4

Apropos discussion at this link: locally isometric is not a symmetric relation.

My instructor told me that a counterexample to the symmetric relation here could be actually even simpler than the complete hyperbolic manifold. For example, a map from a circle to real line or from a torus $\mathbb{T}^2$ to $\mathbb{R}^2$. Is this true? If yes, then how do I go about proving this?

-R.

Community
  • 1
r2d2
  • 45
  • 3
    Do you see why $t\rightarrow e^{it}$ is a local isometry? A local isometry is a local homeomorphism. Does there exist a local homeomorphism $S^1 \rightarrow \mathbb{R}$? – Tim kinsella Oct 04 '14 at 02:26
  • It is worth noting from Tim's comment that both examples the OP gives fail to be locally isometric because of topological properties of the spaces (namely that the two spaces are not homeomorphic), as opposed to metric properties. In contrast, a sphere $\mathbb{S}^2$ and a plane with a spherical bump is an example of two spaces which are homeomorphic but not locally isometric (so the failure is in the metric property). See: https://math.stackexchange.com/questions/1350361/if-m-is-compact-and-n-is-connected-then-locally-isometry-is-symmetric – user676464327 Apr 23 '20 at 21:15

1 Answers1

2

Since $S^1$ is compact, any map $f$ of the circle into $\mathbb{R}$ must have a compact connected image. Thus the image must be a closed interval (possibly degenerating to a point).

Now consider a point $x$ on the circle s.t. $f(x)$ is the least upper bound of the image $f(S^1)$. Removing $x$ from a small connected neighborhood $\mathcal{O}$ (any open arc containing $x$ but properly contained in $S^1$ will do) will separate it into two connected components.

But the same is not true of removing $f(x)$ from $f(\mathcal{O})$. Because $f(x)$ is the upper endpoint of $f(S^1)$, $f(\mathcal{O})$ (a subinterval of $f(S^1)$) remains connected with the removal of $f(x)$.

Thus $f$ is not a homeomorphism on a neighborhood of $x$, and certainly not a local isometry there.

On the other hand a uniform wrapping of the real line around the circle, $g:\mathbb{R}\to S^1$, preserving arc length, is a local isometry.

Added:

Here's a more "metric" way of proving no local isometry $f$ maps $S^1$ continuously into $\mathbb{R}$. Suppose one exists, for the sake of contradiction.

Let $\epsilon: S^1 \to \mathbb{R}^+$ denote the largest radius $\epsilon(x)$ for each point $x\in S^1$ such that on the metric ball $\mathscr{B}(x,\epsilon(x))$ centered on $x$, the map $f$ is an isometry. By assumption $\epsilon(x) \gt 0$ at each point, and $\epsilon$ is easily seen to be continuous. Since $S^1$ is compact, there exists positive greatest lower bound $\epsilon_0$, the minimum of $\epsilon(x)$, such that:

$$ \epsilon(x) \ge \epsilon_0 \gt 0 \;\;\; \forall x \in S^1 $$

For any point $x_0$ in $S^1$, consider that $f$ restricted to the ball (arc) $\mathscr{B}(x_0,\epsilon_0)$ is an isometry and thus has as image the real interval $(f(x_0)-\epsilon_0,f(x_0)+\epsilon_0)$. The same argument applies to the endpoints of the arc $\mathscr{B}(x_0,\epsilon_0)$, so that those neighborhoods are mapped to intervals $(f(x_0)-2\epsilon_0,f(x_0))$ and $(f(x_0),f(x_0)+2\epsilon_0)$ in one order or the other.

Piecing these neighborhood restrictions of $f$ together tells us $f$ is an isometry on $(f(x_0)-2\epsilon_0,f(x_0)+2\epsilon_0)$. Since $x_0$ was arbitrary, this shows every point in $S^1$ has a metric ball of radius $2\epsilon_0$ on which $f$ is an isometry. But this contradicts the minimality of $\epsilon_0$ for $\epsilon:S^1 \to \mathbb{R}^+$.

Therefore no local isometry $f$ exists.

hardmath
  • 37,015
  • Thank you hardmath. So, I understand from your answer that local isometry preserves connectedness. However, I wanted to show this using the definition of Euclidean (or some other) metric so that the metric is not preserved in one of the maps. Do you have any idea how should I go about it? – r2d2 Oct 09 '14 at 20:55
  • You've taken several days to think about it, so I guess there's little use in repeating my argument. Shortest arc length (aka Riemannian) metric gives the usual topology on $S^1$. A continuous map of $S^1$ with this topology into $\mathbb{R}$ cannot be locally one-one, so it cannot be a local isometry. Mapping in the other direction successfully does require showing the metric is locally preserved. – hardmath Oct 09 '14 at 21:54