# How to prove the holonomy group is preserved under Ricci flow?

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.

-
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 – matgaio Apr 30 '12 at 19:36
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. – wspin Sep 5 '14 at 8:39

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 $t \in (0, T)$.