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Apr 13, 2021 at 13:02 comment added PseudoNeo @BillJohnson: is it not? what's wrong with the following construction: you take a finite number of points $F$ in a very small region of the unit sphere with no (nontrivial) symmetry and take the images of this finite set by all elements of G. The convex hull $\mathbf{co}\left(\bigcup_{g\in G} gF\right)$ is the unit ball of a norm whose isometries are G, right?
Feb 4, 2010 at 23:27 comment added Bill Johnson Apparently Gordon & Loewy did not know of Kalyuzhnyi's paper. Their proof is only one page (their paper is devoted mostly to other things) but the result is not obvious IMO.
Feb 2, 2010 at 22:17 comment added Bill Johnson 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.
Feb 1, 2010 at 18:26 vote accept Jason Reed
Feb 1, 2010 at 18:26
Feb 1, 2010 at 14:50 vote accept Jason Reed
Feb 1, 2010 at 14:50
Feb 1, 2010 at 0:33 comment added Konrad Swanepoel @Bill: I missed your comment, which is essentially my answer.
Feb 1, 2010 at 0:22 comment added Bill Johnson 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 $X$.
Jan 31, 2010 at 23:46 comment added Deane Yang It seems to me that someone like Bill Johnson should be given the 500 reputation points automatically.
Jan 31, 2010 at 23:29 comment added Bill Johnson I see Leonid added tags, which is something I don't know how to do.
Jan 31, 2010 at 23:28 history answered Bill Johnson CC BY-SA 2.5