Lagrangian fibrations on hyperkähler manifolds -- question of Beauville (Q2854290)

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scientific article; zbMATH DE number 6216389
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Lagrangian fibrations on hyperkähler manifolds -- question of Beauville
scientific article; zbMATH DE number 6216389

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    18 October 2013
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    hyperkähler manifolds
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    Lagrangian fibration
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    cycle spaces
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    deformation of pairs
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    math.AG
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    math.CV
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    math.DG
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    Lagrangian fibrations on hyperkähler manifolds -- question of Beauville (English)
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    The article under reviewed is concerned with the geometry of (irreducible) hyperkähler manifolds (i.e. compact simply-connected Kähler manifolds whose space of global holomorphic 2-forms is generated by a symplectic form) and in particular with the following question raised by \textit{A. Beauville} [Springer Proc. Math. 8, 49--63 (2011; Zbl 1231.32012)]. NEWLINENEWLINENEWLINE \textit{Question:} If \(X\) is a hyperkähler manifold and \(L\subset X\) a Lagrangian torus, is \(L\) a fibre of an (almost holomorphic) Lagrangian fibration \(f:X\rightarrow B\)?NEWLINENEWLINENEWLINENEWLINE A submanifold \(L\) of \(X\) is said to be Lagrangian if \(\mathrm{dim}(X)=2\,\mathrm{dim}(L)\) and if the restriction of the symplectic form to \(L\) vanishes identically.\newline The main result of this paper is a positive answer to the latter question in non-projective case (building on the earlier work [\textit{F. Campana, K. Oguiso} and \textit{T. Peternell}, J. Differ. Geom. 85, No. 3, 397--424 (2010; Zbl 1232.53042)]).NEWLINENEWLINENEWLINENEWLINE Theorem. (Th. 4.1) If \(X\) is a non-projective hyperkähler manifold and \(L\) a Lagrangian torus, then \(X\) has algebraic dimension \(n\) and \(L\) is a fibre of some algebraic reduction \(f:X\rightarrow B\).NEWLINENEWLINENEWLINENEWLINE The main tools used to prove this result are infinitesial computations in the cycle space (Barlet space) and a criterion of non-simplicity (in addition to the results of Campana et al. [loc. cit.]. In the projective case, the natural idea is to try to deform the pair \((X,L)\) to a non-projective one. This is easily achieved when the absolute situation is considered but the presence of the submanifold \(L\) makes the analysis harder. However the authors are able to derive the following criterion.NEWLINENEWLINENEWLINENEWLINE Theorem. (Th. 5.3) Let \(X\) be a hyperkähler manifold and \(L\) a Lagrangian subtorus. Then \(L\) is a fiber of a Lagrangian fibration if and only if there exists an effective divisor \(D\) in \(X\) such that \(c_1(\mathcal{O}_X(D))\) belongs to NEWLINE\[NEWLINE\mathrm{Ker}\big(\mathrm{H}^{1,1}(X,\mathbb{R})\longrightarrow\mathrm{H}^{1,1}(L,\mathbb{R})\big).NEWLINE\]NEWLINE It is worth noting here that this last assumption has been recently proved in [\textit{J.-M. Hwang} and \textit{R. Weiss}, Invent. Math. 192, No. 1, 83--109 (2013; Zbl 1276.14059)], then providing us with a complete and positive answer to the question of Beauville.\newline The final section is devoted to proving the existence of a nice model of the Lagrangian fibration (when it exists). The abutment of a carefully chosen MMP with scaling is actually a smooth hyperkähler manifold endowed with a holomorphic model of the Lagrangian fibration.
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