3
$\begingroup$

How does one know in general that the toric fibers of a Kahler toric manifold are Lagrangian submanifolds? I can roughly fathom this for e.g. a circle in $\mathbb{C}P^1$, but how about more general toric manifolds of higher dimension?

$\endgroup$

1 Answer 1

3
$\begingroup$

By definition, a toric variety of complex dimension $n$ comes with an action of a complex torus $(\mathbb{C}^{*})^n$. I guess that by Kähler toric manifold, you mean a toric variety $X$ equipped with a Kähler form $\omega$ which is preserved by a compact subgroup $U(1)^n$ of $(\mathbb{C}^{*})^n$. The toric fibration $X \rightarrow \mathbb{R}^n$ is then the moment map of this action of $U(1)^n$ on $(X,\omega)$. More precisely, choosing Hamiltonian functions $H_1$,...,$H_n$ generating the action of $U(1)^n$ (i.e. such that the corresponding Hamiltonian vector fields generate the action of $U(1)^n$), then the toric fibration $X \rightarrow \mathbb{R}^n$ is given by $x \mapsto (H_1(x),...,H_n(x))$. Because $U(1)^n$ is abelian, the vector fields generating the action commute and so the Hamiltonian functions Poisson commute. It is a general fact that if one has a symplectic manifold of real dimension $2n$, with $n$ functions $H_1,...,H_n$ Poisson commuting (and with linearly independent differentials), then the common vanishing locus of these functions is a Lagrangian submanifold (for example, by Darboux lemma, you can locally complete these $n$ functions by $n$ other functions $\theta_1$,...,$\theta_n$ such that $H_i$, $\theta_j$ are local coorrdinates such that $\omega=\sum_{i=1}^n dH_i \wedge d\theta_i$ and the result is then obvious).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.