# Are all $C^1$ arcs tame?

Let $n$ be a positive integer, and let $\mathbf{v}$ be a non-zero vector in $\mathbb{R}^n$.
Let $\; p : [0,1] \to \mathbb{R}^n \;$ be injective and $C^1$ and such that $p'$ is nowhere zero.
Does there always exist $\; f : ([0,1] \times \mathbb{R}^n) \to \mathbb{R}^n \;$ such that $f\hspace{0.01 in}$ is $C^1$ and for all members $t$ of $[0,1]$, $\; (\mathbf{x} \in \mathbb{R}^n) \mapsto f\hspace{0.01 in}(\langle t,\mathbf{x} \rangle) \;$ is a $C^1$ diffeomorphism of $\mathbb{R}^n$ and $\; f\hspace{0.01 in}(\langle 1,p(t)\rangle) = t\cdot \mathbf{v} \;$ ?

(That would be the $C^1$ version of an ambient isotopy.
I already know about the Fox-Artin arc, and I can see that it can 'easily' be
made to be discontinuously differentiable with a nowhere zero derivative.)

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I guess you forgot the assumption that $p$ has no self-intersections. – Sergei Ivanov Sep 20 '11 at 19:44
excellent point – Ricky Demer Sep 20 '11 at 19:46
The answer has to be yes. You can deform $p$ to $p^\prime = p\vert\_{[0,t^\prime]}$ for any $0 < t^\prime < t$. Choosing $t^\prime$ small enough so that the direction of $p^\prime$ changes by less than 180deg, then straighten it out. – George Lowther Sep 20 '11 at 20:05
Alternatively, by deforming locally on sections on which its direction changes by less than 180deg, you can make it $C^\infty$, and constant derivative on the initial segment. Then, extend $p^\prime$ locally to a smooth vector field vanishing away from $p$. Moving along the flow of the vector field, you can shorten $p$ to a straight line segment. – George Lowther Sep 20 '11 at 20:19

## 1 Answer

If $p$ is continuously differentiable up to and including the endpoints, then it has a $C^1$ extension to $(-\varepsilon,1+\varepsilon)$. Then it has a tubular neighborhood $U$ parametrized by a $C^1$ diffeomorphism $i:(-\varepsilon,1+\varepsilon)\times D^{n-1}\to U\subset\mathbb R^n$, where $i(t,0)=p(t)$. In the cylinder $(-\varepsilon,1+\varepsilon)\times D^{n-1}$, it is easy to construct an explicit isotopy which shrinks the segment $[0,1]\times \{0\}$ to a small segment $[0,\delta]\times\{0\}$ and stays identical away from a small neighborhood of $[0,1]\times \{0\}$. Composing this isotopy with $i$ (and extending by the identity on $\mathbb R^n\setminus U$) you get the ambient isotopy which sends $p$ into a small sub-interval of itself.

If $\delta$ is sufficiently small, the resulting curve is contained in a graph of a $C^1$ function $f:\mathbb R\to\mathbb R^{n-1}$ (in a suitable coordinate system in $\mathbb R^n$), and it is easy to straighten it into a segment by a linear isotopy preserving the first coordinate.

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How do you show the existence of $U$ and such a parametrization? – Ricky Demer Sep 20 '11 at 21:07
You need $C^1$ functions $v_1,\dots,v_{n-1}:(-\varepsilon,1+\varepsilon)\to\mathbb R^n$ such that $p'(t),v_1(t),\dots,v_{n-1}(t)$ are linearly independent for every $t$; then you define $i(t,x_1,\dots,x_{n-1}) = p(t)+x_1v_1(t)+\dots+x_{n-1}v_{n-1}$ and by the Inverse Function Theorem this is a diffeomorphism in a neighborhood. To construct smooth $v_i(t)$ first construct continuous ones (e.g. let them be a basis of the orthogonal complement of $p'(t)$ constructed locally by orthogonalization of constant family), then smoothen. – Sergei Ivanov Sep 20 '11 at 21:26
You can always define the parameterization $i$ by making it piecewise linear (approximating $p$ by a polygon), then applying $C^1$ bump on the linear segments to make it line up with $p$, then another $C^1$ bump on each segment to make the derivatives line up at the joins. – George Lowther Sep 20 '11 at 21:27