# What is the minimal length of a “Diagonal” in a Torus?

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.

• 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.
– HJRW
May 19 at 15:03
• 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? May 19 at 15:53
• @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!).
– HJRW
May 19 at 19:35

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 $$L^2(ab)=L^2(a)+L^2(b)+2\sqrt{L^2(a)L^2(b)-A^2}.$$
• 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. May 20 at 3:23
• Could you give examples of non-flat Riemannian (or piecewise Euclidean) tori with $l(ab)$ approaching $l(a)+l(b)$?