# Nonplanar equilateral lattice “pentagons”

It is well-known that no two-dimensional point lattice contains a regular pentagon. (See for example http://mathworld.wolfram.com/LatticePolygon.html.) The same is true for lattices in $\mathbb{R}^n$, simply because any such polygon would lie in a two-dimensional sublattice. If we relax the requirement that the pentagon lie in a plane, we can easily find a closed path of 5 equal length sides: for example, in $\mathbb{R}^5$ with the integer lattice, we can use the path $$(1,0,0,0,0)\to(0,1,0,0,0)\to(0,0,1,0,0)\to(0,0,0,1,0)\to(0,0,0,0,1).$$ (These are the vertices of the standard 4-simplex.) The same construction, of course, works to find a nonplanar equilateral lattice "$n$-gon" for any $n$, as the vertices of the standard $n+1$-simplex in the integer lattice. We can improve this to a lattice in $\mathbb{R}^{n-1}$ by restricting to the plane $x_1+\cdots+x_n=1$. My question is if this is the best we can do, dimension-wise. So is it possible, for example, to find a nonplanar equilateral lattice pentagon in some lattice in $\mathbb{R}^3$? What can be said about the minimal dimension of lattice containing a nonplanar equilateral lattice $n$-gon?

(Note: This was originally posted on math stackexchange here, but nobody answered it.)

• There is an obvious 4dim solution, with three of the sides of a square of even length involved, with those three sides parallel to the axes of the 2dim. subspace they inhabit. – The Masked Avenger Dec 13 '13 at 22:10
• If you don't require that the lattice be square then you can even have a planar equilateral "pentagon": start with an equilateral triangle of side $2$ and remove a corner triangle of side $1$. Here two consecutive sides are collinear but that can be avoided. If $n$ is even then you can even make a planar equilateral $n$-gon in the square lattice; for instance for $n=6$ the sides can be the vectors $(5,0)$, $(3,4)$, $(0,5)$ and their negatives. – Noam D. Elkies Dec 13 '13 at 22:10
• Indeed. A Greek Cross is an example with 12 sides, and that can be extended easily by a multiple of 4 sides. Looks like odd n need at most 4 dimensions. – The Masked Avenger Dec 14 '13 at 1:28

Here are the vertices of an equilateral pentagon of side $\sqrt{2}$ in $\mathbb{Q}^3$ in order:

$(0,0,0), (1,1,0), (1,2,1), (0,1,1), (-\frac{1}{3}, -\frac{1}{3}, \frac{4}{3}).$

The first $4$ points form a rhombus. The last point satisfies $z=1-y, x^2+y^2+(1-y)^2=2.$ This has the rational solution $(1,1,0)$ and lines of any rational slope through $(1,1)$ in the $xy$-plane intersect $x^2+y^2+(1-y)^2=2$ in another rational point. I chose $x=y$. Expanding this by a factor of $3$ gives a nonplanar equilateral pentagon in $\mathbb{Z}^3.$

• This is very clever! May I ask how you found this? – Joseph O'Rourke Dec 14 '13 at 13:55
• @Joseph O'Rourke: If the first $4$ points of the pentagon are in a $2$-dimensional lattice, then adding any fifth point will give a pentagon in a $3$-dimensional lattice. Then the question is whether it is commensurate with a square lattice. Actually, there still seems to be a lot of flexibility in the plane. If there is a rational $\sqrt{-c^4+14c^2+15}$ then there is an equilateral pentagon with rational coordinates $(\pm 1,0),(\pm a,b),(0,c)$. I'm not sure if that is possible, but pentagons of that form are quite restricted among all equilateral pentagons. – Douglas Zare Dec 14 '13 at 21:55

Here is Douglas Zare's pentagon, after scaling by $3$: $$(0, 0, 0),\; (3, 3, 0),\; (3, 6, 3),\; (0, 3, 3),\; (-1, -1, 4)$$ • Ooh, I like pictures! – Rob Silversmith Dec 14 '13 at 14:09