The 'hexagonal tiling honeycomb' is a beautiful structure in 3-dimensional hyperbolic space. I'm trying to figure out something about it.
It contains infinitely many sheets of hexagons, tiling planes in the usual way hexagons do. These are flat Euclidean planes in 3d hyperbolic space, called 'horospheres'. I want to know the coordinates of the vertices of these hexagons. I have some clues.
The hexagonal tiling honeycomb has Schläfli symbol {6,3,3} . The Schläfli symbol is defined in a recursive way. The symbol for the hexagon is {6}. The symbol for the hexagonal tiling of the plane is {6,3} because 3 hexagons meet at each vertex. Finally, the hexagonal tiling honeycomb has symbol {6,3,3} because 3 hexagonal tilings meet at each edge!
The symmetry group of the hexagonal tiling honeycomb is the Coxeter group {6,3,3}. This is a discrete subgroup of the Lorentz group O(3,1), which acts on 3d hyperbolic space because that space is the set of points (t,x,y,z) in Minkowski spacetime with
t² − x² − y² − z² = 1 and t > 0
The Coxeter group {6,3,3} is generated by reflections, but its 'even part', generated by pairs of reflections, is a discrete subgroup of PSL(2,ℂ), because this is the identity component of the Lorentz group. In fact, this Coxeter group is almost PSL(2,𝔼), where 𝔼 is the ring of 'Eisenstein integers'. These are complex numbers of the form
a + bω
where a,b are integers and ω is a nontrivial cube root of 1. So there should be a nice description of the hexagonal tiling honeycomb using Eisenstein integers! And this is what I'm trying to find... quickly, before May 1st because I'm have a column due then. 😧
Here are some clues. The connection to Eisenstein integers must work roughly as follows. We can think of Minkowski spacetime as the space 𝔥₂(ℂ) of 2×2 hermitian matrices, since they're all of the form
t+z x-iy
x+iy t-z
and the determinant of such a matrix is t² − x² − y² − z² = 1. So 3d hyperbolic space consists of 2×2 hermitian matrices with det = 1 and t > 0.
If we look at matrices like this where the entries are Eisenstein integers we should get something closely related to the hexagonal tiling lattice. I'm hoping these matrices give the points sitting in the middle of each hexagon. (Or maybe they give the vertices of the hexagons, but that seems less plausible.)
But I want to check this. I want to see why these matrices lie on planes - more precisely horospheres - and give hexagonal tilings of these horospheres!
I'll say some more. But any help you can give would be appreciated. Someone must have worked this out already, but I'm not finding it. I need to read more references in this:
@johncarlosbaez E.Shulte and P.McMullen went on to write a monograph "Abstract regular polytopes", where these things pop up a lot. But it's connected to even more fun things, such as sporadic simple groups - shameless plug here: https://arxiv.org/abs/1603.01710
One clue: as I mentioned, the hexagons in the hexagonal tiling lattice form sheets that live on 'horospheres': that is, flat Euclidean planes sitting in 3d hyperbolic space. To get these horospheres is easy in the description of hyperbolic space as the hyperboloid t² − x² − y² − z² = 1, t > 0 in Minkowski spacetime. Simply take all the vectors in this hyperboloid that are orthogonal to some lightlike vector in Minkowski spacetime!
Let's think about an example using the description of Minkowski spacetime as the space 𝔥₂(ℂ) of 2×2 hermitian matrices. A nice lightlike vector is t=z=1, x=y=0, which gives the matrix
2 0
0 0
The vectors orthogonal to this (using the Minkowski metric) are
u x+iy
x-iy 0
where u = t = z. I.e. we need t = z, but x and y are arbitrary. To get vectors like this in the hyperboloid we also need u > 0 and det = 1, i.e. x²=y². All that's easy: just pick any t > 0 and any x, and let z = t and y = ±x. So that's our horosphere.
But let's take a look at vectors like that also lie in 𝔥₂(𝔼). Now we need t and x+iy to be Eisenstein integers! That means t is a plain old integer, but x+iy is an Eisenstein integer with y = ±x. Hmm, this is not looking good, since the only Eisenstein integer with y = ±x is 0, I think. So this is no good: this particular horosphere only contains a discrete line of points in 𝔥₂(𝔼). I want the points to form a hexagonal or triangular lattice.
Maybe I picked a bad lightlike vector? All future-pointing lightlike vectors are ''the same' in that they're related by Lorentz transformations and rotations. But now we've broken that symmetry by picking a lattice 𝔥₂(𝔼). So some are better than others!
I don’t know if this helps, but the identity matrix / aka the vector (1,0,0,0), has the following 12 nearest neighbours among 2×2 hermitian matrices with determinant 1, positive trace, and all entries Eisenstein integers.
@gregeganSF - thanks, it may help! I wonder what honeycomb you are starting to explore here (and what part of it you're exploring: maybe the vertices, maybe the midpoints of edges or faces, maybe the centers of cells). There are various closely related honeycombs that have the same symmetry group {6,3,3}, so you're probably seeing one of those:
The dual of the hexagonal tiling honeycomb has all its vertices 'at infinity' - i.e. they are lightlike vectors, not actually in the hyperboloid.
So, one thing I want to do is figure out all the lightlike 2x2 matrices (i.e. those with det = 0) that have Eisenstein entries. I want to use the 'Pythagorean spinors' trick.
@gregeganSF - in the hexagonal tiling honeycomb I think it goes like this:
each vertex has 4 nearest neighbors arrange in a tetrahedron
each edge midpoint has 6 nearest neighbors arranged in 2 equilateral triangles
each face midpoint has, hmm, 12 nearest neighbors?
So maybe the last one is what you're seeing. That would actually accord with there being 'sheets' of hexagons lying on horospheres, with the centers of the hexagons on each sheet forming a copy of the Eisenstein integers. (I don't know if that's true, but I'm hoping it's true.)
Coxeter gives the edge length of this honeycomb to be 2φ where cosh^2 φ = 9/8, which I believe means the dot product between the Lorentzian unit timelike vectors corresponding to adjacent vertices should be:
@gregeganSF - I seem to recall a relevant warning from previous studies of this subject. Namely, if the centers of a bunch of hexagons all lie on some horosphere in hyperbolic space, the vertices of those hexagons don't lie on that horosphere - and vice versa.
It's analogous to this more familiar situation, which works the other way around. In Coxeter theory it's natural to think of a polyhedron like a cube as drawn on a sphere. Then its 'square' faces actually have edges that are geodesics in that sphere - so it's a kind of spherical cube. But if we think of the sphere as sitting in flat 3-space, we can't compute the Euclidean 3d distance between a square's center and its vertices using the metric of the sphere.
Here we are thinking of the hexagonal tiling as drawn on a Euclidean plane. Its hexagonal faces actually have edges that are geodesics in that plane. But if we think of this plane as sitting in hyperbolic 3-space, we can't compute the hyperbolic 3d distance between a hexagon's center and its vertices using the metric of the plane.... as you seem to have been doing.
@gregeganSF - I was extremely confused about this for a while. I guess an efficient way to state it is that while the horocycle is a flat plane in hyperbolic space, it's not a 'totally geodesic submanifold': geodesics in the horocycle are not geodesics in hyperbolic space.
I figured out the dot product between hexagon centres, when all the vertices of the hexagons lie on the same horosphere and the dot product between vertices is –5/4 ... and it is indeed –3/2!
@gregeganSF - I see. Yay! Great! I will try to dream up an approach to directly confirming that the points on the hyperboloid whose 2×2 matrices have Eisenstein integer entries are hexagon centers for a hexagonal lattice tiling. As I sleep.
These are the 6 vertices whose hexagon centre is the identity matrix, multiplied by √6 to give Eisenstein-integer matrices.
The dot product between a hexagon centre and any of its vertices is –√(3/2).
The dot product between adjacent vertices is –5/4.
The dot product between nearest-neighbour hexagon centres is –3/2.
These particular 6 vertices lie on the two horospheres defined by dot products of –√(3/2) with the two null vectors (1,0,0,±1).
Empirically, all the vertices in the honeycomb seem to be matrices with Eisenstein integer entries divided by √6. The diagonal entries are always multiples of 3/√6.
@gregeganSF - this is great; this is just the sort of thing that can lead to a clear description of the hexagonal tiling lattice in terms of Eisenstein integers.
One thing confuses me; you write
"These particular 6 vertices lie on the two horospheres defined by dot products of –√(3/2) with the two null vectors (1,0,0,±1)."
A horosphere is a subset of the hyperboloid consisting of points that have vanishing dot product with some null vector. At least that's what Wikipedia says. And it makes sense to me, because this null vector has vanishing dot product with itself, and it defines a point at infinity in hyperbolic space, so the horosphere approaches one point at infinity, as shown here by @roice3.
The set of points in the hyperboloid having some fixed nonzero dot product with a null vector should be either a sphere in hyperbolic space or a 'hypercycle' (meaning it hits many points at infinity).
I went crazy last night trying to make sense of the claim that a horosphere comes from a vanishing dot product with a null vector … then finally realised that it’s nonsensical.
Suppose the null vector is (1,0,0,1). Its dot product with a generic vector (t,x,y,z) is:
d = –t+z
so the only vectors for which d=0 will take the form:
w = (t,x,y,t)
But the dot product of w with itself is:
x^2+y^2
So w is either null or spacelike! There is no intersection with the hyperboloid, which consists of timelike unit vectors!
[Edited to add: I don’t actually see the claim that the dot product vanishes in Wikipedia.
@gregeganSF@roice3 - I'm sorry to have driven you crazy. On Wikipedia I read
"In the hyperboloid model, a horosphere is represented by a plane whose normal lies in the asymptotic cone."
and thought that meant all points on a horosphere have vanishing dot product with some fixed null vector. After all, the normal vector to a plane has vanishing dot product with all tangent vectors to the plane. This is weird in the case where the normal vector is lightlike and thus also tangent to the plane, but I'm used to that.
Similarly in the article on horocycles they say
"In the hyperboloid model they are represented by intersections of the hyperboloid with planes whose normal lies on the asymptotic cone (i.e., is a null vector in three-dimensional Minkowski space.)"
So what the heck do they mean? What does "normal" mean to them?
Anyway, I can see you're right. I can actually see it in the case of 3d spacetime: the plane you're calling 0 = -t+z is tangent to the lightcone along a line and does not poke into the lightcone where the timelike vectors live. So is Wikipedia just wrong or are they poorly explaining something correct?
More importantly, what actually is a horosphere? Once I know this I should be able to find one horosphere containing infinitely many hexagon vertices.... and then by symmetry generate all such horospheres.
If what you're doing is correct, vectors having a dot product of –√(3/2) with the null vector (1,0,0,1) should form a horosphere. And we should be able to see why this horosphere contains exactly one point at infinity.
As far as I’ve always understood the terminology, when someone talks about the “normal” to a hyperplane, N, that means they’re describing the set of points x such that:
x·N = c
for some constant c.
If the hyperplane passes through the origin, then c=0, but for a general affine hyperplane, c need not be zero.
I just want to add something about the horosphere and ideal points. I was a bit confused about that, but now I think I understand what’s going on.
A given horosphere will have a single limit point on the boundary of the Poincaré ball model. In fact, all the horospheres that are defined via all the affine planes that share the same normal null ray will have the same limit point in the ball model.
For example, all the horospheres defined by planes normal to (1,0,0,1) will have a unique limit point of (0,0,1) in the ball model.
But this does not mean that the defining equation of the horosphere:
g((1,0,0,1),(t,x,y,z)) = α
α < 0
has a unique null vector solution.
That equation is solved by a whole 2-parameter family of null vectors:
(t, x, ±√(–x^2–α(2t+α)), t+α)
But what matters, in terms of the ideal point, is the limit as t→∞ of the projection into the Poincaré ball:
So this lets us embed a Euclidean (A_2) lattice in the horosphere, and getting the vertices of a hexagonal honeycomb is just a matter of omitting some points.
@gregeganSF - I claim, and want to check, that we can arrange things so that all the hexagon centers are exactly all the 2×2 self-adjoint matrices with Eisenstein integer entries and det = 1, tr > 0.
I'll start by checking that your matrices 𝐶(𝑗,𝑛) are of this sort (though obviously not all of them). Starting from these, it may be easy to check my claim using symmetry considerations. I will try to spell this out elsewhere.
One nuance is that I'm using the Minkowski metric with the opposite of your sign convention, namely 𝑡²−𝑥²−𝑦²−𝑧², since this equals the determinant of
@gregeganSF - it's amazing how out of practice I am at simple algebra; it took me way too long to compute the determinant of that matrix and if I hadn't known it should be 1 I'd probably never have caught all my mistakes. I could see the trace was > 0.
Now let me start the symmetry story. SL(2,ℂ) acts on the space 𝔥₂(ℂ) of 2×2 hermitian matrices A by
A ↦ gAg*
in a way that preserves the determinant, and since g= ±1 act trivially this is how we see SL(2,ℂ) double covers the Lorentz group. SL(2,𝔼) acts in the same way on the lattice 𝔥₂(𝔼) of 2×2 hermitian matrices with Eisenstein integer entries. And this is almost the symmetry group of the hexagonal tiling honeycomb, which is the Coxeter group {6,3,3}... but not quite!
First, we don't get reflection symmetries this way. At most we'll get the even part of the Coxeter group, called {6,3,3}^+. But this would still be great because {6,3,3}^+ still acts transitively on hexagon centers.
However, Johnson and Weiss seem to say that to get all of {6,3,3}^+ we need a group slightly bigger than SL(2,𝔼). I'll call this larger group GL(2,𝔼) because their notation is annoying to write. This larger group consists of 2×2 Eisenstein integer matrices whose determinant is an Eisenstein integer of norm 1. I think these are just the 2×2 Eiseinstein integer matrices whose inverse is again an Eisenstein integer matrix! This group should be "3 times as big" as SL(2,𝔼).
So, GL(2,𝔼) should act transitively on hexagon centers, and I guess we use the same formula
A ↦ gAg*
But it's possible that already SL(2,𝔼) acts transitively on hexagon centers, since we don't necessarily need to use the full even Coxeter group {6,3,3}^+ to get transitivity.
@gregeganSF - Anyway, let's say we know GL(2,𝔼) is the even Coxeter group {6,3,3}^+. By general nonsense I think it acts transitively on hexagon centers. But GL(2,𝔼) also maps 2×2 Eisenstein integer matrices with det = 1 and tr > 0 to themselves via
A ↦ gAg*
It follows that if one hexagon center is a 2×2 Eisenstein integer matrix with det = 1 and tr > 0, they all are! And we know that one is.
Next I want to show the converse: every 2×2 Eisenstein integer matrix with det = 1 and tr > 0 is a hexagon center.
For this, it would be enough to show that GL(2,𝔼) acts transitively on 2×2 Eisenstein integer matrices with det = 1 and tr > 0. Then we could reverse the argument I just gave: if one 2×2 Eisenstein integer matrix with det = 1 and tr > 0 is a hexagon center, they all are! And we know that one is.
@gregeganSF - what exactly is the difference between using SL(2,𝔼) and GL(2,𝔼)? I don't understand this deeply, but GL(2,𝔼) is generated by SL(2,𝔼) and this one extra matrix
where (\zeta) is a primitive 6th root of unity, so I can try to figure out how this acts on Minkowski spacetime... it must be a rotation of some sort... and why SL(2,𝔼) transformations aren't able to do this.
Grey lines join all nearest neighbours among these 13 points.
The 6 blue vertices that form a hexagon centred on the black point are all equidistant from the black point, one magenta point and one yellow point. They are:
We then map that central hexagon to 12 others, using Lorentz transformations that take I to each of its 12 nearest neighbours in 𝔥₂(𝔼), and that take one hexagon edge to the opposite edge.
@gregeganSF - this is wonderful! If we can show these 12 neighbors are the closest points with Eisenstein coordinates to the central one (which you claim), and find GL(2,𝔾) transformations mapping the central point to these 12 neighbors (which you've done), I think we've almost proved that all the Eisenstein points are exactly all the hexagon centers. I'll try to fill in some details.
I've been chewing away at the other end of this biscuit:
The elements of 𝔥₂(𝔼) can be partially parameterised as:
[\left(
\begin{array}{cc}
a & c+d \omega \
(c-d)-d \omega & g \
\end{array}
\right)]
For integers a,c,d,g this will have Eisenstein integer entries and will automatically be Hermitian, but we then have to require additionally:
[\begin{array}{rcl}
a+g &>&0 \
a g-c^2+c d-d^2 &=&1
\end{array}]
The nearest neighbour to the identity matrix will have the smallest possible trace that is greater than 2; this is just a restatement of the normal Minkowski dot product in terms of our matrices, and specialising to the dot product with the identity matrix.
The determinant condition gives us:
[\begin{array}{rcl}
a g &=&1+c^2-c d+d^2
\end{array}]
The RHS here can only take positive integer values. If it is 1, then a g = 1 only has the positive-trace solution a=g=1, the identity matrix.
If the RHS is 2, then a g = 2 only has the positive-trace solutions a=1, g=2 and a=2, g=1, with a trace of 3. Since the trace must be an integer, 3 is the smallest possible value greater than 2, and it is achievable with either of those 2 choices for a and g, along with 6 choices for c and d.
So the 12 matrices I listed are precisely the nearest neighbours of the identity matrix.
In the 1990s, J.H. Conway published a combinatorial-geometric method for analyzing integer-valued binary quadratic forms (BQFs). Using a visualization he named the "topograph," Conway revisited the reduction of BQFs and the solution of quadratic Diophantine equations such as Pell's equation. It appears that the crux of his method is the coincidence between the arithmetic group PGL2(ℤ) and the Coxeter group of type (3,∞). There are many arithmetic Coxeter groups, and each may have unforeseen applications to arithmetic. We introduce Conway's topograph, and generalizations to other arithmetic Coxeter groups. This includes a study of "arithmetic flags" and variants of binary quadratic forms.
@gregeganSF - I used to talk to Martin Weissman when I was in Singapore, since he'd gotten a job at Yale-NUS (a college set up by Yale and the National University of Singapore, which later went bust).
Anyway, I'm starting to read "The arithmetic of arithmetic Coxeter groups", and I'm hoping it holds the answers to my remaining puzzles. [EDIT: no, it does not.] The Eisenstein integers show up in Theorem 2.1.1, I guess you've seen, and it's interesting that PSL(2,𝔾) is not the whole even (=orientation-preserving) part of the Coxeter group {6,3,3}⁺ , but only an index-2 subgroup. Johnson and Weiss show PGL(2,𝔾) is all of {6,3,3}⁺. All this is a bit confusing to me since the simplest element in GL(2,𝔾) that's not in SL(2,𝔾) is
ζ 0
0 1
where ζ = exp(iπ/3). This element has order 6, but somehow it makes PGL(2,𝔾) just twice as big as SL(2,𝔾).
This is not a paradox, since all sorts of things are going on here. For example in the definition of PGL we are also modding out by more than we are in the definition of PSL. But I don't clearly see how it works. It should be visible in the symmetries of the hexagonal tiling honeycomb.
@gregeganSF - Btw, here's my current best strategy to show that every element A ∈ 𝔥₂(𝔼) with det(A) = 1 and tr(A) > 0 actually lives in the honeycomb you're building, i.e. is obtained starting from A = I by repeatedly applying transformations
First, we show that by repeatedly applying these 12 transformations however we like, we can take any B ∈ 𝔥₂(𝔼) with det(B) = 1 and tr(B) > 0 and move it close to the identity (in some sense to be worked out). The result will be some element C ∈ 𝔥₂(𝔼) with det(C) = 1 and tr(C) > 0 that's close to the identity. Then, we show that any such element C must be the identity or one of the 12 nearest neighbors you found - you've done that for some definition of "close to the identity". And that means B must be obtainable starting from A = I by repeatedly applying the 12 transformations above.
I something think I see a slicker argument, but so far those keep falling through.
I think you mean for these traces to be positive, rather than equal to 1.
And I guess the simplest sense of how close to the identity you are is just the trace itself, which, if it’s greater than 2, you should always be able to bring closer to 2.
the changes in the trace of the original matrix are linear expressions in either a, c and d or g, c and d. It will only be impossible to reduce the trace if all 12 expressions are non-negative, for parameters where the determinant is 1.
It becomes much easier to see what’s happening on a plot.
We would need to find values of a and g such that the green ellipse (the determinant condition) has a point with integer coordinates inside the smaller of the two hexagons (one of which scales with a, the other with g).
The shape of the ellipse and hexagons are such that if the ellipse passed through any one hexagon vertex it would pass through all of them.
The ratio between the size of the ellipse and the size of the a-hexagon is:
[\frac{\sqrt{3} \sqrt{a g-1}}{a}]
Wlog, we assume a ≤ g, and look for integer parameters where the ratio is less than or equal to 1. The plot below shows that this only happens for a=g=1, the identity matrix.
So, unless we are starting with the identity matrix, we can always use one of those 12 elements of SL(2,𝔼) acting on any given 𝔥₂(𝔼) matrix with determinant 1 and positive trace, and end up with a smaller trace.
I should also show that we can’t end up with a non-positive trace. I’m sure that’s also true, but I haven’t proved it.
[Edited to add: Oh, of course we can’t end up with a non-positive trace, because SL(2,ℂ) is the double cover of the identity component of the Lorentz group, which preserves the direction of time.]
@gregeganSF - wow, this is a great argument, Greg! Of course it leaves a lot of computations under the hood but that's fine with me.
What's cool about it is that ultimately you're proving that every solution of the Diophantine equations
𝑎𝑔 − 𝑐² + 𝑐𝑑 − 𝑑² = 1
can be obtained from the "trivial" solution 𝑎 = 𝑔 = 1 , 𝑐 = 𝑑 = 0 either by acting by SL(2,𝔼) or by negating all four of 𝑎, 𝑐, 𝑑, 𝑔. And this argument has a similar flavor to Minkowski's "geometry of numbers". For example, he would show unique prime factorization for Eisenstein integers by showing that a version of Euclid's algorithm works in this ring - again a kind of recursive "norm-reducing" procedure, which in the end relies on the fact that every complex number has distance ≤1 to some Eisenstein integer.
@johncarlosbaez@gregeganSF Here is a different approach which uses a bit more about Coxeter groups. The key point is that explicit bounding arguments can be replaced by the standard fact that every Coxeter group acts transitively on the chambers of its Coxeter complex.
Let me start with R^4 with elements denoted (t, x, y, z) and equipped with the negative of the Minkowski form, that is, -t^2 + x^2 + y^2 + z^2. Inside here, I choose vectors
e_1 = (0, sqrt(3)/2, 1/2, 0)
e_2 = (0, -1, 0, 0)
e_3 = (1/2, 1/2, -sqrt(3)/2, 1/2)
e_4 = (0, 0, 0, -1)
By checking inner products, we see that these vectors and the aforementioned bilinear form determine a copy of the canonical representation of the rank-4 Coxeter group with diagram
O--(6)--O----O----O
where the (6) means that the edge is labeled '6'. My notation for the canonical representation is consistent with https://en.wikipedia.org/wiki/Coxeter_complex
By the general theory, the canonical representation acts transitively on the set of chambers, where the fundamental chamber is the tetrahedral cone cut out by the hyperplanes (through the origin) which are orthogonal to e_1, e_2, e_3, e_4. Direct computation (e.g. matrix inverse) shows us that the extremal rays of the fundamental chamber are given by the vectors
(3, 0, -sqrt(3), 0)
(4, 1, -sqrt(3), 0)
(1, 0, 0, 0)
(1, 0, 0, 1)
Here I have dropped constant factors, but one can (and perhaps should) normalize to ensure Minkowski norm 1. Now I check explicitly that these vectors match up with Greg's earlier explicit description of "a portion of the honeycomb":
The vector (3, 0, -sqrt(3), 0)/sqrt(6) is one of the blue vertices (k=1) of the hexagon centered on the black point.
The vector (4, 1, -sqrt(3), 0) after normalization becomes the midpoint of the two blue vertices indexed by k=0 and k=1.
The vector (1, 0, 0, 0) is the identity matrix, i.e. the black point.
The vector (1, 0, 0, 1) is the null vector for one of the two horospheres which contains the hexagon centered on the black point.
According to https://en.wikipedia.org/wiki/Hexagonal_tiling_honeycomb , the desired honeycomb is constructed from the aforementioned Coxeter group by applying the Wythoff construction with only the first vertex circled. This confirms that the hexagon centers correspond to the vector (1, 0, 0, 0) and its images under the Coxeter group action.
It remains to show that the hexagon centers coincide with the elements of the Eisenstein lattice with Minkowski norm 1.
To show that the hexagon centers are contained in the Eisenstein lattice, it suffices to show that the Eisenstein lattice is invariant under the Coxeter group action. This follows by checking the action of each of the simple reflections:
v mapsto v - 2 <v, e_i> e_i.
In more detail, for e_1, the inner product <v, e_1> is a half-integer if v is in the Eisenstein lattice, and the claim follows because e_1 is in the Eisenstein lattice. The same statement applies for e_2 and e_4, and 'half-integer' can even replaced by 'integer.' For e_3, the inner product <v, e_3> is an integer multiple of sqrt(3)/2, and the claim follows because e_3 * sqrt(3) is in the Eisenstein lattice.
To show that the elements of the Eisenstein lattice with Minkowski norm 1 are hexagon centers, it suffices to show this statement within the fundamental chamber. Let n := (1, 0, 0, 1) be the null vector from before. Observe the following:
If v lies in the forward light cone, i.e. norm(v) > 0 [EDIT: and t > 0], then <v, n> > 0.
If v lies in the Eisenstein lattice, then <v, n> is an integer.
If v lies in the fundamental chamber and satisfies norm(v) = 1, then <v, n> < 2. (Sketch: Restricting to norm(v) = 1 gives hyperbolic geometry, and the interior-of-horosphere <v, n> < 2 is convex, so it suffices to check the special cases when v is a vertex of the fundamental chamber.)
These observations imply that, if v is an element of the Eisenstein lattice with norm(v) = 1 lying in the fundamental chamber, then <v, n> = 1.
Next, write v = (t, x, y, z). Since <v, n> = 1 rewrites as t - z = 1, the relation norm(v) = 1 rewrites as x^2 + y^2 = 2z. This tells us that the elements of the Eisenstein lattice with norm(v) = 1 and <v, n> = 1 are in bijection with the Eisenstein integers via
x + iy corresponds to ((x^2 + y^2)/2 + 1, x, y, (x^2 + y^2)/2).
It is easy to check that the simple reflections e_2, e_3, e_4 [EDIT: this should say e_1, e_2, e_3] preserve the aforementioned set of vectors v, and this bijection intertwines those simple reflections with the usual reflection symmetries of the Eisenstein integers (viewed as the vertices of the equilateral triangle lattice). Therefore, all of the aforementioned vectors v can be brought to (1, 0, 0, 0) via a Coxeter group action, as desired.
@ai - thanks, this proof looks really interesting! I'm thinking it over. If I want to include your proof in the article I'm writing for the n-Category Cafe, will that be okay with you? If so, should I cite you as "alchemist" and give a link to your post, or do you have some other name you'd like to use?
By the way, where you write
"If v lies in the forward light cone, i.e. norm(v) > 0"
I think you meant to say norm(v) > 0 and also t > 0. It doesn't really affect your argument.
@johncarlosbaez Hi John! Please feel free to use my post however you want, with any attribution or no attribution at all, anything is okay with me. Here I usually go by "mist" (I change my display name from time to time, but it's always a pun on 'mist')
Yes, I did mean to include t > 0, thanks for the correction. I've edited my post accordingly
I also found another error: in the last paragraph, the action on the horosphere slice is given by simple reflections e_1, e_2, e_3, NOT by the simple reflections e_2, e_3, e_4. I used a different numbering on my scratch paper and copied it over incorrectly. I've edited to fix this as well
@gregeganSF wrote: "I think you mean for these traces to be positive, rather than equal to 1."
Yes. I fixed that, though maybe I shouldn't go back and fix things in these posts that aren't "for the world at large".
"And I guess the simplest sense of how close to the identity you are is just the trace itself, which, if it’s greater than 2, you should always be able to bring closer to 2."
Yes! We could use the invariant distance function on hyperbolic space, but the distance from A to 1 is just some simple monotone function of tr(A).
It looks like you are way ahead of me now. I'll try to catch up as I drink coffee and wake up.
By the way, I hope this stuff is fun: I can certainly get a column out by May 1st regardless of whether this gets solved, so please don't feel you need to work on this unless it's fun.
@gregeganSF - wow, this is great! This is exactly what I was yearning for, but I was having trouble getting there and it would have taken me far longer.
This paper:
• C. W. L. Garner, Coordinates for vertices of regular honeycombs in hyperbolic space, Proc. Roy. Soc. London A, 293 (1966), 94–107. https://www.jstor.org/stable/2415373
gives a Diophantine equation whose solutions are all the vertices of the honeycomb we're talking about - that is, the hexagon vertices. But this description doesn't make it obvious that the vertices come in sheets lying on horospheres!
He's also describing the hyperboloid using different coordinates, which somehow hide the Eisenstein integers.
@johncarlosbaez I think his “absolute quadric” eqn (2) might mean he’s rescaled things to hide a square root of 3; he seems to be describing a light cone where time and space are measured in units that differ by that factor.
That is, the hexagon centre I gave as (C(j,n)) above, and the six associated vertices, are all orthogonal to (P(j,n)).
The dot products between the (P(j,n)) for neighbouring hexagons are always ½. So that sort of hints as to how you can use these vectors as kaleidoscope mirrors to extend the construction into the whole honeycomb.
@gregeganSF - Right. The factor of 3 is a giveaway, but it's sort of surprising to me that Garner can hide all the Eisenstein integers by stretching out the time axis by a factor of √3, using integer coordinates for spacetime points, and requiring that these coordinates sum to 0 mod 4. It's an elegant alternative to your approach but it sheds far less light on the structure of the honeycomb than yours!
"Absolute geometry" is a name for the geometry where you can't tell whether you're doing Euclidean, elliptic (I might say spherical but opposite points are identified) or hyperbolic geometry. A better name is "neutral geometry". I'm not sure how the "absolute quadric" is connected to absolute geometry, but it's the lightcone. Searching under "absolute quadric" gives me many copies of a paper on camera design.
@johncarlosbaez@gregeganSF@roice3
Don't know if this helps, but I once asked Vladimir Bulatov "Wat is the horosphere?". He said: "Just a fancy name for the plane.".
@GerardWestendorp - well, a horosphere is a flat Euclidean plane in hyperbolic 3-space. It looks like this in the Poincare ball model of hyperbolic 3-space. (Here it's been decorated with a tiling of the plane by hexagons, but that's irrelevant to the concept of horosphere.)
@johncarlosbaez@gregeganSF One way to see it is that in hyperbolic space, a regular hexagon couldn't have corners with 120-degree angles--because it's a hyperbolic space, the angles have to be smaller than that. So there's no way they could fit into a hexagonal tiling in the horosphere, which is itself planar. If they poke out a bit at the corners, then the tiling can fit into hyperbolic space.
@mattmcirvin@gregeganSF - Thanks, that's a good way to think about it. In theory one should be able actually see the hexagon edges bending up off the horosphere the vertices lie on. I can at least imagine it here:
@johncarlosbaez@gregeganSF Because we're accustomed to living in a very nearly Euclidean spatial geometry, it's hard to imagine a space in which a surface has to extrinsically bulge a bit in order to be flat.
@gregeganSF - great, thanks! I have that paper but didn't extract that clue. My guess is that the matrices you previously listed are hexagon centers. If so, we get some vertices of the hexagonal tiling lattice as 'midpoints' halfway between the identity matrix and the 12 matrices you listed. If so, these would be midpoints in the sense of hyperbolic geometry, so they'd take some work to figure out.
Alas, I don’t think that works out. The dot products between the matrices I found was –3/2, and arccosh(3/2) is not the right distance between the centres of Euclidean hexagons whose edge length is arccosh(5/4).
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