Planes in Lagrangian Grassmannians Let $LG(h,2h)$ be the Lagrangian Grassmannian of subspaces of dimension
$h$ of a complex vector space of dimension $2h$.
For instace, $LG(1,2)=\mathbb{P}^1$, and $LG(2,4)\subset\mathbb{P}^4$ is
a smooth quadratic $3$-fold. So $LG(2,4)$ contains no plane (linearly
embbeded $\mathbb{P}^2$).
Furthermore, by Lemma 2.5.1 here
https://arxiv.org/pdf/math/0209169.pdf
$LG(3,6)$ containes no plane as well.
Is is true that $LG(h,2h)$ containes no plane for all $h\geq 1$.
 A: This is, indeed, true.
To prove this, assume we have an embedding $\mathbb{P}^2 \to \operatorname{LGr}(V)$ (where $V$ is a symplectic vector space). Let $U \subset V \otimes \mathcal{O}$ be the tautological subbundle on $\operatorname{LGr}(V)$. Note that it extends to an exact sequence
$$
0 \to U \to V \otimes \mathcal{O} \to U^\vee \to 0
$$
on $\operatorname{LGr}(V)$ and then, via pullback, also on $\mathbb{P}^2$. So, we consider this sequence on $\mathbb{P}^2$. The condition that the plane is embedded linearly means that
$$
c_1(U) = -h,
$$
where $h$ is the line class on $\mathbb{P}^2$. Therfore, the Chern class of $U$ takes the form
$$
c(U) = 1 - h + xh^2
$$
for some $x \in \mathbb{Z}$. By duality,
$$
c(U^\vee) = 1 + h + xh^2.
$$
Now by Whitney formula we obtain
$$
1 = c(V \otimes \mathcal{O}) = c(U)\cdot c(U^\vee) = (1 - h + xh^2)(1 + h + xh^2) = 1 + (2x - 1)h^2
$$
which implies $2x = 1$. But since $x$ is integer, this is impossible. This contradiction shows that $\operatorname{LGr}(V)$ has no planes.
