What is the obstruction to the existence of a global Kahler potential?

Question

It is a well-know fact that if $(X,\omega)$ is Kahler then about every point $x \in X$ there exists a neighbourhood $U$ and a function $K \in C^{\infty}(U,\mathbb{R})$ such that $\omega|_U = i\partial \bar{\partial} K$. Here $K$ is called a local Kahler potential.

My question is when does this extend globally, i.e. what conditions must be imposed on a Kahler manifold $(X,\omega)$ so that there exists a function $K \in C^{\infty}(X,\mathbb{R})$ such that $\omega = i \partial \bar{\partial} K$? Clearly being non-compact is one condition and there are certainly examples of such manifolds ($\mathbb{C}^n$ etc.). I am guessing there is some vanishing theorem on Dolbeault cohomology but cannot find it in the literature

Answer 1

In the lines of vanishing conditions you can argue as follows. Let $U_\alpha$ be a collection of patches such that $\omega=i\partial\bar\partial f_\alpha$ for some $f_\alpha\in C^{\infty}(M,\mathbb R)$. Then the collection of differences $\{f_\alpha-f_\beta\}$ gives you a Cech cocycle $\phi\in C^1(M,\mathcal P)$, where $\mathcal P$ is a sheaf of real pluriharmonic functions (i.e. satisfying the equation $\partial\bar\partial g=0$). You wish to show that this cocycle is zero in the cohomology group.

Locally any pluriharmonic function is a real part of holomorphic function, so we have a short exact sequence $$ 0 \to \mathbb R\to \mathcal O\xrightarrow{Re}\mathcal P\to 0, $$ We get LES in cohomology $$ \dots\to H^1(M,\mathcal O)\to H^1(M,\mathcal P)\xrightarrow{\delta} H^2(M,\mathbb R)\to\dots $$ Since $\delta([\phi])=[\omega]$, as you have mentioned, the first thing you require is $[\omega]=0$.

Now, given that $[\omega]=0$, it is enough to assume that $H^1(M,\mathcal O)=0$.

Remark. Note that in a non-compact case the vanishing $H^1(M,\mathcal O)=0$ is not implied by $\pi_1(M)=1$, for example, the group $H^1(\mathbb C^2\backslash(0,0),\mathcal O)$ has infinite dimension.

This example allows you to construct a (not necessarily positive) real closed 2-form $\gamma$, which has a local potential, but does not have global. Namely, take a representative $\alpha^{0,1}$ of a nonzero class in $H^1(\mathbb C^2\backslash(0,0),\mathcal O)$ and consider $\gamma=Re(\partial\alpha^{0,1})$ (or $Im(\partial\alpha^{0,1})$).

Answer 2

In this book Geometric and Algebraic Topological Methods in Quantum Mechanics, G. Giachetta,L. Mangiarotti,Gennadiĭ Aleksandrovich Sardanashvili page 181

in remark 2.6.2, it has been written that if $X $ be a simply connected non-compact manifold then then local Kahler potential can be glued into global one, and if $X$ is not simply connected then local potential still exists on an open subset $U$ obtained from $X$ by deleting a real submanifold of lower dimension

Recently A. Loi showed that

Let$ (M, ω) $ be a homogeneous Kahler manifold. Then the following are equivalent:

(a)$ M$ is contractible.

(b)$ (M, ω)$ admits a global Kahler potential.

(c) $(M, ω) $admits a Berezin quantization.

If $\omega$ is a real $d$-exact $(1,1)$-form, then $\omega$ is $\sqrt{-1}∂∂$-exact , i.e. there exits a real function $u$ such that locally $\omega = i∂∂u$

If $ M$ is a compact Kahler manifold then $M$ cannot have a global Kahler potential.see page 6 here

In general Gauduchon showed that for a compact complex manifold $M$, $dd^c$ -lemma for $(1, 1)$-forms is equivalent to the equality $b^1 = 2h^{0,1}$.

Deligne showed that Complex manifolds satisfying the $dd^c$-lemma are formal see New aspects of the ddc -lemma

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If $\omega$ be a Kahler current instead of Kahler form, then still $dd^c $ -lemma holds

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Let $X$ be a compact complex manifold. The equality

$$\sum_{p+q=k}(dim_\mathbb CH_{BC}^{p,q}(X)+dim_A^{p,q}(X))=2dim_\mathbb CH_{dR}^k(X,\mathbb C)$$ holds for every $k \in N$ if and only if $X$ satisfies the $∂\bar∂$-Lemma. here , $H_{BC}^{•,•} (X)$, is Bott-Chern and $H^{•,•}_A(X) $ is Aeppli cohomologies.

In Sasakian cone we have global Kahler potential:

A compact Riemannian manifold $(S,g)$ is Sasakian if and only if its metric cone $(C(S)=R_{>0}\times S, \bar g=dr^2+r^2g)$ is Kahler. which Sasakian manifolds are the odd dimentional view of Kahler manifolds

Then if $S$ be Sasakian, then the cone $C(S)$ has

$$\omega=\frac{1}{2}\sqrt {-1}\partial\bar\partial r^2$$

the function $\frac{1}{2}r^2$ is hence a global Kahler potential for the cone metric

Answer 3

Here's a geometric way to think about the obstruction. Consider a holomorphic line bundle $L$ over a complex manifold $M$ and let $h \in \Gamma\big(\overline{L}\,^*\otimes L^*\big)$ be a Hermitian form on it. With respect to a local holomorphic trivialisation, we may write $h$ as $e^{f}$ for some function $f$ on an open set in $M$. Then the Chern connection is given by $\mathrm d + \partial f = \overline \partial + \partial + \partial f$ in the local holomorphic trivialisation we have chosen and its curvature is $\mathrm d \partial f = \overline \partial \partial f$. We thus see that the Kähler potential $f$ associated to a Kähler form $\omega$ is locally the logarithm of the Hermitian form $h$ on a holomorphic line bundle $L$ whose associated Chern curvature is (up to an overall constant factor) $\omega$ itself.

The function $e^{f}$ is globally defined if and only if (a) $L$ is trivial as a complex line bundle, and (b) the holomorphic structure on $L$ is trivial. These are guaranteed by the conditions $[\omega]=0$ and $H^1(M;\mathscr O^\times)=0$ respectively. Now, keeping in mind the exponential sequence

$$0 \longrightarrow \mathbb Z \longrightarrow \mathscr O \longrightarrow \mathscr O^\times \longrightarrow 0$$

we note that $H^1(M;\mathscr O^\times)=0$ by itself doesn't guarantee that the logarithm $f$ is globally defined. The stronger condition $H^1(M;\mathscr O)=0$ however ensures this.