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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.

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  • $\begingroup$ Is the answer positive for transverse pairs of totally real planes? It seems that the quotient of the space of transverse pairs by $GL(n,\mathbb C)$ has dimension n? I see why it has dimension at least $n$. Indeed, if one takes $L_1$ as the real subspace of vectors $(x_1,..., x_n)$ with $x_i$ real, and takes $n$ complex units $\mu_i$ then one can take as $L_2$ the plane obtained from $L_1$ by applying the diagonal matrix with entries $\mu_i$. $\mu_i$'s are invariants of the pair. For such pairs the desired symplectic form is $\sum dx_i\wedge dy_i$. Are there non equivalent transversal pairs? $\endgroup$ – Dmitri Panov Nov 8 '19 at 20:56
  • $\begingroup$ Dear Dmitri. Thanks for thinking about this question. I agree with you that one can assume $L_1$ to be $(x_1,\dotsc,x_n)$ with all $x_j$ real. For the basis of $L_2$, one can assume that it is $v_j = (a_{j1},\dotsc,a_{jn}) + i e_j$, where $(a_{j1},\dotsc,a_{jn})$ is a real vector, and $e_j$ is the vector $(0,\dotsc,0,1,0,\dotsc,0)$. One way to encode $L_2$ would be as the image of $i \mathbf{1}_n + A$, where $\mathbf{1}_n$ is the unit matrix and $A$ is a real $n\times n$-matrix. I guess, one could identify several cases by conjugating everything with a linear transformation. $\endgroup$ – Klaus Niederkrüger Nov 10 '19 at 12:21
  • $\begingroup$ But then, I don't know what would follow further... $\endgroup$ – Klaus Niederkrüger Nov 10 '19 at 12:25
  • $\begingroup$ This is very interesting (and disappointing). I think I understand that the projection of the totally real planes is a circle and that for the perturbation of the Hopf fibers the images are tiny (and thus disjoint), but how do you see that you need two intersection points between the pair of circles to be able to find the desired 2-form? (or at least that that would be sufficient) $\endgroup$ – Klaus Niederkrüger Nov 10 '19 at 18:55
  • $\begingroup$ @DmitriPanov Thank you very much. I'm very keen on learning more about your findings! $\endgroup$ – Klaus Niederkrüger Nov 12 '19 at 0:45
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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.


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.

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  • $\begingroup$ The solution you give in $\mathbb{C}^2$ is ingenious. Very nice. I need to think about higher dimensions. The problem I have is that I have two totally real submanifolds in an almost complex manifold, and I need locally taming symplectic structures for which both submanifolds are Lagrangian. I guess I'll have to understand the genericity condition of the corresponding tangent planes as a sort of $\det \ne0$ that can be verified explicitly (instead of relying on genericity which would require perturbing the totally real submanifolds). I'll think about it. Thanks again, Dmitri. $\endgroup$ – Klaus Niederkrüger Nov 14 '19 at 22:08
  • $\begingroup$ Klaus, hope this helps you a bit. Do you know anything on how these two sub-manifolds intersect, for example if they are transverse to each other? Because your example shows that sometimes you can not do what you want. Or your desire is to have this condition as explicit as possible? $\endgroup$ – Dmitri Panov Nov 15 '19 at 17:35

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