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Let $r:(a,b)\rightarrow{R^n}$ with $|r^{'}|=1$ is a natural parameter curve in $R^n$.
If $e_1(s)=r'(s),e_2(s),...,e_n(s)$ form an orthonormal frame, then we have Frenet formulae: $e_{i}^{'}=-k_{i-1}e_{i-1}+k_{i}e_{i+1}$ and $k_{0}=k_{n+1}=0,e_{0}=e_{n+1}=0$ ($k_1$ is its curvature).
Above is from the book Geometry 1 written by R.V.Gamkrelidze.

Theorem 1 When $n=2$,
if $k_1=0$, then $r$ is a straight line.
if $k_1=c\not=0$, then $r$ is a sphere.

Theorem 2 When $n=3$,
if $k_1=0$, then $r$ is a straight line.
if $k_1=c\not=0, k_2=0$, then $r$ is a sphere.
if $k_1=c\not=0, k_2=d\not=0$, then $r$ is a spiral line.

Can these theorems be generalized to higher dimension?

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  • $\begingroup$ Maybe we can generalize spiral line. In $R^3$, a spiral is $(acost,asint,bt)$ and its $k_1=\frac{a}{a^2+b^2}$,$k_2=\frac{b}{a^2+b^2}$. From The Uniqueness of Curve, we can get the conclusion. I guess that in $R^n$, we can introduce a "spiral line" like this $(acost,asint,b_1t,b_2t,...,b_{n-2}t)$. I compute its $k_1=\frac{a}{a^2+b_1^2+...+b_{n-2}^2}$. Other $k$, I still have on idea. $\endgroup$ – gaoxinge Jan 2 '14 at 1:38
  • $\begingroup$ $d$ and $c$ are constants, right? $\endgroup$ – Robert Lewis Jan 5 '14 at 5:40
  • $\begingroup$ Yes.@RobertLewis $\endgroup$ – gaoxinge Jan 5 '14 at 5:41
  • $\begingroup$ thanks and thanks for responding so rapidly. Nice question, by the way, +1! $\endgroup$ – Robert Lewis Jan 5 '14 at 5:41
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Check out this reference:http://arxiv.org/abs/math/0412323

The curves you describe have constant curvature ratios in a degenerate sense, but I believe the theorem in the paper still applies, showing that in odd dimensions, such curves are related to curves of constant slope in a generalized torus. It might be true that they can all be described this way, but it's hard to tell from the references.

Edit: From reference 6, it seems that curves with an even number of nonzero curvatues lies on an actual torus with some slope and can be closed, while those with an odd number are helical nature and cannot be closed. It provides exact formulas.

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    $\begingroup$ I would upvote this, but you have $8008$ rep... $\endgroup$ – The Chaz 2.0 Jan 7 '14 at 4:05
  • $\begingroup$ 8008! A powerful processor indeed! $\endgroup$ – Robert Lewis Jan 7 '14 at 5:51

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