MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given a Lagrangian submanifold $L\subset(M,\omega)$ of a symplectic manifold, we have Alan Weinstein's celebrated Lagrangian tubular neighborhood theorem. I now look for the analog on Legendrian submanifolds $K\subset(Y,\lambda)$ of contact manifolds. As the Lagrangian neighborhood construction essentially relies on the Moser method, I don't think it'd be too hard to build a Legendrian neighborhood by using the analogous Gray stability.

However, I am specifically questioning whether I can get a Legendrian neighborhood theorem directly from the Lagrangian neighborhood theorem: By passing to the symplectization $(\mathbb{R}\times Y,d(e^t\lambda)$), a Legendrian submanifold $K$ becomes a Lagrangian submanifold $\mathbb{R}\times K$. I would love to project some Lagrangian tubular neighborhood down into $Y$ to get a desired neighborhood of $K$, but can this actually be done?

(I spoke with Alan, and he said this might be achieved somehow by viewing all of our constructions equivariantly using the translation $\mathbb{R}$-action on our bundle $\mathbb{R}\times Y\to Y$.)

share|cite|improve this question
You're right that Moser's method works well for proving a Legendrian nbhd theorem -- a tubular neighbourhood of $K$ is contactomorphic to a neighbourhood of the zero section of the 1-jet bundle $J^{1}K$ as described in the answers below. A proof can be found in Geiges's book An Introduction to contact topology, Theorem 2.5.8 and Example 2.5.11. You can even make the contactomorphism to be strict, i.e. preserving a given contact form -- a proof of this version is in Kriegl, Michor: The convenient setting of global analysis, Proposition 43.18 (this book is available from Michor's website). – Oldřich Spáčil Sep 10 '13 at 17:13
Also, my feeling is that you'd rather prove the Lagrangian nbhd theorem from the Legendrian one since you supposedly start with a contact manifold and its Legendrian and then consider the symplectization? If I'm not mistaken, the symplectization of $J^{1}K$ is symplectomorphic to $T^{*}(K\times \mathbb{R})$. – Oldřich Spáčil Sep 10 '13 at 17:23
Sorry, my question is focusing on the possibility to pass from Lagrangian neighborhood to Legendrian neighborhood (whereas the answers thus far are only showing me how to get a Legendrian neighborhood). – Chris Gerig Sep 10 '13 at 20:10
Yes, I do understand you're interested in the Lag -> Leg direction... In my second comment I just tried to point out that the other direction seems easier to me, especially if in your setup you start with a Legendrian submanifold and only then consider symplectizations etc. Otherwise I can't come up with anything better than Alan Weinstein's remark on the $\mathbb{R}$-equivariancy. – Oldřich Spáčil Sep 10 '13 at 20:58

Basically, Weinstein's theorem says that you can embed $T^*L$ into $M$ like that: $$ T^*L\cong NL\cong \mathcal{T}_L\subseteq M, $$ where $NL$ is the normal bundle and $\mathcal{T}_L$ is a tubular neighborhood, in such a way that the canonical symplectic form on $T^*L$ is the pull-back of $\omega$. So, the "contact counterpart" of above chain of identifications should read $$ J^1K\cong NK\cong \mathcal{T}_K\subseteq Y $$ where now $J^1K$ is the first-order jet bundle of (smooth) functions on $K$. Notice that all manifolds appearing above are $(2n+1)$-dimensional, and all bundles are over $K$ with $(n+1)$-dimensional fibers. I'm sure this observation can be found in the literature about jet spaces and/or contact manifolds.

My guess is that the above embedding, at least locally, can be always found. Concerning the real question, i.e., whether the canonical contact form on $J^1K$ is the pull-back of $\lambda$, I'm not able to answer, but I feel it is not true!

share|cite|improve this answer
Yes, you can make the contactomorphism between a tubular nbhd of $K$ and the zero section of $J^{1}K$ preserve the contact forms, i.e. the jet contact form and your preferred contact form on your contact manifold. See Kriegl, Michor: The convenient setting of global analysis, Proposition 43.18. – Oldřich Spáčil Sep 10 '13 at 17:16

The closest analog I am aware of to the Weinstein model for Lagrangian submanifolds in the contact setting is the so-called "Lychagin chart:" Every Legendrian submanifold $L$ has a neighborhood contactomorphic to a neighborhood of the zero section of the space of one-jets on $L$. See Banyaga, "The Structure of Classical Diffeomorphism Groups," Section 6.2, also Lychagin, "Local classification of nonlinear first-order partial differential equations," Russ. Math. Surv. 39 (1975), 105-175.

share|cite|improve this answer
You just confirmed what I have guessed in my answer below: but we still do not know whether the contact structure on this neighborhood of $L$ is the restriction of the contact structure on the environment (called $(Y,\lambda)$ in the question). – Giovanni Moreno Sep 10 '13 at 9:48
Just to clarify this a bit, a Lychagin chart is a (smooth) chart on the group of contactomorphisms, while what you have described is the Legendrian neighbourhood theorem. – Oldřich Spáčil Sep 10 '13 at 17:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.