If is irreducible, then roots in 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 and be two roots. We just saw that the -orbit of spans , and so not all the can be perpendicular to . From what we discovered about pairs of roots, we know that if , then the possible ratios of squared lengths are limited. Indeed, this ratio must be one of , , , , or .
If there are three distinct root-lengths, let , , and be samples of each length in increasing order. We must then have and , and so , 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 and 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 . We can replace one root by its negative, if need be, and assume that . Then we may calculate:
We may note, in passing, that the unique maximal root is long. Indeed, it suffices to show that for all . We may, without loss of generality, assume is in the fundamental doman . Since , we must have for any other . In particular, we have and . Putting these together, we conclude
and so must be a long root.
If is irreducible, then the Weyl group acts irreducibly on . That is, we cannot decompose the representation of on as the direct sum of two other representations. Even more explicitly, we cannot write for two nontrivial subspaces and with each one of these subspaces invariant under . If is an invariant subspace, then the orthogonal complement 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 is an invariant subspace under , then it is either trivial or the whole of .
For any root , either or . Indeed, since , we must have . As a reflection, breaks into a one-dimensional eigenspace with eigenvalue and another complementary eigenspace with eigenvalue . If contains the -eigenspace, then . If not, then is perpendicular to or couldn’t be invariant under , and in this case .
So then if isn’t in then it must be in the orthogonal complement . Thus every root is either in or in , and this gives us an orthogonal decomposition of the root system. But since is irreducible, one or the other of these collections must be empty, and thus must be either trivial or the whole of .
Even better, the span of the -orbit of any root spans . Indeed, the subspace spanned by roots of the form is invariant under the action of , and so since is irreducible it must be either trivial (clearly impossible) or the whole of .
First off, remember a root system is reducible if we can write it as the disjoint union of two collections of roots so that each root in is perpendicular to each one in . I assert that, for any base , is reducible if and only if can itself be broken into two collections in just the same way. One direction is easy: if we can decompose , then the roots in are either in or and we can define and .
On the other hand, if we write with each simple root in perpendicular to each one in , then we will find a similar decomposition of . But we know from our study of the Weyl group that every root in can be sent by the Weyl group to some simple root in . So we define to be the collection of roots whose orbit includes a point of , and to be the collection of roots whose orbit includes a point of . So a vector in is of the form for some Weyl group element and some simple root . The Weyl group element can be written as a sequence of simple reflections. A simple reflection corresponding to a root in adds some multiple of that root to the vector, and thus leaves the subspace spanned by invariant; while a simple reflection corresponding to a root in leaves the subspace spanned by fixed point-by-point. Thus any root in must lie in the subspace spanned by , and similarly any root in must lie in the subspace spanned by . This shows that is exactly the sort of decomposition we’re looking for.
If is irreducible, with base , then there is a unique maximal root relative to the partial ordering on roots. In particular, the height of is greater than the height of any other root in , for all simple roots , and in the unique expression
all of the coefficients are strictly positive.
First of all, if is maximal then it’s clearly positive, and so each is either positive or zero. Let be the collection of simple roots so that , and be the collection of simple roots so that . Then is a partition of the base. From here we’ll assume that is nonempty and derive a contradiction.
If , then for any other simple root . From this, we can calculate
Since is irreducible, there must be at least one and so that , and so we must have for this . This proves that must also be a root, which contradicts the maximality of . And so we conclude that is empty, and all the coefficients . In passing, we can use the same fact to show that for all , or else wouldn’t be maximal.
This same argument applies to any other maximal root , giving for all with the inequality strict for at least one . We calculate
which tells us that is a root unless . But if is a root, then either or , contradicting the assumption that both are maximal. Thus must be unique.
When we first discussed Weyl chambers, we defined the fundamental Weyl chamber associated to a base as the collection of all the vectors satisfying for all simple roots . Today, I want to discuss the closure of this set — allowing — and show that it’s a fundamental domain for the action of the Weyl group .
To be more explicit, saying that the fundamental Weyl chamber is a fundamental domain means that each vector is in the orbit of exactly one vector in . That is, there is a unique so that for some .
First, to existence. Given a vector , we consider its orbit — the collection of all the as runs over all elements of . We have to find a vector in this orbit which lies in the fundamental Weyl chamber . To do this, we’ll temporarily extend our partial order to all of by saying that if is a nonnegative -linear combination of simple roots. Relative to this order, pick a maximal vector ; that is, one so that for any we never have . 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 is actually in the fundamental Weyl chamber. Indeed, if it weren’t then there would be some simple root so that . But then we could look at the vector . We calculate
which is a positive -linear combination of simple roots. Thus , which is impossible by assumption. In fact, this gives us a method for constructing a maximal vector in the orbit. Just start with 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 and in the orbit that lie within the fundamental Weyl chamber, then we must have for some . What I’ll show is that if we have for two vectors in the fundamental Weyl chamber, then must be the product of simple reflections which leave fixed, and thus .
We’ll prove this by induction on the length of the Weyl group element . If , then is the identity and the statement is obvious. If then (by the result we proved last time) must send some positive root to a negative root. In particular, cannot send all simple roots to positive roots. So let’s say that is a simple root for which . Then we observe
since and are both in the fundamental Weyl domain. Thus it is forced that , that , and then that . But sends fewer positive roots to negative ones than does, so 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 looks like! It has one point for each vector . If is in the interior of this fundamental domain, then the orbit looks just like a copy of . On the other hand, if lies on one of the boundary hyperplanes the orbit looks like “half” of the Weyl group. That is, if then , so both of the corresponding group elements “collapse” into one point in this orbit. As lies on more and more of the boundary hyperplanes, more and more of the orbit “folds up”, until finally at we have an orbit consisting of exactly one point.
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 can be written as a composition of simple reflections
where all are simple roots for some choice of a base . In general we can do this in many ways, and some will have larger values for than others. But there must be some minimal number of simple reflections it takes to make — some smallest possible value of . This number we call the “length” of the root relative to , and an expression that uses this minimal number of reflections is called “reduced”. By definition we set 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 be the number of positive roots for which — the number of roots that moves from to . I say that for all , and I’ll proceed by induction on . Indeed, the base case is obvious, since the only element of with length zero is the identity, and it sends no positive roots to negative roots.
If this assertion is true for all with , then we write in a reduced form as and set . By one of our lemmas, we see that . By another of our lemmas we know that merely permutes the positive roots other than , and so . On the other hand, , and our inductive hypothesis allows us to conclude that , and thus that .
With our latest lemmas in hand, we’re ready to describe the action of the Weyl group of a root system 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 .
To be a bit more explicit, let be any fixed base of . Then a number of things happen:
- If is any regular vector, then there is some so that for all . That is, sends the Weyl chamber to the fundamental Weyl chamber .
- If is another base, then there is some so that . That is, sends to . We say that the action of the Weyl group is “transitive” on bases and their corresponding Weyl chambers.
- If is any root, then there is some so that .
- The Weyl group is generated by the for .
- If for some , then 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 be the group generated by the for , as in the fourth assertion. We’ll show that this group satisfies the first three assertions, and then show that .
Let be a regular vector and write for the half-sum of the positive roots
Choose some so that is as large as possible. If is simple, then is in too, so we find
which forces for all . None of these inner products can actually equal zero, because if one was then we would have and wouldn’t be regular. Therefore lies in the fundamental Weyl chamber, as desired.
For the second assertion, we know that there must be some regular in the positive half-space for each root , and the first assertion then applies to send to .
For the third assertion, we can invoke the second assertion as long as we know that every root lies in some base . We can find some that’s in no other hyperplane perpendicular to another root (other than ). Then pick some close enough so that , but also for all . The root must then belong to the base .
Okay, now let’s show that . We just need to show that each reflection for (all of which together generate ) is an element of . But using our third assertion we can find some so that . Then
and so .
Finally, suppose that is some non-identity element of so that . Thanks to our fourth assertion we can write as a string of basic reflections, and we can assume that is as small as possible. Then we must have by our final lemma from last time, but we also must have , which gives us a contradiction.
First off, if is positive but not simple, then is a (positive) root for some simple . If for all , then the same argument we used when we showed is linearly independent would show that is linearly independent. But this is impossible because is already a basis.
So for some , and thus . It must be positive, since the height of must be at least . That is, at least one coefficient of with respect to must be positive, and so they all are.
In fact, every can be written (not uniquely) as the sum for a bunch of , and in such a way that each partial sum is itself a positive root. This is a great proof by induction on , for if then is in fact simple itself. If is not simple, then our argument above gives a so that for some with . And so on, by induction.
If is simple, then the reflection permutes the positive roots other than . That is, if , then as well. Indeed, we write
with all nonnegative. Clearly for some (otherwise ). But the coefficient of in must still be . Since this is positive, all the coefficients in the decomposition of are positive, and so . Further, it can’t be itself, because is the image .
In fact, this leads to a particularly useful little trick. Let be the half-sum of all the positive roots. That is,
then for all simple roots . The reflection shuffles around all the positive roots other than itself, which it sends to . This is a difference in the sum of , which the turns into .
Now take a bunch of (not necessarily distinct) and write . If , then there is some index that we can skip. That is,
Write for every from to , and . By our assumption, and . Thus there is some smallest index so that . Then , and we must have . But we know that . In particular,
And then we can write
From this we can conclude that if is an expression in terms of the basic reflections with as small as possible, then . Indeed, if , then
and we’ve just seen that in this case we can leave off as well as some in the expression for .
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 for cannot fill up all of . What’s left over they chop into a bunch of connected components, which we call Weyl chambers. Thus every regular vector belongs to exactly one of these Weyl chambers, denoted .
Saying that two vectors share a Weyl chamber — that — tells us that and lie on the same side of each and every hyperplane for . That is, and are either both positive or both negative. So this means that , and thus the induced bases are equal: . We see, then, that we have a natural bijection between the Weyl chambers of a root system and the bases for .
We write for and call this the fundamental Weyl chamber relative to . Geometrically, is the open convex set consisting of the intersection of all the half-spaces for .
The Weyl group of shuffles Weyl chambers around. Specifically, if and is regular, then .
On the other hand, the Weyl group also sends bases of to each other. If is a base, then is another base. Indeed, since is invertible will still be a basis for . Further, for any we can write , and then use the base property of to write as a nonnegative or nonpositive integral combination of . Hitting everything with makes a nonnegative or nonpositive integral combination of , 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 because preserves the inner product, and so . Thus we write for some regular and find that
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 perpendicular to vectors cannot fill up all of . We call vectors in one of these hyperplanes “singular”, and vectors in none of them “regular”.
When is regular, it divides into two collections. A vector is in if and , and we have a similar definition for . It should be clear that , and that every vector is in one or the other; otherwise would be in . For a regular , we say that is “decomposable” if for . Otherwise, we say that is “indecomposable”.
Now we can state our existence theorem. Given a regular , let be the set of indecomposable roots in . Then is a base of , and every base of arises in this manner. We will prove this in a number of steps.
First off, every vector in is a nonnegative integral linear combination of the vectors in . Otherwise there is some that can’t be written like that, and we can choose so that is as small as possible. itself can’t be indecomposable, so we must have for some two vectors , and so . Each of these two inner products are strictly positive, so to avoid contradicting the minimality of we must be able to write each of and as a nonnegative linear combination of vectors in . But then we can write in this form after all! The assertion follows.
Second, if and are distinct vectors in then . Indeed, by our lemma if then . And so either or lies in . In the first case, we can write , so is decomposable. In the second case, we can similarly show that is decomposable. And thus we have a contradiction and the assertion follows.
Next, is linearly independent. If we have a linear combination
then we can separate out the vectors for which the coefficient and those for which , and write
with all coefficients positive. Call this common sum and calculate
Since each , this whole sum must be nonpositive, which can only happen if . But then
which forces all the . Similarly, all the , and thus the original linear combination must have been trivial. Thus is linearly independent.
Now we can show that is a base. Every vector in is indeed a nonnegative integral linear combination of the vectors in . Since , every vector in this set is a nonpositive integral linear combination of the vectors in . And every vector in is in one or the other of these sets. Also, since spans we find that spans 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 is of the form for some regular . Indeed, we just have to find some for which for each . Then since any is an integral linear combination of we can verify that for all , proving that is regular. and . Then the vectors are clearly indecomposable, showing that . But these sets contain the same number of elements since they’re both bases of , and so .
The only loose end is showing that such a exists. I’ll actually go one better and show that for any basis the intersection of the “half-spaces” is nonempty. To see this, define
This is what’s left of the basis vector after subtracting off its projection onto each of the other basis vectors , leaving its projection onto the line perpendicular to all of them. Then consider the vector where each . It’s a straightforward computation to show that , and so is just such a vector as we’re claiming exists.
We don’t always want to deal with a whole root system . 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 is called a base if first of all is a basis for , and if each vector can be written as a linear combination
where the coefficients are either all nonnegative integers or all nonpositive integers.
Some observations are immediate. Because is a basis, it contains exactly vectors of . It also tells us that the decomposition of each is unique. In fact, as for any basis, every vector in can be written uniquely as a linear combination of the vectors in . What we’re emphasizing here is that for vectors in , 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 on the root system . We say that is a “positive root” with respect to (and write ) if all of its coefficients are nonnegative integers. Similarly, we say that is a “negative root” with respect to (and write ) if all of its coefficients are nonpositive integers. We extend this to a partial order by defining if .
Every root is either positive or negative. We write for the collection of positive roots with respect to a base and for the collection of negative roots. It should be clear that , and also that — the negative roots are exactly the negatives of the positive roots.
We can also define a kind of size of a vector . Given the above (unique) decomposition, we define the “height” of relative to as
This will be useful when it comes to proving statements about all vectors in by induction on their heights.
If are two vectors in a base , then we know that and . Indeed, our lemma tells us that if then would be in . But this is impossible, because every vector in can only be written as a linear combination of vectors in in one way, and that way cannot have some positive signs and some negative signs like does.
What this tells us (among other things) is that must be one end of the root string through . The other end must be , and the root string must be unbroken between these two ends. Every vector with must be in .