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For my class in classical differential geometry I have to solve the following problem:

Let $\alpha$ be a regular curve in $\mathbb{R}^2$. Proof that:

a) $\quad$ $\alpha$ is a straight line if and only if the curvature vanishes.

b) $\quad$ If $\dot{\alpha}$ and $\ddot{\alpha}$ are linearly dependent, then $\alpha$ is a straight line.

Conisder a). At first this problem seemed easy to me. Because if the curvature is defined via

$$\ddot{\alpha}(t) = \kappa(t) \cdot n(t) \qquad \qquad (1)$$

where $n(t)$ is the unit vector perpendicular to the tangent vector, then it is easy to see that $\kappa = 0$ implies $\ddot{\alpha} = 0$ and therefore $\alpha$ is a straight line. However, I realsied, that the above equation is derived by using the fact

$$\big< \dot{\alpha} , \dot{\alpha} \big> = 1 \quad \Longrightarrow \quad \big< \dot{\alpha}, \ddot{\alpha} \big> = 0$$

in the case of $\alpha$ being parameterised by arc length. Since my given $\alpha$ is not parameterised by arc length, I cannot assume equation (1) to hold and therefore cannot assume my $\ddot{\alpha}$ to be perpendicular to the tangent vector. My conclusion would be wrong then. Any ideas on how to solve this problem for general curves $\alpha$?

b) I have no idea on how to show this. Any tips here?

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  • $\begingroup$ If you can prove $(a)$ then $(b)$ follows immediately. If $\alpha'(t)$ and $\alpha''(t)$ are linearly dependent for all $t\in \mathbb{R}$ we know that $\alpha'(t) \times \alpha''(t)=\vec{0}$ for all $t$ and so $\kappa(t)=\frac{\|\alpha'(t) \times \alpha''(t)\|}{\|\alpha'(t)\|}^3=0$. Note the denominator in the previous expression for $\kappa(t)$ never vanishes by our assumption that $\alpha$ is a regular curve. $\endgroup$
    – user801306
    Commented Mar 13, 2021 at 19:39
  • $\begingroup$ I forgot to mention that the curve lies in $\mathbb{R}^2$. I updated my question accordingly. $\endgroup$
    – Octavius
    Commented Mar 13, 2021 at 19:46
  • $\begingroup$ Set $\alpha(t)=\big(x(t),y(t)\big)$. If $\alpha'(t),\alpha''(t)$ are linearly dependent, then $$x'(t)y''(t)-y'(t)x''(t)=\det\Big(\begin{smallmatrix} x'(t) & x''(t)\\ y'(t) & y''(t) \end{smallmatrix}\Big)=0$$ Hence $$\kappa(t)=\frac{x'(t)y''(t)-y'(t)x''(t)}{\Big(\sqrt{\big(x'(t)\big)^2+\big(y'(t)\big)^2}\Big)^3}=0$$ for all $t$. Also, you can assume WLOG that you're parameterizing w.r.t arc length since you're curve $\alpha$ is regular. $\endgroup$
    – user801306
    Commented Mar 13, 2021 at 20:16

2 Answers 2

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Bearing in mind that any regular curve can be parametrized by the arc length then

a) if $\alpha$ is a straight line, i.e., $\alpha(t)=\text{p}+\text{v}t$ and so $k(t)=||\alpha''(t)||=||0||=0$.

If $k(t)=0$ then $\alpha''(t)=0$. If you calculate the antiderivative twice you obtain $\alpha'(t)=\text{v}$ and finally $\alpha(t)=\text{p}+\text{v}t$ for some $\text{p,v}\in\mathbb{R^2}$.

b) As you saw $\langle\alpha',\alpha''\rangle=0$ then $\alpha''$ is null vector or $\alpha'$ is perpendicular to $\alpha''$, but as they are linear dependent then $\alpha''=0$ and by a) it can only be a straight line.

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Hint for $(a)$ : WLOG Assume $\alpha(0)=0$ (Curvature is invariant under isometries) and try looking at $\gamma(s)=(\int_0^scos(\theta(u))du, \int_0^ssin(\theta(u))du)$.

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