6
$\begingroup$

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.)

$\endgroup$
3
  • 2
    $\begingroup$ 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. $\endgroup$ Dec 13, 2013 at 22:10
  • 2
    $\begingroup$ 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. $\endgroup$ Dec 13, 2013 at 22:10
  • 1
    $\begingroup$ 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. $\endgroup$ Dec 14, 2013 at 1:28

5 Answers 5

5
$\begingroup$

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 $\mathbb{Z}^3.$

$\endgroup$
2
  • $\begingroup$ This is very clever! May I ask how you found this? $\endgroup$ Dec 14, 2013 at 13:55
  • 1
    $\begingroup$ @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. $\endgroup$ Dec 14, 2013 at 21:55
6
$\begingroup$

By using integer programming (with the help of Gurobi solver), I found an equilateral pentagon of side $\sqrt{2}$ in $\mathbb{Z}^3$. Its vertices are: $$(0, 0, 0); (1, 1, 0); (1, 0, 1); (0, 0, 2); (0, 1, 1)$$

For those interested in the IP model, below is the code

import gurobipy as gp
from gurobipy import GRB


nSides = 5
model = gp.Model('Equilateral polygon')
model.params.Nonconvex = 2

vars = model.addMVar((nSides, 3), vtype = GRB.INTEGER)
distanceSquared = model.addVar(vtype = GRB.INTEGER, lb = 0, ub = GRB.INFINITY)

# fix a vertex at the origin
model.addConstr(vars[0, 0] == 0)
model.addConstr(vars[0, 1] == 0)
model.addConstr(vars[0, 2] == 0)

# equilateral polygon constraints
model.addConstr(distanceSquared >= 1)
for i in range(nSides):
    xA, yA, zA = vars[i]
    xB, yB, zB = vars[(i+1) % nSides]
    dAB = (xA - xB)**2 + (yA - yB)**2 + (zA - zB)**2
    model.addConstr(dAB == distanceSquared)

# distinct vertices constraints
for i in range(nSides):
    for j in range(i+1, nSides):
        xA, yA, zA = vars[i]
        xB, yB, zB = vars[j]
        dAB = (xA - xB)**2 + (yA - yB)**2 + (zA - zB)**2
        model.addConstr(dAB >= 1)

model.setObjective(distanceSquared, sense = GRB.MINIMIZE)
model.optimize()
print(vars.x)

Picture:

enter image description here

$\endgroup$
1
3
$\begingroup$

Here is Douglas Zare's pentagon, after scaling by $3$: $$(0, 0, 0),\; (3, 3, 0),\; (3, 6, 3),\; (0, 3, 3),\; (-1, -1, 4)$$
     PentagonZare

$\endgroup$
1
  • 1
    $\begingroup$ Ooh, I like pictures! $\endgroup$ Dec 14, 2013 at 14:09
1
$\begingroup$

By using appropriate vector norms it is possible to extract an equilateral triangle from a two-dimensional lattice, and a regular pentagon from a four-dimensional one. By "regular" I mean, in each case, equilateral and equiangular; as we will see in the four-dimensional case the figure (living in four-dimensional space) is not a planar regular pentagon.

The triangular case

The matrix

$M=\begin{pmatrix}0&-1\\1&-1\\\end {pmatrix}$

has the property that if we take successive powers its column vectors cycle among three values:

$M^1=\begin{pmatrix}0&-1\\1&-1\\\end {pmatrix}$

$M^2=\begin{pmatrix}-1&1\\-1&0\\\end {pmatrix}$

$M^3=\begin{pmatrix}1&0\\0&1\\\end {pmatrix}$

and of course the next powers repeat this pattern as $M^3$ is the identity. With that in mind we might suppose that the three points represented by these vectors

$(0,1),(1,0),(-1,-1)$

should somehow form an equilateral triangle.

Plotting on an ordinary Cartesian coordinate plane gives no evidence of any such equality. But suppose that instead of putting the positive $y$ axis at $90°$ from the positive $x$ axis, we were to make that angle $120$°. Using the Law of Cosines for triangles we can render distances from the origin in terms of a modified version of the Euclidean norm:

$||u||=||(u_1,u_2)||^2=u_1^2-u_1u_2+u_2^2 =\frac14(u_1+u_2)^2+\frac34(u_1-u_2)^2$

We see that in this norm all three points are equidistant from the origin, namely all the vectors have norm $1$.

But what about angular displacements? It is not difficult to see that for any vectors $u,v$ the above norm satisfies the relation

$||u+v||^2+||u-v||^2=2(|u|^2+|v|^2)$

and so we can define an associated inner product:

$(u\cdot v)=\frac14(||u+v||^2-||u-v||^2)=u_1v_1+u_2v_2-\frac12(u_1v_2+u_2v_1)$

When this is applied to the three coordinate vectors given above, we find that they all satisfy

$||u||=1,(u\cdot v)=-\frac12.$

In other words, with the right norm all three points are equidistant from the origin and at equal angles relative to each other, thus we have rendered an equilateral triangle using only two dimensions instead of three!

enter image description here Left: The points $(1,0),(0,1),(-1,-1)$ plotted in ordinary Cartesian coordinates do not form an equilateral triangle. Right: Modifying the coordinates so that lengths match a suitable metric renders the triangle equilateral.

The fourth dimension

For the pentagonal case we must start with a matrix whose powers give a 5-cycke of vectors. Such a matrix must have a characteristic polynomial equal to the fifth cyclotomic polynomial, which is degree $4$, and so we move to $\phi(5)=4$ dimensions. Analogously to the two-dimensional case, we have:

$M=\begin{pmatrix}0&0&0&-1\\1&0&0&-1\\0&1&0&-1\\0&0&1&-1\\\end{pmatrix}$

$M^2=\begin{pmatrix}0&0&-1&1\\0&0&-1&0\\1&0&-1&0\\0&1&-1&0\\\end{pmatrix}$

$M^3=\begin{pmatrix}0&-1&1&0\\0&-1&0&1\\0&-1&0&0\\1&-1&0&0\\\end{pmatrix}$

$M^4=\begin{pmatrix}-1&1&0&0\\-1&0&1&0\\-1&0&0&1\\-1&0&0&0\\\end{pmatrix}$

$M^5=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\\end{pmatrix}$

and our proposed cycle of coordinate points

$(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1),(-1,-1,-1,-1).$

As in the two-dimensional case, these points do not form a regular pentagon in the normal Cartesian space; but by using a modified vector norm (which again corresponds to not rendering the axes orthogonally), we gain the symmetry. The appropriate norm and its associated inner product are:

$||u||=||(u_1,u_2,u_3,u_4)||^2=u_1^2+u_2^2+u_3^2+u_4^2-\frac12(u_1u_2+u_1u_3+...+u_3u_4) =\frac1{16}(u_1+u_2+u_3+u_4)^2+\frac5{16}(u_1+u_2-u_3-u_4)^2+\frac5{16}(u_1-u_2-u_3+u_4)^2+\frac5{16}(u_1-u_2+u_3-u_4)^2$

$(u\cdot v)=u_1v_1+u_2v_2+u_3v_3+u_4v_4-\frac14(u_1v_2+u_1v_3+...u_3v_4)$

Here the cross-product terms run through all distinct permutations, including the distinction hetween $u$ and $v$ in the inner product. We note that as in the two-dimensional case, the net effect is to reduce the weight of the "one-signed" coordinate direction along which the point with all coordinates $-1$ lies.

With the correct norm and inner product above, the proposed coordinate vectors all satisfy

$||u||=1, (u\cdot v)=-\frac14.$

This means our five points are the vertices of a $4$-simplex centered at the origin, and any of twelve cycles of five edges embedded in this simplex will be a regular (equilateral, equiangular), nonplanar pentagon.

$\endgroup$
3
  • $\begingroup$ ("acing"?) The original post already had a regular pentagon in four dimensions, didn't it? $\endgroup$ Oct 2, 2022 at 22:17
  • $\begingroup$ @GerryMyerson that depends on definition. The regular pentagon in the original post lies in a four-dimensional cross-section of $\mathbb{Q}^5$; mine is derived directly in $\mathbb{Q}^4$ (with a suitable metric). $\endgroup$ Oct 2, 2022 at 22:43
  • $\begingroup$ @GerryMyerson also it seems that for an object with fivefold symmetry we need at least four dimensions in the lattice. $\endgroup$ Oct 2, 2022 at 22:43
0
$\begingroup$

In order to construct a lattice containing a "regular non-planar $n$-gone" take a lattice $\Lambda$ with an isometry of order exactly $n$. Any orbit with $n$ elements (i.e. with trivial stabilizer for the cyclic subgroup of $n$ elements generated by the above element) defines a non-planar regular $n$-gone. The smallest dimension where this is possible is the smallest dimension $d$ such that the cyclic group $\mathbb Z/n\mathbb Z$ has a faithful linear representation in $\mathrm{GL}_d(\mathbb Q)$ (which is $4$ for the cyclic group of order $5$). There might however exist "non-planar regular $n$-gones" in lattices of smaller dimension. Their vertices do however never form an orbit for an isometry group acting on the lattice (this is of course the case for the $3$-dimensional example given in previous answers). There are therefore three notions of "non-planar regularity": a weak one (just isometric edges) and a stronger one (an isometry subgroup of the lattices acts transitively on vertices) and the strongest one: an isometry subgroup of the containing lattice acts transitively on flags (corresponding to a dihedral group with $2n$ elements). The last one is probably best from a geometric perspective. Minimal possible dimensions for the last two questions are given by minimal dimensions containing rational representations of the underlying groups.

Added: I am not completely sure in the case of dihedral groups, my limited two dimensional intuition might be wrong!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.