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Let $M$ be a compact and connected smooth manifold with a symplectic form $\omega$.

$Ham(M, \omega)$ denotes the space of hamiltonian symplectomorphisms of $(M,\omega)$.

I have the following statement in my lecture notes:

Using Darboux’s theorem one can show that the action of $Ham(M, \omega)$ on M is transitive, that is: for any pair of points $p, q \in M$, there exists $\Phi \in Ham(M, \omega)$ such that $\Phi(p) = q$. The idea is that Darboux's theorem to go from local to global, i.e. to show that points that are close to each other in the symplectic manifold can be mapped to each other via a Hamiltonian diffeomorphism.

...which I unsuccesfully tried to prove. How is it done?

I know that a symplectomorphism of $(M, \omega)$ is a diffeomorphism $\Phi : M \to M$ such that $\Phi ^∗\omega = \omega$. $\Phi$ is Hamiltonian if there exists a Hamiltonian isotopy $\phi_t$ such that $\Phi=\phi_1$.

And that $ Ham(M, \omega)$ is a normal subgroup of $Symp(M, \omega)$ (the space of symplectomorphisms ), if that is useful.

I am not sure what "points close to each other" means here and how to use to arrive to the thesis

Following the suggestion:

Let p, q be in M then By Darboux theorem there are open charts $(U_p,\phi)$ and $(Uq,\psi)$ with $\phi$ and $\psi$ symplectomorphisms from $\Bbb R^{2n}$

But what do I do with this?

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    $\begingroup$ On $(\mathbb{R}^{2n},\omega_{std})$ can you find a hamiltonian symplectomorphism taking $0$ to an arbitrary point $p$? $\endgroup$
    – J.V.Gaiter
    Commented Mar 21 at 15:40
  • $\begingroup$ @J.V.Gaiter Sorry I am very new to all this, how would I do that? $\endgroup$
    – darkside
    Commented Mar 21 at 15:56
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    $\begingroup$ Since $\omega_{std}$ has constant coefficients, any translation $x\mapsto x+p$ is a symplectomorphism. Can you write down a Hamiltonian which generates such a symplectomorphism? $\endgroup$
    – J.V.Gaiter
    Commented Mar 21 at 22:00
  • $\begingroup$ @J.V.Gaiter Not sure what "generates" mean and does Hamiltonian mean Hamiltonian vector field? $\endgroup$
    – darkside
    Commented Mar 21 at 22:05
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    $\begingroup$ Generates as in, is the time 1 flow of. A Hamiltonian is a function which gives a vector field via the symplectic form. Before delving deeper into symplectic geometry, I would recommend reading more on the foundations. The book "symplectic topology" by McDuff and Salamon and the lecture notes of Da Silva are good resource! $\endgroup$
    – J.V.Gaiter
    Commented Mar 21 at 23:10

1 Answer 1

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Translations of $\mathbb{R}^{2n}$ are generated by constant vector fields, i.e. the flow $x\mapsto x+ty$ is generated by the vector field $x\mapsto y$ (under the natural identification $T(\mathbb{R}^{2n})\cong \mathbb{R}^{2n}\times \mathbb{R}^{2n}$). For a constant vector field $X=\sum_{i,j}^nX^i \partial_{q^i}+Y_j\partial_{p_j}$ we have $\iota_{X}\omega=\sum X^idp_i-\sum Y_j dq^j$ and hence $d\iota_{X}\omega_{std}=0$. By the closedness of $\omega_{std}$, $\mathcal{L}_{X}\omega_{std}=0$ and hence the flow $x\mapsto x+ty$ is a symplectomorphism.

There are several ways to see that this flow is Hamiltonian. First off, we can write a Hamiltonian $H_y(p,q)=\sum X^ip_i-Y_jq^j=\omega(X,(p,q))$. But also, the Poincare lemma tells us that $\iota_X\omega$ is exact, i.e. $\iota_X\omega_{std}=dH$ for some function $H$ and hence there exists a Hamiltonian for this flow.

This tells us that $Ham(\mathbb{R}^{2n},\omega_{std})$ acts transitively on $\mathbb{R}^{2n}$ since the time one flow of $x\mapsto x+ty$ sends $0$ to $p$.

Now, to show that $Ham(M,\omega)$ acts transitively, we need to show that given to points $p$ and $q$ there exists a Hamiltonian symplectomorphism taking $p$ to $q$. We must assume that $M$ is connected, since $Ham(M,\omega)\subset Diff_0(M)$ cannot switch connected components and hence the proposition is not true when $M$ is not connected.

Take a path $\gamma:[0,1]\to M$ connecting $p$ to $q$ and cover $\gamma([0,1])$ by some number of Darboux charts $\psi_{\alpha}:U_{\alpha}\to \mathbb{R}^{2n}$. We can take a finite number of these charts since $[0,1]$ is compact. Take $t_1,\cdots t_k\in [0,1]$ be a sequence of times such that $t_i$ and $t_{i+1}$ are contained in a common Darboux chart $U_{\alpha_i}$ for $t_0=0$ and $t_{k+1}=1$ (this is guaranteed by taking sufficiently many). Define $$\kappa_i:[0,1]\to \mathbb{R}^{2n}, t\mapsto (1-t)\psi_{\alpha_i}(\gamma(t_i))+t\psi_{\alpha_i}(\gamma(t_{i+1})).$$ This $\kappa_{i}$ is just the straight line connecting $\psi_{\alpha_i}(\gamma(t_i))$ to $\psi_{\alpha_i}(\gamma(t_{i+1}))$. Now, let $\phi_i: M\to \mathbb{R}^{\geq0}$ be a smooth function which is identically $1$ in a neighborhood of $\psi_{\alpha_i}^{-1}(\kappa_i([0,1]))$ and is identically $0$ outside of $U_{\alpha}$. Define $v_i:=(\psi_{\alpha_i}(\gamma(t_i))-\psi_{\alpha_i}(\gamma(t_{i+1}))$. We can then write down the hamiltonian $H_i=\phi_i\cdot \psi_{\alpha}^*H_{v_i}$ as in my second paragraph. This Hamiltonian makes sense by use of the bump function.

The time one flow of $H_i$ takes $\gamma(t_i)$ to $\gamma(t_{i+1})$ as follows. Since $\phi_i$ is identically $1$ near $\psi_{\alpha_i}^{-1}(\kappa_i([0,1]))$, $\psi_{\alpha_{i}*}X_{H_i}=X_{v_i}$ in a neighborhood of $\kappa_i([0,1])$. The $\kappa_i$ is an integral curve of $X_{H_{v_i}}$ contained in the aforementioned neighborhood and hence $\psi_{\alpha_i}^{-1}\circ \kappa_i$ is an integral curve for $X_{H_i}$. The curve has $\psi_{\alpha_i}^{-1}\circ \kappa_i(0)=\gamma(t_i)$ and $\psi_{\alpha_i}^{-1}\circ \kappa_i(1)=\gamma(t_{i+1})$.

We can then conclude that $\Phi^i_1(\gamma(t_{i}))=\gamma(t_{i+1})$ for $\Phi^i_t$ the flow of the hamiltonian vector field $X_{H_i}$ and hence $(\Phi^k_1\circ \Phi^{k-1}_1\circ \cdots \circ \Phi^1_1)(p)=q$ as desired.

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  • $\begingroup$ Thanks for the answer. About the first paragraph 1)Is the $\omega$ in the first paragraph the standard symplectic form? Shoudn't it be denoted something like $ \omega_0$ or$ \omega_{\text{std}}$ so that it doesn't it get mixed with the generic $\omega$ of my manifold M? I am using $ \omega_{\text{std}}=\sum dq^i\wedge dp_i$ to check the computations in the first paragraph. $\endgroup$
    – darkside
    Commented Mar 25 at 19:25
  • $\begingroup$ 2)Is the point $p$ in the map $x \mapsto x+p$ in $\Bbb R^{2n}$ or in$ \Bbb R^{4n}$ ? First you say it is a translation $\Bbb R^{2n}$, so I deduce $ p \in \Bbb R^{2n}$ but then you say that $x \mapsto p$ is a vector field using the natural identification, then it seems like $ p \in \Bbb R^{4n}$ and then I guess $p = (x, v)$, with $v$ a tangent vector at $x \in \Bbb R^{2n}$ Moreover the notation $p$ surely has nothing to do with a vector of $2n$ components $p=(p_1,...p_n)$ where the $p_i$'s are half of the coordinates of $\Bbb R^{2n}$ (the other half being the $q_i$'s) used in $X$, does it? $\endgroup$
    – darkside
    Commented Mar 25 at 19:32
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    $\begingroup$ Taking $y=q-p$ suffices. The more general point is that if I have a transformation $g$ which takes $0$ to $q$ and h which takes $0$ to $p$ then the transformation $hg^{-1}$ takes $p$ to $q$. $\endgroup$
    – J.V.Gaiter
    Commented Mar 25 at 22:32
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    $\begingroup$ The point about the constant vector fields is that since $T(\mathbb{R}^{2n})\cong \mathbb{R}^{2n}\times \mathbb{R}^{2n}$ the specification of a vector field is equivalent to specifying a function $f: \mathbb{R}^{2n}\to \mathbb{R}^{2n}$ which in this case is the constant function $f(x)=y$ for all $x$. $\endgroup$
    – J.V.Gaiter
    Commented Mar 25 at 22:35
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    $\begingroup$ As an alternative, note that the first part shows that the orbit of any point is open; this implies that every orbit is a connected component (any point in the closure of an orbit O has orbit that intersects O, hence must actually be O). $\endgroup$
    – Max
    Commented Mar 26 at 6:10

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