The Lagrangian neiborhood theorem stated as follows:

Let $(M, \omega)$ be a symplectic manifold and $L \subset M$ be a compact Lagrangian submanifold. Then there exists a neighbourhood $\mathcal{N}\left(L_0\right) \subset T^* L$ of the zero section, a neighbourhood $V \subset M$ of $L$, and a diffeomorphism $\phi: \mathcal{N}\left(L_0\right) \rightarrow V$ such that $$ \phi^* \omega=-d \lambda,\left.\quad \phi\right|_L=\mathrm{id}, $$ where $\lambda$ is the canonical 1 -form on $T^* L$.

Which is essentially the symplectic version of tubular neiborhood theorem.

The proof goes as follows:

Proof: The proof rests on the fact that the normal bundle of $L$ in $M$ is isomorphic to the tangent bundle. To define an explicit isomorphism, one may use a compatible complex structure $J$ on the tangent bundle $T M$. The subspace $J_q T_q L \subset T_q M$ is the orthogonal complement of $T_q L$ with respect to the metric $g_J$ induced by $J$, and is a Lagrangian subspace of $\left(T_q M, \omega\right)$. Let $$ \Phi_q: T_q^* L \rightarrow T_q L $$ be the isomorphism induced by the metric $g_J$, i.e. $$ g_J\left(\Phi_q\left(v^*\right), v\right):=v^*(v), \quad v \in T_q L . $$ Now consider the map $\phi: T^* L \rightarrow M$ given by the exponential map of the Riemannian metric $g_J$ : $$ \phi\left(q, v^*\right):=\exp _q\left(J_q \Phi_q\left(v^*\right)\right) . $$ Then for $v=\left(v_0, v_1^*\right) \in T_q L \oplus T_q^* L=T_{(q, 0)} T^* L$ we have $$ d \phi_{(q, 0)}(v)=v_0+J_q \Phi_q\left(v_1^*\right) \tag{*} $$ .....

I have no idea how to compute the (*) ? What have I done is working out the differential of bundle homomorphism $\Phi$ and $J$, however I don't know how to compute the differential of exponential here.


1 Answer 1


The only things you need are the following (which can be found in any textbook on Riemannian geometry)

(1) The exponential map $\exp:TL\rightarrow M$ fixes every point $(q,0)$ of the zero section.

(2) The derivative $d_{(q,0)}\exp$ at points $(q,0)$ of the zero section is the identity map.

Now take $v=(v_0,v_1^{*})\in T_{(q,0)}T^{*}L$. By linearity, we have $$ d_{(q,0)}\phi(v_0,v_1^{*})=d_{(q,0)}\phi(v_0,0)+d_{(q,0)}\phi(0,v_1^{*}). $$ For the first summand, take a curve $\alpha(t)$ in $L$ with $\alpha(0)=q$ and $\alpha'(0)=v_0$. Then $$ d_{(q,0)}\phi(v_0,0)=\left.\frac{d}{dt}\right|_{t=0}\phi(\alpha(t),0)=\left.\frac{d}{dt}\right|_{t=0}\exp_{\alpha(t)}(0)=\left.\frac{d}{dt}\right|_{t=0}\alpha(t)=v_0, $$ where we used in the third equality that $\exp$ fixes points on the zero section.

For the second summand, we have $$ d_{(q,0)}\phi(0,v_1^{*})=\left.\frac{d}{dt}\right|_{t=0}\phi(q,tv_1^{*})=d_{q}\exp\left(\left.\frac{d}{dt}\right|_{t=0}J_q\Phi_q(tv_1^{*})\right)=d_{q}\exp(J_q\Phi_q(v_1^{*}))=J_q\Phi_q(v_1^{*}). $$ Here we used the chain rule in the second equality, the fact that $J_q\Phi_q$ is linear in the third equality and the last equality holds because the derivative of $\exp$ is the identity along the zero section.

  • $\begingroup$ Thank you studiosus $\endgroup$
    – yi li
    Commented Oct 19, 2022 at 2:29

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