The Geometry of Root Systems. Brian C. Hall

The Geometry of Root Systems A E Z S Brian C. Hall

T G R S

T G R S 1 1. I Root systems arise in the theory of Lie groups and Lie algebras, but can also be studied as mathematical objects in their own right. A root system is a discrete geometric object that captures many of the interesting properties of a Lie group (i.e., a continuous group). In particle physics, for example, the eight-dimensional group SU(3) (consisting of 3 3 matrices with determinant 1) is used to describe the interactions of quarks. Much of the structure of the eight-dimensional, continuous group SU(3) can be captured in the geometry of the two-dimensional, discrete root system called A 2 (Figure 2). These instructions and the accompanying Zometool kit give the user a chance to explore two- and three-dimensional root systems and related ideas. The instructions give the definition of a root system and provide several low-dimensional examples. No attempt is made to prove any theorem about root systems or to explain the connection between root systems and Lie groups and Lie algebras. For that information, the interested reader should consult a textbook on Lie theory, such as my own book, Lie groups, Lie algebras, and representations: An elementary introduction, Second Edition [Springer, Graduate Texts in Mathematics, 2015], hereafter referred to as Hall. 2. D A key idea in the theory of root systems is that of a reflection or mirror symmetry. Suppose E is an n-dimensional Euclidean space and P is a hyperplane in E, that is, a subspace of dimension n 1. If v is any vector in E, the reflection about P should leave the component of v in P unchanged but should flip the sign of the component of v perpendicular to P. Now, if α is a nonzero vector in E, we can consider the hyperplane P α consisting of vectors perpendicular to α. We then let s α denote the reflection about the hyperplane P α. We may compute s α algebraically by the following formula: α, β s α (β) = β 2 α, (1) α, α where α, β denotes the inner product of α and β. Note that if β is perpendicular to α (i.e., if α, β = 0), then s α (β) = β. Thus, s α is simply the identity on P α. On the other hand, we can easily see that s α (α) = α, showing that s α changes the sign of vectors perpendicular to P α. Since the quantity α, β 2 α, α appearing in (1) will play an important role in the definition of a root system, let us pause to consider the geometric meaning of this quantity. The orthogonal projection of β onto α is equal to α, β proj α (β) = α. (2) α, α Thus, the quantity 2 α, β / α, α may be understood as twice the coeffi cient of the projection of β onto α. Definition 1. A root system is a finite collection R of nonzero vectors in a Euclidean space E, satisfying the following axioms. Spanning. The elements of R span E.

T G R S 2 Scaling. For each α in R, the vector α is in R but no other multiple of α belongs to R. Reflection. For each α and β in R, the vector s α (β) also belongs to R. Integrality. For each α and β in R, the quantity is an integer. α, β 2 α, α The dimension of E is called the rank of the root system. In light of our geometric interpretation of the quantity 2 α, β / α, α, we may understand the integrality axiom as follows: Integrality : The orthogonal projection of β onto α must be an integer or half-integer multiple of α. Of course, if α and β are in R, then this condition must also hold with the roles of α and β reversed. As it turns out, there are not very many ways to find a pair of vectors α and β where the projection of β onto α is an integer or half-integer multiple of α, and vice versa. Indeed, if the angle between α and β is acute, all the possibilities are depicted in Figure 1. (We assume there that the longer vector is labeled as α.) The reader is invited to verify that the ratio of the length of α to the length of β is 1, 2, and 3 in the three cases. 2 2 60 2 45 2 30 3 2 Figure 1: Pairs of vectors for which both 2 α, β / α, α and 2 β, α / β, β are integers. A first key result about root systems already suggested by Figure 1 is that only certain angles and length ratios can occur. Proposition 2. Suppose α and β are roots, α is not a multiple of β, and α β. Then one of the following holds: 1. α and β are perpendicular 2. α = β and the angle between α and β is 60 or 120

T G R S 3 3. α = 2 β and the angle between α and β is 45 or 135 4. α = 3 β and the angle between α and β is 30 or 150 We may now consider our first example of a root system, conventionally called A 2 (Figure 2). The root system consists of six equally spaced vectors in the plane. The spanning and scaling axioms are clearly satisfied by A 2. The reflection axiom is satisfied because the collection of vectors is invariant under the reflections about each of the dashed lines in the figures. (Note that the relevant reflections are about the lines perpendicular to the roots, not the lines through the roots.) If β = ±α, the projection of β onto α is ±α; otherwise, the projection of β onto α is ± 1 2 α. Thus, A 2 satisfies the integrality axiom. s Figure 2: The A 2 root system A key aspect of root systems is their symmetry properties. A root system R is invariant under all the reflections s α with α in R, and thus, also, under combinations of such reflections. Definition 3. The Weyl group of a root system (E, R) is the group of orthogonal transformations of E generated by the reflections s α, with α ranging over R. In the case of the A 2 root system, the Weyl group is the symmetry group of an equilateral triangle, as shown in Figure 3. Observe that each reflection about a dashed line maps the triangle to itself. The symmetry group of a triangle has six elements, the reflections about the three dashed lines, together with rotations by multiples of 120. A rotation by ±120 may be obtained by performing two consecutive reflections about two different dashed lines in Figure 3. Note that the Weyl group of A 2 is not the full symmetry group of the A 2 root system, which is the symmetry group of a hexagon. 3. R 1 2 In rank one, there is only one possible root system, called A 1, consisting of two vectors of the form ±α. In rank two, there are four possibilities, shown in Figure 4. The first consists of two perpendicular pairs of vectors {±α, ±β}, with no restriction on the ratio of the length of α to the length of β. This root system is called A 1 A 1 (two copies of A 1 ). We have already encountered the second possibility, A 2. The third possibility, denoted B 2, consists of eight equally spaced vectors that alternate between two different lengths, with the ratio of longer to shorter being 2. The last possibility, denoted G 2, consists of twelve equally spaced vectors that alternate between two different lengths, but with the length ratio now being 3. It is not hard to check that each of these figures is actually a

T G R S 4 Figure 3: The Weyl group of A 2 is the symmetry group of the indicated equilateral triangle. root system; in particular, all angles and length ratios are consistent with Proposition 2. For all the rank-two systems other than A 2, the Weyl group is the full symmetry groups of the figure: the symmetry group of a rectangle (A 1 A 1 ), the symmetry group of a square (B 2 ), and the symmetry group of a hexagon (G 2 ). A 1 A 1 A2 B 2 G 2 Figure 4: The rank-two root systems, with their conventional names. We now turn to the construction of rank-two root systems using Zometool, omitting the uninteresting case of A 1 A 1. It turns out that A 2 and B 2 are fully constructible, whereas for G 2, one can make a convincing fake. Specifically, in Figure 5, the red lines in the second image do not lie in the plane of the blue vectors. Nevertheless, if you project the red lines into the plane of the blue lines, you get a true G 2 root system. Project 1. Construct the A 2 root system using blues or greens and verify that the Weyl group is the symmetry group of the triangle in Figure 3. Project 2. Construct the B 2 root system using a combination of blues and greens. Construct an approximation to the G 2 root system using blues and reds. (See Figures 5 and 6.) In each case, determine the Weyl group of the root system.

T G R S 5 Figure 5: The A 2 root system (left) and an approximation to the G 2 root system (right). Figure 6: The B 2 root system built with green and blue lines. The reader will quickly discover that the green lines in the Zometool system are more complicated to deal with than the other lines, because the same green line can go into the same hole with several different orientations. In constructing the B 2 root system, it will make your life easier if you put in the four blue lines first; then insert the green lines in such a way that they lie in the same plane as the blue lines. For the G 2 approximation, the red lines should be inserted as close as possible to the plane of the blue lines. 4. T A 3 4.1. Different views of A 3. In rank three, we can form a root system by putting one of the rank-two root systems into a plane and then putting a copy of A 1 (i.e., two vectors of the form ±α) in the line perpendicular to the plane. Such root systems may be considered boring, in that they are just rank-two root systems boosted up to three dimensions in a simple way. Thus, we will consider only the irreducible root systems of rank three those that are not combinations of a rank-two root system and A 1 beginning with the system known as A 3. Figure 7 gives three views of the A 3 root system. In the upper left corner, we have the A 3 root system by itself, consisting of 12 vectors emanating from the origin. The vectors in A 3 form the vertices of a polyhedron known as a cuboctahedron, which has six square faces and eight triangular faces, as shown in the upper right portion of the figure. The points in A 3 can also be visualized as the midpoints of the edges of a cube, as in the bottom portion of Figure 7. Algebraically, we can describe A 3 as the set of 12 vectors in R 3 of the form (±1, ±1, 0), (±1, 0, ±1), and (0, ±1, ±1). For α and β in A 3, one may readily calculate that 2 α, β / α, α has the value ±2 (if β = ±α), 0 (if α and β are perpendicular), or ±1 (in all other cases). Geometrically, if β ±α and α and β are not perpendicular, then the angle between α and β will be either 60 or 120, so that α and β fall into Case 2 of Proposition 2. Meanwhile, suppose we orient A 3 so that one of the roots α is vertical, as in Figure 8. Then the plane

T G R S 6 Figure 7: Three views of the A 3 root system. perpendicular to α is the (x, y)-plane and the reflection s α will reverse the sign of the z-component of each vector, without chaning the x- or y-components. (Since you cannot do a reflection physically with the model, you have to simply visualize the map sending (x, y, z) to (x, y, z).) Two of the roots of A 3 lie in the plane perpendicular to α, so that these roots are unchanged by the reflection s α. The remaining eight roots in A 3 (other than ±α and the ones perpendicular to α) line up in pairs directly above and below each other, so that the reflection s α maps each root to another root, as in Figure 8. (No matter which root α you start with, if you orient α vertically, Figure 8 will apply.) Figure 8: The reflection about the horizontal plane interchanges the two shaded roots. Project 3. Construct the A 3 root system using green lines, following the steps below. Optionally, embed A 3 either in a cube or in a cuboctahedron as in Figure 7.

T G R S 7 Step 1: Build two perpendicular pairs of green vectors. To do this, pay close attention to the orientation of the green lines as they go into the ball, as shown in Figure 9. Alternatively, you may start with a B 2 root system and remove the blue lines. Figure 9: Detail of Step 1 in the construction of A 3. Step 2: Add four additional green lines above the ones you constructed in Step 1. The four new green balls should form a square and each of the new green lines should be equally spaced between and above two green lines from Step 1, as shown in Figure 10. Step 3: Add four more green lines below the ones from Step 1. These should also form a square and each of these green lines should come straight out from a line from Step 2. You should now have the twelve vectors making up the A 3 root system, and you should be able to identify the square and triangular faces of the cuboctahedron in the top right portion of Figure 7. Figure 10: Step 2 in the construction of A 3. 4.2. The Weyl group. The Weyl group of A 3 is the symmetry group of a tetrahedron, as shown in Figure 11. As in the A 3 case, the Weyl group is not the full symmetry group of A 3. After all, the map sending α to α is, by definition, a symmetry of any root system, but this map does not send the tetrahedron to itself. Project 4. Construct the A 3 root system and the tetrahedron in Figure 11. Verify that the Weyl group of A 3 is precisely the symmetry group of the tetrahedron. Each yellow line in Figure 11 should be surrounded by three green lines as shown in Figure 12. Hints for determining the Weyl group: First, in Figure 11, if α is one of the vertical roots in the figure, the reflection s α will interchange two of the vertices of the

T G R S 8 tetrahedron and leave the other two vertices unchanged. Second, the symmetry group of a tetrahedron including both reflection symmetries and rotational symmetries is the full permutation group on the four vertices. 4.3. Base and Dynkin diagram. Much of the information about a root system is captured in a special sort of subset called a base. Definition 4. If (E, R) is a root system, a subset of R is called a base for R if the following conditions are satisfied. Basis: The set should form a basis for E as a vector space. Integrality: For each root α, the coeffi cients of α when expanded in terms of are all integers. Sign condition: For each root α, the coeffi cients of α when expanded in terms of are either all greater than or equal to zero or all less than or equal to zero. That is, a given root α cannot have both positive coeffi cients and negative coeffi cients. Figure 11: The Weyl group of A 3 is the symmetry group of the indicated tetrahedron. Figure 12: Detail of the construction of the tetrahedron in Figure 11. Once we fix a base, each vector either has some positive coeffi cients but no negative coeffi cients, or else some negative coeffi cients but no positive coeffi cients. (In both cases,

T G R S 9 some of the coeffi cients may be zero.) If α has positive coeffi cients but no negative ones, we call α a positive root; in the other case, it is a negative root. Of course, the elements of itself are positive roots, sometimes called the positive simple roots. It is shown in Theorem 8.16 and Proposition 8.28 of Hall that every root system has a base and that any base can be mapped to any other base by the action of the Weyl group. Furthermore, it can be shown (Proposition 8.13 of Hall) that if α and β belong to a base, then the angle between α and β must be obtuse, that is, α, β 0. Figure 13 shows a base for the G 2 root system. 2 2 1 2 2 3 1 2 2 1 1 2 3 1 Figure 13: The roots α 1 and α 2 form a base for G 2. Project 5. Find a base for the A 3 root system. Identify the positive roots and express each positive root as a non-negative integer combination of elements of the base. Convince yourself that any base can be mapped to any other base by the action of the Weyl group. Figure 14 shows the base for A 3 (shaded) along with the other positive roots. If the elements of the base are labeled (left to right) as α 1, α 2, and α 3, the positive roots are α 1, α 2, α 3, α 1 + α 2, α 2 + α 3, and α 1 + α 2 + α 3. The remaining six (negative) roots for A 3 are just the negatives of the positive roots. Figure 14: A base (gray) and the associated positive roots for A 3. Associated to a base is an object called a Dynkin diagram. Suppose = {α 1,..., α r } is a base for a root system (E, R) of rank r. The Dynkin diagram for R is then a graph having vertices v 1,..., v r. For each pair of vertices v j and v k, if the corresponding roots α j and α k are perpendicular, we do not put an edge between v j and v k. If the angle between α j and α k is 120, we put one edge between v j and v k. If the angle between α j and α k is 135, we put two edges between α j and α k, and if the angle is 150, we put three edges. Since the angles in a base are always obtuse, these rules cover all possibilities. In light of Proposition 2, we have the following alternative description of the rule for edges:

T G R S 10 The number of edges between non-perpendicular roots in a base is the square of the length ratio. Finally, if α j and α k are non-perpendicular and have different lengths, we put an arrow on the edge(s) between α j and α k pointing from the longer root to the shorter root. (Thinking of the arrow as a greater than sign (>) should make it clear which way the arrow is supposed to go!) Consider, for example, the base for G 2 in Figure 13. Since we have two roots in the base, our Dynkin diagram has two vertices. Since the angle between α 1 and α 2 is 150 (and the length ratio is 3), we put three edges between the two vertices. Finally, since α 2 is longer than α 1, we put an arrow pointing from the second vertex to the first, resulting in the diagram in Figure 15. Figure 15: The Dynkin diagram for G 2. Project 6. Verify that the Dynkin diagram for A 3 is the one shown in Figure 16. Figure 16: The Dynkin diagram for A 3. 4.4. Weights. As we have noted at the beginning of these notes, root systems arise in the study of Lie groups and Lie algebras. In particular, when studying representations of Lie algebras, a key idea is that of a weight of a representation, and the possible weights are described geometrically in terms of the associated root system. Actually, it turns out that all possible weights are expressible as integer linear combinations of certain fundamental weights, which we now describe. 2 2 1 1 Figure 17: The vectors µ 1 and µ 2 are the fundamental weights associated to the base {α 1, α 2 } for A 2. Definition 5. Let (E, R) be a root system and let = {α 1,..., α r } be a base for R. Then the fundamental weights for R (relative to ) are the vectors {µ 1,..., µ r } determined by the following condition: For all j and k, we have 2 α j, µ k α j, α j = δ jk.

T G R S 11 Here δ jk is the Kronecker delta, equal to 1 if j = k and equal to 0 if j k. Recall from (2) that the quantity α j, µ k / α j, α j is the coeffi cient of the projection of µ j onto α j. Thus, we may describe the fundamental weights in words as follows: Each fundamental weight should be perpendicular to all but one of the roots in and should project orthogonally to one-half of the remaining root in. In Figure 17, for example, µ 2 is orthogonal to α 1 and projects orthogonally (along a horizontal line) to half of α 2. Similarly, in Figure 18, the blue line projects orthogonally (along a vertical line) to half of the gree line. Figure 18: The blue line projects orthogonally to the midpoint of the green line. Project 7. Select a base for A 3 and determine the associated fundamental weights. Figure 19: The base (gray) and the fundamental weights for A 3. In the representation theory of semisimple Lie algebras, each irreducible representation has a highest weight µ, which is a non-negative integer combination of the fundamental weights. All the weights of the representation then lie inside the polyhedron whose vertices are the orbit of µ under the action of W. That is to say, we take all vectors of the form w µ with w in W, and then we take the unique convex polyhedron whose vertices are of this form. Project 8. Let µ be the sum of the three fundamental weights for A 3. Construct the orbit of µ under the action of the Weyl group and build the associated polyhedron, as in Figure 20. To find the orbit, start with the sum of the three fundamental weights and then perform repeated reflections. Each edge of the figure should be a green line parallel to one of the roots in the A 3 root system itself. In Figure 21, for example, reflecting the sum of the three fundamental weights (red ball) about each element of the base gives the three dark green edges in the figure.

T G R S 12 Figure 20: The Weyl-group orbit of the sum of the three fundamental weights for A 3. Figure 21: Detail of the model in Figure 20.

T G R S 13 5. T B 3 C 3 The B 3 root system is obtained by starting with the twelve vectors (±1, ±1, 0), (±1, 0, ±1), and (0, ±1, ±1) of A 3 and adding six additional vectors: (±1, 0, 0), (0, ±1, 0), and (0, 0, ±1), as shown in Figure 22. Each of the new roots is shorter than the original roots by a factor of 2. The reader may check algebraically that 2 α, β / α, α has the value of ±1 or ±2 for each pair of roots in B 3. Geometrically, the angles and length ratios are all compatible with Proposition 2; in particular, the angle between a new root (blue) and an old root (green) is always either 45 or 135, with a length ratio of 2. Figure 22: The B 3 root system. In B 3, there are two distinct types of reflections, those about the plane perpendicular to a blue edge and those about the plane perpendicular to a green edge. The reader may easily verify that each of these types of reflections maps B 3 to itself. The Weyl group of B 3 is the symmetry group of the cube in Figure 23. Figure 23: The Weyl group of B 3 is the symmetry group of the indicated cube. Project 9. (1) Construct the B 3 root system and verify that its Weyl group is the symmetry group of a cube. (2) Find a base for B 3, determine the associated positive roots, and express each positive root as a non-negative integer combination of elements of the base. (3) Show that the Dynkin diagram for B 3 as in Figure 24. The C 3 root system similarly starts with the A 3 root system. This time, however, we add the six vectors (±2, 0, 0), (0, ±2, 0), and (0, 0, ±2), which are now longer than the

T G R S 14 B 3 C 3 Figure 24: The Dynkin diagrams for B 3 and C 3. roots of A 3 by a factor of 2. The vectors in C 3 form the vertices of an octahedron, as shown on the right-hand side of Figure 25. The blue nodes in the figure are not part of the C 3 root system but are needed to construct blue edges of length 2. (Alternatively, one can construct C 3 using single blue lines and half-size green lines, as in Figure 8.20 in Hall. Half-green lines are available from zometool.com under the Bulk Parts link.) It is straightforward to check that C 3 satisfies the integrality axiom of a root system. Since each root of C 3 is a multiple of a root for B 3, all the reflections for C 3 are the same as those for B 3. Thus, the Weyl group of C 3 is the same as that for B 3. Figure 25: The C 3 root system. Project 10. (1) Construct the C 3 root system. (2) Find a base for C 3, determine the associated positive roots, and express each positive root as a non-negative integer combination of elements of the base. (3) Show that the Dynkin diagram for C 3 is as in Figure 24. Figure 26: A base (gray) and the associated positive roots for B 3 (left) and C 3 (right).

T G R S 15 Once the fundamental weights have been constructed, we can construct the weight polyhedra associated to some non-negative integer combination of those weights. Project 11. Choose a base for B 3 and for C 3 and construct the associated set of fundamental weights, as in Definition 5. In each case, take some small non-negative integer combination µ of the fundamental weights and construct the polyhedron whose vertices are the Weyl-group orbit of µ, as in Project 8 in the A 3 case. Figure 27: The Weyl-group orbit of the sum of the three fundamental weights for B 3. Each edge of each polyhedron in Project 11 should be an integer multiple of one of the roots in the corresponding root system. In Figure 27, for example, the edges are the same green and blue lines as in the B 3 root system itself. A generic polyhedron for B 3 or C 3 will have 48 vertices (the number of elements in the Weyl group), but if the coeffi cient of one of the fundamental weights is zero, the number of vertices will be smaller. In Figure 27, the faces are rectangles, hexagons, and octagons, with the edges in the octagonal faces alternating between green and blue lines. Figure 28: Detail of the model in Figure 27.

T G R S 16 6. T When studying representations, one encounters the notion of an integral element. If (E, R) is a root system, then an element µ of E is called an integral element if µ satisfies α, µ 2 α, α is an integer for every root α in R. It turns out that if is a base for R and (3) holds for all α in, then (3) also holds for all α in R, showing that µ is integral. (See Proposition 8.35 in Hall.) From this we can easily deduce the following result: An element µ of E is integral if and only if µ can be expressed as an integer linear combination of the fundamental weights {µ j } in Definition 5. The set of all integral elements is called the weight lattice associated to R. The weight lattice is invariant under the Weyl group and also under the map µ µ. If R is A 3, B 3, or C 3, it follows that the weight lattice is invariant under all of the symmetries of a cube. It turns out that there are (up to scaling) only three such lattices, the simple cubic lattice, the body-centered cubic lattice, and the face-centered cubic lattice. We abbreviate these as SC, BCC, and FCC. The simple cubic lattice consists of all the vertices of a family of cubes stacked in the obvious way; a portion of a SC lattice is shown in the top left portion of Figure 29. A body-centered cubic lattice takes a simple cubic lattice and adds vertices at the center of each cube in the SC lattice, as in the top right portion of Figure 29. Finally, the facecentered cubic lattice starts with a simple cubic lattice and adds in vertices a the center of each face in the SC lattice, as in the bottom portion of Figure 29. (3) Figure 29: Simple cubic lattice (top left), body-centered cubic lattice (top right), and face-centered cubic lattice (bottom). There is, in addition to the weight lattice, another important lattice associated to a root system, the root lattice. The root lattice is simply the set of all integer linear combinations of the roots. Note that the integrality axiom says precisely that each root is an integral element. It follows that any integer linear combination of roots is also an

T G R S 17 integral element; that is, the root lattice is contained in the weight lattice. The root lattice for A 3, B 3, and C 3 is also invariant both under the Weyl group of the corresponding root system and under µ µ. Thus, these lattices must also be SC, BCC, or FCC. Although the weight lattices and root lattices for A 3, B 3, and C 3 all fall into one of the three types of lattices with cube symmetry, each lattice can be based either on a cube of edge-length 1 or a cube of edge-length 2. We will denote these differently scaled lattices SC-1 and SC-2, and similarly for the other types. Figure 30: The root lattice for A 3 is FCC-2. The distinction between the weight lattice and the root lattice is important in representation theory. According to Theorem 10.1 in Hall, the weights of a representation with highest weight µ must lie in the polyhedron generated by the Weyl-group orbit of µ. But in addition, each weight λ of the representation must differ from µ by an element of the root lattice. Thus, in most cases, there will exist integral elements inside the polyhedron generated by µ that are not weights of the representation with highest weight µ. Project 12. Build enough of the root lattice and weight lattice for A 3 to verify that the root lattice is FCC-2 and the weight lattice is BCC-1. Figure 31 shows a large model illustrating the two lattices in Project 12. This model is not constructible using only the pieces included with this kit! Note that in Figure 31, the center of the model is a yellow vertex, that is, not in the root lattice. Thus, you should think of the origin in E as being any vertex that has green lines attached, such as the red ball in the figure. (The significance of the red lines in the figure will be explained below.) Although this kit is not large enough to build all of the model in the figure, it should suffi ce to identify the two lattices in question. In calculating the lattices, it may be helpful to insert three perpendicular blue lines to serve as a coordinate system. There is one additional aspect of the root and weight lattices, which we now explore. We may think of both the root lattice and the weight lattice as a group under the operation of vector addition. The root lattice is a subgroup of the weight lattice, and since both groups are commutative, it is actually a normal subgroup. We may therefore construct the (commutative) quotient group G = (weight lattice)/(root lattice). (4)

T G R S 18 Figure 31: The root lattice (green) and the weight lattice (green and yellow) for A 3. It turns out that the group G is isomorphic to the center of the simply connected group associated to a given root system. (This result is obtained from Proposition 13.42 in Hall by dualizing. ) The A 3 root system, for example, comes from the simply connected compact group SU(4) (4 4 unitary matrices with determinant 1), the center of which is Z/4 (integers mod 4). Thus, the group G in (4) should also be Z/4 in the A 3 case. Figure 31 shows the weight lattice (yellow and green) and the root lattice (green) for A 3. If we take the red vertex as our origin, then the four red lines in the figure represent the four elements of the group G. Each yellow vertex should have eight yellow lines attached to it, while each green vertex should have eight yellow lines and twelve green lines (excepting the vertices near the edge of the model). The geometric center of the model is not in the root lattice and, thus, is not connected to any green lines. As we have already noted, the model in Figure 31 requires significantly more Zometool pieces than are included with this kit. Furthermore, it is much less diffi cult but still challenging to build the model in Figure 31 using long green and long yellow pieces, rather than the medium green and medium yellow pieces included with this kit. To build the model, you will need 89 balls optionally divided into 38 green balls and 51 yellow balls 144 long green lines, and 232 long yellow lines. (To order just the parts you need, visit zometool.com and select Bulk Parts at the bottom of the page. Then select Bulk Green Struts or Bulk Yellow Struts and select size 2 (long). ) As the model becomes more interconnected, it will become more diffi cult to insert the pieces. Some flexing of the long green edges will be required! Project 13. Verify that the quotient group G in (4) is isomorphic to Z/4 in the case of the A 3 root system. What the claim means, explicitly, is that there is a vector v in the weight lattice such

T G R S 19 that v, 2v, and 3v are not in the root lattice, but 4v is in the root lattice. In addition, every element of the weight lattice should be expressible as an element of the root lattice plus kv, for k = 0, 1, 2, or 3. Project 14. Identify the root lattice and weight lattice for the B 3 and C 3 root systems, and compute the quotient group G (as in (4)) in each case. Each of the lattices in Project 14 should come from the following list: SC-1, BCC-1, and FCC-2.