Geometry « The Unapologetic Mathematician

Construction of E-Series Root Systems

Today we construct the last of our root systems, following our setup. These correspond to the Dynkin diagrams $E_6$ , $E_7$ , and $E_8$ . But there are transformations of Dynkin diagrams that send $E_6$ into $E_7$ , and $E_7$ on into $E_8$ . Thus all we really have to construct is $E_8$ , and then cut off the right simple roots in order to give $E_7$ , and then $E_6$ .

We start similarly to our construction of the $F_4$ root system; take the eight-dimensional space with the integer-coefficient lattice $I$ , and then build up the set of half-integer coefficient vectors

$\displaystyle I'=I+\frac{1}{2}(\epsilon_1+\epsilon_2+\epsilon_3+\epsilon_4+\epsilon_5+\epsilon_6+\epsilon_7+\epsilon_8)$

Starting from lattice $I\cup I'$ , we can write a generic lattice vector as

$\displaystyle\frac{c^0}{2}(\epsilon_1+\epsilon_2+\epsilon_3+\epsilon_4+\epsilon_5+\epsilon_6+\epsilon_7+\epsilon_8)+c^1\epsilon_1+c^2\epsilon_2+c^3\epsilon_3+c^4\epsilon_4+c^5\epsilon_5+c^6\epsilon_6+c^7\epsilon_7+c^8\epsilon_8$

and we let $J\subseteq I\cup I'$ be the collection of lattice vectors so that the sum of the coefficients $c^i$ is even. This is well-defined even though the coefficients aren’t unique, because the only redundancy is that we can take ${2}$ from $c^0$ and add ${1}$ to each of the other eight coefficients, which preserves the total parity of all the coefficients.

Now let $\Phi$ consist of those vectors $\alpha\in J$ with $\langle\alpha,\alpha\rangle=2$ . The explicit description is similar to that from the $F_4$ root system. From $I$ , we get the vectors $\pm(\epsilon_i\pm\epsilon_j)$ , but not the vectors $\pm\epsilon_i$ because these don’t make it into $J$ . From $I'$ we get some vectors of the form

$\displaystyle\frac{1}{2}(\pm\epsilon_1\pm\epsilon_2\pm\epsilon_3\pm\epsilon_4\pm\epsilon_5\pm\epsilon_6\pm\epsilon_7\pm\epsilon_8)$

Starting with the choice of all minus signs, this vector is not in $J$ because $c^0=-1$ and all the other coefficients are ${0}$ . To flip a sign, we add $\epsilon_i$ , which flips the total parity of the coefficients. Thus the vectors of this form that make it into $\Phi$ are exactly those with an odd number of minus signs.

We need to verify that $\frac{2\langle\beta,\alpha\rangle}{\langle\alpha,\alpha\rangle}=\langle\beta,\alpha\rangle\in\mathbb{Z}$ for all $\alpha$ and $\beta$ in $\Phi$ (technically we should have done this yesterday for $F_4$ , but here it is. If both $\alpha$ and $\beta$ come from $I$ , this is clear since all their coefficients are integers. If $\alpha=\pm\epsilon_i\pm\epsilon_j\in I$ and $\beta\in I'$ , then the inner product is the sum of the $i$ th and $j$ th coefficients of $\beta$ , but with possibly flipped signs. No matter how we choose $\alpha\in I$ and $\beta\in I'$ , the resulting inner product is either $-1$ , ${0}$ , or ${1}$ . Finally, if both $\alpha$ and $\beta$ are chosen from $I'$ , then each one is $c=-\frac{1}{2}(\epsilon_1+\epsilon_2+\epsilon_3+\epsilon_4+\epsilon_5+\epsilon_6+\epsilon_7+\epsilon_8)$ plus an odd number of the $\epsilon_i$ , which we write as $a$ and $b$ , respectively. Thus the inner product is

$\displaystyle\langle\alpha,\beta\rangle=\langle c+a,c+b\rangle=\langle c,c\rangle+\langle c,b\rangle+\langle a,c\rangle+\langle a,b\rangle$

The first term here is $2$ , and the last term is also an integer because the coefficients of $a$ and $b$ are all integers. The middle two terms are each a sum of an odd number of $\pm\frac{1}{2}$ , and so each of them is a half-integer. The whole inner product then is an integer, as we need.

What explicit base $\Delta$ should we pick? We start out as we’ve did for $F_4$ with $\epsilon_2-\epsilon_3$ , $\epsilon_3-\epsilon_4$ , and so on up to $\epsilon_7-\epsilon_8$ . These provide six of our eight vertices, and the last two of them are perfect for cutting off later to make the $E_7$ and $E_6$ root systems. We also throw in $\epsilon_2+\epsilon_3$ , like we did for the $D_n$ series. This provides us with the triple vertex in the $E_8$ Dynkin diagram.

We need one more vertex off to the left. It should be orthogonal to every one of the simple roots we’ve chosen so far except for $\epsilon_2+\epsilon_3$ , with which it should have the inner product $-1$ . It should also be a half-integer root, so that we can get access to the rest of them. For this purpose, we choose the root $\frac{1}{2}(\epsilon_1-\epsilon_2-\epsilon_3-\epsilon_4-\epsilon_5-\epsilon_6-\epsilon_7-\epsilon_8)$ . Establishing that the reflection with respect to this vector preserves the lattice $J$ — and thus the root system $\Phi$ — proceeds as in the $F_4$ case.

The Weyl group of $E_8$ is again the group of symmetries of a polytope. In this case, it turns out that the vectors in $\Phi$ are exactly the vertices of a regular eight-dimensional polytope inscribed in the sphere of radius ${2}$ , and the Weyl group of $E_8$ is exactly the group of symmetries of this polyhedron! Notice that this is actually something interesting; in the $A_2$ case the roots formed the vertices of a hexagon, but the Weyl group wasn’t the whole group of symmetries of the hexagon. This is related to the fact that the $A_2$ diagram possesses a symmetry that flips it end-over-end, and we will explore this behavior further.

The Weyl groups of $E_7$ and $E_6$ are also the symmetries of seven- and six-dimensional polytopes, respectively, but these aren’t quite so nicely apparent from their root systems.

As the most intricate (in a sense) of these root systems, $E_8$ has inspired quite a lot of study and effort to visualize its structure. I’ll leave you with an animation I found on Garrett Lisi’s notewiki, Deferential Geometry (with the help of Sarah Kavassalis).

March 10, 2010 Posted by John Armstrong | Geometry, Root Systems | Leave a Comment

Construction of the F4 Root System

Today we construct the $F_4$ root system starting from our setup.

As we might see, this root system lives in four-dimensional space, and so we start with this space and its integer-component lattice $I$ . However, we now take another copy of $I$ and push it off by the vector $\frac{1}{2}(\epsilon_1+\epsilon_2+\epsilon_3+\epsilon_4)$ . This set $I'$ consists of all vectors each of whose components is half an odd integer (a “half-integer” for short). Together with $I$ , we get a new lattice $J=I\cup I'$ consisting of vectors whose components are either all integers or all half-integers. Within this lattice $J$ , we let $\Phi$ consist of those vectors of squared-length $2$ or $1$ : $\langle\alpha,\alpha\rangle=2$ or $\langle\alpha,\alpha\rangle=1$ ; we want to describe these vectors explicitly.

When we constructed the $B_n$ and $C_n$ series, we saw that the vectors of squared-length $1$ and $2$ in $I$ are those of the form $\pm\epsilon_i$ (squared-length $1$ ) and of the form $\pm(\epsilon_i\pm\epsilon_j)$ for $i\neq j$ (squared-length $2$ ). But what about the vectors in $I'$ ? We definitely have $\left(\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2}\right)$ — with squared-length $1$ — but can we have any others? The next longest vector in $I'$ will have one component $\pm\frac{3}{2}$ and the rest $\pm\frac{1}{2}$ , but this has squared-length $3$ and won’t fit into $\Phi$ ! We thus have twenty-four long roots of squared-length $2$ and twenty-four short roots of squared-length $1$ .

Now, of course we need an explicit base $\Delta$ , and we can guess from the diagram $F_4$ that two must be long and two must be short. In fact, in a similar way to the $B_3$ root system, we start by picking $\epsilon_2-\epsilon_3$ and $\epsilon_3-\epsilon_4$ as two long roots, along with $\epsilon_4$ as one short root. Indeed, we can see a transformation of Dynkin diagrams sending $B_3$ into $F_4$ , and sending the specified base of $B_3$ to these three vectors.

But we need another short root which will both give a component in the direction of $\epsilon_1$ and will give us access to $I'$ . Further, it should be orthogonal to both $\epsilon_2-\epsilon_3$ and $\epsilon_3-\epsilon_4$ , and should have a Cartan integer of $-1$ with $\epsilon_4$ in either order. For this purpose, we pick $\frac{1}{2}(\epsilon_1-\epsilon_2-\epsilon_3-\epsilon_4)$ , which then gives us the last vertex of the $F_4$ Dynkin diagram.

Does the reflection with respect to this last vector preserve the root system, though? What is its effect on vectors in $J$ ? We calculate

$\displaystyle\begin{aligned}\sigma_{\frac{1}{2}(\epsilon_1-\epsilon_2-\epsilon_3-\epsilon_4)}(v)&=v-\frac{2\left\langle v,\frac{1}{2}(\epsilon_1-\epsilon_2-\epsilon_3-\epsilon_4)\right\rangle}{\left\langle\frac{1}{2}(\epsilon_1-\epsilon_2-\epsilon_3-\epsilon_4),\frac{1}{2}(\epsilon_1-\epsilon_2-\epsilon_3-\epsilon_4)\right\rangle}\frac{1}{2}(\epsilon_1-\epsilon_2-\epsilon_3-\epsilon_4)\\&=v-\left\langle v,\frac{1}{2}(\epsilon_1-\epsilon_2-\epsilon_3-\epsilon_4)\right\rangle(\epsilon_1-\epsilon_2-\epsilon_3-\epsilon_4)\\&=v-\frac{v^1-v^2-v^3-v^4}{2}(\epsilon_1-\epsilon_2-\epsilon_3-\epsilon_4)\end{aligned}$

Now the sum $v^1-v^2-v^3-v^4$ is always an integer, whether the components of $v$ are integers or half-integers. If the sum is even, then we are changing each component of $v$ by an integer, which sends $I$ and $I'$ back to themselves. If the sum is off, then we are changing each component of $v$ by a half-integer, which swaps $I$ and $I'$ . In either case, the lattice $J$ is sent back to itself, and so this reflection fixes $\Phi$ .

Like we say for $G_2$ it’s difficult to understand the Weyl group of $F_4$ in terms of its action on the components of $v$ . However, also like $G_2$ , we can understand it geometrically. But instead of a hexagon, now the long and short roots each make up a four-dimensional polytope called the “24-cell”. It’s a shape with 24 vertices, 96 edges, 96 equilateral triangular faces, and 24 three-dimensional “cells”, each of which is a regular octahedron; the Weyl group of $F_4$ is its group of symmetries, just like the Weyl group of $G_2$ was the group of symmetries of the hexagon.

Also like the $G_2$ case, the $F_4$ root system is isomorphic to its own dual. The long roots stay the same length when dualized, while the short roots double in length and become the long roots of the dual root system. Again, a scaling and rotation sends the dual system back to the one we constructed.

March 9, 2010 Posted by John Armstrong | Geometry, Root Systems | 2 Comments

Construction of the G2 Root System

We’ve actually already seen the $G_2$ root system, back when we saw a bunch of two-dimensional root system. But let’s examine how we can construct it in line with our setup.

The $G_2$ root system is, as we can see by looking at it, closely related to the $A_2$ root system. And so we start again with the $2$ -dimensional subspace of $\mathbb{R}^3$ consisting of vectors with coefficients summing to zero, and we use the same lattice $J$ . But now we let $\Phi$ be the vectors $\alpha\in J$ of squared-length $2$ or $6$ : $\langle\alpha,\alpha\rangle=2$ or $\langle\alpha,\alpha\rangle=6$ . Explicitly, we have the six vectors from $A_2$ — $\pm(\epsilon_1-\epsilon_2)$ , $\pm(\epsilon_1-\epsilon_3)$ , and $\pm(\epsilon_2-\epsilon_3)$ — and six new vectors — $\pm(2\epsilon_2-\epsilon_1-\epsilon_3)$ , $\pm(2\epsilon_1-\epsilon_2-\epsilon_3)$ , and $\pm(\epsilon_1+\epsilon_2-2\epsilon_3)$ .

We can pick a base $\Delta=\{\epsilon_1-\epsilon_2,2\epsilon_2-\epsilon_1-\epsilon_3\}$ . These vectors are clearly independent. We can easily write each of the above vectors with a positive sign as a positive sum of the two vectors in $\Delta$ . For example, in accordance with an earlier lemma, we can write

$\displaystyle\begin{aligned}\epsilon_1+\epsilon_2-2\epsilon_3&=(2\epsilon_2-\epsilon_1-\epsilon_3)\\&+(\epsilon_1-\epsilon_2)\\&+(\epsilon_1-\epsilon_2)\\&+(\epsilon_1-\epsilon_2)\\&+(2\epsilon_2-\epsilon_1-\epsilon_3)\end{aligned}$

where after adding each term we have one of the positive roots. In fact, this path hits all but one of the six positive roots on its way to the unique maximal root.

It’s straightforward to calculate the Cartan integers for $\Delta$ .

$\displaystyle\frac{2\langle2\epsilon_2-\epsilon_1-\epsilon_3,\epsilon_1-\epsilon_2\rangle}{\langle\epsilon_1-\epsilon_2,\epsilon_1-\epsilon_2\rangle}=\langle2\epsilon_2-\epsilon_1-\epsilon_3,\epsilon_1-\epsilon_2\rangle=-3$

$\displaystyle\frac{2\langle\epsilon_1-\epsilon_2,2\epsilon_2-\epsilon_1-\epsilon_3\rangle}{\langle2\epsilon_2-\epsilon_1-\epsilon_3,2\epsilon_2-\epsilon_1-\epsilon_3\rangle}=\frac{1}{3}\langle\epsilon_1-\epsilon_2,2\epsilon_2-\epsilon_1-\epsilon_3\rangle=-1$

which shows that we do indeed get the Dynkin diagram $G_2$ .

And, of course, we must consider the reflections with respect to both vectors in $\Delta$ . Unfortunately, computations like those we’ve used before get complicated. However, we can just go back to the picture that we drew before (and that I linked to at the top of this post). It’s a nice, clean, two-dimensional picture, and it’s clear that these reflections send $\Phi$ back to itself, which establishes that $\Phi$ is really a root system.

We can also figure out the Weyl group geometrically from this picture. Draw line segments connecting the tips of either the long or the short roots, and we find a regular hexagon. Then the reflections with respect to the roots generate the symmetry group of this shape. The twelve roots are the twelve axes of symmetry of the polygon, and we can get rotations by first reflecting across one root and then across another. For example, rotating by a sixth of a turn can be effected by reflecting with the basic short root, followed by reflecting with the basic long root.

Finally, we can see that this root system is isomorphic to its own dual. Indeed, if $\alpha$ is a short root, then the dual root is $\alpha$ itself:

$\alpha^\vee=\frac{2}{\langle\alpha,\alpha\rangle}\alpha=\alpha$

On the other hand, if $\alpha$ is a long root, then we find

$\alpha^\vee=\frac{2}{\langle\alpha,\alpha\rangle}\alpha=\frac{1}{3}\alpha$

and so the squared-length of $\alpha^\vee$ is $\frac{2}{3}$ . These are now the short roots of the dual system. Scaling the dual system up by a factor of $9$ and rotating $\frac{1}{12}$ of a turn, we recover the original $G_2$ root system.

March 8, 2010 Posted by John Armstrong | Geometry, Root Systems | 1 Comment

Transformations of Dynkin Diagrams

Before we continue constructing root systems, we want to stop and observe a couple things about transformations of Dynkin diagrams.

First off, I want to be clear about what kinds of transformations I mean. Given Dynkin diagrams $X$ and $Y$ , I want to consider a mapping $\phi$ that sends every vertex of $X$ to a vertex of $Y$ . Further, if $\xi_1$ and $\xi_2$ are vertices of $X$ joined by $n$ edges, then $\phi(\xi_1)$ and $\phi(\xi_2)$ should be joined by $n$ edges in $Y$ as well, and the orientation of double and triple edges should be the same.

But remember that $\xi_1$ and $\xi_2$ , as vertices, really stand for vectors in some base of a root system, and the number of edges connecting them encodes their Cartan integers. If we slightly abuse notation and write $X$ and $Y$ for these bases, then the mapping $\phi$ defines images of the vectors in $X$ , which is a basis of a vector space. Thus $\phi$ extends uniquely to a linear transformation from the vector space spanned by $X$ to that spanned by $Y$ . And our assumption about the number of edges joining two vertices means that $\phi$ preserves the Cartan integers of the base $X$ .

Now, just like we saw when we showed that the Cartan matrix determines the root system up to isomorphism, we can extend $\phi$ to a map from the root system generated by $X$ to the root system generated by $Y$ . That is, a transformation of Dynkin diagrams gives rise to a morphism of root systems.

Unfortunately, the converse doesn’t necessarily hold. Look back at our two-dimensional examples; specifically, consider the $A_2$ and $G_2$ root systems. Even though we haven’t really constructed the latter yet, we can still use what we see. There are linear maps taking the six roots in $A_2$ to either the six long roots or the six short roots in $G_2$ . These maps are all morphisms of root systems, but none of them can be given by transformations of Dynkin diagrams. Indeed, the image of any base for $A_2$ would contain either two long roots in $G_2$ or two short roots, but any base of $G_2$ would need to contain both a long and a short root.

However, not all is lost. If we have an isomorphism of root systems, then it must send a base to a base, and thus it can be seen as a transformation of the Dynkin diagrams. Indeed, an isomorphism of root systems gives rise to an isomorphism of Dynkin diagrams.

The other observation we want to make is that duality of root systems is easily expressed in terms of Dynkin diagrams: just reverse all the oriented edges! Indeed, we’ve already seen this in the case of $B_n$ and $C_n$ root systems. When we get to constructing $G_2$ and $F_4$ , we will see that they are self-dual, in keeping with the fact that reversing the directed edge in each case doesn’t really change the diagram.

March 5, 2010 Posted by John Armstrong | Geometry, Root Systems | 3 Comments

Construction of B- and C-Series Root Systems

Starting from our setup, we construct root systems corresponding to the $B_n$ (for $n\geq2$ ) and $C_n$ (for $n\geq3$ ) Dynkin diagrams. First will be the $B_n$ series.

As we did for the $D_n$ series, we start out with an $n$ dimensional space with the lattice $I$ of integer-coefficient vectors. This time, though, we let $\Phi$ be the collection of vectors $\alpha\in I$ of squared-length ${2}$ or ${1}$ : either $\langle\alpha,\alpha\rangle=2$ or $\langle\alpha,\alpha\rangle=1$ . Explicitly, this is the collection of vectors $\pm(\epsilon_i\pm\epsilon_j)$ for $i\neq j$ (signs chosen independently) from the $D_n$ root system, plus all the vectors $\pm\epsilon_i$ .

Similarly to the $A_n$ series, and exactly as in the $D_n$ series, we define $\alpha_i=\epsilon_i-\epsilon_{i+1}$ for $1\leq i\leq n-1$ . This time, though, to get vectors whose coefficients don’t sum to zero we can just define $\alpha_n=\epsilon_n$ , which is independent of the other vectors. Since it has $n$ vectors, the independent set $\Delta=\{\alpha_i\}$ is a basis for our vector space.

As in the $A_n$ and $D_n$ cases, any vector $\epsilon_i-\epsilon_j$ with $i<j$ can be written

$\epsilon_i-\epsilon_j=(\epsilon_i-\epsilon_{i+1})+\dots+(\epsilon_{j-1}-\epsilon_j)$

This time, any of the $\epsilon_i$ can be written

$\epsilon_i=(\epsilon_i-\epsilon_{i+1})+\dots+(\epsilon_{n-1}-\epsilon_n)+\epsilon_n$

Thus any vector $\epsilon_i+\epsilon_j$ can be written as the sum of two of these vectors. And so $\Delta$ is a base for $\Phi$ .

We calculate the Cartan integers. For $i$ and $j$ less than $n$ , we again have the same calculation as in the $A_n$ case, which gives a simple chain of length $n-1$ vertices. But when we involve $\alpha_n$ things are a little different.

$\displaystyle\frac{2\langle\epsilon_i-\epsilon_{i+1},\epsilon_n\rangle}{\langle\epsilon_n,\epsilon_n\rangle}=2\langle\epsilon_i-\epsilon_{i+1},\epsilon_n\rangle$

$\displaystyle\frac{2\langle\epsilon_n,\epsilon_i-\epsilon_{i+1}\rangle}{\langle\epsilon_i-\epsilon_{i+1},\epsilon_i-\epsilon_{i+1}\rangle}=\langle\epsilon_n,\epsilon_i-\epsilon_{i+1}\rangle$

If $1\leq i<n-1$ , then both of these are zero. On the other hand, if $i=n-1$ , then the first is $-2$ and the second is $-1$ . Thus we get a double edge from $\alpha_{n-1}$ to $\alpha_n$ , and $\alpha_{n-1}$ is the longer root. And so we obtain the $B_n$ Dynkin diagram.

Considering the reflections with respect to the $\alpha_i$ , we find that $\sigma_{\alpha_i}$ swaps the coefficients of $\epsilon_i$ and $\epsilon_{i+1}$ for $1\leq i\leq n-1$ . But what about $\alpha_n$ ? We calculate

$\displaystyle\begin{aligned}\sigma_{\alpha_n}(v)&=v-\frac{2\langle v,\alpha_n\rangle}{\langle\alpha_n,\alpha_n\rangle}\alpha_n\\&=v-2\langle v,\alpha_n\rangle\alpha_n\\&=v-2v^n\epsilon_n\end{aligned}$

which flips the sign of the last coefficient of $v$ . As we did in the $D_n$ case, we can use this to flip the signs of whichever coefficients we want. Since these transformations send the lattice $I$ back into itself, they send $\Phi$ to itself and we do have a root system.

Finally, since we don’t have any restrictions on how many signs we can flip, the Weyl group for $B_n$ is exactly the wreath product $S_n\wr\mathbb{Z}_2$ .

So, what about $C_n$ ? This is just the dual root system to $B_n$ ! The roots of squared-length ${2}$ are left unchanged, but the roots of squared-length ${1}$ are doubled. The Weyl group is the same — $S_n\wr\mathbb{Z}_2$ — but now the short root in the base $\Delta$ is the long root, and so we flip the direction of the double arrow in the Dynkin diagram, giving the $C_n$ diagram.

March 4, 2010 Posted by John Armstrong | Geometry, Root Systems | 2 Comments

Construction of D-Series Root Systems

Starting from our setup, we construct root systems corresponding to the $D_n$ Dynkin diagrams (for $n\geq4$ ).

The construction is similar to that of the $A_n$ series, but instead of starting with a hyperplane in $n+1$ -dimensional space, we just start with $n$ -dimensional space itself with the lattice $I$ of integer-coefficient vectors. We again take $\Phi$ to be the collection of vectors $\alpha\in I$ of squared-length $2$ : $\langle\alpha,\alpha\rangle=2$ . Explicitly, this is the collection of vectors $\pm(\epsilon_i\pm\epsilon_j)$ for $i\neq j$ , where we can choose the two signs independently.

Similarly to the $A_n$ case, we define $\alpha_i=\epsilon_i-\epsilon_{i+1}$ for $1\leq i\leq n-1$ , but these can only give vectors whose coefficients sum to ${0}$ . To get other vectors, we throw in $\alpha_n=\epsilon_{n-1}+\epsilon_n$ , which is independent of the others. The linearly independent collection $\Delta=\{\alpha_i\}$ has $n$ vectors, and so must be a basis of the $n$ -dimensional space.

As before, any vector in $\phi$ of the form $\epsilon_i-\epsilon_j$ for $i<j$ can be written as

$\displaystyle\epsilon_i-\epsilon_j=(\epsilon_i-\epsilon_{i+1})+\dots+(\epsilon_{j-1}-\epsilon_j)$

while vectors of the form $\epsilon_i+\epsilon_j$ are a little more complicated. We can start with

$\displaystyle\epsilon_i+\epsilon_n=(\epsilon_i-\epsilon_{i+1})+\dots+(\epsilon_{j-1}+\epsilon_j)$

and from this we can always build $2\epsilon_j=(\epsilon_j-\epsilon_n)+(\epsilon_j+\epsilon_n)$ for $1\leq j\leq n-1$ . Then if $i<j\leq n-1$ we can write $\epsilon_i+\epsilon_j=(\epsilon_i-\epsilon_j)+2\epsilon_j$ . This proves that $\Delta$ is a base for $\Phi$ .

Again, we calculate the Cartan integers. The calculation for $i$ and $j$ both less than $n$ is exactly as before, showing that these vectors form a simple chain in the Dynkin diagram of length $n-1$ . However, when we involve $\alpha_n$ we find

$\displaystyle\frac{2\langle\epsilon_i-\epsilon_{i+1},\epsilon_{n-1}+\epsilon_n\rangle}{\langle\epsilon_i-\epsilon_{i+1},\epsilon_i-\epsilon_{i+1}\rangle}=\langle\epsilon_i-\epsilon_{i+1},\epsilon_{n-1}+\epsilon_n\rangle$

For $i<n-2$ , this is automatically ${0}$ ; for $i=n-2$ , we get the value $-1$ ; and for $i=n-1$ we again get ${0}$ . This shows that the Dynkin diagram of $\Delta$ is $D_n$ .

Finally, we consider the reflections with respect to the $\alpha_i$ . As in the $A_n$ case, we find that $\sigma_{\alpha_i}$ swaps the coefficients of $\epsilon_i$ and $\epsilon_{i+1}$ for $1\leq i\leq n-1$ . But what about $\alpha_n$ ?

$\displaystyle\begin{aligned}\sigma_{\alpha_n}(v)&=v-\frac{2\langle v,\alpha_n\rangle}{\langle\alpha_n,\alpha_n\rangle}\alpha_n\\&=v-\langle v,\alpha_n\rangle\alpha_n\\&=v-(v^{n-1}+v^n)(\epsilon_{n-1}+\epsilon_n)\\&=v-\langle v,\alpha_n\rangle\alpha_n\\&=v-(v^{n-1}\epsilon_{n-1}+v^n\epsilon_n)-(v^n\epsilon_{n-1}+v^{n-1}\epsilon_n)\end{aligned}$

This swaps the last two coefficients of $v$ and flips their sign. Clearly, this sends the lattice $I$ back to itself, showing that $\Phi$ is indeed a root system.

Now we can use $\sigma_{\alpha_n}\sigma_{\alpha_{n-1}}$ to flip the signs of coefficients of $v$ , two at a time. We use whatever of the $\sigma_{\alpha_i}$ we need to get the two coefficients we want into the last two slots, hit it with $\sigma_{\alpha_n}\sigma_{\alpha_{n-1}}$ to flip them, and then invert the first permutation to move everything back where it started from. In fact, this is a lot like what we saw way back with the Rubik’s cube, when dealing with the edge group. We can effect whatever permutation we want on the coefficients, and we can flip any even number of them.

The Weyl group of $D_n$ is then the subgroup of the wreath product $S_n\wr\mathbb{Z}_2$ consisting of those transformations with an even number of flips coming from the $\mathbb{Z}_2$ components. Explicitly, we can write $\mathbb{Z}_2^{n-1}$ as the subgroup of $\mathbb{Z}_2^n$ with sum zero. Then we can let $S_n$ act on $\mathbb{Z}_2^n$ by permuting the components, and use this to give an action of $S_n$ on $\mathbb{Z}_2^{n-1}$ , and thus form the semidirect product $S_n\ltimes\mathbb{Z}_2^{n-1}$ .

March 3, 2010 Posted by John Armstrong | Geometry, Root Systems | 2 Comments

Construction of A-Series Root Systems

Starting from our setup, we construct root systems corresponding to the $A_n$ Dynkin diagrams.

We start with the $n+1$ -dimensional space $\mathbb{R}^{n+1}$ with orthonormal basis $\{\epsilon_0,\dots,\epsilon_n\}$ , and cut out the $n$ -dimensional subspace $E$ orthogonal to the vector $\epsilon_0+\dots+\epsilon_n$ . This consists of those vectors $v=\sum_{k=0}^nv^k\epsilon_k$ for which the coefficients sum to zero: $\sum_{k=0}^nv^k=0$ . We let $J=I\cap E$ , consisting of the lattice vectors whose (integer) coefficients sum to zero. Finally, we define our root system $\Phi$ to consist of those vectors $\alpha\in J$ such that $\langle\alpha,\alpha\rangle=2$ .

From this construction it should be clear that $\Phi$ consists of the vectors $\{\epsilon_i-\epsilon_j\vert i\neq j\}$ . The $n$ vectors $\Delta=\{\alpha_i=\epsilon_{i-1}-\epsilon_i\}$ are independent, and thus form a basis of the $n$ -dimensional space $E$ . This establishes that $\Phi$ spans $E$ . In particular, if $i<j$ we can write

$\displaystyle(\epsilon_i-\epsilon_j)=(\epsilon_i-\epsilon_{i+1})+(\epsilon_{i+1}-\epsilon_{i+2})+\dots+(\epsilon_{j-1}-\epsilon_j)$

showing that $\Delta$ forms a base for $\Phi$ .

We calculate the Cartan integers for this base

$\displaystyle\frac{2\langle\epsilon_{j-1}-\epsilon_j,\epsilon_{i-1}-\epsilon_i\rangle}{\langle\epsilon_{i-1}-\epsilon_i,\epsilon_{i-1}-\epsilon_i\rangle}=\langle\epsilon_{j-1}-\epsilon_j,\epsilon_{i-1}-\epsilon_i\rangle$

For $i=j$ we get the value ${2}$ ; for $j+1$ or $j-1$ we get the value $-1$ ; otherwise we get the value ${0}$ . This clearly gives us the Dynkin diagram $A_n$ .

Finally, the reflections with respect to the $\alpha_i$ should generate the entire Weyl group. We must verify that these leave the lattice $J$ invariant to be sure that we have a root system. We calculate

$\displaystyle\begin{aligned}\sigma_{\alpha_i}(v)&=v-\frac{2\langle v,\alpha_i\rangle}{\langle\alpha_i,\alpha_i\rangle}\alpha_i\\&=v-\langle v,\alpha_i\rangle\alpha_i\\&=v-(v^{i-1}-v^i)(\epsilon_{i-1}-\epsilon_i)\\&=v-(v^{i-1}\epsilon_{i-1}+v^i\epsilon_i)+(v^i\epsilon_{i-1}+v^{i-1}\epsilon_i)\end{aligned}$

That is, it swaps the coefficients of $\epsilon_{i-1}$ and $\epsilon_i$ , and thus sends the lattice $J$ back to itself, as we need.

We can also see from this effect that any combination of the $\sigma_{\alpha_i}$ serves to permute the $n+1$ coefficients of a given vector. That is, the Weyl group of the $A_n$ system is naturally isomorphic to the symmetric group $S_{n+1}$ .

March 2, 2010 Posted by John Armstrong | Geometry, Root Systems | 4 Comments

Construction of Root Systems (setup)

Now that we’ve proven the classification theorem, we know all about root systems, right? No! All we know is which Dynkin diagrams could possibly arise from root systems. We don’t know whether there actually exists a root system for any given one of them. The situation is sort of like what we found way back when we solved Rubik’s magic cube: first we established some restrictions on allowable moves, and then we showed that everything else actually happened.

And so we must construct some actual root systems. For this task, we let $E$ stand for a finite-dimensional real vector space $\mathbb{R}^n$ for various $n$ , equipped with its usual inner product. We pick an orthonormal basis $\{\epsilon_1,\dots,\epsilon_n\}$ and let the integral linear combinations of these basis vectors form the lattice $I$ . Here, I do not mean “lattice” in the order-theory sense. I mean that this is a discrete collection of points in the vector space that is closed under addition.

In every case we’re going to take either the lattice $I$ , or a slightly modified lattice $J$ . We’ll define our root system $\Phi$ to be the collection of vectors in the lattice of either one or two specified lengths (since there can be at most two root lengths). That is, we’re considering the intersection of a discrete collection of points with one or two spheres. These spheres are closed and bounded, and thus compact. The collection $\Phi$ must be finite or else it would have an accumulation point by Bolzano-Weierstrass, and thus wouldn’t be discrete!

Any one of our constructed collections will span $E$ , and in fact an explicit basis will be shown in each case, in case it’s not clear. It should also be clear that none of them can contain the vector ${0}$ , and so the first condition of being a root system will hold. Our choice of lengths will make it clear that there are no possible scalar multiples of a root besides itself and its negative. On the other hand, it should be clear that if $\alpha$ is in a lattice $I$ and on a sphere $\lVert\alpha\rVert^2=r^2$ , then $-\alpha$ is also in both, and thus the second condition holds.

The reflection $\sigma_\alpha$ preserves lengths, and so it sends the spheres back to themselves. We’ll have to check in each case that $\sigma_\alpha$ sends every vector in our collection back into the lattice, which will establish the third condition.

As to the fourth condition, the inner product $\langle\alpha,\beta\rangle$ is automatically going to be in $\mathbb{Z}$ when we pick $\alpha$ and $\beta$ from a lattice, and so picking the squared radii of our spheres to divide $2$ should be enough to guarantee that $\frac{2\langle\beta,\alpha\rangle}{\langle\alpha,\alpha\rangle}\in\mathbb{Z}$ .

Tomorrow we start in constructing our root systems, towards the theorem: For each Dynkin diagam allowed by the classification theorem, there exists an irreducible root system having that diagram.

March 1, 2010 Posted by John Armstrong | Geometry, Root Systems | 7 Comments

Proving the Classification Theorem V

Today we conclude the proof of the classification theorem. The first four parts of the proof are here, here, here, and here.

The only possible Coxeter graphs with a triple vertex are those of the forms $D_n$ , $E_6$ , $E_7$ , and $E_8$ .

From step 8 we have the labelled graph

Like we did in step 9, we define

$\displaystyle\epsilon=\sum\limits_{i=1}^{p-1}i\epsilon_i$

$\displaystyle\eta=\sum\limits_{j=1}^{q-1}i\eta_j$

$\displaystyle\zeta=\sum\limits_{k=1}^{r-1}i\zeta_k$

These vectors $\epsilon$ , $\eta$ , and $\zeta$ are mutually orthogonal, linearly independent vectors, and that $\psi$ is not in the subspace that they span.

We look back at our proof of step 4 to determine that $\cos(\theta_1)^2+\cos(\theta_2)^2+\cos(\theta_3)^3<0$ ; where $\theta_1$ , $\theta_2$ , and $\theta_3$ are the angles between $\psi$ and $\epsilon$ , $\eta$ , and $\zeta$ , respectively. We look back at our proof of step 9 to determine that $\langle\epsilon,\epsilon\rangle=\frac{p(p-1)}{2}$ , $\langle\eta,\eta\rangle=\frac{q(q-1)}{2}$ , and $\langle\zeta,\zeta\rangle=\frac{r(r-1)}{2}$ . Thus we can calculate the cosine

$\displaystyle\begin{aligned}\cos(\theta_1)^2&=\frac{\langle\epsilon,\psi\rangle^2}{\langle\epsilon,\epsilon\rangle\langle\psi,\psi\rangle}\\&=\frac{(p-1)^2\langle\epsilon_{p-1},\psi\rangle^2}{\langle\epsilon,\epsilon\rangle}\\&=\frac{(p-1)^2\frac{1}{4}}{\frac{p(p-1)}{2}}\\&=\frac{p-1}{2p}\\&=\frac{1}{2}\left(1-\frac{1}{p}\right)\end{aligned}$

And similarly we find $\cos(\theta_2)^2=\frac{1}{2}(1-\frac{1}{q})$ , and $\cos(\theta_3)^2=\frac{1}{2}(1-\frac{1}{r})$ . Adding up, we find

$\displaystyle\begin{aligned}\frac{1}{2}\left(3-\frac{1}{p}-\frac{1}{q}-\frac{1}{r}\right)&<1\\3-\frac{1}{p}-\frac{1}{q}-\frac{1}{r}&<2\\\frac{1}{p}+\frac{1}{q}+\frac{1}{r}&>1\end{aligned}$

This last inequality, by the way, is hugely important in many areas of mathematics, and it’s really interesting to find it cropping up here.

Anyway, now none of $p$ , $q$ , or $r$ can be $1$ or we don’t have a triple vertex at all. We can also choose which strand is which so that

$\displaystyle\frac{1}{p}\leq\frac{1}{q}\leq\frac{1}{r}\leq\frac{1}{2}$

We can determine from here that

$\displaystyle\frac{3}{2}\geq\frac{3}{r}\geq\frac{1}{p}+\frac{1}{q}+\frac{1}{r}>1$

and so we must have $r=2$ , and the shortest leg must be one edge long. Now we have $\frac{1}{p}+\frac{1}{q}>\frac{1}{2}$ , and so $\frac{2}{q}>\frac{1}{2}$ , and $q$ must be either $2$ or $3$ .

If $q=3$ , then the second shortest leg is two edges long. In this case, $\frac{1}{p}>\frac{1}{6}$ and $p<6$ . The possibilities for the triple $(p,q,r)$ are $(3,3,2)$ , $(4,3,2)$ , and $(5,3,2)$ ; giving graphs $E_6$ , $E_7$ , and $E_8$ , respectively.

On the other hand, if $q=2$ , then the second shortest leg is also one edge long. In this case, there is no more restriction on $p$ , and so the remaining leg can be as long as we like. This gives us the $D_n$ family of graphs.

And we’re done! If we have one triple edge, we must have the graph $G_2$ . If we have a double edge or a triple vertex, we can have only one, and we can’t have one of each. Step 9 narrows down graphs with a double edge to $F_4$ and the families $B_n$ and $C_n$ , while step 10 narrows down graphs with a triple vertex to $E_6$ , $E_7$ , and $E_8$ , and the family $D_n$ . Finally, if there are no triple vertices or double edges, we’re left with a single simple chain of type $A_n$ .

February 26, 2010 Posted by John Armstrong | Geometry, Root Systems | Leave a Comment

Proving the Classification Theorem IV

We continue proving the classification theorem. The first three parts are here, here, and here.

Any connected graph $\Gamma$ takes one of the four following forms: a simple chain, the $G_2$ graph, three simple chains joined at a central vertex, or a chain with exactly one double edge.

This step largely consolidates what we’ve done to this point. Here are the four possible graphs:

The labels will help with later steps.

Step 5 told us that there’s only one connected graph that contains a triple edge. Similarly, if we had more than one double edge or triple vertex, then we must be able to find two of them connected by a simple chain. But that will violate step 7, and so we can only have one of these features either.

The only possible Coxeter graphs with a double edge are those underlying the Dynkin diagrams $B_n$ , $C_n$ , and $F_4$ .

Here we’ll use the labels on the above graph. We define

$\displaystyle\epsilon=\sum\limits_{i=1}^pi\epsilon_i$

$\displaystyle\eta=\sum\limits_{j=1}^qj\eta_j$

As in step 6, we find that $2\langle\epsilon_i,\epsilon_{i+1}\rangle=-1=2\langle\eta_j,\eta_{j+1}\rangle$ and all other pairs of vectors are orthogonal. And so we calculate

$\displaystyle\begin{aligned}\langle\epsilon,\epsilon\rangle&=\sum\limits_{i=1}^pi^2\langle\epsilon_i,\epsilon_i\rangle+2\sum\limits_{i<j}ij\langle\epsilon_i,\epsilon_j\rangle\\&=\sum\limits_{i=1}^pi^2+\sum\limits_{i=1}^{p-1}i(i+1)2\langle\epsilon_i,\epsilon_{i+1}\rangle\\&=\sum\limits_{i=1}^pi^2-\sum\limits_{i=1}^{p-1}(i^2+i)\\&=p^2-\sum\limits_{i=1}^{p-1}i\\&=p^2-\frac{p(p-1)}{2}\\&=\frac{p(p+1)}{2}\end{aligned}$

And similarly, $\langle\eta,\eta\rangle=\frac{q(q+1)}{2}$ . We also know that $4\langle\epsilon_p,\eta_q\rangle^2=2$ , and so we find

$\displaystyle\langle\epsilon,\eta\rangle^2=p^2q^2\langle\epsilon_p,\eta_q\rangle^2=\frac{p^2q^2}{2}$

Now we can use the Cauchy-Schwarz inequality to conclude that

$\frac{p^2q^2}{2}=\langle\epsilon,\eta\rangle^2<\langle\epsilon,\epsilon\rangle\langle\eta,\eta\rangle=\frac{p(p+1)q(q+1)}{4}$

where the inequality is strict, since $\epsilon$ and $\eta$ are linearly independent. And so we find

$\displaystyle\begin{aligned}\frac{p^2q^2}{2}&<\frac{p(p+1)q(q+1)}{4}\\2p^2q^2&<p(p+1)q(q+1)\\2pq&<(p+1)(q+1)\\2pq&<pq+p+q+1\\pq-p-q+1&<2\\(p-1)(q-1)&<2\end{aligned}$

We thus must have either $p=q=2$ , which gives us the $F_4$ diagram, or $p=1$ or $q=1$ with the other arbitary, which give rise the the $B_n$ and $C_n$ Coxeter graphs.

February 25, 2010 Posted by John Armstrong | Geometry, Root Systems | 5 Comments

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