Hallo,


Let $M$ be a compact real-analytic Riemannian manifold with Riemannian metric $g$. Let $U \subset T^{\*}M$ be a open neighbourhood of the zero section. On $U$ there exists a complex structure $J$ and a Kähler form $-i\partial \overline{\partial} \phi$ such that $M$ is a Lagrangian manifold and $\phi = d\phi =0$ on $M$. This shows the existence. Now lets assume that there are two symplectic forms $\omega\_{1} = -i\partial \overline{\partial} \phi\_{1}$ and $\omega\_{2} = -i\partial \overline{\partial} \phi\_{2}$ in the same complex structure and such that $M$ is a Lagrangian manifold with respect to both and $\omega\_{1}^{n} = \omega\_{2}^{n}$ (they have the same volume), where $n$ is the dimension of $M$. My question is: are these Kähler forms the same? If not, what assumptions does one need to make in order that they are equal? Or, is there any chance, at all, to make them equal?


hapchiu



Uniqueness of Kähler form with same volume
Your question doesn't really refer to the metric $g$ as part of the data, so I'll ignore it. What you really seem to be asking is this: Let $X$ be a complex $n$-manifold and let $M\subset X$ be a compact, analytic, totally real $n$-dimensional submanifold of $X$. Suppose you have two Kähler forms $\Omega\_1$ and $\Omega\_2$ on an open $M$-neighborhood $U\subset X$ that have the same volume form and for which $M$ is Lagrangian with respect to each. What can you say about the relationship between $\Omega\_1$ and $\Omega\_2$? Are they equal in some neighborhood of $M$?


I claim that the answer is 'no' when $n>1$, as can be seen by example: Let $\Lambda\subset \mathbb{R}^n$ be a lattice (i.e., a discrete co-compact subgroup), let $X = \mathbb{C}^n/\Lambda$ and let $M=\mathbb{R}^n/\Lambda$. Let $z = (z\_k) = (x\_k+iy\_k)$ be the standard coordinates on $\mathbb{C}^n$. Then a function $\phi:\mathbb{C}^n\to\mathbb{R}$ is well-defined on $X$ if it is $\Lambda$-periodic in the $x$-variables. One easily computes that $\phi\_0:X\to\mathbb{R}$ defined by 
$$
\phi\_0 = {y\_1}^2+\cdots+{y\_n}^2
$$
has the property that $\Omega\_0 = i\partial\bar\partial\phi\_0$ is a Kähler form on $X$ for which $M$ is Lagrangian. For any $\phi:X\to\mathbb{R}$, the equation
$$
(i\partial\bar\partial\phi)^n = {\Omega\_0}^n \tag{1}
$$
is an elliptic second order PDE for $\phi$ as long as $i\partial\bar\partial\phi > 0$. Moreover, $M$ will be Lagrangian for $i\partial\bar\partial\phi$ as long as the equations
$$
\frac{\partial^2\phi}{\partial x\_k\partial y\_j} 
- \frac{\partial^2\phi}{\partial x\_j\partial y\_k} = 0\tag{2}
$$
hold along the locus $y\_1 = \cdots y\_n = 0$ (which defines $M$). Now, since $(1)$ is a real-analytic elliptic equation, one has the following existence theorem: Let $\psi$ be any real-analytic function on an open $M$-neighborhood $U$ such that $i\partial\bar\partial\psi>0$. Then there exists a unique, real-analytic function $\phi$ on an open $M$-neighborhood $V\subset U$ that satisfies $(1)$ and
$$
\phi(x,y\_1,\ldots,y\_{n-1},0) = \psi(x,y\_1,\ldots,y\_{n-1},0)
$$
and
$$
\frac{\partial\phi}{\partial y\_n}(x,y\_1,\ldots,y\_{n-1},0) 
= \frac{\partial\psi}{\partial y\_n}(x,y\_1,\ldots,y\_{n-1},0).
$$
In particular, the space of solutions to $(1)$ defined on an open $M$-neighborhood depends essentially on two arbitrary analytic functions of $2n{-}1$ variables (subject to some open conditions that say that $i\partial\bar\partial\phi$ should be positive on a neighborhood of $M$). The Lagrangian condition only imposes a very small set of restrictions on the $\psi$s that you can choose. For example, if you just require that the $y$-partials of $\psi$ vanish along $M$, i.e., 
that $\frac{\partial\psi}{\partial y\_k}(x,0)=0$, then $(2)$ will be satisfied, and $M$ will be Lagrangian with respect to $\Omega\_\phi = i\partial\bar\partial\phi$. 


Thus, there are many, many $M$-local solutions to your problem and, without further assumptions, you can't hope to get uniqueness.


Here is another way to see that you won't generally, get uniqueness: Start with the given data of $M\subset X$ as above and choose a real-analytic density $\nu$ on $M$. (I'm using densities rather than volume forms because $M$ may not be orientable; think, for example, of $\mathbb{RP}^n\subset\mathbb{CP}^n$ for $n$ even.) Then, on an $M$-neighborhood $U\subset X$ there will exist a canonical, nonvanishing $2n$-form $\hat\nu$ defined as follows: Near each point of $M$, there will exist a holomorphic coordinate system $z= (z\_k) = (x\_k+iy\_k)$ in which $M$ is defined by the equations $y\_k=0$ and such that $|dx\_1\wedge\cdots\wedge dx\_n| = \nu$. Then set $\hat\nu = dx\_1\wedge dy\_1\wedge\cdots dx\_n\wedge dy\_n$. Now note that any real-analytic tangent vector field $xi$ on $M$ whose flow preserves $\nu$ will extend uniquely to a holomorphic vector field $\zeta$ on $X$ whose (local) flow preserves $\hat\nu$. (When $n>1$, there will be lots of these vector fields $\xi$.) Then, if you take any $M$-local function $\phi:U\to\mathbb{R}$ such that $i\partial\bar\partial\phi>0$ and such that $(i\partial\bar\partial\phi)^n = \hat\nu$, then the local flow of a vector field $\zeta$ as above will carry $\phi$ to another solution $\tilde\phi$ of $(i\partial\bar\partial\tilde\phi)^n = \hat\nu$. Moreover, if $M$ is Lagrangian with respect to $i\partial\bar\partial\phi$, it will be Lagrangian with respect to $i\partial\bar\partial\tilde\phi$.


The end result is that this construction will generate lots and lots of unequal solutions to your problem. Now, you might ask whether all the solutions to your problem might be 'equivalent' (though not equal) under this 'reparametrization' diffeomorphism pseudogroup, but the answer to that is still 'no' because the 'reparametrizations' depend only on arbitrary functions of $n$ variables while the general solutions as above depend on arbitrary functions of $2n{-}1$ variables, and, as long as $n>1$, there is no hope that you'll be able to get all of the solutions in this way.


Finally, you may note that I have been assuming $n>1$ throughout all of this. The reason is that, when $n=1$, you do indeed have uniqueness, but this is trivial.


