Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
Last update: 11 August 2012
This is a typed version of I.G. Macdonald's lecture notes from lectures at the University of California San Diego from January to March of 1991.
Notation etc. as in the previous chapter. If
denote the constant term of (the coefficient of ). I am going to define various scalar products on
and the number of terms in this sum is
If this is 0, and if it is
where as before
is the Weyl denominator.
which is zero if because the orbits
do not intersect. If we have
(because ). Hence we get
Let be a real number, and let
As explained earlier, we may regard as a continuous function on the torus
For we define
Notice that when we have and when we have
so that when we get the two previous scalar products.
Since is invariant it follows that
a product of convergent series, since Hence we can multiply them all together and get say
and in particular Hence we have
where and then
with equality if and only if Hence
Now let We cannot have and hence (since the scalar product is symmetric) we may assume We have
If the series (*) and (**) have just one term in common, namely and hence
(So are dual bases.)
Proof of (4.18).
From (†) it follows that
which proves (4.18)(ii). Now
If is singular, then
if nonsingular, then
for a unique and and then
So from (††) we have
In fact the are polynomials in with positive coefficients (Lusztig: he shows that they are Kazhdan-Lusztig polynomials for the affine Weyl group). Elementary proof?
is Kostant's formula for weight multiplicities.
Now let be independent indeterminants [or real variables, ] and let
In particular, if where then
We now define a new scalar product on if then
Once again, this is symmetric and positive definite. Notice that when we have
Also if (i.e., ) we have
Finally if (i.e., ) we have
both independent of
For each there exists
This will be a consequence of the following proposition:
There exists a linear operator
satisfying the following three conditions:
( is self adjoint),
( is triangular relative to the orbit-sums ),
( has distinct eigenvalues).
Let us first show that Proposition 1.6 implies Proposition 1.5, and then address the problem of constructing a suitable for each root system For each define
and then let
Then the so defined have the desired properties. Indeed, it is clear from the definition and (b) above that satisfies Proposition 1.5 (i).
Let be the (finite-dimensional) subspace of spanned by the such that (and ). Then is stable under and the minimal (characteristic polynomial of
since the are all distinct. Hence
on and therefore
(so the are the eigenvectors of ). Hence
Since if it follows that
So it remains to construct a linear operator on for each reduced irreducible root system satisfying the conditions (a), (b), (c) of Proposition 1.6. In fact we need two constructions, according as or
Minuscule weights (of )
Assume reduced, irreducible. Let
is a minuscule weight of if
Clearly 0 is always a minuscule weight.
be the basis of dual to
Then is a basis of the lattice
be the highest root of and let
(It might be empty.)
The minuscule weights of are 0 and
is a minuscule weight. Then
or 1 for each and
If this is zero, then all the are 0, and If it is 1, then for some we have
(whence ) and
for (whence since ). So
for all because hence
The minuscule weights have other nice properties:
They are coset representatives of in hence there are of them (i.e. nonzero minuscule weights).
Let Then is a minuscule weight (for not ) if and only if
(i.e. the weights form a single orbit).
They are the minimal elements of with respect to:
So there are nonzero weights except in types
all fundamental weights are minuscule and property (2) above translate into the fact that
PICTURE OF DYNKIN DIAGRAMS
Suppose (i.e. that is not of type or ) and let be a nonzero minuscule weight of so that
extended by linearity. Clearly
so that is an automorphism of the algebra Moreover, is self-adjoint for the scalar product
which is symmetrical in and hence equal to
If we interpret as a function on by the rule — different from the previous one —
and hence by linearity
for all i.e. is a translation operator.
From above we have
(so that ) and let
We now define, for
(conjugate by and then sum over the Weyl group). We shall show that maps into and satisfies conditions (a) and (b) (self adjointness and triangularity) of Proposition 1.6, but not always (c) (distinct eigenvalues)).
Let us first verify that is self-adjoint. Since
as we saw earlier, it follows that
Next we have to compute
We have from (OP 5)
where as before
which shows that maps into
Now let then
is singular (i.e. not regular) then
is regular, then there exists and such that
for some hence
and it follows that is a linear combination of the
and hence also a linear combination of the
Moreover in the above calculations we shall have if and only if and i.e.
which by (4.11) forces
So the coefficient of (or equivalently of ) in is
It remains to examine whether satisfies condition (c) of Proposition 1.6, i.e. whether the
are all distinct. In fact this is true in all cases except
(and where there is no nonzero minuscule weight). With as above let
Clearly the are invariant polynomial functions on i.e.
in the notation of Chapter III. (Also )
Suppose that is not of type or and that if is of type the minuscule weight is the fundamental weight corresponding to an end node of the Dynkin diagram. Then the generate the algebra
This is easily checked for of type or For and we refer to M.L. Mehta (Communications in Algebra16 (1988) 1083-1098).
Suppose now ( excluded) are such that i.e.
Operate on both sides with
and then set We obtain
Hence by Proposition 3.2
for all from which we can conclude that thanks to the following general lemma:
Let be a finite group of automorphisms of and let
be the algebra of invariant polynomial functions on Suppose are such that for all Then are in the same orbit in (i.e. separates the orbits in )
Let be distinct points of and let Then I claim that there exists
To prove this it is enough to construct
for we can then take
For each ordered pair with and we can find
(we may take linear, nonhomogeneous). Then
does the trick.
be distinct orbits in From above there exists such that
Then and takes the value 1 (resp. 0) at each (resp. ). So separates the orbits in which is equivalent to the result stated.
What about ? Here there are 3 minuscule weights, hence 3 operators It turns out that no one of them has distinct eigenvalues, but that a suitable linear combination of all three does. We omit the details. So we have constructed for each with an operator satisfying the three conditions of Proposition 1.6.
This time we take highest root of so that (2.4)
As before, let
and define now
Then it turns out, as a result of calculations analogous to those we have already done, that has the 3 desired properties in the cases not covered by the previous construction. So the existence of the
claimed by Proposition 1.5 is established in all cases.
Here I can only conjecture
(C2q) Let Then
True if of type
Consider the special case of (C2q). Since we have
and hence (C2q) would give
I claim that (OP 6) is equivalent to (C1q). Let
and therefore (since is invariant)
(because have the same constant term). It follows from (OP 6) that
Consider then the polynomial
it follows that
Hence (OP 6) becomes
So (C2q) (C1q). (Tip of the iceberg!)
Consider as (i.e. ). We have
(continuous function on )
True when (From the denominator formula for affine root systems.)
The limiting case
As we have
as kernel for the scalar product. So we have
When is of type these are Jack's symmetric functions (suitably normalized) with parameter When (C2q) is in fact true: (Heckman, Opdam)
Proof is now elementary — depends on their notion of shift operators which enables one to compute the quotient
and hence prove (OP 7) (for a positive integer) by induction on (when
contains no cycles (hence is a tree if is irreducible).
Suppose contains a cycle of length Then with a suitable numbering of the simple roots we have
Let angle between and Then is hence and therefore
Consider the vector
So and hence impossible as the are linearly independent.
Let be an irreducible root system; If
then for some
Since is irreducible, is an irreducible module (2.1). Hence spans hence there exists such that
Replace by we may assume
If we are done; if we have
Replacing by if necessary, we may assume that
Then we have
Can use this, together with (1.21a), to find the orders of the Weyl groups of the exceptional root systems. (Use .)
72 roots, forming a single orbit. The extended Dynkin diagram shows that the highest root is orthogonal to all except hence
126 roots, forming a single orbit. Here
24 long roots,
Let (reduced), say
The support of is the subgraph of (= Dynkin diagram of ) obtained by deleting the vertices of for which
Show that is connected, for each
Let be such that is connected ( delete verticies outside ). Show that
May assume that (because clearly ). Induction on
If then and consists of one point. So assume then has at least two elements. We have
say and hence
for some and hence (1.4)
indeed because the coefficients of in terms of the are
If then and
hence by the inductive hypothesis is connected.
If then hence the string through is
say, where By (1.5)
it follows that
for some Now is connected, by the induction hypothesis; and is joined to some Hence is connected.
Induction on OK if so assume Since contains no cycles, neither does i.e. is a tree. Let be an end vertex of Then is connected, hence and
for some so that
and therefore (1.4)
of type then all positive roots are as in (ii). For the other root systems this is not so.
(Parabolic subgroups of ). Let be a subset of Let
is a root system in with basis and Weyl group if
The subgroup of generated by and is
Thus the form a lattice of subgroups of (parabolic subgroups).
Proof of (i).
(i) Check the axioms: clearly satisfies (R1), also (R2) because if then
is a basis of hence every is a linear combination of the The coefficients must be integers all of the same sign, hence is a basis of Hence is the Weyl group of
Let be the basis of dual to the basis so that
Proof of claim.
⇒) Let If then (because ). Since is generated by the it follows that
If fixes for all then fixes hence (1.21a) is a product of reflections fixing So suppose say
If fixes then
hence for each so that
Now do (ii)-(iv):
Proof of (ii)-(iv).
(ii) If we may assume there exists We have but (Otherwise would fix hence ) So
(iv) Suppose then fixes whenever or hence whenever Hence
Descriptions of the exceptional root systems
Let be a reduced irreducible root system with basis the highest root or
Let be the extended Dynkin diagram of with vertices corresponding to
and labels where
If we remove the vertex of we get a (usually disconnected) diagram of a root system in with basis
and Weyl group say. (In particular ) Since and it follows that
From (OP 8) we have
Consider the root lattices
Since we have and is a finite group. Let
be the image of in Then
if and (OP 9) shows that
hence is a linear combination of the But (OP 8) is the only linear dependence relation among the
hence is a multiple of It follows that is a cyclic group of order generated by
Assume now that for each so that and hence ( is simply laced.) Then the dual lattices of are
we have and is cyclic of order (4.1), (4.2). So we have the tower of lattices
and hence if is the index of connection of we have
of type the extended Dynkin diagram is
and hence the root lattice of lies between the root lattice
and the weight lattice
This underlies the construction of that I gave earlier:
Exercise: Figure out the roots of in this representation.
So we have 3 pairwise orthogonal copies of sitting inside we can think of as the sixth roots of unity in So we may work in and take ( basis of as vector space)
What about We must have
In this representation the roots of are
For these are certainly roots, and there are
of them. So they are all the roots of
Exercise: Figure out the action of the Weyl group in this description.
of the type
is easily described; it is the lattice of all
Exercise: Figure out the roots of in this representation
What are the roots of in this description?
So certainly all of the vectors
are roots. There are
of these vectors, hence they are the lot.
Again take So we may take
So certainly the following vectors are roots:
So these are all roots.
Orders of the Weyl groups of
Recall (Exercise): irreducible,
Then are in the same orbit. Apply this when
The roots form a single orbit. Consider in particular the highest root Since the stabilizer is generated by the such that So we can read off from the extended Dynkin diagram, and we have
24 long roots, hence
The Weyl groups of
For we have
it follows that has order hence is a 5-dimensional vector space over the field of 3 elements. If the scalar product is normalized so that
for all it induces a nonsingular scalar product on So acts on as a group of orthogonal transformations, i.e.
In fact has index 2 in this group (but it is not
but another subgroup of index 2). Now in general for a Chevalley group we have
( the rank, the number of positive roots). So
then is of type so that
For we have
it follows that has order hence is a 6 dimensional space over Normalize the scalar product so that
for all then it induces a nonsingular alternating bilinear form on acts and we have an isomorphism
For the root system is so that here
Again for is a vector space of dimension 7 over Normalize the scalar product so that this time, for all Then we have a nondegenerate quadratic form on and acts as a group of orthogonal transformations: we have an isomorphism
Repeat Exercise 3 for Again we find
(check the orders). For the root system is so that the are and
Arun: The next section appears to be out of order/without context. Have typed and left to be fixed/moved.
Let number of roots of height Then
Hence number of degrees equal to so
is a partition and
Can now check (3) directly case by case. Probably as quick, if not quicker, than a uniform proof
Issac Newton Institute for the Mathematical Sciences
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