Radius of comparison equals half the mean dimension

The explanation

Cover the interval [0, 1] by two open pieces, [0, 0.6) and (0.4, 1]. At most two pieces overlap at any point. The order of the cover is one less than that, so it is 1. No cover of the interval by small pieces can do better, which is why its dimension is 1. A square needs points where three pieces meet, so its dimension is 2.

Now let a map h move the points of a space X, one step per unit of time. Mean dimension asks how much dimension the system uses per stepSource 1. Refine a cover by following each point for N steps, take the order of that refined cover, and divide by N. Gromov introduced mean dimension, and Lindenstrauss and Weiss developed itSource 2. It stays finite for some systems on infinite-dimensional spaces. A homeomorphism is minimal when every orbit comes close to every point.

Each such system gives a C*-algebra, the crossed product, built from functions on X and one operator for the map. In nice algebras, one positive element sits below another whenever its size under every trace is smaller. Toms's radius of comparison measures the extra margin needed for that to workSource 3. The Phillips–Toms conjecture says the radius equals half the mean dimension. The half is suggested by how complex rank relates to real dimension: one complex direction is two real ones. Phillips proved the radius is at most 1 + 36 times the mean dimensionSource 4. Niu proved the sharp upper bound of one halfSource 5. For mean dimension 2, Phillips's bound gives 73, and Niu's gives 1. The obstacle was the matching lower bound.

The paper claims the conjecture for every minimal homeomorphism of an infinite compact metrizable spaceSource 6. This includes the case where both sides are infinite. It follows that zero mean dimension is equivalent to Jiang–Su stability and to finite nuclear dimension, two key regularity propertiesSource 7. The proof assumes the cover rate is more than twice a comparison bound, then turns comparison into a bundle embedding. A bundle is a family of vector spaces over a space. The companion paper supplies the step that pushes maps into cubes onto cube boundariesSource 8. It needs the bundle's target rank to be less than twice the source rankSource 9. An example at equality shows this strict inequality cannot be removedSource 10. The compressed maps give a smaller cover, which is a contradiction.

How sure is this? There is no Lean check. A computer proof checker has not looked at any part of it. The proof spans two long papers, one in stable homotopy theory. No specialist review is published.

What it doesn't show. This does not cover actions of groups other than the integers, which the broader form of the conjecture includes.