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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 $\mathbb{R}\times Y\to Y$.)

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    $\begingroup$ 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). $\endgroup$
    – Oldřich Spáčil
    Commented Sep 10, 2013 at 17:13
  • $\begingroup$ 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 $T^{*}(K\times \mathbb{R})$. $\endgroup$
    – Oldřich Spáčil
    Commented Sep 10, 2013 at 17:23
  • $\begingroup$ 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). $\endgroup$
    – Chris Gerig
    Commented Sep 10, 2013 at 20:10
  • $\begingroup$ 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. $\endgroup$
    – Oldřich Spáčil
    Commented Sep 10, 2013 at 20:58

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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!

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    $\begingroup$ 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. $\endgroup$
    – Oldřich Spáčil
    Commented Sep 10, 2013 at 17:16
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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.

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  • $\begingroup$ 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). $\endgroup$
    – Giovanni Moreno
    Commented Sep 10, 2013 at 9:48
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    $\begingroup$ 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. $\endgroup$
    – Oldřich Spáčil
    Commented Sep 10, 2013 at 17:07

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