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If $[\sigma ]\ne 0$ in de Rham cohomology, a twisted cotangent symplectic form has no Lagrangian submanifold whose projection is a diffeomorphism onto the entire base. Such a submanifold would be a global one-form graph, and the twisted Lagrangian graph criterion would make $\sigma$ exact.

A differential one-form $\theta$ is a smooth section of a vector bundle $T^{\ast }X\to X$. Its graph is its image as a section, and the projection restricts to a diffeomorphism onto $X$. Pullback of the Liouville one-form by that section equals $\theta$, making the twisted Lagrangian graph criterion immediate.

The differential one-form $\theta$ defines a smooth section of a vector bundle, and its one-form graph is an embedded copy of $X$ because $\pi \circ \theta ={\mathrm{id}}_X$. The defining property of the Liouville one-form gives ${\theta }^{\ast }\alpha =\theta$. ConsequentlyThe one-form graph has dimension $n$, half the dimension of the cotangent bundle. It is therefore a Lagrangian submanifold precisely when this pullback of a differential form vanishes:This is the twisted Lagrangian graph criterion.

Now suppose the projection restricts to a diffeomorphism on a Lagrangian submanifold $L$. Its inverse followed by the inclusion defines a smooth section $s:X\to T^{\ast }X$, hence a global differential one-form $\theta$ with image $L$. The criterion forces $\sigma =d\theta$, so its class in de Rham cohomology is zero. Thus $[\sigma ]\ne 0$ rules out every such Lagrangian section. This cohomological obstruction to a Lagrangian section concerns graphs over the entire base; it does not claim that all Lagrangian submanifolds are absent.