Let $n>0$, $\mathbb S^n$ be $n$-sphere and $1\in \mathbb S^n$ be its north pole.

A am looking for an example of compact manifold $M$ with a continuous $n$-parameter family of maps $h_x\colon M\to M$, $x\in \mathbb S^n$ such that

  • $h_1=\mathop{\rm id}_M$
  • For any Riemannian metric $g$ on $M$ and any family of maps $h'_x\sim h_x$ (i.e., such that the maps $h,h'\colon \mathbb S^n\times M\to M$ are homotopic) there is $x\in\mathbb S^n$ such that the Lipschitz constant of $h'_x\colon (M,g)\to(M,g)$ is $\ge 10000$.


  • I think about this problem for few years; I do not think it is an easy one, but I want to make sure that I did not miss a well known trick. A positive answer might have interesting consequences for collapsing with lower curvature bound.
  • If $n=0$, so $\mathbb S^0=\{+1,-1\}$ then $M=\mathbb T^k$ and $h_{-1}=$ a hyperbolic linear map with a big eigenvalue will do.
  • Denote by $F$ the space of all maps $M\to M$. If $\pi_n(F,\mathop{\rm id}_M)$ is finitely generated then for every $h_x\colon (M,g)\to (M,g)$ there is $h'_x$ with a fixed Lipschitz constant $L=L(M,g)$ for all $x$.
  • I know few examples of families $h_x$ for which one can not find a metric $g$ and homotopic family $h'_x$ with Lipschitz constant $=1$ (in other words the family can not be made isometric).
  • I do not know an answer even if instead of 10000 I would have $1+\varepsilon$ for a fixed $\varepsilon> 0$.
  • 3
    $\begingroup$ So is the following rephrasing correct: The question is whether any element in $\pi_k(Map(M,M))$ lies in the image of $\pi_k(Lip_{10000}(M,g))$ for any Riemannian metric $g$ on $M$, where $Lip_C(M,g)$ denotes the space of self maps of $(M,g)$ Lipschitz constant $C$. $\endgroup$ – HenrikRüping Jun 11 '12 at 11:01
  • $\begingroup$ The paper "Homotopical effects of dilation" by Gromov considers similar question, but I haven't checked the exact relationship. Links: projecteuclid.org/euclid.jdg/1214434601 , or (non-paywalled) seven.ihes.fr/~gromov/PDF/4%5B19%5D.pdf $\endgroup$ – Lennart Meier Jun 11 '12 at 12:00
  • $\begingroup$ @Henrik: it is almost correct, you should change "any" to "some". $\endgroup$ – Anton Petrunin Jun 11 '12 at 16:19
  • 1
    $\begingroup$ If it were me, I'd try to start : Let $\Sigma$ be an exotic $n$-sphere. By conjugacy through the round sphere, this carries a very-transitive action by homeomorphisms of $S O (n+1)$; this action cannot be by diffeomorphisms, or else one could use it to give $\Sigma$ an $S O (n+1)$-equivariant Riemann metric... But I can't see whether this is helpful after perturbing away from the subspace of group actions; still, I do expect exotic spheres to be useful somewhere, especially since it is known that some carry obstructions to global curvature positivities. $\endgroup$ – some guy on the street Jul 5 '12 at 16:09

Your Answer

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

Browse other questions tagged or ask your own question.