$\varphi_t$ 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 .


1 Answer 1


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 $\partial_t g_{ab}$.

I do not know about Ricci flow in particular, but I suggest you get a good grounding in differential and Riemannian geometry first. For a general introduction to differential geometry I would look at Tu’s Introduction to Manifolds or Lee’s Smooth manifolds. For Riemannian geometry therea re good books by John Lee again or Do Carmo.

  • $\begingroup$ How compute $\mathcal{L}_vg_{ij}=g_{ij}\nabla_jV^k+g_{jk}\nabla_iV^k$? $\endgroup$
    – lanse2pty
    Oct 6, 2015 at 13:38
  • $\begingroup$ 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. $\endgroup$
    – lanse2pty
    Oct 6, 2015 at 13:44

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