Uniform equivalence of norms induced by Riemannian metrics.

Question

Let $g$ and $h$ be Riemannian metrics on $M$. Since $M$ is finite dimensional, at every point $x\in M$ there are constants $c_x$ and $C_x$, both depending on $x$, such that $$c_xh(v,v)\leq g(v,v)\leq C_xh(v,v),\quad\text{for any $v\in T_xM$}.$$ Assume now that $M$ is compact. Can we find constants $c$ and $C$ independent of $x\in M$ such that $$ch(V,V)\leq g(V,V)\leq Ch(V,V)$$ for any vector field $V$? In other words, are the metrics $g$ and $h$ uniformly equivalent when assuming $M$ is compact?

Answer 1

Let $SM$ be the unit sphere bundle of $TM$ with respect to $h$. As $M$ is compact, $SM$ is compact. Now $g$ induces a continuous map $SM \to \mathbb{R}$ given by $g \mapsto g(v, v)$. As $SM$ is compact, the image is compact and hence contained in a closed interval $[c, C]$. That is, $c \leq g(v, v) \leq C$ for all $v \in SM$ (equivalently, for all $v \in TM$ with $h(v, v) = 1$).

For $V \in T_pM\setminus\{0\}$, set

$$v = \frac{1}{\sqrt{h(V, V)}}V.$$

Note that $h(v, v) = 1$, and

$$g(v, v) = g\left(\frac{1}{\sqrt{h(V, V)}}V, \frac{1}{\sqrt{h(V, V)}}V\right) = \frac{1}{h(V, V)}g(V, V)$$

so multiplying $c \leq g(v, v) \leq C$ through by $h(V, V)$ gives

$$ch(V, V) \leq g(V, V) \leq Ch(V, V).$$

Note, this formula also holds when $V = 0 \in T_pM$.