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Recall Wolpert's lemma:

Let X,Y be hyperbolic surfaces and $f:X\to Y$ a $K$-quasiconformal homeomorphism. For any homotopy class of curves $c$ let $\ell(c)$ denote the length of the geodesic in the class. Then $$\frac{\ell(c)}{K}\leq \ell(f(c))\leq K\ell(c)$$

I am wondering: if $f$ is a $C^1$ diffeomorphism between closed hyperbolic surfaces that has this property, is it $K$-quasiconformal? Of course it is quasiconformal for some other constant, by nature of being $C^1$.

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    $\begingroup$ Since you only ask about homotopy classes of curves, you should be asking instead about the homotopy (or equivalently isotopy) class of $f$, or there are trivial counterexamples. $\endgroup$ Commented Apr 11, 2022 at 12:25

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The answer is no, but it's actually a deep question and leads to another metric on the Teichmüller space of surfaces. The minimum quasi-conformal constant of a map in a given homotopy class is the Teichmüller metric on surfaces; it is also equal to the maximal ratio of extremal lengths of any curves. If you look at the ratio of hyperbolic lengths, as in your question, you get a quite different metric on the space of surfaces, studied by William Thurston: Minimal stretch maps between hyperbolic surfaces. That paper effectively uses an asymmetric version of the inequality ($\ell(f(C)) \le K \ell(C)$), but the symmetrized version is also a very different metric from the standard one given by quasiconformal stretch. Both the asymmetric stretch metric was studied recently in detail by Dumas, Lenzhen, Rafi, and Tao.

The paper The converse of the Schwarz Lemma is false by Maxime Fortier-Borque gives many relevant examples, although that paper is concerned with the slightly different situation of embeddings between non-compact surfaces and maps that decrease length, effectively the case $K=1$ in your questions.

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No. The issue is that quasiconformality constants are much more sensitive to local distortion than lengths. This is morally a $L^\infty$ (quasiconformal constants) to $L^1$ (lengths) comparison: $L^\infty$ bounds on a function on a finite measure space imply $L^1$ bounds, but in general the converse fails.

To sketch a counter-example, take a hyperbolic surface with an $\epsilon$-short curve for $\epsilon > 0$ very small. Such a curve has a long collar neighborhood that is close to a standard form. Leaving the core curve fixed, stretch a small annular neighborhood yielding a dilatation of $2$ at the core curve. Compensate for the stretch in a slightly larger neighborhood to produce a $C^1$ homeomorphism supported in a small annular neighborhood $V_\delta$ of the core curve. Call this map from our hyperbolic surface to itself $f_\delta$. By construction, $K(f_\delta) \geq 2$.

One sees that for any geodesic $c$, the length of $f(c)$ is close to the length of $c$. Taking $\epsilon, \delta$ close to $0$, the numbers $\sup\limits_{c} l(f_\delta(c))/l(c)$ and $\inf\limits_c l(f_\delta(c))/l(c)$ can be made to be arbitrarily close to $1$. The key thing to do here is to bound the distortion of geodesic segments in $V_\delta$ in terms of the homotopy class (rel $\partial V_\delta$) of the segment.

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  • $\begingroup$ Does your contstruction work if one looks at the homotopy class of $f_\delta$, as the question probably should have asked? $\endgroup$ Commented Apr 11, 2022 at 12:28
  • $\begingroup$ @DylanThurston It won't -- this is very much a construction of a bad representative of the isotopy class. One will need to drop the $C^1$ hypothesis to work with extremizers, but after doing so this becomes an interesting question of asymptotic comparison of Teichmuller's metric and W. Thurston's asymmetric metric. $\endgroup$
    – Alex Nolte
    Commented Apr 11, 2022 at 16:03
  • $\begingroup$ Are you sure you can't tweak it? This situation with thin curves is also the place you see that the stretch metric is very asymmetric, and I believe also should show you that the symmetrized stretch metric is different from the standard Teichmüller metric. (Also, I think the $C^1$ hypothesis is harmless, you only need to drop it if you want actual extremizers.) $\endgroup$ Commented Apr 11, 2022 at 17:10

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