# Why $-2Ric=\mathcal{L}_vg$?

$\varphi_t$ is a one-parameter group of diffeomorphisms generated by a vector field $V$ on $M$,

$$g_{ij}=\varphi_t^*g_{ij}(x,0) ~~~ \frac{\partial g_{ij}}{\partial t}=-2R_{ij}$$

How to show that $-2Ric=\mathcal{L}_vg$ ? The $\mathcal{L}_vg$ is Lie derivative

I only have a little knowledge of Riemannian geometry and PDE.But I must read some paper about Ricci flow .What should do? What book I should read before read Ricci flow? Thanks for any helping .

You already have it! I assume $v$ in your Lie derivative is the vector field tangent to the orbits of $\phi_t$. If that is the case then $v=\partial/\partial_t$ and the Lie derivative of $g_{ab}$ is just $\partial_t g_{ab}$.
• How compute $\mathcal{L}_vg_{ij}=g_{ij}\nabla_jV^k+g_{jk}\nabla_iV^k$? Commented Oct 6, 2015 at 13:38
• I think it should be $\mathcal{L}_vg_{ij}=g_{ij}\nabla_jV^k+g_{jk}\nabla_iV^k+V^k\frac{\partial g_{ij}}{\partial x^k}$, but the it is above one on the book. Commented Oct 6, 2015 at 13:44