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Let $S$ be a torus with one boundary component. The group $H_1(S)$ equals $\mathbb{Z}^2$. Fixing some such identification, $(a,b) \in \mathbb{Z}^2 = H_1(S)$ can be represented by a simple closed curve iff and only if $a$ and $b$ are relatively prime.

Let $v_0$ be a point on $\partial S$, and let $\alpha, \beta \in \pi_1(S,v_0)$ be elements with the following properties.

  1. $\{\alpha,\beta\}$ is a free basis for $\pi_1(S,v_0)$.
  2. $[\alpha] = (1,0) \in H_1(S)$ and $[\beta] = (0,1) \in H_1(S)$.
  3. $\alpha$ and $\beta$ can be represented by based simple closed curves that only intersect at the basepoint $v_0$.

If you draw the picture, this is just the standard way of drawing a basis for $\pi_1(S,v_0)$.

This brings me to my question. Fix $(a,b) \in H_1(S)$ such that $a$ and $b$ are relatively prime. What is the shortest word $w$ in $\{\alpha,\beta\}$ such that $[w] = (a,b)$ and $w$ can be represented by a based simple closed curve?

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Take the straight line from $0$ to $(a, b)$ in the usual integer lattice (think of ruled paper). Every time you intersect a horizontal line, write $\alpha,$ every time you intersect a vertical line, write $\beta$ When you are done, you will have your shortest word -- I might be off by one, so you should look at the paper by Osborne and Zieschang from 1981. For much more on this subject, look at papers by G. McShane and I. Rivin.

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It's been years since I've thought about this, but since the torus is punctured I was expecting the answer to be "take the unique representative which is a geodesic for the hyperbolic metric". If it makes sense, can you explain why you are doing something Euclidean rather than hyperbolic here? (Not that I'm doubting your answer...) – Pete L. Clark Apr 17 '11 at 8:11
@Pete: the point is that there are infinitely many hyperbolic geodesics in a given homology class, but only one of them is a simple closed curve. It happens to be the shortest geodesic in the homology class (see the paper of McShane and Rivin in IMRN a few years back). The other geodesics, when drawn in the Euclidean metric, will loop around the lattice points (which are thought of as punctures). – Igor Rivin Apr 17 '11 at 13:36

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