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

differential-geometrymanifoldssymplectic-geometry

  • 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.)

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.

Best Answer

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 $M$.)

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