Compactness of a certain set coming from the Ricci Flow I was wondering about the following situation:
Suppose we have a solution to the Ricci Flow on a compact manifold $M$, $g(t)=g_t$, on a compact time interval $[0, \delta]$, and we consider $\{(t,v)\in [0, \delta] \times TM \mid g_t(v,v)=1\}$. The question is very simply, "Is this set compact?"
It seems like it should be, since if you fix a time $t$, then we're just looking at the sphere bundle of $M$ (which is compact), so more or less the set is like a product of the sphere bundle with a compact interval. I'm just having an issue trying to pin down that more or less part so that's it's precise.
 A: Your intuition is spot on. The most direct way to formalize it is probably to consider the map $F : [0,\delta]\times UTM \to [0,\delta] \times TM$ given by $F(t,u) = (t,(g_t(u,u))^{-1/2}u)$ where $UTM = \{ u \in TM : g_0(u,u)=1 \}$ is the unit sphere bundle in the initial metric.  Since all the metrics $g_t$ are non-degenerate and they vary smoothly in time, this is a continuous map; and its image is the set you are interested in. Thus the compactness of $[0,\delta] \times UTM$ gives you your desired result.
A: I will record a solution I obtained earlier for posterity.
I will write $g_t(v)$ for the length of $v$ at time t. Consider $UTM=\{v\in TM \mid g_0(v)=1\}$, and the quantities $\displaystyle\max_{v\in UTM} g_t(v)$ and $\displaystyle\min_{v\in UTM} g_t(v)$. These are continuous functions of the variable $t$, as $UTM$ is compact. Let $M=\displaystyle \max_{t\in [0,\delta]} \max_{v\in UTM} g_t(v)$ and $\mu=\displaystyle \min_{t\in [0,\delta]} \min_{v\in UTM} g_t(v)$, both of which are necessarily positive since the metric is positive definite. What we're doing is fixing the sphere bundle at the initial time and ascertaining how extreme possible distortions of length are becoming as we move forward in time. A key point is the continuity of the "spatial" maximums as a function of time, in order that they have extrema which are obtained.
Thus, for any $v\in UTM$, $\mu \le g_t(v) \le M$ for all $t$. Now we want to see how "far away" the unit tangent bundle at later times is from $UTM$, so suppose $w$ is such that $g_t(w)=1$. Scale $w$ by $\lambda>0$ so that $\lambda w=v$ where $v\in UTM$. Then by the inequalities above, $\mu\le g_t(v)=g_t(\lambda w)=\lambda g_t(w)=\lambda \le M$. Thus $w=\frac{1}{\lambda}v$ where $\frac{1}{M} \le \frac{1}{\lambda} \le \frac{1}{\mu}$. The conclusion is that, for any $(t,w)$ in the set of interest (call it $\Sigma$), $(t,w)$ must lie in $[0, \delta] \times \{v\in TM \mid \frac{1}{M} \le g_0(v) \le \frac{1}{\mu} \}$; the second factor is clearly compact, so $\Sigma$ is contained in a compact set. However, $\Sigma$ is manifestly closed, and closed subsets of compact sets are compact, so this finishes the proof.
