7
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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
    May 13, 2017 at 19:04
  • 1
    $\begingroup$ 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
    May 13, 2017 at 19:21

3 Answers 3

6
$\begingroup$

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).

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Aug 14, 2014 at 12:41
8
$\begingroup$

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

$\endgroup$
3
  • 4
    $\begingroup$ 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
    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
    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
    Oct 24, 2017 at 21:00
6
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.