# Does an immersed curve have to be “curve-like” somewhere?

Let $$C$$ be a smooth immersed curve, i.e. an image of a smooth immersion $$\varphi\colon \mathbb{R} \to \mathbb{R}^n$$. We will say that $$C$$ is curve-like at a point $$p \in C$$ if, for every two smooth immersions $$\alpha, \beta\colon \mathbb{R} \to C$$ with $$\alpha(0) = \beta(0) = p$$, we have neighborhoods $$V_\alpha, V_\beta$$ of 0 so that $$\alpha(V_a) = \beta(V_b)$$.

An embedded curve is curve-like at every point. However, a curve with "self-intersections", like the $$\infty$$ symbol, fails to be curve-like at those self-intersections.

The existence of space-filling curves shows that the image of $$\mathbb{R}$$ by a continuous map does not have to be curve-like anywhere. When the parameterization $$\varphi$$ is guaranteed to be a smooth immersion, does $$C$$ need to have at least one curve-like point?

• Do you assume $\alpha$ and $\beta$ to be smooth too, or can they be merely continous? – hmakholm left over Monica Jun 1 '19 at 13:34
• It is good enough for my purposes to assume they are smooth. Edited to clarify this. – Christopher Gadzinski Jun 1 '19 at 13:53
• @user10354138: That is true, but not being a self-intersection is not a sufficient criterion for being curve-like. – hmakholm left over Monica Jun 1 '19 at 14:48

No -- it is possible for $$C$$ to be non-curvelike everywhere.

Let $$g$$ be a fixed smooth bump funcion $$\mathbb R\to\mathbb R$$ supported on the open unit interval and define $$\hat g(x)=g(x-\lfloor x\rfloor)$$, an infinite row of bumps.

To warm up, let us first consider this set: $$D = \{(x,0)\mid x\in\mathbb R\} \cup \bigcup_{n=1}^\infty\{(x,e^{-n}\hat g(nx))\mid x\in\mathbb R\}$$ It consists of the $$x$$ axis overlaid by a series of bumps that get narrower or narrower, with smaller and smaller amplitude. The amplitudes decrease vastly faster than the widths.

Even thought the wiggles only touch the $$x$$-axis at rational coordinates, $$D$$ is not curve-like at any point on the $$x$$-axis. If we have an irrational point we can still let $$\alpha$$ be the $$x$$-axis itself, and let $$\beta$$ include a sequence of non-overlapping bumps that collectively approach $$p$$. If $$\beta$$ is parameterized by the $$x$$-coordinate, it will be smooth, and it doesn't coincide with $$\alpha$$ in any neighborhood of $$p$$.

First construction: With this construction in mind, we can construct a smooth curve $$\varphi$$ whose image is nowhere curvelike:

Start by choosing an surjection $$B: \mathbb N\to \mathbb Z\times \mathbb N$$ such that the first component of $$B(m)$$ is always smaller than $$m$$.

• For $$t\le 0$$ we have $$\varphi(t)=(t,0)$$. Ho hum.
• For each $$m\in\mathbb N$$, let $$(k,n)=B(m)$$ and construct $$\varphi(t)$$ for $$6m as follows:
• On the interval $$6m+5\le t\le 6m+6$$, let $$\varphi(t)$$ follow the already chosen path for $$\varphi$$ on $$[k,k+1]$$, but offset by the wiggle function $$e^{-n}\hat g(nx)$$ added either to the $$y$$ or the $$x$$ coordinate depending on whether the $$[k,k+1]$$ interval is considered "horizontal-ish" or "vertical-ish" (and this classification is then taken over for $$[6m+5,6m+6]$$).
• On the interval $$6m < t < 6m+5$$ choose a smooth connector curve from where the previous step left off to the location and direction we need to reach at $$t=6m+5$$. If necessary, we can construct the connector such that each of its five unit intervals is clearly either horizontal-ish or vertical-ish for when we add bumps to it later.

Now each $$p\in C$$ will either correspond to an integer $$t$$ (and then $$C$$ is clearly not curve-like at $$p$$, since differently bumpy segments of $$\varphi$$ join up there by construction), or else there's a neighborhood of $$p$$ that has a subset that looks locally like $$D$$ (modulo a diffeomorphism of $$\mathbb R^2$$).

A more explicit construction thought up later. Let $$\psi$$ be a surjection from $$\mathbb Z$$ to the set of finite subsets of $$\mathbb N$$. For example $$\psi(n)$$ could be $$\varnothing$$ for negative $$n$$ and consist of the positions of $$1$$ bits in the binary representation of $$n$$ when $$n\ge 0$$. Then set $$\phi(k+x) = \biggl[ 1 + \sum_{n\in \psi(k)} e^{-n}\hat g(nx)\biggr]\cdot(\cos 2\pi t,\sin 2\pi t)$$ for all $$k\in\mathbb Z, x\in[0,1)$$.

Now $$C$$ is bounded, and each of the derivatives of $$\varphi$$ is globally bounded too.

• By the way, I think you mean $\require{color}({\color{red}x},e^{-1/n}\hat{g}(nx))$. – user10354138 Jun 1 '19 at 14:49
• @user10354138: Typo fixed, thanks. I'm not sure I understand what you're getting at in your first comment. There's a neighborhood of $p$ in $C$ that contains something that looks like $D$, and then we can use this to transfer the curves $\alpha$ and $\beta$ I made in $D$ previously, to $C$ such that they differ on every $t$-neighborhood of $0$. – hmakholm left over Monica Jun 1 '19 at 14:52
• Am I mistaken in thinking you need more than that? You need bumps on the bumps, then bumps on the bumps on the bumps, etc, so you should biject $\mathbb{N}\to\bigcup_{n\geq 1}\mathbb{N}^n$ first? Also, there are no surjections $\mathbb{N}\to\mathbb{Z}\times\mathbb{R}$. – user10354138 Jun 1 '19 at 15:03
• @user10354138: That $\mathbb R$ was another embarassing typo; I meant $\mathbb N$. You do in fact get bumps on the bumps, ad infinitum, because the bumpy interval $[6m+5,6m+6]$ will become the $[k,k+1]$ in a sequence of later steps of the construction to have its own bumps added to it. – hmakholm left over Monica Jun 1 '19 at 15:08
• It is much clearer now, thanks. – user10354138 Jun 1 '19 at 15:11