We defined a Ricci soliton as a Riemannian manifold $(M,g_0)$ such that $$\mathrm{Ric}_{g_0}+\frac{1}{2}\mathcal{L}_{X}g_0=\lambda g_0$$ for some constant $\lambda$ and some vector field $X\in TM$, where $\mathcal{L}_X$ denotes the Lie derivative. For each $p\in M$, let $\phi_t(p)$ solve the ODE $$\frac{\partial}{\partial t}\phi_t(p)=\frac{1}{1-2\lambda t}X|_{\phi_t(p)}, \qquad \phi_0(p)=p$$I would like to verify that the family of metrics $$g(t)=(1-2\lambda t)\phi^{\ast}_t(g_0)$$ solves the Ricci flow, i.e. that $\frac{\partial}{\partial t}g(t)=-2\mathrm{Ric}_{g(t)}$. So I try differentiating: $$\frac{\partial}{\partial t}g(t)=-2\lambda \phi^{\ast}_t(g_0)+(1-2\lambda t)\frac{\partial}{\partial t}\phi^{\ast}_t(g_0).$$ How do I calculate $\frac{\partial }{\partial t}\phi^{\ast}_t(g_0)$? Do I need to use metric compatibility?


1 Answer 1


The easy case is when $\lambda = 0$, since then $\phi_t$ is the flow of $X$ and we have $\partial_t \phi_t^* g_0 = \phi_t^*\mathcal L_X g_0$ by the definition of the Lie derivative. (At $t=0$ this is literally the definition; at other times you also need the group law $\phi_{t+s} = \phi_t\phi_s$)

In order to tackle the general case, we can look for a reparametrization $t(s)$ that turns the ODE into the flow of some vector field, and then apply the same formula for the Lie derivative. By the chain rule we have \begin{align} \partial_s \phi_{t(s)}(p) &= \frac{dt}{ds}(\partial_t \phi)_{t(s)}(p) \\ &= \frac{dt}{ds}\frac{1}{1-2 \lambda t(s)} X(\phi_{t(s)} p); \end{align}

so the solution to the ODE

$$\begin{cases} t'(s) = 1 - 2 \lambda t(s) \\ t(0) = 0 \end{cases}$$

lets us write $\phi_t = \psi_{s(t)}$ where $\psi_s$ is the flow of $X.$ (You can solve this ODE by hand to confirm the solution is invertible on its maximal domain of existence.)

We can now compute the derivative of the pullback using the chain rule: we have $$\partial_t \phi_t^* g_0= \partial_t \psi_{s}^* g_0 = s'(t)\partial_s \psi^*_{s} g_0 = \frac{1}{1-2 \lambda t}\phi_t^*\mathcal L_X g_0.$$


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