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.



Symplectic structure vanishing simultaneously on two totally real subspaces
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. 


