# Lagrangian Kleinian bottles

I remember some talks some time ago about proofs of nonexistence of Lagrangian Kleinian bottles in C^2 for the standard symplectic structure, mentioning that this were the only compact surface for which the problem were open. Question 1: Where to find references for the other surfaces? Question 2: What is the current status for Kleinian bottles? Do there exist written proofs now?

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Just googling "Lagrangian klein bottles" returns the following :

Stefan Nemirovski, Lagrangian Klein Bottles in R2n Geometric And Functional Analysis Volume 19, Number 3 (2009), 902-909

the freely accessible arxiv version is at http://arxiv.org/abs/0712.1760 abstract : "It is shown that the n-dimensional Klein bottle admits a Lagrangian embedding into R^{2n} if and only if n is odd."

There was a previous result by the same author about the $n=2$ case, but it had a flaw initially, and was completely rewritten. See http://arxiv.org/abs/math/0106122

There is also another proof of this result by V. Schevchishin http://front.math.ucdavis.edu/0707.2085

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While working on my thesis, I (by accident) constructed a Lagrangian Klein bottle in $(S^2\times S^2, \omega\oplus\omega)$. The construction works in $S^2\times D^2$ as long as the area of $D^2$ is strictly greater than half the area of $S^2$. The method is to pick a Hamiltonian on $S^2$ that rotates $S^2$ halfway around, and realize it as the monodromy around $\partial D^2$, of a symplectic fibration over $D^2$ with fiber $S^2$. There is a circle that goes through the two fixed points of the rotation, which then traces out the Klein bottle. See McDuff's Geometric Variants of the Hofer Norm'' for the precise version of how to realize such a monodromy and the constraints: the area of the base is an easy way to think of it, but it's really a ratio of volume to fiber area.

As I understand it the paper by Nemirovskii called homology class of a Lagrangian Klein bottle'', is flawed, because it predates the accepted proofs of non-existence as described by others. I would love to know if its main claim is true, that a Lagrangian Klein bottle in a symplectic 4-manifold must realize a nontrivial second homology class with mod 2 coefficients. This is true for the example above. (Might need to ask this as a question.)

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May be this is relevant:

http://arxiv.org/abs/0712.1760

Lagrangian Klein bottles in R^{2n}

Stefan Nemirovski

It is shown that the n-dimensional Klein bottle admits a Lagrangian embedding into R^{2n} if and only if n is odd.

Comments: V.2 - explicit formula for the Luttinger-type surgery; 6 pages

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Just to explicitly answer the first part of your question, the original version of Nemirovski's first paper (http://arxiv.org/abs/math/0106122v1) surveys what is known about the other surfaces. Namely:

• You can easily deduce that orientable surfaces must be tori: pick a compatible almost complex structure J. Applying J to a vector tangent to the Lagrangian will get you a normal vector, hence the normal bundle and the tangent bundle are isomorphic and because the Lagrangian is nullhomologous its normal bundle has vanishing Euler characteristic (the Euler characteristic equals the self-intersection, which is where we use orientability). Hence it's a torus. You usually think of the cotangent and normal bundles as being isomorphic because of Weinstein's neighbourhood theorem, but of course the tangent and cotangent bundles are musically isomorphic via the metric obtained from the symplectic form and J.

• For nonorientable Lagrangians a formula of Audin from this paper: M. Audin, "Quelques remarques sur les surfaces lagrangiennes de Givental", Journal of Geometry and Physics 7, 1990 p.583--598 tells you that the Euler characteristic must be divisible by 4. You can also deduce the orientable case from this formula.

• In this paper: A. B. Givental, "Lagrangian imbeddings of surfaces and unfolded Whitney umbrella", Funkts. Anal. Prilozh., 20:3 (1986), 35–41, Givental constructs examples of nonorientable surfaces when the Euler characteristic is -4k for k at least 1.

That leaves k=0 for Nemirovski/Schevchishin.

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