Last update: 27 August 2012
This is a typed version of I.G. Macdonald's lecture notes on Kac-Moody Lie algebras from 1983.
In this chapter, is a Cartan matrix until further notice.
Recall that is the lattice generated by in , and that is a root of if and . The multiplicity of is
(and may well be ).
Let denote the set of roots. By (1.7), each root is either positive or negative . Moreover the involution interchanges and , hence if is a root, so is with the same multiplicity.
Let denote the set of positive roots. We have
(i), (ii) done above; (iii) because ; (iv) is clear.
(v) Let be a nonzero element of , so that (and ). Then and lies in , where . Thus and
Recall from Chapter I that if is a minimal realization of the matrix , then is a minimal realization of the transposed matrix . Let
be the lattice in spanned by . Then plays the role of for the Lie algebra (note that is also a Cartan matrix). The root system of is called the dual of the root system . The simple roots of are , etc.
For define by
Let denote the group of automorphisms of generated by . By (1.1) it depends only on the matrix if we need to make the dependance explicit. It is called the Weyl group of , or of .
acts contragradiently on :
In particular we have
for all .
From (1) and (2) it follows that acting on is the Weyl group of :
We want next to show that permutes the roots, ie that for each . For this purpose we recall (1.19) that are locally nilpotent derivations of , so that are automorphisms. Let us compute their effect on the generators of :
If we have , whence and therefore
Next, we have , whence
(2.3) for all (which justifies the choice of notation)
|Proof is calculation.|
(2.4) Let , then .
Let . Then we have
This calculation shows that , by (1.7) again, hence that . It follows that , i.e. that is a root; and then, replacing by we have (because ). Hence
so we have equality throughout, and hence .
(2.5) If and , then and .
Enough to prove this when is a generator of , and then it follows from (2.4).
(2.6) If , then . Thus permutes the set .
Say , so that the are and some , is . Then still has as coefficient of (for this ), hence is a positive root.
(2.7) Let be such that . Then
We shall next prove that is a Coxeter group, i.e. that it is generated by subject only to the relations of the form
where are positive integers (or ). The proof will depend only on the last two propositions (2.6), (2.7).
Each element can be written (in many ways) as a word in the generators (recall that ), say
For a given , the least value of is called the length of , and if , is called a reduced word for .
(2.8) Let be a reduced word for . Then the set of positive roots such that is negative is
(so that ).
By induction or . For this is (2.6). Assume and write (for convenience of notation) for , so that .
As a corollary of (2.8) we have
(2.9) If is such that , then .
For then , hence is empty, so and therefore .
In particular, it follows that acts faithfully as a group of permutations of :
(2.11) Exchange Lemma Let be a reduced expression, and suppose that . Then for some we have
i.e. we can "exchange" with some .
Induction on . If then and there is nothing to prove. Assume that , and let .
(2.12) Theorem. is a Coxeter group, generated by subject to the relations
where are given in terms of the Cartan matrix by the following table:
The exchange lemma (2.11) implies that is a Coxeter group (proof in Bourbaki, LG + LA, Chapter IV). It remains to compute the order of in . For the moment, exclude the case .
Let . Then together with and span . For iff , i.e. iff
and these equations have only the solution (since we are assuming ).
Now fixes pointwise, and on the 2-plane spanned by and it acts as follows:
and therefore the matrix of relative to the basis is
the eigenvalues of which are the roots of the quadratic equation
so we have the following table, in which :
When the eigenvalues satisfy , hence do not lie on the unit circle in , and therefore cannot be roots of unity; consequently has infinite order in this case.
Finally, when , the eigenvalues are both 1, but , hence is unipotent and therefore again of infinite order.
Remark. We may summarize the table in (2.12) as follows:
In particular, is "crystallographic" – it acts faithfully on the lattice hence embeds in .
Let be a (not necessarily minimal) realization of the Cartan matrix over . The fundamental chamber is
the transforms of under the elements of the Weyl group are the chambers; and the union of all the chambers
is called the Tits cone.
Define a partial ordering on :
(i) and (ii) will both follow from the
Lemma. Let , and let be a reduced word for . If and , then .
(2.14) The following conditions are equivalent:
(i) (ii) Let . The orbit of is finite; Let be a maximal element of this orbit for the ordering on . I claim that ; for if not, then for some , whence , impossible. Thus and therefore . Thus .
(ii) (iii) Choose such that . Then for all ; but , hence for almost all . It follows that is finite, hence so is .
(iii) (i) because acts faithfully on : (remark following (2.9)).
(i) (iv) because have the same Weyl group.
Let be any subset of the index set . Then if , is a realization of the principal submatrix . Define in the obvious way:
We have then
(2.16) Let . Then iff is finite. ( interior of )
We may assume that , because . Let , then by (2.13)(ii), and .
Suppose . Let , then for some , hence by (2.15). It follows that and hence by (2.14) that is finite.
Conversely, suppose that is finite, so that (2.14) . Let
Hence is an interior point of .
Let be a finite-dimensional real vector space. Recall that a non-empty subset of is a convex cone if is closed under addition and multiplication by non-negative scalars:
Let be a closed convex cone in , and let be the dual of . The dual cone is defined by
Clearly is a closed convex cone (it is an intersection of closed half-spaces). The basic fact we shall need is the
Dualilty theorem is the dual of , i.e.
Let denote the r.h.s. above. Clearly . The nontrivial part is to prove that , i.e. that if there exists such that ; or equivalently that if there exists a hyperplane in which seperates and .
Let be a positive definite scalar product on : write and , the usual Euclidean metric. We shall show that there is a unique point for which is minimal, and that the hyperplane (through 0) perpendicular to seperates from .
Let be the unit sphere in . For each , let denote the shortest distance from to the ray , so that by Pythagoras
Clearly is a continuous function on . Now is compact (because is compact and is closed), hence attains its lower bound on , i.e. there exists for which is a minimum (, since ). Moreover this is unique, for if and are both minimal, and then is strictly smaller, where .
Let , then . I claim that for all . For if is such that , then the angle is acute; hence if is the foot of the perpendicular form to the segment , we have (convexity) and , contradiction.
Now define by ; then and .
We shall require the following consequence of the duality theorem:
Lemma 1 Let be any real matrix. Consider the systems of linear inequalities ,
Then either (1) has a solution or (2) has a nontrivial solution.
Let be the standard basis of and let be the closed convex cone generated by the vectors of . We have
If on the other hand , then by the duality theorem there is a linear form on which is negative at and at all points of . But then satisfy the inequalities (1).
Until further notice, will be a real matrix satisfying
If satisfies these conditions, so does its transpose .
Assume also that is indecomposable.
Let be the space of column vectors . Write (resp. ) to mean (resp. ) for . Define the closed convex cones
The interior of is
Lemma 2 .
Let ; suppose for , for (where are complementary subsets of ). Then we have
for all , in particular for . But and , whence for all . Since is indecomposable, it follows that either , in which case ; or , in which case .
As regards and , there are two possibilities:
Case (a). By Lemma 2 we have . Now is connected (because convex) and we have
where is the complement of in , so that
as a union of disjoint relatively open sets, of which is not empty (by Lemma 2). Hence there are two possibilities:
is connected, in which case , i.e. so that (Lemma 2)
is not connected, in which case is a line in (i.e. a subspace). (For if lie in different components of , the line segment is contained in , hence must pass through 0, whence for some . If now , then either and lie in different components, whence (as above) ; or and lie in different components, in which case . Thus .)
Let be the null-space of . Clearly .
In case , contains no vector subspace of , because clearly doesn't. Hence , i.e. is nonsingular. In this case we say that is of positive type.
In case we have . For if , then ; but also , whence and therefore , i.e. . So and therefore . Hence is singular, of rank . We say that is of zero type.
In either case or , the inequalities
have no solution. For if then , which in case implies either or , and in case implies , i.e. . By Lemma 1, therefore, the inequalities
have a nontrivial solution, i.e. we have
where . It follows that the matrix satisfies (a), and is therefore of positive or zero type according as is (because ). Hence conditions (a), and therefore also (b) are stable under transposition.
Case (b).In this case and therefore also , i.e. the inequalities
have only the trivial solution. Hence, by Lemma 1, the inequalities
have a solution. We say that (and ) is of negative type.
(2.17) Theorem (Vinberg) Each indecomposable real matrix satisfying the conditions
belongs to exactly one of the following there categories:
|is nonsingular, and or||(positive type)|
|(0)||, and||(zero type)|
|such that ; and||(negative type)|
The transposed matrix belongs to the same category as .
Moreover, is of positive (resp. zero, negative) type iff such that (resp. , ).
Only the last sentence requires comment. Observe that is a finite intersection of closed half-spaces and therefore its interior is the intersection of the corresponding open half-spaces, i.e. , and is non empty if is nonsingular (because the columns of are then a basis of ).
If is of positive type, we have , hence , and is nonempty, whence with . If is of zero type, then intersects , hence with . Finally, if is of negative type, then with , from above.
Conversely, let . If , then cannot be of zero or negative type, hence is of positive type. If , then cannot be of negative type or of positive type (because ), hence is of zero type. Finally, if , the vector satisfies , whence is not of positive or zero type, hence of negative type.
(2.18) If is indecomposable and of positive or zero type, then for every proper subset of the principal submatrix has all its components of positive type.
We may assume that is indecomposable. By (2.17) such that and . Let ; then , and has components
The first sum on the right is , and so is the second, since for , , and . Moreover the second sum is 0 iff for all . Since is indecomposable, and such that , hence at least one of the components of the vector is . Thus we have
whence by (2.17) is of positive type.
A matrix satisfying the conditions is said to be symmetrizable if there exists a diagonal matrix with such that is symmetric.
Hence is symmetrizable iff can be made symmetric by multiplication (on the left or on the right) by a positive diagonal matrix;
(b) If is symmetrizable and is symmetric, then is uniquely determined by . For
and therefore (as if )
Call the symmetrization of : notation .
(c) If is indecomposable and symmetrizable, the diagonal matrix above is unique up to a positive scalar multiple. For if , then with we have , i.e. and therefore whenever , i.e. whenever and are joined by an edge in the graph of . Since is connected (1.11) it follows that all the are equal, i.e. .
Let be a matrix satisfying , its graph. Let
be a path in , i.e. are vertices of such that is an edge, so that and . Define
so that and has the sign of . If is the reverse path, then is positive.
(2.19) is symmetrizable for each loop in .
In particular, is symmetrizable if is a tree.
diagonal matrix such that is symmetric: whence for any loop in (the middle equality because is symmetric).
: Fix an index . For each there is at least one path from to in (we may assume is indecomposable, i.e. connected). If is another such path, then is a loop, hence
Hence we may define unambiguously
(in particular, ).
I claim that . This is trivially true if or if and are not lined by an edge in , for then both sides are 0. If is an edge in , and if is a path from to in , then is a path from to , and so that
Hence the matrix is symmetric, i.e. is symmetrizable.
Terminology. For a Cartan matrix (indecomposable)
Assume from now on that is an indecomposable Cartan matrix. Then all the previous results are applicable.
(2.20) Let be an indecomposable Cartan matrix of finite or affine type. Then either the graph of is a tree, or else is a cycle with vertices and
(i.e. if : indices mod ). This is of affine/zero type.
Suppose contains a cycle. By considering a cycle in with the least possible number of vertices, we see that has a principal submatrix of the form
for some (after a permutation of the indices) with when . By (2.18), is of finite or affine type, hence such that : i.e. such that
(where suffixes are read modulo ). Add these inequalities:
Now the 's are integers , hence
Since the are we must have equality throughout; so the 's are all , and , whence is of affine/zero type and therefore (2.18) .
(2.21) Let be an indecomposable Cartan matrix of finite/positive or affine/zero type. Then is symmetrizable, and its symmetrization is positive definite or positive semidefinite (of corank 1) according as is of finite/positive or affine/zero type.
The first assertion follows from (2.19) and (2.20). By (2.17) such that , hence if we have and , more precisely, if is of positive type, if is of zero type. It follows that for all scalars , hence is of positive type for all , hence is nonsingular (2.17). Thus the eigenvalue of (which are real, because is a real symmetric matrix) are all . If is of positive type, they are all , whence is positive definite; and if is of zero type, 0 is an eigenvalue of with multiplicity 1, and all other eigenvalues of are .
Remark. If is a symmetrizable indecomposable Cartan matrix of negative type, then is indefinite. For since the quadratic form takes positive values, and since by (2.17) with , the vector satisfies and , whence . Hence, for symmetrizable
(2.23) Let be an indecomposable Cartan matrix in which all proper principal minors (i.e. for all ) are positive. Then
(i) Let be the cofactor of in . We may assume that . The expansion of is
summed over all permutations of . Write as a product of disjoint cycles, say . Now is the coefficient of in the expansion (1), hence we have to consider all cycles such that : say , and correspondingly the product . Thus product will be zero unless is a simple path (i.e. with no repeated vertices) from to in the graph of ; and then it will be equal to , in the notation of (2.20). Thus it follows from (1) that
where the sum is over all simple paths from to in , and is the complement of the set of vertices of . Each term in the sum (2) is , and the sum is not empty, because is connected. Hence .
(ii) By (2.17), is of positive (zero, negative) type according as with . Let , then premultiplication by the matrix of cofactors gives
Since and , this gives the result.
(2.24) Let be an indecomposable Cartan matrix. Then the following are equivalent:
(a) (b) If is of positive type, its symmetrization is positive definite. Hence each principal minor of , being equal to the corresponding principal minor of , is positive. In particular, .
If is of zero type, we know (2.17) that and (2.19) that each indecomposable proper principal submatrix is of positive type, hence from above.
(b) (a) follows from (2.23).
If is an indecomposable Cartan matrix of positive or zero type, then for each pair we have by (2.18) and (2.24)
(with equality only when and is of zero type). Thus
This is a fancier version of the graph of that we defined earlier. In the Dynkin diagram of the vertices are connected by lines, with an arrow ponting towards if . Thus the possibilites are given in the following table:
We do not attempt to define a Dynkin diagram for Cartan matrices in which for some pairs .
The table above shows that determines uniquely. Observe also that the Dynkin diagram of is obtained from that of by reversing all arrows; also that is indecomposable iff is connected.
We shall say that is of finite type (resp. affine type) according as is.
We begin by verifying the last statement. The equation , i.e.
can be rewritten as followss:
where the sum is over all the vertices j in joined directly to , and
[except if .]
Then (ii) is easily checked diagram by diagram. It follows from (2.17) that all the diagrams in Table A are of affine type. Since each diagram in Table F occurs as a subdiagram of one in table A, it follows from (2.18) that all diagrams in Table F are of finite type.
(The number of vertices is .)
|(The number of vertices is ).|
If is of (finite) type , then is the number of 1's in the diagram . The 1's form a single orbit under the group .
We now have to prove the converse, namely that every connected diagram of finite or affine type occurs in Table F or Table A. Let be the number of vertices in .
If , the only possibility is .
If , we have already enumerated the possibilities: of finite type, and , of affine type.
If , either is a tree or is (by (2.21)). If is a tree then
where are positive integers; by (2.19) and , so that ; more precisely, or 3 if is of finite type, if is of affine type. So the possibilities are
From (2.19) we have:
If a subdiagram of occurs in Table A, then it is the whole of . Hence to show that Table A it is enough to show that some subdiagram of is in Table A.
If is not a tree then by (2.20) . So we may assume that is a tree, and that .
If contains multiple bonds, they are double bonds . For otherwise contains either or , in which case by ; or contains a connected proper subdiagram of 3 nodes containing a triple bond, which (see above) can only be or , and therefore , by again.
If contains two or more double bonds, it contains a subdiagram of type or or , hence by this subdiagram is the whole of .
So we may assume that contains at most one double bond.
Suppose now that has at least one branch point. If it contains a double bond as well, then it contains a subdiagram of type or , and again by this subdiagram is the whole of . So assume now that is simple-laced (i.e. no multiple bonds). If there is more then one branch point, then contains as a subdiagram, hence again this is the whole of . So we may assume that has only one branch point. If there are edges issuing from this point, then contains as a subdiagram, which is therefore the whole of . If there are 3 edges issuing from the branch point, let be the number of vertices of on the three arms, where (and ).
If then contains
, hence by
If then contains , hence .
If then contains , hence .
If then respectively.
If then .
Finally assume has no branch points, hence is a chain. If is simply-laced, then . The remaining possibility is that contains just one double bond. Suppose there are nodes on one side of the double bond, on the other, where .
If then contains
as a proper subdiagram, contradicting
If then contains or as a subdiagram. Now apply .
If we obtain ; and if we obtain or .
Explanation of notation: Let be any of the symbols a (finite) root system of type ; the simple roots, the lowest root (i.e. is minimum). Let
Then is the Cartan matrix of type (finite type) and is the Cartan matrix of type (affine type).
We have say, with the positive integers, i.e.
if we define ; but then
showing that is of affine type and the are the labels attached to the nodes of the diagram of type in Table A.
Remark. The classification theorem (2.22) gives a complete list of the indecomposable Cartan matrices of finite or affine type. Any Cartan matrix not in this list is therefore of indefinite (i.e. negative) type, so the Cartan matrices of indefinite type are the pink-heap. However, there are two subclasses which can be explicitly classified: an indecomposable Cartan matrix of indefinite type is said to be hyperbolic (resp. strictly hyperbolic) if every proper principal submatrix has all its components of finite or affine type (resp. of finite type). All the Cartan matrices with are of strictly hyperbolic type; apart from these there are only finitely many, and they all have Dynkin diagrams.
Exercise. If is an indecomposable Cartan matrix, then
Here are two with : (probably the only two)
and (I think) the only strictly hyperbolic matrix with has diagram
If is hyperbolic and is affine ( connected) then . For otherwise we should have with connected and both inclusions strict, and then would not be either affine or finite type.
(2.26) Let be a Cartan matrix. Then is symmetrizable iff there exists a –invariant symmetric bilinear form on (with values in ), such that is positive rational for all . Moreover such a form is nondegenerate.
Suppose is symmetrizable, then such that for all . Since the are integers we may assume that the are rational (or even positive integers).
As before, let , and let be a vector space complement of in . Define as follows:
(Notice that and also in particular, therefore, so that the above definition is unambiguous.) Now we have
which is symmetrical in and , so that
from which it follows that for all , by induction on . This is –invariant.
Conversely, if we have such a –invariant form on , then from (1) it follows that
for all : taking , we obtain
and therefore, putting ,
which shows that is symmetrizable.
Finally, if for all then in particular , so that (1.10); but then say, and
so that in and therefore .
From (2.26) it follows that the mapping
is an isomorphism, and we can therefore transport the scalar product
Let . Then
so that ; taking we obtain (since )
is the coroot of .
Finally note that
Action of :
For each , let be the hyperplane in . Then is reflection in this hyperplane: because if is the midpoint of then
and is a scalar multiple of .
So is realized as a group generated by reflections.
We can now characterize the algebras for which is a Cartan matrix of finite type:
(2.27) Let be an indecomposable Cartan matrix. Then the following conditions are equivalent:
(i) (ii) by (2.26) and (2.21)
(ii) (iii) (Here ). The matrix is nonsingular, hence is a basis of and therefore is a lattice in . Consequently is a lattice in the real vector space .
Let be the orthogonal group of the form , acting on . is compact, hence a bounded subset of ; is a subgroup of and preserves the lattice . i.e. , which is finite.
(iii) (ii) Let be any positive definite scalar product on . Then
is –invariant and positive definite. Hence is symmetrizable, by (2.23).
(iii) (iv) proved earlier (2.14)
(iv) (v) because (direct sum) (1.7); and is simpe by (1.13).
Now suppose that is an indecomposable Cartan matrix of affine type. By (2.17) there is a unique vector
with components which are mutually prime positive integers, such that
Dually, the matrix is also of affine type, hence there is a unique vector
with components which are mutually prime positive integers, such that
Now is symmetrizable (2.21), hence there is a diagonal matrix (with positive diagonal entries ) such that is symmetric, i.e.
But then by (2) and (3), so that from (1), for some scalar . Replacing by we have then , i.e.,
Then we have
(We shall see later that is a root) for by (1). Hence fixes . Dually define
then we have
for by (2). Hence fixes .
Recall that is the centre of , and that he. Thus (we are taking here): is the canonical central element.
Next we shall construct the scalar product on as in (2.26). We have , so we can take as a basis of the elements and say, where . (By (5) we must have ).
Remark. this of course does not determine uniquely: we could add on any linear combination of . At this stage, however, that doesn't matter.
We have then
From (6) and (7), therefore,
and in particular
As in (2.26) let be the isomorphism defined by the scalar product, so that . Then
because , so that
Now consider the action of the Weyl group . We shall show that acts as a group generated by reflections in a real Euclidean space of dimension . Since , we have to cut down the number of dimensions by 2. We do this in two stages.
(1) Since fixes it follows that acts (faithfully) on . Each such that defines a linear form on , which we denote by . Thus we have . Notice that (in its action on ) fixes . If is the projection, the image of the fundamental chamber is the set of such that and the image of the Tits cone is the union of the , .
(2) For each real number let
If this is an affine hyperplane in , hence of dimension . If , , where as usual is the subspace of spanned by . (For .) Each is stable under the action of because fixes . Moreover acts faithfully on . For if fixes pointwise, it fixes all points of not in , hence fixes all points of (for the fixed point set of is in any case a vector subspace of ), whence is the identity. Thus we have realized as a group of affine-linear transformations of (any ). These actions of are all essentially the same, so we may as well take and concentrate attention on the affine space .
Now the restriction to of the scalar product is the positive semidefinite, or rank (because , and the matrix is positive semidefinite of rank ); also for all , by (8). Hence we have a positive definite scalar product on , and therefore has the structure of a Euclidean space of dimension . The restriction of to is an affine-linear function on this Euclidean space, and is an affine hyperplane in ; these hyperplanes are the focus of an –simplex in , namely . The generator of acts on as reflection in the hyperplane . Thus is realized as a group generated by reflections in the Euclidean space subscript?.
The transforms of the "fundamental alcove" therefore fill up the space . Consequently the union of the chambers for all is the open half space together with the origin. Pulling back to , we see that the Tits cone is
By (2.1) and (2.5) each is a root of multiplicity . Kac calls these the real roots. In the classical case, where the Cartan matrix is of finite type, all the roots are real (proof later). In general however there will be other roots as well, whcih Kac calls imaginary roots. (The justification for this terminology will be apparent shortly.)
Let denote the sets of real and imaginary roots, respectively. Also put
So by definition
Likewise for the dual root system , with simple roots :
Consider the real roots first. If define the coroot of to be
This definition is justified by (2.7), because if also , then we have and therefore , i.e., .
We have then
Next, for a real root we define (acting on and on ) by the formula
and we verify easily that
(2.28) The mapping is a bijection of into such that
(i), (ii) are clear. As to (iii), let , then since we have , hence by (2.10) , hence again by (2.10) (applied this time to ) , i.e. .
Next we consider the imaginary roots.
The next proposition justifies the names "real" and "imaginary".
(2.30) Assume that the Cartan matrix is symmetrizable, and let be a –invariant symmetric bilinear form on , as in (2.26). If is a root then
If is real, say , then (by our choice of scalar product).
Conversely, suppose . By (2.29) we have for all , and . Hence we may assume that has minimum height in its orbit , i.e. that for all .
Since , we have
and therefore , i.e. and therefore also . But , say, with coefficients , hence
(2.31) Let and let be a simple root such that . Then the set of integers such that is a root is a finite interval in , where and . (If ) is positive all these roots are positive.
Without loss of generality we can assume , say with coefficients . If is a root, we must have (because some is , since ), i.e. . It follows that the set is bounded below (and is not empty, because ).
is in any case a disjoint union of intervals in . Suppose is one of these intervals, and consider the vector space
Then is stable under and – for example, which is either contained in or is zero. Hence is stable under . From (2.5) it follows that the set is stable under . But now, since
it follows that the mapping maps the interval onto itself. Since is bounded below (because is), must be a finite interval with midpoint . Hence all the component intervals of have the same midpoint, hence there is only one component, i.e. is an interval with midpoint , i.e. (and because ).
The set is the –string through .
Corollary. If is not a root, then . (For in (2.31).)
(2.31) valid for any real root : if , , then the –string through is
where and .
For ; now apply (2.31) to and : , .
The following results should have occurred in Chapter I: they are valid for any matrix (satisfying the condition iff , so that the graph of is defined). If is any subset of we have the principal submatrix and its graph , which is the full subgraph of obtained by deleting the vertices of not belonging to . If is connected we shall say simply that is connected.
Now let be any element of . The support of , denoted by , is defined to be the set
(2.32) Let . Then is connected.
Let , then , the root system of . If is not connected then is decomposable, say and hence (1.12) . Hence the root space lies in either or ; in either case, : contradiction.
(2.33) Suppose is infinite. Then for each such that .
Suppose not, then such that . Let , . Then (because ), from which it follows that and therefore the ideal generated by in is
Hence unless . But by (1.13) we have (because clearly ), hence in particular , i.e. for all positive roots . Hence all satisfy , whence (and therefore ) is finite.
Let be a positive imaginary root. By (2.29) is positive for all . We shall say that is minimal if for all .
This implies in particular that ; since , it follows that , i.e. that , the dual fundamental chamber. If also is minimal, then , and therefore by (2.13)(i) (applied in the dual situation, i.e. to ) we have . Thus each –orbit of positive imaginary roots has a unique minimal element.
Conversely, if is such that , then and hence by (2.13)(iii) we have for all , i.e. ; hence , and therefore is a minimal positive imaginary root.
If , then . Thus for al minimal positive imaginary root the vector satisfies
from which it follows from (2.17) that the Cartan matrix cannot be of finite type. Thus if is of finite type there are no imaginary roots.
From these remarks and (2.32) it follows that the minimal imaginary root satisfies
(i) is connected; (ii) .
In fact these necessary conditions are also sufficient:
(2.34) Let . Then the following conditions are equivalent:
We have just observed that (ii) (i).
(i) (ii). From the remarks above, it is enough to prove that is a root.
Suppose then that is not a root. Let , so that say
Since for all it follows as above that the matrix is not of finite type.
Choose a positive root of maximal height, say
Let , say
An immediate corollary of (2.34) is the last part of (2.29): if is an imaginary root, then so is for all integers .
For we may assume that is positive and minimal, and then satisfies the conditions of (2.34) for any integer .
(2.35) Let be an indecomposable Cartan matrix.
Let be a symmetrizable Cartan matrix. The standard bilinear form on may clearly be chosen so that the scalar products are integers. It follows that the number
exists and is . (Notation: .)
(2.36) Let be an indecomposable Cartan matrix of finite or affine or symmetrizable hyperbolic type. If and , then or .
Suppose not, then , where , and the support of respectively are disjoint. We have
Suppose first that all components of and of are of finite type. Then and are , hence ; also (because if ). But this contradicts (1).
Suppose then that (say) has a component of affine type. Then this component is the whole of , and consists of a single vertex , by virtue of (2.18); moreover by connectedness we have for some , whence . But this time we have and , whence again (1) is contradicted.
(2.37) Let be as in (2.36).
Let be such that . Then is a real root, and hence
(i) We have for all , hence by (2.36). Replacing by if necessary, we may assume that . Let be of minimal height in , say . Then we have
so that for some index , i.e. . But then has height less than , hence and therefore for , i.e. ; but then , because . Hence and therefore is a real root, and .
(ii) Suppose is a long (real) root of . Then is a short real root of ; but , so that for all . Conversely if this contradiction is satisfied, then and hence and therefore .
(iii) If , then by (2.30). Consequently, suppose and . By (2.36) we may assume . Again choose of minimal height in , say . This time
Suppose for some index ; then , whence , which as before implies that , whence . Consequently , i.e. , for . To complete the proof, by (2.34) it is enough to show that is connected. If not, let be a component of , and , and put . Then are of finite or affine type, whence and ; also (because for all ). Hence , contradiction.
Hence is a minimal imaginary root, hence .
We can now describe the affine root systems explicitly.
Assume that the Dynkin diagram is not of type . Then there exists an index such that , by inspection of Table A. Denote this index by 0 and the others by , where . Let the subdiagram of obtained by erasing the vertex 0 from . Then is of finite type and is a root system with as its Dynkin diagram, hence finite by (2.27). As before let
Then or 3, and or for all real roots (again by inspection of Table A). (This is not true for the excluded case , where there are roots of 3 lengths.) Let ; likewise , (short roots and long roots). (If then .) Finally let
so that if either is simply-laced or is a long root.
(2.38) If is of affine type (Table A) but not of type , then
Let , then (because ). We have and , where
Finally, if is of type , choose (as before, but this time ); then is of type , so that explicitly the roots in are
where , so that or 2.
In we have or 4 (for a real root): short, medium and long. One finds
(2.39) Let be an indecomposable Cartan matrix, the Tits cone (here ). Then the closure of (in the usual topology of ) is given by
This is clear if is of finite type, for then , and . If is of affine type we have seen earlier that iff either or , so that iff ; and the positive imaginary roots are positive integer multiples of (2.33).
So assume that is of indefinite type, and let
Clearly the fundamental chamber , and is –stable by virtue of (2.29)(i or ii?). Hence for all , and therefore ; moreover is closed, because it is an intersection of closed half-spaces, so that .
Conversely, let and assume first that . Choose a positive imaginary root such that (2.35)(iii). To show that it is enough by (2.13)(iv) to show that there are only finitely many positive real roots such that , i.e. such that .
For such an we have
where (because ). Since is a positive imaginary root and , we have and therefore
So has height , and therefore there are only finitely many possibilities for . Hence . By replacing by a rational scalar multiple of , it follows that
But these are dense in , hence .
(2.33) has the following geometrical interpretation. Let be the positive imaginary cone, i.e. the cone in generated by the positive imaginary roots, i.e., is the set of all finite linear combinations with and . Then
(2.40) The cones in and in are duals of each other.
When the Cartan matrix is of indefinite type, the only case (to my knowledge) in which the closure of of the Tits cone can be explicitly described is that in which is hyperbolic and symmetrizable.