11
$\begingroup$

Let $(M,g)$ be a closed simply-connected Riemannian manifold. Can we find a constant $C$, which depends on $M$, such that for any closed curve $\alpha_0$ with $L(\alpha_0) \le 1$, there exists a homotopy $\{\alpha_s:0\le s \le 1\}$ satisfying

(1) $\alpha_1$ is a point;

(2) For any $0 \le s \le 1$, $L(\alpha_s) \le C$?

Here $L$ denotes the length of a curve.

$\endgroup$
2
  • 4
    $\begingroup$ This is a very nice and natural question. The result is a special case of a much more general result due to Lang and Schlichenmaier and I will write details later when I have time. $\endgroup$ Commented Jan 29, 2019 at 17:49
  • 4
    $\begingroup$ If you want $C$ to depend only on $M$ and not on the metric $g$, then such $C$ can't exit. For example for $M=S^2$ take a double-ball movementfirst.sg/self-myofascial-release/… It has a shortish geodesic that can not be contracted without increasing its length significantly $\endgroup$ Commented Jan 29, 2019 at 23:45

1 Answer 1

17
$\begingroup$

The result is true and in fact we do not need the condition $L(\alpha_0)\leq 1$ since a stronger result is true:

Theorem 1. If $(M,g)$ is a closed simply-connected Riemannian manifold, then there is a constant $C\geq 1$ such that for every closed curve $\alpha_0$ of finite length $L<\infty$, there is a homotopy $\alpha_s$, $0\leq s\leq 1$ between $\alpha_0$ and a constant curve $\alpha_1$ such that $L(\alpha_s)\leq CL$ for all $0\leq s\leq 1$.

Since a reparametrization of a curve does not change its length, we can assume that the curve of length $L$ is parametrized by the constant speed which makes the curve $L/(2\pi)$-Lipschitz as a mapping from $\mathbb{S}^1$ (with arc-length distance) to $(M,g)$. Now the following result is a straightforward consequence of Theorem 5.1 in [1].

Theorem 2. If $M$ is a compact connected and simply-connected Riemannian manifold, then there is $\gamma>0$ such that every $L$-Lipschitz mapping $\alpha:\mathbb{S}^1\to M$ admits a $\gamma L$-Lipschitz extension $A:\overline{\mathbb{B}}^2(0,1)\to M$.

Indeed, using terminology from [1], $M$ is $1$-Lipschitz connected in the small, because it has a finite covering by balls that are bi-Lipschitz homeomorphic to Euclidean balls.

Now, as we pointed out at the beginning, we can assume that $\alpha_0$ is $L/(2\pi)$-Lipschitz. Therefore the extension $A$ is $\gamma L/(2\pi)$ Lipschitz and restrictions of $A$ to circles of radii $0\leq t\leq 1$ will give us a desired homotopy: the restriction to the circle of radius $t$ will have length at most $\gamma Lt\leq \gamma L$. This completes the proof of Theorem 1.

[1] Lang, U., Schlichenmaier, T., Nagata dimension, quasisymmetric embeddings, and Lipschitz extensions. Int. Math. Res. Not. 2005, no. 58, 3625-3655.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .