This is a note on parts of Gromov’s paper ‘width and related invariants of Riemannian manifolds’ (1988).

For a compact subset of , we define the **k-codimensional width** (or simply **k-width**) to be the smallest possible number where there exists a k-dimensional affine subspace s.t. all points of is no more than away from .

i.e.

where is the length of the orthogonal segment from to .

It’s easy to see that, for any ,

.

At the first glance it may seems that . However it is not the case since for example the equilateral triangle of side length in has diameter but 0-width . In fact, by a theorem of Jung, this is indeed the optimum case, i.e. we have:

At this point one might wonder (at least I did), if we want to invent a notion that captures the ‘diameter’ after we ‘forget the longest k-dimensions’, a more direct way seem to be taking the smallest possible number where there is an orthogonal projection of onto a dimensional subspace where any point has pre-image with diameter .

i.e.

Now we easily have .

However, the disadvantage of this notion is, for example, there is no reason for a semicircle arc to have 1-width 0 but a three-quarters circular arc having positive 1-width.

Since we are measuring how far is the set from being linear, taking convex hull should not make the set ‘wider’ , unlike is not invariant under taking convex hulls. Note that for convex sets we do have

iff is contained in a -plane.

We now generalize this notion to general metric spaces:

**Definition:** The **Uryson k-width** of a compact metric space is the smallest number where there exists dimensional topological space and a continuous map where any point has pre-image with diameter .

i.e.

**Note:** Here dimension is the usual covering dimension for topological spaces: i.e. a topological space is dimensional if any finite cover of has a finite refinement s.t. no point of is contained in more than sets in the cover and is the smallest number with this property.

For compact subsets of with induced metric, we obviously we have since the pair is clearly among the pairs we are minimizing over.

Speaking of topological dimensions, one of the classical results is the following:

**Lebesgue’s lemma:** Let be the solid n-dimensional cube, then for any topological space with and any continuous map , we have image of at least one pair of opposite -faces intersect.

Since the conclusion is purely topological, this applies equally well to rectangles. i.e. for , , we have ; furthermore, for all .

(If the later statement does not hold, we write as , being the product of the first coordinates. Now ).

In light of the earlier post about minimax inequality, we should note that if we restrict to be a homeomorphic copy of then the notion is the same as the minimax length of fibres. In particular as proved in the post the minimax length of the unit disc to is 2.

**Exercise:** Check that for the unit -disk, , i.e. the optimum is obtained by contracting the disc onto a triod.

Hence it can indeed be strictly smaller than merely taking as the targeting space, even for simply connected sets. This gives a better measurement of ‘width’ in the sense that, for example, the neighborhood of a tree will have about .

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