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Let $M$ be a complex manifold of complex dimension 2. What do we know about the set all Kähler metrics on $M$ in general and in the case of 4-torus $C^2/Z^4$?

For the case of surfaces ($dim_C=1$), any compatible metric is Kähler and by the uniformization theorem, the answer is that every two such metrics are conformally equivalent and the set all Kähler metrics is nonempty.

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  • $\begingroup$ In general, Kahler metrics in $[\omega_0]$ can also be parametrised as metrics of the same volume conformally equivalent to $\omega_0$ by $$\{\varphi\in C^\infty(X,\mathbb R)|\; \int_Xe^\varphi\omega_0^n=\int_X\omega_0^n=vol(X,[\omega_0]) \}$$ $\endgroup$
    – user21574
    Commented May 13, 2017 at 19:04
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    $\begingroup$ Moreover if two metric be comformally equivalent $\omega_\varphi^n=e^u\omega_0^n$ then conformal factor and Kahler potential are related by $(1+\Delta_{\omega_0}\varphi)=e^u$ $\endgroup$
    – user21574
    Commented May 13, 2017 at 19:21

3 Answers 3

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Let $M$ be a compact complex manifold. The set of Kahler metric on $M$ in a given Kahler class is parametrized by the set of positive volume forms with given integral, because (by Calabi-Yau theorem) any given positive volume form is a volume form of a Kahler metric in a given cohomology class, assuming their integrals agree. This is actually used when they put a structure of an infinite-dimensional symmetric space on the space of all Kahler metrics. See for example here: http://www.emis.de/journals/NYJM/JDG/p/2000/56-2-1.pdf (THE SPACE OF KAHLER METRICS, by XIUXIONG CHEN).

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  • $\begingroup$ Ziegler's answer and yours complements each other in the following way. If our compact complex manifold admits any Kahler metric ever, then on can find all the metrics by first fixing a Kahler class in the Kahler cone and then changing the metric in the same class by choosing a smooth function. $\endgroup$
    – Asghar Ghorbanpour
    Commented Aug 14, 2014 at 12:41
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adding my comment as an answer

In general, Kahler metrics in $[ω_0]$ can also be parametrised as metrics of the same volume conformally equivalent to $ω_0$ by

$$\{\varphi\in C^\infty(X,\mathbb R)|\; \int_Xe^\varphi\omega_0^n=\int_X\omega_0^n=vol(X,[\omega_0]) \}$$

Moreover if two metric be comformally equivalent $\omega_\varphi^n=e^u\omega_0^n$ then conformal factor and Kahler potential are related by $(1+\Delta_{\omega_0}\varphi)=e^u$

In Mirror symmetric language

If $X$ and $\hat X$ be mirror to each other and be CY, then the Kahler moduli space of $\hat X$, denoted by $\mathcal M_{kah}(\hat X)$ can be identified with $K_\mathbb C(\hat X)/Aut(\hat X)$ where $$K_\mathbb C(\hat X)=\{\omega\in H^2(\hat X,\mathbb C)|Im(\omega)\in K(\hat X)\}/im H^2(\hat X,\mathbb Z)$$

We can identify the moduli space of Kahler spaces of $\hat X$ with moduli space of complex space of $X$ via Yukawa couplings

See Moduli Spaces of Hyperkahler Manifolds and Mirror Symmetry by Daniel Huybrechts

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    $\begingroup$ Given a family $f:X→S$ with singular central fibre $X_0$ and with generic fibre $X_s$ a Calabi-Yau variety, mirror symmetry produces, in some cases, a dual family $\tilde f:\tilde X→S$ satisfying certain properties. For instance, the moduli space of complex structures on $X_s$ is isomorphic to the complexified moduli space of Kähler structures on $\tilde X_s$ and vice versa. $\endgroup$
    – user21574
    Commented Jul 26, 2017 at 4:19
  • $\begingroup$ Let $X$ be a Kahler manifolds and $L$ be an ample line bundle then if we denote $\mathcal H_L$ be the space of all Kahler metrics $\omega\in c_1(L)$, then we know $\mathcal H(L)\cong \frac{GL(N,\mathbb C)}{U(N)}$ $\endgroup$
    – user21574
    Commented Oct 24, 2017 at 21:00
  • $\begingroup$ ....Now I give relative version of pervious comment: suppose that $\mathcal X\to S$ be a holomorphic surjective map of Kahler manifolds and $\mathcal L$ be a relatively ample line bundle over $\mathcal X$. Now denote $\mathcal K_{\mathcal L/S}$ be the space of all Kahler metrics $\omega$ on $\mathcal L$ with positive curvature such that its restriction on each fiber $\mathcal X_s$, we have $\omega_s\in c_1(\mathcal L_s)$ then $\mathcal K_{\mathcal L/S}$ is an infinite dimensional fiber bundle over $S$ whose fibers are of the form $\mathcal H_L$ $\endgroup$
    – user21574
    Commented Oct 24, 2017 at 21:00
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Compact case (since you mention $\mathbf C^2/\mathbf Z^4$):

  • For $M$ to be Kähler its first Betti number must be even, and conversely every compact complex surface with even $b_1(M)$ is Kähler (Kodaira's conjecture, proved by Siu, Lamari, Buchdahl).

  • Lamari and Buchdahl also describe "how many" Kähler metrics then exist, i.e. the so-called "Kähler cone" of classes in $H^{1,1}_{\mathbf R}(M)$ which can be represented by positive closed $(1,1)$-forms.

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