Smooth curve with no Frenet frame

Solution 1:

I presume that implicit in the definition is that the frame varies smoothly (as Elliott pointed out).

Here is an example in $\mathbb R^2$:

\begin{equation} \gamma(t) = \begin{cases} (\mathrm{e}^{1/t}, 0) &\text{ for $t < 0$} \\\\ (0,0) &\text{ for $t = 0$} \\\\ (0, \mathrm{e}^{-1/t}) &\text{ for $t > 0$} \end{cases} \end{equation}

EDIT:

To give an embedded curve, I will modify the example to work in $\mathbb R^3$. \begin{equation} \gamma(t) = \begin{cases} (\mathrm{e}^{1/t}, 0, t) &\text{ for $t < 0$} \\\\ (0,0, 0) &\text{ for $t = 0$} \\\\ (0, \mathrm{e}^{-1/t}, t) &\text{ for $t > 0$} \end{cases} \end{equation}

Observe that this curve is embedded with $\dot \gamma \ne 0$. This means the first vector field in the Frenet frame has to be $\frac{1}{|| \dot \gamma ||} \dot \gamma$. In particular then, at $t=0$, $X_1 = (1, 0, 0)$.

We've now moved the problem to the second derivative. For $t < 0$, $\ddot \gamma$ is in the span of $(0, 1, 0)$ and for $t > 0$, $\ddot \gamma$ is in the span of $(0, 0, 1)$. This means that $X_2$ can't exist.

Let me say that we have a partial Frenet frame if we can find such vector fields $X_1, \dots, X_m$, where $m < n$. If the first $m$ derivatives $\dot \gamma, \ddot \gamma, \dots, \gamma^{(m)}$ are linearly independent, we have the existence of a partial frame by applying Gram-Schmidt.

The way for the Frenet frame to fail to exist is for there to be an $m$ such that at some point, the span of $\{ \dot \gamma, \ddot \gamma, \dots, \gamma^{(m)} \}$ has dimension less than $m$. In both examples, at the image of $t=0$, the span dropped (for $m=1$ and $m=2$ respectively). Obviously, this is not a sufficient condition, such we can easily construct examples where things work out.

This construction seems very similar to the one for obtaining a smooth family of matrices without a corresponding smooth frame of eigenvectors. I don't immediately see how to transform this question to that one. The question about eigenvectors has been discussed MO before. (I just provide the link for curiosity's sake, unless someone can explain how to relate these two questions.)