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

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

• The "standard textbook" proof shows that if $[\omega]=0$ in $H^2(X,\mathbb{R})$ and if $H^{0,1}_{\bar\partial}(X)$ vanishes, then you can write $\omega=i\partial\bar\partial K$. You can easily reconstruct the proof by yourself. – YangMills Oct 24 '15 at 5:55

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 $Im(\partial\alpha^{0,1})$).

• Nice example, Here you mean Kahler current or Kahler forms? Can you specify – user21574 Oct 24 '15 at 3:30
• No, I do not consider currents here, though after certain modifications the argument might work. In the example $\gamma$ is merely a closed real 2-form and is not necessarily Kahler. By the way, having this example in mind, I find the statement in the first quote in your answer a bit confusing. Do you have a proof in mind? – Yury Ustinovskiy Oct 24 '15 at 3:38
• I know that for non-compact homogeneous manifolds global potential is related to topological candition that manifold must be contractible but I don't have any proof for my first statement. My conjecture is that global potential must be related to some correction in geometric quantization. – user21574 Oct 24 '15 at 4:26
• In Sasakian cone which is kahler manifold, we have global Kahler potential also and these spaces are good examples for global Kahler potential. – user21574 Oct 24 '15 at 5:56
• Thanks, this appears to be the answer I am after. Could you expand (i.e. spell out the argument) on why $\phi$ must be zero in cohomology. I only know a bit of sheaf cohomology and used the top answer here mathoverflow.net/questions/125200/… for an intuitive idea of the first cohomology group of the kernel of a surjective morphism $F \to F''$. Hence $\phi$ being a coboundary means that we can lift a global section of $F''$ to one of $F$, but fail to see what $F$ and $F''$ are in this case (although guess $F$ is the sheaf of potentials). – user78400 Oct 28 '15 at 17:03

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

• Thanks for your reply, but that doesn't really answer the question. As we both said, compact Kahler manifolds cannot admit a global Kahler potential and so I am interested in the case when $(X,\omega)$ is exact. I don't know whether your first statement applies in this case. – user78400 Oct 23 '15 at 11:51
• I edited it , I added some references – user21574 Oct 23 '15 at 12:50
• The statement of the result of Gauduchon is wrong, it should say $b^1=2h^{0,1}$. – YangMills Oct 24 '15 at 5:58
• arxiv.org/pdf/1506.07421.pdf Theorem 4.1, can you give a reason, I don't see it – user21574 Oct 24 '15 at 6:06
• There is a typo in the paper. $\dim H^1(X,\mathcal O)$ is $h^{0,1}$. – Yury Ustinovskiy Oct 24 '15 at 18:48