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

Hallo,

It is a known fact that any real-analytic Riemannian manifold $M$ admits a isometric embedding in a Kähler manifold $\Omega$, where $M$ is totally real in $\Omega$. Of $\Omega$ can be taught of as some open neighbourhood of the zero section of the cotangent bundle $T^{*}M$. This complex manifold is far from being compact. My question is now: Can one choose the complex manifold $\Omega$ not to be just some open neighbourhood of $M$ but to be a compact Kähler manifold? In other words: Can one embed any real-analytic compact Riemannian manifold isometrically in a compact Kähler manifold, such that the real-analytic Riemannian manifold is totally real in the Kähler manifold? Are there any references on this topic? What furter assumptions does one need in order that this works? Or is it entirely impossible? If so, why? By compact I mean compact without boundary!

hapchiu

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

I think the answer to your question should be positive and below is a sketch of what should work (I think).

Any real analytic manifold can be realised as the real part of a complex projective manifold. I.e. one should embedd the real analytic manifold smoothly in $\mathbb R^n$ and then approximate by a real algebraic submanifold, whose complexification will be the Kahler manifold you are thinking about. Now, the emdedding can be done almost isometrically. I think, using this one should be able to show that the standard (say Fubini-Study) metric on the complexification can be adjusted a bit to make the embedding in the complexification an isometry.

share|improve this answer
    
Thanks for the answer Dmitri. Is there any reference on this? –  hapchiu Nov 29 '12 at 9:12
    
Hapchiu, I don't know a reference that would tell all this. The fact that every Riemannian manifold can be embedded isometrically in $\mathbb R^n$ of some dimension is I guess due to Nash. The fact that you can get your real analytic guy as a real part of complex projective manifold is something like "Weierstrass approximation". Finally the fact that you can perturb the metric to make everything an isometry this is my guess, but I don't see how this guess can be wrong... –  Dmitri Nov 29 '12 at 9:21
    
If all this works is the resulting complex manifold then compact (compact without boundary)? –  hapchiu Nov 29 '12 at 9:27
    
Yes of course it will be compact without boundary. You should think of $\mathbb R^n$ as sitting in $\mathbb RP^n$ and then the complexification of an arbitrary algebraic submanifold in $\mathbb RP^n$ will be a complex projective subvariety in $\mathbb CP^n$ –  Dmitri Nov 29 '12 at 10:21
1  
Dear Ian, the real analytic structure on a smooth manifold is unique. So I am not sure if I understand your question. –  Dmitri Nov 29 '12 at 18:59
show 1 more 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.