symplectic geometry: help showing the cotangent lift of an action to a symplectic manifold is a symplectic action

Question

I am following da Silva's lectures on symplectic geometry.

She defines the lift of a diffeomorphism as follows:

Let $X_1$ and $X_2$ be $n$-dimensional manifolds with cotangent bundles $M_1=T^*X_1$ and $M_2=T^*X_2$ and suppose $f:X_1 \rightarrow X_2$ is a diffeomorphism.

Then the lift $f_\#:M_1 \rightarrow M_2$ is defined

$$f_\#(p_1) = p_2 = (x_2, \xi_2) = (f(x_1), \xi_2) $$ where $\xi_1 = (f_{x_1})^*\xi_2 = \xi_{2}\circ df_{x_1}$ and $\xi_2 \in (T^*X_2)_{f(x_1)}$

$f_\#\rvert_{T^*X_1}$ is therefore the inverse map of $(f_{x_1})^*$

and so $\xi_2 = [(f_{x_1})^*]^{-1}\xi_1 = (f^{-1}_{x_1})^*$

Now let $(M, \omega)$ be the symplectic manifold obtained by equipping the tangent bundle $M=T^*N$ with the canonical 2-form $\omega = \sum_i dx^i \wedge d\xi^i$ and let the Lie Group G act on N: $$\psi: G \rightarrow \text{Diff(M)}, \quad g \rightarrow \psi_g \\ $$ I am trying to prove the contangent lift of the action is symplectic (and hamiltonian)

We must have $$(\psi_g)_\#(p) = (\psi_g)_\#(x, \xi) = (\psi_g(x), (\psi_g^{-1})^*(\xi)) = (\psi_g(x), (\psi_{g^{-1}})^*(\xi)) $$

This must preserve the symplectic form, i.e. $$((f_\#)^*\omega)_p (u, v) = \omega_{f_\#(p)}(df_p(u), df_p(v)) $$

I am unsure what to do from here. I can see the symplectic form takes in two coordinates that transform in seemingly inverse ways. Could someone point me in the right direction ? It would also help if anyone spots any errors in the way I state the problem, I have studied physics so far and am slightly out my depth with the mathematics.

EDIT: Solution using tautological one-form (this method is in da Silva's notes)

We want to show $(f_\#)^*\alpha_2 = \alpha_1$

The tautological form on $M_1$, $M_2$ is defined: $$(\alpha_1)_{p_1} = \pi^*_{p_1}\xi_1, \quad \quad (\alpha_2)_{p_2} = \pi^*_{p_2}\xi_2$$

So we have $$\begin{align}(f_\#)^*_{p_1}(\alpha_2)_{p_2} &= (f_\#)^*_{p_1}\pi^*_{p_2}(\xi_2)] \\ &= (\pi_2 \circ f_\#)^*_{p_1} \xi_2 \\ &= (f \circ \pi_1)^*_{p_1} \xi_2 \quad\quad \text{(The lift is constructed as such)} \\ &= (\pi_1)^*_{p_1}f^*_{x_1}\xi_2 \\ &= (\pi_1)^*_{p_1}\xi_1 \quad\quad \text{(By definition of the lift)}\\ &= (\alpha_1)_{p_1} \end{align} $$

Therefore $f_\#$ preserves the tautological form as well as $d\alpha_1$, the canonical 2-form which is symplectic on $M_1$

Therefore the lift of the diffeomorphism $\psi_g:X_1 \rightarrow X_2$ is:

$(\psi_\#)_g:M_1 \rightarrow M_2$

and it is a symplectomorphism.

The lift of the action $\psi_\#$ is therefore symplectic.

It can also be shown the action is hamiltonian. It seems obvious since we are preserving the tautological one-form, although I'm not sure on the proof.

Answer 1

My own attempt attempt at an answer after reading the comments:

Proving that the lift of the flow of a vector field $V$ on $X_1$ is a hamiltonian vector field.

The flow of a vector field $V$ on a manifold $M$ is an action of the real numbers on the manifold: $f: \mathbb{R} \times M \rightarrow M$

There is a bijection between symplectic actions of $\mathbb{R}$ on $M$ and complete vector fields on $M$ (given by the flow of the vector field).

We have shown the lift of the flow $f_\#$ is a symplectic action on $M_1$. Let its associated vector field be $V_\#$.

$V_\#$ is hamiltonian if $ \iota_{V_\#} \omega$ is exact.

$$\iota_{V_\#} \omega = -\iota_{V_\#} d\alpha $$ By Cartan's magic formula: $$\iota_{V_\#} \omega = d\iota_{V_\#}\alpha - \mathcal{L}_{V_\#}\alpha = d\iota_{V_\#}\alpha $$ Since $f_\#$ preserves the tautological form $\alpha$

$V_\#$ is therefore hamiltonian with hamiltonian function $H = \iota_{V_\#}\alpha$

Answer 2

Here is the general proof of your statement that actions of Lie groups lift to Hamiltonian actions (in your answer you only showed this for $\mathbb R$-actions).

If $G$ acts on $X$ from the left, there is an induced left action of $G$ on $M=T^*X$, given by $$ g\cdot\eta=(\psi_{g^{-1}}){}^*\eta=\eta\circ d(\psi_{g^{-1}})_{gx} $$ for all $\eta\in T_x^*X$; here we have denoted $\psi_g(x)=g\cdot x$. It's easy to see this is a left action, and we will denote it as $\bar\psi_g(\eta)=g\cdot \eta$. Now this induced action preserves the tautological form, i.e. $(\bar\psi_g)^*\tau=\tau$ or more explicitly $$ \tau_{g\cdot \eta}\circ d(\bar\psi_g)_\eta=\tau_\eta $$ which stems from the fact that the following diagram commutes (you've basically shown this in your question): $\require{AMScd}$ \begin{CD} T^*X @>{\bar\psi_g}>> T^*X\\ @VVV @VVV\\ X @>{\psi_g}>> X \end{CD}

To show that the action is moreover Hamiltonian, we need to come up with a momentum map, and the natural candidate is given by the dual of the "fundamental map" of the action of $G$ on $X$: recall that we have a linear map $\sigma\colon \mathfrak g\rightarrow \mathfrak X(X)$ which is given pointwise as $$ \sigma(\xi)_x=\frac{d}{dt}\bigg|_{t=0}\exp(t\xi)\cdot x. $$ Taking the dual of this map, we obtain $\mu_x\colon T^*_xX\rightarrow \mathfrak g^*$, explicitly given as $\mu_x(\eta)(\xi)=\eta(\sigma(\xi)_x)$. These maps altogether define a map $\mu\colon T^*X\rightarrow \mathfrak g^*$. It is customary to denote for any $\xi\in\mathfrak g$ the function $\mu^\xi\in C^\infty(T^*X)$, $\mu^\xi(\eta)=\mu(\eta)(\xi).$

To check that this is indeed a moment map for the action, you now need to show two things:

• $d(\mu^\xi)_\eta(v)=\omega(\rho(\xi),v)$ for all $v\in T_\eta(T^*X)$. Here, $\rho\colon \mathfrak g\rightarrow \mathfrak X(T^*X)$ is the fundamental map of the induced action.
• Equivariance with respect to the coadjoint representation, i.e. $$ \mu(g\cdot \eta)(\xi)=\mu(\eta)(\mathrm{Ad}_{g^{-1}}\xi). $$

The first one is a consequence of the fact that the fundamental vector fields $\rho(\xi)$ and $\sigma(\xi)$ are $\pi$-related for any $\xi\in\mathfrak g$, i.e. there holds $d\pi_\eta(\rho(\xi)_\eta)=\sigma(\xi)_{\pi(\eta)}$ for all $\eta\in T^*X$. Here, $\pi\colon T^*X\rightarrow X$ is the canonical projection. Hence: $$ \tau_\eta(\rho(\xi)_\eta)=\eta\circ d\pi_\eta(\rho(\xi)_\eta)=\eta(\sigma(\xi)_{\pi(\eta)})=\mu^\xi(\eta). $$ Using this with the magic formula as in your answer, gives the first point: $$ \require{cancel} \iota_{\rho(\xi)}\omega=-\iota_{\rho(\xi)}d\tau=d\iota_{\rho(\xi)}\tau-\cancel{\mathcal L_{\rho(\xi)}\tau}=d(\tau(\rho(\xi)))=d(\mu^\xi) $$

The second one follows by a simple computation using the naturality of the exponential map $C_g\circ \exp=\exp\circ \mathrm{Ad}_g$ and our definition of the momentum map $\mu$.