Let denote the -dimensional real space with the norm or . In two-dimensional case it is easy to plot the unit notice the pattern: from squareness to roundness back to squareness, as increases from to .

Unit spheres, from http://en.wikipedia.org/wiki/Unit_ball

In particular, and are isometric spaces. But is an exception: already in three dimensions we see that the unit balls are different. In it is an octahedron

Octahedron, from http://en.wikipedia.org/wiki/File:Octahedron.svg

and in it is a hexahedron aka cube:

Cube, from http://en.wikipedia.org/wiki/File:Hexahedron.svg

They differ in the number of faces and of vertices. But how can we use this discrepancy to show that and are not isometric for ?

One possibility it to invoke the Mazur-Ulam theorem which states that any surjective isometry between normed spaces is affine. Then the proof is finished by observing that the unit balls differ in the number of extreme points: for but only for . The high-dimensional versions of the octahedron, called cross-polytopes, have very few vertices which are, in some sense, extremely pointy (a lot of curvature must concentrate at each vertex since there are so few).

But the following proof does not rely on the Mazur-Ulam theorem and gives a stronger result:

does not admit an isometric embedding into for

Note that the Mazur-Ulam theorem does not apply here, since non-surjective isometries may well be non-linear. For example, the map defined by is an isometric embedding.

Proof: Suppose is an isometric embedding. We may assume . Let be the set of vertices of the unit cube, i.e., the set of all vectors with entries . Note three properties of :

1. all points are at distance from the origin
2. all points are at distance from one another
3. the cardinality of the set is .

All these properties must be shared by . For each vector consider its positive and negative supports and . Since , we may assume that at least vectors in have nonempty positive supports. It follows that at least two of them, say and , have overlapping positive supports. This creates cancellation when we compute the distance between them: , a contradiction. QED

After some tweaks the argument can be applied to -biLipschitz embeddings, defined by the double inequality , where and are the norms of the domain and the target space. (I used to write in such inequalities but recently picked notation from a paper by Gromov.) I give only the most straightforward modification: define and . Observe that with each vector of has nonempty support of one or of the other kind. Repeating the same proof brings us to the point where . This yields , and we conclude that for there is no -biLipschitz embedding with . The estimate is wasteful since the bound and the support-counting argument cannot be sharp at the same time. But in any case this proof is just a toy; the real stuff can be found, for example, in the book by Milman and Schechtman.

Bonus content: there is a simple (and well-known) isometric embedding in the opposite direction, from into with . It is best illustrated by an example with : the vector is mapped into the vector with coordinates , , , . At least one of the sums will have all terms of the same sign, and therefore have absolute value equal to .