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Schoen's Calabi-Yau 3-fold is the fiber product $X=Y_1\times_{\mathbb{P}^1}Y_2$ of two rational elliptic surfaces $Y_1\rightarrow\mathbb{P}^1$ and $Y_2\rightarrow\mathbb{P}^1$ with $\chi(X)=0$ and $h^{1,1}(X)=h^{1,2}(X)=19$, see: http://link.springer.com/article/10.1007%2FBF01215188#page-1.

It is claimed by Kovalev many years ago that a special Lagrangian torus fibration on $X$ can be constructed by decomposing $X$ into two pieces, doing constructions separately, and then gluing them together. He also did a similar construction: http://arxiv.org/pdf/math/0511150.pdf, which is a coassociative $K3$ fibration on a $G_2$ manifold.

The work of Gross via toric degeneration (http://arxiv.org/abs/math/0406171) shows that the discriminant $\Delta$ of such a special Lagrangian torus fibration $f:X\rightarrow S^3$ should be a disjoint union of 24 circles. If we treat two sets of 12 parallel circles respectively as one single circle, then this looks like a Hopf link.

My question is how to explicitly construct such a Lagrangian torus fibration?

We may use the method of Bernard-Matessi (http://arxiv.org/abs/math/0611139) to glue singular fibers of a generic local model over $S^3\setminus\Delta$, but then the resulting total space $X'\rightarrow S^3$ is only homeomorphic to $X$. There should be an explicit construction for such a Lagrangian fibration as all the singular Lagrangian fibers are expected to be generic (locally $X$ just looks like $T^\ast S^3$), but I can't find any reference.

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  • $\begingroup$ There is not much known about the existence of smooth Special Lagrangian fibrations. See 2013 or 2008 survey of Gross on SYZ. $\endgroup$ Sep 24, 2014 at 2:46
  • $\begingroup$ @MohammadF.Tehrani I'm not asking about special Lag fibrations, I only seek for Lagrangian fibration with good behavior. There is no need for the fibration to be smooth, of course when the Lagrangian fibration is complicated enough, then that of its mirror is generally believed to be only piecewise smooth. $\endgroup$
    – YHBKJ
    Sep 24, 2014 at 15:53

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I realise this is an old question (and by now you may already know the answer) but here's a way I think you can construct this fibration.

Suppose that $E\stackrel{f}{\to}\mathbf{P}^1$ and $E'\stackrel{f'}{\to}\mathbf{P}^1$ are your elliptically fibred rational surfaces. In other words, they're blow ups of the plane at the nine basepoints of a cubic pencil. Parametrise the base of $f$ so that $\infty$ is not one of the (twelve) critical points. Let $\phi_t\colon\mathbf{P}^1\to\mathbf{P}^1$ be the map $z\mapsto tz$. Pullback $f$ along $\phi_t$ to get a new elliptically fibred surface $E_t\stackrel{f_t}{\to}\mathbf{P}^1$ whose critical points have been squished towards zero (if $|t|<1$).

The family of 3-folds $E_t\times_{\mathbf{P}^1} E'$ will now have a limit point as $t\to 0$ which I claim will be the union of $E\times e'$ and $e\times E'$ (where $e'=(f')^{-1}(0)$ and $e=f^{-1}(\infty)$) along the complex torus $e\times e'$. To see this, just think about the family of projection maps to $\mathbf{P}^1$: in the limit the base should degenerate to a chain of two $\mathbf{P}^1$s glued together at a point; the critical points for $f$ should end up in one $\mathbf{P}^1$ and for $f'$ should end up in the other.

Now observe that $E$ and $E'$ individually admit almost toric fibrations $L\colon E\to B$ and $L'\colon E'\to B'$ over the disc with 12 focus-focus fibres each (make three nodal trades to the standard moment triangle of $\mathbf{CP}^2$ and then perform three blow-ups at interior points of each of the three edges; each of these operations introduces a nodal fibre) and where the elliptic curve $e$ (respectively $e'$) is the preimage of the almost toric boundary. The bases of these almost toric fibrations are discs, and when you multiply $E$ with $e'$ and take the product of the almost toric fibration $L$ with $L'|_{e'}$ you get a Lagrangian torus fibration of $E\times e'$ over the solid torus (disc times circle); similarly for the other piece. This corresponds to the usual Heegaard decomposition of the sphere because you're matching the fibres up so that $L(e)$ is the meridian (boundary of a disc) when considered in one solid torus and is the longitude when considered in the other. The discriminant locus is what you expect because it's just (12 points in the disc) times $S^1$ and $S^1$ times (12 points in the disc).

Now, following Ruan, take the composition of the symplectic parallel transport in the degenerating family of varieties with this Lagrangian torus fibration on the (normal crossing) singular fibre and you get a Lagrangian torus fibration on the smooth guy, which is the Schoen 3-fold.

Incidentally, here there is no issue with smoothness or need to appeal to Castano-Bernard--Matessi because there are no negative vertices (or indeed any vertices) in the discriminant locus.

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