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Let $J_0$ be the standard complex structure on ${\mathbb{R}}^{2n}$, let ${\omega }_0$ be the standard symplectic form, and let $g_0$ be the Euclidean inner product. They satisfyThe two-out-of-three property for unitary structures says that a real linear map preserving any two of these structures preserves the third. In terms of the corresponding matrix groups,

For example, if $A$ preserves $J_0$ and ${\omega }_0$, thenso $A$ preserves $g_0$. If it preserves $J_0$ and $g_0$, the second displayed identity shows that it preserves ${\omega }_0$. Finally, $J_0$ is uniquely determined by $g_0(J_{0}u,v)={\omega }_0(u,v)$, so preservation of $g_0$ and ${\omega }_0$ implies $A{J}_0=J_{0}A$. The common intersection is therefore the unitary group.

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,is an oriented circle bundle.

If a transition function of $T{S}^2$ 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 givesBy the Poincaré-Hopf theorem,for the chosen orientation, up to changing both signs. Hence the Euler number of $\mathcal{L}Gr(T{S}^2)$ is $\pm 4$, which is nonzero. A smoothly trivial oriented circle bundle has zero Euler class, so $\mathcal{L}Gr(T{S}^2)\to S^2$ cannot be smoothly trivial for any choice of symplectic form.

On ${\mathbb{R}}^{2n}$ let $J_0$ be the standard complex structure, let ${\omega }_0$ be the standard symplectic form, and let $g_0(u,v)={\omega }_0(u,J_{0}v)$ be the Euclidean inner product. A linear map preserving any two of $J_0$, ${\omega }_0$, and $g_0$ preserves the third. Equivalently,