Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
Last update: 01 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.
Let be a root system in (not necessarily reduced), spanning Let be subspaces of such that
In this situation the following are equivalent:
and (resp. ) is a root system in (resp. ),
the hyperplane perpendicular to Since
then there exists such that
We have (1.1)
hence So either or i.e.,
Replace by and we see that
But spans hence (resp. ) spans (resp. ). So
Clearly each satisfy (R1) (integrality). As to (R2): if
So are root systems.
Let be the Weyl groups of respectively. Then
(direct product). is stable under and fixed under hence stable under Likewise for
These equivalent conditions mean that is reducible.
a basis of Let
is called the Cartan matrix of It is independent of the choice of basis because (1.11) any other basis is of the form
for some and
because is an isometry.
The matrix satisfies the following conditions:
For is the matrix
is a positive integer.
The Cartan matrix determines the reduced root system up to isomorphism. More precisely:
Let be another root system spanning a vector space and let be a basis of Suppose
is a bijection such that
Then there is a unique isomorphism
(i.e. extends ).
Since (resp. ) is a basis of (resp. ) there is just one isomorphism
that extends We have to show that maps onto
Then we have
[Don't confuse with the Coxeter matrix
This is a graph that encodes the information given by the Cartan matrix. Let be a basis of (reduced) as before and let be the following graph: vertex set = (so vertices labelled ).
Edges: the vertices and are joined by
edges (or bonds), together with an arrow pointing from to if and
From (1.2) we see that or So if
the possibilities are
Dynkin diagrams of
This table shows that the Cartan matrix can be reconstructed unambiguously from the Dynkin diagram hence by (1.22) determines the reduced root system up to isomorphism.
[Coxeter diagram: describes the Weyl group
The beauty of this description of is that
is irreducible is connected.
Suppose reducible: Let be a basis of then
is a basis of and
Hence no vertex of (the diagram of ) is joined to any vertex of the diagram of So is not connected.
Conversely, suppose is not connected. Then
(and neither nor empty). Let be the subspace spanned by Then
(orthogonal direct sum) and each of is stable and hence (Prop. 1.1) is reducible.
So to classify irreducible root systems it suffices to list all possible connected Dynkin diagrams. To achieve this we shall first classify another class of Dynkin diagrams, called affine Dynkin diagrams (because this classification is easier). To do this we need the notion of the highest root of an (irreducible, reduced) root system.
(reduced), as usual.
Suppose irreducible. Then has a "highest root" such that for all We have
or for all
Let be a maximal element of for the partial order If
for some then by (1.4) we should have either
contradicting the maximality of or
Suppose then there exists and such that
(otherwise the Dynkin diagram of would be disconnected, contrary to (Prop. 3.1)). But then
(because each summand is and at least one is ) contradicting (*). So i.e., for
Now let be another maximal element of As above we have
for all moreover
for some (if
for all then
for all so that ). Hence
Suppose From (1.4) we infer that
in which case or
in which case Both of these are impossible, so that
we may replace by and hence (1.21) assume that (fundamental domain for ). We have hence
for all in particular
Let From (*) we have
just proved. By (1.2) it follows that
Note that (from the proof)
with each a positive integer.
Let be the lowest root of (relative to ). Then the relation (**) takes the form
where the are positive integers, and We define the extended Cartan matrix (with rows and columns)
as before, and the extended Dynkin diagram by the same rules as before. Each determines the other. is connected, because from the proof of (Prop. 4.1) we have
positive and hence
negative for at least one
So we have
(with no arrow, because
In all other cases the bond strengths
are as in
(every positive root is
where ), so
Come back to the relation
Taking scalar products with we have
Since and for we can rewrite this in the form
gives the set of vertices of directly linked to the vertex and
(either no arrow, or arrow pointing away from ) whereas
Suppose first no multiple bonds issuing from the vertex Then all
are equal to 1, and (Cla 3) takes the form
Next suppose that there are multiple bonds issuing from the vertex of If the arrow points outwards, treat it as a single bond. If the arrow points inwards, replace
So the upshot is that the diagram can be labeled with positive integers in such a way that the conditions (Cla 3), (Cla 4) and (Cla 5) are satisfied.
is a labeled diagram (also called an affine Dynkin diagram). So each reduced irreducible root system gives rise to an affine Dynkin diagram.
We shall now temporarily forget about the root system that gave rise to and and define a generalized Cartan matrix to be any matrix of integers
satisfying the conditions
We associate with a diagram (or graph) by the same rules as before: has vertices numbered and vertices
are joined by
edges, with an arrow pointing towards if
As before, each of determines the other.
The diagram will be connected if and only if is irreducible, i.e. if we cannot partition the index set into two proper subsets, such that
A generalized Cartan matrix is of affine type if
There exists a vector
of positive integers such that
(which implies ). Correspondingly, the diagram is of affine type if
can be labeled according to the previous rules.
Finally, a subdiagram of is any diagram obtained from by either deleting some vertices (and the edges issuing from them) or deleting some edges (or both) (i.e. it is subgraph of ). Then the basic fact is that
Let be a diagram of affine type. Then no proper subdiagram of is of affine type.
The moral of (Prop. 4.2) is that affine diagrams are rigid.
We shall use (Prop. 4.2) to classify all possible diagrams of affine type. We know that each extended diagram must occur in our list, and that (the diagram of ) is obtained from by deleting one vertex. So we shall also obtain a list of all possible diagrams
The diagrams do not satisfy the rigidity property (Prop. 4.2) (e.g. is a subdiagram of or of is a subdiagram of ). This is why the affine diagrams are easier to classify.
The proof of (Prop. 4.2) rests on two lemmas about real matrices. We consider matrices
satisfying the condition
i.e. the off-diagonal elements of are If the set can be partitioned into two nonempty subsets such that
we say that is reducible.
is a column vector, we shall write (resp. ) to mean that
(resp. ) for all
Let be an irreducible matrix satisfying (Cla 7). Suppose that there exists such that and
Then either is nonsingular or else
Suppose is singular (). Then there exists
such that We may assume (replacing by if necessary) that some Let
is for all and for some Let
If we have
because the are and the are From (***) it follows that
for all i.e.,
and hence for all
Since is irreducible and it follows that
so that and hence from (***)
as above, and let
These are the principal submatrices of and their determinants are the principal minors.
be two irreducible matrices satisfying (Cla 7) (off-diagonal elements ) and such that (i.e.,
for all ), Suppose there exists
such that and Then all principal minors of (including itself) are nonzero.
Consider first itself. We have
(since ) and
(because ). Hence
by Lemma 4.4.
Now let be a proper nonempty subset of If is reducible, there are complementary subsets of such that
So we may assume irreducible.
If we have
which is (because ). Hence
we should have
for all whence
for and and this is impossible because is irreducible. So we have
by Lemma 4.4.
Proof of Proposition 4.2.
Let be a diagram of affine type, the corresponding Cartan matrix, and let be a proper subdiagram of Deleting edges from corresponds to increasing off diagonal entries of (i.e. decreasing ); deleting vertices of corresponds to passing to a principal submatrix. Hence the Cartan matrix of is of the form where (take outside ). By Lemma 4.5, is nonsingular, hence is not of affine type.
Classification of diagrams of affine type
We shall make use of Proposition 4.2 in the following form: if is an affine diagram and is an affine subdiagram of then is the whole of
Let be an affine diagram.
Suppose contains a cycle (with vertices). Then has a subdiagram of the form
which can be labeled, hence is affine, hence is the whole of So assume henceforth that is a tree (no cycles).
Suppose has a branch point of order (joined to 4 or more vertices). Then has a subdiagram of the form
which is affine, hence is the whole of
Suppose that has 2 or more branch points of order 3. Then has a subdiagram of the form
Suppose that has one branch point of order 3 and at least one multiple bond. Then has a subdiagram of one of the forms
Suppose that has one branch point of order 3 and no multiple bonds. Let be the number of vertices on each branch (including the centre). Suppose labels at the ends of the branches are and that the label at the center is Then
the solutions of which (with ) are
We have now exhausted all the possibilities where has a branch point. Assume now that is a chain. If has at least two multiple bonds it will contain one of the following diagrams,
each of which is labeled, hence affine. So must be one of these.
Assume now that is a chain with just one multiple bond.
strength 2: the multiple bond can't be at an end (subdiagram of (6)). Can't contain properly either of
and can't be a subdiagram of either.
(No multiple bonds). Finally
is not affine, because it is a proper subdiagram of the necklace (1).
So our list of affine (i.e. labeled) diagrams is complete.
Finally, the Dynkin diagrams of irreducible reduced root systems must all occur as connected subdiagrams of these obtained by removing a vertex with label 1. Looking through our collection we obtain the list
Of these we have already encountered
and As to the other exceptional diagrams, we still have to establish that they do correspond to root systems.
Not all the affine diagrams that we have encountered are of the form (extended Dynkin diagrams): about half of them are. The affine diagrams classify (irreducible, reduced) affine root systems.
Affine root systems
as before a real vector space of finite dimension equipped with a positive definite inner product. An affine linear function on is a function of the form
where and is the gradient of (in the usual sense of elementary calculus). We may write
Let be the space of all these functions: it is a vector space of dimension isomorphic to For we define
this scalar product on is positive semidefinite (if is constant, i.e. then clearly
for all ). ( the constant functions.) If is an affine isometry then acts on
In particular, let be non-constant. Then
is an affine hyperplane in Let
be orthogonal reflection in this hyperplane. Then the action of on is given by
just as in (1.1).
We can now define an affine root system. It is a subset of spanning and consisting of functions
and satisfying the following axioms:
(AR3) (with the discrete topology) acts properly on
This means that the mapping
of into is a proper map, which (since is locally compact) means that for each compact subset of
is compact. Since is discrete, this in turn means that
(AR3') Given any two compact subsets in the set of such that meets is finite.
It follows that the set
of hyperplanes in is locally finite, i.e. each compact (closed, bounded) subset of is intersected by only finitely many of them.
From (AR1)-(AR3) it is clear that if is an affine root system, so is
an irreducible reduced root system in Then
is an affine root system.
Hence so is
Warning: this is not necessarily isomorphic to
In fact these two examples are almost exhaustive of the reduced irreducible affine root systems; there is just one other family
constructed by a variant of Ex. 1.
All the main features of the finite theory (chambers, bases, Cartan matrix, Dynkin diagram) have their counterparts for affine root systems. The Dynkin diagrams are precisely the "affine" Dynkin diagrams that we have just been classifying. (In the table, the left hand column consists of the diagrams for reduced and irreducible; the right hand column consists of the others.
(Affine root systems and Dedekind's function, Inv. Math. 14 (1972) 91-143.)
Affine (Kac-Moody) Lie algebras (see Kac's book).
The Weyl groups are affine Weyl groups (infinite discrete groups of isometries of Euclidean spaces generated by reflections). Draw pictures of
Let be the standard basis of with the usual inner product
so that Let
be the orthogonal projection, given by
a lattice in generated by the nine vectors
is a root system in of type
First check the axioms. spans because e.g.
for we have and hence (R1) becomes
where Then by (Cla 8) we have
and in particular
Hence is a rational number whose square is an integer, hence and therefore (ClaE8 1) shows that
Next, if we have
proving that (R2) holds. So is a root system in clearly reduced.
Next let us list the elements of Suppose
we may choose
If it follows from (ClaE8 2) that
we obtain likewise from (ClaE8 2) that
To recapitulate, the elements of are
I claim that
is a basis of This will complete the proof, since the Dynkin diagram is
are in as defined by So are
because this is the sum of three terms of the form
is the sum of six of So these are the positive roots, and is indeed a basis of
To obtain (resp. ) delete (resp. ).
For the positive roots are
So 63 positive roots, 126 roots altogether.
For the positive roots are
So 36 positive roots, 72 roots altogether.
Alternate description of
This time we work in with standard basis Let
and is even (because ). Let
is a root system of type In this description the elements of are
The Dynkin diagram is
Let be the lattice in generated by and
belongs to if and only if the are all integers or all half-integers. Let
is a root system of type
If the coordinates of can take only the values
So consists only of the vectors
Check (R1), (R2). Let
is a basis of The positive roots are
and the Dynkin diagram is
(highest root ).
Non-reduced root systems
We have already encountered
In fact these are the only non-reduced irreducible root systems. I shall omit the proof, which is not difficult but not particularly instructive (Bourbaki, Groupes et Algèbres de Lie, Ch. VI §1, Props. 13 and 14 (p. 152)).
Show is reduced, 2 root lengths
elements of pairwise orthogonal ⇒ type hence
Coxeter groups and Weyl groups
A Coxeter group is any group with generators say and relations
where is an integer or Graphically it may be described by a Coxeter diagram, which is a graph with one vertex for each generator the vertices
being joined as follows:
So it is not very different from a Dynkin diagram. irreducible ⇔ diagram connected.
Classification of Coxeter groups
finite, irreducible. We have seen that Weyl groups are Coxeter groups. To convert the Dynkin diagram of into the Coxeter diagram of we have only to replace PICTURE by PICTURE and
This however does not exhaust the list of finite irreducible Coxeter groups: there are also the dihedral groups (order )
for all (which are Weyl groups only for
( respectively)) and two more with diagrams
has order 120, has order 1440. is the symmetry group of an icosahedron: it contains the alternating group as a subgroup of index 2, but it is not isomorphic to likewise is the symmetry group of a 4-dimensional regular polytope.
affine, i.e. having a faithful representation in which the are reflections in hyperplanes in Euclidean space. The irreducible affine Coxeter groups are precisely the "affine Weyl groups", in one to one correspondence with the finite (reduced, irreducible) root systems. So their Coxeter diagrams are obtained from the affine Dynkin diagrams (left hand column) by replacing
Geometric representation of a Coxeter group
standard basis define a scalar product by
and let be the group of automorphisms of generated by the Then is isomorphic to (the isomorphism being
). (Proof omitted.)
The affine Weyl group
Let be a reduced irreducible root system, spanning For each and let
Let be the orthogonal reflection in the affine hyperplane
Let be the group of affine isometries of generated by the reflections is the affine Weyl group of Let
is the group consisting of all pairs with multiplication
From (A1)(ii) we have
for all Define
Claim that is an isomorphism. We have
so that is a homomorphism. It is injective because
for all hence
and so Also is surjective because (A1)(ii)
be the group of all with By (A1)(iv), induces an isomorphism
So has finite index in
is an open subset of and its connected component are therefore also open: they are called the alcoves of Since by (A1) permutes the hyperplanes it acts on as a group of homomorphisms, hence permutes the alcoves.
Fix a basis of and let
Let be the highest root.
is an alcove.
Let Then if and only if
Certainly also is convex (if then the line segment is contained in ), hence connected. Suppose Then for some either
so that is separated from by one of the hyperplanes It follows that is a connected component of i.e. an alcove.
Conversely if satisfies these inequalities, then
for all also for all hence is a sum of simple roots and therefore
From (A3)(ii) it follows that is an open Euclidean simplex with faces
These hyperplanes are the walls of Let reflection in
permules the alcoves transitively.
is generated by
Let be the subgroup of generated by the We shall show that permutes the alcoves transitively. So let be an alcove and let
The orbit of is a discrete subset of because by (A2) it consists of the points
hence so is Choose in this orbit at minimum distance from If then for some
and are on opposite sides of the hyperplane But then (see fig.) is nearer to than to contradiction. Hence but also hence So each alcove has walls
To complete the proof it is enough to show that for each Choose closer to than any other reflecting hyperplane and let be the alcove containing Then is a wall of and
for some and therefore (A1)
Let denote the set of hyperplanes
For each let
By (A4)(ii) we can write
Call this a reduced expression for if is as small as possible, and define the length of to be
be a reduced expression. Then
and these hyperplanes are all distinct. Hence
We proceed by induction on If then and is empty. Assume (A5) true for of length and consider of length Then we have to show that
Consider (b) first. If
for some From (A1) it follows that
and therefore that
Now consider (a). From above,
i.e. does not separate and hence does not separate and But separates and hence separates and So
(because is the only hyperplane separating and hence does not separate and hence does not separate and ).
From (A5), just as in Chapter I (Root Systems), we may deduce the exchange Lemma (1.8) for and hence that is a Coxeter group with generators
where is the order (possibly infinite) of in
permutes the alcoves simply transitively.
is a fundamental domain for the action of on
and hence is not empty, by (A5). Hence there is at least one hyperplane separating and Hence
for some alcove
So meets every orbit in and it remains to show that it meets it exactly once. So suppose and are both in We must show that If this is clear, so we may proceed by induction on If
there exists such that
i.e. separates and say
are the two open half spaces with boundary But then
we conclude that
Let be the basis of dual to
Then is a basis of
be the highest root of so that (Prop. 4.1) all the are
(A7) The elements of
are 0 and the such that
Then if and only if
So either all the are zero, i.e.
which means that
for some such that
For uniformity let us define
Then (A7) says that the elements of
are the Recall that is the group of all
it contains as a normal subgroup of finite index, and
Let be the subgroup of such that To describe the elements of we introduce the following notation: let
and let be the longest element of
and we have
(A8) The elements of are
so that number of labels in the extended Dynkin diagram of
If so certainly maps into Now let
let and let
We must check that i.e. that
Suppose first Then
because the coefficient of in is
We have therefore checked that the do belong to Conversely, let say
for some So
fixes 0, i.e. and Since (the Weyl chamber) it follows that
whence (because are disjoint if (1.21)). Hence
Then is an alcove, hence
for a unique by (A6), and hence
So factors uniquely as
and under this isomorphism the element corresponds to the coset of in
are coset representatives of
number of labels in the extended Dynkin diagram of
Follows from the remarks above.
are called the minuscule weights (or ).
(A10) Let The is minuscule if and only if
or for all
⇒) Let so that
for each If now is minuscule, for some such that so that
⇐) We have
or for each and
for each so that If on the other hand
then for some
Finite Dynkin diagrams
Affine Dynkin diagrams
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