1
$\begingroup$

Let $\gamma_1,\gamma_2: \mathbb{S}^1 \to \mathbb{R}^2$ be two smooth, closed, convex curves that their (special)affine curvature, $\mu_1,\mu_2$ are equal, that is $\mu_1(\theta)=\mu_2(\theta)$, for any $\theta\in \mathbb{S}^1$. Are$\gamma_1,\gamma_2$ related to each other by an affine transformation and reparameterization? Are these two curves, $\gamma_1(\mathbb{S}^1 ),\gamma_2(\mathbb{S}^1) $, are related by an affine transformation?

$\endgroup$
0
$\begingroup$

Note that closed curves made by two arcs of ellipses of the given areas, say $a$ and $b$ may be not related to each other by an affine transformation. They have constant affine curvatures on two arcs, so for suitable parametrization the curvatures are the same.

The example is $C^{1,1}$-smooth, but it can be $C^\infty$-smoothed.

$\endgroup$
  • 1
    $\begingroup$ If we apply your construction to curvature, instead of affine curvature, using two circles with areas a,b, then does not your construction suggest that there are two closed, convex curves, that their Euclidean curvatures are equal, although they are not related by Euclidean motions. This would contradict the fact that the curvature in the plane determines the curve up Euclidean motions. $\endgroup$ – xyz abcd Mar 13 '15 at 0:18
  • 1
    $\begingroup$ @xyzabcd: Actually, you do not have the Euclidean theorem stated correctly. What is true is that the curvature plus the element of arc determines the curve up to Euclidean motions; thus, you need two things to determine the curve geometrically. Moreover, when the common curvature of the two curves $\gamma_i:\mathbb{S}^1\to\mathbb{R}^2$ that you want to compare is not constant, you can't usually use reparametrization to 'line things up' if they have different elements of arc, because most reparametrizations that line up the elements of arc won't preserve the curvature. $\endgroup$ – Robert Bryant Mar 13 '15 at 14:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.