The Mahler conjectures, functional inequalities and polar-product symplectic width

The explanation

Start with a square whose corners are (±1, ±1). Its area is 4. Now collect every point y with x·y ≤ 1 for all points x of the square. Here x·y means x₁y₁ + x₂y₂. Checking the four corners is enough, and they carve out a diamond with corners at (±1, 0) and (0, ±1). Its area is 2. This diamond is the square's polar, a partner shape that every symmetric shape has. Multiply the two areas and you get 8. That number is the volume product.

The product ignores size. Double the square's side and its area becomes 16, while its diamond shrinks to area 1/2. The product is still 8. Stretching or tilting the square does not change it either, so the product measures shape alone. Now try a disc of radius 1. Its polar is the same disc, so the product is π × π, about 9.87. The disc scores higher than the square.

In 1939, Kurt Mahler asked how low this product can go. He meant convex bodies, solid shapes with no dents, that are symmetric about their centre. He settled flat shapes himself. The expected answer in n dimensions is 4^n/n!, the score of a cube and its polar. In two dimensions that is 16/2 = 8, our square again. In three dimensions, a cube with corners (±1, ±1, ±1) has volume 8, and its polar octahedron has volume 4/3. Their product is 32/3, which matches 4³/3!.

The single difficulty is that the lowest score is shared. The octahedron also scores 32/3. In four dimensions, so does a prism built on an octahedron. No stretch or tilt turns it into a cube or the cube's polar. All of these are Hanner shapes. They are built from line segments by two moves: taking products, and joining shapes at the centre. The family grows with dimension, and a proof must let every member tie at once. Earlier methods reached the sharp bound only for special classes. Jean Saint-Raymond did it in 1981 for shapes unchanged by flipping the sign of any coordinate. Shlomo Reisner did it for zonoids. In 2020, Hiroshi Iriyeh and Masataka Shibata settled three dimensions. In 1987, Jean Bourgain and Vitali Milman proved a floor of a^n/n! in every dimension, for some fixed a above zero. Raising a all the way to 4 stayed open.

This paper proves the floor 4^n/n! in every dimension and finds every shape that reaches itSource 1. The idea is to count one total in two ways. First replace the shape by one cut out by flat slabs. Its polar is then the smallest convex shape holding a finite list of points. For each point of the original shape, the paper picks out small pointed pieces. Each has one corner at the centre and the rest at listed points. All of them sit inside the polarSource 2. So adding up their volumes, across the whole shape, gives at most the shape's volume times the polar's volume.

The new step shows the same total is at least 4^n/n!Source 3. It comes from a random experiment. Pick angles on a circle at random. Send them through a special map from a disc onto a lens shape. Then ask how often the result lands in the right place. A calculation from complex analysis shows these chances add up to at least 1Source 4. That is exactly the floor. Approximation then carries the result from slab shapes to every symmetric convex bodySource 5.

For the shapes that tie, the paper finds a hidden property. Measure distance using the polar as the unit ball. Then any three balls that overlap in pairs share a common pointSource 6. A known theorem of Hansen and Lima says only Hanner shapes, and their stretched or tilted copies, have this propertySource 7. So the square, the cube and their relatives are the only shapes on the floor, up to linear changesSource 1.

The floor needs symmetry. A companion paper drops it and measures each shape from its best interior point. There the floor is (n+1)^(n+1)/(n!)², reached exactly by simplices such as trianglesSource 8. For a triangle it is 27/4 = 6.75, below the square's 8. A second companion paper proves a statement in symplectic geometry, the geometry of position and momentumSource 9. That statement implies the symmetric floor by itselfSource 10. The paper also derives a sharp version for functions instead of shapes.

How sure is this? A computer proof checker called Lean has checked formal versions of four results. They are the symmetric floor, its equality cases, the general simplex floor and the symplectic statement. The functional inequalities are outside that check. No independent expert review has been published.

What it doesn't show. This does not compute the volume product of any specific body.