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Given a Riemannian torus $(T,d)$ with fundamental group $\pi_1(T)=\langle a,b \mid ab=ba \rangle$. Denote for any $\gamma \in \pi_1(T)$ the infimum length of all representatives of $\gamma$ by $L(\gamma)$.

Is there a way to get a general upper bound on $L(ab)$ in terms of $L(a),L(b)$ and the volume of $T$?

An obvious one is $L(ab) \leq L(a) + L(b)$, but i guess, there are better results.

This is inspired by Loewner's torus inequality.

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    $\begingroup$ Can't $l(ab)$ be computed exactly using trigonometry? Lift everything to the universal cover, the Euclidean plane. The volume of the torus is the area of the fundamental parallelogram, which is $l(a)l(b)\sin\theta$ where $\theta$ is the angle between $a$ and $b$. Now the geodesic representative of $ab$ is the diagonal of the parallelogram opposite $\theta$, and its length can be computed using the cosine rule. $\endgroup$
    – HJRW
    Commented May 19, 2021 at 15:03
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    $\begingroup$ Is the metric on the torus meant to be flat, in which case HJRW's remark would be my answer? Or do we have varying curvature, in which case I'll be quiet and see what the actual differential geometers have to say? $\endgroup$
    – David E Speyer
    Commented May 19, 2021 at 15:53
  • $\begingroup$ @DavidESpeyer: you're quite right, I misread the question and thought the torus was assumed to be flat. My guess is that this is that the Euclidean case is also the extremal one, but it would be interesting to see a proof (or counterexample!). $\endgroup$
    – HJRW
    Commented May 19, 2021 at 19:35

1 Answer 1

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There is no better upper bound independent of the area. Suppose that your torus is the factor of the plane by integer lattice, and the metric is Euclidean. Let $a=m$ and $b=m+1$. These generate the lattice. And length of $ab$ is $\sqrt{4m^2+1}\sim 2m, m\to\infty$.

If you want to include area, you can obtain an exact formula, rather than an inequality. Denote the angle between $a$ and $b$ by $\phi$, then $L^2(ab)=L^2(a)+L^2(b)+2L(a)L(b)\cos\phi,$ while the area $A=2L(a)L(b)\sin\phi$. Eliminating $\phi$ we obtain $$L^2(ab)=L^2(a)+L^2(b)+2\sqrt{L^2(a)L^2(b)-A^2}.$$

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  • $\begingroup$ About the second part. The metric is not assumed to be flat, so for sure you will not get equality there. $\endgroup$
    – Anton Petrunin
    Commented May 19, 2021 at 20:16
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    $\begingroup$ If you assume nothing about the curvature of the metric, then there is no inequality involving area, and $L(ab)\leq L(a)+L(b)$ is the only thing you can say. $\endgroup$
    – Alexandre Eremenko
    Commented May 20, 2021 at 3:23
  • $\begingroup$ Thank you for your answer. Is there a possibility if we consider a triangulation with a piecewise flat metric? Or instead of triangles use quadrilaterals with piecewise euclidean metric. $\endgroup$
    – Sebastian
    Commented May 20, 2021 at 6:31
  • $\begingroup$ Could you give examples of non-flat Riemannian (or piecewise Euclidean) tori with $l(ab)$ approaching $l(a)+l(b)$? $\endgroup$
    – HJRW
    Commented May 20, 2021 at 7:52
  • $\begingroup$ Yes, of course. Consider my example 1. Now triangle with sides a,b,ab lies entirely inside a fundamental region. Change the metric inside this triangle to achieve arbitrary area. You can make it arbitrarily large by gluing in a bubble, or arbitrarily small by choosing a metric of very negative curvature, without affecting the lengths of a,b,ab at all. My exact claim is that you can choose the three lengths as on a Euclidean torus, and obtain arbitrary area. $\endgroup$
    – Alexandre Eremenko
    Commented May 20, 2021 at 12:47

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