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Can anyone point me to a proof of the Frenet-Serret formulas for arbitrary (i.e. $N>3$) dimensions?

Michael Hardy
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bob.sacamento
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1 Answers1

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Assuming appropriate genericity of the curve (linear independence of all derivatives), it's really the same proof. Assuming an arclength parametrization $\alpha(s)$, set $\alpha'(s)=e_1(s)$. Now, assuming that you've defined $e_1,\dots,e_k$ with $e_1' = \kappa_1e_2$ $(\kappa_1>0), \dots, e_j'=-\kappa_{j-1}e_{j-1}+\kappa_{j}e_{j+1}$ for $2\le j\le k-1$, you write $e_k'=-\kappa_{k-1}e_{k-1}+\kappa_{k}e_{k+1}$ for $\kappa_{k}>0$ and an appropriate unit vector $e_{k+1}$. When you get to $k=N-1$, you can ordain that $e_1,\dots,e_N$ be a positively oriented orthonormal basis, and then $\kappa_{N-1}$ will have a sign. The skew-symmetry follows, as in the $\Bbb R^3$ case, by differentiating $e_i\cdot e_j = 0$, $i\ne j$, and $e_i\cdot e_i=1$.

I don't know your background, but a beautiful exposition of this and far more (if you are familiar with differential forms and Lie groups) is in Phillip Griffiths's paper "On Cartan's Method of Lie Groups and Moving Frames as Applied to Existence and Uniqueness Questions in Differential Geometry," Duke Math J. {\bf 41}, 1974. (However, I think his subscripts may get a bit messed up on this.)

Ted Shifrin
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    OK, your expression for $e^{\prime}j$ is exactly where my problem is. My understanding of the frame is such that you start with the various derivatives of the motion, and then perform a Gramm-Schmidt process to get the various $e$'s. And I don't see how that process implies $e^{\prime}_j=−κ{j−1}e_{j−1}+κje{j+1}$. I mean, I don't see a priori why $e^{\prime}_j$ for the general case can't have components of any number of the other vectors. Am I missing something obvious? Thanks for your help. – bob.sacamento Mar 14 '16 at 19:00
  • I'm new at this. The paper you mention is beyond me right now. – bob.sacamento Mar 14 '16 at 19:06
  • It's easier to work with the frame itself, rather than with various higher derivatives. The reason that $e_j'$ can't involve vectors other than $e_{j-1}$ and $e_{j+1}$ ($2\le j\le N-1$) is the skew-symmetry result I mentioned. For example, since $e_1'\cdot e_j = 0$ for $j>2$, it follows that $e_j'\cdot e_1=0$ for $j>2$. Similarly, $e_2'\cdot e_4=0 \implies e_4'\cdot e_2=0$. – Ted Shifrin Mar 14 '16 at 19:47
  • Sorry, I don't mean to be obtuse. And if you want to give up, I don't blame you. I understand $e^{\prime}_1\cdot e_j=0, j>2$, and I understand $e^{\prime}_2\cdot e_4=-e^{\prime}_4\cdot e_2$. For the life of me, I can't see why $e^{\prime}_2\cdot e_4 = 0$. – bob.sacamento Mar 14 '16 at 20:35
  • Oh ... Because we define $\kappa_2 e_3$ by being what's left when we remove $-\kappa_1e_1$ from $e_2'$. This was the step I wrote above, writing $e_k' = -\kappa_{k-1}e_{k-1}+\kappa_ke_{k+1}$ for the appropriate choice of $e_{k+1}$ ... for the case $k=2$. ... I guess this is harder to discern in the 3-dimensional case, because once we have $\mathbf T$ and $\mathbf N$ ($e_1$ and $e_2$), $\mathbf B$ ($e_3$) is obtained by taking the cross product. But you could think of $\mathbf B$ as (in this case, $\pm$) the direction that's left when you look at $\mathbf N' +\kappa\mathbf T$. – Ted Shifrin Mar 14 '16 at 21:20
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    Wouldn't "arbitrary dimensions" include countably and uncountably infinite dimensions? You've answered for what appears to be the finite cases. – Galen Jun 30 '21 at 14:54
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    @Galen Not in the universe in which I do differential geometry. – Ted Shifrin Jun 30 '21 at 15:08
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    @TedShifrin That's understandable, and the finite case does cover most use cases. – Galen Jun 30 '21 at 15:19