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I've heard that on a Kähler manifold $(M,g_0)$, if you evolve the metric $g$ by Ricci flow $\partial g_{ij}(t)/\partial t=-2R_{ij}$, and $g(0)=g_0$, then you always have $g(t)$ is a Kähler metric on $M$.

All the references I saw refer this fact to that the holonomy group of $(M,g(t))$ is preserved under Ricci flow, but I don't know how to prove it.

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  • $\begingroup$ I know it is not too much, but I can suggest Chow's "The Ricci Flow: an introduction". You can find it for download here: 4shared.com/office/xU-k3g2N/the_ricci_flowan_introduction_.html $\endgroup$
    – matgaio
    Commented Apr 30, 2012 at 19:36
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    $\begingroup$ In the case $M$ is simply connected this should follow since the Holonomy group changes continously with $t$ and by the Berger Holonomy classification this means that it must be constant. $\endgroup$ Commented Sep 5, 2014 at 8:39

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Hamilton outlines the proof that the reduced holonomy group does not expand under Ricci flow in theorem 4.1 of "The Formation of Singularities in the Ricci Flow." The argument uses a maximum principle applied to the curvature operator. Kotschwar's "Ricci Flow and the Holonomy Group" shows that the reduced holonomy group is preserved in backwards time as well. Consequently, the reduced holonomy is exactly preserved by Ricci flow.

An alternative argument applies in the particular case of an initially Kahler manifold evolving by Ricci flow. Namely, the associated 2-form $\omega_t = g_t(J \cdot, \cdot)$ satisfies $$\frac{\partial}{\partial t} d \omega_t = d \rho_t, \quad d\omega_0 = 0, \text{ and} \quad d\rho_0 = 0,$$ where $\rho = Ric(J \cdot, \cdot)$ is the Ricci form. With a suitable uniqueness result on solutions to this differential equation, one can then conclude that $d\omega_t = 0$ for all $t \in (0, T)$. That is, $(M, g_t, J, \omega_t)$ is Kahler for all $t \in (0, T)$.

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