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.

$\endgroup$
  • 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. $\endgroup$ – Piotr Hajlasz Jan 29 at 17:49
  • 3
    $\begingroup$ 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 $\endgroup$ – Dmitri Panov Jan 29 at 23:45
16
$\begingroup$

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. https://arxiv.org/abs/math/0410048

$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.