How big can the Hausdorff dimension of the closure of a smooth curve be?

Question

Consider curves in $\mathbb{R}^n$. Smooth curves have Hausdorff dimension $1$.

The closure of a smooth curve can have Hausdorff dimension $> 1$. (For example, a curve dense in a torus.)

(Can it be $> 2$? For instance, can there be a smooth curve dense in a solid torus?)

How are you defining a smooth curve? Does it just have to be a smooth mapping $\gamma: [0,1] \to \mathbb{R}^n$, or must it be regular as well ($\gamma' \neq 0$)? See my comment on Vasily Mitch's answer below. — Michael Seifert, Aug 16 '19 at 18:20

Travis Willse · Accepted Answer · 2020-01-09T10:23:33.860

6

We can extend the example you mention, sometimes called an irrational winding of a torus, to higher-dimensional tori $\Bbb T_n$:

Pick a constants $\alpha_1, \ldots, \alpha_n \in \Bbb R$ such that $1, \alpha_1, \ldots, \alpha_n$ are linearly independent over $\Bbb Q$, and denote $$\alpha := (\alpha_1, \ldots, \alpha_n) \in \Bbb R^n .$$ Then, the image of the curve $$\gamma : \Bbb R \to \Bbb T_n \cong \Bbb R^n / \Bbb Z^n, \qquad\gamma(t) := t \alpha \pmod {\Bbb Z^n}$$ is dense in $\Bbb T_n$.

(For $n = 2$ the linear independence condition is just that the ratio $\frac{\alpha_2}{\alpha_1}$ is irrational.)

edited Jan 09 '20 at 10:23

answered Aug 16 '19 at 17:50

Travis Willse

99,363

Thanks! You might also like: Hausdorff dimension of Hamiltonian orbit closure and symplectic leaves – Ricardo Buring Aug 16 '19 at 18:05
You're welcome, I hope you found it useful. I looked at your other question, but I'm too rusty on my symplectic geometry to answer it without doing some review. But it's a good question, so with luck it will get the attention of someone who can. – Travis Willse Aug 16 '19 at 18:11
How to prove the image of $\gamma$ is dense in $\Bbb T^n$? – math112358 Jan 09 '20 at 00:57
See https://math.stackexchange.com/questions/1210717/an-irrational-flow-on-a-torus-is-dense for the $n = 2$ case. The general case is similar. – Travis Willse Jan 09 '20 at 10:22

score 2 · Answer 2 · answered Aug 16 '19 at 17:50

2

Consider all points in $[0,1]^n$ with rational coordinates. We can enumerate them with natural numbers, so they produce a sequence $P_i$. We can also enumerate them in such a way so $|P_{i+1}-P_i|>\epsilon$ for every $i$ and fixed $\epsilon$. Now you can assign a spline that goes through all points in a succession. By construction, the closure of the curve will include $[0,1]^n$ and thus will have the dimension $n$

answered Aug 16 '19 at 17:50

Vasily Mitch

10,129

1

It is not at all clear to me that a spline through countably many point is smooth (or even differentiable, for that matter). Could you expand on that point, please? – Xander Henderson Aug 16 '19 at 17:53
Spline is a piecewise function with matching $k$-derivatives at nodes. So the curve between nodes $P_i$ and $P_{i+1}$ doesn't depend on the position of other points. – Vasily Mitch Aug 16 '19 at 17:58
If the curve only needs to be smooth (and not regular), then you could also just use a "smooth step function" (a function $f:[a,b] \to [0,1]$ for which $f(a) = 0$, $f(b) = 1$, and all the derivatives vanish at either end) to travel along the straight-line segments between each pair of successive points. – Michael Seifert Aug 16 '19 at 18:23

How big can the Hausdorff dimension of the closure of a smooth curve be?

2 Answers2