# About the proof of the Lagrangian neiborhood theorem

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) 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.