Normalized Hamiltonian holomorphic vector fields on Sasakian manifolds - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T15:29:38Z http://mathoverflow.net/feeds/question/119590 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/119590/normalized-hamiltonian-holomorphic-vector-fields-on-sasakian-manifolds Normalized Hamiltonian holomorphic vector fields on Sasakian manifolds David Petrecca 2013-01-22T18:26:40Z 2013-04-04T01:14:28Z <p>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. </p> <p>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 $M$.</p> <p>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 $\iota_X \frac 1 2 d\eta = i \bar \partial_B u_X$.</p> <p>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).</p> <p>They consider normalized vector fields. They are the ones such that $\int_M u_X e^h (1/2 d\eta)^n \wedge \eta = 0$.</p> <p>It can be checked that the set of Hamiltonian holomorphic vector fields is closed under the Lie bracket.</p> <p>Is it true also for normalized ones?</p> <p>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.</p> <p>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?</p> <p>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 <em>not</em> 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.</p> <p>thank you</p> <p>David</p> http://mathoverflow.net/questions/119590/normalized-hamiltonian-holomorphic-vector-fields-on-sasakian-manifolds/126465#126465 Answer by Craig for Normalized Hamiltonian holomorphic vector fields on Sasakian manifolds Craig 2013-04-04T01:14:28Z 2013-04-04T01:14:28Z <p>The equation </p> <p>$$\int (\Delta^h u \cdot v - u \Delta^h v)e^h (1/2 d\eta)^n \wedge \eta$$</p> <p>already gives a proof. $u$ and $v$ are eigenvalues of $\Delta^h$, so the integrand is zero.</p> <p>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</p> <p>$$(\operatorname{Ham},[,]) \rightarrow (\chi,[,])$$</p> <p>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.</p>