Last update: 10 September 2012
This is a typed version of I.G. Macdonald's lecture notes on Kac-Moody Lie algebras from 1983.
To begin with, let be any matrix over the field . Eventually will have to be a symmetrizable Cartan matrix, but we shall bring in that assumption only when it becomes necessary.
and that each root space is a finite-dimensional (1.7).
Let be a –module, i.e. a –vector space on which acts, so that we are given a Lie algebra homomorphism which extends to i.e is a –module. Notation when I want to be pedantic. More often then not I shall suppress and write or for .
For any –module and any we define
If we say that is a weight of , that is the weight space and that the elements of are the weight vectors for the weight . We have
where is the –dimensional –module defined by that is to say where for all The isomorphism associates to each the homomorphism which takes to From it follows that (for a fixed ) is a left exact functor (from –modules to –modules).
Example: The weight spaces are and the and the set of weights is
For and we have
The sum is direct, and is a –submodule of
If is a –module homomorphism, then for all
If is any –module, let denote the set of weights of (It might be empty.) Also, for each let
and for any subset of let
We shall use this notation only for finite subsets of
Let denote the category of –modules which satisfy the following two conditions:
is –diagonalizable with finite dimensional weight spaces, i.e.
(direct sum, by (3.1)), with each finite-dimensional;
for some finite
The morphisms in are –module homomorphisms.
(3.2) Let be a short exact sequences of –modules, with . Then
For each the sequence is exact;
Since is –diagonalizable we have by (1.5), and by (3.1), so that is finite-dimensional and Hence
Next, we have by (3.1), for all hence
consequently we have equality throughout, whence for each and the sequence is therefore exact. Finally, is finite-dimensional and so that and (iii) is now obvious.
Recall the partial order on iff with each
(3.3) Each module has at least one maximal weight.
Suppose has no maximal weight. Then contains an infinite strictly increasing sequence For each the form a subsequence. Since is finite, at least one of these subsequences is infinite, say in Each hence is a nonnegative integer. It follows that the sequence is an infinite strictly decreasing sequence of integers which is absurd.
If has a unique maximal weight then is called the highest weight of
We shall say that a –module is a highest weight (h.w.) module if
has a highest weight, say
is generated (as –module) by some
(3.4) Let be a h.w. –module, with highest weight Then
has a unique maximal submodule, hence a unique simple quotient;
If is a nonzero homomorphic image of then is h.w. with h.w.
We shall now show how to construct all h.w. –modules.
Let and let as before denote the 1–dimensional –module corresponding to where for all
Let be the subalgebra of generated by and The subalgebra is a semidirect product because is an ideal in and We may regard as a –module by making act trivially, i.e. The Verma module is defined to be the induced –module
Let Clearly generates and since we have showing that is a h.w. –module with highest weight and that it is free of rank 1 as a –module.
Alternative description of let denote the left ideal in generated by something and all Then
For if is the representation of on then is such that and all hence is the left ideal of codimension 1 in generated by and all tensoring the exact sequence (of left –modules)
with (over ) gives
and the image of in is
The Verma modules are the "universal" h.w. –modules:
is a h.w. –module with highest weight
Every h.w. –module with highest weight is a homomorphic image of
By (3.4)(iv) it follows that has a unique simple quotient by (3.2) and (3.4)(i), we have Moreover, the are precisely the simple objects in the category
(3.6) If is simple, then for a unique
By (3.3), has at least one maximal weight, say Let Then (because hence is killed by and therefore the submodule generated by is a quotient of Since and is simple, we have hence is a simple quotient of hence
Suppose also that Then we have a –isomorphism under which weight spaces correspond (3.1). Hence is a weight of whence similarly and therefore
(3.7) Example. When with trivial –action and The maximal submodule of is the augmentation ideal of hence is the trivial 1–dimensional –module.
(3.8) Ket be a h.w. module. Then
Let be a h.w. vector which generates If is a –module homomorphism something have hence for some (because (3.4)). The Kernel of is a submodule of which contains hence is the whole of i.e.
Let be the set of all functions such that for some finite subset of i.e. unless for some Clearly is closed under addition and subtraction of functions; define multiplication by convolution:
If then Thus is a commutative ring.
A family of functions in is summable if finite such that
for each we have for almost all
In that case the function defined by
is well defined and belongs to and we write
For each let denote the characteristic function of Then from the rule (1) defining multiplication. For any the family is summable, and we have
Now let and define the formal character of to be the function defined by
thus for some finite so that and by (2) we have
Thus is nothing but the generating function for the multiplicities of the weights
(3.9) ch is an additive function on the category i.e. if is an exact sequence in then
This follows from (3.2)(ii) by counting dimensions
More generally, if is an exact sequence in we have
by breaking up the exact sequence into short exact sequences and applying (3.9).
We shall first compute the character of a Verma module
(3.10) Let Then
where as usual (The product on the right is a unit in
We saw earlier that is a free –module of rank 1. As before, let be a –basis of consisting of root vectors, say Then has a –basis consisting of weight vectors where each (and the vector just written being of weight
As before, is any matrix and Suppose that there exists a symmetric (–valued) bilinear form on such that
for all (invariance)
the restriction of to is nondegenerate.
Then for any we have
Let Condition (ii) ensures that taking in the calculation above we have
and therefore the matrix is symmetric, i.e. is symmetrizable ( is a nonsingular diagonal matrix).
This proves the first part of
(3.11) Let be an invariant symmetric bilinear form on whose restriction to is nondegenerate. Then
the matrix is symmetrizable
is nondegenerate on
restricted to (where ) is
If and then
where is defined by for all
Proposition (3.11) has a converse:
(3.12) Suppose that the matrix is symmetrizable. Then there exists a nondegenerate symmetric invariant bilinear form on (which therefore has the properties listed in (3.11)).
By assumption, there exist non-zero scalars such that We shall first construct the form on (of (2.23)) and then extend it to
Choose a vector space complement of in (where as usual ) and define on by
To see that this form is nondegenerate, suppose that is such that Then in particular we have whence thus say, and then
for all so that in hence and so something.
Recall the principal –grading of
The extension of to is unique, for by (3.11)(iii) we must have if and also It is straightforward to verify that
whenever all 5 terms lie in
We shall now extend to a symmetric bilinear form on by induction on such that (1) holds whenever all 5 terms are in and such that
So assume and defined on satisfying (1) and (2). To extend the form to we have, in view of (2), only to define on Write
where (resp. are homogeneous of positive (resp. negative) degree, hence lie in Define now
The whole point is to show that this is well defined, i.e. that it does not depend on the expression (4) for For this purpose we make the following calculation: dropping the suffixes,
i.e. we have
From (3), (4) and (6) it follows that
and therefore (as defined by (5)) is well-defined, and satisfies the invariance condition (1) by our definition (5).
Notation. In we have in particular
and an isomorphism definted by
for all whence
We use to transport the scalar product from to thus
In the classical situation, where is finite-dimensional, the Casimir operator plays an important role in representation theory. The invariant bilinear form may be regarded as an element since it is non-degenerate it induces an isomorphism of with hence determines an element of The image of this in (which is a quotient of the tensor algebra is the Casimir element . Since is invariant it follows that is in the centre of hence acts as a scalar on any simple –module. Explicitly, if is any –basis of let be the dual basis (so that then
In the present situation, where is any symmetrizable matrix, we proceed as follows. Let by (3.11), the bilinear form is nondegenerate on choose a basis of let be the dual basis of and define
Then is independent of the choice of dual bases, for if is another pair of dual bases, we have
If is not a root (or zero) we define (the sum is empty).
Example: We have hence
Let then we have
where, for the same reason as before, and are independent of the choice of dual bases. Since and we have Let be a basis of the dual basis of Then
i.e. we have the formula
In particular, unless both and are positive roots (or 0); and likewise unless both
(in some completion of ...)
(3.13) We have
where image of under the isomorphism induced by the bilinear form. (i.e. so that
But unless and therefore
Similarly we have
Finally by (2).
Choose an element such that
(thus if is a Cartan matrix). Then we have
Now let and define a –linear map
as follows: if then
is a finite sum, because for almost all
(3.14) is a –module homomorphism.
Since is generated by the the and it is enough to verify that commutes with the action of each of these elements. So we calculate:
Remark: is functorial, i.e. if is a –module homomorphism (with then the diagram
commutes. For commutes with the action of and preserves weight spaces.
(3.15) Example. Let be a h.w. module with h. wt. If is a generator of (3.4), we have for all hence and therefore Hence by (3.14) (since generates
(3.16) Let be such that for some scalar Let be a finite subset of such that and let
Then there exist integers such that
If we have for some hence is an integer Define the depth of (relative to to be
so that also define
Since is finite there are only finitely many of given depth; in particular, has only finitely many weights of least depth and they are all maximal weights. Call them
We shall kill the weight spaces Let and let
Choose a –basis of each and let be the –homomorphism which maps the generators of the summands of to the chosen basis elements of the Let be the kernel and cokernel of so that we have an exact sequence
Then because it is a submodule of and because it is a quotient of Now acts as scalar multiplication by on (3.15), and hence also on the image which is a non zero submodule of Since by hypothesis acts as scalar multiplication by on it follows that i.e. Hence acts as on and hence on also on By construction we have and and by additivity of ch (3.9)
Now repeat the same procedure on and After we have done it times we shall have say
where is some finite subset of and for all Now let and we have
Remark. Suppose in particular that is a h.w. module, with highest weight Then
with and in particular
From now on, is a Cartan matrix.
Let be a h.w. module, with highest weight Then each is a locally nilpotent endomorphism of For if say and if then if
If also each is locally nilpotent on we shall say that is a quasi-simple –module. (Later we shall see that quasi-simple simple).
(3.17) Let be a h.w. module with highest weight and generator If such that for then is quasi-simple.
Recall the formula (1.16)
associative ring Let so that for some and apply (1.16) with
Now is locally nilpotent on (1.19), hence also on so that for some Hence if is large enough would do) either or for each and so something
(3.18) Let be a quasi-simple –module with highest weight Then:
is –invariant (as a function on
If and then the set of integers such that is a finite interval in where and
We shall make use of the following formula:
for elements of an associative –algebra, with nilpotent (so that is defined). The proof is very simple: we have and commute, hence
A linear form is integral if dominant integral if for
Let (resp. denote the set of all integral (resp. dominant integral) Notice that each because thus (Warning: Clearly the dual fundamental chamber).
(3.19) Let be a quasi-simple –module with highest weight Then Conversely, if then is quasi-simple.
Let be a generator of Since is quasi-simple, such that also whence
and therefore Thus (Notice that this gives another proof of (3.18)(c), namely that for if then
For the second part, let be the generator of and let I claim that For now we have from (1) that if i.e. also if (because then commute). Hence generates a proper submodule of Since is simple, we must have By (3.17), it follows that is quasi-simple.
Recall that was chosen such that Since is now a Cartan matrix, this condition now becomes
For let (sign character of
(3.20) is –skew, i.e.
It is enough to verify this when is a generator of We have
On the other hand (2.6), sends to and permutes the positive roots . Thus
Now assume that the Cartan matrix is symmetrizable. Then the scalar product on and is –invariant (2.23), and we have
For the same reason, if we have
(3.21) Theorem (V. Kac) Let be a symmetrizable Cartan matrix and let be a quasi-simple –module with highest weight Then
From (3.16) we have, writing
summed over such that with coefficients and, in particular,
Now is –invariant (3.18) and is –skew (3.20). Hence there product is –skew, and therefore for each we have
so that Hence if we have for all choose so that is minimal and put Then i.e. and therefore or equivalently
These three conditions force for we have
But by (2) because (3.19); (1); and ((i) above). Hence and therefore all coefficients are 0, hence and therefore
Recall (3.19) that is quasi-simple if The character formula (3.21) shows that if is a quasi-simple –module with highest weight then depends only on It follows that
i.e. for all But is in any case a homomorphic image of and so we conclude that
(3.22) Every quasi-simple –module is simple.
Another corollary of (3.21) is the "denominator formula":
(3.23) For any symmetrizable Cartan matrix we have
Take in (3.21) and observe that is the trivial 1–dimensional –module (3.7), so that
We can write (3.23) in another form, as follows. Recall that for
is a finite set (2.10). Define
a finite sum of positive roots. We then have the formula (for any Cartan matrix A)
If is a reduced word for then (2.9)
from which it follows that if
i.e. if with then
To prove (3.24), we proceed by induction on The result is clearly true when for then Assume and write as above, then
By virtue of (3.24) we can rewrite (3.23) in the form
Also form (3.23) we can rewrite the character formula (3.21) in the form
(3.21) Let then
(3.23) is a statement about the root system and the Weyl group it may be formally inverted to give a formula for the multiplicities (recall that all real roots have multiplicity 1; the imaginary roots may have multiplicities
Suppose is of finite type, so that is finite-dimensional and is finite. In that case
For if is the sum of the positive roots then by (2.6)
but on the other hand so that for all Since is nonsingular, is spanned by and therefore
The Denominator formula (3.23) now reads
by virtue of where is this from? It is a polynomial identity in the group ring
For a specific example, take of type Let be the standard basis of then the roots may be taken to be and the positive roots Thus
Put the Weyl group is here the symmetric group acting by permuting the (or, equivalently, the We have
where is the Vandermonde determinant. On the other side,
and therefore the "denominator formula" in this case reduces to the familiar factorization of the Vanermonde determinant:
So it is an essentially trivial polynomial identity in this case.
Let of affine type. The Weyl group is infinite dihedral, generated by reflections So its elements are 1 and
We have and likewise So we get the following picture:
Now if to terms then
so that if we put we have
On the other hand, the positive roots are
is an imaginary root; in fact (as we shall see later) it has multiplicity 1. So we obtain the identity
in If we put it takes the form
and is due to Jacobi (and earlier, unpublished, to Gauss): it is called Jacobi's triple product identity and it can be specialized in various ways, at least two of which are worth notice:
Put and we get
Euler's pentagonal number theorem.
Divide both sides by and then set On the product side we get and on the sum side
Thus we obtain another famous identity due to Jacobi:
Finally, if is symmetrizable and of indefinite type, the denominator formula can be used to compute the multiplicities of the imaginary roots. For a simple example take The Weyl group is infinite dihedral:
So the real roots are
the coefficients of which we recognise as Fibonacci numbers: the real roots are
From this we calculate easily
so that the series is
i.e. it is a 'sparse' power series in By factorizing it as a product of factors we compute the multiplicities of the roots: