pull-back of the tautological 1-form to a section

Question

• Suppose $M$ is a differentiable manifold of dimension $m < \infty$.

• Let $T^*M$ be its cotangent bundle. Let $\alpha \in T^*T^*M$ be the tautological $1$-form given by $\alpha_{(p, \lambda)}(\xi) := \lambda(D\pi(p, \lambda)\xi)$ where $\pi(p, \lambda):= p$ is the projection $T^*M\rightarrow M$ and therefore $D\pi(p, \lambda): T_{(p, \lambda)}T^*M\rightarrow T_pM$.

• Let $\omega := -\mathrm{d}\alpha$ the symplectic 2-form associated to it. Let $\mu$ be a closed $1$-form on $M$. (In fancy notation $\mu \in \Lambda^1(T^*M) = \Gamma(T^*M)$.)

There is this classical fact that if $\iota:\mu(M)\rightarrow T^*M$ be the inclusion, then $\iota^*\omega = 0$ and therefore, $\mu(M)$ is a Lagrangian submanifold of $T^*M$. Now for some reason I have trouble to show this. Could you help me to complete my attempt? (I don't want to identify objects that do not belong to the same set in my proof.)

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Attempt:

Firstly, $\iota^*\omega = \iota^* (-\mathrm{d}\alpha) = -\mathrm{d}(\iota^*\alpha)$. It is therefore enough to show that $\mathrm{d}(\iota^*\alpha) = 0$. Let $\underset{\theta}{\underbrace{\left.\iota\right|_{\mu(M)}}}:= \iota^*\alpha$. Then for $(p, \lambda) \in T^*M$ (i.e. $p \in M$ and $\lambda \in T_p^*M$): \begin{align*} (\mathrm{d}\theta)_{(p, \lambda)}(v_1, v_2) &= \theta_{(p, \lambda)}(v_2) - \theta_{(p, \lambda)}(v_1) - \theta_{(p, \lambda)}([v_1, v_2]) \\ &= \iota_{(p, \lambda)}(D\iota(p, \lambda)v_2) - \iota_{(p, \lambda)}(D\iota(p, \lambda)v_1) - \iota_{(p, \lambda)}(D\iota(p, \lambda)[v_1, v_2]). \end{align*}

I am stuck here. I think I have to somehow invlove $\mathrm{d}\mu = 0$ into my computations. (The classic argument says that the pull back of the tautological $1$-form is $\mu$ and the rest is therefore obvious. I'm exactly unsure of how to show the first part of this argument.) Thanks again for your help in advance.

Answer 1

Sometimes local coordinates make the formalism clearer. In the standard "physicists' notation," let $(q,p)$ be the usual local coordinates on $T^*M$. Then $\alpha = p\,dq = \sum p_i dq^i$. If $\mu = \sum f_i\,dq^i = f\,dq$, then $s_\mu(q)=(q,f(q))$, and so $s_\mu^*\alpha = f\,dq = \mu$, and so $d(s_\mu^*\alpha)=d\mu = 0$. (Indeed, $\alpha$ is called the tautological $1$-form because $s_\mu^*\alpha = \mu$. Note, also, that since $s_\mu$ is a diffeomorphism to its image, it is natural to think of $\iota$ precisely as $s_\mu$, with domain $M$.)

Answer 2

I just realized what is going on. I read a proof but it seems a little bit strange to me why everything just works perfectly. (Maybe because that is the right way to write everything clearly? Anyways, below I try to explain Da Silva's proof.)

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Just to be clear, we consider a difference between $\displaystyle\bigsqcup_{x \in M} T_x^*M$ and $\displaystyle\bigcup_{x \in M} T_x^*M$ and what we consider as the tangent bundle is the second one. Then

$$ T^*M = \bigsqcup_{x \in M} T_x^*M := \bigcup_{x \in M} \Big(\{x\} \times T_x^*M\Big) \subseteq M\times \bigcup_{x \in M} T_x^*M $$ and \begin{align*} \lambda:M\longrightarrow &\bigcup_{x \in M} T_x^*M; \quad x \mapsto \lambda(x) \\[2ex] \lambda_\mu: M \longrightarrow &T^*M; \quad x \mapsto (x, \lambda(x)). \end{align*}

We also recall that if $L:V\rightarrow W$ is a linear map, then $L^*:W^* \rightarrow V^*$ is another linear map associated to it given by $L^*(\lambda):= \lambda \circ L$ which completes the following diagram:

With these out of the way, we can do the following series of computation:

\begin{align*} (s_\lambda^* \alpha)(x) &= \alpha_{s_\lambda(x)} \circ Ds_\lambda(x) \tag{definition of pull back} \\ &= (Ds_\lambda(x))^*\alpha_{s_\lambda(x)} \\ &= (DS_\lambda(x))^*D\pi(S_\lambda(x))^*(\lambda(x)) \tag{definition of $\alpha$} \\ &= (D\pi(S_\lambda(x)\circ DS_\lambda(x))^*(\lambda(x)) \\ &= (D(\pi\circ S_\lambda)(x))^*\mu(x) = (\mathrm{id})^*\lambda(x) = \lambda(x). \end{align*}

This shows that $s_\lambda^* \alpha = \lambda$. Therefore

\begin{align*} s_\lambda^* \omega &= s_\lambda^* (-\mathrm{d}\alpha) \\ &= -\mathrm{d}(s_\lambda^*{\alpha}) \\ &= -\mathrm{d}\lambda \\ &= 0. \end{align*}