4
$\begingroup$

The group of isometries of a circle in Euclidean space with the Euclidean metric is uncountable, because every rotational symmetry is an isometry. However if we change to any unequally weighted Euclidean-metric the circle is now an ellipse and has a finite number of symmetries with respect to the Euclidean norm. By this I mean that after we draw the circle, we forget about the norm that we used to create it. This happens with other metrics as well such as those induced by p-norms, the max norm, and the taxicab metric.

My question is why does this happen? Is this because of the space is Euclidean or is it because we're considering the isometries with respect to the Euclidean norm? For example if I started with a circle in the taxicab metric will the Euclidean norm have a finite group of symmetries with respect to it? I can see how a loop of string pulled tight around a square could have a group structure it's just unclear to me what the Euclidean metric might look like with respect to it.

$\endgroup$
3
  • 4
    $\begingroup$ What is a "symmetry with respect to Euclidean norm"? Note also that the metric is derived from the norm, so if you switch one but not the other, you can't expect them to play well with each other... $\endgroup$ Aug 8 at 19:54
  • 2
    $\begingroup$ If you draw a circle using the Euclidean norm and then decide to use the taxicab norm for isometries, you'll have limited isometries of the circle because it has just four points that are at a minimum distance from the origin and four that are at a maximum distance. $\endgroup$
    – David K
    Aug 9 at 2:05
  • $\begingroup$ In fact, even if you create the circle using the taxicab norm and look at its isometries under the taxicab norm, I think there are only finitely many. There are two pairs of points on a taxicab circle of radius $1$ such that the entire circle of radius $1$ lies on just two circles of radius $2$ about those two points. No other pairs of points have that property, so I think the isometries can only map those pairs to themselves or to each other. $\endgroup$
    – David K
    Aug 9 at 18:56

1 Answer 1

4
$\begingroup$

Consider the space $\mathbb R^2$ with some inner product $\phi(x,y)$. Use the norm corresponding to that inner product: $\|x\|_\phi = \big(\phi(x,x)\big)^{1/2}$. Of course the set $\{x \in \mathbb R^2 : \|x\|_\phi = 1\}$ is an ellipse. (And every ellipse centered at the origin is obtained in this way.) Then there are uncountably many linear maps $T : \mathbb R^2 \to \mathbb R^2$ that preserve the norm. So your "ellipse" is not a very good example for what you ask.

It is true that the $l^p$ norm, $p \ne 2$, has far fewer isometries.

$\endgroup$
1
  • $\begingroup$ Thanks for the response, this seems to solve the elliptical case. If impose the restriction that it fixes the circle and not just the norm it seems to give the same group, $D_8$. Also, if we call the circle $C$ and index the transformations as $\varphi_i$ is the set $\{\varphi_i(x): x \in C\}$ a traditional circle? $\endgroup$ Aug 9 at 1:38

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.