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In dimensions 1 and 2 there is only one, respectively 2, compact Kaehler manifolds with zero first Chern class, up to diffeomorphism. However, it is an open problem whether or not the number of topological types of such manifolds of dimension 3 (Calabi-Yau threefolds) is bounded. I would like to ask what is known in this direction. In particular, is it known that the Euler characteristic or the total Betti number of Calabi-Yau threefolds can't be arbitrarily large? Are there any mathematical (or physical?) reasons to expect either answer?

As a side question: I remember having heard several times something like "Calabi-Yau 3-folds parametrize (some kind of) vacua in string theory" but was never able to make precise sense of this. So any comments on this point or references accessible to mathematicians would be very welcome.

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BTW, one of the principal themes of Peter Woit's blog at math.columbia.edu/~woit/wordpress is that string theory is fundamentally not capable of prediction because of the landscape. –  Steve Huntsman Jan 19 '10 at 5:43
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I seem to recall that Batyrev has done some work in these directions. This paper might be of some interest to you: arxiv.org/abs/math.AG/0505432 –  Kevin H. Lin Jan 19 '10 at 8:32
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"Complex analytic" is not a correct terminology, this just means a a manifold with holomorphic strucutre planetmath.org/encyclopedia/ComplexAnalyticSubmanifold.html In dimension 2 you have infinite number of topological types of Kodaira surfaces, they have $c_1=0$. So you should ask this question about Kahler manfiolds (or complex algebraic) instead of complex analytic –  Dmitri Jan 19 '10 at 9:18
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@Steve: most physical theories have "landscape". Quantum field theories like the ones the predictive standard model are based, have landscape: take any compact Lie group with any ad-invariant inner product, then take any unitary representation of that group,... This is already an infinite choice, and nothing intrinsic to the theory will pick one. To make a choice one needs experimental input and that is the crucial difference with string theory. Woit understands this, but it is not nearly as controversial. –  José Figueroa-O'Farrill Jan 19 '10 at 9:58

3 Answers 3

up vote 28 down vote accepted

This is a very good question, and I would really love to know the answer since its current state seems to be quite obscure. Below is just a collection of remarks, surely not the full answer by any means. I would like to argue that for the moment there is no any deep mathematical reason to think that the Euler number of CY 3-folds is bounded. I don't believe either that there is any physical intuition on this matter. But there is some empirical information, and I will describe it now, starting by speaking about "how many topological types of CY manifolds we know for the moment".

As far as I understand for today the construction of Calabi-Yau 3-folds, that brought by far the largest amount of examples is the construction of Batyrev. He starts with a reflexive polytope in dimension 4, takes the corresponding toric 4-fold, takes a generic anti-canonical section and obtains this way a Calabi-Yau orbifold. There is always a crepant resolution. So you get a smooth Calabi-Yau. Reflexive polytopes in dimension 4 are classified the number is 473,800,776. I guess, this number let Miles Reid to say in his article "Updates on 3-folds" in 2002 http://arxiv.org/PS_cache/math/pdf/0206/0206157v3.pdf , page 519

"This gives some 500,000,000 families of CY 3-folds, so much more impressive than a mere infinity (see the website http://tph16.tuwien.ac.at/~kreuzer/CY/). There are certainly many more; I believe there are infinitely many families, but the contrary opinion is widespread"

A problem with the number 500,000,000 in this phrase is that it seems more related to the number of CY orbifolds, rather than to the number of CY manifolds obtained by resolving them. Namely, the singularities that appear in these CY orbifolds can be quite involved and they have a lot of resolutions (I guess at least thousands sometimes), so the meaning of 500,000,000 is not very clear here.

This summer I asked Maximillian Kreuzer (one of the persons who actually got this number 473,800,776 of polytopes), a question similar to what you ask here. And he said that he can guarantee that there exist at least 30108 topological types of CY 3-folds. Why? Because for all these examples you can calculate Hodge numbers $h^{1,1}$ and $h^{2,1}$, and you get 30108 different values. Much less that 473,800,776. As for more refined topological invariants (like multiplication in cohomology) according to him, this was not really studied, so unfortunately 30108 seems to be the maximal number guarantied for today. But I would really love to know that I am making a mistake here, and there is some other information.

Now, it seems to me that the reason, that some people say, that the Euler characteristics of CY 3-folds could be bounded is purely empirical. Namely, the search for CY 3-folds is going for 20 years already. Since then a lot of new families were found. We know that mirror symmetry started with this symmetric table of numbers "($h^{1,1}, h^{2,1})$", and the curious fact is that, according to Maximillian, what happened to this table in 20 years -- it has not got any wider in 20 years, it just got denser. The famous picture can be found on page 9 of the following notes of Dominic Joyce http://people.maths.ox.ac.uk/~joyce/SympGeom2009/SGlect13+14.pdf . So, this means that we do find new families of CY manifolds, all the time. But the values of their Hodge numbers for some reason stay in the same region. Of course this could easily mean that we are just lacking a good construction.

Final remark is that in the first version of this question it was proposed to consider complex analytic manifolds with $c_1=0$. If we don't impose condition of been Kahler, then already in complex dimension 2 there is infinite number of topological types, given by Kodaira surfaces, they are elliptic bundles over elliptic curve. In complex dimension 3 Tian have shown that for every $n>1$ there is a holomorphic structure on the connected sum of n copies of $S^3\times S^3$, with a non-vanishing holomorphic form. Surely these manifolds are non-Kahler. So if you want to speak about any finiteness, you need to discuss say, Kahler 3-folds with non-vanishing holomorphic volume form, but not all complex analytic ones.

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This is a great answer! Fascinating, even to a non-specialist like myself. Could you run it through a rudimentary spell checker? It would be much easier to read with fewer glaring typos. –  HJRW Jan 19 '10 at 19:29
    
All the typos are corrected :) At least those that my computer finds –  Dmitri Jan 19 '10 at 19:46
    
PS Dima, I just realized that you're you! :) Thanks! –  HJRW Jan 19 '10 at 19:51

The smallest known Euler characteristic of a Calabi-Yau threefold is -960. The threefold is a hypersurface in the weighted projective space P_(1,1,12,28,42).

There are some ideas from type IIA-heterotic duality suggesting that this is extremal, but isn't even a physics proof.

Note that 42 occurs as the largest possible denominator in writing 1 as the sum of Egyptian fractions 1 = 1/2+1/3+1/7+1/42.

(1) A. Degeratu, K. Wendland: Friendly giant meets pointlike instantons? On a new conjecture by John McKay http://www.opus-bayern.de/uni-augsburg/volltexte/2007/700/pdf/mpreprint_07_037.pdf

(2) Kachru and Vafa, hep-th/9505105

For 4-folds see: (3) http://arxiv.org/abs/hep-th/9701023v2

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42 is the answer to Ultimate Question of Life, the Universe and Everything, as everybody knows! There is no need to involve the egyptians :P –  Mariano Suárez-Alvarez Jan 19 '10 at 19:52

I am going to give a rough account of the physical interpretation of Calabi-Yau manifolds in string theory.

We are studying particles moving around in 10 spacetime dimensions. We want to pick these 10 dimensions so that spacetime looks realistic, so we choose our Lorentzian 10-manifold to be $R^{1,3}\times X$ where $X$ is a compact 6-manifold. If we choose $X$ to be sufficiently small then at low energies $X$ itself is not directly observable. Thus to an observer who studies the world with low energy probes spacetime will have just the familiar 4 dimensions.

Nevertheless, one cannot chose $X$ in a completely arbitrary way. Einstein's theory of general relativity tells us that the metric on $X$ carries energy measured by its Ricci curvature. In the absence of matter, that is $in \ the \ vacuum,$ the spacetime must be in a lowest energy Ricci flat configuration. So $X$ must be such that it admits a metric of vanishing Ricci curvature.

To get to the Calabi-Yau condition we need to add one more physical criterion and this is called supersymmetry. Supersymmetry is an extremely attractive conjectural symmetry of nature that says that bosons (e.g. photons), and fermions (e.g. electrons) are paired together and related. Mathematically fermions are described by spinor valued fields on spacetime, while bosons are described by functions or one-forms on spacetime. To relate these objects, and thus have a theory with supersymmetry, what is required is a covariantly constant spinor field on $X$. This constraint on $X$ reduces its holonomy from that of a general Riemanian 6-manifold, $SO(6)$, to $SU(3)$. Indeed $SO(6)$ is locally isomorphic to $SU(4)$ and the spinor representation is $4 \oplus \bar{4}$ where 4 denotes the fundamental representation of $SU(4)$. Since $X$ admits a covairiantly constant spinor its holonomy must be contained in the subgroup of $SU(4)$ that stabilizes a spinor, and it is clear form the decomposition above that this subgroup is $SU(3)$. Hence for supersymmetry, $X$ must be Calabi-Yau.

Thus we can now clearly state exacty what it is that Calabi-Yau threefolds describe physically. They are lowest energy vacuum configurations of string theory where the physics is supersymmetric. For more details one might want to take a look at the book "Mirror Symmetry" by Vafa and Zaslow.

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Thanks, Clay, this was very helpful. But are there reasons to expect finitely of infinitely many topologically distinct CY's? –  algori Apr 8 '10 at 17:18
    
As far as we know now there is no physical argument that suggests that there are finitely many Calabi-Yau n-folds. –  Clay Cordova Apr 8 '10 at 18:02

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