Continuous automorphism groups of normed vector spaces? - MathOverflow

Continuous automorphism groups of normed vector spaces?

2010-01-31T22:57:08Z

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?

Answer by Bill Johnson for Continuous automorphism groups of normed vector spaces?

2010-01-31T23:28:16Z

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.

Answer by Konrad Swanepoel for Continuous automorphism groups of normed vector spaces?

2010-02-01T00:31:55Z

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.

Answer by Ady for Continuous automorphism groups of normed vector spaces?

2010-02-05T02:32:52Z

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\ $}.