Let $(M,\omega)$ be a manifold endowed with symplectic form. Then the product manifold $M\times M$ with symplectic form $\omega\times -\omega$ is symplectic, and the diagonal submanifold $\Delta\equiv\{(x,x)|x\in M\}$ is Lagrangian. Does the product manifold in fact admit a Lagrangian fibration over $\Delta$?



Does $(M\times M, \omega\times -\omega)$ admit a Lagrangian fibration?
No (independent of the full meaning of the question) in general it does not.


Any manifold which admits a Lagrangian fibration to a half dimensional base, also admits a Lagrangian foliation, and hence admits a global continuous choice of Lagrangian at each point. 


If this were the case then it means that the classifying map of the tangent bundle (with an almost complex structure) is defined as the tensor product of $\mathbb{C}$ with a real vector bundle.


More, precisely; the classifying map of the tangent bundle of $M$ is a map $M \to BO(2n)$, this is lifted (canonically up to homotopy) to a map $M\to BU(n)$ by choosing an almost complex structure, and having a Lagrangian fibration implies that this lifts to $BO(n)\to BU(n)$. Since the composed map $BO(n) \to BO(2n)$ is simply direct sum with itself (or multiplication with 2) we can say something about the Stiefel-Whitney classes of the image.


This leads to the following counter example: $M=\mathbb{C} P^2$. Indeed, $M$ is not spinable because its second Stiefel-Whitney class is non-zero. Hence so is true for $M\times M$, now we see that $T(M\times M)$ is never 2 times a real half dimensional vector bundle. Indeed, to get the non-zero second Stiefel-Whitney class you need a non-zero first Stiefel Whitney class on the bundle that you double, and this can't happen since $M\times M$ is simply connected.


