I understand this statement from the physics side. Consider an $n-$dimensional manifold $\cal M$ ("configuration space") and its cotangent bundle ${\cal P} = T^*\cal M$ ("phase space"), a symplectic manifold. Now consider a scalar density $f:{\cal P}\to\mathbb{R}$ ("phase space distribution function").

An operation which a physicist is interested in is going from the phase space distribution function to the configuration space distribution function. Let's choose coordinates $q^i:{\cal M}\to \mathbb{R}$, $i=1\ldots n$ for ${\cal M}$, and in the usual way, take the canonical coordinates $p_i:{\cal P}\to \mathbb{R}$ so the collection of $q$'s (lifted) and $p$'s are coordinates for ${\cal P}$. Then we define the configuration space distribution function as $$ F(q) \equiv \int_{\pi^{-1}(q)} d^n p \ \ f(q,p) \,. \qquad\qquad (1) $$ This yields a scalar density of weight 1 on ${\cal M}$.

Above I have done everything in terms of specific coordinate functions, but I feel that it must be possible to state this all geometrically. Over each point $q\in{\cal M}$ there is a fiber $\pi^{-1}(q)$, which is a Lagrangian submanifold ($\cal P$ has a natural Lagrangian fibration).

Question 1: Is there a natural volume form for this fiber? What is the geometric way to define the integral in (1)?

Question 2: Is this possible only for the cotangent bundle of some manifold, or for any symplectic manifold which admits a Lagrangian fibration?

EDIT: As best as I can tell, $d^np$ is best defined as $$ d^np \equiv \iota_{e_1}\iota_{e_2}\cdots\iota_{e_n}\mathrm{dVol} $$ where $e_i\equiv \partial/\partial q^i$, and where $$ \mathrm{dVol} \equiv \omega^{\wedge n}\\ = dq^1 \wedge \cdots \wedge dq^n \wedge dp^1 \wedge \cdots \wedge dp^n $$ the symplectic 2-form wedged with itself $n$ times to make the top-rank volume form. It's still not clear to me how to define this all in a geometric way without regard to coordinates, or at least to show that it is coordinate independent; and if this can be done on any symplectic manifold with Lagrangian fibration.

  • $\begingroup$ You can consider the 1-dimensional case, when symplectic form coincides with the volume form. The thing you obtained by integrating along the fibre is a section of the dual bundle of the density bundle of the base. $\endgroup$
    – Guangbo Xu
    Nov 22, 2014 at 0:13
  • $\begingroup$ @GuangboXu, why do you say it's a section of the dual density bundle, and not just a section of the density bundle? And do you know if this is specific to the cotangent bundle of a manifold, or any symplectic manifold admitting a Lagrangian fibration? $\endgroup$ Nov 22, 2014 at 18:57

1 Answer 1


This is just a special case of Integration along fibers (http://en.wikipedia.org/wiki/Integration_along_fibers). However, notice that integrating of $f \omega^n$ along the natural fibration results in an $n$-form on the base manifold. In your coordinate language, you then have contracted this $n$-form with the global vector fields $e_i = \partial_{q^i}$.


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