# Every Riemannian metric is conformally related to a complete metric

If $(M,g)$ is a Riemannian manifold, there is a metric $\tilde g=hg$ on $M$ which is complete, where $h$ is a positive smooth function. I've been given the hint to let $f:M\to \Bbb R$ be a smooth exhaustion function, which means that $f^{-1}(-\infty,c]$ is compact for all $c\in\Bbb R$. Then one should pick $h$ so that $f$ is bounded on $\tilde g$-bounded sets ($\tilde g$-bounded means bounded wrt. the distance function induced by $\tilde g$). (I don't want to repeat the classical proof by Nomizu and Ozeki.)

Assuming such an $h$ can be found, I understand how to complete the proof. I think that $h$ should be of the form $\mathrm e^{-2f}$ or $f^{-2}$ so that distances wrt. $\tilde g$ are shorter than wrt. $g$, so that the bounded sets are "shrunk."

I tried to assume the opposite, namely I chose one of those $h$s and assumed $f$ is unbounded on a $\tilde g$-bounded set $B$. Letting $(y_n)$ be a sequence in $B$ such that $f(y_n)>n$ for all $n$, I tried to contradict that $B$ is bounded, but nothing seems to work. (I just end up with trivial inequalities and the like.)

I'd appreciate a hint on how to start. My $h$ is probably wrong, but even if I had the right $h$, I'm not sure how to link $\tilde g$-boundedness to the boundedness of $f$.

One way to solve this problem is to use Whitney embedding theorem: Embed $M$ in $R^{2n+1}$ via a proper map $i: M\to R^{2n+1}$ so that $0\notin i(M)$. Then the function $f: x\mapsto ||i(x)||$ is a smooth exhaustion function on $M$. (Properness of $i$ translates to the property that $f$ is an exhaustion function.)

Edit. In the above I explained how to construct an exhaustion function: It appears that you already know how to do this. I will now explain how to find $h$. Let $(\bar{M}, \bar{d})$ denote the metric completion of $(M, d)$ where $d$ is the distance function defined via the metric $g$. If $\bar{M}=M$, you are done, hence, I will assume that the complement $\bar{M}-M$ is nonempty. Define the function $\delta(x)$ on $M$ to be the distance function from $x$ to the closed subset $\bar{M}- M$ in the metric space $(\bar{M}, \bar{d})$. This is a positive 1-Lipschitz function, hence, it is continuous. Now, set $$h_o(x)= \delta^{-1}(x).$$ I will leave you to verify that the continuous Riemannian metric $g_o=h_o^2g$ yields a complete distance function on $M$. (This amounts to computing an elementary integral.) In order to find a smooth metric $\tilde{g}=h^2g$ conformal to $g$ you approximate $h_o$ by a smooth function $h$ so that $$||\tilde{g} -g||<1.$$ (This can be done for instance by using a smooth proper embedding of $M$ into $R^{2n+1}$ and then applying the classical Weierstrass approximation theorem.) What does it have to do with the exhaustion function $f$ I do not know (I think, none). I also do not know how similar is this proof to the classical one.

• Allow me to bother you with a few questions, please. 1) Why is $\bar M \setminus M$ closed in $\bar M$? I am thinking of $M = \mathbb Q$, but it is true that this is not a Riemannian manifold. 2) What is the elementary integral? I cannot see it. 3) How do you define the norm on the space of tensors? 4) Is it $\| \tilde g - g \|<1$ or $\| \tilde g - g_0 \|<1$? 5) Why is $\tilde g$ complete? Thank you for the clarifications. Jun 9 '18 at 14:54

I, too, got stuck on this problem, even though I was able to prove it in other ways (as in the classical proof, and the one using the Whitney embedding theorem).

Here's my progress so far. If you look at the construction of the smooth exhaustion function (Prop. 2.28 in Lee's "Introduction to Smooth Manifolds", where this problem is from), it actually guarantees that $f \ge 1$, which may be useful.

Given a $\tilde{g}$-bounded set $B$, and two points $p$ and $q$ such that $d(p, q) = D \le K$, let $\gamma$ be the $\tilde{g}$-geodesic of unit speed from $p$ to $q$. (EDIT: Never mind the rest; such a geodesic isn't guaranteed to exist!) Then we have $$\int_0^D \sqrt{h(\gamma(t)) g_{\gamma(t)}(\gamma'(t), \gamma'(t))} \, dt = D\text{,}$$ where the integrand is equal to $\sqrt{\tilde{g}_{\gamma(t)}(\gamma'(t), \gamma'(t))}$, and is thus equal to $1$. Therefore, $$h(\gamma(t)) g_{\gamma(t)}(\gamma'(t), \gamma'(t)) = 1\text{.}$$ However, without a way to bound $g$ also, I don't see how to define $h$ purely in terms of $f$ to get a bound on $f$.

• Since you yourself noticed that $\gamma$ might not exist, and since even this mistaken approach was incomplete, does it make sense to keep this answer? MSE is not suposed to be a draft of failed attempts. Jun 9 '18 at 14:14

For some reason I am finding it easier to work directly with an exhaustion by compact sets rather than an exhaustion function, but these tools are equivalent anyway.

Let $K_1 \subset \mathring K_2 \subset K_2 \subset \ldots \bigcup_{n\in \mathbb{N}}K_n = M$ be an exhaustion by compact sets. Let $\delta_n:= d_{g}(\partial K_n, \partial K_{n+1})$ be the distance between boundaries of successive compact sets in the original metric, and define a piecewise-constant function $h':= \sum_{n\in \mathbb{N}} \frac{1}{n \cdot \delta_n} \cdot \chi_{(K_n \setminus K_{n-1})}$ , with $\chi$ denoting an indicator function. Now if $(x_n)_{n\in \mathbb{N}}$ is a sequence in $M$ with $x_n \to \infty$ (i.e. for any compact $K$, there is $N > 0$ such that $\forall n > N: x_n \not \in K$), it follows that $d_{h'g}(x_1,x_n) \to \infty$ since the sequence $d_{h'g}(x_1,x_n)$ coincides with the tails of partial sums of the harmonic series.

Now approximate $h'$ from above by a smooth function $h$ (first construct by hand a continuous approximation from above, then approximate it from above by a smooth function). Since $h > h'$, it again follows that for any sequence $(x_n)_{n\in \mathbb{N}}$ with $x_n \to \infty$, $d_{hg}(x_1,x_n) \to \infty$.

Define the Riemannian metric $\tilde{g}:=h g$. Now supposing that $(x_n)_{n\in \mathbb{N}}$ is a $\tilde{g}$-Cauchy sequence, it follows that this sequence is $\tilde{g}$-bounded. The contrapositive of the last paragraph's claim shows that $(x_n)_{n \in \mathbb{N}} \subset K$ for some compact $K$, so $(x_n)_{n\in \mathbb{N}}$ has a convergent subsequence to a point $p \in K$ and hence $x_n \to p$, since the sequence was $\tilde{g}$-Cauchy and the manifold topology is compatible with the $\tilde{g}$ metric topology.

Here's one approach using an exhaustion function.

Let $$f:M\to \mathbb R$$ be a smooth exhaustion function for $$M$$. Define $$h:M\to \mathbb{R}$$ by $$h=\sqrt{df(\operatorname{grad}f)+1}=\sqrt{|\operatorname{grad}f|_g^2+1}$$. (We are adding $$1$$ so that $$h$$ is smooth and positive everywhere). Note that the Schwarz inequality implies that $$h(p)\cdot|v|_g>|(\operatorname{grad}f)_p|_g\cdot|v|_g\geq \langle(\operatorname{grad}f)_p,v\rangle_g=df_p(v)$$ for all $$p\in M$$ and $$v\in T_pM.$$

Now let $$\tilde{g}=hg$$. For any piecewise smooth curve segment $$\gamma:[a,b]\to M$$, we have $$\begin{eqnarray}\text{L}_{\tilde{g}}(\gamma)&=&\int_{a}^{b}|\gamma'(t)|_{\tilde{g}}dt\\&=&\int_{a}^{b}h(\gamma(t))|\gamma'(t)|_{g}dt\\&\geq&\int_{a}^{b}df_{\gamma(t)}(\gamma'(t))dt\\&=&f(\gamma(b))-f(\gamma(a)),\end{eqnarray}$$ which implies that $$f$$ is bounded on every $$\tilde{g}$$-bounded sets.