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Suppose $(V^{2g}, g, \omega, J)$ is an almost Kahler manifold. ie. $(V,\omega)$ is a symplectic manifold with $\omega$-compatible almost complex structure $J$ ($J$ is a symplectomorphism) and such that $\omega(\cdot, J\cdot)$ coincides with a riemannian metric $g$ on $V$.

Suppose further that $L$ is a lagrangian submanifold of $V$. Then $L$ has a volume, as defined with respect to the riemannian metric $g$.

Nonetheless I want to believe that there still is an alternative expression of the metric quantity $vol(L,g)$ which exploits both the (almost) Kahler structure and the fact that the submanifold $L$ is lagrangian (ie. totally isotropic) and that we have a canonical decompostion $TM|L=TL \oplus T JL$, ie. the normal bundle of $L$ is precisely $JL$.

Can anybody testify or provide some evidence or enlightenment on this matter?

Alternatively, an even simpler question: if we suppose $V$ is closed, then can we give an expression to the volume of $V$ (wrt g) in terms of $vol(L,g)$ which depends essentially on the circumstances of $L$ being lagrangian and $g=\omega(\cdot, J\cdot)$?

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Your definition of a Kahler manifold should also include the assumption that $J$ is integrable. What you defined is usually called an "almost Kahler" structure. –  YangMills Jul 18 '12 at 21:49
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Also, $\omega^g$ is a $2g$-form so it cannot yield a volume form on any $g$-dimensional submanifold, Lagrangian or not. –  YangMills Jul 18 '12 at 21:51
    
sure, forgot to add integrable. And of course $\omega^g$ is not a volume form on any $g$-dimensional submanifold. I was hasty and should have given $g/2$ (when $g$ even) to emphasize just that $\omega$ vanishes on $L$, so it ''appears'' to give to no data on $J$. I'll correct this typos. –  J. Martel Jul 18 '12 at 23:03
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This is probably closer in spirit to what you're looking for than what you've received in the comments. If $(V^{2m}, J, \omega, g)$ is Calabi-Yau (which for me means that $J$ is integrable, and the first Chern class $c_1(V) = 0$), then one can say much more. In this case there exists a holomorphic nowhere vanishing $(m,0)$ form $\Omega$, called the "holomorphic volume form." The form $\Omega$ is unique up to multiplication by a nowhere vanishing holomorphic function. We don't need to assume that $\Omega$ is parallel, but Yau's theorem does tell us (if $V$ is compact) that we can change the metric, keeping the Kaehler class $[\omega]$ unchanged, to make $\Omega$ parallel (and consequently also the new metric will be Ricci-flat.) But I have digressed.

In such a situation, if $L$ is Lagrangian, then it is well known that the restriction of $\Omega$ to $L$ is equal to $e^{i \theta} \mathrm{Vol_L}$, where $\mathrm{Vol}_L$ is the volume form of $L$ (with the induced metric) and $e^{i \theta}$ is the "phase" of the Lagrangian, where $\theta : L \to \mathbb R/ (2 \pi \mathbb Z)$ is a smooth, multivalued function on $L$. In addition, the mean curvature $H$ of $L$ in $V$ is given by $H = J \nabla \theta$. So the minimal Lagrangian submanifolds (vanishing mean curvature) correspond to those with constant phase function $\theta = \theta_0$. In this case, if $L$ is compact, then the volume of $L$ is given by

$\mathrm{Vol} (L) = \int_L \mathrm{Vol}_L = \int_L e^{-i \theta_0} \Omega = e^{- i \theta_0} [\Omega] \cdot [L],$

which is topological. (It looks complex, but it's actually real, because $[\Omega]$ is a class in $H^g(V, \mathbb C)$.) If you prefer, you can just replace $\Omega$ by $e^{- i \theta_0} \Omega$ to get rid of the phase factor.

Such "minimal Lagrangian" submanifolds, whose volume is purely topological, are also called special Lagrangian submanifolds, and are widely studied in calibrated geometry and differential geometric approaches to mirror symmetry. The best place to start looking is the text "Riemannian Holonomy Groups and Calibrated Geometry" by Dominic Joyce and its multiple references.

I'm not sure how much of this will extend to the case of $J$ non-integrable and $c_1(V) \neq 0$. I'd have to think about it.

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I appreciate your answer very much Spiro. This formula for the volume is impressive, although i do not really know what it means. In my own case, I am looking at the volumes of lagrangians which are rational inside a given symplectic lattice. More precisely, let's fix the standard Kahler structure $(\mathbb{R}^{2g}, \omega, J)$. We say $\Lambda$ is a \emph{symplectic lattice} if $\Lambda=A\mathbb{Z}^{2g}$ for $A \in Sp{2g}\mathbb{R}$. Notice that $\omega$ will be integral and unimodular on $\Lambda$. Let's say a lagrangian $\ell$ in $\mathbb{R}^{2g}$ is $\Lambda$-\emph{rational} if –  J. Martel Aug 23 '12 at 4:20
    
$\ell \cap \Lambda$ is a cocompact lattice in $\ell$. In this sense we have a volume $vol(\ell, \Lambda):=vol(\ell / \ell \cap \Lambda)$. Hence the volume of $\ell$ in $\Lambda$ is the volume of the compact embedded lagrangian $\ell / \ell \cap \Lambda$ in the torus $\mathbb{R}^{2g} / \Lambda$ equipped with its flat Kahler structure. Now we have a problem here, right? Since my assumptions are exactly that this torus is actually a principally polarized abelian variety. Now whatever... –  J. Martel Aug 23 '12 at 4:27
    
that means', it at least says that the first Chern class of the torus is nonzero (since $\omega$ is nontrivial on the torus). But we do have of course $c_1 \mathbb{R}^{2g}=0$, and i am computing volumes of lagrangian paralellipipeds. Now at this point i should like to know precisely what' $\Omega$ is, what are the phases $\theta$ of these lagrangian parallelipipeds, and what is the computation that yields $[\Omega] \cdot [L]$? What confuses me is why the formula does not yield $Vol(L)=0$ for all lagrangian submanifolds in $\mathbb{R}^{2g}$? I'm admitting my own ignorance –  J. Martel Aug 23 '12 at 4:32
    
because i truly do not understand why $[L] \neq 0$. ANyway, i need to think more on this. I'll look at your reference. Thank you again for your answer, it is tremendously motivating. –  J. Martel Aug 23 '12 at 4:33
    
I'll try to clear up some of your confusions, but I am also confused by some of your comments. First of all, $c_1$ of a torus is zero. (I'm talking about the first Chern class (with complex or real coefficients) of its tangent bundle. I don't know anything about principally polarized abelian varieties. You might be saying that $c_1$ has torsion? I am pretty sure that doesn't matter here. On $\mathbb R^{2g} \cong \mathbb C^g$, the form $\Omega$ is just $dz^1 \wedge \cdots \wedge dz^g$. This descends to the quotient $\mathbb C^g / \Lambda$. –  Spiro Karigiannis Aug 23 '12 at 13:27
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Take an open rectangle in $\mathbb R^2$ with side lengths $a$ and $b$. This is a Riemannian manifold, thus symplectic, and has an almost complex structure. The center line is a Lagrangian submanifold and has volume $a$.

Two versions of this whole ensemble are symplectomorphic if $a_1b_1=a_2b_2$. So the volume of the center line is not invariant under symplectomorphisms.

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Ok, so what? This is no answer to my question. If we fix a Kahler structure on $(V,g)$, then every lagrangian submanifold has a well-defined volume (with respect to the Kahler structure). Your rectangle has varying Kahler structure, so of course lagrangians have different length. –  J. Martel Jul 18 '12 at 23:17
    
Rereading my question, I see that the point of your example is that one cannot expect volume to be a measure of only $\omega$, ie. at some point we have to refer to the metric $\omega(\cdot, J\cdot)$. –  J. Martel Jul 18 '12 at 23:21
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Do you, after rereading, feel that there is a question still to be answered? (If it's the second question, then could you please clarify what you are looking for? You seem to request, if $vol(L,g)$ is not invariant under symplectomorphisms, an alternate measure which is not invariant under symplectomorphisms.) –  Will Sawin Jul 18 '12 at 23:26
    
I just want to know $vol(L,g)$ in another way. There may be possible another expression of $vol(L,g)$ which depends essentially on the circumstances of $L$ lagrangian, and $g=\omega(\cdot, J\cdot)$. OR does a Kahler structure on a symplectic manifold yield a different functional on lagrangians? –  J. Martel Jul 18 '12 at 23:32
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If I'm given, say, a region in $\mathbb R^n$ I want to compute the volume of, I can give you a Lagrangian submanfiold of $\mathbb R^{2n}$ with the same volume. So if you have some easy way to compute the volumes of Lagrangian submanifolds, then I have an easy way to compute the volume of everything else. –  Will Sawin Jul 19 '12 at 13:48
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