# The Unapologetic Mathematician

## Properties of Irreducible Root Systems III

Today we conclude with our series of lemmas on irreducible root systems.

If $\Phi$ is irreducible, then roots in $\Phi$ have at most two different lengths. Here I mean actual geometric lengths, as measured by the inner product, not the “length” of a Weyl group element. Further, any two roots of the same length can be sent to each other by the action of the Weyl group.

Let $\alpha$ and $\beta$ be two roots. We just saw that the $\mathcal{W}$-orbit of $\alpha$ spans $V$, and so not all the $\sigma(\alpha)$ can be perpendicular to $\beta$. From what we discovered about pairs of roots, we know that if $\langle\alpha,\beta\rangle\neq0$, then the possible ratios of squared lengths $\frac{\lVert\beta\rVert^2}{\lVert\alpha\rVert^2}$ are limited. Indeed, this ratio must be one of $\frac{1}{3}$, $\frac{1}{2}$, ${1}$, ${2}$, or ${3}$.

If there are three distinct root-lengths, let $\alpha$, $\beta$, and $\gamma$ be samples of each length in increasing order. We must then have $\frac{\lVert\beta\rVert^2}{\lVert\alpha\rVert^2}=2$ and $\frac{\lVert\gamma\rVert^2}{\lVert\alpha\rVert^2}=3$, and so $\frac{\lVert\gamma\rVert^2}{\lVert\beta\rVert^2}=\frac{3}{2}$, which clearly violates our conditions. Thus there can be at most two root lengths, as asserted. We call those of the smaller length “short roots”, and the others “long roots”. If there is only one length, we call all the roots long, by convention.

Now let $\alpha$ and $\beta$ have the same length. By using the Weyl group as above, we may assume that these roots are non-orthogonal. We may also assume that they’re distinct, or else we’re already done! By the same data as before, we conclude that $\alpha\rtimes\beta=\beta\rtimes\alpha=\pm1$. We can replace one root by its negative, if need be, and assume that $\alpha\rtimes\beta=1$. Then we may calculate:

$\displaystyle[\sigma_\alpha\sigma_\beta\sigma_\alpha](\beta)=[\sigma_\alpha\sigma_\beta](\beta-\alpha)=\sigma_\alpha(-\beta-\alpha+\beta)=\alpha$.

We may note, in passing, that the unique maximal root $\beta$ is long. Indeed, it suffices to show that $\langle\beta,\beta\rangle\geq\langle\alpha,\alpha\rangle$ for all $\alpha\in\Phi$. We may, without loss of generality, assume $\alpha$ is in the fundamental doman $\overline{\mathfrak{C}(\Delta)}$. Since $\beta-\alpha\succ0$, we must have $\langle\gamma,\beta-\alpha\rangle\geq0$ for any other $\gamma\in\overline{\mathfrak{C}(\Delta)}$. In particular, we have $\langle\beta,\beta-\alpha\rangle\geq0$ and $\langle\alpha,\beta-\alpha\rangle\geq0$. Putting these together, we conclude

$\displaystyle\langle\beta,\beta\rangle\geq\langle\alpha,\beta\rangle\geq\langle\alpha,\alpha\rangle$

and so $\beta$ must be a long root.

February 12, 2010 Posted by | Geometry, Root Systems | 4 Comments

## Properties of Irreducible Root Systems II

We continue with our series of lemmas on irreducible root systems.

If $\Phi$ is irreducible, then the Weyl group $\mathcal{W}$ acts irreducibly on $V$. That is, we cannot decompose the representation of $\mathcal{W}$ on $V$ as the direct sum of two other representations. Even more explicitly, we cannot write $V=W\oplus W'$ for two nontrivial subspaces $W$ and $W'$ with each one of these subspaces invariant under $\mathcal{W}$. If $W$ is an invariant subspace, then the orthogonal complement $W'$ will also be invariant. This is a basic fact about the representation theory of finite groups, which I will simply quote for now, since I haven’t covered that in detail. Thus my assertion is that if $W$ is an invariant subspace under $\mathcal{W}$, then it is either trivial or the whole of $V$.

For any root $\alpha\in\Phi$, either $\alpha\in W$ or $W\subseteq P_\alpha$. Indeed, since $\sigma_\alpha\in\mathcal{W}$, we must have $\sigma_\alpha(W)=W$. As a reflection, $\sigma_\alpha$ breaks $V$ into a one-dimensional eigenspace with eigenvalue $-1$ and another complementary eigenspace with eigenvalue ${1}$. If $W$ contains the $-1$-eigenspace, then $\alpha\in W$. If not, then $\alpha$ is perpendicular to $W$ or $W$ couldn’t be invariant under $\sigma_\alpha$, and in this case $W\subseteq P_\alpha$.

So then if $\alpha$ isn’t in $W$ then it must be in the orthogonal complement $W'$. Thus every root is either in $W$ or in $W'$, and this gives us an orthogonal decomposition of the root system. But since $\Phi$ is irreducible, one or the other of these collections must be empty, and thus $W$ must be either trivial or the whole of $V$.

Even better, the span of the $\mathcal{W}$-orbit of any root $\alpha\in\Phi$ spans $V$. Indeed, the subspace spanned by roots of the form $\sigma(\alpha)$ is invariant under the action of $\mathcal{W}$, and so since $V$ is irreducible it must be either trivial (clearly impossible) or the whole of $V$.

February 11, 2010 Posted by | Geometry, Root Systems | 4 Comments

## Properties of Irreducible Root Systems I

Now we can turn towards the project of classifying irreducible root systems up to isomorphism. And we start with some properties of irreducible root systems.

First off, remember a root system $\Phi$ is reducible if we can write it as the disjoint union of two collections of roots $\Phi=\Psi\uplus\Psi'$ so that each root in $\Psi$ is perpendicular to each one in $\Psi'$. I assert that, for any base $\Delta\subseteq\Phi$, $\Phi$ is reducible if and only if $\Delta$ can itself be broken into two collections $\Delta=\Gamma\uplus\Gamma'$ in just the same way. One direction is easy: if we can decompose $\Phi$, then the roots in $\Delta$ are either in $\Psi$ or $\Psi'$ and we can define $\Gamma=\Delta\cap\Psi$ and $\Gamma'=\Delta\cap\Psi'$.

On the other hand, if we write $\Delta=\Gamma\uplus\Gamma'$ with each simple root in $\Gamma$ perpendicular to each one in $\Gamma'$, then we will find a similar decomposition of $\Phi$. But we know from our study of the Weyl group that every root in $\Phi$ can be sent by the Weyl group to some simple root in $\Delta$. So we define $\Psi$ to be the collection of roots whose orbit includes a point of $\Gamma$, and $\Psi'$ to be the collection of roots whose orbit includes a point of $\Gamma'$. So a vector in $\Psi$ is of the form $\sigma(\alpha)$ for some Weyl group element $\sigma\in\mathcal{W}$ and some simple root $\alpha\in\Gamma$. The Weyl group element $\sigma$ can be written as a sequence of simple reflections. A simple reflection corresponding to a root in $\Gamma$ adds some multiple of that root to the vector, and thus leaves the subspace spanned by $\Gamma$ invariant; while a simple reflection corresponding to a root in $\Gamma'$ leaves the subspace spanned by $\Gamma$ fixed point-by-point. Thus any root in $\Psi$ must lie in the subspace spanned by $\Gamma$, and similarly any root in $\Psi'$ must lie in the subspace spanned by $\Gamma'$. This shows that $\Phi=\Psi\uplus\Psi'$ is exactly the sort of decomposition we’re looking for.

If $\Phi$ is irreducible, with base $\Delta$, then there is a unique maximal root $\beta\in\Phi$ relative to the partial ordering $\prec$ on roots. In particular, the height of $\beta$ is greater than the height of any other root in $\Phi$, $\langle\beta,\alpha\rangle\geq0$ for all simple roots $\alpha\in\Delta$, and in the unique expression

$\displaystyle\beta=\sum\limits_{\alpha\in\Delta}k_\alpha\alpha$

all of the coefficients $k_\alpha$ are strictly positive.

First of all, if $\beta$ is maximal then it’s clearly positive, and so each $k_\alpha$ is either positive or zero. Let $\Gamma$ be the collection of simple roots $\alpha$ so that $k_\alpha>0$, and $\Gamma'$ be the collection of simple roots $\alpha$ so that $k_\alpha=0$. Then $\Delta=\Gamma\uplus\Gamma'$ is a partition of the base. From here we’ll assume that $\Gamma'$ is nonempty and derive a contradiction.

If $\alpha'\in\Gamma'$, then $\langle\alpha,\alpha'\rangle\leq0$ for any other simple root $\alpha\in\Delta$. From this, we can calculate

$\displaystyle\langle\beta,\alpha'\rangle=\left\langle\sum\limits_{\substack{\alpha\in\Delta\\\alpha\neq\alpha'}}k_\alpha\alpha,\alpha'\right\rangle=\sum\limits_{\substack{\alpha\in\Delta\\\alpha\neq\alpha'}}k_\alpha\langle\alpha,\alpha'\rangle\leq0$

Since $\Phi$ is irreducible, there must be at least one $\alpha\in\Gamma$ and $\alpha'\in\Gamma'$ so that $\langle\alpha,\alpha'\rangle<0$, and so we must have $\langle\beta,\alpha'\rangle<0$ for this $\alpha'$. This proves that $\beta+\alpha'$ must also be a root, which contradicts the maximality of $\beta$. And so we conclude that $\Gamma'$ is empty, and all the coefficients $k_\alpha>0$. In passing, we can use the same fact to show that $\langle\beta,\alpha\rangle\geq0$ for all $\alpha\in\Delta$, or else $\beta$ wouldn’t be maximal.

This same argument applies to any other maximal root $\beta'$, giving $\langle\alpha,\beta'\rangle$ for all $\alpha\in\Delta$ with the inequality strict for at least one $\alpha$. We calculate

$\displaystyle\langle\beta,\beta'\rangle=\sum\limits_{\alpha\in\Delta}k_\alpha\langle\alpha,\beta'\rangle>0$

which tells us that $\beta-\beta'$ is a root unless $\beta=\beta'$. But if $\beta-\beta'$ is a root, then either $\beta\prec\beta'$ or $\beta\succ\beta'$, contradicting the assumption that both are maximal. Thus $\beta$ must be unique.

February 10, 2010 Posted by | Geometry, Root Systems | 4 Comments

## The Fundamental Weyl Chamber

When we first discussed Weyl chambers, we defined the fundamental Weyl chamber $\mathfrak{C}(\Delta)$ associated to a base $\Delta$ as the collection of all the vectors $\lambda\in V$ satisfying $\langle\lambda,\alpha\rangle>0$ for all simple roots $\alpha\in\Delta$. Today, I want to discuss the closure $\overline{\mathfrak{C}(\Delta)}$ of this set — allowing $\langle\lambda,\alpha\rangle=0$ — and show that it’s a fundamental domain for the action of the Weyl group $\mathcal{W}$.

To be more explicit, saying that the fundamental Weyl chamber $\overline{\mathfrak{C}(\Delta)}$ is a fundamental domain means that each vector $\mu\in V$ is in the orbit of exactly one vector in $\overline{\mathfrak{C}(\Delta)}$. That is, there is a unique $\lambda\in\overline{\mathfrak{C}(\Delta)}$ so that $\lambda=\sigma(\mu)$ for some $\sigma\in\mathcal{W}$.

First, to existence. Given a vector $\mu\in V$, we consider its orbit — the collection of all the $\sigma(\mu)$ as $\sigma$ runs over all elements of $\mathcal{W}$. We have to find a vector in this orbit which lies in the fundamental Weyl chamber $\overline{\mathfrak{C}(\Delta)}$. To do this, we’ll temporarily extend our partial order to all of $V$ by saying that $\mu\prec\nu$ if $\nu-\mu$ is a nonnegative $\mathbb{R}$-linear combination of simple roots. Relative to this order, pick a maximal vector $\lambda=\sigma(\mu)$; that is, one so that for any $\nu=\tau(\mu)$ we never have $\nu\succ\lambda$. There may well be more than one such maximal vector, given what we’ve said so far, but there will always be at least one.

I say that this $\lambda=\sigma(\mu)$ is actually in the fundamental Weyl chamber. Indeed, if it weren’t then there would be some simple root $\alpha\in\Delta$ so that $\langle\lambda,\alpha,\rangle<0$. But then we could look at the vector $\sigma_\alpha(\lambda)=[\sigma_\alpha\sigma](\mu)\in\mathcal{W}\mu$. We calculate

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

which is a positive $\mathbb{R}$-linear combination of simple roots. Thus $\sigma_\alpha(\lambda)\succ\lambda$, which is impossible by assumption. In fact, this gives us a method for constructing a maximal vector in the orbit. Just start with $\mu$ and form its inner product with all the simple roots. If we find one for which the inner product is negative, reflect the vector through the plane perpendicular to that simple root. Eventually, you’ll end up with a vector in the fundamental Weyl chamber!

Now for uniqueness: if there are two vectors $\lambda_1$ and $\lambda_2$ in the orbit $\mathcal{W}\mu$ that lie within the fundamental Weyl chamber, then we must have $\lambda_1=\sigma(\lambda_2)$ for some $\sigma\in\mathcal{W}$. What I’ll show is that if we have $\lambda_1=\sigma(\lambda_2)$ for two vectors in the fundamental Weyl chamber, then $\sigma$ must be the product of simple reflections which leave $\lambda_2$ fixed, and thus $\lambda_1=\lambda_2$.

We’ll prove this by induction on the length of the Weyl group element $\sigma$. If $l(\sigma)=0$, then $\sigma$ is the identity and the statement is obvious. If $l(\sigma)>0$ then (by the result we proved last time) $\sigma$ must send some positive root to a negative root. In particular, $\sigma$ cannot send all simple roots to positive roots. So let’s say that $\alpha\in\Delta$ is a simple root for which $\sigma(\alpha)\prec0$. Then we observe

$\displaystyle0\geq\langle\lambda_1,\sigma(\alpha)\rangle=\langle\sigma^{-1}(\lambda_1),\alpha\rangle=\langle\lambda_2,\alpha\rangle\geq0$

since $\lambda_1$ and $\lambda_2$ are both in the fundamental Weyl domain. Thus it is forced that $\langle\lambda_2,\alpha\rangle=0$, that $\sigma_\alpha(\lambda_2)=\lambda_2$, and then that $[\sigma\sigma_\alpha](\lambda_2)=\lambda_1$. But $\sigma\sigma_\alpha$ sends fewer positive roots to negative ones than $\sigma$ does, so $l(\sigma\sigma_\alpha) and we can invoke the inductive hypothesis to finish the job.

The upshot of all this is that we know what the space of orbits of $\mathcal{W}$ looks like! It has one point for each vector $\lambda\in\overline{\mathfrak{C}(\Delta)}$. If $\lambda$ is in the interior of this fundamental domain, then the orbit looks just like a copy of $\mathcal{W}$. On the other hand, if $\lambda$ lies on one of the boundary hyperplanes the orbit looks like “half” of the Weyl group. That is, if $\langle\lambda,\alpha\rangle=0$ then $\sigma(\lambda)=[\sigma\sigma_\alpha](\lambda)$, so both of the corresponding group elements “collapse” into one point in this orbit. As $\lambda$ lies on more and more of the boundary hyperplanes, more and more of the orbit “folds up”, until finally at $\lambda=0$ we have an orbit consisting of exactly one point.

February 9, 2010 Posted by | Geometry, Root Systems | 2 Comments

## Lengths of Weyl Group Elements

With our theorem from last time about the Weyl group action, and the lemmas from earlier about simple roots and reflections, we can define a few notions that make discussing Weyl groups easier. Any Weyl group element $\sigma\in\mathcal{W}$ can be written as a composition of simple reflections

$\displaystyle\sigma=\sigma_{\alpha_1}\dots\sigma_{\alpha_t}$

where all $\alpha_k\in\Delta$ are simple roots for some choice of a base $\Delta\subseteq\Phi$. In general we can do this in many ways, and some will have larger values for $t$ than others. But there must be some minimal number of simple reflections it takes to make $\sigma$ — some smallest possible value of $t$. This number we call the “length” $l(\sigma)$ of the root $\sigma$ relative to $\Delta$, and an expression that uses this minimal number of reflections is called “reduced”. By definition we set $l(1)=0$ for the identity element, since we can write it with no reflections at all.

Now, we also have another characterization of the length of a root. Let $n(\sigma)$ be the number of positive roots $\alpha\in\Phi^+$ for which $\sigma(\alpha)\prec0$ — the number of roots that $\sigma$ moves from $\Phi^+$ to $\Phi^-$. I say that $l(\sigma)=n(\sigma)$ for all $\sigma\in\mathcal{W}$, and I’ll proceed by induction on $l(\sigma)$. Indeed, the base case is obvious, since the only element of $\mathcal{W}$ with length zero is the identity, and it sends no positive roots to negative roots.

If this assertion is true for all $\tau\in\mathcal{W}$ with $l(\tau), then we write $\sigma$ in a reduced form as $\sigma_{\alpha_1}\dots\sigma_{\alpha_t}$ and set $\alpha=\alpha_t$. By one of our lemmas, we see that $\sigma(\alpha)\prec0$. By another of our lemmas we know that $\sigma_\alpha$ merely permutes the positive roots other than $\alpha$, and so $n(\sigma\sigma_\alpha)=n(\sigma)-1$. On the other hand, $l(\sigma\sigma_\alpha)=l(\sigma)-1, and our inductive hypothesis allows us to conclude that $l(\sigma\sigma_\alpha)=n(\sigma\sigma_\alpha)$, and thus that $l(\sigma)=n(\sigma)$.

February 8, 2010 Posted by | Geometry, Root Systems | 2 Comments

## The Action of the Weyl Group on Weyl Chambers

With our latest lemmas in hand, we’re ready to describe the action of the Weyl group $\mathcal{W}$ of a root system $\Phi$ on the set of its Weyl chambers. Specifically, the action is “simply transitive”, and the group itself is generated by the reflections corresponding to the simple roots in any given base $\Delta$.

To be a bit more explicit, let $\Delta$ be any fixed base of $\Phi$. Then a number of things happen:

• If $\gamma$ is any regular vector, then there is some $\sigma\in\mathcal{W}$ so that $\langle\sigma(\gamma),\alpha\rangle>0$ for all $\alpha\in\Delta$. That is, $\sigma$ sends the Weyl chamber $\mathfrak{C}(\gamma)$ to the fundamental Weyl chamber $\mathfrak{C}(\Delta)$.
• If $\Delta'$ is another base, then there is some $\sigma\in\mathcal{W}$ so that $\sigma(\Delta')=\Delta$. That is, $\sigma$ sends $\mathfrak{C}(\Delta')$ to $\mathfrak{C}(\Delta)$. We say that the action of the Weyl group is “transitive” on bases and their corresponding Weyl chambers.
• If $\alpha\in\Phi$ is any root, then there is some $\sigma\in\mathcal{W}$ so that $\sigma(\alpha)\in\Delta$.
• The Weyl group $\mathcal{W}$ is generated by the $\sigma_\alpha$ for $\alpha\in\Delta$.
• If $\sigma(\Delta)=\Delta$ for some $\sigma\in\mathcal{W}$, then $\sigma$ is the identity transformation. That is, the only transformation in the Weyl group that sends a base back to itself is the trivial one. We say that the action of the Weyl group is “simple” on bases and their corresponding Weyl chambers.

What we’ll do is let $\mathcal{W}'$ be the group generated by the $\sigma_\alpha$ for $\alpha\in\Delta$, as in the fourth assertion. We’ll show that this group satisfies the first three assertions, and then show that $\mathcal{W}'=\mathcal{W}$.

Let $\gamma$ be a regular vector and write $\delta$ for the half-sum of the positive roots

$\displaystyle\delta=\frac{1}{2}\sum\limits_{\beta\in\Phi^+}\beta$

Choose some $\sigma\in\mathcal{W}'$ so that $\langle\sigma(\gamma),\delta\rangle$ is as large as possible. If $\sigma_\alpha$ is simple, then $\sigma_\alpha\sigma$ is in $\mathcal{W}'$ too, so we find

\displaystyle\begin{aligned}\langle\sigma(\gamma),\delta\rangle&\geq\langle\sigma_\alpha\sigma(\gamma),\delta\rangle\\&=\langle\sigma(\gamma),\sigma_\alpha(\delta)\rangle\\&=\langle\sigma(\gamma),\delta-\alpha\rangle\\&=\langle\sigma(\gamma),\delta\rangle-\langle\sigma(\gamma),\alpha\rangle\end{aligned}

which forces $\langle\sigma(\gamma),\alpha\rangle\geq0$ for all $\alpha\in\Delta$. None of these inner products can actually equal zero, because if one was then we would have $\gamma\in P_\alpha$ and $\gamma$ wouldn’t be regular. Therefore $\sigma(\gamma)$ lies in the fundamental Weyl chamber, as desired.

For the second assertion, we know that there must be some regular $\gamma$ in the positive half-space for each root $\alpha'\in\Delta'$, and the first assertion then applies to send $\Delta'$ to $\Delta$.

For the third assertion, we can invoke the second assertion as long as we know that every root $\alpha\in\Phi$ lies in some base $\Delta'$. We can find some $\gamma\in P_\alpha$ that’s in no other hyperplane perpendicular to another root (other than $-\alpha$). Then pick some close enough $\gamma'$ so that $\langle\gamma',\alpha\rangle=\epsilon>0$, but also $\lvert\langle\gamma',\beta\rangle\rvert>0$ for all $\beta\neq\pm\alpha$. The root $\alpha$ must then belong to the base $\Delta(\gamma')$.

Okay, now let’s show that $\mathcal{W}'=\mathcal{W}$. We just need to show that each reflection $\sigma_\alpha$ for $\alpha\in\Phi$ (all of which together generate $\mathcal{W}$) is an element of $\mathcal{W}'$. But using our third assertion we can find some $\tau\in\mathcal{W}'$ so that $\beta=\tau(\alpha)\in\Delta$. Then

$\displaystyle\sigma_\beta=\sigma_{\tau(\alpha)}=\tau\sigma_\alpha\tau^{-1}$

and so $\sigma_\alpha=\tau^{-1}\sigma_\beta\tau\in\mathcal{W}'$.

Finally, suppose that $\sigma$ is some non-identity element of $\mathcal{W}$ so that $\sigma(\Delta)=\Delta$. Thanks to our fourth assertion we can write $\sigma$ as a string $\sigma_1\dots\sigma_t$ of basic reflections, and we can assume that $t$ is as small as possible. Then we must have $\sigma(\alpha_t)\prec0$ by our final lemma from last time, but we also must have $\sigma(\alpha_t)\in\Delta\subseteq\Phi^+$, which gives us a contradiction.

February 5, 2010 Posted by | Geometry, Root Systems | 7 Comments

## Some Lemmas on Simple Roots

If $\Delta$ is some fixed base of a root system $\Phi$, we call the roots $\alpha\in\Delta$ “simple”. Simple roots have a number of nice properties, some of which we’ll run through now.

First off, if $\alpha\in\Phi^+$ is positive but not simple, then $\alpha-\beta$ is a (positive) root for some simple $\beta\in\Delta$. If $\langle\alpha,\beta\rangle\leq0$ for all $\beta\in\Delta$, then the same argument we used when we showed $\Delta(\gamma)$ is linearly independent would show that $\Delta\cup\{\alpha\}$ is linearly independent. But this is impossible because $\Delta$ is already a basis.

So $\langle\alpha,\beta\rangle>0$ for some $\beta\in\Delta$, and thus $\alpha-\beta\in\Phi$. It must be positive, since the height of $\alpha$ must be at least $2$. That is, at least one coefficient of $\alpha-\beta$ with respect to $\Delta$ must be positive, and so they all are.

In fact, every $\alpha\in\Phi^+$ can be written (not uniquely) as the sum $\beta_1+\dots+\beta_{\mathrm{ht}(\alpha)}$ for a bunch of $\beta_i\in\Delta$, and in such a way that each partial sum $\beta_1+\dots+\beta_k$ is itself a positive root. This is a great proof by induction on $\mathrm{ht}(\alpha)$, for if $\mathrm{ht}(\alpha)=1$ then $\alpha$ is in fact simple itself. If $\alpha$ is not simple, then our argument above gives a $\beta_{\mathrm{ht}(\alpha)}$ so that $\alpha=\alpha'+\beta_{\mathrm{ht}(\alpha)}$ for some $\alpha'\in\Phi^+$ with $\mathrm{ht}(\alpha')=\mathrm{ht}(\alpha)-1$. And so on, by induction.

If $\alpha$ is simple, then the reflection $\sigma_\alpha$ permutes the positive roots other than $\alpha$. That is, if $\alpha\neq\beta\in\Phi^+$, then $\sigma_\alpha(\beta)\in\Phi^+$ as well. Indeed, we write

$\displaystyle\beta=\sum\limits_{\gamma\in\Delta}k_\gamma\gamma$

with all $k_\gamma$ nonnegative. Clearly $k_\gamma\neq0$ for some $\gamma\neq\alpha$ (otherwise $\beta=\alpha$). But the coefficient of $\gamma$ in $\sigma_\alpha(\beta)=\beta-(\beta\rtimes\alpha)\alpha$ must still be $k_\gamma$. Since this is positive, all the coefficients in the decomposition of $\sigma_\alpha(\beta)$ are positive, and so $\sigma_\alpha(\beta)\in\Phi^+$. Further, it can’t be $\alpha$ itself, because $\alpha$ is the image $\sigma_\alpha(-\alpha)$.

In fact, this leads to a particularly useful little trick. Let $\delta$ be the half-sum of all the positive roots. That is,

$\displaystyle\delta=\frac{1}{2}\sum\limits_{\beta\in\Phi^+}\beta$

then $\sigma_\alpha(\delta)=\delta-\alpha$ for all simple roots $\alpha$. The reflection shuffles around all the positive roots other than $\alpha$ itself, which it sends to $-\alpha$. This is a difference in the sum of $-2\alpha$, which the $\frac{1}{2}$ turns into $-\alpha$.

Now take a bunch of $\alpha_1,\dots,\alpha_t\in\Delta$ (not necessarily distinct) and write $\sigma_i=\sigma_{\alpha_i}$. If $\sigma_1\dots\sigma_{t-1}(\alpha_t)\prec0$, then there is some index $1\leq s that we can skip. That is,

$\displaystyle\sigma_1\dots\sigma_t=\sigma_1\dots\sigma_{s-1}\sigma_{s+1}\dots\sigma_{t-1}$

Write $\beta_i=\sigma_{i+1}\dots\sigma_{t-1}(\alpha_t)$ for every $i$ from ${0}$ to $t-2$, and $\beta_{t-1}=\alpha_t$. By our assumption, $\beta_0\prec0$ and $\beta_{t-1}\succ0$. Thus there is some smallest index $s$ so that $\beta_s\succ0$. Then $\sigma_s(\beta_s)\prec0$, and we must have $\beta_s=\alpha_s$. But we know that $\sigma_{\tau(\alpha)}=\tau\sigma_\alpha\tau^{-1}$. In particular,

$\displaystyle\sigma_s=\left(\sigma_{s+1}\dots\sigma_{t-1}\right)\sigma_t\left(\sigma_{s+1}\dots\sigma_{t-1}\right)^{-1}$

And then we can write

\displaystyle\begin{aligned}\sigma_1\dots\sigma_{s-1}(\sigma_s\sigma_{s+1}\dots\sigma_{t-1})\sigma_t&=\sigma_1\dots\sigma_{s-1}(\sigma_{s+1}\dots\sigma_{t-1}\sigma_t)\sigma_t\\&=\sigma_1\dots\sigma_{s-1}\sigma_{s+1}\dots\sigma_{t-1}\end{aligned}

From this we can conclude that if $\sigma=\sigma_1\dots\sigma_t$ is an expression in terms of the basic reflections with $t$ as small as possible, then $\sigma(\alpha_t)\prec0$. Indeed, if $\sigma(\alpha_t)\succ0$, then

$\displaystyle\sigma_1\dots\sigma_{t-1}(\alpha_t)=\sigma_1\dots\sigma_{t-1}(\sigma_t(-\alpha_t))=-\sigma(\alpha_t)\prec0$

and we’ve just seen that in this case we can leave off $\sigma_t$ as well as some $\sigma_s$ in the expression for $\sigma$.

February 4, 2010 Posted by | Geometry, Root Systems | 8 Comments

## Weyl Chambers

A very useful concept in our study of root systems will be that of a Weyl chamber. As we showed at the beginning of last time, the hyperplanes $P_\alpha$ for $\alpha\in\Phi$ cannot fill up all of $V$. What’s left over they chop into a bunch of connected components, which we call Weyl chambers. Thus every regular vector $\gamma$ belongs to exactly one of these Weyl chambers, denoted $\mathfrak{C}(\gamma)$.

Saying that two vectors share a Weyl chamber — that $\mathfrak{C}(\gamma)=\mathfrak{C}(\gamma')$ — tells us that $\gamma$ and $\gamma'$ lie on the same side of each and every hyperplane $P_\alpha$ for $\alpha\in\Phi$. That is, $\langle\gamma,\alpha\rangle$ and $\langle\gamma',\alpha\rangle$ are either both positive or both negative. So this means that $\Phi^+(\gamma)=\Phi^+(\gamma')$, and thus the induced bases are equal: $\Delta(\gamma)=\Delta(\gamma')$. We see, then, that we have a natural bijection between the Weyl chambers of a root system $\Phi$ and the bases for $\Phi$.

We write $\mathfrak{C}(\Delta)=\mathfrak{C}(\gamma)$ for $\Delta=\Delta(\gamma)$ and call this the fundamental Weyl chamber relative to $\Delta$. Geometrically, $\mathfrak{C}(\Delta)$ is the open convex set consisting of the intersection of all the half-spaces $\{\gamma\vert\langle\gamma,\alpha\rangle>0\}$ for $\alpha\in\Delta$.

The Weyl group $\mathcal{W}$ of $\Phi$ shuffles Weyl chambers around. Specifically, if $\sigma\in\mathcal{W}$ and $\gamma$ is regular, then $\sigma(\mathfrak{C}(\gamma))=\mathfrak{C}(\sigma(\gamma))$.

On the other hand, the Weyl group also sends bases of $\Phi$ to each other. If $\Delta\subseteq\Phi$ is a base, then $\sigma(\Delta)$ is another base. Indeed, since $\sigma$ is invertible $\sigma(\Delta)$ will still be a basis for $V$. Further, for any $\beta\in\Phi$ we can write $\beta=\sigma(\beta')$, and then use the base property of $\Delta$ to write $\beta'$ as a nonnegative or nonpositive integral combination of $\Delta$. Hitting everything with $\sigma$ makes $\beta$ a nonnegative or nonpositive integral combination of $\sigma(\Delta)$, and so this is indeed a base.

And, just as we’d hope, these two actions of the Weyl group are equivalent by the bijection above. We have $\sigma(\Delta(\gamma))=\Delta(\sigma(\gamma))$ because $\sigma$ preserves the inner product, and so $\langle\sigma(\gamma),\sigma(\alpha)\rangle=\langle\gamma,\alpha\rangle$. Thus we write $\Delta=\Delta(\gamma)$ for some regular $\gamma$ and find that

\displaystyle\begin{aligned}\sigma(\mathfrak{C}(\Delta))&=\sigma(\mathfrak{C}(\Delta(\gamma)))\\&=\sigma(\mathfrak{C}(\gamma))\\&=\mathfrak{C}(\sigma(\gamma))\\&=\mathfrak{C}(\Delta(\sigma(\gamma)))\\&=\mathfrak{C}(\sigma(\Delta(\gamma)))\\&=\mathfrak{C}(\sigma(\Delta))\end{aligned}

February 3, 2010 Posted by | Geometry, Root Systems | 3 Comments

## The Existence of Bases for Root Systems

We’ve defined what a base for a root system is, but we haven’t provided any evidence yet that they even exist. Today we’ll not only see that every root system has a base, but we’ll show how all possible bases arise. This will be sort of a long and dense one.

First of all, we observe that any hyperplane has measure zero, and so any finite collection of them will too. Thus the collection of all the hyperplanes $P_\alpha$ perpendicular to vectors $\alpha\in\Phi$ cannot fill up all of $V$. We call vectors in one of these hyperplanes “singular”, and vectors in none of them “regular”.

When $\gamma$ is regular, it divides $\Phi$ into two collections. A vector $\alpha$ is in $\Phi^+(\gamma)$ if $\alpha\in\Phi$ and $\langle\alpha,\gamma\rangle>0$, and we have a similar definition for $\Phi^-(\gamma)$. It should be clear that $\Phi^-(\gamma)=-\Phi^+(\gamma)$, and that every vector $\alpha\in\Phi$ is in one or the other; otherwise $\gamma$ would be in $P_\alpha$. For a regular $\gamma$, we say that $\alpha\in\Phi^+(\gamma)$ is “decomposable” if $\alpha=\beta_1+\beta_2$ for $\beta_1,\beta_2\in\Phi^+(\gamma)$. Otherwise, we say that $\alpha$ is “indecomposable”.

Now we can state our existence theorem. Given a regular $\gamma$, let $\Delta(\gamma)$ be the set of indecomposable roots in $\Phi^+(\gamma)$. Then $\Delta(\gamma)$ is a base of $\Phi$, and every base of $\Phi$ arises in this manner. We will prove this in a number of steps.

First off, every vector in $\Phi^+(\gamma)$ is a nonnegative integral linear combination of the vectors in $\Delta(\gamma)$. Otherwise there is some $\alpha\in\Phi^+(\gamma)$ that can’t be written like that, and we can choose $\alpha$ so that $\langle\gamma,\alpha\rangle$ is as small as possible. $\alpha$ itself can’t be indecomposable, so we must have $\alpha=\beta_1+\beta_2$ for some two vectors $\beta_1,\beta_2\in\Phi^+(\gamma)$, and so $\langle\gamma,\alpha\rangle=\langle\gamma,\beta_1\rangle+\langle\gamma,\beta_2\rangle$. Each of these two inner products are strictly positive, so to avoid contradicting the minimality of $\langle\gamma,\alpha\rangle$ we must be able to write each of $\beta_1$ and $\beta_2$ as a nonnegative linear combination of vectors in $\Delta(\gamma)$. But then we can write $\alpha$ in this form after all! The assertion follows.

Second, if $\alpha$ and $\beta$ are distinct vectors in $\Delta(\gamma)$ then $\langle\alpha,\beta\rangle\leq0$. Indeed, by our lemma if $\langle\alpha,\beta\rangle>0$ then $\alpha-\beta\in\Phi$. And so either $\alpha-\beta$ or $\beta-\alpha$ lies in $\Phi^+(\gamma)$. In the first case, we can write $\alpha=\beta+(\alpha-\beta)$, so $\alpha$ is decomposable. In the second case, we can similarly show that $\beta$ is decomposable. And thus we have a contradiction and the assertion follows.

Next, $\Delta(\gamma)$ is linearly independent. If we have a linear combination

$\displaystyle\sum\limits_{\alpha\in\Delta(\gamma)}r_\alpha\alpha=0$

then we can separate out the vectors $\alpha$ for which the coefficient $r_\alpha>0$ and those $\beta$ for which $r_\beta<0$, and write

$\displaystyle\sum\limits_\alpha s_\alpha\alpha=\sum\limits_\beta t_\beta\beta$

with all coefficients positive. Call this common sum $\epsilon$ and calculate

$\displaystyle\langle\epsilon,\epsilon\rangle=\sum\limits_{\alpha,\beta}s_\alpha t_\beta\langle\alpha,\beta\rangle$

Since each $\langle\alpha,\beta\rangle\leq0$, this whole sum must be nonpositive, which can only happen if $\epsilon=0$. But then

$\displaystyle0=\langle\gamma,\epsilon\rangle=\sum\limits_\alpha s_\alpha\langle\gamma,\alpha\rangle$

which forces all the $s_\alpha=0$. Similarly, all the $t_\beta=0$, and thus the original linear combination must have been trivial. Thus $\Delta(\gamma)$ is linearly independent.

Now we can show that $\Delta(\gamma)$ is a base. Every vector in $\Phi^+(\gamma)$ is indeed a nonnegative integral linear combination of the vectors in $\Delta(\gamma)$. Since $\Phi^-(\gamma)=-\Phi^+(\gamma)$, every vector in this set is a nonpositive integral linear combination of the vectors in $\Delta(\gamma)$. And every vector in $\Phi$ is in one or the other of these sets. Also, since $\Phi$ spans $V$ we find that $\Delta(\gamma)$ spans $V$ as well. But since it’s linearly independent, it must be a basis. And so it satisfies both of the criteria to be a base.

Finally, every base $\Delta$ is of the form $\Delta(\gamma)$ for some regular $\gamma$. Indeed, we just have to find some $\gamma$ for which $\langle\gamma,\alpha\rangle>0$ for each $\alpha\in\Delta$. Then since any $\beta\in\Phi$ is an integral linear combination of $\alpha\in\Delta$ we can verify that $\langle\gamma,\beta\rangle\neq0$ for all $\beta\in\Phi$, proving that $\gamma$ is regular. and $\Phi^+=\Phi^+(\gamma)$. Then the vectors $\alpha\in\Delta$ are clearly indecomposable, showing that $\Delta\subseteq\Delta(\gamma)$. But these sets contain the same number of elements since they’re both bases of $V$, and so $\Delta=\Delta(\gamma)$.

The only loose end is showing that such a $\gamma$ exists. I’ll actually go one better and show that for any basis $\{\eta_i\}_{i=1}^{\dim(V)}$ the intersection of the “half-spaces” $\{\gamma\vert\langle\gamma,\eta_i\rangle\}$ is nonempty. To see this, define

$\displaystyle\delta_i=\eta_i-\sum\limits_{\substack{1\leq j\leq\dim(V)\\j\neq i}}\frac{\langle\eta_i,\eta_j\rangle}{\langle\eta_j,\eta_j\rangle}\eta_j$

This is what’s left of the basis vector $\eta_i$ after subtracting off its projection onto each of the other basis vectors $\eta_j$, leaving its projection onto the line perpendicular to all of them. Then consider the vector $\gamma=r^i\delta_i$ where each $r^i>0$. It’s a straightforward computation to show that $\langle\gamma,\eta_k\rangle=r^i\langle\delta_k,\eta_k\rangle>0$, and so $\gamma$ is just such a vector as we’re claiming exists.

February 2, 2010 Posted by | Geometry, Root Systems | 5 Comments

## Bases for Root Systems

We don’t always want to deal with a whole root system $\Phi\subseteq V$. Indeed, that’s sort of like using a whole group when all the information is contained in some much smaller generating set. For a vector space we call such a small generating set a basis. For a root system, we call it a base. Specifically, a subset $\Delta\subseteq\Phi$ is called a base if first of all $\Delta$ is a basis for $V$, and if each vector $\beta\in\Phi$ can be written as a linear combination

$\displaystyle\beta=\sum\limits_{\alpha\in\Delta}k_\alpha\alpha$

where the coefficients $k_\alpha$ are either all nonnegative integers or all nonpositive integers.

Some observations are immediate. Because $\Delta$ is a basis, it contains exactly $n=\dim(V)$ vectors of $\Phi$. It also tells us that the decomposition of each $\beta$ is unique. In fact, as for any basis, every vector in $V$ can be written uniquely as a linear combination of the vectors in $\Delta$. What we’re emphasizing here is that for vectors in $\Phi$, the coefficients are all integers, and they’re either all nonnegative or all nonpositive.

Another thing a choice of base gives us is a partial order $\preceq$ on the root system $\Phi$. We say that $\beta$ is a “positive root” with respect to $\Delta$ (and write $\beta\succeq0$) if all of its coefficients are nonnegative integers. Similarly, we say that $\beta$ is a “negative root” with respect to $\Delta$ (and write $\beta\preceq0$) if all of its coefficients are nonpositive integers. We extend this to a partial order by defining $\beta\preceq\alpha$ if $\beta-\alpha\preceq0$.

Every root is either positive or negative. We write $\Phi^+$ for the collection of positive roots with respect to a base $\Delta$ and $\Phi^-$ for the collection of negative roots. It should be clear that $\Delta\subseteq\Phi^+$, and also that $\Phi^-=-\Phi^+$ — the negative roots are exactly the negatives of the positive roots.

We can also define a kind of size of a vector $\beta\in\Phi$. Given the above (unique) decomposition, we define the “height” of $\beta$ relative to $\Delta$ as

$\displaystyle\mathrm{ht}(\beta)=\sum\limits_{\alpha\in\Delta}k_\alpha$

This will be useful when it comes to proving statements about all vectors in $\Phi^+$ by induction on their heights.

If $\alpha\neq\beta$ are two vectors in a base $\Delta\subseteq\Phi$, then we know that $\langle\alpha,\beta\rangle\leq0$ and $\alpha-\beta\notin\Phi$. Indeed, our lemma tells us that if $\langle\alpha,\beta\rangle>0$ then $\alpha-\beta$ would be in $\Phi$. But this is impossible, because every vector in $\Phi$ can only be written as a linear combination of vectors in $\Delta$ in one way, and that way cannot have some positive signs and some negative signs like $\alpha-\beta$ does.

What this tells us (among other things) is that $\beta$ must be one end of the $\alpha$ root string through $\beta$. The other end must be $\sigma_\alpha(\beta)$, and the root string must be unbroken between these two ends. Every vector $\beta+k\alpha$ with $0\leq k\leq-\beta\rtimes\alpha$ must be in $\Phi^+$.

February 1, 2010 Posted by | Geometry, Root Systems | 17 Comments