First, a lemma: given a root system , let and be nonproportional roots. That is, . Then if — if the angle between the vectors is strictly acute — then is also a root. On the other hand, if then is also a root. We know that is a root. If we replace with we can see that the second of these assertions follows from the first, so we just need to prove that one.
We know that is positive if and only if is. So let’s look at the table we worked up last time. We see that when is positive, then either that or equals . If it’s , then
is a root. On the other hand, if then is a root.
So given two nonproportional and nonorthogonal roots and we’re guaranteed to have more than one vector of the form for some integer in the root system . We call the collection of all such vectors in the -string through .
Let be the largest integer so that , and let be the largest integer so that . I say that the root string is unbroken. That is, for all .
Indeed, if there’s some integer so that , then we can find and with so that
But then the above lemma tells us that , while . Subtracting, we find
The two inequalities tell us that this difference should be positive, but is positive and is negative. Thus we have a contradiction, and the root string must be unbroken from to .
We can also tell that just adds a positive or negative multiple of to any root. Then it’s clear from the geometry that just reverses a root string end to end. That is, . But we can also calculate
Thus . And so the length of the string through can be no more than .
When we look at a root system, the integrality condition 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
Then we can consider the product
This has to be an integer between and . And it can only hit if is a multiple of , in which case we already know that must be . So some simple counting shows the possibilities. Without loss of generality we’ll assume that .
We see that if and 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 and span some two-dimensional space. We can choose an orthogonal basis (and thus coordinates) that makes , and then . We calculate
The first of these conditions says that for some integer . We can manipulate the second to tell us that for some integer . Let’s plot some of these!
The vector pointing to is . The blue vertical lines correspond to the first condition for , while the red circles correspond to the second condition for . Any larger values of will put lines further out than any of the circles can touch, while any larger values of will put smaller circles further in than any line can touch. Thus, all possible values of 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 . We can see that if we project any of the examples onto , the projection is a half-integral multiple of . Conversely, projecting 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 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.
Given a root system in an inner product space we may be able to partition it into two collections so that each root in is orthogonal to every root in and vice versa. In this case, the subspace spanned by and the subspace spanned by are orthogonal to each other, and together they span the whole of . Thus we can write , and thus is the coproduct .
In this situation, we say that is a “reducible” root system. On the other hand, if there is no way of writing 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 , and this is definitely irreducible. On the other hand, given an -dimensional root system , either it’s irreducible or not. If it is, we’re done. But if not, then we can break it into and , each of which has dimension strictly less than . 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 . Suppose that we have two different decompositions, and . In each case, every root in is actually in exactly one of the subspaces and exactly one of the subspaces , thus partitioning the root system on the one hand as and .
All we have to do is define . From here we can see that
Further, each root in is in exactly one of the , 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!
Given a root system , there’s a very interesting related root system , called the “dual” or “inverse” root system. It’s made up of the “duals” , defined by
This is the vector that represents the linear functional . That is, .
The dual root is proportional to , and so . The dual reflections are the same as the original reflections, and so they generate the same subgroup of . That is, the Weyl group of is the same as the Weyl group of .
As we should hope, dualizing twice gives back the original root system. That is, . We can even show that . Indeed, we calculate
It turns out that passing to duals reverses the roles of roots, in a way, just as we might expect from a dualization. Specifically, . Indeed, we calculate
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 and in inner-product spaces and , we take the direct sum of the vector spaces, which makes vectors from each vector space orthogonal to vectors from the other one.
The coproduct root system consists of the vectors of the form for and for . Indeed, this collection is finite, spans , and does not contain . The only multiples of any given vector in are that vector and its negative. The reflection sends vectors coming from to each other, and leaves vectors coming from fixed, and similarly for the reflection . Finally,
All this goes to show that 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 and , which are inherited from the direct sum of and . This satisfies the universal condition of a coproduct, since the direct sum does. Indeed, if is another root system, and if and are linear transformations sending and into , respectively, then sends into , and is the unique such transformation compatible with the inclusions.
Interestingly, the Weyl group of the coproduct is the product of the Weyl groups. Indeed, for every generator of and every generator of we get a generator . 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 that would be compatible with the projections and . It’s made up of the points for and . 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
and similarly for the other projection. In other words
But this isn’t a reflection! Indeed, each reflection has determinant , and this is the composition of two reflections (one for each component) so it has determinant . 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 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.
As with so many of the objects we study, root systems form a category. If is a root system in the inner product space , and is a root system in the inner product space , then a morphism from to will be a linear map so that if then . Further, we’ll require that for all roots .
Immediately from this, we find that the Weyl group of not only acts on itself, but on . Indeed, induces a homomorphism that sends the generator to the generator . Even better, actually intertwines these actions! That is, . Indeed, we can calculate
In particular, we can say that two root systems are isomorphic if there’s an invertible linear transformation sending to , and whose inverse sends back onto . In this case, the intertwining property can be written as an isomorphism of Weyl groups sending to .
Even more particularly, an automorphism of is an isomorphism from to itself. That is, it’s an invertible linear transformation from to itself that leaves invariant. And so we see that itself is a subgroup of . In fact, the Weyl group is a normal subgroup of the automorphism group. That is, given an element of and an automorphism of , the conjugation 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.
Let’s take a root system in the inner product space . Each vector in gives rise to a reflection in , the group of transformations preserving the inner product on . So what sorts of transformations can we build up from these reflections? The subgroup of generated by the reflections for all is called the Weyl group 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 as a subgroup of the group of permutations on the vectors in . Indeed, by definition each sends each vector in back to another vector in , and so shuffles them around. So if has vectors, the Weyl group can be realized as a subgroup of .
In particular, is a finite group, as a subgroup of another finite group. In fact, we even know that the number of transformations in divides . It may well (and usually does) have elements which are not of the form , but there are still only a finite number of them.
The first thing we want to take note of is how certain transformations in act on by conjugation. Specifically, if leaves invariant, then it induces an automorphism on that sends the generator to — which (it turns out) is the generator — for all . Further, it turns out that for all .
Indeed, we can calculate
Now, every vector in is of the form for some , and so sends it to the vector , which is again in , so it leaves invariant. The transformation also fixes every vector in the hyperplane , for if is orthogonal to , then the above formula shows that is left unchanged by the transformation. Finally, sends to .
This is all the data we need to invoke our lemma, and conclude that is actually equal to . Specifying the action on the generators of is enough to determine the whole automorphism. Of course, we can also just let act on each element of 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
Comparing this with the equation above, we find that , as asserted.
Okay, now to lay out the actual objects of our current interest. These are basically collections of vectors in some inner product space , 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 . We also may as well assume that spans , because it certainly spans some subspace of and anything that happens off of this subspace is pretty uninteresting as far as goes. These are the things that would just be silly not to ask for.
Now, the core requirement is that if , then the reflection should leave invariant. That is, if is any vector in , then
is also a vector in . In particular, this means that we have to have . But we don’t want any other scalar multiples of to be in , 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 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 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 and in , we have . In other words, the length of the projection of onto must be a half-integral multiple of the length of . This makes it so that the displacement from to is some integral multiple of . This provides a certain rigidity to our discussion.
So, let’s recap:
- is a finite, spanning set of vectors in which does not contain .
- If then the only scalar multiples of in are .
- If then the reflection leaves invariant.
- If and are in , then is an integer.
A collection of vectors satisfying all of these conditions is called a “root system”, and the vectors in 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.
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 equipped with an inner product .
Now, let’s say is some finite collection of vectors which span (it doesn’t matter if they’re linearly independent or not). Let be a linear transformation which leaves invariant. That is, if we pick any vector then the image will be another vector in . Let’s also assume that there is some -dimensional subspace which leaves completely untouched. That is, for every . Finally, say that there’s some so that (clearly ) and also that is invariant under . Then I say that and .
We’ll proceed by actually considering the transformation , and showing that this is the identity. First off, definitely fixes , since
so acts as the identity on the line . In fact, I assert that also acts as the identity on the quotient space . Indeed, acts trivially on , and every vector in has a unique representative in . And then acts trivially on , and every vector in has a unique representative in .
This does not, however, mean that acts trivially on any given complement of . All we really know at this point is that for every the difference between and is some scalar multiple of . 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 -dimensional space. This tells us that all of our eigenvalues are , and the characteristic polynomial is , where . We can evaluate this on the transformation to find that
Now let’s try to use the collection of vectors . We assumed that both and send vectors in back to other vectors in , and so the same must be true of . But there are only finitely many vectors (say of them) in to begin with, so must act as some sort of permutation of the vectors in . But every permutation in has an order that divides . That is, applying times must send every vector in back to itself. But since is a spanning set for , this means that , or that
So we have two polynomial relations satisfied by , and 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 must satisfy the greatest common divisor of and . It’s not hard to show that this greatest common divisor is , which means that we must have or .
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.
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 equipped with an inner product . If you want a specific example you can think of the space consisting of -tuples of real numbers . Remember that we’re writing our indices as superscripts, so we shouldn’t think of these as powers of some number , but as the components of a vector. For the inner product, you can think of the regular “dot product” .
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 -dimensional “hyperplane” that acts like a mirror. Every vector in itself is just left where it is, and a vector on the line that points perpendicularly to will be sent to its negative — “reflecting” through the “mirror” of .
Any nonzero vector spans a line , and the orthogonal complement — all the vectors perpendicular to — forms an -dimensional subspace , which we can use to make just such a reflection. We’ll write for the reflection determined in this way by . We can easily write down a formula for this reflection:
It’s easy to check that if then , while if is perpendicular to — if — then , leaving the vector fixed. Thus this formula does satisfy the definition of a reflection through .
The amount that reflection moves in the above formula will come up a lot in the near future; enough so we’ll want to give it the notation . That is, we define:
Notice that this is only linear in , not in . You might also notice that this is exactly twice the length of the projection of the vector onto the vector . 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.