Hello, I am reading the paper Futaki; Ono; Wang Transverse Kähler geometry of Sasaki manifolds and toric Sasaki-Einstein manifolds. J. Differential Geom. 83 (2009), no. 3, 585–635.

For your convenience I repeat the setting. Let $(M^{2n+1}, \eta, \xi, g)$ be a compact Sasakian manifold and $\pi_\alpha: U_\alpha \rightarrow C^n$ the submersions defining the characteristic foliation on $M$.

Hamiltonian holomorphic vector fields are introduced to be complex vector fields $X$ on $M$ such that the field $d\pi_\alpha X$ is holomorphic and the function $u_X := i \eta(X)$ is such that $\iota_X \frac 1 2 d\eta = i \bar \partial_B u_X$.

Let $h$ be a real basic function such that $\rho^T - (2n+2)\omega^T = i \partial_B \bar \partial_B h$ (transverse Ricci form and transverse Kaehler form).

They consider normalized vector fields. They are the ones such that $\int_M u_X e^h (1/2 d\eta)^n \wedge \eta = 0$.

It can be checked that the set of Hamiltonian holomorphic vector fields is closed under the Lie bracket.

Is it true also for normalized ones?

After using Theorem 5.1 of the paper which states that the space of normalized ham holo flds is in correpspondance with the $2n+1$ eigenspace of the operator on complex valued basic functions $$ \Delta^h u = \Delta u - \nabla^i u \nabla_i h $$ where $\Delta$ is the basic laplacian and $\nabla$ the transverse LC connection.

The correspondence is given by $u \mapsto u \xi + \nabla^i u (\partial_{z_i} - \eta(\partial_{z_i})\xi)$. Btw this is where I have another question. Shouldn't it be $u \mapsto -iu \xi + ...$ in order to have a hamiltonian holomorphic vector field in the above sense?

Anyway I computed the bracket of two fields $X, Y$ image of basic functions $u,v$ and found out that the Hamiltonian function of $[X,Y]$ is $Xv - Yu = \nabla^i u \nabla_i v - \nabla^i v \nabla_i u$. Its itegral with respect to the weighted measure is then $$ \int (\Delta^h u \cdot v - u \Delta^h v)e^h (1/2 d\eta)^n \wedge \eta $$ which is not the self-adjoint relation for $\Delta^h$, which would contain some conjugates, being $\Delta^h$ the $\bar \partial$-Laplacian wrt the weighted measure. This is where I am stuck.

thank you



1 Answer 1


The equation

$$\int (\Delta^h u \cdot v - u \Delta^h v)e^h (1/2 d\eta)^n \wedge \eta$$

already gives a proof. $u$ and $v$ are eigenvalues of $\Delta^h$, so the integrand is zero.

This covers the case of $c_1(\mathcal{F})=ad\eta,\ a>0,$ but I'm pretty sure the proof works for general Sasakian manifolds. We have a Lie algebra homomorphism

$$ (\operatorname{Ham},[,]) \rightarrow (\chi,[,]) $$

taking $u_X$ to $X$ with poisson bracket $[u ,v]= \nabla^i u \nabla_i v - \nabla^i v \nabla_i u$ which has intgral zero. So there is always of splitting of the Lie algebra map when $M$ is compact.

  • $\begingroup$ Thank you, Craig! When you talk about the splitting of the Lie algebra you mean a sasakian analog of the decomposition of the Lie algebra of holomorphic fields that holds in the Kahler case? In this case for the algebra of normalized ones ofc. Anyway without using it is a subalgebra I already managed to write a decomposition by using that it is a set of representatives for the quotient {Ham. holo flds}/ $\xi$, in analogy with other splittings holding in the Sasaki-extremal case and more in general for transvesely Kahler foliations (Tondeur-Nishikawa) $\endgroup$
    – David P
    Commented Apr 4, 2013 at 7:58
  • $\begingroup$ I may have not been very clear. $(\chi,[,])$ denotes the transversely holomorphic vector fields with Hamiltonian potentials. The kernel of the above map is just $C$, constant functions. So the splitting is just to Hamiltonian potentials of integral zero. $\endgroup$
    – Craig
    Commented Oct 27, 2013 at 8:39

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