# Smooth curve with no Frenet frame

Let $\gamma: I \rightarrow \mathbb{R}^n$ be a smooth curve. We define a Frenet frame to be an orthonormal frame $X_1, \ldots X_n$ such that for all $1 \leq k \leq n$, $\gamma^{(k)}(t)$ is contained in the linear span of $X_1(t), \ldots, X_k(t)$, $t \in I$.

What is an example of a smooth curve $\gamma$ with no Frenet frame?

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Just for reference, this is an exercise in Unit 2 of Balázs Csikós's "Budapest Semesters in Mathematics" differential geometry notes, page 7 of the following pdf: cs.elte.hu/geometry/csikos/dif/dif2.pdf –  Jonas Meyer Feb 22 '11 at 22:31
Are you sure that this can be done with a smooth curve? Also, I think it might be important to mention that $X_1, \ldots, X_n$ are smooth vector fields along $\gamma$... –  Elliott Feb 26 '11 at 6:16

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

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

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

EDIT:

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

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.)

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Argh. I utterly despise this word game of having a smooth curve that is not regular. (Not addressed at @Sam, but at Csikós.) One of those cases where I think the terminology (a curve refers to the parametrisation and not that actual image) actually harms the intuition. (@Sam: +1 for giving the simple answer.) –  Willie Wong Apr 19 '11 at 12:56
Just to elaborate on @Willie's comment: if one looks at the image of this smooth curve, one sees that it is the union of the positive $x$-axis and the positive $y$-axis, which is not so smooth! How does a smooth curve trace out this non-smooth path? Because it uses a bump function to come to a complete stop at $(0,0)$ so that it can then change its direction from one axis to the other. As Willie wrote: Argh! –  Matt E Apr 19 '11 at 15:07
@WillieWong, @MattE : You are absolutely right that this example is taking advantage of the difference between a smooth map and an immersed submanifold. I am updating my answer with what gives a hint at a more general construction. –  Sam Lisi Apr 19 '11 at 17:05