I am trying to solve the excercise 1.5.18 from Do Carmo's curves and surfaces. Although hints are included in the book I have difficulties with the details.
Part of the Exercise is:
Let $\alpha: I \mapsto \mathbb{R}^3$ be a parametrized regular curve (not necessarily by arc length) with $k(t) \neq 0, \tau(t) \neq 0, \forall t \in I$. The curve $\alpha$ is called a Bertrand curve if there exists a curve $\tilde{\alpha}: I \mapsto \mathbb{R}^3$ such that the normal lines of $\alpha$, and $\tilde{\alpha}$ at $t \in I$ are equal. In this case, $\tilde{\alpha}$ is called a Bertrand mate of $\alpha$ and we can write $$\tilde{\alpha}(t) = \alpha(t) + r n(t).$$ Prove that
$r$ is constant.
$\alpha$ is a Bertrand curve if and only if there exists a linear relation $$Ak(t) + B\tau(t) = 1, \forall t \in I,$$
where A, B are nonzero constants and $k$ and $\tau$ are the curvature and torsion of $\alpha$ respectively.
My Problems: I am fine with statement 1.
Question 1For statement 2.
One can consider the tangent vector for $t$ for $\alpha$ and the tangent vector $\tilde{t}$ for $\tilde{\alpha}$. Using frenet formulas one can show that $(t \cdot \tilde{t})'=0$. Now if $(t \cdot \tilde{t}) \neq 0$ everything is fine, but if $(t \cdot \tilde{t})=0$ one has that the curvature $k$ of $\alpha$ is constant and given by $k=1/r$.
In this case the only way to find a nonzero constant $B$ is to show that the torsion $\tau$ is also constant.
Is there a way to prove this? If not is there a counterexample? Meaning a curve with constant curvature and nonconstant, nonzero torsion having a bertrand mate?
Question 2
For the other direction assume a curve $\alpha$ satisfies $A k + B \tau =1$ with nonzero constants $A,B$.
One then can consider the curve $\tilde{\alpha}:= \alpha + A \cdot n$. In this case computations using the frenet formulas yield
$$ \tilde{n}= \frac{B k - A \tau}{\sqrt{A^2 +B^2}} \cdot n,$$
where $\tilde{n}$ is the normal vector for $\tilde{\alpha}$. If $B k - A \tau \neq 0$ everything is fine. How to exclude the case $B k - A \tau=0$?
For this I have an example: One can consider the helix $(a \cdot \cos(s) , a \cdot \sin(s) , b \cdot s )$ with $a= b= 1/\sqrt{2}$. Then the relation above is satisfied with $A=B=1/\sqrt{2}$. This implies that the curve $\tilde{\alpha}$ is given by $(0,0,b \cdot s)$ such a curve has no normal vector and thus no good definition of normal line.
So given the relation $A k + B \tau=1$ if $B k - A \tau=0$ for some time $t$ is there another curve, which is a (well defined) bertrand mate?
If no is the statement of the excercise unprecise and if so how to make it rigorous?
I don't think you should be doing special cases. You've shown there is a constant angle, say $\theta$, between $t$ and $\tilde t$, so write $\tilde t =\cos\theta t + \sin\theta b$. Now what do you get when you differentiate? Remember, of course, that even if $\alpha$ is arclength-parametrized, $\tilde\alpha$ is almost surely not.
With regard to your question about the converse, if $A\kappa+B\tau = 1$ and $B\kappa=A\tau$, then $B=B(A\kappa+B\tau) = A^2\tau + B^2\tau = (A^2+B^2)\tau \ne 0$ unless $\tau = 0$. But we're told that the curvature and torsion never vanish.