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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.

In cotangent bundle coordinates $(q^1,\dots ,q^n,p_1,\dots ,p_n)$, the Liouville one-form and the canonical symplectic form areThe twisted cotangent symplectic form is closed: pullback of a differential form commutes with the exterior derivative, soTo prove the form is nondegenerate, write a tangent vector in these coordinates as $(u,v)$ and test its pairing with an arbitrary vertical vector $(0,w)$. The base differential two-form has zero pairing with vertical vectors, henceA vector in the kernel of a linear map therefore has $u=0$. Pairing it next with every $(z,0)$ gives $-z^i{v}_i=0$, so $v=0$. Thus the kernel is zero at every point, independently of the size or rank of $\sigma$. Together with closedness, this proves ${\omega }_{\sigma }$ is a symplectic form. The argument is local only for convenience; the nondegenerate property is intrinsic and the coordinate charts cover the whole cotangent bundle.

The area form $\sigma$ on $S^2$ is closed. For any closed two-form $B$ on a base, the twisted cotangent symplectic form $\omega +{\pi }^{\ast }B$ is closed and nondegenerate: pairing a putative kernel vector first with vertical vectors kills its horizontal component, and then pairing with horizontal vectors kills its vertical component. Thus both ${\omega }_{+}$ and ${\omega }_{-}$ are symplectic.

The cotangent bundle deformation-retracts onto its zero section. Their cohomology classes satisfyThey are therefore not strongly isotopic.

Let $f:S^2\to S^2$ be an orientation-reversing isometry. Then $f^{\ast }\sigma =-\sigma$, while its cotangent lift preserves the canonical form and satisfies $\pi \circ f_{\#}=f\circ \pi$. Consequentlyso the two forms are symplectomorphic.