# The Unapologetic Mathematician

## Root Strings

First, a lemma: given a root system $\Phi$, let $\alpha$ and $\beta$ be nonproportional roots. That is, $\beta\neq\pm\alpha$. Then if $\langle\alpha,\beta\rangle>0$ — if the angle between the vectors is strictly acute — then $\alpha-\beta$ is also a root. On the other hand, if $\langle\alpha,\beta\rangle<0$ then $\alpha+\beta$ is also a root. We know that $-\beta$ is a root. If we replace $\beta$ with $-\beta$ we can see that the second of these assertions follows from the first, so we just need to prove that one.

We know that $\langle\alpha,\beta\rangle$ is positive if and only if $\alpha\rtimes\beta$ is. So let’s look at the table we worked up last time. We see that when $\alpha\rtimes\beta$ is positive, then either that or $\beta\rtimes\alpha$ equals ${1}$. If it’s $\alpha\rtimes\beta$, then

$\sigma_\beta(\alpha)=\alpha-(\alpha\rtimes\beta)\beta=\alpha-\beta$

is a root. On the other hand, if $\beta\rtimes\alpha=1$ then $\alpha-\beta=-\sigma_\alpha(\beta)$ is a root.

So given two nonproportional and nonorthogonal roots $\alpha$ and $\beta$ we’re guaranteed to have more than one vector of the form $\beta+k\alpha$ for some integer $k$ in the root system $\Phi$. We call the collection of all such vectors in $\Phi$ the $\alpha$-string through $\beta$.

Let $r$ be the largest integer so that $\beta-r\alpha\in\Phi$, and let $q$ be the largest integer so that $\beta+q\alpha\in\Phi$. I say that the root string is unbroken. That is, $\beta+k\alpha\in\Phi$ for all $-r\leq k\leq q$.

Indeed, if there’s some integer $k$ so that $\beta+k\alpha\notin\Phi$, then we can find $p$ and $s$ with $-r\leq p so that

\displaystyle\begin{aligned}\beta+p\alpha&\in\Phi\\\beta+(p+1)\alpha&\notin\Phi\\\beta+(s-1)\alpha&\notin\Phi\\\beta+s\alpha&\in\Phi\end{aligned}

But then the above lemma tells us that $\langle\alpha,\beta+p\alpha\rangle\geq0$, while $\langle\alpha,\beta+s\alpha\rangle\leq0$. Subtracting, we find

\displaystyle\begin{aligned}\langle\alpha,\beta+p\alpha\rangle-\langle\alpha,\beta+s\alpha\rangle&=\langle\alpha,(\beta+p\alpha)-(\beta+s\alpha)\rangle\\&=\langle\alpha,(p-s)\alpha\\&=(p-s)\langle\alpha,\alpha\rangle\end{aligned}

The two inequalities tell us that this difference should be positive, but $\langle\alpha,\alpha\rangle$ is positive and $p-s$ is negative. Thus we have a contradiction, and the root string must be unbroken from $\beta-r\alpha$ to $\beta+q\alpha$.

We can also tell that $\sigma_\alpha$ just adds a positive or negative multiple of $\alpha$ to any root. Then it’s clear from the geometry that $\sigma_\alpha$ just reverses a root string end to end. That is, $\sigma_\alpha(\beta+q\alpha)=\beta-r\alpha$. But we can also calculate

\displaystyle\begin{aligned}\sigma_\alpha(\beta+q\alpha)&=\sigma_\alpha(\beta)+q\sigma_\alpha(\alpha)\\&=\beta-(\beta\rtimes\alpha)\alpha-q\alpha\\&=\beta-((\beta\rtimes\alpha)+q)\alpha\\&=\beta-r\alpha\end{aligned}

Thus $r-q=\beta\rtimes\alpha$. And so the length of the $\alpha$ string through $\beta$ can be no more than $4$.

January 29, 2010 Posted by | Geometry, Root Systems | 6 Comments

## Pairs of Roots

When we look at a root system, the integrality condition $\beta\rtimes\alpha\in\mathbb{Z}$ puts strong restrictions on the relationship between any two vectors in the root system. Today we’ll look at this in two ways: one more formal and the other more visual.

First off, let’s look a bit deeper at the condition

$\displaystyle\beta\rtimes\alpha=\frac{2\langle\beta,\alpha\rangle}{\langle\alpha,\alpha\rangle}=k\in\mathbb{Z}$

We know a lot about the relation between the inner product and the lengths of vectors and the angle between them. Specifically, we can write

$\displaystyle\frac{2\langle\beta,\alpha\rangle}{\langle\alpha,\alpha\rangle}=\frac{2\lVert\beta\rVert\lVert\alpha\rVert\cos(\theta)}{\lVert\alpha\rVert^2}=2\frac{\lVert\beta\rVert}{\lVert\alpha\rVert}\cos(\theta)$

Then we can consider the product

$\displaystyle(\beta\rtimes\alpha)(\alpha\rtimes\beta)=\left(2\frac{\lVert\beta\rVert}{\lVert\alpha\rVert}\cos(\theta)\right)\left(2\frac{\lVert\alpha\rVert}{\lVert\beta\rVert}\cos(\theta)\right)=4\cos(\theta)^2$

This has to be an integer between ${0}$ and $4$. And it can only hit $4$ if $\theta$ is a multiple of $\pi$, in which case we already know that $\beta$ must be $\pm\alpha$. So some simple counting shows the possibilities. Without loss of generality we’ll assume that $\lVert\beta\rVert\geq\lVert\alpha\rVert$.

$\begin{tabular}{rrcc}\hline$$\alpha\rtimes\beta$$&$$\beta\rtimes\alpha$$&$$\theta$$&$$\frac{\lVert\beta\rVert^2}{\lVert\alpha\rVert^2}$$\\\hline0&0&$$\frac{\pi}{2}$$&arbitrary\\1&1&$$\frac{\pi}{3}$$&1\\-1&-1&$$\frac{2\pi}{3}$$&1\\1&2&$$\frac{\pi}{4}$$&2\\-1&-2&$$\frac{3\pi}{4}$$&2\\1&3&$$\frac{\pi}{6}$$&3\\-1&-3&$$\frac{5\pi}{6}$$&3\\\hline\end{tabular}$

We see that if $\alpha$ and $\beta$ are orthogonal then we have no information about their relative lengths. This is to be expected, since when we form the coproduct of two root systems the roots from each side are orthogonal to each other, and we should have no restriction on their relative lengths. On the other hand, if they aren’t orthogonal, then there are only a handful of possibilities.

Let’s try to look at this more visually. The vectors $\alpha$ and $\beta$ span some two-dimensional space. We can choose an orthogonal basis (and thus coordinates) that makes $\alpha=(1,0)$, and then $\beta=(x,y)$. We calculate

\displaystyle\begin{aligned}\beta\rtimes\alpha=\frac{2\langle(x,y),(1,0)\rangle}{\langle(1,0),(1,0)\rangle}=2x&=k\in\mathbb{Z}\\\alpha\rtimes\beta=\frac{2\langle(1,0),(x,y)\rangle}{\langle(x,y),(x,y)\rangle}=\frac{2x}{x^2+y^2}&=l\in\mathbb{Z}\end{aligned}

The first of these conditions says that $x=\frac{k}{2}$ for some integer $k$. We can manipulate the second to tell us that $\left(x-\frac{1}{l}\right)^2+y^2=\left(\frac{1}{l}\right)^2$ for some integer $l$. Let’s plot some of these!

The vector pointing to $(1,0)$ is $\alpha$. The blue vertical lines correspond to the first condition for $k=\pm1,\pm2,\pm3,\pm4$, while the red circles correspond to the second condition for $l=\pm1,\pm2,\pm3,\pm4$. Any larger values of $k$ will put lines further out than any of the circles can touch, while any larger values of $l$ will put smaller circles further in than any line can touch. Thus, all possible values of $\beta$ which satisfy both conditions are the twenty highlighted points in this image, six examples of which are indicated in the diagram.

The three examples towards the top of the diagram correspond to the second, fourth, and sixth lines in the above table. The three examples toward the bottom are similar, but have $\lVert\beta\rVert\leq\lVert\alpha\rVert$. We can see that if we project any of the examples onto $\alpha$, the projection is a half-integral multiple of $\alpha$. Conversely, projecting $\alpha$ onto any of the examples gives a half-integral multiple of that vector.

So we can say a lot about how any two vectors in $\Phi$ are related if they aren’t parallel or orthogonal. The question now is how we can pick a whole collection of vectors so that each pair relates to each other in one of these very particular ways.

January 28, 2010 Posted by | Geometry, Root Systems | 5 Comments

## Irreducible Root Systems

Given a root system $\Phi$ in an inner product space $U$ we may be able to partition it into two collections $\Phi=\Psi\uplus\Psi'$ so that each root in $\Psi$ is orthogonal to every root in $\Psi'$ and vice versa. In this case, the subspace $V$ spanned by $\Psi$ and the subspace $V'$ spanned by $\Psi'$ are orthogonal to each other, and together they span the whole of $U$. Thus we can write $U=V\oplus V'$, and thus $\Phi$ is the coproduct $\Psi\amalg\Psi'$.

In this situation, we say that $\Phi$ is a “reducible” root system. On the other hand, if there is no way of writing $\Phi$ as the coproduct of two other root systems we say it’s “irreducible”. We will show that every root system is the coproduct of a finite number of irreducible root systems in a unique way (up to reordering the irreducibles).

Okay, first off we can show that there’s at least one decomposition into irreducibles, and we’ll proceed by induction on the dimension of the root system. To start with, any one-dimensional root system is of the form $\{\alpha,-\alpha\}$, and this is definitely irreducible. On the other hand, given an $n$-dimensional root system $\Phi$, either it’s irreducible or not. If it is, we’re done. But if not, then we can break it into $\Psi$ and $\Psi'$, each of which has dimension strictly less than $n$. By the inductive hypothesis, each of these can be written as the coproduct of a bunch of irreducibles, and so we just take the coproduct of these two collections.

On the other hand, how do we know that the decomposition is essentially unique? This essentially comes down to what the guys at the Secret Blogging Seminar call the diamond lemma: if we have two different ways of partitioning a root system, then there is some common way of partitioning each of those, thus “completing the diamond”. Thus, any choices we may have made in showing that a decomposition exists ultimately don’t matter.

So, let’s say we have a root system $\Phi\subseteq U$. Suppose that we have two different decompositions, $U=V_1\oplus\dots\oplus V_m$ and $U=V'_1\oplus\dots\oplus V'_n$. In each case, every root in $\Phi$ is actually in exactly one of the subspaces $V_i$ and exactly one of the subspaces $V'_j$, thus partitioning the root system on the one hand as $\Phi=\Psi_1\amalg\dots\amalg\Psi_m$ and $\Phi=\Psi'_1\amalg\dots\amalg\Psi'_n$.

All we have to do is define $W_{ij}=V_i\cap V'_j$. From here we can see that

\displaystyle\begin{aligned}V_i&=\bigoplus\limits_{j=1}^nW_{ij}\\V'_j&=\bigoplus\limits_{i=1}^mW_{ij}\\U&=\bigoplus\limits_{\substack{1\leq i\leq m\\1\leq j\leq n}}W_{ij}\end{aligned}

Further, each root in $\Psi$ is in exactly one of the $W_{ij}$, so we have a decomposition of the root system.

This result does a lot towards advancing our goal. Any root system can be written as the coproduct of a bunch of irreducible root systems. Now all we have to do is classify the irreducible root systems (up to isomorphism) and we’re done!

January 27, 2010 Posted by | Geometry, Root Systems | 6 Comments

## Dual Root Systems

Given a root system $\Phi$, there’s a very interesting related root system $\Phi^\vee$, called the “dual” or “inverse” root system. It’s made up of the “duals” $\alpha^\vee$, defined by

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

This is the vector that represents the linear functional $\underline{\hphantom{X}}\rtimes\alpha$. That is, $\beta\rtimes\alpha=\langle\beta,\alpha^\vee\rangle$.

The dual root $\alpha^\vee$ is proportional to $\alpha$, and so $\sigma_{\alpha^\vee}=\sigma_\alpha$. The dual reflections are the same as the original reflections, and so they generate the same subgroup of $\mathrm{O}(V)$. That is, the Weyl group of $\Phi^\vee$ is the same as the Weyl group of $\Phi$.

As we should hope, dualizing twice gives back the original root system. That is, $\left(\Phi^\vee\right)^\vee=\Phi$. We can even show that $\left(\alpha^\vee\right)^\vee=\alpha$. Indeed, we calculate

\displaystyle\begin{aligned}\left(\alpha^\vee\right)^\vee&=\frac{2}{\langle\alpha^\vee,\alpha^\vee\rangle}\alpha^\vee\\&=\frac{2}{\langle\frac{2}{\langle\alpha,\alpha\rangle}\alpha,\frac{2}{\langle\alpha,\alpha\rangle}\alpha\rangle}\frac{2}{\langle\alpha,\alpha\rangle}\alpha\\&=\frac{4}{\frac{4}{\langle\alpha,\alpha\rangle\langle\alpha,\alpha\rangle}\langle\alpha,\alpha\rangle\langle\alpha,\alpha\rangle}\alpha\\&=\alpha\end{aligned}

It turns out that passing to duals reverses the roles of roots, in a way, just as we might expect from a dualization. Specifically, $\alpha^\vee\rtimes\beta^\vee=\beta\rtimes\alpha$. Indeed, we calculate

\displaystyle\begin{aligned}\alpha^\vee\rtimes\beta^\vee&=\frac{2\langle\alpha^\vee,\beta^\vee\rangle}{\langle\beta^\vee,\beta^\vee\rangle}\\&=\frac{2\langle\frac{2}{\langle\alpha,\alpha\rangle}\alpha,\frac{2}{\langle\beta,\beta\rangle}\beta\rangle}{\langle\frac{2}{\langle\beta,\beta\rangle}\beta,\frac{2}{\langle\beta,\beta\rangle}\beta\rangle}\\&=\frac{\frac{8}{\langle\alpha,\alpha\rangle\langle\beta,\beta\rangle}\langle\alpha,\beta\rangle}{\frac{4}{\langle\beta,\beta\rangle\langle\beta,\beta\rangle}\langle\beta,\beta\rangle}\\&=\frac{2\langle\beta,\alpha\rangle}{\langle\alpha,\alpha\rangle}\\&=\beta\rtimes\alpha\end{aligned}

January 26, 2010 Posted by | Geometry, Root Systems | 4 Comments

## Coproduct Root Systems

We should also note that the category of root systems has binary (and thus finite) coproducts. They both start the same way: given root systems $\Phi$ and $\Phi'$ in inner-product spaces $V$ and $V'$, we take the direct sum $V\oplus V'$ of the vector spaces, which makes vectors from each vector space orthogonal to vectors from the other one.

The coproduct $\Phi\amalg\Phi'$ root system consists of the vectors of the form $(\alpha,0)$ for $\alpha\in\Phi$ and $(0,\alpha')$ for $\alpha'\in\Phi'$. Indeed, this collection is finite, spans $V\oplus V'$, and does not contain $(0,0)$. The only multiples of any given vector in $\Phi\amalg\Phi'$ are that vector and its negative. The reflection $\sigma_{(\alpha,0)}$ sends vectors coming from $\Phi$ to each other, and leaves vectors coming from $\Phi'$ fixed, and similarly for the reflection $\sigma_{(0,\alpha')}$. Finally,

\displaystyle\begin{aligned}(\beta,0)\rtimes(\alpha,0)=\beta\rtimes\alpha&\in\mathbb{Z}\\(0,\beta')\rtimes(0,\alpha')=\beta'\rtimes\alpha'&\in\mathbb{Z}\\(\beta,0)\rtimes(0,\alpha')=(0,\beta')\rtimes(\alpha,0)=0&\in\mathbb{Z}\end{aligned}

All this goes to show that $\Phi\amalg\Phi'$ actually is a root system. As a set, it’s the disjoint union of the two sets of roots.

As a coproduct, we do have the inclusion morphisms $\iota_1:\Phi\rightarrow\Phi\amalg\Phi'$ and $\iota_2:\Phi'\rightarrow\Phi\amalg\Phi'$, which are inherited from the direct sum of $V$ and $V'$. This satisfies the universal condition of a coproduct, since the direct sum does. Indeed, if $\Psi\subseteq U$ is another root system, and if $f:V\rightarrow U$ and $f':V'\rightarrow U$ are linear transformations sending $\Phi$ and $\Phi'$ into $\Psi$, respectively, then $(a,b)\mapsto f(a)+g(b)$ sends $\Phi\amalg\Phi'$ into $\Psi$, and is the unique such transformation compatible with the inclusions.

Interestingly, the Weyl group of the coproduct is the product $\mathcal{W}\times\mathcal{W}'$ of the Weyl groups. Indeed, for every generator $\sigma_\alpha$ of $\mathcal{W}$ and every generator $\sigma_{\alpha'}$ of $\mathcal{W}'$ we get a generator $\sigma_{(\alpha,0)}$. And the two families of generators commute with each other, because each one only acts on the one summand.

On the other hand, there are no product root systems in general! There is only one natural candidate for $\Phi\times\Phi'$ that would be compatible with the projections $\pi_1:V\oplus V'\rightarrow V$ and $\pi_2:V\oplus V'\rightarrow V'$. It’s made up of the points $(\alpha,\alpha')$ for $\alpha\in\Phi$ and $\alpha'\in\Phi'$. But now we must consider how the projections interact with reflections, and it isn’t very pretty.

The projections should act as intertwinors. Specifically, we should have

$\displaystyle\pi_1(\sigma_{(\alpha,\alpha')}(\beta,\beta'))=\sigma_{\pi_1(\alpha,\alpha')}(\pi_1(\beta,\beta'))=\sigma_\alpha(\beta)$

and similarly for the other projection. In other words

$\displaystyle\sigma_{(\alpha,\alpha')}(\beta,\beta')=(\sigma_\alpha(\beta),\sigma_{\alpha'}(\beta'))$

But this isn’t a reflection! Indeed, each reflection has determinant $-1$, and this is the composition of two reflections (one for each component) so it has determinant ${1}$. Thus it cannot be a reflection, and everything comes crashing down.

That all said, the Weyl group of the coproduct root system is the product of the two Weyl groups, and many people are mostly concerned with the Weyl group of symmetries anyway. And besides, the direct sum is just as much a product as it is a coproduct. And so people will often write $\Phi\times\Phi'$ even though it’s really not a product. I won’t write it like that here, but be warned that that notation is out there, lurking.

January 25, 2010

## The Category of Root Systems

As with so many of the objects we study, root systems form a category. If $\Phi$ is a root system in the inner product space $V$, and $\Phi'$ is a root system in the inner product space $V'$, then a morphism from $\Phi$ to $\Phi'$ will be a linear map $\tau:V\rightarrow V'$ so that if $\alpha\in\Phi$ then $\tau(\alpha)\in\Phi'$. Further, we’ll require that $\tau(\beta)\rtimes\tau(\alpha)=\beta\rtimes\alpha$ for all roots $\alpha,\beta\in\Phi$.

Immediately from this, we find that the Weyl group $\mathcal{W}$ of $\Phi$ not only acts on $\Phi$ itself, but on $\Phi'$. Indeed, $\tau$ induces a homomorphism $\mathcal{W}\rightarrow\mathcal{W}'$ that sends the generator $\sigma_\alpha$ to the generator $\sigma_{\tau(\alpha)}$. Even better, $\tau$ actually intertwines these actions! That is, $\sigma_{\tau(\alpha)}(\tau(\beta))=\tau(\sigma_\alpha(\beta))$. Indeed, we can calculate

\displaystyle\begin{aligned}\sigma_{\tau(\alpha)}(\tau(\beta))&=\tau(\beta)-(\tau(\beta)\rtimes\tau(\alpha))\tau(\alpha)\\&=\tau(\beta-(\tau(\beta)\rtimes\tau(\alpha))\alpha)\\&=\tau(\beta-(\beta\rtimes\alpha)\alpha)\\&=\tau(\sigma_\alpha(\beta))\end{aligned}

In particular, we can say that two root systems are isomorphic if there’s an invertible linear transformation $\tau$ sending $\Phi$ to $\Phi'$, and whose inverse $\tau^{-1}$ sends $\Phi'$ back onto $\Phi$. In this case, the intertwining property can be written as an isomorphism of Weyl groups sending $\sigma\in\mathcal{W}$ to $\tau\sigma\tau^{-1}\in\mathcal{W}'$.

Even more particularly, an automorphism of $\Phi$ is an isomorphism from $\Phi$ to itself. That is, it’s an invertible linear transformation from $V$ to itself that leaves $\Phi$ invariant. And so we see that $\mathcal{W}$ itself is a subgroup of $\mathrm{Aut}(\Phi)$. In fact, the Weyl group is a normal subgroup of the automorphism group. That is, given an element $\sigma$ of $\mathcal{W}$ and an automorphism $\tau$ of $\Phi$, the conjugation $\tau\sigma\tau^{-1}$ is again in the Weyl group. And this is exactly what we proved last time!

We can now revise our goal: we want to classify all possible root systems up to isomorphism.

January 22, 2010

## The Weyl Group of a Root System

Let’s take a root system $\Phi$ in the inner product space $V$. Each vector $\alpha$ in $\Phi$ gives rise to a reflection in $\sigma_\alpha\in\mathrm{O}(V)$, the group of transformations preserving the inner product on $V$. So what sorts of transformations can we build up from these reflections? The subgroup of $\mathrm{O}(V)$ generated by the reflections $\sigma_\alpha$ for all $\alpha\in\Phi$ is called the Weyl group $\mathcal{W}$ of the root system. It’s pronounced “vile”, but we don’t mean that as any sort of value judgement.

Anyway, we can also realize $\mathcal{W}$ as a subgroup of the group of permutations on the vectors in $\Phi$. Indeed, by definition each $\sigma_\alpha$ sends each vector in $\Phi$ back to another vector in $\Phi$, and so shuffles them around. So if $\Phi$ has $k$ vectors, the Weyl group can be realized as a subgroup of $S_k$.

In particular, $\mathcal{W}$ is a finite group, as a subgroup of another finite group. In fact, we even know that the number of transformations in $\mathcal{W}$ divides $k!$. It may well (and usually does) have elements which are not of the form $\sigma_\alpha$, but there are still only a finite number of them.

The first thing we want to take note of is how certain transformations in $\mathrm{GL}(V)$ act on $\mathcal{W}$ by conjugation. Specifically, if $\tau$ leaves $\Phi$ invariant, then it induces an automorphism on $\mathcal{W}$ that sends the generator $\sigma_\alpha$ to $\tau\sigma_\alpha\tau^{-1}$ — which (it turns out) is the generator $\sigma_{\tau(\alpha)}$ — for all $\alpha\in\Phi$. Further, it turns out that $\beta\rtimes\alpha=\tau(\beta)\rtimes\tau(\alpha)$ for all $\alpha,\beta\in\Phi$.

Indeed, we can calculate

$\displaystyle\left[\tau\sigma_\alpha\tau^{-1}\right](\tau(\beta))=\tau(\sigma_\alpha(\beta))=\tau(\beta-(\beta\rtimes\alpha)\alpha)=\tau(\beta)-(\beta\rtimes\alpha)\tau(\alpha)$

Now, every vector in $\Phi$ is of the form $\tau(\beta)$ for some $\beta$, and so $\tau\sigma_\alpha\tau^{-1}$ sends it to the vector $\tau(\sigma_\alpha(\beta))$, which is again in $\Phi$, so it leaves $\Phi$ invariant. The transformation $\tau\sigma_\alpha\tau^{-1}$ also fixes every vector in the hyperplane $\tau(P_\alpha)$, for if $\beta$ is orthogonal to $\alpha$, then the above formula shows that $\tau(\beta)$ is left unchanged by the transformation. Finally, $\tau\sigma_\alpha\tau^{-1}$ sends $\tau(\alpha)$ to $-\tau(\alpha)$.

This is all the data we need to invoke our lemma, and conclude that $\tau\sigma_\alpha\tau^{-1}$ is actually equal to $\sigma_{\tau(\alpha)}$. Specifying the action on the generators of $\mathcal{W}$ is enough to determine the whole automorphism. Of course, we can also just let $\sigma$ act on each element of $\mathcal{W}$ by conjugation, but it’s useful to know that the generating reflections are sent to each other exactly as their corresponding vectors are.

Now we can calculate from the definition of a reflection

$\displaystyle\left[\tau\sigma_\alpha\tau^{-1}\right](\tau(\beta))=\sigma_{\tau(\alpha)}(\tau(\beta))=\tau(\beta)-(\tau(\beta)\rtimes\tau(\alpha))\tau(\alpha)$

Comparing this with the equation above, we find that $\tau(\beta)\rtimes\tau(\alpha)=\beta\rtimes\alpha$, as asserted.

January 21, 2010

## Root Systems

Okay, now to lay out the actual objects of our current interest. These are basically collections $\Phi$ of vectors in some inner product space $V$, but with each vector comes a reflection and we want these reflections to play nicely with the vectors themselves. In a way, each point acts as both “program” — an operation to be performed — and “data” — an object to which operations can be applied — and the interplay between these two roles leads to some very interesting structure.

First off, only nonzero vectors give rise to reflections, so we don’t really want zero to be in our collection $\Phi$. We also may as well assume that $\Phi$ spans $V$, because it certainly spans some subspace of $V$ and anything that happens off of this subspace is pretty uninteresting as far as $\Phi$ goes. These are the things that would just be silly not to ask for.

Now, the core requirement is that if $\alpha\in\Phi$, then the reflection $\sigma_\alpha$ should leave $\Phi$ invariant. That is, if $\beta$ is any vector in $\Phi$, then

$\sigma_\alpha(\beta)=\beta-(\beta\rtimes\alpha)\alpha=\beta-\frac{2\langle\beta,\alpha\rangle}{\langle\alpha,\alpha\rangle}\alpha$

is also a vector in $\Phi$. In particular, this means that we have to have $\sigma_\alpha(\alpha)=-\alpha\in\Phi$. But we don’t want any other scalar multiples of $\alpha$ to be in $\Phi$, because they’d just give the same reflection again and that would be redundant.

Of course, we could just throw in more and more vectors as we need to make $\Phi$ invariant under all of its reflections, and each new vector introduces not only new images under the existing reflections, but whole new reflections we have to handle. We want this process to stop after a while, so we’ll insist that $\Phi$ is a finite collection of vectors. This is probably the biggest constraint on our collections.

We have one last condition to add: we want to ask that for every pair of vectors $\alpha$ and $\beta$ in $\Phi$, we have $\beta\rtimes\alpha\in\mathbb{Z}$. In other words, the length of the projection of $\beta$ onto $\alpha$ must be a half-integral multiple of the length of $\alpha$. This makes it so that the displacement from $\beta$ to $\sigma_\alpha(\beta)$ is some integral multiple of $\alpha$. This provides a certain rigidity to our discussion.

So, let’s recap:

• $\Phi$ is a finite, spanning set of vectors in $V$ which does not contain $0\in V$.
• If $\alpha\in\Phi$ then the only scalar multiples of $\alpha$ in $\Phi$ are $\pm\alpha$.
• If $\alpha\in\Phi$ then the reflection $\sigma_\alpha$ leaves $\Phi$ invariant.
• If $\alpha$ and $\beta$ are in $\Phi$, then $\displaystyle\beta\rtimes\alpha=\frac{2\langle\beta,\alpha\rangle}{\langle\alpha,\alpha\rangle}$ is an integer.

A collection $\Phi$ of vectors satisfying all of these conditions is called a “root system”, and the vectors in $\Phi$ are called “roots” for ABSOLUTELY ARBITRARY REASONS THAT HAVE ABSOLUTELY NOTHING TO DO WITH ANYTHING. As far as we’re concerned for now.

So yeah: “root system”. Just ’cause…

Our lofty goal, for the immediate future, is to classify all the possible root systems.

January 20, 2010 Posted by | Geometry, Root Systems | 29 Comments

## A Lemma on Reflections

Here’s a fact we’ll find useful soon enough as we talk about reflections. Hopefully it will also help get back into thinking about linear transformations and inner product spaces. However, if the linear algebra gets a little hairy (or if you’re just joining us) you can just take this fact as given. Remember that we’re looking at a real vector space $V$ equipped with an inner product $\langle\underline{\hphantom{X}},\underline{\hphantom{X}}\rangle$.

Now, let’s say $\Phi$ is some finite collection of vectors which span $V$ (it doesn’t matter if they’re linearly independent or not). Let $\sigma$ be a linear transformation which leaves $\Phi$ invariant. That is, if we pick any vector $\phi\in\Phi$ then the image $\sigma(\phi)$ will be another vector in $\Phi$. Let’s also assume that there is some $n-1$-dimensional subspace $P$ which $\sigma$ leaves completely untouched. That is, $\sigma(v)=v$ for every $v\in P$. Finally, say that there’s some $\alpha\in\Phi$ so that $\sigma(\alpha)=-\alpha$ (clearly $\alpha\notin P$) and also that $\Phi$ is invariant under $\sigma_\alpha$. Then I say that $\sigma=\sigma_\alpha$ and $P=P_\alpha$.

We’ll proceed by actually considering the transformation $\tau=\sigma\sigma_\alpha$, and showing that this is the identity. First off, $\tau$ definitely fixes $\alpha$, since

$\displaystyle\tau(\alpha)=\sigma\left(\sigma_\alpha(\alpha)\right)=\sigma(-\alpha)=-(-\alpha)=\alpha$

so $\tau$ acts as the identity on the line $\mathbb{R}\alpha$. In fact, I assert that $\tau$ also acts as the identity on the quotient space $V/\mathbb{R}\alpha$. Indeed, $\sigma_\alpha$ acts trivially on $P_\alpha$, and every vector in $V/\mathbb{R}\alpha$ has a unique representative in $P_\alpha$. And then $\sigma$ acts trivially on $P$, and every vector in $V/\mathbb{R}\alpha$ has a unique representative in $P$.

This does not, however, mean that $\tau$ acts trivially on any given complement of $\mathbb{R}\alpha$. All we really know at this point is that for every $v\in V$ the difference between $v$ and $\tau(v)$ is some scalar multiple of $\alpha$. On the other hand, remember how we found upper-triangular matrices before. This time we peeled off one vector and the remaining transformation was the identity on the remaining $n-1$-dimensional space. This tells us that all of our eigenvalues are ${1}$, and the characteristic polynomial is $(T-1)^n$, where $n=\dim(V)$. We can evaluate this on the transformation $\tau$ to find that $(\tau-1)^n=0$

Now let’s try to use the collection of vectors $\Phi$. We assumed that both $\sigma$ and $\sigma_\alpha$ send vectors in $\Phi$ back to other vectors in $\Phi$, and so the same must be true of $\tau$. But there are only finitely many vectors (say $k$ of them) in $\Phi$ to begin with, so $\tau$ must act as some sort of permutation of the $k$ vectors in $\Phi$. But every permutation in $S_k$ has an order that divides $k!$. That is, applying $\tau$ $k!$ times must send every vector in $\Phi$ back to itself. But since $\Phi$ is a spanning set for $V$, this means that $\tau^{k!}=1$, or that $\tau^{k!}-1=0$

So we have two polynomial relations satisfied by $\tau$, and $\tau$ will clearly satisfy any linear combination of these relations. But Euclid’s algorithm shows us that we can write the greatest common divisor of these relations as a linear combination, and so $\tau$ must satisfy the greatest common divisor of $T^{k!}-1$ and $(T-1)^n$. It’s not hard to show that this greatest common divisor is $T-1$, which means that we must have $\tau-1=0$ or $\tau=1$.

It’s sort of convoluted, but there are some neat tricks along the way, and we’ll be able to put this result to good use soon.

January 19, 2010

## Reflections

Before introducing my main question for the next series of posts, I’d like to talk a bit about reflections in a real vector space $V$ equipped with an inner product $\langle\underline{\hphantom{X}},\underline{\hphantom{X}}\rangle$. If you want a specific example you can think of the space $\mathbb{R}^n$ consisting of $n$-tuples of real numbers $v=(v^1,\dots,v^n)$. Remember that we’re writing our indices as superscripts, so we shouldn’t think of these as powers of some number $v$, but as the components of a vector. For the inner product, $\langle u,v\rangle$ you can think of the regular “dot product” $\langle u,v\rangle=u^1v^1+\dots+u^nv^n$.

Everybody with me? Good. Now that we’ve got our playing field down, we need to define a reflection. This will be an orthogonal transformation, which is just a fancy way of saying “preserves lengths and angles”. What makes it a reflection is that there’s some $n-1$-dimensional “hyperplane” $P$ that acts like a mirror. Every vector in $P$ itself is just left where it is, and a vector on the line that points perpendicularly to $P$ will be sent to its negative — “reflecting” through the “mirror” of $P$.

Any nonzero vector $\alpha$ spans a line $\mathbb{R}\alpha$, and the orthogonal complement — all the vectors perpendicular to $\alpha$ — forms an $n-1$-dimensional subspace $P_\alpha$, which we can use to make just such a reflection. We’ll write $\sigma_\alpha$ for the reflection determined in this way by $\alpha$. We can easily write down a formula for this reflection:

$\displaystyle\sigma_\alpha(\beta)=\beta-\frac{2\langle\beta,\alpha\rangle}{\langle\alpha,\alpha\rangle}\alpha$

It’s easy to check that if $\beta=c\alpha$ then $\sigma_\alpha(\beta)=-\beta$, while if $\beta$ is perpendicular to $\alpha$ — if $\langle\beta,\alpha\rangle=0$ — then $\sigma_\alpha(\beta)=\beta$, leaving the vector fixed. Thus this formula does satisfy the definition of a reflection through $P_\alpha$.

The amount that reflection moves $\beta$ in the above formula will come up a lot in the near future; enough so we’ll want to give it the notation $\beta\rtimes\alpha$. That is, we define:

$\displaystyle\beta\rtimes\alpha=\frac{2\langle\beta,\alpha\rangle}{\langle\alpha,\alpha\rangle}$

Notice that this is only linear in $\beta$, not in $\alpha$. You might also notice that this is exactly twice the length of the projection of the vector $\beta$ onto the vector $\alpha$. This notation isn’t standard, but the more common notation conflicts with other notational choices we’ve made on this weblog, so I’ve made an executive decision to try it this way.

January 18, 2010