Weinstein's neighborhood theorem says that every Lagrangian has a standard neighborhood. The more precise statement goes like this.

Theorem 1: (Lagrangian Neighborhood Theorem) Let $(X,\omega)$ be a symplectic manifold and $L \subset X$ be a closed Lagrangian. Then there exists a neighborhood $U$ of $L$ in $X$ and a symplectomorphism $\varphi:U \simeq V \subset T^*L$ taking $L$ identically to the zero-section $L \subset T^*L$.

Now let $(W,\lambda)$ be a Liouville domain. That is, $W$ is a compact manifold with boundary, and $\lambda$ is a $1$-form on $W$ such that $d\lambda$ is symplectic and $\lambda|_{\partial W}$ is a contact form. Furthermore, let $L \subset W$ be a compact Lagrangian sub-manifold with Legendrian boundary $\partial L \subset \partial W$.

My question is whether the following version of the neighborhood theorem holds in this setting. It seems to me that if it is true, then it should be standard, but I can't find a reference.

Theorem 2 (Maybe?): There exists a neighborhood $U$ of $L$ in $W$ and a symplectomorphism of manifolds with boundary $\varphi:U \simeq V \subset T^*L$ taking $L$ identically to the zero-section $L \subset T^*L$.

Remark On Proof Of Theorem 1: The basic result that the usual Lagrangian neighborhood theorem depends on is the following lemma (see [1] or McDuff-Salamon).

Lemma: Let $X$ be a manifold with closed sub-manifold $S \subset X$, and let $\omega_0, \omega_1$ be two symplectic forms on $X$. Suppose that $\omega_0 = \omega_1$ on the fiber $T_sX$ for any $s \in S$.

Then there exists neighborhoods $N_0$ and $N_1$ of $S$ and a symplectomorphism $\varphi:N_0 \to N_1$ with $\varphi|_S = \text{Id}$ and $\varphi^*\omega_1 = \omega_0$.

The proof is a version of the usual Moser trick. You find a $1$-form $\sigma$ in a neighborhood with $d\sigma = \omega_1 - \omega_0$ and then you integrate the vector-field $Z_t$ satisfying: $$\iota_{Z_t}\omega_t = -\sigma\quad\text{where}\quad\omega_t = (1-t)\omega_0 + t\omega_1$$ This gives you a family of diffeomorphisms with $\varphi^*_t\omega_t = \omega_0$ and you're done. If you try to run this proof on a sub-manifold $S \subset X$ with $\partial S \subset \partial X$, you run into the issue that $Z_t$ needs to be parallel to the boundary $\partial X$ in order for the flow to be well-defined. If I'm not mistaken, the criterion for this to be the case is: $$ T(\partial X)^{\omega_t} \subset \ker(\sigma) \text{ on }\partial X $$ Here $T(\partial X)^{\omega_t}$ is the characteristic foliation on $\partial X$ with respect to $\omega_t$, i.e. the symplectic perp to the tangent space to $\partial X$. It isn't clear to me that you can even accomplish the above inclusion for $\sigma$, or that you can upgrade $\sigma$ to a family $\sigma_t$ with this property.

Speculation On Validity Of Theorem 2: On a conceptual level, I can't decide whether or not Theorem 2 is too optimistic. Here is what makes me skeptical about it.

Theorem 2 would imply not only that the boundaries $\partial U \simeq \partial V$ of $U$ and $V$ were contactomorphic, but also that the characteristic foliations $T(\partial U)^{d\lambda}$ and $T(\partial V)^{d\lambda}$ on $U$ and $V$ near $\partial L$ were the same. The characteristic foliation of a contact hypersurface is generally very sensitive to the embedding of said hypersurface, and from that perspective a standard neighborhood in the vein of Theorem 2 would be a bit surprising.

I haven't pursued this idea enough to produce a counter-example unfortunately.

  • $\begingroup$ If there is a symplectic cap, or a "symplectic collar", to $W$ in which the Lagrangian $L$ extends as a Lagrangian (now with the original Legendrian $\partial L$ as a hypersurface), you could just take the restriction of the Weinstein tubular neighborhood. $\leftarrow$ Might be something stupid. $\endgroup$ Commented Oct 30, 2018 at 16:31
  • $\begingroup$ @ChrisGerig the issue with this approach is that the Weinstein neighborhood map $\varphi:U' \to TL'$ of the closed extension $L'$ into the capped symplectic manifold $W'$ may not send $U' \cap W$ to $T^*L \subset T^*L'$. In other words, the domain of the restricted chart may not be correct. $\endgroup$ Commented Nov 6, 2018 at 19:53

1 Answer 1


Theorem 2 is true, verbatim. I will give an outline of the proof here, since the details make it kind of long. If you would like a detailed write-up and you don't want to do it yourself, DM or email me.

The actual statement that is true is more general: you do not need the boundary $\partial X$ to be contact or for the boundary $\partial L$ to be Legendrian.

Theorem: (Weinstein Neighborhood With Boundary) Let $(X,\omega)$ be a symplectic manifold with boundary $\partial X$ and let $L \subset X$ be a properly embedded, Lagrangain sub-manifold with boundary $\partial L \subset \partial X$ transverse to $T(\partial X)^\omega$.

Then there exists a neighborhood $U \subset T^*L$ of $L$ (as the zero section), a neighborhood $V \subset X$ of $L$ and a diffeomorphism $f:U \simeq V$ such that $\varphi^*(\omega|_V) = \omega_{\text{std}}|_U$.

Proof: The proof has two steps. First, we construct neighborhoods $U \subset T^*L$ and $V \subset X$ of $L$, and a diffeomorphism $\varphi:U \simeq V$ such that: \begin{equation} \varphi|_L = \text{Id} \qquad \varphi^*(\omega|_V)|_L = \omega_{\text{std}}|_L \qquad T(\partial U)^{\omega_{\text{std}}} = T(\partial U)^{\varphi^*\omega} \end{equation} Here $T(\partial U)^{\omega_{\text{std}}} \subset T(\partial U)$ is the symplectic perpendicular to $T(\partial U)$ with respect to $\omega_{\text{std}}$ (and similarly for $T(\partial U)^{\varphi^*\omega}$. Second, we apply Lemma 1 (below) and a Moser-type argument to conclude the result.

For the first part, the proof proceeds like this. First you pick a metric on $L$ and use the exponential map in the usual way, to get a diffeomorphism $\varphi:U \simeq V$ with $U \subset T^*L$, $V \subset X$ and $\varphi^*\omega_{\text{std}} = \omega$ along $L$. Then we use Lemma 2 below to modify $\varphi$ in a collar neighborhood of $\partial U$ to satisdy $T(\partial U)^{\omega_{\text{std}}} = T(\partial U)^{\varphi^*\omega}$.

The second part is basically identical to the usual Moser argument.

Lemma 1: (Fiber Integration With Boundary) Let $X$ be a compact manifold with boundary, $\pi:E \to X$ be a rank $k$ vector-bundle with metric and $\pi:U \to X$ be the closed disk bundle of $E$. Let $\kappa \subset T(\partial U)$ be a distribution on $\partial U$ invariant under fiber-wise scaling. Finally, suppose that $\tau \in \Omega^{k+1}(U)$ is a $(k+1)$-form such that: \begin{equation} \label{eqn:fiber_integration_sigma} d\tau = 0 \qquad \tau|_X = 0 \qquad (\iota^*_{\partial X}\tau)|_\kappa = 0\end{equation}

Then there exists a $k$-form $\sigma \in \Omega^k(U)$ with the following properties. \begin{equation} \label{eqn:fiber_integration_tau} d\sigma = \tau \qquad \sigma|_X = 0 \qquad (\iota^*_{\partial X}\sigma)|_\kappa = 0 \end{equation}

The proof of Lemma 1 just involves examining the proof of the version of the Poincare Lemma in McDuff-Salamon, and checking that the primitive constructed there satisfies the 3rd property. Note that to apply this lemma, you need to show that the characteristic foliation of $\partial(T^*L) \subset T^*L$ is invariant under fiber-scaling, but this is a quite easy Lemma.

Lemma 2: Let $U$ be a manifold and $L \subset U$ be a closed sub-manifold. Let $\kappa_0,\kappa_1$ be rank $1$ orientable distributions in $TU$ such that $\kappa_i|_L \cap TL = \{0\}$ and $\kappa_0|_L = \kappa_1|_L$.

Then there exists a neighborhood $U' \subset U$ of $L$ and a family of smooth embeddings $\psi:U' s\times I \to U$ with the following four properties. $$ \psi_t|_{\partial L} = \text{Id} \qquad d(\psi_t)_u = \text{Id} \text{ for }u \in L \qquad \psi_0 = \text{Id} \qquad [\psi_1]_*(\kappa_0) = \kappa_1 $$ Furthermore, we can take $\psi_t$ to be $t$-independent for $t$ near $0$ and $1$.

The proof of Lemma 2 is straight-forward.

  • $\begingroup$ It seems from the proof that we don't even need the submanifold to be Lagrangian. In particular, if we consider generally a coisotropic submanifold with boundary that is transverse to the Reeb foliation(for instance, if it's a hypersurface such that the boundary is convex), then we will have a version of Weinstein neighbourhood theorem. $\endgroup$
    – TheWildCat
    Commented Aug 23, 2023 at 15:59

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