Coxeter groups
Part I. Geometry and combinatorics

There are naturally occurring cases where S is infinite, but in these notes S will be assumed to be finite. We shall see eventually that this is not a serious restriction.

Associated to a Coxeter matrix is a Coxeter diagram. Its nodes are indexed by S, and there is an edge between two nodes s and t if m_s,t is 3 or more. This edge is labeled by m_s,t , usually only implicitly if m_s,t = 3 . The Coxeter matrix is said to be irreducible if its Coxeter diagram is connected.

For example, if the Coxeter matrix is then its Coxeter diagram is .

The Coxeter group W = W_S associated to a Coxeter matrix m is the group with generators S and relations

• s² = 1
• st ... = ts ... (with m_s,t factors on both sides) whenever s and t are distinct and m_s,t is finite

This simple definition conveys, without elaboration, no idea of how interesting such groups are. They are among the most intriguing of all mathematical structures.

• deleting a pair ss;
• inserting a pair ss;
• replacing one side of a braid relation by the other.

Exercise 1. Verify that if w = s₁ ... s_n then w^-1 = s_n ... s₁.

Exercise 2. Verify that W is always a group.

Exercise 3. Verify in general that for all s and t the product st satisfies (st)^m_s,t = 1.

Exercise 4. Verify that we could have stipulated W to be defined by relations (st)^m_s,t = 1.

If a Coxeter diagram decomposes into two components S₁ and S₂, then the corresponding group is a direct product of the groups parametrized by S₁ and S₂.

Exercise 5. Verify this last claim.

All words in an equivalence class have the same parity - either even or odd. The sign of w is defined to be 1 if its parity is even, -1 if odd. The sign is a homomorphism from W to the multiplicative group of 1, -1.

A word is said to be reduced if it is of minimal length in its equivalence class. The length l(w) is the common length of all the reduced words of w. The following is an immediate consequence of this definition:

Proposition 1. The length has these properties:

• l(xy) l(x) + l(y);
• l(w^-1) = l(w);
• l(w) = 0 if and only if w is the identity;
• l(w) = 1 if and only if w lies in S (more properly, has a representative in S);
• sign(w) = (-1) ^l(w);
• l(ws) = l(w) + 1 or l(w) - 1;
• l(sw) = l(w) + 1 or l(w) - 1;

Exercise 6. Verify this claim.

Exercise 7. Suppose that S has two elements s and t. Let m = m_s,t . (a) If m is finite, verify directly from the definition that W has 2m elements, and that exactly one of them has two expressions as reduced words. (Hint. It is straightforward to see there are no more than 2m elements. To see that there are exactly that number, represent the group by permutations of 2m elements, effectively that corresponding to left multiplication on the group itself. ) (b) If m is infinite, verify that every element of W has exactly one reduced expression, and that elements of W - other than the identity - correspond bijectively to sequences st ... and ts ... with no repetitions.

If T is a subset of S, then the inclusion of T in S induces a canonical map from W_T to W_S. It is not clear a priori that this is an embedding, although it will turn out that this is so, or that a shortest expression of an element in the image by elements of T will also be one by elements of S. For the moment, all we can see easily is that

• if w is an element of W_T with image w_* in W_S, then l(w_*) l(w).

Recall that a linear reflection is a linear transformation fixing a hyperplane passing through the origin and acting as multiplication by -1 on a line through the origin transverse to this hyperplane.

There are other kinds of reflections, too, but they reduce to linear ones. They are discussed at the end of this section.

Single reflections

The hyperplane fixed by the reflection is where a = 0, and a^v is mapped to its negative as long as < a , a^v > = 2.

The function a and vector a^v are unique up to non-zero scalar multiples. If a is replaced by ca then a^v is replaced by c^-1a^v.

Often a reflection will be orthogonal with respect to some inner product , when the reflection takes v to

v - 2 (a^v v) / (a^v a^v) .

In this situation, the dot product induces a linear map from V to the linear dual space V^*, and a is the image of 2 a^v / (a^v a^v).

The group GL(V) acts by its definition on V, and on V^* according to the prescription < ga, gv > = < a, v >. (If vectors are column vectors, then linear functions are row vectors and the canonical pairing is the matrix product a v. If g is represented by the matrix M then g takes v to Mv and a to a M^-1.) At any rate, the group GL(V) acts transitively on pairs (a, a^v), hence all reflections are conjugate to each other.

Pairs of reflections

Let s and t be a pair of reflections, say corresponding to a, a^v and b, b^v. Define associated real constants

cs, t = < a, bv > ct, s = < b, av > .

Let (s₁, t₁) and (s₂, t₂) be pairs of reflections, conjugate under the linear transformation g. Then

ga1 = ca a2 ga1v = ca-1 a2v gb1 = cb b2 gb1v = cb-1 b2v

for some non-zero constants c_a and c_b, and consequently c_{s2, t2} = (c_b/c_a) c_{s1, t1}, c_{t2, s2} = (c_a/c_b) c_{t1, s1}. The constants c_{s, t} and c_{t, s} are therefore not conjugation-invariant, but the product n_s,t = c_{s, t}c_{t, s} is. For generic pairs it will turn out to possess a simple interpretation.

One way to understand conjugation-invariant phenomena is to consider geometrical configurations - for example, how the various hyperplanes and lines relate qualititatively to each other. One possibility is that in which the reflection hyperplanes of s and t are the same. In this case we may take a = b. Then

cs, t = < a, bv > = < b, bv > = 2 ct, s = < b, av > = < a, av > = 2

so that n_{s, t} = 4. Similarly n_{s, t} = 4 when a^v and b^v are linearly dependent.

Proposition 2. When n_{s, t} is not equal to 4, then a and b are linearly independent, as are a^v and b^v. The plane spanned by a^v and b^v is transverse to the linear subspace of codimension 2 where a and b are both equal to 0.

Proof. Only the last claim remains to be proven.

Exercise 8. Verify the last claim.

We are especially interested in the case where n_{s, t} = 4 but a and b are linearly independent. If L is the intersection of the kernels of a and b, then s and t induce reflections on the two-dimensional quotient V / L. In this plane a^v and b^v are linearly dependent. We have c_{s, t}c_{t, s} = < a, b^v > < b, a^v > = 4, and we can scale a so that in fact c_{s, t} = c_{t, s}, both being equal to 2 or -2. In either case, a^v and b^v lie on a single line through the origin, and both s and t transform a vector v in a direction parallel to that line. In particular the products st and ts are shears along that line.

s	t

The region on one side of that line and in between the two lines a=0 and b=0 will be a fundamental domain for the group acting on the corresponding open half-plane. If we choose the signs of a and b correctly we may assume that this region is that where a > 0 and b > 0. In that case, with our normalization, c_{s , t} = c_{t, s} = -2.

Exercise 9. In the situation described above, explain how s and t act on the space V^* dual to V. This is of importance in the theory of Kac-Moody Lie algebras, since V^* contains the roots of the algebra.

Exercise 10. Show that in dimension 2 there are three conjugacy classes of pairs (s, t) with n_{s, t} = 4 and in dimensions greater than 2 there are four.

From now on, suppose that n_{s, t} is not equal to 4. Then the vectors a^v and b^v span a plane transversal to the intersection of the kernels of a and b, and in effect we may restrict ourselves to dimension 2. In these circumstances, reflections fix lines.

There is one other exceptional collection of cases to deal with - when n_{s, t} = 0. This happens only when either c_{s, t} = < a, b^v > = 0 or c_{t, s} = < b, a^v > = 0. If both are equal to 0, then s and t are a commuting pair of reflections whose product amounts to multiplication by -1. The group generated by s and t has four elements, and any one of the four quadrants is a fundamental domain.

Otherwise, suppose that < a, b^v > = 0, but < b, a^v > is not equal to 0. We may arrange that < b, a^v > is in fact positive.

Then the line a = 0 is fixed by both s and t. The reflection s interchanges the two sides of this line, but t preserves it. The pair of reflections t and sts now both stabilize each side of the line a = 0, and the n-invariant of this pair is 4, since they share the eigenvector b^v. Therefore from what have already learned in this case, we know that with a suitable choice of sign for b the region C where b > 0 and sb > 0 on the side of the line where a > 0 is a fundamental domain for the group generated by t and sts. Furthermore, the reflection s takes C to -C. Note also that the vector a^v lies in its interior, as shown in the figure. Later on we shall need a simple consequence of this clear picture of how the pair s and t act: in this case the region where a > 0, b > 0 is not a fundamental domain for the group generated by s and t. Or, in other words, if n_{s, t} = 0 and this region is a fundamental domain then s and t are a commuting pair of reflections, and c_{s, t} and c_{t, s} both vanish.

From now on, we assume n_{s, t} to be neither 0 nor 4.

Proposition 3. If n be neither 0 nor 4, there is a unique conjugacy class of pairs (s, t) of reflections with n_{s, t} = n.

Proof. Let s be any reflection, say corresponding to a, a^v. All choices of s are certainly conjugate. We may as well take a to be the function x and the vector a^v to be (1, 0). The matrices commuting with s are then the diagonal matrices. We now look for b and b^v with < b, b^v > = 2, < a, b^v > < b, a^v > = n. Let b be any linear function independent of a and not vanishing on a^v. All choices are conjugate by a linear transformation preserving a and a^v. Suppose c = < b, a^v >. It remains to choose b^v independent of a^v with < a, b^v > = n/c. This requires that n differ from 4, and then all choices are conjugate by a linear transformation fixing both the line a = 0 and that through a^v.

Proposition 4. Take a^v and b^v as a basis of V. The matrices of s, t, and st are then

,

,

As one consequence, -2 + n_s,t is the trace of st, hence clearly conjugation-invariant.

Proposition 5. Suppose n_{s, t} to be neither 0 nor 4. Then there exists a non-degenerate quadratic form invariant under both s and t, unique up to scalar multiple. It will be definite if and only if 0 < n_{s, t} < 4.

Proof. Suppose that u v is an inner product invariant under s and t. This happens if and only if a^v is perpendicular to the line a = 0, and similarly for b^v and b. The space of vectors fixed by s is spanned by c_{s, t} a^v - 2b^v, and that fixed by t is spanned by 2a^v - c_{t, s} b^v. For invariance, therefore, we must have

(cs, t av - 2bv) av = 0 (2av - ct, s bv) bv = 0 cs, t (av av) - 2 (bv av) = 0 cs, t (av av) = 2 (bv av) 2 (av bv) - ct, s (bv bv) = 0 ct, s (bv bv) = 2 (av bv)

Since n_{s, t} is not 0, neither c_{s, t} nor c_{t, s} vanish. Therefore the single number a^v b^v determines the inner product on all of V.

The matrix of the quadratic form is

In order for the form to be definite, it is necessary and sufficient that c_{s, t} and c_{t, s} have the same sign, and that the determinant be positive. Therefore we must have 0 < n_{s, t} < 4. QED

Proposition 6. In order for the group generated by s and t to be finite, it is necessary that one of these hold:

• s = t (when n = 4);
• s and t commute (when n = 0);
• n = 4 cos² p / m for some positive integers m and p < m.

The picture accompanying the last assertion is this:

Exercise 11. Prove this in detail.

Suppose now that n is not 4, and that s and t generate an infinite group. We want again to see when the region a > 0, b > 0 is a fundamental domain. From one of the Propositions just proven, we must have n > 4 or n < 0.

Proposition 7. The elements s and t generate an infinite group with fundamental domain a > 0, b > 0 if and only if n is greater than or equal to 4, and both c_{s, t} and c_{t, s} are negative.

Proof. The case n = 4 has been dealt with. It remains to show that n cannot be negative. This case can be eliminated by using the invariance of the pair of lines where the indefinite metric vanishes. The following images portray what things look like (the red lines are where the invariant indefinite quadratic form vanishes):

n > 4	n < 0

Exercise 12. Finish the proof. (Hint: the reflection s preserves the `cones' through which a = 0 passes. What does it do to the other two? This situation is somewhat like the one we saw where n = 0.)

In summary:

Theorem 8. Assume that s and t are a pair of reflections in space with reflection data a, a^v and b, b^v. Assume also that the region a > 0, b > 0 is a fundamental domain for the group they generate. Exactly one of the following cases occurs:

• s and t commute, and c_{s, t} = c_{t, s} = 0;
• st has finite order m > 2, and n_{s, t} = 4 cos² /m;
• st has infinite order and n 4.

In the last two cases, c_{s, t} and c_{t, s} are both negative.

Affine reflections

for some linear function a and constant c. The linear function a is the gradient of f, and for any point x in V and vector v we have f(x + v) = f(x) + < a, v >. The reflection then takes a point x to x - f(x) a^v for some vector a^v in V with < a, a^v > = 2.

An affine reflection is a special case of a linear one, through the familiar trick of embedding an affine space of dimension n into a vector space of dimension n+1. Thus v in V maps to (v, 1) in a larger space V_#. Let be the function on V_# taking (v, x) to x. The affine reflection on V corresponding to a, a^v, and c is the restriction to V of the reflection defined by the linear function a + c on U. In the discussion of pairs of reflections with n = 4 we have already seen an implicit example of this.

Non-Euclidean reflections

The non-Euclidean space H_n of n dimension is the component Q(x) = x_n+1² - x₁² - ... - x_n² = 1, x_n+1 > 0 of the hyperbolic sphere, which can be parametrized by Euclidean space E_n according to the formula

It is taken into itself by the connected component of the orthogonal group of Q and there is a unique Riemannian metric on it invariant under this group and restricting to the Euclidean metric at (0, 0, ... , 0, 1). There are two models of this in Euclidean space of n dimensions, the familiar Poincaré model and the slightly less familiar Klein model. (There is also, for n=2, the upper half-plane, but that is another story.) Both models identify H with the interior of the unit ball in E_n centred at the origin.

For the Klein model, this ball is identified with the intersection of the slice x_n+1 = 1 of the region Q(x) > 0 (the interior of a homogeneous cone). A point x on H maps to x_* = x/x_n+1. It happens that a non-Euclidean geodesic line between two points x and y is the intersection with H of the plane through the origin containing x and y, and this intersects the slice in a line between x_* and y_*. Thus in the Klein model. points of H are identified with points of the interior of a unit ball, and geodesics between such points are line segments in the ball.

The Poincaré model is a transformation of the Klein model. A point x in the interior of the unit ball in E_n maps to the point above it on the upper hemisphere in E_n+1, then by South pole stereographic projection back onto the equator. In this transformation, geodesics become arcs of circles perpendicular to the boundary of the unit ball.

The point for us is that non-Euclidean reflections are those induced on the hyperbolic sphere, or either of its models, by linear reflections in E_n+1. For both models, we start with a hyperplane a = 0 which intersects the region Q(x) > 0 and a vector a^v transverse to it. In the Klein model, we take a point x in the slice Q(x) > 0, x_n+1 = 1 to x - < a, x > a^v and then project it back onto the slice. We have seen examples of this also in the discussion above of pairs of linear reflections with n > 4. In the Poincaré model we unravel by stereographic projection, reflect, and ravel again.

Exercise 13. Design an algorithm to draw the geodesic between two points of the unit disc in the Poincaré model. There are two ways to approach this problem, depending on the tools available for drawing. (1) Suppose all you can do is produce line segments and circles, then you must first decide whether to draw a line or a circle; if it is a circle, you must find the centre, radius, and arc. This becomes delicate only for points nearly lying on a diameter of the unit disc, and suffers to some extent from stability. Discuss how serious this is, analyzing how continuously your drawing depends on the two points. (2) if you can draw Bezier curves, then you should probably use them to approximate circles. Use one or two bisections to allow a reasonable approximation; then use velocity vectors to find the control points. The best way to do all this is likely through stereographic projection, since the geodesics in the disc correspond to simple latitude circles on the sphere.

Finite dihedral groups

2 / m

Rotations are not its only symmetries, since any line through the center of any of its sides and the origin, or any of its corners and the origin, is an axis of mirror symmetry. Since any symmetry must take a corner into some other corner, and can either preserve or reverse orientation, there are 2m symmetries altogether, in which the rotations form a subgroup of order 2.

Proposition 9. The symmetry group of a regular polygon of m sides is a Coxeter group with two generators, say s and t. We have m_{s, t} = m.

This should be clear from the picture. The generators s and t should be chosen to be reflections in neighbouring axes of symmetry, as the red lines in the figure.

They are orthogonal reflections, with the angle between their lines of reflection equal to /m. Choosing the signs of a^v and b^v correctly, we have a^v b^v = -2 cos (/m). The region where a > 0 and b > 0 is a fundamental domain C for the symmetry group.

The 2m elements of the group can be expressed as

We can see how these elements match up with transforms of C in this picture:

Affine dihedral groups

It is an infinite Coxeter group with two generators, say s and t, and with m_{s, t} infinite. As we have already seen, the realization by affine reflections is the restriction to a line of a linear Coxeter group in dimension 2. Again, if a and b are chosen correctly, the region where a > 0 and b > 0 is a fundamental domain for the group.

The elements of the group can be expressed as

Hyperbolic dihedral groups

An observation about groups of rank two

Proposition 10. In these circumstances, if s is a generator in S then sw > w if and only if C and wC lie on the same side of the line a_s = 0. .

In brief, the argument for this is that if sw > w then w = ts .. , and the shortest gallery (sequence of contiguous chambers) from C to swC through chambers is the reflection of the one from C to wC, except that it starts with an extra transition from C to sC. To be more explicit, the shortest gallery from C to wC is the sequence C, tC, stC, ... and is reflected into the longer gallery C, sC, stC, stsC, ...

Groups of rank 3

We know that the standard representation preserves a metric in which |a_i|² = 1 for all i and a_i a_j = -cos (/m_{i, j}) for i distinct from j. If we assume that the indexing is chosen so that m_{1, 2} differs from 2, then by choosing coordinates suitably (so that the finite Coxeter group generated by the first two reflections stabilizes the (x, y)-plane) we can arrange this metric to be x² + y² + c z² with c = 1, 0, or -1, and a₁ = (1, 0, 0), a₂ = (-cos /m_1,2, sin /m_1,2, 0). In order to find a₃ we have to take into account the conditions

• |a₃|² = 1
• a₃ a₁ = - cos /m_{1, 3}
• a₃ a₂ = - cos /m_{2, 3}

• x² + y² + c x₃² = 1
• (x, y, z) (1, 0, 0) = - cos /m_1,3
• (x, y, z) (-cos /m_{1, 2}, sin /m_{1, 2}, 0) = - cos /m_{2, 3}

• x = - cos /m_1,3
• cos /m_{1, 3} cos /m_{1, 2} + y sin /m_1,2 = - cos /m_{2, 3}
• cz² = 1 - x² - y²

Proposition 11. Let r = 1 - 1/m_{1, 2} + 1/m_{1, 3} + 1/m_{2, 3}. Then

• if r < 0 then c > 0 and W is finite;
• if r = 0 then c = 0 and W stabilizes a half-plane, and it acts by affine reflections on a plane parallel to its boundary;
• if r > 0 then c < 0 and the group acts by non-Euclidean reflections.

- /m_{2, 3}

/m_{1, 2} + /m_{1, 3}

y = sin /m_{1, 3}

Here is a table of cases of possible values of the m_{i, j} in weakly increasing order:

The regular solids

Exercise 14. Verify that the Coxeter subgroup with values of m equal to 2, 3, 3 is the symmetry group of the regular tetrahedron.

Exercise 15. Verify that the Coxeter subgroup with values of m equal to 2, 3, 4 is the symmetry group of the cube and the regular octahedron. 2, 3, 4

Exercise 16. Verify that the Coxeter subgroup with values of m equal to 2, 3, 5 is the symmetry group of the icosahedron and the dodecahedron.

Exercise 17. Prove directly that the symmetry group of any regular polyhedron is a Coxeter group.

Exercise 18. Verify that the symmetry group of the regular simplex in n dimensions is a Coxeter group.

Exercise 19. Verify that the symmetry group of the cube in n dimensions is a Coxeter group.

Affine Coxeter groups of rank 3

Affine A2

Affine B2

Affine G2

A hyperbolic Coxeter group of rank 3

Poincaré model

Klein model

A polyhedral realization of a Coxeter group is a linear representation in which

• The group possesses a fundamental domain C which is a polyhedral cone;
• the generators in S are represented by reflections in the walls of this cone.

Every Coxeter group possesses at least one realization, as we shall see in a moment. Geometric properties of realizations translate naturally to combinatorial properties of the group. From the geometry of the simplices neighbouring a fundamental domain, for example, you can read off the Coxeter matrix. This is because if L the intersection of two walls, then the configuration in the neighbourhood of L is essentially that in a realization of the group generated by the two reflections in those walls. This is a special case of a very general result proven in most generality by MacBeath around 1964.

Given a realization, make a choice for each s in S of a pair a_s, a_s^v defining reflection in the wall of the fundamental domain parametrized by s. The sign of each function a_s can (and always will) be made so that a_s > 0 in the interior of the given fundamental domain. Such a linear function a_s is determined up to a positive scalar multiple, and its equivalence class under such multiplications will be called a basic root of the realization (and implicitly also of the choice of fundamental domain). The half-space A_s where a_s > 0 is determined by this class, and will be called a basic geometric root of the realization. The hyperplane a_s = 0 will be called a basic root hyperplane. This terminology is not common, but for general Coxeter groups this notion of root is entirely natural, since the particular choice of a_s has no intrinsic significance. This is not the case if the Coxeter group is the Weyl group of a Kac-Moody Lie algebra, since in that case the roots themselves are part of the structure of the Lie algebra.

The Cartan matrix associated to a choice of functions a_s is the matrix c_{s, t} = < a_s, a_t^v >. If the a_s are chamnged to d_sa_s then c_{s, t} is cahmnged to d_s c_{s, t} d_t^-1. This gives rise to a new matrix DCD^-1 where D is a diagonal matrix with positive entries. But the numbers n_s,t = c_{s, t}c_{t, s} depend only on the realization itself. The next result is a consequence of observations made earlier about conjugacy classes of pairs of reflections:

Theorem 12. In any realization, the Cartan matrix satisfies these conditions:

1. c_{s, s} = 2;
2. c_{s, t} = 0 if and only if c_t,s = 0;
3. c_{s, t} is either 0 or a negative number;
4. if n_{s, t} lies between 0 and 4 then it is equal to 4 cos² (/m_{s, t}).

Any Cartan matrix clearly gives rise to a representation of the associated Coxeter group. In fact:

Theorem 13. The representation of a Coxeter group determined by any abstract Cartan matrix is a realization of the associated Coxeter group.

This will be proven in the next section. One consequence is that every Coxeter group has at least one realization, since there exists always the standard Cartan matrix

c_s,t = -2 cos(/m_s,t) .

Cartan matrices with integral matrices determine Kac-Moody Lie algebras. In this case the representation of its Weyl group on the lattice of roots is the one associated to this Cartan matrix. Coxeter groups which occur as the Weyl groups of Kac-Moody algebras are called crystallographic, and are distinguished by the property that for them the numbers m_{s, t} are either 2, 3, 4, 6 or infinite.

Two Cartan matrices C₁ and C₂ will give rise to isomorphic representations of a Coxeter group if and only if there exists a positive diagonal matrix D with C₂ = D C₁ D^-1. In particular, those Cartan matrices giving rise to realizations equivalent to the standard one are symmetrizable.

Proposition 14. If each m_{s, t} is finite, then the isomorphism classes of Cartan realizations are parametrized by H¹( , R) (where is the Coxeter diagram).

Proof. If all m_{s, t} are finite, then all Cartan matrices are of the form c_{s, t} d_{s, t} where c_{s, t} = -2 cos²/m_{s, t} and d_{s, t} is an arbitrary matrix of positive real numbers with d_{t, s} = 1/d_{s, t}. Two of these will give isomorphic represntations of W when the entries differ by factors d_s/d_t, with all d_s > 0. But the assignment of d_{s, t} to (s, t) defines a cocycle on the Coxeter graph with values in the multiplicative group of positive real numbers, and assignments d_s/d_t are coboundaries. Conclude by applying the logarithm.

Corollary. If W is finite, then the Coxeter diagram has no circuits.

Proof. Because if the diagram were not a tree, the group would possess a continuous family of non-isomorphic representations of dimension r.

Distinct classes can give rise to realizations with very different geometric properties. We have seen this already in the case of the infinite dihedral group, and here are the pictures for two different realizations of the Coxeter group whose Coxeter diagram is :

The Klein model of the standard realization

The Kac-Vinberg realization

The first of these is associated to the standard Cartan matrix, and the second to the integral matrix

which is that of a certain hyperbolic Kac-Moody Lie algebra. It is the second, therefore, which is likely to have intrinsic significance.

Exercise 20. Does the Coxeter group of rank 3 with all m_{s, t} = 3 have non-equivalent Cartan matrices?

Exercise 21. (This is a research problem! ) Prove that the boundary of the second is non-smooth everywhere (i.e. even though it does have tangent lines everywhere, it will not likely be C²). (This conjecture is consistent with, but not directly related to, a curious result of Kac & Vinberg, which asserts that if the boundary of the slice we are looking at is smooth, it is an ellipse. This problem will appear more reasonable after we have looked at the role of finite automata in the geometry of Coxeter groups.)

The principal result connecting combinatorics and the geometry of a Coxeter group is this:

Theorem 15. Suppose w in W, s in S. Suppose that a_s is a basic root in a Cartan representation of W. Then sw > w if and only if wC lies entirely in the region a_s > 0.

This generalizes what we have already seen for groups of rank two. The proof is somewhat intricate.

Proof. It begins with

Lemma. If T is a subset of S and w is any element of W, then there exists u in W_T and x in W such that (a) tx > x for all t in T; (b) l(w) = l(u) + l(x).

This result will be made more precise later on, where we discuss the cosets W_T\W in more detail.

Proof of the Lemma. The following algorithm computes u and x:

	x := w
	u := 1
	while tx < x for some t in T 
		x := tx
		u := ut

We now prove Theorem 3 by induction on l(w). If w = 1 there is no problem. Suppose l(w) > 1. If x = sw < w it must be shown that a_s is negative on wC. But then wC = sxC; by induction a_s is positive on xC, hence sxC lies in the region where a_s < 0.

Now suppose sw > 0. It must be shown that a_s > 0 on wC. Choose t such that tw < w. Find u in W_s,t and x in W satisfying the conditions of the Lemma. Since tw < w, l(x) < l(w). Since sx > x and tx > x, induction lets us see that xC is contained in the region C_s,t where a_s > 0 and a_t > 0. Since l(w) = l(u) + l(x), l(su) = l(u) + 1, and this is still valid if l is the length in W_s,t. From the discussion on groups of rank two, we see that a_s > 0 on the region uC_s,t, hence on wC = uxC as well. QED

Proposition 16. Every root is either positive or negative.

The point is that the chamber C is not intersected by a root hyperplane. This result follows immediately from Theorem 3. A reformulation:

Proposition 17. If wC intersects C then w = 1.

Proof. If w is not equal to 1, then there exists s with sw < w. But then C and wC lie on opposite sides of the hyperplane a_s = 0.

In other words, every Cartan representation is a realization of the group as a subgroup of GL(V). If we restrict the realization to the subgroup generated by T, we see that W_T embeds into W, too.

For a subset T of S, let C_T be the region in the boundary of C where the basic roots a_t = 0 for t in T, a_s > 0 for s not in T. Every point of C_T is fixed by t in T, hence all elements of the subgroup W_T. Conversely:

Proposition 18. 3. Let be the closure of C. If w intersects then the intersection equals the closure of C_T for some T, and w lies in W_T.

Proof. By induction on l(w). If w = 0 or 1, no problem. Otherwise, say sw < w and w = sx for l(x) < l(w). Then a_s 0 on x while a_a 0 on w. Therefore the intersection of C and wC lies in a_s = 0. But then the intersection of x and is non-empty, so we can apply the induction hypothesis. We deduce that the intersection of x and is equal to the the closure of some C_T, and x lies in W_T. But then sC intersects C_T as well, and s lies in T also. Hence w lies in W_T.

Corollary. Any face of a chamber of a realization is the W-transform of a unique face of C.

Proof. If xC_U = yC_T, then y^-1xC_U = C_T. But then y^-1x must lie in W_T.

In other words, each face of a chamber is labeled by a unique subset T.

A gallery is a sequence of chambers C_i with each successive pair sharing a face of codimension one. If w = s₁ ... s_n then the sequence C, s₁C, s₁s₂C, ... , s₁s₂ ... s_nC is a gallery. The chambers C and sC share a face labeled by s, and hence wC and wsC do, too. Minimal galleries correspond to reduced words.

For w in W, let L_w be the set of positive roots r such that w^-1r is negative - i.e. r > 0 on C but r < 0 on wC. This means that the root hyperplane r = 0 separates C from wC. In any gallery linking C to wC it must be one of the walls between two successive chambers in the gallery. Therefore:

Proposition 19. The root hyperplanes for r in L_w are those separating C from w^-1C. A minimal gallery from C through xC to xyC can be split into two disjoint pieces, one from C to xC, the other from xC to xyC. The second is the x-transform by x of a gallery from C to yC. Therefore:

Proposition 20. If l(xy) = l(x) + l(y) then L_xy is the disjoint union of L_x and xL_y.

Corollary. The length l(w) is the cardinality of L_w.

Corollary. The length of w is the number of root hyperplans separating C from wC.

Proposition 21. An element w lies in W_T if and only if L_w is contained in T.

Let R⁺ be the set of positive roots, R_T those generated by W_T from the a_t with t in T.

Proposition 22. An element of W_T permutes the roots in R⁺ - R_T⁺.

Proposition 23. A reduced word for w in W_T is one also in W.

Proposition 24. In every coset W_T \ W there exists a unique element x such that tx > x for all t in T. For any w in this coset, l(w) = l(wx^-1) + l(x).

Let W^T be the set of these distinguished coset representatives.

Proposition 25. Any element w can be expressed as w = xy with x in W_T and y in W^T.

These last will be left as exercises.

The following pictures illustrate how this works on the affine Weyl group of A₂. The chambers are really three dimensional simplicial cones, and we are looking at a slice through them, in which the generators are represented by affine reflections.

s2C	s1s2C
s3s1s2C	s2s3s1s2C

If w = s₁s₂ ... s_n, then this word representation of w corresponds in a simple fashion to a gallery from the fundamental chamber C to wC - namely, the chain of chambers C, s₁C, s₁s₂C, ... , s₁s₂ ... s_nC, in that order. We read the path from left to right.

Proposition 26. A vector v lies in if and only if the set of positive roots b with < b, v > < 0 is finite.

Proof. It is to be shown that if the set of roots b with < b, v > < 0 is finite, then v = wu for some u in the closure of C. If the set of such roots is empty then already v lies in C. Otherwise, < a_s, v > < 0 for some s. Now s permutes the set R⁺ - { a_s } , hence L_sv = s(L_v - { a_s }), which has size one less than L_v. Apply induction.

Proposition 27. The region is convex.

It is called the Tits cone.

Proposition 28. The following are equivalent:

• The group W is finite;
• The set of roots is finite;
• The Tits cone is all of V.

Exercise 22. Prove that in every double coset W_X\ W / W_Y there is a unique element of least length.

Proposition 29. The face C_T lies in the interior of if and only if W_T is finite. Every finite subgroup of W lies in the conjugate of some finite W_T.

An ( n 1):
Bn = Cn (n 2) :
Dn (n 4) :
E6 :
E7 :
E8 :
F4 :
G2 :
H3 :
H4 :
Ip (p = 5, 7) :

This is justified in VI.4 of the book by Bourbaki. The basic idea is to check when the standard realization preserves a positive definite quadratic form. These are the cases when the Tits `cone' is the whole vector space. The starting point is that the Coxeter diagram cannot contain any circuits. Another easy remark is that the number of branches from any point can be at most 3. But from there the argument is complicated. The argument of Bourbaki uses the criterion that W is finite if and only if the quadratic form left invariant by W in the standard realization is definite. An argument along different lines can presumably be put together using ideas of Kac and Vinberg.

Exercise 23. How large are these groups?

These are the Coxeter diagrams for those irreducible Coxeter groups which can be interpreted as affine reflections:

A1~:
An~ ( n 2):
B2~ :
Bn~ (n 2) :
Cn~ (n 2) :
Dn~ (n 4) :
E6~ :
E7~ :
E8~ :
F4~ :
G2~ :

This also is in the book by Bourbaki. These are the cases when the Tits `cone' is a half-space.

Exercise 24. The aim of this exercise and the next two is to explain how the regular polyhedra in all dimensions of Euclidean space are classified in terms of Coxeter diagrams. Recall that a regular polyhedron is a polyhedron in Euclidean space whose symmetry group acts transitively on the faces of any given dimension. Prove that the symmetry group of any regular polyhedron in n dimensions is a Coxeter group, and that the fundamental domain of the action is a slice through a fundamental domain in a realization of the group. (Hint: Start with dimensions 2 and 3. )

Exercise 25. Suppose that v is a vector in a realization of a finite Coxeter group. We may assume that v lies in the closure of a fundamental domain for the group. Let T be the complement of the set of generators fixing v. Let P be the convex hull of the W orbit of V. Show that the faces of P meeting the fundamental domain are parametrized by the connected components of the Coxeter diagram containing T (Hint: if X is such aset, the face will be the orbit of v under W_X.) (c) Show that P will be a regular polyhedron if and only if T is a single generator and the Coxeter diagram has no branches.

Exercise 26. Prove that the regular Euclidean polyhedra are classified by isomorphism classes of (a) a Coxeter diagram associated to a finite Coxeter group together with (b) a single node of the diagram on its boundary. List explicitly all the ones occurring in dimension 4.

• N. Bourbaki, Chapitres IV, V, VI of Groupes et algebres de Lie, Hermann, Paris, 1968.
• James Humphreys, Reflection groups and Coxeter groups, Cambridge Press, 1990.
• V. Kac, Infinite dimensional Lie algebras, Cambridge University Press.
• E. B. Vinberg, Discrete linear groups generated by reflections, Math. U. S. S. R. Izvestia 5 (1971).
• V. Kac and E. B. Vinberg, Quasi-homogeneous cones (in Russian), Math. Zametki 3 (1967), pp. 347-354.
• R. Howlett, lecture notes available on the Internet (http://www.maths.usyd.edu.au/): Miscellaneous facts about Coxeter groups (June, 1993) and Introduction to Coxeter groups (February, 1996).
• A. M. MacBeath, Annals of Mathematics 79 (1964).