Three sides and four are proved

The most area a fence can hold asked which shape of given perimeter holds the most area, and the answer was the circle. Every chord slid to the middle gave the tool that proves it — Steiner symmetrisation, which slides every chord across a line until the line cuts it in half, keeps the area and never lengthens the boundary — and ended on a question the tool cannot answer. It concerned a different quantity, one that behaves like perimeter under symmetrisation, and a restriction that symmetrisation does not respect: the number of corners.

The quantity is the lowest eigenvalue of the region. Take a region in the plane and look for a function that is zero on its boundary and satisfies $-\Delta u = \lambda u$ inside it, where $\Delta u$ is the sum of the second derivatives of $u$ in the two directions. Nonzero solutions exist only for particular values of $\lambda$, and the smallest is the region’s lowest eigenvalue. It measures how much room the region has: shrink a region by a factor of two and $\lambda$ quadruples, so the number that describes the shape alone is $\lambda$ times the area. Physically it is the lowest frequency, squared, of a membrane stretched over the region, or the slowest rate at which heat drains out of it through a cold boundary; mathematically it is the least value of the energy of a function relative to its size, as in the highest point on the sphere is an eigenvalue.

The disc is lowest, and symmetrising proves it #

Among all regions of a given area, the disc has the smallest lowest eigenvalue. That is the theorem of Georg Faber and Edgar Krahn from the 1920s, and its value is $\lambda \cdot \text{area} = \pi j^2 \approx 18.168$, where $j \approx 2.4048$ is the first zero of the Bessel function $J_0$ — the function that the lowest mode of a disc is made of, as the sine is for an interval, which a plucked string keeps its corners builds every motion of a string from.

The proof is symmetrisation, applied to the function rather than only the region. Take the lowest mode of any region and replace each of its level sets by a disc of the same area, centred at one point; the resulting function is zero on the boundary of a disc of the same area as the region, and a rearrangement inequality of Pólya and Szegő says its energy is no larger, while its size is unchanged. The disc’s lowest eigenvalue is the least energy over all functions on it, so it is at most the region’s. Every step is a version of the chord-sliding of every chord slid to the middle, done on the level sets of a function instead of on one shape.

The regular polygons give the obvious picture. Their lowest modes have every symmetry of the polygon — a lowest mode is unique up to a constant, so any symmetry of the region must carry it to itself — and so each could be computed on a single slice, a triangle from the centre to half a side, with the cut edges acting as mirrors, the device that half a circle against a wall used to turn a fence against a wall into a whole circle. The computation here does the whole polygon and lets the symmetry appear on its own, which is a check that the method is not quietly breaking it. The equilateral triangle has $\lambda \cdot \text{area} = 4\pi^2/\sqrt3 \approx 22.79$, the square $2\pi^2 \approx 19.74$, the pentagon 18.92, the hexagon 18.59, and the values fall towards the disc’s 18.168 as the number of sides grows. The computation behind the figure is linear finite elements: each polygon is cut into a fan of triangles from its centre, each triangle into hundreds of smaller ones, and the eigenvalue problem becomes a matrix problem whose smallest eigenvalue is found by inverse iteration. The mesh fits the polygon exactly, so the only error is the size of the small triangles, and computing at two sizes and extrapolating removes most of it — the triangle and square agree with their exact values to the places shown.

The eigenvalue is an energy, and the energy is perimeter #

The reason symmetrisation lowers the eigenvalue is that the eigenvalue is a perimeter in disguise. The lowest eigenvalue is the least possible ratio of a function’s energy, the integral of the square of its gradient, to its size, the integral of its square, over functions that vanish on the boundary. The lowest mode is the function that achieves the least ratio, and any other function gives an upper bound.

Slice a function into its level sets, the regions where it exceeds each value. Its size depends only on the areas of those regions. Its energy depends on how steeply it climbs from one level to the next, and, roughly speaking, a function climbing through level sets of given areas spends the least energy when the boundaries of those sets are as short as possible — the gradient is concentrated along the boundaries, and a shorter boundary to cross means less gradient spread over less length. The isoperimetric inequality says the shortest boundary for a given area is a circle. So replacing every level set by a disc of the same area keeps the size, does not raise the energy, and gives a function on a disc: the disc’s eigenvalue is no larger.

That is the Faber–Krahn theorem in one paragraph, and it shows exactly what the polygon problem lacks. The rearranged function lives on a disc, not a polygon, because the shortest boundary for each level set is a circle. To stay among $n$-gons the level sets would have to be replaced by the best $n$-gons of the same area, and the regular $n$-gon is indeed the best $n$-gon for perimeter — but the level sets of a function on a pentagon are not pentagons, so there is nothing to replace them with.

Triangles: the equilateral is lowest #

Polygons raise a question that the disc does not. Fix the number of sides. Which $n$-gon of given area has the smallest lowest eigenvalue? The natural guess is the regular one, and George Pólya and Gábor Szegő proved it for triangles and quadrilaterals in 1951.

The figure measures every triangle on a base by where its apex sits. The values rise in every direction from the equilateral apex — slowly along the line of isosceles triangles above the middle of the base, quickly for lopsided or flattened triangles, reaching 44.5 at the bottom corners of the grid, where the triangles are long and thin and the lowest mode has little room.

The proof uses the chord-sliding exactly as it is. Symmetrise a triangle across the perpendicular bisector of one of its sides, and every chord parallel to that side is slid to the bisector: the result is again a triangle, isosceles on that side, with the same area and — by the Faber–Krahn argument applied to symmetrisation across a line — a lowest eigenvalue no larger. Doing it to each side in turn drives any triangle towards the equilateral one, and since the eigenvalue can only fall along the way, the equilateral triangle is lowest.

Quadrilaterals: the square is lowest #

For rectangles the answer can be written down: a rectangle with sides $a$ and $b$ has lowest mode $\sin(\pi x/a)\sin(\pi y/b)$ and $\lambda = \pi^2(1/a^2 + 1/b^2)$, so $\lambda \cdot \text{area} = \pi^2(r + 1/r)$ for side ratio $r$, least at $r = 1$. The computed values in the figure agree with the formula, which is a second check on the method. For rhombi there is no formula, and the computation shows the value rising as the angle moves away from a right angle, slowly at first — a rhombus of 80° is only 1% above the square — and then quickly.

Pólya and Szegő’s proof for quadrilaterals symmetrises three times. Symmetrising a quadrilateral across a line perpendicular to one diagonal slides the chords parallel to that diagonal, and since two of the corners lie on one such chord, the result has only four corners again: a kite, symmetric about the line. Symmetrising the kite across the perpendicular to its other diagonal makes it a rhombus. And symmetrising a rhombus across a line perpendicular to two of its sides makes it a rectangle, since every chord parallel to those sides has the same length. Among rectangles the formula settles it. At each step the number of corners stays at four and the eigenvalue does not rise.

Pentagons: a minimum, by computation #

For five sides the same move fails. Symmetrising a pentagon across a line produces a shape whose boundary has a corner wherever an old corner’s chord lands, and a pentagon generally has corners at five different heights, which become up to eight corners in the result. The output is not a pentagon, and the argument has nothing to compare it with. That obstruction was drawn in every chord slid to the middle, and it is the whole reason the question is open from five sides up.

What can be done is to test the regular pentagon against its neighbours.

Every one of the forty nudged pentagons has a higher value than the regular one, from moves of one hundredth of the pentagon’s radius to moves of a sixth. And the excess grows as the square of the distance moved — the fitted slope on the logarithmic plot is 2.01 — which is how a smooth function behaves near a strict minimum: no first-order change in any direction, a positive second-order change in every direction. The regular pentagon is a local minimum, on this evidence and on a careful version of it. Work in the 2020s turned exactly this kind of computation into a proof of local minimality for the regular pentagon and hexagon, with interval arithmetic in place of floating-point numbers and with the second-order terms bounded rigorously.

Local is the operative word. A local minimum need not be the global one, and a computer can check the regular pentagon against its neighbours more easily than against every pentagon of the same area. Proving that it beats pentagons far from regular — a long thin one, a nearly triangular one with two short sides — needs either a different idea or a way to bound the eigenvalue of every pentagon in a whole region of shapes at once, and the second is the route the recent computer-assisted work has taken.

Perimeter and eigenvalue are not the same measure #

It is tempting to think that the eigenvalue question is the perimeter question in disguise, since both are minimised by the round shape and both are decreased by symmetrisation. Among pentagons the perimeter question is settled: the regular pentagon has the least perimeter for its area, a theorem going back to Zenodorus in antiquity and proved properly by the nineteenth century. If the eigenvalue were a function of the perimeter and area alone, the pentagon question would be answered.

It is not. The figure places random convex pentagons by both measures, and while they rise together and the regular pentagon sits at the bottom of both, the cloud has width: the vertical line joins two pentagons whose isoperimetric quotients differ by less than 0.005 and whose eigenvalues differ by more than four. The eigenvalue depends on how the area is distributed — a pentagon with a long thin spike can have a modest perimeter for its area and still give the lowest mode little room in the spike — and no inequality between perimeter and eigenvalue can transfer Zenodorus’s theorem to the eigenvalue. Bounds in one direction exist, and they lose too much near the regular pentagon to decide anything.

Why the corners matter #

The difference between three or four sides and five is a fact about the tool rather than about the shapes. Symmetrisation across a line through the midpoints of opposite sides keeps a rectangle a rectangle; across the perpendicular bisector of a side it keeps a triangle a triangle. For a pentagon no line has that property in general, because a pentagon has no pair of parallel sides and no axis that its corners are arranged around. The tool that works for the disc, and for all shapes when the number of corners is allowed to grow, breaks exactly at the restriction.

That puts the polygon question beside the least wall for equal rooms, where the hexagonal honeycomb’s optimality among all partitions took until 1999 to prove although among partitions into convex polygons it had been known for decades: restricting the class of shapes changes which tools apply, and the restricted question can be harder than the unrestricted one rather than easier. It is also a cousin of nearly the most means nearly round, which made the perimeter inequality quantitative. The quantitative Faber–Krahn inequality — a shape whose eigenvalue is close to the disc’s must be close to a disc — is known, and it says nothing about which pentagon is lowest, since all pentagons are far from discs.

Corners cost less than perimeter suggests #

The regular polygons in the second figure approach the disc, and the rate at which they do is a measurement of how much the corners cost. The excess of $\lambda \cdot \text{area}$ over the disc’s value is 1.57 for the square, 0.42 for the hexagon, 0.17 for the octagon and 0.05 for the twelve-sided polygon. Each doubling of the number of sides divides the excess by about eight, and the known expansion of the eigenvalue of the regular $n$-gon makes that exact in the limit: the excess is $4\zeta(3)/n^3$ of the disc’s value to leading order, where $\zeta(3) = 1 + 1/8 + 1/27 + \cdots \approx 1.202$. The computed values agree with it to within a few per cent from eight sides on.

The perimeter behaves differently. A regular $n$-gon’s isoperimetric quotient exceeds the circle’s by about $\pi^2/(3n^2)$, so each doubling of the sides divides the perimeter’s excess by four, not eight. The eigenvalue converges to the disc’s value faster than the perimeter does, by a whole power of $n$. Its lowest mode, as the first figure showed, is nearly round in the middle and feels the corners only near the edge, where it is already small; the perimeter has no such interior to hide them in. It is one more sign that the two measures are not one, and that a proof for the eigenvalue cannot simply be borrowed from the perimeter.

What the computation does not show #

The figures use finite elements with an error that falls as the square of the mesh size, extrapolated from two meshes. For the comparisons drawn — differences between nearby pentagons of a few thousandths of the value — the error is smaller than the differences, which is why the smallest nudges still show the slope of two cleanly. But a computation of finitely many pentagons is evidence, not a proof that none of the uncountably many others is lower, and the proofs that exist for the pentagon and hexagon cover neighbourhoods of the regular polygon.

The figures also concern convex polygons only. The conjecture is stated for every $n$-gon, convex or not, and a non-convex pentagon, with one corner pointing inward, has not been drawn here; symmetrisation makes any polygon convex, which is part of why it is the natural tool, and part of why losing it costs so much.

Still open: every number of sides #

The conjecture that the regular $n$-gon has the smallest lowest eigenvalue among $n$-gons of given area is proved for $n = 3$ and $n = 4$ by symmetrisation, supported for $n = 5$ and $n = 6$ by computer-assisted proofs of local minimality and extensive numerical search, and unproved for every $n$ from five upward. A proof for one more value of $n$ by a new method, or a replacement for symmetrisation that respects the number of corners, would settle the matter for that $n$ and probably for all of them.

The same pattern holds for other quantities. Pólya and Szegő’s book lists the electrostatic capacity, the torsional rigidity of a beam of polygonal cross-section and several others, each extremal for the circle among all shapes and conjectured extremal for the regular polygon among $n$-gons, and each proved only where symmetrisation keeps the number of sides. The polygonal versions stand or fall together on whether a tool that keeps corners can be found.