1.Fundamentals of the Ricci Flow Equation

Hamilton’s Ricci flow#

Hamilton’s Ricci flow equation is

• EX 1: Let $M^{n}=S^{n}$ and let $g_{S^{n}}$ denote the standard metric on the unit $n$-sphere in Euclidean space. If $g_{0}=r_{0}^{2} g_{S^{n}}$ for some $r_{0}>0$ ( $r_{0}$ is the radius), then $$\begin{equation*} g(t) \doteqdot\left(r_{0}^{2}-2(n-1) t\right) g_{S^{n}} \tag{2.1} \end{equation*}$$ is a solution to the Ricci flow with $g(0)=g_{0}$ defined on the maximal time interval $(-\infty, T)$, where $T \doteqdot r_{0}^{2} / 2(n-1)$. That is, under the Ricci flow, the sphere stays round and shrinks at a steady rate.

Proof:

• EX 2: (Homothetic Einstein solutions). Suppose that $g_{0}$ is an Einstein metric, i.e., $\operatorname{Rc}\left(g_{0}\right) \equiv c g_{0}$ for some $c \in \mathbb{R}$. Derive the explicit formula for the solution $g(t)$ of the Ricci flow with $g(0)=g_{0}$. Observe that $g(t)$ is homothetic to the initial metric $g_{0}$ and that it shrinks, is stationary, or expands depending on whether $c$ is positive, zero, or negative, respectively.

Proof:

• EX 3: Let $\left(M_{1}, g_{1}(t)\right)$ and $\left(M_{2}, g_{2}(t)\right)$ be solutions of the Ricci flow on a common time interval I. Show that $$\left(M_{1} \times M_{2}, g_{1}(t)+g_{2}(t)\right)$$ is a solution of the Ricci flow. In particular, if $\left(M^{n}, g(t)\right)$ is a solution of the Ricci flow, then so is $\left(M^{n} \times \mathbb{R}, g(t)+d r^{2}\right)$ (we can replace $\left(\mathbb{R}, d r^{2}\right)$ by any static flat manifold).

Proof:

the evolution equation for the scalar curvature#

the evolution equation for the scalar curvature

When $n=2$, since then $\mathrm{Rc}=\frac{1}{2} R g$, we have

Note the similarity to the heat equation $\frac{\partial u}{\partial t}=\Delta u$.

Lemma (Variation of scalar curvature). If $\frac{\partial}{\partial s} g_{i j}=v_{i j}$, then

where $V=g^{i j} v_{i j}=\operatorname{trace}(v)$ is the trace of $v$.

Proof:

Proof of formula (2.2):

• Ex 4: . Let $\left(M^{2}, h\right)$ be a Riemannian surface. Recall that if $g=u \cdot h$ for some function $u$ on $M$, then $$R_{g}=u^{-1}\left(R_{h}-\Delta_{h} \log u\right)$$ Using this equation and the fact that $R_{i j}=\frac{1}{2} R g_{i j}$ when $n=2$, show that $g(t)=u(t) \cdot h$ is a solution of the Ricci flow if and only if $$\begin{equation*} \frac{\partial u}{\partial t}=\Delta_{h} \log u-R_{h} \tag{2.5} \end{equation*}$$