Consider the metric space on, say, ℝ2 induced by the various $L^p$ norms, and the group of isometries from that space into itself that preserve the origin. When $p=2$ I get the continuous group of rotations, but when $p\in\{1,3,4,5,...\infty\}$ it looks like I just get $D_8$, the symmetry group of the square. Question: what's going on here? Why is 2 so special? Are there other natural norms on ℝ2 (or on ℝn) besides the euclidean one that give interesting isometry groups?

  • $\begingroup$ Fair enough. I was wondering about whether there was something more exotic than pasting together norms I already knew about. $\endgroup$ – Jason Reed Feb 1 '10 at 18:28

The following answer gives a partial description of the isometry groups of finite-dimensional normed spaces.

I assume that an isometry is a bijection preserving the distance function. By the Mazur-Ulam theorem it then follows that an isometry is a linear transformation composed with a translation. Thus we may assume without loss of generality that an isometry fixes the origin, so the isometry group is a subgroup of $GL(n)$.

Then the isometry group of any (real) finite-dimensional normed space is conjugate in $GL(n)$ to a closed subgroup of $O(n)$ that contain $-id$. This is seen as follows.

Consider the John ellipsoid $E$ of the unit ball $B$ of some $n$-dimensional normed space. This is the ellipsoid of largest volume contained in $B$ and, crucially, it is the unique such ellipsoid.

After some choice of basis we may assume that $E$ is the Euclidean ball. An isometry maps $B$ onto $B$, so it must map the John ellipsoid to the John ellipsoid. It follows that the isometry group is a subgroup of $O(n)$ containing $-id$. This subgroup is clearly closed, hence compact.

The converse is surely false. The following is an attempt at constructing a norm from such a subgroup. Fix a Euclidean unit vector $v$. Then its $Gv$ is a compact set of Euclidean unit vectors, symmetric with respect to the origin. Its convex hull $Gv$ is still compact and symmetric, so gives a unit ball $B_0$ of some norm on the linear span of $Gv$. If this linear span is not all of $\mathbb{R}^n$, then the unit ball has to be made full-dimensional in a sufficiently rough way, so as not to add any more isometries.

However, as pointed out by Leonid Kovalev in the comments, there are closed subgroups of $O(n)$, such as $U(n)$, where this construction gives a norm with a strictly larger isometry group (in the case of $U(n)$, the Euclidean norm).

As pointed out by Bill Johnson in a comment to his answer, it was shown by Gordon and Loewy that any $finite$ subgroup of $O(n)$ that contains $-id$ is the isometry group of some norm on $\mathbb{R}^n$. It's still my guess that the only way you can get infinite isometry groups (in the finite-dimensional case) is by having Euclidean subspaces, and for the norm to be so symmetric that it shares all the symmetries of this subspace.

  • $\begingroup$ There is no problem making the unit ball full dimensional, since you can include the orbit under $G$ of the unit vector basis. This is no loss of generality by Auerbach's lemma. Also, this construction, if it works would give a unit ball that is inside the Euclidean ball. The Euclidean ball would be the ellipsoid of minimal volume containing the unit ball (i.e., the polar of the John ellipsoid). $\endgroup$ – Bill Johnson Feb 1 '10 at 15:54
  • $\begingroup$ Why was this answer accepted? Konrad suggested an approach but did not give an answer. $\endgroup$ – Bill Johnson Feb 1 '10 at 16:53
  • $\begingroup$ Sorry, I may have misunderstood the norms of what "accepting" an answer is supposed to mean. $\endgroup$ – Jason Reed Feb 1 '10 at 18:29
  • $\begingroup$ Could somebody give a good description of the closed subgroups of $O(n)$? Perhaps this deserves to be a new question. In the two-dimensional case, the group is either finite (and dihedral) or the whole $O(2)$. For general $n$, perhaps there is an orthogonal decomposition of the space such that the orbit of a unit vector in each component is either finite or the whole unit sphere. $\endgroup$ – Konrad Swanepoel Feb 1 '10 at 19:00

I think what groups can be the isometry group of a finite dimensional normed space are classified, maybe by Y. Gordon and/or D.R. Lewis. I don't have access to emath from home but will check the reference tomorrow if no one has answered by then.

BTW: Banach-spaces would be a more appropriate tag IMO.

  • 1
    $\begingroup$ I see Leonid added tags, which is something I don't know how to do. $\endgroup$ – Bill Johnson Jan 31 '10 at 23:29
  • 2
    $\begingroup$ It seems to me that someone like Bill Johnson should be given the 500 reputation points automatically. $\endgroup$ – Deane Yang Jan 31 '10 at 23:46
  • 3
    $\begingroup$ Thanks, Leonid (and thanks for the vote of confidence, Deane). One other comment: Any group which is the group of isometries for some n dimensional normed space $X$ must be a (necessarily compact) subgroup of the orthogonal group because isometries of it preserve the ellipsoid of maximal volume inside the unit ball of $X$. $\endgroup$ – Bill Johnson Feb 1 '10 at 0:22
  • $\begingroup$ @Bill: I missed your comment, which is essentially my answer. $\endgroup$ – Konrad Swanepoel Feb 1 '10 at 0:33
  • $\begingroup$ Gordon and Loewy in Math. Annalen 241, 159-180 (1979) consider the question: If $G$ is a group of linear operators on $R^n$ which contains $I$ and $-I$, is it the group of isometries of some norm on $R^n$? Among other results, they prove that the answer is yes if $G$ is finite. $\endgroup$ – Bill Johnson Feb 2 '10 at 22:17

Consider the following norm on $\mathbb{R}^{2}$: $||(x,y)||$ := $|x|+|y|$ if $xy\leq0$; $||(x,y)||$ := $|y|$ if $xy\geq0$ and $|y|$ $\geq3|x|$; $||(x,y)||$ := $|x|+\frac{2}{3}|y|$ if $xy$ $>0$ and $|y|$ $\leq3|x|$. Then the group of isometries is { $\pm I\ $}.

  • $\begingroup$ Bill Davis proved in the 1970s that any (I think separable) Banach space can be equivalently renormed so that the only isometries are $\pm I$. $\endgroup$ – Bill Johnson Feb 5 '10 at 3:21
  • $\begingroup$ Obviously I should not have relied on my memory. Thanks for the correction, Leonid. $\endgroup$ – Bill Johnson Feb 9 '10 at 21:35
  • 1
    $\begingroup$ This is true for all [real] Banach spaces (separable or not), due to K.Jarosz siue.edu/MATH/kj_papers/AnyBanach.pdf . $\endgroup$ – Ady Feb 10 '10 at 22:09

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.