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Let $n \ge 2$ be some positive integer. Given a norm $p : \mathbb{R}^n \to \mathbb{R}$, one can inquire about the structure and properties of its isometry group, i.e. the group of all bijections $F:\mathbb{R}^{n}\to\mathbb{R}^{n}$ such that $p\left(v\right)=p\left(F\left(v\right)\right)$ for all $v \in \mathbb{R}^n$. By the Mazur-Ulam theorem, the isometries of $p$ are affine transformations, so the subgroup $Iso_{0}\left(p\right)$ of all isometries which fix the origin, consists of linear transformations, and is therefore a closed subgroup of $GL_n (\mathbb{R})$. Essentially, $Iso_{0}\left(p\right)$ is what we get after modding out the obvious distance-preserving functions (the translations), which have no relation to $p$ whatsoever.

For some norms, $Iso_{0}\left(p\right)$ is a very small group. For instance, if $p$ is either the $\ell^\infty$ norm or the $\ell^1$ norm, then the group of $Iso_{0}\left(p\right)$ is finite, and this seems to be the case for all $\ell^q$ norms other than $q=2$. On the other hand, for the euclidean norm ($p = \ell^2$), $Iso_{0}\left(p\right)$ is quite a rich group: it is a Lie group of positive dimension (which increases significantly with $n$) - the orthogonal group $O(n)$.

Obviously, a significant difference between this norm and other possible norms is that it arises from an inner product on $\mathbb{R}^n$. Since there is only one inner product structure on $\mathbb{R}^n$ (up to conjugacy) this essentially gives us just one example of a norm with a rich group of linear isometries. So my questions are:

  1. Are there any other norms (not induced by an inner product) on $\mathbb{R}^n$ which have a rich group of linear isometries? Here let me define "rich" as having positive dimension as a Lie subgroup of $GL_n (\mathbb{R})$ (every closed linear group is canonically a submanifold of $GL_n (\mathbb{R})$ compatible with the group structure).
  2. If not, why is the property of being induced from an inner product the "right" condition for having many symmetries? That is, from a geometric point of view, why does this property make the norm especially "symmetric" or "smooth"?
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For question 1 can't you do something silly like the direct sum of a norm which is induced from an inner product on $\mathbb{R}^{n-1}$ and some other $p$-power norm on $\mathbb{R}$? –  Qiaochu Yuan May 10 '11 at 1:34
    
@Qiaochu Yuan: That does indeed work. Define the norm $\|(x,x_{n+1})\|:=max\lbrace \|x\|_2,|x_{n+1}|\rbrace$ on $\mathbb{R}^{n+1}$. Then $O(n)$ is a subgroup of the isometry group of this norm. –  Johannes Hahn May 10 '11 at 11:32
    
That's very nice. Thanks to both of you. –  Mark May 10 '11 at 11:59
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6 Answers 6

up vote 11 down vote accepted

As an answer to (1), consider any compact subgroup $G\subset O(n)$ and look at the $G$-invariant functions on $\mathbb{R}^n$ that have homogeneity $1$. As long as $G$ does not act transitively on the space of lines in $\mathbb{R}^n$, that set of functions will be properly larger than the $O(n)$-invariant functions of homogeneity $1$, and any one of them that is sufficiently near the standard norm will be a norm.

As a specific example, consider the irreducible action of $SO(3)$ on the $5$-dimensional vector space $S$ of traceless symmetric $3$-by-$3$ matrices. There are two $SO(3)$-invariant polynomials on $S$: $p_2(m) = trace(m^2)$ and $p_3(m) = det(m)$. (They generate the ring of $SO(3)$-invariant polynomials on $S$.) Now let $$ |m| = \bigl(p_2(m)^3 + \epsilon\ p_3(m)^2\bigr)^{1/6}. $$ for some small $\epsilon$. When $\epsilon>0$ is sufficiently small, this will be a strictly convex norm on $S$, and its symmetry group will be $SO(3)$, which acts irreducibly on $S$.

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Little contribution: The linear isometries in the plan R2, for a p-norm, is discrete and finite, excepted for p=2, the euclidean plan. This is demonstrated with elemantary maths in http://lucas.borboleta.blog.free.fr/index.php?trackback/2219894.

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A source of examples for question (1): If two finite-dimensional normed spaces have "rich" symmetry groups, then the space of linear maps between these with the induced operator norm will also have a "rich" symmetry group.

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As implicit in Bill's answer and prooved in Konrad's the answer to Continuous automorphism groups of normed vector spaces?, any isometry group of a normed space is a subgroup $G$ of $O(n)$ that contains $-id$.

It may be a good starting point.

Does the positive dimension imply that the space contains an Euclidean plane?

Forget the original norm and consider Euclidean $n$-space with a closed subgroup of $O(n)$ acting on it.

Let $A$ be any non zero element of the Lie algebra tangent to $G$. Since $exp(\epsilon A)$ is a rotation, and every rotation is equivalent to a matrix that has $2\times 2$ rotation blocks in the diagonal and zeros elsewhere. Expressed in the same (or other?) coordinates, $A$ probably had $2\times 2$ antisymmetric blocks in the diagonal and zeros elsewhere (although I don't know right now how to prove this fact without abandoning the algebra). The point having first coordinate $1$ and then zeros moves in a circle around the origin. Hence the spaces contains an Euclidean plane.

Related stuff:

Easy proof of the fact that isotropic spaces are Euclidean Maximal Ellipsoid Towards a metric characterization of Euclidean spaces (not too related but I'm trying to promote it).

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Thanks! The links were helpful. –  Mark May 10 '11 at 11:59
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Related problems have been studied by people in Banach space theory. Why don't you talk with Yehoram (Joram) Gordon, who did deep work in this direction?

As for (2), the answer is easy: If $C$ is the John's maximal volume ellipsoid on the space, any linear isometry maps $C$ onto itself. This means that you get the "largest" group of isometries on the space from a Euclidean structure.

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An even easier way to answer (2) is to use the Legendre ellipsoid in place of John's -- no need to refer to an existence and uniqueness result. –  Mark Meckes May 23 '11 at 19:28
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