Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
    
Asking for isometry is too much, A better question is if it can be embedded as a smooth manifold. –  Mohammad F. Tehrani Mar 10 '12 at 15:23
    
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 ? –  user22057 Mar 11 '12 at 7:53
    
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 –  hapchiu Mar 12 '12 at 9:49
add comment

2 Answers

  1. 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".

  2. 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".

  3. 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.

share|improve this answer
    
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? –  user22057 Mar 10 '12 at 16:22
    
maybe in higher dimensions ??? –  user22057 Mar 10 '12 at 16:23
    
Do you want complete CY manifold ? –  BS. Mar 10 '12 at 17:07
    
or eaven in dimension 2 for the 2-sphere ? –  user22057 Mar 10 '12 at 17:09
    
no not necessary, like in bryant's paper like a "germ" around the real manifold. is this possible for the 2-sphere, for example ??? –  user22057 Mar 10 '12 at 17:11
show 2 more comments

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).

share|improve this answer
1  
Only if you consider noncompact Calabi-Yau manifolds; see page 3. The real question is about compact Calabi-Yau manifolds. –  Ben McKay Mar 10 '12 at 12:00
    
I agree this is the real question, but since the OP didn't mention compactness I assumed he didn't know R. Bryant's result. –  BS. Mar 10 '12 at 16:19
    
And I realize that even dimension 3 isn't in the OP (nor completeness of the CY, by the way). –  BS. Mar 10 '12 at 17:04
add comment

Your Answer

 
discard

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.