example of special lagrangian submanifold

2012-03-10T09:48:34Z

are there any examples of a real analytic riemannian manifold that cannot be isometrically embedded as a special lagrangian submanifold of a calabi-yau manifold ?

peter hara

Answer by BS for example of special lagrangian submanifold

2012-03-10T11:09:40Z

On the contrary, R. Bryant has shown that any closed oriented real analytic 3-dimensional riemannian manifold is the real locus of an antiholomorphic, isometric involution of a Calabi-Yau 3-fold (see http://arxiv.org/abs/math/9912246).

Answer by Robert Bryant for example of special lagrangian submanifold

2012-03-10T14:43:56Z

If the question is "Are there examples of compact real-analytic Riemannian manifolds that cannot be isometrically embedded as a special Lagrangian submanifold of a compact Calabi-Yau manifold?", then the answer is "yes".
If the question is "Are there known, explicit examples of compact real-analytic Riemannian manifolds that cannot be isometrically embedded as a special Lagrangian submanifold of a compact Calabi-Yau manifold?", then the answer is "probably".
If the question is "Are there known, explicit examples of compact real-analytic Riemannian manifolds for which a proof is known that they cannot be isometrically embedded as a special Lagrangian submanifold of a compact Calabi-Yau manifold?", then the answer is "no" (to my knowledge).

For the first question, just note that, already for dimension 2, the space of compact Calabi-Yau surfaces is a finite-dimensional space, and the metrics that can be realized on compact complex curves in such a Calabi-Yau fall into a countable union of finite dimensional families. (Remember that special Lagrangian surfaces in a Calabi-Yau are complex curves in a different Calabi-Yau metric in the canonical $S^2$-family of Calabi-Yau metrics.) Thus, the set of such realizable metrics, even on the $2$-sphere, constitutes a countable union of finite dimensional families. This could never account for all of the real-analytic metrics on the $2$-sphere. Thus, some example exists, though we don't know one explicitly.

For the second question, consider the fact that it is highly unlikely that the induced metric on any complex curve in a Calabi-Yau surface has constant Gaussian curvature. The 'reason' is that most (non-flat) Ricci-flat Kahler metrics contain no complex curves with constant Gaussian curvature. It would be remarkable indeed if one of the Ricci-flat Kahler metrics on a (non-flat) compact 4-manifold had such a curve. In particular, I regard it as highly likely that the standard round metric on the $2$-sphere cannot be isometrically embedded as a complex curve in any compact Calabi-Yau surface.

My answer to the third question is just an affirmation of my ignorance.

A remark about the local story: peter h asked about what I would call the 'local case', i.e., whether a real analytic Riemannian manifold can be isometrically embedded as a special Lagrangian submanifold in some Calabi-Yau, with no assumptions about completeness of the ambient manifold. In particular, he raised the question for surfaces.

Now, in the case of a real-analytic metric on a Riemann surface, the answer would be 'yes', according to a paper in 2000 by D. Kaledin, "Hyperkaehler structures on total spaces of holomorphic cotangent bundles", which is available on the arXive (arXiv:alg-geom/9710026v1). (It's 100 pages, and I don't claim that I have read it, I'm just pointing out that it is there.) The main theorem of this paper is that, given any real-analytic Kahler manifold $M$, there exists a hyperKahler metric on a neighborhood of the $0$-section of the cotangent bundle $T^\ast M$ that is compatible with the natural complex and holomorphic structures on $T^\ast M$ and that induces the original metric on the $0$-section.

When the (real) dimension of $M$ is $2$, this would apply to show that $M$ is isometrically imbedded as a complex curve in a Calabi-Yau (complex) surface, and then one can apply the 'rotation trick' to turn this into a special Lagrangian surface when the ambient $4$-manifold is regarded as a complex surface with respect to one of the orthogonal complex structures. Thus, the case of surfaces would be covered by this theorem.

In fact, this would work in any even dimension when the given real-analytic metric is actually Kahler.

There would remain the question (which I raised in my original paper) of whether every real-analytic metric on $S^4$ can be realized by an embedding as a special Lagrangian submanifold of a $4$-dimensional Calabi-Yau.

Answer by peter h for example of special lagrangian submanifold

2012-03-10T16:22:22Z

@Robert Bryant: Actually I am considering the question: Are there examples of compact real analytic Riemannian manifolds that cannot be isometrically embedded as special lagrangian submanifolds of a (not necessary compact) Calabi-Yau manifold?

Answer by peter h for example of special lagrangian submanifold

2012-03-11T07:53:46Z

Actually I am intrested in the following: Are there examples of compact real analytic Riemannian manifolds that cannot be isometrically embedded as special lagrangian submanifolds of a Calabi-Yau manifold? Here the Calabi-Yau manifold doesn't have to be compact or complete, it should be like a "germ" around the Riemannian manifold (like in Bryant's paper). Are there counterexamples ? What about the two-sphere, that Robert Bryant mentioned ?

Answer by hapchiu for example of special lagrangian submanifold

2012-03-12T09:49:10Z

I think, eaven in this case you mentioned there counterexamples, but cannot be given explicitely. See the post of Bryant (above), part 1. Am I right?

hapchiu

example of special lagrangian submanifold - MathOverflow

example of special lagrangian submanifold

Answer by BS for example of special lagrangian submanifold

Answer by Robert Bryant for example of special lagrangian submanifold

Answer by peter h for example of special lagrangian submanifold

Answer by peter h for example of special lagrangian submanifold

Answer by hapchiu for example of special lagrangian submanifold