
In its full 8dimensional glory, this shape has:
and
Remember, a 3simplex is a tetrahedron:
while a 3orthoplex is an octahedron:
Anyone can look up these crazy numbers on Wikipedia. But how can we see that they're right?
That's our challenge for today. It's easiest if we define the \(\mathrm{E}_8\) root polytope starting from this picture:
and use the theory of simplylaced Dynkin diagrams, which are Dynkin diagrams without any multiple edges between dots. (Most of what I'm about to say generalizes to other Dynkin diagrams, but I'll save a bit of time by focusing on this case.)
Suppose we've got such a diagram with \(n\) dots. At an elementary level, this diagram tells us to take a bunch of unit vectors in \(n\)dimensional Euclidean space, and make sure they lie at a \(90^\circ\) angle if there's no edge connecting them, and a \(120^\circ\) angle if there is an edge connecting them.
We get a lattice called the root lattice by taking all integer linear combinations of these vectors. The nonzero vectors closest to the origin are called roots. These roots are the vertices of an \(n\)dimensional convex polytope called the root polytope.
The roots include the vectors we started with, one for each dot in our Dynkin diagram: these are called the simple roots. But there are also a lot more.
Starting with the \(\mathrm{E}_8\) Dynkin diagram, we can in principle figure out everything we want to know about the \(\mathrm{E}_8\) root polytope, including how many faces of each dimension it has, and the size of its symmetry group. But to do such things elegantly, it helps to use some tricks.
The main trick is to use the theory of Coxeter groups. Any simplylaced Dynkin diagram gives such a group, and it's very easy to describe. It has one generator \(s_i\) obeying
$$s_i^2 = 1$$for each dot in the Dynkin diagram, one relation
$$ s_i s_j s_i = s_i s_j s_i $$for each pair of dots connected by an edge, and one relation
$$ s_i s_j = s_j s_i $$for each pair of dots not connected by an edge. Each generator \(s_i\) corresponds to a reflection: the reflection that flips the \(i\)th simple root to its negative.
For example, starting with the Dynkin diagram called \(\mathrm{A}_3\):
we get a group with 3 generators, which happens to be the symmetry group of a regular tetrahedron:
So, you should think of the three dots in the Dynkin diagram as standing for 'vertex', 'edge' and 'face'. Why? Because the corresponding generators of the Coxeter group are reflections that:
• switch two neighboring vertices,
• switch two neighboring edges, and
• switch two neigboring faces,
respectively!
The \(\mathrm{A}_n\) diagram gives the symmetry group of a regular \(n\)simplex in just the same way. A similar story applies to the symmetry group of the regular \(n\)dimensional orthoplex, but the Dynkin diagram for this is not simply laced, so I won't get into the details. The cases we really need now are the \(\mathrm{E}_8\) Dynkin diagram and the subdiagrams of this.
For example, if we pull off the end dot of \(\mathrm{E}_8\) we get a subdiagram called \(\mathrm{D}_7\), which looks like this:
or, after prettying it up, this:
Reading from left to right, can think of the dots here as standing for 'vertex', 'edge', '2d face', '3d face' and so on... but when we get to the right end of the diagram, obviously something funny must happen!
There are two dots that mean 'topdimensional face'. The reason is that this Coxeter group is the symmetry group of a 7orthoplex that has its 6dimensional faces colored alternately white and black. We only consider symmetries that carry faces to faces of the same color. Two 6d faces intersect in at most one 5d face, so there's no dot for '5d face'. We can summarize the story like this:
All the \(\mathrm{D}_n\) Dynkin diagrams look similar, with some number of dots in a row followed by two at the end... and they all work the same way. Their Coxeter groups are the symmetry groups of orthoplexes with their topdimensional faces colored alternately white and black.
In the case of \(\mathrm{E}_8\), something even stranger happens! The topdimensional faces are of two kinds: 7simplex and 7orthoplex:
Other subtleties arise for the subdiagrams \(\mathrm{E}_7\) and \(\mathrm{E}_6\). So, you're probably thinking that this subject is an elaborate quagmire. But it's not.
To deal with all Coxeter groups in a systematic way, it's better to think of them as symmetry groups of certain simplicial complexes called 'Coxeter complexes'. Roughly speaking, a simplicial complex is a gadget made of 0simplexes, 1simplexes, 2simplexes, 3simplexes, and so on — all stuck together in a nice way.
If you have a Coxeter diagram with \(n\) dots, the highest dimension of the simplexes in its Coxeter complex is \(n1\). There is one of these topdimensional simplexes for each element of the Coxeter group. For example, I've already said this Dynkin diagram:
gives the Coxeter group consisting of symmetries of a regular tetrahedron — by which I mean all reflections and rotations. This group has 4! = 24 elements, so the Coxeter complex is built from 24 triangles. And in fact, we get it by barycentrically subdividing the surface of a tetrahedron! We can draw it on a sphere, like this:
You can see there are 24 triangles. And, you can see that if you fix a given triangle, there's a unique symmetry of the Coxeter group mapping it to any other triangle.
In general, the Coxeter group of a Dynkin diagram with \(n\) dots always acts as linear transformations of \(\mathbb{R}^n\). Each root gives a reflection that flips that root to its negative. So, this group also acts on the \((n1)\)sphere. If we take this sphere and chop it up along the hyperplanes corresponding to all the reflections in the Coxeter group, we get the Coxeter complex. See if you can visualize this in the picture above.
Even better, if you pick any topdimensional simplex in the Coxeter complex, there always exactly one element of the Coxeter group that maps it to any other topdimensional simplex. So the Coxeter complex is the best possible thing made out of simplexes on which the Coxeter group acts as symmetries!
The size of the Coxeter group corresponding to a Dynkin diagram is always the product of some integers, one for each dot in the Dynkin diagram. These integers have lots of important properties, and calculating them is really the key to all the problems we've set out to solve. To start with, let me just show you these numbers and what we can do with them.
Here they are. For any Dynkin diagram \(D\), let \(W(D)\) be its Coxeter group, also known as its Weyl group. Here are the sizes of these groups for all the simplylaced Dynkin diagrams:
$$ \begin{array}{ccll} W(\mathrm{A}_n) &=& 2 \cdot 3 \cdot 4 \cdot \cdots \cdot (n+1) &=& (n+1)! \\ W(\mathrm{D}_n) &=& 2 \cdot 4 \cdot 6 \cdot \cdots \cdot 2(n1) \cdot n &=& (2n)?! \\ W(\mathrm{E}_6) &=& 2 \cdot 5 \cdot 6 \cdot 8 \cdot 9 \cdot 12 &=& 51,840 \\ \\ W(\mathrm{E}_7) &=& 2 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdot 14 \cdot 18 &=& 2,903,040 \\ W(\mathrm{E}_8) &=& 2 \cdot 8 \cdot 12 \cdot 14 \cdot 18 \cdot 20 \cdot 24 \cdot 30 &=& 696,729,600 \end{array} $$It's easy to see that \(W(A_n)\), the symmetry group of the \(n\)dimensional simplex, has size equal to
$$ W(\mathrm{A}_n) = (n+1)! $$since the \(n\)simplex has symmetries permuting its vertices any way you want. While it's not a simplylaced example, it's also easy to see that \(W(B_n)\), the symmetry group of the \(n\)dimensional orthoplex, has size equal to
$$ W(\mathrm{B}_n) = 2^n n! $$This number is equal to the double factorial
$$ (2n)!! = 2 \cdot 4 \cdot \cdots \cdot 2n $$Finally, it's also easy to see that \(W(D_n)\), the symmetry group of the orthoplex with its topdimensional faces colored alternately white and black, is half as big. So, its size is the half double factorial of \(2n\):
$$ (2n)?! = 2 \cdot 4 \cdots \cdot 2(n1) \cdot n $$So, the only problem is computing the size of \(W(\mathrm{E}_6)\), \(W(\mathrm{E}_7)\), and \(W(\mathrm{E}_8)\). I'll talk about this later.
You might think it's somewhat arbitrary how we wrote the size of these Coxeter groups as products of numbers—but it's not! I'll talk about this more at the end of the article. But here's a little taste of the fun:
Puzzle 1. If you double each of these numbers and subtract one, then add up the results, you get the dimension of the Lie group corresponding to this Dynkin diagram. Why?
Let's check that it works for \(\mathrm{E}_8\):
$$ 2 \cdot (2 + 8 + 12 + 14 +18 + 20 + 24 + 30)  8 = 2 \cdot 128  8 = 248 $$Yes, this is the dimension of the Lie group \(\mathrm{E}_8\)!
Assuming we know the size of the Coxeter groups for all the simplylaced Dynkin diagrams, let's see if we can calculate how many faces of each kind the \(\mathrm{E}_8\) root polytope has.
I explained the method in week187. Each Dynkin diagram \(D\) describes a kind of 'incidence geometry' with different kinds of figures—points, edges, triangles, tetrahedra and so on—one for each dot in the Dynkin diagram. These figures can be 'incident' to each other—e.g., a triangle can lie on a tetrahedron—and there's one basic incidence relation for each edge in the Dynkin diagram.
The Coxeter group \(W(D)\) acts transitively on the figures of any given kind. We can also work out the subgroup of \(W(D)\) that preserves a figure of a given kind. To do this, we just remove the corresponding dot from the Dynkin diagram, and get a subdiagram \(E \subset D\). Then \(W(E) \subseteq W(D)\) is the subgroup we want!
So, the set of the figures of a given kind is \(W(D)/W(E)\). Thus, the number of figures of this kind is the ratio
$$ \frac{W(D)}{W(E)} $$To see if you understand this, let's do an easy example! How many edges does a tetrahedron have? Here our Dynkin diagram \(D\) is \(\mathrm{A}_3\):
It describes a geometry, namely a tetrahedron, with figures of three kinds:
Suppose we want to count the number of 1simplexes, or edges. We take the '1simplex' dot:
and we remove it, obtaining this subdiagram:
Hmm, this isn't connected! But that's okay. This diagram is the disjoint union of two copies of the onedot Dynkin diagram, \(\mathrm{A}_1\). If you follow the rules, you'll see its Coxeter group is the product
$$ W(\mathrm{A}_1) \times W(\mathrm{A}_1) $$So, the number of edges in a tetrahedron is
$$ \frac{W(\mathrm{A}_3)}{W(\mathrm{A}_1) \times W(\mathrm{A}_1)} = \frac{4!}{2! \times 2!} = 6 $$which is right!
If you've never done this sort of stuff, I encourage you to play around with more examples. It's lots of fun! But having spent years warming up with exercises like that, I'm now going to climb Mount Everest and count the various figures in the \(\mathrm{E}_8\) root polytope.
It starts out easy, but near the end, as you might imagine from this picture, it gets tricky—and I'll slip and fall. Hanging from a icy cliff, I'll ask for your help!
If we remove the 'vertex' dot from the \(\mathrm{E}_8\) Dynkin diagram:
we're left with \(\mathrm{E}_7\)
So, the number of vertices of the \(\mathrm{E}_8\) root polytope is
$$ \begin{array}{ccl} \frac{W(\mathrm{E}_8)}{W(\mathrm{E}_7)} &=& \frac{2 \cdot 8 \cdot 12 \cdot 14 \cdot 18 \cdot 20 \cdot 24 \cdot 30}{2 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdot 14 \cdot 18} \\ \\ &=& \frac{ 20 \cdot 24 \cdot 30}{6 \cdot 10} \\ \\ &=& 240 \end{array} $$as we knew!
Just for fun, in the next to last step I cancelled all the numbers I could from the big fraction in the top line. This is worth doing because a \(q\)deformed version of the same calculation lets us count the number of points in a certain space called a 'Grassmannian' for the version of the group \(\mathrm{E}_8\) defined over the finite field with \(q\) elements, where \(q\) is any prime power. I explained how this works in week186 and week187.
In the \(q\)deformed count we replace integers with \(q\)integers, defined as follows:
\( [n] = 1 + q + q^2 + \cdots + q^{n1} \)So, for this particular Grassmannian the number of points is
$$ \frac{[2] \cdot [8] \cdot [12] \cdot [14] \cdot [18] \cdot [20] \cdot [24] \cdot [30]}{[2] \cdot [6] \cdot [8] \cdot [10] \cdot [12] \cdot [14] \cdot [18]} $$or, doing all the cancellations we can,
$$ \frac{ [20] \cdot [24] \cdot [30]}{[6] \cdot [10]} $$This, believe it or not, is a polynomial in \(q\). But we can't say it's equal to \([240]\), because in general \([n m] \ne [n][m]\). Only when we set \(q = 1\) does this equation hold, and then we get 240 points. So, our problem now, involving Coxeter groups, concerns the special case of the legendary but so far still mythical 'field with one element'.
But if we wanted, we could look at the group \(\mathrm{E}_8\) defined over the field with 2 elements. It acts in a transitive way on a set with
$$ \frac{ [20] \cdot [24] \cdot [30]}{[6] \cdot [10]} $$points. How many points is that? In this case
$$ [n] = 1 + 2 + 4 + \cdots + 2^{n1} = 2^n  1 $$so the number of points is
$$ \frac{(2^{20}  1)(2^{24}  1)(2^{30}  1)}{(2^6  1)(2^{10}  1)} = 293,091,386,578,365,375 $$If we remove the 'edge' dot from the \(\mathrm{E}_8\) Dynkin diagram:
we're left with \( \mathrm{A}_1 \times \mathrm{E}_6\):
So, the number of edges of the \(\mathrm{E}_8\) root polytope is
$$ \begin{array}{ccc} \frac{W(\mathrm{E}_8)}{W(\mathrm{A}_1) \times W(\mathrm{E}_6)} &=& \frac{2 \cdot 8 \cdot 12 \cdot 14 \cdot 18 \cdot 20 \cdot 24 \cdot 30}{2 \times (2 \cdot 5 \cdot 6 \cdot 8 \cdot 9 \cdot 12)} \\ \\ &=& \frac{14 \cdot 18 \cdot 20 \cdot 24 \cdot 30}{2 \cdot 5 \cdot 6 \cdot 9} \\ \\ &=& 6,720 \end{array} $$A 2simplex is just a triangle. If we remove the '2simplex' dot:
we're left with \(\mathrm{A}_2 \times \mathrm{D}_5\):
so the number of 2d faces is
$$ \begin{array}{ccc} \frac{W(\mathrm{E}_8)}{W(\mathrm{A}_2) \times W(\mathrm{D}_5)} &=& \frac{2 \cdot 8 \cdot 12 \cdot 14 \cdot 18 \cdot 20 \cdot 24 \cdot 30}{(2 \cdot 3) \times (2 \cdot 4 \cdot 6 \cdot 8 \cdot 5)} \\ \\ &=& \frac{12 \cdot 14 \cdot 18 \cdot 20 \cdot 24 \cdot 30}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 8}\\ \\ &=& 60,480 \end{array} $$This is a funny picture of a 3simplex, or tetrahedron. If we remove the '3simplex' dot:
we're left with \(\mathrm{A}_3 \times \mathrm{A}_4\):
so the number of 3d faces is
$$ \begin{array}{ccc} \frac{W(\mathrm{E}_8)}{W(\mathrm{A}_3) \times W(\mathrm{A}_4)} &=& \frac{2 \cdot 8 \cdot 12 \cdot 14 \cdot 18 \cdot 20 \cdot 24 \cdot 30}{(2 \cdot 3 \cdot 4) \times (2 \cdot 3 \cdot 4 \cdot 5)} \\ \\ &=& \frac{8 \cdot 12 \cdot 14 \cdot 18 \cdot 20 \cdot 24 \cdot 30}{2 \cdot 3 \cdot 3 \cdot 4 \cdot 4 \cdot 5}\\ \\ &=& 241,920 \end{array} $$This is a picture of a 4simplex. If we remove the '4simplex' dot from the \(\mathrm{E}_8\) Dynkin diagram:
we're left with \(\mathrm{A}_4 \times \mathrm{A}_1 \times \mathrm{A}_2\):
so the number of 4d faces is
$$ \begin{array}{ccc} \frac{W(\mathrm{E}_8)}{W(\mathrm{A}_4) \times W(\mathrm{A}_1) \times W(\mathrm{A}_4)} &=& \frac{2 \cdot 8 \cdot 12 \cdot 14 \cdot 18 \cdot 20 \cdot 24 \cdot 30}{(2 \cdot 3 \cdot 4 \cdot 5) \times 2 \times (2 \cdot 3)} \\ \\ &=& \frac{8 \cdot 12 \cdot 14 \cdot 18 \cdot 20 \cdot 24 \cdot 30}{2 \cdot 2 \cdot 3 \cdot 3 \cdot 4 \cdot 5} \\ \\ &=& 483,840 \end{array} $$At this point we meet a mystery: there is no '5simplex' dot in the \(\mathrm{E}_8\) Dynkin diagram!
This is a bit like how there is no 5simplex dot in the \(\mathrm{D}_7\) Dynkin diagram:
The resolution there is that if we have 'complete flag' consisting of one figure of each kind shown in this picture, all incident, there is a unique 5simplex that's incident to all the other figures. Indeed, there's at most one 5simplex touching any pair consisting of a white and a black 6simplex. So, the 5simplex is 'redundant'.
The same sort of resolution must, I think, apply to \(\mathrm{E}_8\). Figuring this out in detail would make a good test of our understanding. But let's move on!
If we remove the '6simplex' dot:
we're left with \(\mathrm{A}_6 \times \mathrm{A}_1\):
so the number of 6d faces is apparently:
$$ \begin{array}{ccc} \frac{W(\mathrm{E}_8)}{W(\mathrm{A}_6) \times W(\mathrm{A}_1) } &=& \frac{2 \cdot 8 \cdot 12 \cdot 14 \cdot 18 \cdot 20 \cdot 24 \cdot 30}{(2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7) \times 2} \\ \\ &=& \frac{8 \cdot 12 \cdot 14 \cdot 18 \cdot 20 \cdot 24 \cdot 30}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7} \\ \\ &=& 69,120 \end{array} $$But this does not agree with Wikipedia's answer! It says our polytope has 207,360 6dimensional faces, which are all regular 6simplexes. So, we have another mystery on our hands here, which we will return to.
If we remove the '7simplex' dot:
we are left with \(\mathrm{A}_7\):
so the number of 7dimensional simplex faces is:
$$ \begin{array}{ccc} \frac{W(\mathrm{E}_8)}{W(\mathrm{A}_7) } &=& \frac{2 \cdot 8 \cdot 12 \cdot 14 \cdot 18 \cdot 20 \cdot 24 \cdot 30}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8} \\ \\ &=& \frac{12 \cdot 14 \cdot 18 \cdot 20 \cdot 24 \cdot 30}{3 \cdot 4 \cdot 5 \cdot 6 \cdot 7} \\ \\ &=& 17,280 \end{array} $$If we remove the '7orthoplex' dot:
we are left with \(\mathrm{D}_7\):
so the number of 7dimensional orthoplex faces is:
$$ \begin{array}{ccc} \frac{W(\mathrm{E}_8)}{W(\mathrm{D}_7) } &=& \frac{2 \cdot 8 \cdot 12 \cdot 14 \cdot 18 \cdot 20 \cdot 24 \cdot 30}{2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdot 7} \\ \\ &=& \frac{12 \cdot 14 \cdot 18 \cdot 20 \cdot 24 \cdot 30}{4 \cdot 6 \cdot 7 \cdot 10 \cdot 12} \\ \\ &=& 2,160 \end{array} $$ But we've got some further mysteries to solve.First, our calculations seemed to show the number of 6simplex faces of root polytope of \(\mathrm{E}_8\) is 69,120, while Wikipedia says the answer is 207,360. Luckily we can compute the answer another way, which sheds a little light on this problem.
Our polytope has 17,280 7simplex faces and 2,160 7orthoplex faces. Each 7simplex has 8 faces, all of which are 6simplexes. Each 7orthoplex has 2^{7} faces, all of which are 6simplexes. So, we could naively count the total number of 6simplexes and get
$$ 17,280 \cdot 8 + 2,160 \cdot 2^7 = 138,240 + 276,480 = 414,720 $$But this is doublecounting, since each 6simplex touches two 7simplexes. (The topdimensional faces of a convex polytope always meet in pairs along faces of the next highest dimension.) So the real count is half this:
$$ \frac{138,240}{2} + \frac{276,480}{2} = 69,120 + 138,240 = 207,360 $$This matches the answer on Wikipedia. But it also sheds some light on our wrong answer. Our wrong answer, 69,120, is exactly the total number of faces of all the 7simplexes. It's also half the total number of faces of the 7orthoplexes. It's also exactly 1/3 of the correct answer!
All this suggests that the 6simplex faces of our polytope come in two kinds, which cannot be interchanged by symmetries of the polytope. 1/3 of them are the ones we counted, the kind mentioned in this diagram:
2/3 of them are 6simplexes of some other kind, which do not appear in the above diagram.
If this seems weird, remember that this diagram also does not include the 5simplex faces of our polytope! Also remember that the \(\mathrm{D}_7\) diagram, corresponding to the 7orthoplex, gives another example where some simplexes come in two kinds, while others do not appear at all, even though they exist as faces of the 7orthoplex:
Even if this is true, it's clearly not enough for a full understanding. If our polytope has two kinds of 6simplex as faces, how do they differ? And where is the number 1/3 coming from?
Here's my guess. The 7orthoplex faces of our polytope have 6simplex faces that can be colored alternately white and black, and all the symmetries of our polytope preserve this twocoloring. As evidence, note that the symmetries preserving a 7orthoplex form the group \(\mathrm{D}_7\), which preserves such a twocoloring. But I'm guessing we can do this twocoloring in a consistent way, so every 6simplex in our polytope is either white or black.
And I'm guessing we can do this so that each 7orthoplex touches another 7orthoplex on each of its black faces, and a 7simplex on each of its white faces! On the other hand, every 7simplex is completely surrounded by 7orthoplexes.
If we can do this:
These are the 6simplexes we counted.
where we get a factor of \(1/2\) because only half the 6simplex faces of each 7orthoplex are white. This is a consistency check!
as it says on Wikipedia.
All this fits together nicely... so while I haven't proved the existence of this 'consistent twocoloring' and this arrangement of 6simplexes and 6orthoplexes, I believe it.
Puzzle 2. Prove or disprove my guesses here!
Why does our polytope have \(483,840\) 5d faces, which are all regular 5simplexes? I haven't figured out how to count them. With work I should be able to do it using my guesses in the previous section, but I made these guess just a few minutes ago.
Notably, \(483,840\) is also the number of 4d faces.
Puzzle 3. Show how to count the 5d faces of the \(\mathrm{E}_8\) root polytope.
All my calculations rely on general results that I understand, together with some numbers I don't understand:
$$ \begin{array}{ccll} W(\mathrm{E}_6) &=& 2 \cdot 5 \cdot 6 \cdot 8 \cdot 9 \cdot 12 &=& 51,840 \\ \\ W(\mathrm{E}_7) &=& 2 \cdot 6 \cdot 8 \cdot 10 \cdot 12 \cdot 14 \cdot 18 &=& 2,903,040 \\ \\ W(\mathrm{E}_8) &=& 2 \cdot 8 \cdot 12 \cdot 14 \cdot 18 \cdot 20 \cdot 24 \cdot 30 &=& 696,729,600 \end{array} $$In fact for my calculations I only need to know the sizes of these three Coxeter groups, but it's irresistible to try understanding the 'magic numbers' that you can multiply to get their sizes.
Puzzle 4. What's the best way to calculate these magic numbers?
Back in week186 and week187, I gave many characterizations of these numbers. Unfortunately, they don't seem to make it easy to calculate these numbers for \(\mathrm{E}_6, \mathrm{E}_7\) and \(\mathrm{E}_8\).
I bet somebody knows an easy way. But to help you with this puzzle, I'll tell you eight hard ways!
Start with any Dynkin diagram \(D\), not necessarily simplylaced. Let \(G(\mathbb{F})\) be the corresponding simple algebraic group over a field \(\mathbb{F}\). In particular,
$$ G = G(\mathbb{C}) $$is a complex Lie group. Suppose the diagram \(D\) has \(n\) dots. Then the magic numbers \(d_1, \dots, d_n\) are natural numbers such that
$$ W(D) = d_1 \cdot \cdots \cdot d_n $$Moreover:
If you know enough stuff you may have fun proving all these descriptions of the magic numbers are equivalent.
I have not managed to use any of these descriptions of the magic numbers to solve this problem:
Puzzle 5. Why is these sequence of magic numbers for any Dynkin diagram 'symmetrical'?
To explain what I mean here, look at the pattern of spacings between magic numbers for \(\mathrm{E}_6\):
$$ \begin{array}{cccccccccccc} 2 & & & 5 & 6 & & 8 & 9 & & & 12 \\ \bullet & \circ & \circ & \bullet & \bullet & \circ & \bullet & \bullet & \circ & \circ &\bullet \end{array} $$It doesn't change when you reflect it! The same is true for \(\mathrm{E}_7, \mathrm{E}_8\) and also the nonexceptional cases if you write the magic numbers in increasing order. Why? It's reminiscent of Poincaré duality, but I don't see how it is that.
Puzzle 1. If you double each of the magic numbers and subtract one, then add up the results, you get the dimension of the Lie group corresponding to this Dynkin diagram. Why?
Answer. A selfcontained proof can be found in Humphrey's book Reflection Groups and Coxeter Groups; see Theorem 3.9. Personally I prefer using the fact that the real cohomology of the Lie group \(G\) corresponding to the Dynkin diagram is an exterior algebra on generators \(x_i\) of degrees \(2d_i  1\). Thus, the highestdegree element in the cohomology of \(G\) is the product \(x_1 \cdots x_n\), which has degree \( \sum_{i=1}^n (2d_i  1) \). But the highest degree possible in the cohomology of \(G\) is just the dimension of \(G\).
Puzzle 2. Show that the faces of the 7orthoplexes in the \(\mathrm{E}_8\) polytope can be 2colored in an alternating way so that their black faces touch other 7orthoplexes, while their white faces touch 7simplexes.
Answer. Layra Idrani answered:
You can prove the blackwhite thing by example, or at least I did.
Letting our vertices be
$$ (\pm 1,\pm 1,0,0,0,0,0,0) $$and permutations, and
$$ (\pm 1/2,\pm 1/2,\pm 1/2,\pm 1/2,\pm 1/2,\pm 1/2,\pm 1/2,\pm 1/2) $$with an even number of \(1/2\)s, we get that a pair of vertices are connected by an edge iff their inner product is 1.
We can look at the following orthoplex:
$$(1,1,0,0,0,0,0,0) $$ $$ (1,0,1,0,0,0,0,0) $$ $$ (1,0,0,1,0,0,0,0) $$ $$ (1,0,0,0,1,0,0,0) $$ $$ (1,0,0,0,0,1,0,0) $$ $$(1,0,0,0,0,0,1,0) $$ $$ (1,0,0,0,0,0,0,1) $$ $$ (1,1,0,0,0,0,0,0) $$ $$ (1,0,1,0,0,0,0,0) $$ $$ (1,0,0,1,0,0,0,0) $$ $$ (1,0,0,0,1,0,0,0) $$ $$ (1,0,0,0,0,1,0,0) $$ $$ (1,0,0,0,0,0,1,0) $$ $$ (1,0,0,0,0,0,0,1) $$Now we figure out which sets of seven can be extended to a simplex via adding another vertex.
Consider a vertex \((1,1,0,0,0,0,0,0)\). For another vertex to have an edge to it, i.e. to have inner product 1, its first entry must be positive or 0, and its second entry must be negative or 0, since otherwise there would be a negative term in the inner product; both entries can't be 0, and if both are nonzero then they must be 1/2 and 1/2 respectively.
Similarly, for any other vertex \(v\) in the orthoplex, any vertex \(w\) connected to \(v\) must either match the sign of \(v\) in precisely one of the two nonzero entries of \(v\), or \(w\) must match both and all of its coordinates must have magnitude 1/2. We take a set of seven vertices in our orthoplex such that except for the first entry, if \(v\) has a nonzero entry somewhere then the other 6 vertices have 0 there. These are all our 6dimensional simplices in our orthoplex, since any other set would contain a pair of vertices whose inner product is 0. Suppose that of these, \( \{v_i\}\) for \(i\) from 1 to \(k\) each have a 1 in position \(i\). The point that has coordinate 1/2 in position \(i\) and 1/2 in all other positions is of distance 2 from all the vertices given. Such a point is a vertex iff \(k\) is even. So that gives us 64 6simplices that extend to 7simplices. Moreover, these 6simplices alternate with the remaining 6simplices, since given a 6simplex with an even number of vertices that have a 1 in them, changing any of the vertices such that we still get a 6simplex in the orthoplex yields a 6simplex with an odd number of vertices with a 1 in them, and vice versa.Now we just need to show that the other 64 6simplices don't extend to 7simplices. Consider a set of seven vertices in the orthoplex such that an odd number have an entry of 1 and the seven vertices form a 6simplex.
Suppose we have a vertex \(w\) with a 0 entry. If it has 1 as its first entry, then it must have an entry of 1 or 1 somewhere else, and thus either matches or is orthogonal to a vertex in our 6simplex. If \(w\) has 0 as its first entry, then it must have nonzero entries in two other places, but as there are then five remaining entries, there is a vertex in our 6simplex that is 0 in those two places and hence is orthogonal to \(w\). If \(w\) has only entries of \(\pm 1/2\), then it must match each vertex in our simplex wherever that simplex is nonzero, but doing so yields a point with an odd number of entries of 1/2, which is not a vertex. So these 6simplices cannot be extended to 7simplices, and hence must be connected to orthoplexes (orthoplices?).
So this particular orthoplex has precisely half of its 6faces connected to simplices, and the other half to other orthoplexes, and such faces alternate. By symmetry, this applies to all the orthoplexes.
Puzzle 3. Show how to count the 5d faces of the \(\mathrm{E}_8\) root polytope.
Answer. Layra Idrani give an explanation of the answer in Wikipedia:
In terms of the 5simplices, the fastest way I can think to do it is to count the number of 5simplices per 7simplex (28) and then multiply by the number of 7simplices (17280). This gives the right answer (483,840). Now we just have to show that this is the correct calculation to do.
We know that we only have to worry about 5simplices that are contained in 7simplices, since for any 5simplex in an orthoplex, it can extend to a white 6simplex, and uniquely so.
To show that each 5simplex is in a unique 7simplex, we look at 4simplices. According to the diagram, each 4simplex A is contained in exactly two 7simplices, denoted B and C. Hence if a 5simplex D contains A, then it can only be contained in B or C, no other 7simplices, and indeed must be contained in at least one of B or C. Since all 5simplices are identical, if any 5simplex is in two 7simplices, each 5simplex must be in two 7simplices.
So we assume that D is in both B and C. Since any other D' that contains A must be in B or C, and D is in two simplices, D' must also be in both B and C. We know that D' exists since B has more than one vertex that isn't in A.
D and D' share 5 vertices and are both in the same 7simplex B, and so thus together define a unique 6simplex, and that 6simplex is contained in B. But by assumption, they're also both in C, and so that same 6simplex is also in C. Thus we get that B and C share a 6simplex, which is impossible.
Hence D is in exactly one of B and C, and similarly every 5simplex is in precisely one 7simplex.
The only part of this answer I didn't undertand is why every 4simplex lies in exactly two 7simplexes. However, I was able to confirm this using a slight extension of the technique already used in this article.
First I'll count the pairs consisting of a 4simplex and a 7simplex that contains it. Then I'll count the 4simplexes. Then I'll divide the first number by the second and get 2.
To count the pairs consisting of a 4simplex and a 7simplex that contains it, I mark those dots in the Dynkin diagram:
and then remove them:
leaving me with the Dynkin diagram for \(\mathrm{A}_4 \times \mathrm{A}_2\). So, the number of these pairs is $$ \begin{array}{ccl} \frac{W(\mathrm{E}_8)}{W(\mathrm{A}_4) \times W(\mathrm{A}_2) \times W(\mathrm{A}_4)} &=& \frac{2 \cdot 8 \cdot 12 \cdot 14 \cdot 18 \cdot 20 \cdot 24 \cdot 30}{(2 \cdot 3 \cdot 4 \cdot 5) \times (2 \times 3)}\end{array} $$ On the other hand, I already counted the number of 4simplexes. But just as a reminder: to do this, we remove just the 4simplex dot from the Dynkin diagram:
leaving us with \(\mathrm{A}_4 \times \mathrm{A}_1 \times \mathrm{A}_2\):
so the number of 4d faces is $$ \begin{array}{ccl} \frac{W(\mathrm{E}_8)}{W(\mathrm{A}_4) \times W(\mathrm{A}_1) \times W(\mathrm{A}_4)} &=& \frac{2 \cdot 8 \cdot 12 \cdot 14 \cdot 18 \cdot 20 \cdot 24 \cdot 30}{(2 \cdot 3 \cdot 4 \cdot 5) \times 2 \times (2 \cdot 3)} \end{array} $$ You can see this is half our previous answer. So, yes: there are two 7simplexes containing each 4simplex!
Puzzle 4. What's the best way to calculate the magic numbers for a given Dynkin diagram?
Answer. Two methods, both rather astounding, can be found in Humphrey's book Reflection Groups and Coxeter Groups. The first method relies on hisProposition 3.20. If \(h\) is the Coxeter number of the Dynkin diagram \(D\) and \(1 \le m \le h\) is relatively prime to \(h\), then \(m+1\) is a magic number for \(D\).
Recall that the Coxeter number is the number of roots divided by the number of dots in the Dynkin diagram. (For other more conceptual descriptions, click the link!)
To use this proposition, it may help to know that the number of roots is the dimension of the Lie algebra corresponding to \(D\) minus the number of dots in \(D\). For example, the dimension of the Lie algebra \(\mathrm{E}_8\) is 248, and its Dynkin diagram has 8 dots, so the number of roots is \(248  8 = 240\) (which we already knew), and its Coxeter number is \(240/8 = 30\). The numbers between 1 and 30 that are relatively prime to 30 are: $$ 1, 7, 11, 13, 17, 19, 23, 29 $$
So, adding one, we get some magic numbers for \(\mathrm{E}_8\):
$$ 2, 8, 12, 14, 18, 20, 24, 30 $$But there are eight of them, and there's always just one magic number for each dot in the Dynkin diagram, so these must be all the magic numbers!
This is black magic, and even the normally imperturbable Humphreys gives it an exclamation mark.
By the way, did you notice something funny about these numbers:
$$ 1, 7, 11, 13, 17, 19, 23, 29 ? $$Except for 1, they're all prime! 30 just happens to be the largest number for which every number besides 1 that's smaller than it and relatively prime to it is actually prime. I don't know if this is important here, but it's darn suspicious.
For \(\mathrm{E}_7\) we have to work harder. The dimension of the Lie algebra \(\mathrm{E}_7\) is 133, so the number of roots is \(133  7 = 126\), and the Coxeter number is \(126/7 = 18\). Here are the numbers smaller than 18 that are relatively prime to 18: $$ 1, 5, 7, 11, 13, 17 $$ Adding one, we get magic numbers for \(\mathrm{E}_7\): $$ 2, 6, 8, 12, 14, 18 $$ Unfortunately we only get six of the seven magic numbers this way. Luckily, Humphreys also proved that the sequence of magic numbers must be 'symmetrical' as described in Puzzle 5. The only possibility is that the seventh magic number is 10: $$ \begin{array}{ccccccccccccccccc} 2 &&&& 6 &&8 && \mathbf{10} && 12 &&14 &&&& 18 \\ \bullet &\circ &\circ &\circ & \bullet &\circ &\bullet &\circ &\bullet &\circ &\bullet &\circ &\bullet &\circ&\circ &\circ& \bullet \end{array} $$ By the way, did you notice anything funny about these numbers: $$ 1, 5, 7, 11, 13, 17 ?$$
Yes, again they're all prime. Like 30, 18 is a very round number, meaning that every nature number relatively prime to it and smaller than it is either 1 or prime.
For \(\mathrm{E}_6\) we must work even harder for our supper, but I'll leave that as a puzzle for you! To get you started, I'll remind you that the dimension of the \(\mathrm{E}_6\) is 78, so the number of roots is \(78  6 = 72\), so the Coxeter number is \(72/6 = 12\). Also: puzzle 1 may come in handy.
(By the way, the only very round numbers are $$ 1, 2, 3, 4, 6, 8, 12, 18, 24, 30 $$ and it is a mystery to me that three of them have come up in this discussion. But it is easy to see that there's no very round number bigger than \(30\). First, it would have to be divisible by \(2, 3\), and \(5\), or else some product of two of these primes would be relatively prime to this number and smaller than it, but not prime. So the first option is \(60\). \(60\) is not very round because \(7^2\) is relatively prime to it but not prime. In short, the problem is that there's a prime that does not divide \(60\) whose square is less than \(60\). So, any round number bigger than \(30\) actually needs to be divisible by \(7\), too, which means it must be bigger than \(2 \times 3 \times 5 \times 7 = 210\). But \(11^2 < 210\) and \(13^2 < 210\), so any round number bigger than \(30\) must actually be bigger than \(2 \times 3 \times 5 \times 7 \times 11 \times 13\). And so on: this is clearly a losing battle. At this stage, knowing there's always a prime between \(n\) and \(2n\) is enough to guarantee that we can keep up this argument by induction.)
The second method for computing magic numbers is more systematic.
This method uses the concept of the height of a positive root, meaning the sum of its coefficients when expanded as a linear combination of simple roots. To calculate the magic numbers, you calculate how many positive roots of each height there are, and use this to get a partition of the total number of positive roots. For example, \(\mathrm{B}_2\) has 2 positive roots of height one, 1 of height two and 1 of height three, so we get the partition (2,1,1).
Then you write this partition as a Young diagram:
Then you reflect this Young diagram along the diagonal and get another Young diagram:XX X X
Then you read this as a partition, namely (3,1). And finally, you add one to each of these numbers, and you get the magic numbers! For \(\mathrm{B}_2\) you get (4,2). But the procedure works in general for any Dynkin diagram.XXX X
This sounds insane, but it's true! It's Theorem 3.20 in Humphreys' book — the last theorem in the book.
Puzzle 5. Why is the sequence of magic numbers for any Dynkin diagram 'symmetrical'?
Answer. Allen Knutson suggested consulting Humphreys' book Reflection Groups and Coxeter Groups, and Ben Webster said:Allen is correct; this symmetry is Lemma 3.16 in Humphreys. The quick explanation is that \(\zeta^{d_i1}\) for \(\zeta\) a particular \(h\)th root of unity is the set of eigenvalues of the Coxeter element. Since this is a real matrix, its eigenvalues come in complex conjugate pairs. Here \(h\) is the Coxeter number, the order of the Coxeter element.You can also read comments on the nCategory Café, and make your own comments or ask questions there!
