I am currently studying for an exam in differential geometry. There's a problem which I am not able to solve and do not even know where to start (although I think it has to do with the Frenet equations) :Let $\alpha(s)$ be a curve parametrised by arc length of non zero constant torsion $\frac{1}{a}$. Show that there exists a (vector valued) function $f$ satisfying : $\alpha(s) = a \int (f(s) \times f'(s)) ds$ $\vert f(s) \vert =1$ and $(f(s) \times f'(s)) \cdot f''(s) \not =0$ Any help/hints?

  • $\begingroup$ I take your equation $\alpha(s) = \int f(s) \times f'(s) ds$ to mean $\alpha(s) = \int ((s) \times f'(s)) ds$. (Note parenthesis.) Am I right? $\endgroup$ – Robert Lewis Apr 11 '15 at 22:27
  • $\begingroup$ Jup! I edited it! Thanks! $\endgroup$ – mich95 Apr 11 '15 at 22:35
  • $\begingroup$ I suspected as much; glad to help out! $\endgroup$ – Robert Lewis Apr 11 '15 at 22:46

Let's start with the Frenet-Serret equations for the unit-speed, arc-length parametrized curve $\alpha(s)$. We have the unit tangent vector $T(s)$ along $\alpha(s)$, given by

$T(s) = \alpha'(s); \tag{1}$

the first Frenet-Serret equation is then

$T'(s) = \kappa(s) N(s), \tag{2}$

where $\kappa$ is the curvature of $\alpha(s)$ and $N(s)$ is the unit normal vector along the curve. The second such equation is

$N'(s) = -\kappa(s) T(s) + \tau B(s), \tag{3}$

where $\tau$ is the torsion of and

$B(s) = T(s) \times N(s) \tag{4}$

is the so-called binormal field along $\alpha(s)$. We note that

$\Vert B(s) \Vert = 1 \tag{5}$

for all $s$ by virtue of the definition (4), since $\Vert T \Vert = \Vert N \Vert = 1$ and $\langle T(s), N(s) \rangle = 0$ everywhere.

Finally, the third Frenet-Serret equation is given by

$B'(s) = -\tau N(s); \tag{6}$

in light of the hypothesis $\tau = 1/a$, a constant, (3) and (5) may also be written

$N'(s) = -\kappa(s) T(s) + \dfrac{1}{a} B(s), \tag{7}$

$B'(s) = -\dfrac{1}{a} N(s). \tag{8}$

At this point we need to introduce a hypothesis which was not mentioned by OP mich95 in the text of the question, but which (as far as I can see) is essential to forward progress on this problem; this is that $\kappa(s) = \Vert T'(s) \Vert \ne 0$ for all $s$ for which $\alpha(s)$ is defined; otherwise, $N(s)$ can't easily be defined by an equation such as (2) and the remainder of the Frenet-Serret apparatus breaks down. Bearing these remarks in mind, we proceed.

I claim the unit binormal vector $B(s)$ satisfies the requirements placed upon $f(s)$ in the statement of the question. We have already seen the $\Vert B(s) \Vert = 1$; we can compute $B(s) \times B'(s)$ using (4), (6), (8); we have

$B(s) \times B'(s) = B(s) \times (-\dfrac{1}{a} N(s)) = -\dfrac{1}{a} B(s) \times N(s); \tag{9}$

now from (4) we compute $B(s) \times N(s)$, using the standard identity

$X \times (Y \times Z) = Y(X \cdot Z) - Z (X \cdot Y); \tag{10}$

(10) holds for any three vectors $X, Y, Z \in \Bbb R^3$; see this wikipedia entry. Thus, from (4),

$B(s) \times N(s) = - N(s) \times B(s) = -N(s) \times (T(s) \times N(s)) = -(T(s)(N(s) \cdot N(s)) - N(s)(N(s) \cdot T(s))) = - T(s); \tag{11}$

using (11) in (9) yields

$B(s) \times B'(s) = \dfrac{1}{a}T(s). \tag{12}$

We next evaluate $(B(s) \times B'(s)) \cdot B''(s)$; by (8),

$B''(s) = -\dfrac{1}{a} N'(s) \tag{13}$

since $a$ is constant; inserting (7) into (13) we find

$B''(s) = -\dfrac{1}{a} N'(s) = -\dfrac{1}{a}(-\kappa(s) T(s) + \dfrac{1}{a} B(s)) = \dfrac{\kappa(s)}{a}T(s) - \dfrac{1}{a^2}B(s); \tag{14}$

from this and (12) we obtain

$(B(s) \times B'(s)) \cdot B''(s) = \dfrac{1}{a}T(s) \cdot (\dfrac{\kappa(s)}{a}T(s) - \dfrac{1}{a^2}B(s)) = \dfrac{\kappa(s)}{a^2} \ne 0 \tag{15}$

since $T(s) \cdot T(s) = 1$ and $T(s) \cdot B(s) = 0$. We observe that the assumption $\kappa(s) \ne 0$ made earlier is essential for the validity of (15). Finally, we have from (1) and (12) that

$\alpha'(s) = T(s) = a(B(s) \times B'(s)). \tag{16}$

This shows that

$\alpha(s) = a\int (B(s)\times B'(s))ds \tag{17}$

in the sense that $a(B(s) \times B'(s))$ is a (vector) primitive or anitderivative for $\alpha(s)$; we can also phrase (17) in terms of definite integrals

$\alpha(s) - \alpha(s_0) = a \int_{s_0}^s (B(u) \times B'(u))du \tag{18}$

if we so desire.

We have seen that $B(s)$ satisfies all the requirements imposed upon $f(s)$ in the problem statement.

Finally, a few remarks on the assumption $\kappa(s) \ne 0$. We see in (15) that this additional hypothesis is essential to attain $(B(s) \times B'(s)) \cdot B''(s) \ne 0$, though it was not mentioned in the statement of the problem. Stipulating that $\kappa(s) \ne 0$ is not, however, a very restrictive assumption; indeed, it is "more likely than not" to be true of an arbitrary curve $\alpha(s)$, since the assertion $\kappa(s) = 0$ for some $s$ is really a constraint on the family of regular curves $\alpha(s)$. My phrase "more likely than not" is an attempt to express intuitively, with brevity, the usual mumbo-jumbo about open dense sets etc. etc. etc. Hopefully my readers will get the idea, be tolerant of my lack of rigor here, and be grateful for the omission of the verbiage to which I am often inclined. And I thank you for your attention to these matters.

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    $\begingroup$ Thank you for your answer! That's really ingenious! $\endgroup$ – mich95 Apr 12 '15 at 6:00
  • $\begingroup$ Thank you for the kind words, and the "acceptance"! Ingenuity straight from the old plug and grind method! I'm working on your other one, math.stackexchange.com/questions/1229366/…; nice problems, we'll see what happens . . . thanks again and, as ever, Cheers! $\endgroup$ – Robert Lewis Apr 12 '15 at 6:11
  • $\begingroup$ Thanks again! I am sure I did something wrong in the other one (as I got that those are only part of circles ) $\endgroup$ – mich95 Apr 12 '15 at 6:17
  • $\begingroup$ Don't think they are circles; I'm guessing a kind of spiral. Still trying to figure in out! We'll see what happens . . . thanks again! $\endgroup$ – Robert Lewis Apr 12 '15 at 6:21

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