# Half-dimensional torus fibration vs Lagrangian torus fibration

Assume we have a closed symplectic manifold $$M$$ which is the total space of a smooth fibration by half-dimensional tori. Can we infer that $$M$$ is the total space of a smooth fibration by Lagrangian tori?

This doesn't need to hold. For example, if one takes a $$(T^4,\omega)$$ with a constant symplectic structure $$\omega$$, in order for it to have a fibration by Lagrangian tori one should be able to find a homologically non-trivial $$T^2\subset T^4$$ such that $$\int_{\omega} T^2=0$$ which is impossible for general $$\omega$$.
One can also give counter-examples when a half-dimensional torus bundle exists on the symplectic manifold but no Lagrangian torus bundle exists for any symplectic structure on the manifold. To construct such an example consider $$M^4$$ that fibers over $$B=T^2$$ with a fiber $$F=T^2$$ such that the action of $$\mathbb Z^2=\pi_1(B)$$ on $$H_1(F)=\mathbb Z^2$$ contains a hyperbolic element. Such a manifold is symplectic by Thurston, but it is not a total space of a Lagrangian torus fibration by the classification of such fibrations in dimension $$4$$.
Just to add, that the classification of Lagrangian $$T^2$$ fibrations was done in https://ac.els-cdn.com/S0926224596000241/1-s2.0-S0926224596000241-main.pdf?_tid=824690ea-affb-4cd2-98a6-6d7ed7f08d9f&acdnat=1544694026_3c6783fa389437767cd9c8246616d4dd
And there is as well a very nice paper classifying Lagrangian $$T^2$$ fibrations over the Klein bottle. As you can see from Theorem 2.1 stated in the paper, the representation of $$\pi_1(B)$$ in $$H^1(F)=\mathbb R^2$$ (for Lagragian $$T^2$$ fibrations over $$T^2$$) is unipotent. https://arxiv.org/pdf/0909.2982.pdf