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Is there an intuitional way to understand what the Kahler class of $T^2$ actually is? It would be extremely useful to me if you could provide me some intuition behind it!

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    $\begingroup$ There's information missing from the question: How is the torus constructed, and what's the metric? For example, are you talking about a flat metric on $\mathbf{C}$ modulo a specific lattice? A surface of rotation in space? (There's only one torus topologically, but there's a $1$-dimensional complex family of tori holomorphically, and even after a holomorphic structure is specified, there's an infinite-dimensional space of Kähler metrics....) $\endgroup$ Apr 24, 2015 at 17:11
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    $\begingroup$ Yes, I understand that topologically there is only one $T^2$. Let us assume the fundamental domain of the lattice. Let us construct the torus out of two unit cycles for example, $\alpha$ and $\beta$. I am not sure how to answer to the other questions though. But if you can expand towards the infinite class of Kahler metric maybe this would answer my question. $\endgroup$
    – Marion
    Apr 24, 2015 at 17:46
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    $\begingroup$ The point is, speaking of a Kähler class presumes you have a fixed holomorphic structure and a Kähler metric; having a topological torus isn't enough. (In case it helps, once those data are specified, the Kähler class is the $2$-dimensional de Rham cohomology class represented by the area form of the metric. "Fixing a Kähler class" amounts to normalizing the area of the torus.) $\endgroup$ Apr 24, 2015 at 18:30
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    $\begingroup$ Ok, thus, is it ok if I think of the Kahler class as a parameter parametrizing the area of the torus? Actually, this seems to make some sense in the graphs of the toric non-compact CY threefolds I am using. $\endgroup$
    – Marion
    Apr 24, 2015 at 18:35
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    $\begingroup$ Yes, overall area ( as a parameter) is a reasonable interpretation (with the proviso that there's also a holomorphic structure lurking in the background; the holomorphic structure singles out a unique flat Kähler metric up to overall scaling). :) $\endgroup$ Apr 24, 2015 at 20:27

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Just so this has an answer: Fixing a Kähler class on a compact Riemann surface amounts to fixing the overall area, with some fine print. (Similarly, fixing a Kähler class on a manifold of higher dimension fixes the areas of $2$-dimensional homology classes.)


In more detail, a torus $T$[1] admits a unique flat Kähler metric $g$ of unit area.[2] Let $\omega$ denote the associated Kähler form.

If $g'$ is an arbitrary Kähler metric of area $\alpha > 0$ on $T$, then the Kähler form $\omega'$ is cohomologous to $\alpha\omega$; that is, $[\omega'] = [\alpha\omega]$ in the de Rham space $H^{2}(T, \mathbf{R}) \simeq \mathbf{R}$.

Thus, a choice of Kähler class on $T$ amounts to fixing the area of $T$.[3]

[1] Namely, a compact Riemann surface of genus one, a.k.a a holomorphic quotient $\mathbf{C}/\Lambda$ for some rank-$2$ integer lattice $\Lambda$.

[2] This metric is induced (up to overall scale) by the Euclidean metric on $\mathbf{C}$, and is a Riemannian product of circles if and only if $\Lambda$ is rectangular. ("Usually not.")

[3] With the understanding that speaking of $T$ itself entails fixing a holomorphic structure. Two flat tori of equal area are not isometric unless their underlying holomorphic structures are the same. ("Usually not.")

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    $\begingroup$ "(Similarly, fixing a Kähler class on a manifold of higher dimension fixes the areas of 2-dimensional homology classes.)" Actually, it fixes the lengths/areas/volumes of all classes; a Kahler class defines an inner product on the whole cohomology ring. $\endgroup$ Apr 24, 2015 at 21:37
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See my answer In MO

https://mathoverflow.net/questions/178494/all-kähler-metrics-on-a-complex-manifold/273696#273696

I give a general view of the space of Kahler metrics In the Kahler class $[\omega]$ on a compact Kahler manifold by using Semme's construction which is less known

The Kahler metrics In the Kahler class $[\omega]$ on a compact Kahler manifold $M$ is one-to-one correspondance with exact Lagrangian symplectic submanifolds In the space $\mathcal W_{[\omega]}$.

Now I define the space $\mathcal W_{[\omega]}$.

Let $\{ U_i , i\in I\}$ be a covering of $M$ such that $ \omega|_{U_i}=\sqrt{-1}\partial\bar\partial \rho_i $ . For any $x=y\in U_i \cap U_j$ we identify $(x,v_i)\in T^*U_i $ with $(y,v_j)\in T^*U_j$ if $v_i=v_j+\partial(\rho_i-\rho_j)$ . Then $\mathcal W_{[\omega]}$ consists of all these équivalence classes of $[x, v_i]$

This is knows as Semme's construction , see p.12 of the Well written paper of Tian with Chen

Chen, X. X.; Tian, G. Geometry of Kähler metrics and foliations by holomorphic discs Publications Mathématiques de l'IHÉS, Tome 107 (2008) , p. 1-107 http://www.numdam.org/item/PMIHES_2008__107__1_0

The space of Kahler forms in the class of $[\omega]$, can be written as $$\mathcal K=\{\omega_\varphi=\omega+dd^c\varphi, \;\omega_\varphi>0 \}$$ By using Moser lemma, for every $\omega_\varphi\in \mathcal K$, we have a diffeomorphism $F_\varphi^*\omega_\varphi=\omega$, as an exercise we can show that maximal leaf of the involutive distribution of $F$ passing through the complex structure $J$ is the image of the map $$\mathcal K\times G\to F, \; \; \; (\omega_\varphi,\sigma)\to \sigma^* F_{\varphi}^* J$$ where here $G$ is the group of symplectomorphisms. Note that in Fano Kahler-Einstein manifolds the space of Kahler metrics $\mathcal K$, corresponds to symmetric space $G^\mathbb C/G$.

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  • $\begingroup$ See p.64 of this book also springer.com/us/book/9780817641030 $\endgroup$
    – user61135
    Feb 15, 2018 at 13:10
  • $\begingroup$ Note that when $G$ is the group of symplectomorphisms , then the complexification of $G$, i.e., $ G^\mathbb C$ is not a Lie group In général and we see it as set theoritic. $\endgroup$
    – user61135
    Feb 20, 2018 at 0:55

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