Following the notation of Moroianu's Lectures on Kähler Geometry, we let $(M,g,\Omega,J)$ denote the metric $g$, symplectic form $\Omega$, and complex structure $J$ of a Kähler manifold $M$, satisfying the compatibility condition $g(X,Y) = g(JX,JY) = \Omega(X,JY)$ for vector fields $X,Y$.

Supposing further that $V\lrcorner\ \Omega = dH$ for $H$ a real-valued (explicitly biholomorphic) Hamiltonian potential $H:M\to\mathbb R$, such that V is real-holomorphic (is this correct?) then we immediately have the following Lie derivative relations:

$$ \mathcal{L}_V \Omega = \mathcal{L}_V J = 0\quad\Rightarrow\quad \mathcal{L}_V g = 0 $$

or equivalently this proposition:

Proposition Every real-holomorphic Hamiltonian vector field on a Kähler manifold is Killing.

This proposition is (in essence) a Hamiltonian converse to the following proposition of Moroianu's

Proposition 9.5 (Moroianu) Every Killing vector field on a compact Kähler manifold is real holomorphic.

Three specific questions are asked:

Q1  Is the proposition $\mathcal{L}_V g = 0$ correct (for the assumptions given)?

Q2  Where can it be found in the literature?

Q3  Does it "trivially" imply $\mathcal{L}_V \mathcal{R} = 0$, where $\mathcal{R}$ is the scalar curvature?

Note: My numerical calculations suggest $\mathcal{L}_V \mathcal{R} \ne 0$. The practical question is simply which is buggy: the formal reasoning associated to the proposition, or the code, or the "trivial" expectation that Q3's answer is "yes"?

Deficiencies in my understanding of terms like "real-holomorphic" are plausible (and even likely). Answers/references/general advice are very welcome!

  • $\begingroup$ The current title of this question brings certain bits of pop culture to mind for some people, yes? en.wikipedia.org/wiki/Death%27s_Head $\endgroup$
    – Yemon Choi
    Jun 14, 2013 at 0:13
  • $\begingroup$ On further reflection, it appears that the requiring the Hamiltonian potential H to be biholomorphic and real may not (in general) suffice to ensure that the flow is real-holomorphic. I will grind out some (index-heavy) calculations to check this, but if anyone wants to post an answer, please do so! $\endgroup$ Jun 14, 2013 at 2:01
  • $\begingroup$ @Yemon Choi: yes. $\endgroup$
    – johndoe
    Jun 14, 2013 at 9:15

2 Answers 2


If $\mathcal{L}_v J= 0$ and $\mathcal{L}_v \Omega= 0$, then $\mathcal{L}_v g=0$ so $v$ is a Killing vector field. Indeed, the property $\mathcal{L}_v J= 0$ is equivalent to the condition that the (local) flow of $v$ preserves $J$ and the property $\mathcal{L}_v \Omega= 0$ is equivalent to the condition that the (local) flow of $v$ preserves $\Omega$. Since $g$ is costructable by $J$ and $\Omega$, if both $J$ and $\Omega$ are preserved by the flow, then $g$ is preserved by the flow as well and the vector field is Killing.

Hamiltonian vector fields have the property $\mathcal{L}_v \Omega= 0$. If your definition of ``real-holomorphic'' implies $\mathcal{L}_v J= 0$, then the answer on your question is yes, yes, yes, ya, ya, ya.


Late to the party, but here's some extra nugget of information for the record. If $(M,\omega, J,g)$ is Kähler, $\omega(X,Y) = g(JX,Y)$ implies that $$(\mathcal{L}_X\omega)(Y,Z) = (\mathcal{L}_Xg)(Y,Z) + g((\mathcal{L}_XJ)Y,Z),$$for all fields $X,Y,Z \in \mathfrak{X}(M)$. It follows from this that any two of the following conditions together imply the remaining one: $\mathcal{L}_X\omega = 0$, $\mathcal{L}_Xg = 0$, $\mathcal{L}_XJ = 0$. In other words, any two of the following conditions together imply the remaining one: $X$ is symplectic, $X$ is Killing, $X$ is real-holomorphic.

If we also assume, in addition, that $M$ is compact, then every Killing field is Hamiltonian and real-holomorphic, but while $X$ real-holomorphic implies $JX$ real-holomorphic, $X$ and $JX$ are not both Killing unless $X$ is parallel (with respect to the Levi-Civita connection of $g$). See the second chapter of Einstein Manifolds by Arthur Besse for more details.


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