If S is any set, a Coxeter matrix indexed by
S is a function m_{s,t} on S × S with values in
either the positive integers or infinity, such that
m_{t,s} = m_{s,t} > 1 for distinct s and t. 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 $\left[ \begin{array} { l l l } { 1 } & { 3 } & { 4 } \\ { 3 } & { 1 } & { 3 } \\ { 4 } & { 3 } & { 1 } \end{array} \right]$ then its Coxeter diagram is Error! Click to view log. . The Coxeter group W = W_{S} associated to a Coxeter matrix m is the group with generators S and relations
This simple definition conveys, without elaboration, no idea of how interesting such groups are. They are among the most intriguing of all mathematical structures. 
More precisely, the definition sets W to be the set of
words in S (i.e. finite sequences s_{1} ... s_{n} of elements of S)
modulo an equivalence relation.
Words x and y
are equivalent if x is obtained from y by a chain of these
elementary transformations:
stssts=stts, stts=ss, ss=1. Exercise 1. Verify that if w = s_{1} ... s_{n} then w^{−1} = s_{n} ... s_{1}. 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)^{ms,t} = 1. Exercise 4. Verify that we could have stipulated W to be defined by relations (st)^{ms,t} = 1. If a Coxeter diagram decomposes into two components S_{1} and S_{2}, then the corresponding group is a direct product of the groups parametrized by S_{1} and S_{2}. 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:
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

It is possible to develop the subject of Coxeter groups entirely
in combinatorial terms (this is done  well, at least thoroughly attempted  in
the book by Bourbaki), but certain geometric representations
of Coxeter groups, in which the group acts discretely
on a certain domain, and in which the generators are
represented by reflections, allow one to visualize nicely
what is going on.
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 reflectionsIf s is a reflection in a space V, there exists a linear functional a and a vector a^{v} so that the reflection takes v to
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 nonzero scalar multiples. If a is replaced by ca then a^{v} is replaced by c^{−1}a^{v}. Often a reflection will be orthogonal with respect to some inner product •, when the reflection takes v to
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 reflectionsSuppose s and t to be a pair of reflections. Such pairs turn out to make up essentially a oneparameter family of conjugacy classes.Let s and t be a pair of reflections, say corresponding to a, a^{v} and b, b^{v}. Define associated real constants
Let (s_{1}, t_{1}) and (s_{2}, t_{2}) be pairs of reflections, conjugate under the linear transformation g. Then
for some nonzero 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 conjugationinvariant, 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 conjugationinvariant 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
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 twodimensional 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.
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 halfplane. 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 KacMoody 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 ninvariant 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 conjugationinvariant. Proposition 5. Suppose n_{s, t} to be neither 0 nor 4. Then there exists a nondegenerate 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
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:
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):
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:
In the last two cases, c_{s, t} and c_{t, s} are both negative. Affine reflectionsAn affine reflection reflects points in an affine hyperplane f(x) = 0 for some affine function f(x). The function f can be expressed as
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 δ 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 δ on U. In the discussion of pairs of reflections with n = 4 we have already seen an implicit example of this. NonEuclidean reflectionsA nonEuclidean reflection reflects points of a nonEuclidean space in a nonEuclidean hyperplane. This also can be explained in terms of linear reflections.The nonEuclidean space H_{n} of n dimension is the component Q(x) = x_{n+1}^{2} − x_{1}^{2} − ... − x_{n}^{2} = 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 halfplane, 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 nonEuclidean 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 nonEuclidean 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. 
In this section we shall look at some Coxeter groups defined geometrically.
Finite dihedral groupsSuppose P to be a regular polygon in the plane. That is to say, it is a polygon of m sides for some m > 2, centred at the origin, and invariant under rotations by 2 π/ 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 π/m. Choosing the signs of a^{v} and b^{v} correctly, we have a^{v} • b^{v} = −2 cos (π/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 groupsNow we look at the group generated by affine reflections in the points at the end of a line segment in one dimension.
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 groupsNow we look at the group generated by two hyperbolic reflections in the ends of a segment on the hyperbola Q(x, y) = 1, where Q is an indefinite nondegenerate quadratic form. It, too, is an infinite Coxeter group with two generators. And again, if a and b are chosen correctly, the region where a > 0 and b > 0 is a fundamental domain for the group.
An observation about groups of rank twoThe examples we have just examined exhaust the possible representations of rank two Coxeter groups with the property that the region C where a > 0, b > 0 is a fundamental domain. An observation we shall need later, based on observation of these cases, is thatProposition 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 3Suppose now that s_{0}, s_{1}, and s_{2} generate a Coxeter group W of rank 3 with Coxeter matrix (m_{i, j}) for i, j = 1, 2, 3. I assume that all the m_{i, j} are finite, so that all the dihedral groups W_{i, j} are finite. How can we tell whether W is finite?We know that the standard representation preserves a metric in which a_{i}^{2} = 1 for all i and a_{i} • a_{j} = −cos (π/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 x^{2} + y^{2} + c z^{2} with c = 1, 0, or −1, and a_{1} = (1, 0, 0), a_{2} = (−cos π/m_{1,2}, sin π/m_{1,2}, 0). In order to find a_{3} we have to take into account the conditions
Proposition 11. Let r = 1 − 1/m_{1, 2} + 1/m_{1, 3} + 1/m_{2, 3}. Then
Here is a table of cases of possible values of the m_{i, j} in weakly increasing order:
The regular solidsThe classification of the finite Coxeter groups in three dimensions is intimately related to the classification of 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
A hyperbolic Coxeter group of rank 3

A polyhedral realization of a Coxeter group is a linear representation in which
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 halfspace 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 KacMoody 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_{s}a_{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:
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
Cartan matrices with integral matrices determine KacMoody 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 KacMoody 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 C_{1} and C_{2} will give rise to isomorphic representations of a Coxeter group if and only if there exists a positive diagonal matrix D with C_{2} = D C_{1} 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 H^{1}( Γ, R) (where Γ 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 cos^{2}π/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 nonisomorphic 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 Error! Click to view log.:
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 KacMoody 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 nonequivalent Cartan matrices? Exercise 21. (This is a research problem! ) Prove that the boundary of the second is nonsmooth everywhere (i.e. even though it does have tangent lines everywhere, it will not likely be C^{2}). (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 chambers are parametrized by elements of W,
and the geometrical structure of
the complex they make up mirrors the structure of W.
These chambers are simplicial cones embedded in the
vector space V, and the left action of
W on V is compatible
with the left action of W on itself.
But there is also a right action of W on itself,
and this corresponds to a right action of W on the
set of chambers. If C_{*} = xC is a chamber then
C_{*}w = xwC defines the right transform of C_{*} by w.
Thus xC = Cx. The right action of
generators is particularly simple: C_{*} and C_{*}s share
a wall of codimension one labeled by s.
The following pictures illustrate how this works on the affine Weyl group of A_{2}. 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.
If w = s_{1}s_{2} ... 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, s_{1}C, s_{1}s_{2}C, ... , s_{1}s_{2} ... s_{n}C, in that order. We read the path from left to right. 
Define $\cal C$ to be the union of the closed chambers of a realization.
Proposition 26. A vector v lies in $\cal C$ 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 $\cal C$ is convex. Proposition 28. The following are equivalent:
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 $\cal C$ if and only if W_{T} is finite. Every finite subgroup of W lies in the conjugate of some finite W_{T}. 
These are the Coxeter diagrams for those irreducible
Coxeter groups which are finite:
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:
This also is in the book by Bourbaki. These are the cases when the Tits `cone' is a halfspace. 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. 
