let $(M,\omega)$ be a compact Kähler manifold. Let $\mathfrak{g}=H^{0}(M,T_{M})$ be the Lie algebra of holomorphic vector fields on $M$.We can decompose $\mathfrak{g}$ as $$\mathfrak{g}=\mathfrak{h}\oplus\mathfrak{a}$$ where $\mathfrak{h}$ is the space of holomorphic vector fields vanishing somewhere on $M$. Suppose $X\in \mathfrak{h}$ has a Hamiltonian potential $\varphi_{X}\in C^{\infty}(M,\mathbb{R})$ $$\overline{\partial}\varphi_{X}=-\frac{1}{2}i_{X}\omega$$ Now we blow up $M$ at some points $p_{1},\ldots,p_{n}$ and we denote with $\tilde{M}$ this blow up and with $\pi$ the canonical surjection on $M$. Suppose $X$ lifts to $\tilde{M}$ i.e. $X$ vanishes at $p_{1},\ldots,p_{n}$. Now we pick a Kähler metric $\omega_{\varepsilon}\in[\omega_{\varepsilon}]$

$$[\omega_{\varepsilon}]:=\pi^{*}[\omega]+\varepsilon\sum_{i=1}^{n}c_{1}(\mathcal{O}(-E_{i}))$$ with $E_{i}$ the exceptional divisors at points. We DO NOT assume that $\omega_{\varepsilon}$ is invariant for the flow of the lift of $X$.

The question is the following: is there always a Hamiltonian potential $\tilde{\varphi}_{X}\in C^{\infty}(M,\mathbb{R})$ such that
$$\overline{\partial}\tilde{\varphi}_{X}=-\frac{1}{2}i_{X}\omega_{\varepsilon}$$ or there is some obstruction?


1 Answer 1


I guess there will not exist a $\tilde \varphi_X \in C^{\infty}(\tilde M, \mathbb{R})$ if $\omega_\epsilon$ is not invariant under $Re(X)$. But if you allow $C^{\infty}(\tilde M, \mathbb{C})$, then the following general fact is true :

$\textbf{Proposition}:$ A holomorphic vector field has a hamiltonian (possibly complex valued) (with respect to any Kahler form) if and only if it vanishes at some point.

This fact is an exercise in "An introduction to extremal Kahler metrics" by Gabor Szekelyhidi. He attributes this fact to LeBrun-Simanca.

One possible way of proving the Proposition is the following - Let $(M,\omega)$ be a Kahler manifold and $X$ a holomorphic vector field such that $X(p)=0$ for some $p\in M$. Since $X$ is holomorphic, $i_X\omega$ is $\bar\partial$-closed $(0,1)$ form, and represents a co-homology class. So there must exist a unique harmonic $(0,1)$ form $\alpha \in [i_X\omega]$. If we can show that $\alpha = 0$, we are done, since then $i_X\omega$ would have to be $\bar \partial$-exact which is equivalent to saying that $X$ is Hamiltonian.

$\textbf{Claim}$ : $\bar \alpha (X) = 0$. To see this, we first notice that since we are working on Kahler manifold and $\alpha$ is $\bar\partial$-harmonic $(0,1)$ form, $\bar\alpha$ is a $\bar\partial$-harmonic $(1,0)$ form. So $\bar\partial \bar\alpha = 0$. But then this implies that $\bar\alpha(X)$ is a holomorphic function which vanishes at $p\in M$, and so must be identically zero. We can now complete the proof -

$$\int_{M}|\alpha|^2\frac{\omega^n}{n!} = \int_{M}\bar\alpha\wedge\alpha\wedge\frac{\omega^{n-1}}{(n-1)!} = \int_{M}\bar\alpha \wedge i_{X}\omega \wedge \frac{\omega^{n-1}}{(n-1)!} = \int_{M}\bar\alpha(X)\frac{\omega^n}{n!}=0.$$

The second equality follows from integration by parts, since $\alpha = i_X\omega + \bar\partial f$ for some function $f$, and $\omega$ and $\bar\alpha$ are $\bar\partial$-closed. Hence $\alpha = 0$, and this completes the proof of the Proposition.

So in your problem since the lifted vector does vanish (on all exceptional divisors), it will have a (possibly complex valued) hamiltonian with respect to any Kahler form on $\tilde M$.

  • 1
    $\begingroup$ The point of the question is that he asks for a real potential. $\endgroup$ Nov 8, 2014 at 15:15

Your Answer

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

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