Signed curvatures

Question

Let $\textbf{$\gamma$}$ and $\textbf{$\tilde{\gamma}$}$ be two plane curves.

Show that, if $\textbf{$\tilde{\gamma}$}$ is obtained from $\gamma$ by applying an isometry $M$ of $\mathbb{R}^2$, the signed curvatures $κ_s$ and $\tilde{κ}_s$ of $\textbf{$\gamma$}$ and $\textbf{$\tilde{\gamma}$}$ are equal if $M$ is direct but that $\tilde{κ}_s = −κ_s$ if $M$ is opposite (in particular, $\textbf{$\gamma$}$ and $\textbf{$\tilde{\gamma}$}$ have the same curvature).

Show, conversely, that if $\textbf{$\gamma$}$ and $\textbf{$\tilde{\gamma}$}$ have the same nowhere-vanishing curvature, then $\textbf{$\tilde{\gamma}$}$ can be obtained from $\textbf{$\gamma$}$ by applying an isometry of $\mathbb{R}^2$.

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Could you give me some hints how we could show that? I don't really have an idea...

Answer 1

This is precisely exercise 2.2.3 on page 43 of "Elementary Differential Geometry" by A.N. Pressley. The chapter relevant to this problem has been made freely available by the publisher.

Definions and notations

As usual, we shall assume our curves to be unit-speed, since this makes formulae simpler without reducing generality.

If the tangent vector at time $t$ is $\dot \gamma (t)$ (having length $1$), the signed normal $\Bbb n (t)$ is the unique vector of length $1$ perpendicular to $\dot \gamma (t)$ that makes the frame $\{ \dot \gamma (t), \Bbb n (t) \}$ positive-oriented. Note that since $\| \dot \gamma \| ^2 = 1$, deriving with respect to $t$ gives $\langle \dot \gamma, \ddot \gamma \rangle = 0$, so $\ddot \gamma (t)$ is a vector on the line perpendicular to $\dot \gamma (t)$, therefore it must be a multiple of $\Bbb n (t)$; let the signed curvature $k(t)$ be the coefficient giving this proportionality, i.e. $\ddot \gamma (t) = k(t) \Bbb n (t)$. (Note that, unlike in Pressley's notation, I have dropped the lower index ${}_s$ from the signed normal and the signed curvature, for simplicity.)

The first part of the question

Remember that an isometry preserves orthogonality and length, therefore the image through $M$ of the positive-oriented orthonormal frame $\{ \dot \gamma, \Bbb n \}$ will be the orthonormal frame $\{ M \dot \gamma, M \Bbb n \}$, which is $\{ \dot {\tilde \gamma}, M \Bbb n \}$.

If $M$ is direct, then this frame will also be positive-oriented. On the other hand, $\{ \dot {\tilde \gamma}, \tilde {\Bbb n} \}$ is another positive-oriented orthonormal frame, having the same first vector, therefore $M \Bbb n = \tilde {\Bbb n}$.

If $M$ reverses orientation, then $\{ \dot {\tilde \gamma}, M \Bbb n \}$ will be negative-oriented, so $\{ \dot {\tilde \gamma}, - M \Bbb n \}$ will be positive-oriented and with the argument in the above paragraph, $- M \Bbb n = \tilde {\Bbb n}$.

Since the determinant of an isometry is $\pm 1$ (understood as the Jacobian, because translations are not linear), we can summarize the above paragraphs in a single formula: $M \Bbb n = (\det M) \tilde {\Bbb n}$.

Finally,

$$\ddot {\tilde \gamma} = \ddot {(M \gamma)} = M \ddot \gamma = M (k \Bbb n) = k (M \Bbb n) = k (\det M) \tilde {\Bbb n} ;$$

on the other hand, $\ddot {\tilde \gamma} = \tilde k \tilde {\Bbb n}$, whence $\tilde k = (\det M) k$. If $M$ is direct, then $\tilde k = k$; is it is not, then $\tilde k = -k$.

The second part of the question

A similar statement but using the signed curvature and direct isometries instead of the unsigned curvature and isometries is made in theorem 2.2.6. Please note that the statement of the problem is wrong: consider the curves $t \mapsto (t, 0)$ for $t \in [0,1]$ and $t \mapsto (t, 0)$ for $t \in [0,2]$; being line segments, they both have curvature $0$, yet they cannot be transformed into one another by an isometry because they have different lengths. One way to correct it is to impose that they be speed-unit curves defined on the same interval $[a.b]$. These assumptions are made in theorem 2.2.6 too, therefore it is in this context that I shall work.

I shall denote the unsigned curvature by $c$. By definition, $c = \| \ddot \gamma \|$. By hypothesis, $\tilde c = c$.

One approach is to mimic the proof given to theorem 2.2.6 (since $c$ is a continuous function and the problem says that it doesn't vanish, you may assume $c>0$, the proof for $c<0$ being analogous). I don't like it, I consider it ugly. I suggest the following, more conceptually elegant approach.

For each $s \in [a,b]$, consider the orthonormal bases $\{ \dot \gamma, \Bbb n \}$ and $\{ \dot {\tilde \gamma}, \tilde {\Bbb n} \}$ in $\Bbb R ^2$. They will be related by an orthogonal transformation $M(s)$ of $\Bbb R ^2$. Since the dependence of $\dot \gamma (s), \Bbb n (s), \dot {\tilde \gamma} (s), \tilde {\Bbb n} (s)$ on $s$ is smooth, so will be the map $s \mapsto M(s)$, therefore we may speak of its derivative with respect to $s$, denoted $\dot M$.

The fact that $M(s)$ connects the two frames above can be encoded into the formulae $\dot {\tilde \gamma} = M \dot \gamma, \ \tilde {\Bbb n} = M \Bbb n$. Deriving the first of them produces

$$\ddot {\tilde \gamma} = \dot M \dot \gamma + M \ddot \gamma = \dot M \dot \gamma + M (c \Bbb n) = \dot M \dot \gamma + c \tilde {\Bbb n} .$$

On the other hand, $\ddot {\tilde \gamma} = \tilde c \tilde {\Bbb n} = c \tilde {\Bbb n}$. Equating the two we get $\dot M \dot \gamma = 0$. Since $\dot \gamma \ne 0$ (by hypothesis, $\| \dot \gamma \| = 1$), this implies $\dot M = 0$, which means that $s \mapsto M(s)$ is a constant map, i.e. $M(s) = M_0 \ \forall s \in [a,b]$. This $M_0$ is an orthogonal transformation (therefore an isometry, but not yet the required one).

Finally, let's integrate the equality $\dot {\tilde \gamma} = M_0 \dot \gamma$. We get

$$\tilde \gamma (s) = \int \limits _a ^s \dot {\tilde \gamma} (t) \ \Bbb d t + \tilde \gamma (a) = \int \limits _a ^s M_0 \dot \gamma (t) \ \Bbb d t + \tilde \gamma (a) = M_0 \big( \gamma (s) - \gamma (a) \big) + \tilde \gamma (a) = \\ M_0 \gamma (s) + \big( \tilde \gamma (a) - M_0 \gamma (a) \big) .$$

Therefore, $\tilde \gamma (s)$ is obtaind by applying the orthogonal transformation $M_0$ to $\gamma (s)$, followed by a translation by the constant vector $\tilde \gamma (a) - M_0 \gamma (a)$. Since orthogonal transformations and translations are isometries, their composition will also be so, and this is the desired isometry.

(Please note that nowhere have we used that $c(s) \ne 0$, therefore this approach is better than Pressley's proof, making this assumption unnecessary.)

Answer 2

$\large \textbf{First Part}$

The orthogonal $2\times 2$ matrices are of the two forms

$$ \begin{pmatrix} \cos \theta & - \sin \theta\\ \sin \theta & \cos \theta \end{pmatrix}, \qquad \begin{pmatrix} \cos \theta & \sin \theta\\ \sin \theta & -\cos \theta \end{pmatrix} $$

The first kind corresponds to $\textit{direct isometries: }$ a rotation (which is the composition of two reflections). The second kind corresponds to $\textit{opposite isometries: }$ a reflection.

If $M$ is a reflection on a line $\ell$ then $M(v) = Pv + a$ where

$$P = \begin{pmatrix} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta\end{pmatrix}$$

for some angle $\theta$.

Now, $\tilde{\gamma} = P \gamma + a$ and we can write $\dot{\gamma} = (\cos \varphi, \sin \varphi)$ and $\dot{\tilde{\gamma}} = (\cos \tilde{\varphi}, \sin \tilde{\varphi})$; furthermore we $$ \dot{\tilde{\gamma}} = P \dot{\gamma} = \begin{pmatrix} \cos \theta & \sin \theta\\ \sin \theta & -\cos \theta \end{pmatrix} \begin{pmatrix} \cos \varphi\\ \sin \varphi \end{pmatrix} = (\cos (\theta - \varphi), \sin (\theta - \varphi)). $$ Then we have the equality $$(\cos \tilde{\varphi}, \sin \tilde{\varphi}) = (\cos (\theta - \phi), \sin (\theta - \phi))$$

which implies that $$ \tilde{\varphi} = \theta - \varphi + 2\pi n, \quad \text{for some integer } n, $$

and then

$$ \begin{align*} \frac{d\tilde{\varphi}}{ds} &= - \frac{d\varphi}{ds}\\ \tilde{k}_{s} &= -k_{s}. \end{align*} $$

If $M$ is a rotation followed by a translation, then $M = P v + a$, where $$ P = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} $$ and by the previous reasoning we have

$$ (\cos \tilde{\varphi}, \sin \tilde{\varphi}) = (\cos (\theta + \varphi), \sin(\theta + \varphi)) $$

which implies that $$ \tilde{\varphi} = \theta + \varphi + 2\pi n, \text{ for some integer } n, $$

hence, $\tilde{k}_{s} = k_{s}$.

$\large \textbf{Second Part}$

If $\gamma$ and $\tilde{\gamma}$ have the same non-zero curvature then their signed curvature are the same or differ by the sign. If the signed curvature are the same, theorem 2.2.6 from the book by Andrew Presly ensures that $\gamma$ and $\tilde{\gamma}$ differ by a direct isometry. If the signed curvatures differ in sign, applying a reflection to one curve give us two curves with the same signed curvature, then these two last curves differ by a direct isometry and hence the original curves differ by an opposite isometry.