= Solution
Any <symplectic form> on $S^2$ orients its <tangent bundle>. Choose a compatible <complex structure> and <inner product>; this reduces the structure group to $SO(2)$. Since every line in an oriented symplectic plane is Lagrangian,
$$
\mathcal LGr(TS^2)=\mathbb P(TS^2)
$$
is an oriented <circle bundle>.
If a transition function of $TS^2$ rotates vectors through an angle $\alpha$, its action on unoriented lines rotates the coordinate $e^{2i\theta}$ through $2\alpha$. The <Euler class of the Lagrangian-line bundle of an oriented plane bundle> therefore gives
$$
e(\mathcal LGr(TS^2))=2e(TS^2).
$$
By the <Poincaré-Hopf theorem>,
$$
\langle e(TS^2),[S^2]\rangle=\chi(S^2)=2
$$
for the chosen orientation, up to changing both signs. Hence the Euler number of $\mathcal LGr(TS^2)$ is $\pm4$, which is nonzero. A smoothly trivial oriented circle bundle has zero <Euler class>, so $\mathcal LGr(TS^2)\to S^2$ cannot be smoothly trivial for any choice of symplectic form.
Back to article page
