# Differentiable Manifolds, Metric Space Structure, Lipschitz Continuity

I would like to prove that if $f:M\rightarrow M$ is a continuously differentiable mapping on a compact manifold, then $f$ must be Lipschitz continuous. However, I do not know how to properly interpret what a Lipschitz continuous function on a manifold should be. To define this, I would need a metric space structure on the manifold.

Really, I would like to know how I should formulate and state that proposition. But as a concrete question, is it possible to define a metric $d:M\times M\rightarrow \mathbb{R}$, inducing the topology on $M$, such that we may write $M$ as a finite union of charts so that the coordinate functions are isometries between the Euclidean metric on $\mathbb{R}^n$ and the metric $d$ on $M$?

I know very little beyond the definition of a differentiable manifold.

The short answer to the question in your second paragraph is: No.

If you know very little beyond the definition of differentiable manifold, then it's hard to explain everything but...

In a literal sense, you can't talk about Lipschitz functions on $M$ without a metric on $M$, but to explain the answer to your question: If you have a smooth, compact manifold whose topology arises from a metric (in the sense of a function $d : M \times M \to \mathbb{R}$) then you can use that metric to define a Riemannian metric on the manifold and then it is a well-known fact that the manifold can only be locally isometric to Euclidean space if it is flat (has no Riemannian curvature). But there are lots of manifolds that have Riemannian curvature.

You can introduce a metric on $M$ wich induces the topology. This is a standard result from differential geometry. One way of doing this is to introduce a scalar product (usually denoted $g$, also called a Riemannian metric) on each tangent space in a smooth way. It is not difficult to show that this is possible by patching it together from local charts with a partition of unity. Once you have a scalar product you can define the lengths of smooth curves $\gamma:[0,1]\rightarrow M$ by defining $$\ell(\gamma) := \int_0^1 \sqrt{g_{\gamma(t)}(\gamma^\prime(t), \gamma^\prime(t))}d\, t$$

Once you have a notion of length you can define $$d(p,q):= \inf\{\ell(\gamma):\gamma \text{ is a smooth curve from p to q}\}$$ (Assume $M$ to connected)

It can then be shown that this is a metric on $M$ which, in fact, is compatible with the topology. And of course you can then define Lipshitz continuity for $f:M\rightarrow M$ in the usual way.

Alternatively you can embed $M$ into some $\mathbb{R}^d$ -- according to a theorem of Whitney this is always possible. Then you can define a metric by requiring this to be an isometry.

In general $M$ will, however, not be locally isometric to some $\mathbb{R}^n$ (regardless of how you choose the metric).

For details you should consult textbooks on Differential Geometry.

• Thanks for the response. There are metrics on R that induce the same topology but disagree about which functions are Lipschitz continuous. Will the metrics obtained from different choices of a Riemannian metric have some relationship with one another? Will they be strongly equivalent, for example. Commented May 25, 2017 at 14:24
• @user3281410 You assumed $M$ to be compact. In general ($M$ not necessarily compact) you may get different classes of (globally) Lipshitz continuous functions, but not locally. So on compact manifolds there is no such phenomenon. On a compact manifold you can always estimate two Riemannian metrics against each other, which implies that the derived quantities (e.g. distance on $M$ or norm of a derivative w.r.t either Riemannian metric can be also esitmated against each other. With the norm of $Df$ you can then find a Lipshitz constant for $f$ (which may in addition depend on the geometriy of M) Commented May 25, 2017 at 14:46