Symplectic structure vanishing simultaneously on two totally real subspaces

Question

Let $L_1,L_2$ be two $\mathbb{R}$-linear subspaces of $\mathbb{C}^n$ that are both totally real, that is, $$L_j \cap \bigl(i\cdot L_j\bigr) = \{0\}$$ and $$\dim_{\mathbb{R}} L_j = \dim_{\mathbb{C}} \mathbb{C}^n = n$$ or equivalently $$L_j \oplus (i\cdot L_j) = \mathbb{C}^n . $$

I would like to know under what conditions I can find a symplectic structure $\omega$ on $\mathbb{C}^n$ that tames $i$ and that vanishes when restricted to $L_1$ and $L_2$.

Taming means $$\omega\bigl(v,i\cdot v\bigr) > 0$$ for every $v\in \mathbb{C}^n$ different from $0$.

Remark: In general such $\omega$ might not exist, suppose for example that $L_1$ is spanned by $(1,0)$, and $(0,1)$ in $\mathbb{C}^2$ and that $L_2$ is spanned by $(1,0)$ and $(i,1)$. Then we can easily check that $\omega$ cannot vanish on $L_2$ as $$ \omega\bigl((1,0), (i,1)\bigr) = \omega\bigl((1,0), (i,0)\bigr) + \omega\bigl((1,0), (0,1)\bigr) = \omega\bigl((1,0), i\cdot (1,0)\bigr) > 0 ,$$ where we only have used that $\omega$ vanishes on $L_1$, and that $\omega$ tames $i$.

It deduce that a vector $v\in L_2$ that is transverse to $L_1\cap L_2$ must not lie in $L_1 + \mathbb{C}\cdot (L_1\cap L_2)$, but I have not managed to prove that this condition is sufficient.

Answer 1

I'll give a positive answer for two generic totally real planes in $\mathbb C^2$. I believe this generalises to larger $n$, though I don't prove it - just give a possible plan of a proof with one step completed.

Lemma 1. Suppose $L_1$ and $L_2$ are two generic totally real $2$-planes in $\mathbb C^2$ then the desired form exists.

Proof. Let $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$ be the complex coordinates in $\mathbb C^2$. Let $$\omega_1=dx_1\wedge dx_2-dy_1\wedge dy_2,\;\;\omega_2=dx_1\wedge dy_2+dy_1\wedge dx_2.$$

Note that $\omega_1=\mathrm{Re}(dz_1\wedge dz_2)$, $\omega_2=\mathrm{Im}(dz_1\wedge dz_2)$. In particular, both $\omega_1$ and $\omega_2$ restrict to zero on all complex lines in $\mathbb C^2$.

Now, let $W$ be the two-dimensional vector space in $\Lambda^2({\mathbb R^4}^*)$ spanned by $\omega_1$ and $\omega_2$. Consider the evaluation map given by restricting a form $\omega\in W$ to $L_1$ and $L_2$ $$e: W\to \Lambda^2 L_1^*\oplus \Lambda^2 L_2^*.$$

I claim that for generic $L_1$ and $L_2$ this map is an isomorphism. Indeed, it is enough to prove this claim for at least one pair of real planes. So let $L_1$ be the plane spanned by vectors $e_{x_1},e_{x_2}$ and $L_2$ to be spanned by $e_{x_1}, e_{y_2}$. In this case the map is clearly an isomorphsim.

Now suppose that $L_1$ and $L_2$ are generic, and such that the map $e$ is an isomorphism. Let us take $\omega=dx_1\wedge dx_2+dy_1\wedge dy_2$. Then its restriction to $L_1$ and $L_2$ gives an element $v\in \Lambda^2 L_1^*\oplus \Lambda^2 L_2^*$. Let $\tilde\omega\in W$ be such that $e(\tilde \omega)=v$. Then it is easy to see that the symplectic form $\omega-\tilde \omega$ is doing the job. Indeed $\omega$ restricts positively to all complex lines and $\tilde \omega$ is zero on all complex lines.

END OF Proof.

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It seems to me that the same idea could work in higher dimensions as well. One takes again $W$ to be the space spanned by all real and imaginary parts of holomorphic $2$-forms on $\mathbb C^n$. The real dimension of this space is $n(n-1)=2*\frac{n(n-1)}{2}$. So it might be that the eavaluation map $e$ is still generically an isomorphism from $W$ to $\Lambda^2 L_1^*\oplus \Lambda^2 L_2^*$. But one has to check whether this is indeed so.

So, to check that the map $e$ is injective it would be enough to show that there exist two real $n$-planes $L_1, L_2\subset \mathbb C^n$ such that there is no holomorphic $2$-form $\omega$ such that ${\mathrm Re}(\omega)$ restricts as zero to both $L_1$ and $L_2$. Here one can consider different cases when ${\mathrm Re}(\omega)$ has different ranks, but assume for simplicity that ${\mathrm Re}(\omega)$ is symplectic. Then the space of couples of $L_1, L_2$ on which it vanishes has dimension $n(n+1)$ which is twice the dimension of the Lagrangian grassmanian. Now, recall the the space of all forms in $W$ has dimension $n(n-1)$, however two proportional sympelctic forms define the same Lagrangian grassmanian. So we get $n(n-1)-1$. So we conclude that the dimension of pairs of $\mathbb R^n$'s in $\mathbb C^n$ on which a non-degenerate form from $W$ can vanish is $n(n+1)+n(n-1)-1=2n^2-1$. However, the dimension of the space of all pairs of real $n$-planes in $\mathbb C^n$ is $2n^2$. Since $2n^2>2n^2-1$ we conclude that there is a pair on which none of non-degenerate forms from $W$ vanish simultaneously.

I would guess that one can treat the cases of degenerate forms from $W$ in a similar way.