2D slice of a 6D Calabi-Yau manifold , and other?

[Spinnor]

"2D slice of a 6D Calabi-Yau manifold", and other?

Mathematically what does it mean to take a "2D slice of a 6D Calabi-Yau manifold"?

Part of quote taken from the top of,

http://en.wikipedia.org/wiki/Calabi–Yau_manifold

Is there a finite number of slices of a 6D Calabi-Yau manifold that in principle could define that Calabi-Yau manifold?

How are 6D Calabi-Yau manifolds classified?

Does it make sense to take some 6D Calabi-Yau manifold and distort its shape a little and still keep it the "same" topologically?

Can we distort one 6D Calabi-Yau manifold into another?

Does it make sense to say that there is a minimum number of "flat" complex dimensions a typical 6D Calabi-Yau manifold can be embedded?

Is there a 6D Calabi-Yau manifold that could be considered most simple?

What cool things "happen" when we allow for complex manifolds that don't happen for real manifolds?

In string theory do the vibrating strings interact with the Calabi-Yau manifold to change the Calabi-Yau manifold even if a little bit?

If some questions don't make sense, sorry, hopefully they can be modified so that they do make sense.

Thanks for any help!

 

[jedishrfu]

This may help with visualizing a Calabi-yau figure in 2D by specifying the projection angle of interest:

http://demonstrations.wolfram.com/CalabiYauSpace/

and here's an article that may answer some of your questions:

http://www.dummies.com/how-to/content/string-theory-and-calabiyau-manifolds.html

 

[Ben Niehoff]

Most of your questions are very vague and open-ended, and you should do some reading. However, this one is simple to answer:

Does it make sense to say that there is a minimum number of "flat" complex dimensions a typical 6D Calabi-Yau manifold can be embedded?

No. In real geometry it is easy to show that every real manifold is isometrically embeddable in some $\mathbb{R}^n$. However, the analogous situation with $\mathbb{C}^n$ is not true in complex geometry. The catch is that complex embeddings have to be holomorphic, and this severely restricts the kinds of embeddings you can have.

It is true, however, that every compact complex manifold can be embedded in some $\mathbb{CP}^n$ (i.e. complex projective space).

 

[Spinnor]

Most of your questions are very vague and open-ended, and you should do some reading. However, this one is simple to answer:
No. In real geometry it is easy to show that every real manifold is isometrically embeddable in some $\mathbb{R}^n$. However, the analogous situation with $\mathbb{C}^n$ is not true in complex geometry. The catch is that complex embeddings have to be holomorphic, and this severely restricts the kinds of embeddings you can have.

It is true, however, that every compact complex manifold can be embedded in some $\mathbb{CP}^n$ (i.e. complex projective space).

That is neat, thanks! Do we know what the smallest n is in CP^n that would handle all 6D Calabi-Yau manifolds?

Will do some reading.

 

[Spinnor]

This may help with visualizing a Calabi-yau figure in 2D by specifying the projection angle of interest:

http://demonstrations.wolfram.com/CalabiYauSpace/

and here's an article that may answer some of your questions:

http://www.dummies.com/how-to/content/string-theory-and-calabiyau-manifolds.html

Nice demo! Amazon let's me take a pretty good look at the book,

https://www.amazon.com/dp/047046724X/?tag=pfamazon01-20

Thanks!

 

[Therodre]

It is true, however, that every compact complex manifold can be embedded in some $\mathbb{CP}^n$ (i.e. complex projective space).

Actually no, to be projective you need at least to be Kähler, and some compact complex surfaces are not even Kähler, like Hopf surface.
There are cohomological obstructions to being projective, for instance, all your H^p for p odd must be of even rank, this is not the case for Hopf surfaces, whose H^1 is easily seen to be Z.

 

[Ben Niehoff]

Actually no, to be projective you need at least to be Kähler, and some compact complex surfaces are not even Kähler, like Hopf surface.
There are cohomological obstructions to being projective, for instance, all your H^p for p odd must be of even rank, this is not the case for Hopf surfaces, whose H^1 is easily seen to be Z.

You're right. I was thinking Calabi-Yau and I wrote "complex".

 

[mathwonk]

although there exist non projective complex manifolds, if a compact complex manifold of dimension n is projective, then it embeds holomorphically in CP^(2n+1), by the usual method of projecting it down from wherever you were able to embed it. I.e.a projection map from a point off the manifold is always holomorphic, and is an embedding if it does not lie on a secant of the embedded manifold.

Since the bundle of secants has dimension 2n+1, it does not fill up CP^k when k > 2n+1, so you can always choose a point to project from to a lower dimension. In particular all complex 3 dimensional projective Calabi Yau's embed in CP^7. So the quintic threefold in CP^4 is rather special, and in some ways is one of the most simple, or at least one of the most familiar such animals. The other examples given in the wikipedia article of degree k+2 in CP^(k+1), seem also rather special, being hyper surfaces.

I cannot find many reference to 2 dimensional slices of a 6d Calabi Yau in your link. Complkex manifolds do often vary in families, in which the complex structure but not the topology varies. In fact in any smooth family of complex compact manifolds the topology is constant, so the question of "deforming" one manifold into another is one of determining how many different families of such manifolds exist, and whether the different families are somehow connected, all deep questions.

As an example, all quoin tic three folds in CP^4 form one family, obtained by varying the quintic polynomial, so all of these are deformations of one another and all are thus homeomorphic. (In fact diffeomorphic, cf "theorem of Ehresmann".)

Taking a complex 1 diml slice of an embedded cx 3diml manifold is done by intersecting it with a general linear space of complex codimension 2.