# Why is the base of SLAG fibration of CY3 expected to be $S^3$?

The SYZ conjecture roughly says that any Calabi-Yau threefold $X$ has a special Lagrnagian fibration $\pi:X\rightarrow B$ by 3-tous with section and one of its mirror partners $\check{X}$ is obtained by dualizing each smooth fiber.

I often come across a statement claiming that the base $B$ is expected to be a 3-sphere $S^3$. If I remember correctly, one of explanation is given by homological mirror symmetry conjecture. Could anyone give me a reference for this or kindly explain why this is true?

I am a math graduate student with some background in some algebraic and symplectic geometry. Thank you.

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If your CY manifold is simply connected, the base of the torus-fibration will have to be simply connected too, since a homotopically non-trivial loop downstairs would lift to a loop upstairs which does not bound a disc. In 3 dimensions, that's the end of the story by the Poincaré conjecture.

I'll try to explain via homological mirror symmetry (HMS) why, even in higher dimensions, the base of the SYZ fibration should be a rational homology-sphere. This only applies to "strict" CY manifolds.

Say we have a special Lagrangian torus-fibration $\check{X}\to B$, and we would like to understand the Fukaya category as the derived category of a mirror $X$, defined over some field $K$ of characteristic zero (depending on the formulation of the Fukaya category, $K$ might be the field of rational or complex Novikov series; optimists think that $\mathbb{C}$ could also be a possibility for $K$).

A basic aspect of HMS is the prediction that the mirror to a smooth torus-fiber $F_b$ will be a skyscraper sheaf $\mathcal{O}_{X,x}$ on $X$. That prediction gives rise to another: that the mirror $L$ to the structure sheaf $\mathcal{O}_X$ should be a Lagrangian section of $\check{X}\to B$. The reason is that $\mathrm{Ext}^\ast(\mathcal{O}_X,\mathcal{O}_{X,x})=H^\ast(\mathcal{O}_{X,x})=K$, so by HMS one should have $HF(L,F_b)=K$ for each fibre $F_b$. Taking Euler characteristics of the latter isomorphism, we get $[L]\cdot [F_b]=1$. So $L$ is at least a homology-section, and we guess that it should be a true section. In particular, $H^\ast(L;K)=H^\ast(B;K)$.

By a "strict" CY $n$-manifold I mean that as well as trivial canonical bundle, one has $H^i(\mathcal{O}_X)=0$ for $0<i<n$. (In the setting of complex manifolds, this means that the holonomy is exactly $SU(n)$.) By Serre duality, $H^n(\mathcal{O}_X)=K$. Hence $\mathrm{Ext}^*(\mathcal{O}_X,\mathcal{O}_X)= H^\ast(\mathcal{O}_X)$ is isomorphic as a graded $K$-algebra to $H^\ast(S^n;K)$. On the other hand, $\mathrm{Ext}^*(\mathcal{O}_X,\mathcal{O}_X)\cong HF(L,L)$ by HMS. One makes the reasonable guess that $HF(L,L)\cong H^\ast(L;K)$, and infers that $H^*(B;K) = H^*(L;K)\cong H^*(S^n;K)$.

Edit: Ah, I think we don't need to guess at the last stage! The DGA of cochains computing $\mathrm{Ext}^*(\mathcal{O}_X,\mathcal{O}_X)$ is formal - over $\mathbb{C}$, we get that from Deligne-Griffiths-Morgan-Sullivan plus Hodge by using a Dolbeault model. By HMS, $CF(L,L)$ is then formal as an $A_\infty$-algebra. The Oh spectral sequence $H^\ast(L;K) \Rightarrow HF(L,L)$ must then surely degenerate at $E_1$, so $H^\ast(L)\cong HF(L,L)$.

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Not convinced by your first paragraph. Why couldn't the long exact sequence of homotopy groups go $0=\pi_1(X) \to \pi_1(B) \to \pi_2(F) = \mathbb{Z}^{\binom{n}{2}}$ with $\pi_1(B)$ nontrivial? Some nice intuition in the later part, thank you! – David Speyer Sep 7 '12 at 6:42
David: thanks for the comments. In the first para, the LES (to the extent that it applies here) sandwiches $\pi_1(B)$ between $\pi_1(X)$ and $\pi_0(F)$. – Tim Perutz Sep 7 '12 at 13:39
You conclude in your first paragraph that in $3d$ the base should be simply connected, and so a homotopy $S^3$. Is there a way to get $S^3$ itself without appealing to Perelman-Poincare? – Chris Brav Sep 7 '12 at 14:02
Chris: I don't know another way. However, $B$ does carry a singular affine structure induced by the torus-fibration, and if the singularities are sufficiently nice that ought to be a strong constraint. – Tim Perutz Sep 7 '12 at 14:11
Incidentally, can anyboydy think of a simply connected rational homology $n$-sphere that's not a homotopy-sphere? – Tim Perutz Sep 7 '12 at 14:11

There are topological conditions for a manifold to be a base of lagrangian fibration.

For example, in the real 4 dimensional case, if you want a smooth lagrangian fibration without singular fiber, then the base have to be the unique integral affine surface, that is T^2. If you allow certain singular fibres, then the base has to be some special ones, here is a good reference: http://arxiv.org/abs/math/0312165

Similarly, in the real 6 dimensional case, there are affine structure condition on the base.

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