)$, where $\mu^X$ is a hamiltonian function for $X^* $ chosen so that $\mu$ is G-equivariant with respect to the coadjoint action on $\operatorname{Lie}(G)^* $. (Note that in the case where $G$ is commutative such as $G=\mathbb T^n$, this last condition boils down to $\mu$ being $G$-invariant.)

The question: I am trying to prove that the ad-hoc definition implies the general definition in the $\mathbb T^n$ 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\mathbb S^1\times\ldots\times 1 \subset\mathbb T^n $) are hamiltonian. Knowing that the $X_i$'s form a basis of $\operatorname{Lie}(\mathbb T^n)$, does this imply that every $X\in\operatorname{Lie}(\mathbb T^n)$ induces a hamiltonian $X^* $. Does something like $\color{red}{(X+Y)^* =X^* +Y^*}$ hold?

More generally, given a Lie group $G$ acting on a symplectic manifold $M$, is it necessary to check that each $X\in\operatorname{Lie}(G) $ induce a hamiltonian vector field on $M$, or is it sufficient to check this for a basis of $\operatorname{Lie}(G) $ ?



3 Answers 3


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.

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.

  • $\begingroup$ Ah yes, I was missing this insights. Thank you Paul. $\endgroup$
    – Gigou
    Sep 21, 2011 at 23:50
  • $\begingroup$ Glad it helped :) $\endgroup$
    – user17945
    Sep 21, 2011 at 23:52

Your belief is only partially correct. The existence of a momentum map requires two conditions:

  1. 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.

  2. But also you need equivariance, which boils down to $\lbrace \mu^X,\mu^Y \rbrace = \mu^{[X,Y]}$.

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.

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.

  • $\begingroup$ Hello José, thanks for answering my question. I am not saying that 1. implies 2., but I am saying that the locally constant function in 1. can be chosen so that 2. is satisfied. Is this false? (Sorry, I don't understand the Lie algebra cohomology argument.) Now concerning my actual question, which you answer by the affirmative (i.e. (X+Y)*=X*+Y*), can you provide an explanation or a reference? Can you read this property just from properties of the exponential map? $\endgroup$
    – Gigou
    Sep 21, 2011 at 20:45
  • $\begingroup$ By the way, just so we're clear, I agree that X-->X* is a Lie algebra homomorphism PROVIDED the action is hamiltonian (i.e., provided there exists a moment map for it, and in particular, provided X* is hamiltonian for each X in Lie(G)). But of course this is irrelevant for my question because I work with the hypothesis that I have only a basis of X's whose X*'s are hamiltonian. And using only this hypothesis, I want to show that every Y in Lie(G) induces a hamiltonian Y*. One way to do this is if we had (X+Y)*=X*+Y*, but without a moment map, how do we prove this? $\endgroup$
    – Gigou
    Sep 21, 2011 at 21:27
  • $\begingroup$ The locally function in 1. cannot always be chosen so that 2. is satisfied. The obstruction is the class of the cocycle $c$ in the Lie algebra cohomology and that need not be zero. In a way, of course, this is just a tautology, but phrasing it in terms of Lie algebra cohomology allows you to conclude that in many cases (e.g., semisimple Lie algebras) there is no obstruction. $$ $$ The map $X \to X^*$ is a Lie algebra homomorphism independently of whether the action is hamiltonian or even symplectic. It's simply the fact that you have an action of a Lie group on a manifold. $\endgroup$ Sep 21, 2011 at 22:32
  • $\begingroup$ "locally function" of course means "locally constant function". (I wish I could edit comments...) $\endgroup$ Sep 21, 2011 at 22:32
  • $\begingroup$ Hey. I thought I was done with this but it came back to haunt me! You said the map X-->X* is a Lie algebra homomorphism, and I thought I could prove this using Paul Skerritt's insight that $X^* =\Phi^p_*(X)$ by using the fact that if $V_i$, $W_i$ (i=1,2) are F-related at p, then so are $[V_1,V_2]$ and $[W_1,W_2]$. Hence, $[X,Y]^*_p=\Phi^p_*([X,Y]_e)=[\Phi^p_*(X),\Phi^p_*(Y)]_p=[X^*,Y^*]_p$. BUT, Silva says that X-->X* is supposed to be a Lie-algebra anti-homomorphism. And if I assume the existence of a moment map (as per your definition), then using that $f\rightarrow X_f$ is a $\endgroup$
    – Gigou
    Sep 26, 2011 at 21:50

Check out:


  • $\begingroup$ Thanks Igor, unfortunately, the information I am after is dismissed with "The moment map property is again easy to check." on page 8. $\endgroup$
    – Gigou
    Sep 21, 2011 at 20:32

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