Can a lagrangian submanifold of a 4-dimensional real space be diffeomorphic to sphere or real projective plane?

Question

Is it possible to find a lagrangian submanifold of $\mathbb{R}^4$, which is diffeomorphic to $S^2$ or $\mathbb{R}P^2$? As I have found, the answer is no for $S^2$, because $S^2$ is simply connected, while each lagrangian submanifold of $\mathbb{R}^4$ is not simply connected. But I do not know how to prove it, I think it has to be simplier methods to show it. The answer for $\mathbb{R}P^2$ I still do not know.

Answer 1

As a consequence of Weinstein's neighborhood Theorem, one can rule out many closed surfaces in $\mathbb{R}^4$ from being Lagrangian submanifolds (with respect to any symplectic form on $\mathbb{R}^4$). In fact, using it, I can rule out all oriented surfaces except $T^2$, and all non-orientable surfaces with odd Euler characteristic. In particular, $S^2$ and $\mathbb{R}P^2$ cannot occur.

To see this, suppose $L$ is a closed surfaced embedded into $\mathbb{R}^4$ as a Lagrangian submanifold with respect to some symplectic form on $\mathbb{R}^4$. Weinstein's theorem gives an embedding of $T^\ast L \cong TL$ into $\mathbb{R}^4$.

Now, the point is that $TL$ is open in $\mathbb{R}^4$, so the normal bundle of $L$ in $TL$ (which is isomorphic to $TL$), must be isomorphic to the normal bundle of $L$ in $\mathbb{R}^4$.

If $L$ is orientable, then the normal bundle in $\mathbb{R}^4$ has trivial Euler class (Milnor-Stasheff, Corollary 11.4), while the Euler class of $L$ in $TL$ measures the Euler class. Hence, if $L$ is orientable, $L$ must be a torus.

If $L$ is non-orientable, then the normal bundle in $\mathbb{R}^4$ has trivial top Stiefel-Whitney class. But this only happens in the tangent bundle $TL$ if $L$ has even Euler characteristic.