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Let $M$ be a smooth manifold, $\dim M=n$ and $\gamma:[0;1]\to M$ be continuous. Is it true that there exists local coordinates $(y^0,\ldots,y^n)$ in a neighborhood $V$ of the graph $\{(t,\gamma(t)),t\in[0;1]\}\subset\mathbb{R}\times M$ such that $y^0(t,x)=t$ and $\frac{\partial}{\partial t}y^k(t,x)=0$ for any $(t,x)\in V$?

This question arises in calculus of variations and optimal control: when you are trying to apply a small variation to a curve, it is very convenient to do it in local coordinates and these coordinates must be defined along the entire curve (for example, for Jacobi-type conditions).

If (i) $\gamma\in C^\infty$, (ii) $\dot\gamma(t)\ne 0$, and (iii) $\gamma$ has no self-intersections, then the answer is trivially positive (a tubular neighborhood of $\gamma$ in $M$ gives $y^k$ for $k\ge 1$).

If the original question has negative answer, I wonder which of the three previous properties are essential? For example, what if $\gamma\in \mathrm{Lip}[0;1]$ and $|\dot\gamma(t)|\ge\mathrm{const}$ (in a Riemannian metric on $M$)?

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    $\begingroup$ Suppose $M=\mathbb{R}^n$ with Euclidean coordinates $(x^k)$. Denote by $V$ the Euclidean space $\mathbb{R}\times \mathbb{R}^n$. It is an open neighborhood of the graph. The obvious Euclidean coordinates $(y^0, y^1, \dotsc y^n)$ on $V$ seem to do the trick. Above $y^k=x^k$ for $k=1,\dotsc, y^n$. $\endgroup$
    – Liviu Nicolaescu
    Commented Sep 20, 2022 at 12:23
  • $\begingroup$ Yes. Obviously, if there exist local coordinates near $\gamma[0;1]\subset M$, then the answer is trivially yes. I'm interested in the non-trivial case when there are no local coordinates near $\gamma[0;1]$. For example, if $M=S^1$ and $\gamma(t)=2\pi t$, then there are no local coordinates near $\gamma[0;1]$, but there exist fiberwise local coordinates near the graph $\{(t,\gamma(t))\}$. $\endgroup$
    – Lev
    Commented Sep 20, 2022 at 13:32

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