Why is the base of SLAG fibration of CY3 expected to be $S^3$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T14:21:23Z http://mathoverflow.net/feeds/question/106553 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106553/why-is-the-base-of-slag-fibration-of-cy3-expected-to-be-s3 Why is the base of SLAG fibration of CY3 expected to be $S^3$? C Chan 2012-09-06T22:10:09Z 2012-11-28T04:17:46Z <p>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. </p> <p>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? </p> <p>I am a math graduate student with some background in some algebraic and symplectic geometry. Thank you. </p> http://mathoverflow.net/questions/106553/why-is-the-base-of-slag-fibration-of-cy3-expected-to-be-s3/106569#106569 Answer by Tim Perutz for Why is the base of SLAG fibration of CY3 expected to be $S^3$? Tim Perutz 2012-09-07T04:56:07Z 2012-09-07T15:15:20Z <p>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 Poincar&eacute; conjecture.</p> <p>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.</p> <p>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 $K$). </p> <p>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 <code>$\mathrm{Ext}^\ast(\mathcal{O}_X,\mathcal{O}_{X,x})=H^\ast(\mathcal{O}_{X,x})=K$</code>, 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, $H^\ast(L;K)=H^\ast(B;K)$.</p> <p>By a "strict" CY $n$-manifold I mean that as well as trivial canonical bundle, one has $H^i(\mathcal{O}_X)=0$ for <code>$0&lt;i&lt;n$</code>. (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 <code>$H^*(B;K) = H^*(L;K)\cong H^*(S^n;K)$</code>.</p> <p><i>Edit:</i> Ah, I think we don't need to guess at the last stage! The DGA of cochains computing <code>$\mathrm{Ext}^*(\mathcal{O}_X,\mathcal{O}_X)$</code> is <i>formal</i> - 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 $H^\ast(L)\cong HF(L,L)$.</p> http://mathoverflow.net/questions/106553/why-is-the-base-of-slag-fibration-of-cy3-expected-to-be-s3/114726#114726 Answer by Jay for Why is the base of SLAG fibration of CY3 expected to be $S^3$? Jay 2012-11-28T04:17:46Z 2012-11-28T04:17:46Z <p>There are topological conditions for a manifold to be a base of lagrangian fibration.</p> <p>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: <a href="http://arxiv.org/abs/math/0312165" rel="nofollow">http://arxiv.org/abs/math/0312165</a></p> <p>Similarly, in the real 6 dimensional case, there are affine structure condition on the base.</p>