# rational rotation vector and closed curves

On $\mathbb{T}^n$ with a Riemannian metric, the stable norm is defined as $$\Vert h\Vert=\inf \sum |r_i| \cdot \mathrm{length}(\sigma_i),$$

where $h\in H_1(\mathbb{T}^n,\mathbb R)$ and $\sum_i r_i\sigma_i$ is a Lipschitz cycle representing $h$. The $\inf$ is taken over all such cycles.

The question is, if we consider

(1)$h$ is completely rational, i.e. there are n-1 linearly independent integer vectors perpendicular to $h$,

(2) The $\inf$ is achieved on a set supporting a measure that is uniquely ergodic. (This excludes Hedlund's counterexample)

then is it correct that the $\inf$ is always achieved by a cycle, i.e. the measure is supported on a closed orbit?

This is true in 2-dim case, but not sure about the high dim case.

## 1 Answer

Let $\mathbb T^n=\mathbb R^n\pmod {2{\cdot}\pi}$. Consider the embedding $f\colon\mathbb T^2\hookrightarrow \mathbb T^3$ defined as $$f\colon(u,v)\mapsto (\sin u,\cos u, v).$$

It is easy to find a metric on $\mathbb T^3$ so that for $h=(0,0,1)$, the infimum is achieved on a geodesic which runs in the embedded $\mathbb T^2$ in an irrational direction.

Is that what you wanted?

• Thanks very much. This is a nice example. I am also trying to exclude the possibility of Denjoy minimal set, i.e. if we take a section to the measure, we get a Cantor set. In general, orbits in Denjoy minimal set has hyperbolicity, so it is different from your example. Is it possible to have a Denjoy minimal set as support of measure with rational rotation vector? – John Galt Aug 13 '13 at 14:39