Continuous automorphism groups of normed vector spaces? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T22:53:27Z http://mathoverflow.net/feeds/question/13596 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/13596/continuous-automorphism-groups-of-normed-vector-spaces Continuous automorphism groups of normed vector spaces? Jason Reed 2010-01-31T22:57:08Z 2010-02-05T02:32:52Z <p>Consider the metric space on, say, &#8477;<sup>2</sup> 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 &#8477;<sup>2</sup> (or on &#8477;<sup>n</sup>) besides the euclidean one that give interesting isometry groups?</p> http://mathoverflow.net/questions/13596/continuous-automorphism-groups-of-normed-vector-spaces/13600#13600 Answer by Bill Johnson for Continuous automorphism groups of normed vector spaces? Bill Johnson 2010-01-31T23:28:16Z 2010-01-31T23:28:16Z <p>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.</p> <p>BTW: Banach-spaces would be a more appropriate tag IMO.</p> http://mathoverflow.net/questions/13596/continuous-automorphism-groups-of-normed-vector-spaces/13608#13608 Answer by Konrad Swanepoel for Continuous automorphism groups of normed vector spaces? Konrad Swanepoel 2010-02-01T00:31:55Z 2010-02-03T12:12:44Z <p>The following answer gives a partial description of the isometry groups of finite-dimensional normed spaces.</p> <p>I assume that an isometry is a <em>bijection</em> 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 $GL(n)$.</p> <p>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.</p> <p>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.</p> <p>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.</p> <p>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.</p> <p>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).</p> <p>As pointed out by Bill Johnson in a comment to his answer, it was shown by <a href="http://gdz.sub.uni-goettingen.de/en/dms/load/img/?PPN=PPN235181684%5F0241&amp;DMDID=dmdlog32" rel="nofollow">Gordon and Loewy</a> 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.</p> http://mathoverflow.net/questions/13596/continuous-automorphism-groups-of-normed-vector-spaces/14217#14217 Answer by Ady for Continuous automorphism groups of normed vector spaces? Ady 2010-02-05T02:32:52Z 2010-02-05T02:32:52Z <p>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\$}.</p>