# What exactly is the Kahler class of a torus?

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!

• 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....) Apr 24, 2015 at 17:11
• 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. Apr 24, 2015 at 17:46
• 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.) Apr 24, 2015 at 18:30
• 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. Apr 24, 2015 at 18:35
• 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). :) Apr 24, 2015 at 20:27

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.")

• "(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. Apr 24, 2015 at 21:37

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 $G^\mathbb C/G$.

• See p.64 of this book also springer.com/us/book/9780817641030
– user61135
Feb 15, 2018 at 13:10
• 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.
– user61135
Feb 20, 2018 at 0:55