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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?

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

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.

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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). –  Bill Johnson Feb 1 '10 at 15:54
    
Why was this answer accepted? Konrad suggested an approach but did not give an answer. –  Bill Johnson Feb 1 '10 at 16:53
    
Sorry, I may have misunderstood the norms of what "accepting" an answer is supposed to mean. –  Jason Reed Feb 1 '10 at 18:29
    
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. –  Konrad Swanepoel Feb 1 '10 at 19:00
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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.

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I see Leonid added tags, which is something I don't know how to do. –  Bill Johnson Jan 31 '10 at 23:29
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It seems to me that someone like Bill Johnson should be given the 500 reputation points automatically. –  Deane Yang Jan 31 '10 at 23:46
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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 $X$. –  Bill Johnson Feb 1 '10 at 0:22
    
@Bill: I missed your comment, which is essentially my answer. –  Konrad Swanepoel Feb 1 '10 at 0:33
    
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. –  Bill Johnson Feb 2 '10 at 22:17
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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\ $}.

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