A trace formula for G2

An n-dimensional matrix representation of a group G on a vector space V is a homomorphism from G to GL(V). For our purposes, we consider an irreducible representation to be a representation which cannot be decomposed into the direct sum of smaller-dimensional representations. Let H be a subgroup of G. The way in which irreducible representations of G decompose into irreducible representations of H is called branching. In order to calculate such branching, one must first obtain a trace formula for each conjugacy class of H in irreducible representations of G. In this thesis, we calculate this trace formula for each of the sixteen conjugacy classes of G2(2) as a subgroup of the real compact form of G2. The smallest of the five exceptional Lie groups G2 is realized as the group of automorphisms of the real octonians or Cayley numbers. Inside the Cayley numbers, there exists a nice order or in other words a ring of integers much like the ring of Gaussian integers in C. This order, known as the Coxeter order, gives an integral structure or lattice inside the octonions. We define G2(Z) to be the subgroup of G2 preserving the lattice. We define the group G2(p) to be the automorphism group of the octonions taken modulo p. G2(Z) acts on the Coxeter order, and therefore G2(Z) acts on the Coxeter order modulo p. In the case of p = 2, this action yields an isomorphism, thereby establishing that G2(2) is a finite subgroup of G2. The special unitary group SU(3) is also abstractly a subgroup of G2. We show that each of the 16 conjugacy classes of G2(2) is conjugate to a diagonal element of SU(3) inside of G2. Using the character table and power table for G2(2) along with the properties of SU(3), we find a diagonal representative element in SU(3) for each of the 16 conjugacy classes of G2(2). We use theWeyl character formula to write the trace formulas for each of the regular conjugacy classes. Since the Weyl character formula is undefined for irregular elements, we use a formal differential operator together with l'Hospital's rule to develop a general formula for irregular elements. This allows us to calculate the traces of the irregular conjugacy classes of G2(2). This process gives a closed-form formula for the character of each of the 16 conjugacy classes of G2(2). In other words, given only the highest weight l of an irreducible representation Vl of G2, one may use our formulas to calculate the multiplicity of each of the irreducible representations of G2(2) in the direct sum decomposition of Vl . We conclude with Matlab code which implements our formulas along with examples of calculation and verification of our formulas.

A TRACE FORMULA FOR G2 . . . . . . by . Steven M. Sullivan . . . A Senior Honors Thesis Submitted to the Faculty of The University of Utah In Partial Fulfillment of the Requirements for the . Honors Degree in Bachelor of Arts . . . In . Mathematics . Approved: . . TTTTTTTTTTTTTTTTT Dr. Peter Trapa Chair, Department of Mathematics TTTTTTTTTTTTTTTTT Dr. Gordan Savin Supervisor . . . . . TTTTTTTTTTTTTTTTT Dr. Fernando Guevara Vasquez Department Honors Advisor TTTTTTTTTTTTTTTTT Dr. Sylvia D. Torti Dean, Honors College May 2013 ABSTRACT . An n-dimensional matrix representation of a group G on a vector space V is a homomorphism from G to GL(V ). For our purposes, we consider an irreducible representation to be a representation which cannot be decomposed into the direct sum of smaller-dimensional representations. Let H be a subgroup of G. The way in which irreducible representations of G decompose into irreducible representations of H is called branching. In order to calculate such branching, one must first obtain a trace formula for each conjugacy class of H in irreducible representations of G. In this thesis, we calculate this trace formula for each of the sixteen conjugacy classes of G2 (2) as a subgroup of the real compact form of G2 . The smallest of the five exceptional Lie groups G2 is realized as the group of automorphisms of the real octonians or Cayley numbers. Inside the Cayley numbers, there exists a nice order or in other words a ring of integers much like the ring of Gaussian integers in C. This order, known as the Coxeter order, gives an integral structure or lattice inside the octonions. We define G2 (Z) to be the subgroup of G2 preserving the lattice. We define the group G2 (p) to be the automorphism group of the octonions taken modulo p. G2 (Z) acts on the Coxeter order, and therefore G2 (Z) acts on the Coxeter order modulo p. In the case of p = 2, this action yields an isomorphism, thereby establishing that G2 (2) is a finite subgroup of G2 . The special unitary group SU(3) is also abstractly a subgroup of G2 . We show that each of the 16 conjugacy classes of G2 (2) is conjugate to a diagonal element of SU(3) inside of G2 . Using the character table and power table for G2 (2) along with the properties of SU(3), we find a diagonal representative element in SU(3) for each of the 16 conjugacy classes of G2 (2). We use the Weyl character formula to write the trace formulas for each of the regular conjugacy classes. Since the Weyl character formula is undefined for irregular elements, we use a formal differential operator together with l’Hospital’s rule to develop a ii general formula for irregular elements. This allows us to calculate the traces of the irregular conjugacy classes of G2 (2). This process gives a closed-form formula for the character of each of the 16 conjugacy classes of G2 (2). In other words, given only the highest weight λ of an irreducible representation Vλ of G2 , one may use our formulas to calculate the multiplicity of each of the irreducible representations of G2 (2) in the direct sum decomposition of Vλ . We conclude with Matlab code which implements our formulas along with examples of calculation and verification of our formulas. iii TABLE OF CONTENTS ABSTRACT ii 1 INTRODUCTION 1 1.1 REVIEW OF REPRESENTATION THEORY . . . . . . . . . 1 1.2 EXAMPLE OF BRANCHING . . . . . . . . . . . . . . . . . 3 1.3 THE OCTONIONS AND G2 . . . . . . . . . . . . . . . . . . 8 1.4 G2 (2) AS A SUBGROUP OF G2 . . . . . . . . . . . . . . . . 10 2 THE CONJUGACY CLASSES OF G2 (2) 13 2.1 SU(3) AS A SUBGROUP OF G2 . . . . . . . . . . . . . . . . 14 2.2 G2 (2) AND SU(3) . . . . . . . . . . . . . . . . . . . . . . . 20 2.3 FUSING OF CLASSES INSIDE G2 . . . . . . . . . . . . . . 21 2.4 REPRESENTATIVE ELEMENTS IN SU(3) . . . . . . . . . . 24 3 TRACE FORMULA FOR G2 (2) 31 3.1 THE WEYL CHARACTER FORMULA . . . . . . . . . . . . 31 3.2 A TRACE FORMULA FOR REGULAR ELEMENTS . . . . 36 3.3 A TRACE FORMULA FOR IRREGULAR ELEMENTS . . . 43 4 BRANCHING FROM G2 TO G2 (2) 54 4.1 IMPLEMENTATION . . . . . . . . . . . . . . . . . . . . . . 54 4.2 AGREEMENT WITH FORMER RESULTS . . . . . . . . . . 60 5 REFERENCES 62 iv 1 1. INTRODUCTION 1.1. REVIEW OF REPRESENTATION THEORY We begin with a few basic definitions from representation theory. Definition 1.1. A representation ρ of a group G on an n-dimensional vector space V is a group homomorphism ρ : G → GL(V ). By choosing a basis for V , one obtains a map known as a matrix representation. In general, computations are very difficult in abstract groups. Representations give us a way to better understand the group G and simplify computation. Once we find a homomorphism to some matrix group, the abstract composition of elements of G becomes matrix multiplication. Example 1.2. Consider the rotational symmetries of a regular triangle. If we denote the 120◦ counterclockwise rotation by the element x, then abstractly this group is the cyclic group of order 3 generated by x: G = {1, x, x2 }. We obtain a matrix representation of G by considering the regular triangle to be centered at the origin in R2 and letting u, v, and w be the vectors defined by the vertices of the triangle labeled in counterclockwise order as shown in Figure 1. We choose u and v as a basis for R2 . Then counterclockwise rotation by 120◦ sends u to v and v to w = −u − v, and 240◦ rotation sends u to w = −u − v and v to u. Thus we define a matrix representation R : G → SL2 (Z) in the basis {u, v} as follows: " # 1 0 R(1) = , 0 1 " # −1 −1 R(x) = , 1 0 " 0 1 R(x2 ) = −1 −1 # . 2 u v 0 w Figure 1: Regular triangle centered at the origin in R2 with vectors u, v, and w defined by the vertices The beauty of this representation is that it is faithful, meaning that R is an isomorphism of G onto R(G). This allows us to do any computation in G using matrix multiplication in SL2 (Z). Definition 1.3. In general, a representation ρ of a group G on a vector space V is called irreducible if V has no proper G-invariant subspace. Otherwise ρ is called reducible. Instead of explaining what this means, we simply give an alternative definition. We call a representation irreducible if it is not isomorphic to the direct sum of representations of G of smaller dimension. These two definitions are not equivalent in general, but they are for the applications in this thesis. The idea of irreducibility together with Theorem 1.4 gives us a way to describe the representations of finite groups. The proof of Theorem 1.4 is found in chapter 10 of [2]. Theorem 1.4. A finite group always has the same number of irreducible representations as conjugacy classes. Another tool which is indispensible in classifying the representations of a finite group is the character of a representation. Definition 1.5. The character χR of a representation R of a group G is the map χR : G → C defined by χR (g) = tr(R(g)) for all g ∈ G, where tr: G → C is the usual trace map. 3 The character has many important properties which make it a powerful tool in classifying representations of finite groups. One such property is that if a representation R can be written as the direct sum of smaller representations R∼ = R1 ⊕ R2 ⊕ · · · ⊕ Rk , then the character χR can be written as the sum χR = χR1 + χR2 + · · · + χRk . We may also define a Hermitian form on the characters of a group G: hχi , χ j i = 1 ∑ χi(g)χ j (g). |G| g∈G (1) We use the following properties of the character throughout this thesis. 1. The character is constant on conjugacy classes. 2. Isomorphic representations have the same character. 3. Irreducible characters are orthonormal with respect to the Hermitian form (1). Property 1 is familiar from linear algebra as the statement that similar matrices have the same trace. This tells us that when dealing with the character, all we need are a representative element of each conjugacy class and the order of each conjugacy class. Property 2 tells us that we can use the characters of representations to classify the representations. Property 3 is useful in the calculation of branching. 1.2. EXAMPLE OF BRANCHING Clearly, restriction of a representation R of a group G to a finite subgroup H ≤ G defines a representation of H. However, even if R is an irreducible representation of G, the restriction R|H is not necessarily an irreducible representation of H. The restriction of R to H is 4 always isomorphic to a direct sum of irreducible representations of H. Let R1 , . . . , Rk be the irreducible representations of H. Then R restricted to H is the direct sum R|H ∼ = d1 R1 ⊕ d2 R2 ⊕ · · · ⊕ dk Rk , where d1 , . . . , dk are the multiplicities of the irreducible representations of H in the direct sum. A general formula for computing these multiplicities is called the branching rule for the restriction of irreducible representations of G to H. Such a general branching rule may be quite difficult to obtain. However, if we know the traces of all the elements of H in irreducible representations of G, and if we have the character table of H, then given any irreducible representation of G, we may calculate d1 , . . . , dk . We demonstrate a method for doing this in the following example. Example 1.6. Branching for the restriction of SO(3) to A4 There are only five families of finite subgroups of SO(3), namely 1. Rotational symmetries of a regular polygon (cyclic groups Cn ). 2. Rotational and reflectional symmetries of a regular polygon (dihedral groups D2n ). 3. Rotational symmetries of a tetrahedron (abstractly A4 ). 4. Rotational symmetries of a cube (abstractly S4 ). 5. Rotational symmetries of a dodecahedron (abstractly A5 ). For details, see [2]. Thus abstractly A4 ≤ SO(3). If we think of SO(3) as the subgroup of SU(3) of matrices with real coefficients, then we easily see that we have the inclusion SO(3) ≤ SU(3). Thus A4 ≤ SO(3) ≤ SU(3). Since the trace is constant on conjugacy classes, we only need one representative from each of the conjugacy classes of A4 . The following theorem gives us a nice way to write representatives for each class. 5 Theorem 1.7. Every matrix in SO(3) becomes conjugate to a diagonal matrix of the form  α  A= 1  α −1   (2) in SU(3) where α = eiθ for some θ ∈ [0, 2π). Proof. Euler’s Theorem, found in Section 5.2 of [2], states that the 3 × 3 rotation matrices are precisely the elements of SO(3). So we may think of SO(3) as the group of rotations of a sphere. Also in Section 5.2 of [2], Artin proves that each matrix in SO(3) has 1 as an eigenvalue. In other words, every rotation of a sphere is equivalent to the rotation about some fixed axis by a certain angle. Let B ∈ SO(3). Then abstractly B is the rotation about some axis β by an angle θ . Let T be an element of SO(3) that sends the vertical axis z to β . The matrix  cos θ − sin θ 0   Zθ =  sin θ cos θ 0  0 0 1  is the matrix of rotation about z by the angle θ . Then B is conjugate to Zθ since B = T −1 Zθ T. Since conjugate matrices have the same trace, it follows that tr(B) = tr(Zθ ). Finally, in SU(3), Zθ can be diagonalized to a matrix of the form (2). Thus B is conjugate to A in SU(3). In order to be brief, we make the claim without proof that there is exactly one irreducible representation of SO(3) of each dimension 2n + 1. Also, a matrix of the form given in (2) is represented in R2n+1 by a matrix which is conjugate in SU(2n + 1) to a diagonal matrix with diagonal entries (α n , α n−1 , . . . , α, 1, α −1 , . . . , α −n+1 , α −n ), which has trace n S2n+1 = ∑ j=−n α j. 6 Note that when θ = 0 the matrix Zθ is the identity and this formula simply gives the dimension of the representation. Otherwise we write αS2n+1 − S2n+1 = α n+1 − α −n . By solving for S2n+1 , we obtain the formula S2n+1 = χ2n+1 (A) = α n+1 − α −n . α −1 (3) Theorem 1.7 gives us a way to calculate the trace of any element of SO(3) in any irreducible representation of SO(3). In particular, we can find the traces of the conjugacy classes of A4 in irreducible representations of SO(3). We need to know the conjugacy classes of A4 . The character table for A4 which we have included as Table 1 can be found in Section 7.5 of [2]. χ1 χ2 χ3 χ4 (1) 1 (4) x (4) x2 (3) z 1 1 1 3 1 ω ω2 0 1 ω2 ω 0 1 1 1 -1 2πi/3 Table 1: Character Table for A4 ω = e The columns of Table 1 represent the conjugacy classes of A4 . In parentheses, we have the order of each class. Below the order is the representative element. The element 1 represents the class of the identity. The element x represents rotations of a face of the tetrahedron by 120◦ , the element x2 represents rotations of a face of the tetrahedron by 240◦ , and the element z represents rotation by 180◦ about an axis through the centers of opposing edges. We use (3) to find the character of each of these conjugacy classes in a given representation: 7 χ2n+1 (I) = dim(R2n+1 ) = 2n + 1. χ2n+1 (x) = χ2n+1 (x2 ) =    1, n ≡ 0 mod 3; = 0, n ≡ 1 mod 3;   −1, n ≡ 2 mod 3. (e2πi/3 )n+1 − (e2πi/3 )−n e2πi/3 − 1 (−1)n+1 − (−1)−n χ2n+1 (z) = = −1 − 1 ( 1, n ≡ 0 mod 2; −1, n ≡ 1 mod 2. To illustrate how we compute branching, we use the example of the irreducible 7dimensional representation of SO(3), which we obtain by letting n = 3. We want to decompose the restriction of R7 to A4 . We let χ1 , χ2 , χ3 , and χ4 be the characters of the four irreducible representations of A4 as in the table. We write R7 |A4 as a direct sum R7 |A4 = d1 χ1 + d2 χ2 + d3 χ3 + d4 χ4 . Since the irreducible characters of A4 are orthonormal with respect to the Hermitian form (1), we have hχ j , χ7 i = d1 hχ j , χ1 i + d2 hχ j , χ2 i + d3 hχ j , χ3 i + d4 hχ j , χ4 i = d j . (4) This gives us a way to calculate the multiplicities d1 , . . . , d4 : d1 = hχ1 , χ7 i = 1 1 χ1 (a)χ7 (a) = (7 + 4(1) + 4(1) + 3(−1)) = 1, ∑ 12 a∈A4 12 d2 = hχ2 , χ7 i = 1 1 χ2 (a)χ7 (a) = (7 + 4ω(1) + 4ω 2 (1) + 3(−1)) = 0, ∑ 12 a∈A4 12 d3 = hχ3 , χ7 i = 1 1 χ3 (a)χ7 (a) = (7 + 4ω 2 (1) + 4ω(1) + 3(−1)) = 0, ∑ 12 a∈A4 12 d4 = hχ4 , χ7 i = 1 1 χ4 (a)χ7 (a) = (3(7) + 3(−1)(−1)) = 2. ∑ 12 a∈A4 12 So the irreducible 7-dimensional representation of SO(3) decomposes as R7 |A4 ∼ = R1 ⊕ 2R2 8 when restricted to A4 . In this example, we have the good fortune that the group SO(3) is a subgroup of SU(3), for which any element of finite order is diagonalizable. This gives us a nice way to find representative elements for each of the conjugacy classes of A4 . In the remainder of this thesis, we apply this same sort of method to calculate the branching for the restriction of the group G2 to the finite subgroup G2 (2). For the restriction of G2 to G2 (2), we do not have a simple inclusion of G2 into a nice group like SU(3). However, we show that the group SU(3) is abstractly a subgroup of G2 and that every element of G2 is conjugate to a diagonal element of SU(3). This allows us to represent the conjugacy classes of G2 (2) in G2 using diagonal elements of SU(3) just as we did for the conjugacy classes of A4 in the example. 1.3. THE OCTONIONS AND G2 In this section, we give a brief overview of the group G2 and the finite subgroup G2 (2). In the interest of brevity, we omit many details. However, we attempt to give references to where the details may be found. Killing first discovered the exceptional Lie algebras. However, his classification of these algebras was incomplete. His list was redundant and his proofs were not completely correct. Élie Cartan was the first to give rigorous proofs of the complete classification of simple Lie algebras over the field of complex numbers. This classification included the five exceptional Lie algebras of dimensions 14, 52, 78, 133, and 248. Cartan also gave constructions of the Lie groups corresponding to these algebras. In 1914 he showed that the smallest of the exceptional Lie groups, known as G2 , is the automorphism group of the octonions [3]. In order to define G2 as the automorphism group of the octonions, first we should say a little bit about the octonions themselves. The octonions are the largest of the composition algebras. For a detailed definition of composition algebras, see Chapter 6 of [4]. In brief, a 9 composition algebra is an algebra which has a norm for which the norm of a product equals the product of the norms. A well-known theorem of Hurwitz found in Section 6.4 of [4] states that the only algebras with such a composition law are R, C, H, and O, where H and O are the quaternions and the octionions respectively. We define the octonions as Conway and Smith do in [4], namely as the formal expressions x∞ + x0 i0 + x1 i1 + x2 i2 + x3 i3 + x4 i4 + x5 i5 + x6 i6 (5) with xt real, which form the algebra over R generated by i0 , . . . , i6 satisfying the the following multiplication rules: i2n = −1, in+1 in+2 = in+4 = −in+2 in+1 , in+2 in+4 = in+1 = −in+4 in+2 , in+4 in+1 = in+2 = −in+1 in+4 . where we read the subscripts modulo 7. Now we may officially define the group G2 . Definition 1.8. Throughout this thesis, when we write G2 we mean the real compact form of the smallest of the exceptional Lie groups which is realized as the automorphism group of the octonions, i.e., G2 = Aut(O). Though G2 is the smallest of the exceptional Lie groups, this is by no means a reason to celebrate its simplicity. In fact, the following observation summarizing Sections 6.7 and 6.8 of [4] allows us to see intuitively that G2 is 14-dimensional. Observation 1.9. If we take two units i, j ∈ O which are orthogonal to 1 and to each other along with their product k = i j, then i, j, and k satisfy Hamilton’s relations. In other words 10 they form a basis for a quaternion subalgebra of O. We relabel these elements as i = j1 , j = j2 , k = j4 , and then choose a unit j0 such that j0 is orthogonal to the quaternion subalgebra with basis 1, j1 , j2 , and j4 . Then by defining j0 j1 = j3 , j0 j2 = j6 , j0 j4 = j5 , we obtain a basis {1, j0 , j1 , . . . , j6 } for O. This analysis provides information about the automorphism group of O. By the above argument, we obtain a unique element of G2 in the following way: we choose any unit j1 orthogonal to 1 and send i1 7→ j1 . Then we choose j2 to be any unit orthogonal to 1 and j1 and send i2 7→ j2 . Finally, we choose j4 to be any unit orthogonal to 1, j1 , j2 , and j1 j2 and send i4 7→ j4 . We choose j1 from the 6-dimensional unit sphere orthogonal to 1. We choose j2 from the 5-dimensional unit sphere orthogonal to 1 and j1 . Finally, we choose j4 from the 3-dimensional unit sphere orthogonal to 1, j1 , j2 , and j1 j2 . These three spaces are manifolds of dimensions 6, 5, and 3. Thus the dimension of G2 is 6 + 5 + 3 = 14. 1.4. G2 (2) AS A SUBGROUP OF G2 We call Q(x0 , x1 , x2 , x3 , x4 , x5 , x6 ) the ring of rational octonions. Though it is important to realize that this ring does not have associativity. An element of this ring takes the form c = c∞ + c0 x0 + c1 x1 + c2 x2 + c3 x3 + c4 x4 + c5 x5 + c6 x6 (6) and satisfies what Coxeter calls the rank equation x2 − 2c∞ x + (c2∞ + c20 + c21 · · · + c26 ) (7) 11 with rational coefficients. For details see [5]. We are interested in the octonions which satisfy (7) with integer coefficients, i.e., such that 2c∞ = c + c̄ ∈ Z, N(c) = c̄c ∈ Z. These elements, which we denote by O, are referred to by Coxeter as the integral Cayley numbers. He gives a detailed construction of O in [5]. In [4], Conway and Smith call O the integral octavians and so do we. We note that O is to the octonions as the Gaussian integers Z[i] are to the complex numbers. Just as the Gaussian integers form a lattice inside C, the integer octavians form a lattice inside O. For C we may ask: what are the automorphisms of C which preserve Z[i]? We know that each automorphism of C must send elements of norm 1 to elements of norm 1. The four units in Z[i] are ±1, ±i. Since the only nontrivial automorphism of C preserving the units of Z[i] must send i to −i and fix ±1, this is just a group with two elements, namely the identity and complex conjugation. So we may write Aut(Z[i]) ∼ = Z/2. We may naturally ask the same question for the octonions and the lattice O. We define G2 (Z) to be the subgroup of G2 = Aut(O) which preserves the octavian integers O. Since G2 (Z) is the automorphism group of a lattice, G2 (Z) is a discrete subgroup of G2 . But G2 is compact, so we conclude that G2 (Z) is finite. In fact, in [4] the order of G2 (Z) given to be 12096. Since the integral octonions form a lattice inside G2 , it makes sense to look at the lattice modulo p. We say that two elements a, b ∈ O are congruent modulo p if they differ by p times an octavian integer. We define G2 (p) for each prime p to be the automorphism group of the octonions taken modulo p. Given any automorphism in G2 (Z), we take the automorphism modulo p to yield an automorphism in G2 (p). This gives a group homomorphism G2 (Z) → G2 (p). 12 This map is injective, though this is not immediately apparent in the case of p = 2. The order of the group G2 (p) is p6 (p2 − 1)(p6 − 1). So in the case of p = 2 we have |G2 (2)| = 12096. Since |G2 (2)|=|G2 (Z)|, the injective homomorphism yields an isomorphism: G2 (2) ∼ = G2 (Z) ≤ G2 . 13 2. THE CONJUGACY CLASSES OF G2 (2) Recall that we aim to develop a trace formula for all elements of G2 (2) in arbitrary irreducible representations of G2 . In order to simplify this task, we make use of the first property of the character which we listed in Section 1.1, namely that the character is constant on conjugacy classes. Because of this property, we only need to find one representative element for each of the conjugacy classes of G2 (2). Then the trace formula we compute for that element is the formula for the entire conjugacy class of that element. First, we require knowledge of the conjugacy classes of G2 (2). Following the labeling in [7], we name the 16 conjugacy classes of G2 (2) according to the order of the elements in the conjugacy class. For instance, we label the three classes of elements of order 12 as 12a, 12b, and 12c. Table 2 gives the characters of each irreducible representation of G2 (2) on each of the conjugacy classes. Table 3 is the power table for the conjugacy classes. These tables are included in the GAP system and are also found in [7]. Table 2 tells us that G2 (2) has 16 conjugacy classes, so we need to develop 16 trace formulas. Recall that in Example 1.6 we use the inclusion of SO(3) into SU(3) in order to write the representative elements of the conjugacy classes of A4 as diagonal matrices in SU(3). We use a similar method here. However, instead of using an inclusion of G2 into some larger group, we show that abstractly SU(3) is a subgroup of G2 and further that every conjugacy class of G2 (2) contains an element which is conjugate to a diagonal element of this SU(3). Then we use the properties of SU(3) along with the character table and power table for G2 (2) to produce such a diagonal element of SU(3) to represent each of the conjugacy classes of G2 (2). 14 (1) (63) (56) (672) (126) (378) (504) (1728) (1512) (1008) (252) (252) (2016) (1512) (1008) (1008) x 1a 2a 3a 3b 4a 4b 6a 7a 8a 12a 2b 4c 6b 8b 12b 12c χ1 (x) χ2 (x) χ3 (x) χ4 (x) χ5 (x) χ6 (x) χ7 (x) χ8 (x) χ9 (x) χ10 (x) χ11 (x) χ12 (x) χ13 (x) χ14 (x) χ15 (x) χ16 (x) 1 1 6 6 7 7 14 14 14 21 21 42 27 27 56 64 1 1 -2 -2 -1 -1 6 -2 -2 5 5 2 3 3 -8 0 1 1 -3 -3 -2 -2 -4 5 5 3 3 6 0 0 2 -8 1 1 0 0 1 1 2 -1 -1 0 0 0 0 0 2 -2 1 1 -2 -2 3 3 -2 2 2 1 1 -6 3 3 0 0 1 1 2 2 -1 -1 2 2 2 1 1 -2 -1 -1 0 0 1 1 1 1 2 2 0 1 1 -1 -1 2 0 0 -2 0 1 1 -1 -1 0 0 0 0 0 0 0 0 -1 -1 0 1 1 1 0 0 -1 -1 0 0 0 -1 -1 0 1 1 0 0 1 1 1 1 0 0 -2 -1 -1 1 1 0 0 0 0 0 1 -1 0 0 1 -1 0 2 -2 3 -3 0 3 -3 0 0 1 -1 0 0 -3 3 0 -2 2 -1 1 0 3 -3 0 0 1 -1 0 0 1 -1 0 -1 1 0 0 0 0 0 0 0 1 -1 0 0 -1 1 0 0 0 1 -1 0 -1 1 0 0 1 -1 A -A 0 0 0 1 -1 -1 1 0 0 0 0 0 1 -1 -A A 0 0 0 1 -1 -1 1 0 0 0 0 0 √ A=i 3 Table 2: Character Table for G2 (2) x∈ 1a 2a 3a 3b 4a 4b 6a 7a 8a 12a 2b 4c 6b 8b 12b 12c x2 ∈ x3 ∈ x7 ∈ 1a 1a 1a 1a 2a 2a 3a 1a 3a 3b 1a 3b 2a 4a 4a 2a 4b 4b 3a 2a 6a 7a 7a 1a 4a 8a 8a 6a 4a 12a 1a 2b 2b 2a 4c 4c 3b 2b 6b 4b 8b 8b 6a 4c 12b 6a 4c 12c Table 3: Power Table for G2 (2) 2.1. SU(3) AS A SUBGROUP OF G2 Much of this section follows arguments given by Reeder in [8]. We previously introduced G2 as the automorphism group of the octonions. Recall that the octonions are formal expressions of type (6), which is the algebra over the reals generated by the units i0 , . . . , i6 that satisfy the necessary multiplication properties given in Section 2.1. For j = 0, . . . , 6, we define Gi j to be the stabilizer of i j in Aut(O) ∼ = G2 . We denote the traceless octonions 15 by O0 . Clearly O decomposes as the direct sum O∼ = R ⊕ O0 . We fix i j , one of the generators of O0 . Define V j to be the 6-dimensional orthogonal complement of i j in O0 . Then V j consists of formal expressions x0 i0 + x1 i1 + x2 i2 + x3 i3 + x4 i4 + x5 i5 + x6 i6 where the coefficient of i j is 0. We show that V j is closed under multiplication by i j using the multiplication rules of O given in Section 1.3 to compute i j i j+1 = i j+3 , i j i j+2 = i j+6 , i j i j+3 = −i j+1 , i j i j+4 = i j+5 , i j i j+5 = −i j+4 , i j i j+6 = −i j+2 . where we take subscripts modulo 7. Thus multiplication by i j gives i j (x0 i0 + x1 i1 + · · · + x6 i6 ) = x j+1 i j+3 + x j+2 i j+6 − x j+3 i j+1 + x j+4 i j+5 − x j+5 i j+4 − x j+6 i j+2 ∈ V j . Hence we write O0 as the direct sum O0 ∼ = Ri j ⊕V j . Therefore we may write O as the direct sum O∼ = R ⊕ Ri j ⊕V j . There is an isomorphism R ⊕ Ri j ∼ = C defined by i j 7→ i. Thus O∼ = C ⊕V j . 16 We note that Gi j is precisely the subgroup of automorphisms of O that fix this C. Theorem 2.2 shows that Gi j ∼ = SU(3). Before we can prove the theorem, we need to understand a little bit about how calculations are done in O. Since O is not associative, calculation can be quite difficult. In O, we take the norm to be N(x) = xx̄. Then we define the norm form on O to be (x, y) = N(x + y) − N(x) − N(y) . 2 We let w, x, y, z ∈ O0 be elements of norm 1. Then we have (xy, xz) = N(x) · (y, z) = (y, z) · N(x) = (yx, zx). (8) By replacing z with w and x with x + z in (8) we obtain (xy, zw) + (xw, zy) = 2(x, y)(z, w). (9) Therefore, when (x, y) = 0, we obtain the equation (xy, zw) = −(xw, zy). (10) (y, x̄z) = (xy, z) = (x, zȳ). (11) By setting w = 1 in (9) we obtain We obtain the next equation by noting that the form (·, ·) is symmetric and (a, a) > 0 whenever a 6= 0. Therefore (a,t) = 0 for all t if and only if a = 0. By replacing a with a − b, we may conclude that a = b if and only if (a,t) = (b,t) for all t. We apply (11) repeatedly to obtain (xy,t) = (ȳx̄,t), 17 which implies that xy = ȳx̄. (12) A simple calculation shows that 1 (x, y) = (xȳ + yx̄). 2 It follows immediately that if (x, y) = 0, then xy = ȳx = −yx. (13) Now we have all the tools necessary to prove the three equations which will allow us to prove the theorem. Proposition 2.1. Let x, y, and z be elements of O of norm 1 and orthogonal to 1. Also let x be orthogonal to the other two. Then the following three equations hold x(yz) = y(x̄z), (xy)z = x(zy), (xy)(xz) = −zȳ. Proof. For the first equation we have (11) (10) (11) (x(yz),t) = (yz, x̄t) = −(x̄z, yt) = (y(x̄z),t). For the second equation we have (11) (13) (10) (11) (12) (13) ((xy)z,t) = (xy,t z̄) = (ȳx,t z̄) = (ȳz̄,tx) = (zy,tx) = ((zy)x,t) = (x(zy),t). For the third equation we have (11) (13) (10) (11) ((xy)(xz),t) = (xy,t(xz)) = −(xy,t(xz)) = (x(xz),ty) = −(xz, x(ty)) (8) (11) = −(z,ty) = (−zȳ,t). 18 Theorem 2.2. The stabilizer of i j in G2 = Aut(O) is SU(V j ) ∼ = SU(3), and therefore abstractly SU(3) ≤ G2 . Proof. We have shown that V j is invariant under multiplication by i j , and that therefore O breaks up as the direct sum O = C ⊕V j . The stabilizer subgroup Gi j is a Lie subgroup of G2 which acts on V j by C-linear transformations. We define a Hermitian product on V j by [u, v] = (u, v) + i j (i j u, v) for all u, v ∈ V j . We need to show that the action of Gi j preserves this Hermitian product. To do this, we take σ ∈ Gi j . Since σ is an automorphism of O, it preserves the inner product (·, ·), so we have [σ (u), σ (v)] = (σ (u), σ (v)) + i j (i j σ (u), σ (v)) = (σ (u), σ (v)) + i j (σ (ı j u), σ (v)) = (u, v) + i j (i j u, v) = [u, v]. Thus Gi j ≤ U(V j ) ∼ = U(3). Now, since σ is a C-linear transormation on V j , which can be viewed as 3-dimensional complex space. Then σ has eigenvalues u, v, w ∈ V j which are orthonormal with respect to this Hermitian product. We may as well assume w = uv. Since σ is an automorphism, it must preserve the norm. Therefore the eigenvalues corresponding to these three eigenvectors have norm 1. Therefore they lie in the unit circle of C. So we write eθ i j and eφ i j for the eigenvalues of u and v respectively. Then by applying Proposition 19 (2.1) where needed we obtain σ (uv) = σ (u)σ (v) = (eθ i j u)(eφ i j v) = (cos(θ )u + i j sin(θ )u)(cos(φ )v + i j sin(φ )v) = cos θ cos φ (uv) + sin θ sin φ (i j u)(i j v) + cos θ sin φ u(i j v) + sin θ cos φ (i j u)v = cos θ cos φ (uv) + sin θ sin φ (−vū) + cos θ sin φ i j (ūv) + sin θ cos φ (i j vu) = cos θ cos φ (uv) − sin θ sin φ (uv) − cos θ sin φ i j (uv) − sin θ cos φ i j (uv) = (cos(θ + φ ) − sin(θ + φ )i j )(uv) = e−(θ +φ )i j (uv). Since u, v, and uv are three orthonormal eigenvectors of V6 , in the basis {u, v, uv}, the Clinear transformation φ can be written as a diagonal matrix with the eigenvalues for u, v, and uv on the diagonal. But then det(σ ) = eθ i j eφ i j e−(θ +φ )i j = e0 = 1. So we have Gi j ≤ SU(V j ) ∼ = SU(3). We recall that SU(3) is a connected Lie group of dimension 8. So a proper Lie subgroup of SU(3) must have dimension less than 8. So if Gi j has dimension 8, then Gi j = SU(V6 ) ∼ = SU(3). To see that Gi j has dimension 8 we revisit Observation 1.9. Recall that for any automorphism of O, i1 can be sent to anywhere on the 6-dimensional unit sphere orthogonal to 1. So the stabilizer of i1 in G2 has dimension 14 − 6 = 8. Since i1 may be chosen to be any unit orthogonal to 1, the stabilizer of i j in G2 has dimension 8. Throughout the remainder of this thesis, when we refer to SU(3) as a subgroup of G2 we mean SU(V j ). 20 2.2. G2 (2) AND SU(3) Our reason for showing that G2 contains SU(3) as a subgroup is to have a way to find representatives for the conjugacy classes of G2 (2) in G2 . The following theorem shows why every element of G2 (2) is conjugate to some diagonal element of SU(3) in G2 . Theorem 2.3. Inside G2 , every element of G2 (2) is conjugate to a diagonal element of SU(3). Proof. By following the proof of Theorem 2.2 replacing u with i, v with j, and w with k, we obtain a torus T j ⊂ SU(V6 ) of all automorphisms fixing i j and acting diagonally on the basis {i, j, k} by sending j 7→ eφ i j j, i 7→ eθ i j i, k 7→ eψi j k, where θ + φ + ψ = 0. Adams gives an argument which shows that T j is a maximal torus in both SU(3) and G2 . We do not include his argument, but it is available in Chapter 5 of [1]. In Chapter 7 of [8], Reeder proves that a maximal torus in a compact Lie group meets every conjugacy class in the Lie group. We apply this result to our case to say that there is at least one element of every conjugacy class of G2 contained in T j . Since G2 (2) is a subgroup of G2 , every conjugacy class of G2 (2) is contained in some conjugacy class of G2 . This is clear because if two elements are conjugate in G2 (2) they are conjugate in the larger group G2 . We conclude that each conjugacy class of G2 (2) is conjugate to an element of T j . As we deal almost exclusively with elements of T j throughout the remainder of this thesis, we introduce the notation (a, b, c) for the diagonal matrices of type     a   ∈ Tj . b c 21 2.3. FUSING OF CLASSES INSIDE G2 We recall that our goal is to find the traces of the elements of the 16 conjugacy classes of G2 (2) in irreducible representations of G2 . We have no reason to believe there could not be elements of G2 \ G2 (2) by which some of the conjugacy classes of G2 (2) become conjugates in G2 . Theorem 2.4 explains to what extent this fusing of conjugacy classes of G2 (2) occurs in G2 . In order to prove Theorem 2.4, we need to know how to calculate the characters χ6 and χ9 . The characters χ6 and χ9 correspond to the restrictions from G2 to G2 (2) of the 7-dimensional standard and 14-dimensional adjoint representations of G2 respectively. We call these two representations V(1,0) and V(0,1) . We explain the notation in a later section. We label the restrictions of these two representations to G2 (2) as V6 and V9 . Thanks to the branching formula developed by Gordan Savin in [9], we know that for the restriction of G2 to SU(3) we have V(1,0) |SU(3) = W(1,0) ⊕ 1 ⊕W(0,1) , where W(1,0) and W(0,1) are the standard representation of SU(3) and its dual representation. Therefore the character χ6 of an element γ ∈ T j can be calculated as χ6 (γ) = tr(γ) + 1 + tr(γ̄) = a + b + c + 1 + a−1 + b−1 + c−1 . (14) The restriction from G2 to SU(3) also yields V(0,1) |SU(3) = W(1,0) ⊕W(1,1) ⊕W(0,1) , where W(1,1) is the 8-dimensional adjoint representation of SU(3). Thus the character χ9 of γ can be calculated as χ9 (γ) = tr(γ) + (3 + ab + ac + bc + ab−1 + ac−1 + ba−1 + bc−1 + ca−1 + cb−1 + a + b + c) + tr(γ̄) 22 = 3 + 2(a + b + c + a−1 + b−1 + c−1 ) + ab + ac + bc + ab−1 + ac−1 + ba−1 + bc−1 . Theorem 2.4. Two conjugacy classes C1 and C2 of G2 (2) are conjugate to each other inside G2 if and only if both χ6 and χ9 agree on C1 and C2 . Proof. First assume that two conjugacy classes C1 and C2 of G2 (2) become conjugated inside G2 . Recall that χ6 and χ9 are not only the characters of irreducible representations of G2 (2), but also of V(1,0) and V(0,1) . Since C1 and C2 are conjugate in G2 , they must have the same character on all representations of G2 including V(1,0) and V(0,1) . Therefore χ6 and χ9 agree on C1 and C2 . For the the other implication, we recall that by Theorem 2.3 any element in G2 (2) is conjugate to some element γ = (a, b, c) of the torus T j . Further, we have already calculated the trace of γ in V(1,0) to be χ6 (γ) = a + b + c + 1 + a−1 + b−1 + c−1 . Therefore in V(1,0) the matrix representative of γ is conjugate to the matrix        Γ=        a 0 b c 1 a−1 0 b−1 c−1       .       We show that the characteristic polynomial of Γ is completely determined by χ6 (γ) and χ9 (γ). Then given two elements α, β ∈ G2 such that χ6 (α) = χ6 (β ) and χ9 (α) = χ9 (β ), the matrices for α and β in V(1,0) have the same characteristic polynomial and are therefore conjugate in V(1,0) . Since they are conjugate in a representation of G2 , they are conjugate 23 in G2 . This will complete the proof. The roots of pΓ (x) are precisely 1, a±1 , b±1 , and c±1 . Therefore pΓ (x) is either palindromic or antipalindromic. Then the coefficients of pΓ (x) are given by the exterior powers of Γ in the following manner: 7 pΓ (x) = ∑ (−1)mtr(∧mΓ)x7−m. (15) m=0 From equation (15) we calculate the constant term of pΓ (x) to be − ∧7 (Γ) = − det(Γ) = −1. This, allows us to conclude that pΓ (x) is antipalindromic. This fact together with equation (15) allow us to conclude that Γ is of the form pΓ (x) = x7 − Ax6 + Bx5 −Cx4 +Cx3 − Bx2 + Ax − 1, where A = tr(∧1 Γ), B = tr(∧2 Γ), C = tr(∧3 Γ). We clearly have A = tr(∧1 Γ) = tr(Γ) = χ6 (γ). The calculations for B and C are straightforward, but in the interest of brevity we omit them. We know that 2 dim ∧ V6 7 = = 21, 2 and in fact the calculation of B yields B = χ6 (γ) + χ9 (γ), which is the character of V6 ⊕V9 , a 21-dimensional representation of G2 . For C we have 3 dim ∧ V6 7 = = 35, 3 24 and the calculation of C yields C = (χ6 (γ))2 − χ9 (γ), which is the character of a 35-dimensional representation of G2 . We have shown that pΓ (x) is completely determined by χ6 (γ) and χ9 (γ), which completes the proof. Now we use Theorem 2.4 along with Table 2 to find out precisely which conjugacy classes of G2 (2) become conjugates in G2 . Corollary 2.5. The following three sets of conjugacy classes of G2 (2) are conjugate to each other inside G2 : 1. 2a and 2b 2. 4a and 4c 3. 12a, 12b, and 12c Proof. The proof of this corollary just involves checking in Table 2 that χ6 and χ9 agree on these three sets and on no others. By Corollary 2.5, we only need to find one representative element for each of the sets 1, 2, and 3. Therefore, we need to find a total of 12 representative elements. 2.4. REPRESENTATIVE ELEMENTS IN SU(3) As we have shown, T j is a maximal torus of G2 (2) and every conjugacy class of G2 (2) is conjugate to an element of T j . In this section, we find such representative elements for each of the conjugacy classes of G2 (2). Clearly the class of the identity is represented by the identity in SU(3), which we denote by (1, 1, 1). Now we use the properties of SU(3) along with the power table and character table for G2 (2) to identify representatives for the remaining conjugacy classes. In order to reduce the amount of work involved in finding a representative for each class, we make the following observation: 25 Observation 2.6. Two elements (a1 , b1 , c1 ) and (a2 , b2 , c2 ) of T j are conjugates in G2 if −1 −1 {a1 , b1 , c1 } = {a2 , b2 , c2 } as sets or {a1 , b1 , c1 } = {a−1 2 , b2 , c2 } as sets, i.e., they differ by a permutation of entries or the inverse of a permutation of entries. This has to do with the fact that the Weyl group of G2 acts on T j by precisely these permutations of entries and inverses of permutations of entries. We discuss the Weyl group along with the Weyl character formula in Section 3.1. Because of Observation 2.6, we do not distinguish between elements of T j which differ by a permutation of entries or the inverse of a permutation of entries. So when we say that a given representative is unique, we mean that it is unique up to permutations and their inverses. Since G2 (2) is a finite group, all of the representative elements have finite order. Let γ = (a, b, c) be an element of T j of finite order n. Then γ n = 1, which implies that a, b, and c are nth roots of 1. Since γ ∈ SU(3), we also have abc = 1. We denote by ωn a general primitive nth root of unity. The following theorem gives a representative for each of the 16 conjugacy classes of elements in G2 (2). Theorem 2.7. Table 4 gives the unique diagonal representative element in SU(3) for each of the 16 conjugacy classes of G2 (2). We give a proof of Theorem 2.7 for each conjugacy class in the following seven subsections. The following proposition allows us to further simplify our search for representatives. Proposition 2.8. Every element of T j of order n = 1, 2, 3, 4, 6, 7, 8 or 12 contains a primitive nth root as an entry. Proof. Let γ = (ωa , ωb , ωc ) be an element of T j of order n. Since the order of γ is n, it follows that lcm(a, b, c) = n. We call an element γ nontrivial if none of a, b, or c is equal to n and trivial otherwise. Clearly every trivial element contains a primitive nth root. We easily see that all elements of T j of orders 1, 2, 3, 4, 7, and 8 are trivial, and therefore contain a primitive root. 26 Representative 1a 2a and 2b 3a 3b 4a and 4c 4b 6a 7a 8a 12a, 12b and 12c 6b 8b (1, 1, 1) (−1, −1, 1) (ζ3 , ζ3 , ζ3 ) (ζ3 , 1, ζ3−1 ) (i, 1, −i) (i, i, −1) (ζ6 , ζ6 , ζ6−2 ) (ζ7 , ζ72 , ζ74 ) (ζ8 , ζ83 , ζ84 ) 4 ,ζ7 ) (ζ12 , ζ12 12 (ζ6 , ζ62 , ζ63 ) (ζ8 , ζ82 , ζ85 ) Table 4: Table of Representatives for G2 (2) in SU(3) The only way there could possibly be a nontrivial element of order 6 is if a = 2 and b = 3. But then since γ ∈ T j ⊂ SU(3), we must have ωc = ω2−1 ω3−1 = (ω2 ω3 )−1 , which is a primitive 6th root of unity. So there are no nontrivial elements of order 6. Similarly, the only there could possibly be an element of order 12 which is nontrivial is if a = 3 and b = 4. But then since γ ∈ SU(3), we must have ωc = ω3−1 ω4−1 = (ω3 ω4 )−1 , which is a primitive 12th root of unity. So there are no nontrivial elements of order 12. Let (a, b, c) be a representative for a conjugacy class of G2 (2) of order n. As a consequence of Proposition 2.8 and Observation 2.6, we may assume that a = ωn , where ωn is some primitive nth root of unity. This reduces the number of possible representatives we need to check for each class. Throughout the remainder of this thesis we denote by ζn the primitive nth root of unity e2πi/n . 27 2.4.1. Classes of Elements of Order 2 (2a and 2b) In this case we have a = ω2 . The only primitive second root of unity is −1, so a = −1. Thus bc = −1. Therefore the representative for the conjugacy classes 2a and 2b is (−1, −1, 1). 2.4.2. Classes of Elements of Order 3 (3a and 3b) Representative elements of order 3 have a = ω3 . So there are two cases, namely (ω3 , ω3 , ω3 ) and (ω3 , 1, ω3−1 ). Since there are two primitive third roots of unity, namely ζ3 and ζ32 , we have the two possible representatives (ζ3 , ζ3 , ζ3 ) and (ζ3 , 1, ζ3−1 ). By Theorem 2.4, we know that the classes 3a and 3b do not agree on the irreducible character χ6 . So χ6 differentiates between them. We use (14) to calculate: χ6 (ζ3 , ζ3 , ζ3 ) = −2, χ6 (ζ3 , 1, ζ3−1 ) = 1. By checking Table 2 we immediately see that (ζ3 , ζ3 , ζ3 ) is the representative in T j for the conjugacy class 3a, and (ζ3 , 1, ζ3−1 ) is the representative for 3b. 2.4.3. Classes of Elements of Order 4 (4a, 4b, and 4c) We move on to representative elements of order 4. Here we have a = ω4 . The two possible choices for b and c yield (ω4 , ω4 , ω4−2 ) and (ω4 , 1, ω4−1 ). Since i and −i are the only primitive fourth roots of unity, we only have the two possibilities (i, i, −1) and (i, 1, −i). By Theorem 2.4, one of these must be the representative for 4a and 4c and the other must be the representative for 4b. We compute the character χ6 of both elements: χ6 (i, i, −1) = −1, χ6 (i, 1, −i) = 3. By consulting Table 2, we see that (i, i, −1) is the representative element for 4b and (i, 1, −i) is the representative element for 4a and 4c. 28 2.4.4. Classes of Elements of Order 6 (6a and 6b) Next, we find representative elements of order 6. In this case, we have a = ω6 . The three possible choices for b and c yield: (ω6 , 1, ω6−1 ), (ω6 , ω6 , ω6−2 ), (ω6 , ω62 , ω63 ). Since ζ6 and ζ6−1 are the only primitive sixth roots of unity, the possible representatives are (ζ6 , 1, ζ6−1 ), (ζ6 , ζ6 , ζ6−2 ), or (ζ6 , ζ62 , ζ63 ). By Theorem 2.4, one of these is the representative for 6a, and a different is the representative for 6b. We calculate the character χ6 of each element: χ6 (ζ6 , 1, ζ6−1 ) = 5, χ6 (ζ6 , ζ6 , ζ6−2 ) = 2, χ6 (ζ6 , ζ62 , ζ63 ) = −1. So (ζ6 , ζ6 , ζ6−2 ) is the representative for 6a and (ζ6 , ζ62 , ζ63 ) is the representative for 6b. 2.4.5. Class of Elements of Order 7 (7a) Now we find the representative for the class of elements of order 7. We have a = ω7 . The four possible choices for b, and c yield (ω7 , 1, ω7−1 ), (ω7 , ω7 , ω7−2 ), (ω7 , ω72 , ω7−3 ), and (ω7 , ω73 , ω73 ). The primitive seventh roots of unity are ζ7 , ζ7−1 , ζ72 , ζ7−2 , ζ73 , ζ7−3 . So the possible representatives are (ζ7 , 1, ζ7−1 ), (ζ72 , 1, ζ7−2 ), (ζ73 , 1, ζ7−3 ), (ζ7 , ζ7 , ζ7−2 ), (ζ72 , ζ72 , ζ73 ), (ζ7 , ζ72 , ζ7−3 ), or (ζ7 , ζ73 , ζ73 ). We calculate χ6 of each element: χ6 (ζ7 , 1, ζ7−1 ) = 3 + 2ζ7 + 2ζ7−1 , χ6 (ζ72 , 1, ζ7−2 ) = 3 + 2ζ72 + 2ζ7−2 , χ6 (ζ73 , 1, ζ7−3 ) = 3 + 2ζ73 + 2ζ7−3 , χ6 (ζ7 , ζ7 , ζ7−2 ) = 1 + 2ζ7 + 2ζ7−1 + ζ72 + ζ7−2 , χ6 (ζ72 , ζ72 , ζ73 ) = 1 + 2ζ72 + 2ζ7−2 + ζ73 + ζ7−3 , χ6 (ζ7 , ζ72 , ζ7−3 ) = 1, χ6 (ζ7 , ζ73 , ζ73 ) = 1 + ζ7 + ζ7−1 + 2ζ73 + 2ζ7−3 . We consult Table 2 and see that (ζ7 , ζ72 , ζ7−3 ) must be the representative element for 7a. 29 2.4.6. Classes of Elements of Order 8 (8a and 8b) Next, we find representative elements for the classes of elements of order 8. We have a = ω8 . So the four choices for b and c yield (ω8 , 1, ω8−1 ), (ω8 , ω8 , ω8−2 ), (ω8 , ω82 , ω8−3 ), and (ω8 , ω83 , ω84 ). The primitive eighth roots of unity are ζ8 , ζ8−1 , ζ83 , and ζ84 . So the six possible representatives are (ζ8 , 1, ζ8−1 ), (ζ83 , 1, ζ8−3 ), (ζ8 , ζ8 , ζ8−2 ), (ζ83 , ζ83 , ζ82 ), (ζ8 , ζ82 , ζ8−3 ), and (ζ8 , ζ83 , ζ84 ). We calculate the character χ6 for each element: χ6 (ζ8 , 1, ζ8−1 ) = 3 + 2ζ8 + 2ζ8−1 , χ6 (ζ83 , 1, ζ8−3 ) = 3 + 2ζ83 + 2ζ8−3 , χ6 (ζ8 , ζ8 , ζ8−2 ) = 1 + 2ζ8 + 2ζ8−1 + ζ82 + ζ8−2 , χ6 (ζ83 , ζ83 , ζ82 ) = 1 + 2ζ83 + 2ζ8−3 + ζ82 + ζ8−2 , χ6 (ζ8 , ζ82 , ζ8−3 ) = 1, χ6 (ζ8 , ζ83 , ζ84 ) = −1. Then Table 2 tells us that (ζ8 , ζ83 , ζ84 ) is the representative for 8a and (ζ8 , ζ82 , ζ8−3 ) is the representative for 8b. 2.4.7. Classes of Elements of Order 12 (12a, 12b, and 12c) We find a representative for the classes of elements of order 12. We have a = ω12 , so the six −1 −2 2 , ω −3 ), (ω , ω 3 , ω −4 ), choices for b and c yield (ω12 , 1, ω12 ), (ω12 , ω12 , ω12 ), (ω12 , ω12 12 12 12 12 4 , ω −5 ), and (ω , ω 5 , ω 6 ). The power table, included as Table 3, tells us that (ω12 , ω12 12 12 12 12 squaring an element of 12a, 12b, or 12c gives an element in 6a. This allows us to eliminate −1 2 , ω −3 ), (ω , ω 3 , ω −4 )and (ω , ω 5 , ω 6 ) as possibilities. Cubing (ω12 , 1, ω12 ), (ω12 , ω12 12 12 12 12 12 12 12 an element of 12a gives an element of 4a and cubing an element of 12b or 12c gives an −2 element of 4c, allowing us to eliminate (ω12 , ω12 , ω12 ) as a possibility. This leaves only 4 , ω −5 ). The primitive 12th roots of unity are ζ , ζ 5 , ζ 7 , and ζ 11 . So the only (ω12 , ω12 12 12 12 12 12 30 4 , ζ −5 ). As expected we calculate possible representative for 12a, 12b, and 12c is (ζ12 , ζ12 12 4 −5 χ6 (ζ12 , ζ12 , ζ12 ) = 0, which agrees with Table 2. This completes the proof of Theorem 2.7. 31 3. TRACE FORMULA FOR G2 (2) 3.1. THE WEYL CHARACTER FORMULA The Weyl character formula was first proved in 1925 by Hermann Weyl in [10], [11], and [12]. Before introducing the formula, we need talk a little bit about the language in which the formula is given. The purpose of this thesis is not to give an introduction to Lie algebras, so we only introduce enough information in order to use the Weyl character formula to calculate characters in irreducible representations of G2 . One may consult [6] for a thorough introduction. Let g2 be the Lie algebra corresponding to G2 , and let h be the maximal cartan subalgebra corresponding to the maximal torus T j . If g is n-dimensional, then elements of h are diagonal elements of su3 , i.e., of the form  x  y     z where x + y + z = 0. The roots of g2 are the weights which occur in the adjoint representation. They are the twelve functionals on the matrices in h defined by ±(x − y), ±(x − z), ±(y − z), ±x, ±y, ±z. The roots lie in a 2-dimensional hyperplane as pictured in Figure 2, where we give the following labels to the roots: α1 = y, α2 = x − y, β1 = x, β2 = y − z, ω1 = −z, ω2 = x − z. The weights occurring in any irreducible representation of G2 differ by integral combinations of these roots. For each root ρ = (r1 , r2 , r3 ) and each element γ = (a, b, c) ∈ T j , the map eα : G2 → C 32 C+ ω2 β1 β2 ω1 α2 −α1 −β2 α1 −ω1 −β1 −α2 −ω2 Figure 2: Root system of G2 along with the positive Weyl chamber C + is defined by eα (γ) = ar1 br2 cr3 . When we actually do calculations involving these exponentiation maps, we identify the roots of g2 with the plane x + y + z = 0 as follows: 1 2 1 α1 7→ − , , − , 3 3 3 β2 7→ (0, 1 − 1), α2 7→ (1, −1, 0), 1 1 2 ω1 7→ , ,− , 3 3 3 β1 7→ 2 1 1 ,− ,− , 3 3 3 ω2 7→ (1, 0, −1). An example of how this identification works is eα1 (γ) = a−1/3 b2/3 c−1/3 = √ √ 3 3 a−1 b2 c−1 = b3 = b. This gives us a more concrete way to complete the calculations involved when using the Weyl character formula. We choose the positive Weyl chamber C + to be the Weyl chamber which contains all 33 positive integer combinations of ω1 and ω2 as shown in Figure 2. Then the positive roots are Φ+ = {α1 , α2 , β1 , β2 , ω1 , ω2 }, and every irreducible representation of G2 is determined by its highest weight, which is of the form λ = mω1 + nω2 , where m, n ∈ N. We denote the irreducible representation with highest weight λ as Vλ or V(m,n) . The roots ω1 and ω2 correspond to the 7-dimensional standard representation V(1,0) and the 14-dimensional adjoint representation V(0,1) which we have already encountered. When we identify the root system of g2 with the plane x + y + z = 0, we can calculate λ in this identification: 1 1 1 2 , ,− + n(1, 0, −1) = (m + 3n, m, −2m − 3n). λ =m 3 3 3 3 (16) We define ρ= 1 ∑+ α = ω1 + ω2. 2 α∈Φ This ρ is added to λ in the Weyl character formula to ensure that the highest weight does not lie on the walls of C + . We calculate the image of ρ in the identification with the plane by setting m = n = 1 in Equation (16) to obtain 1 ρ = (4, 1, −5). 3 The Weyl group W of g2 has twelve elements. We denote them by their action on the roots of g2 . They are the positive and negative permutations w1 = 1, w2 = −(13), w3 = −(321), w7 = −1, w8 = (13), w9 = (321), w4 = (12), w5 = (123), w6 = −(23), w10 = −(12), w11 = −(123), w12 = (23). When we write a negative sign in front of a permutation, this means the action of permuting 34 and then taking negatives. We define a notion of parity of the elements of the Weyl group. For each w ∈ W , let ε(w) = (−1)k , where k is the minimal number of reflections in the plane x + y + z = 0 and perpendicular to the simple roots α1 and α2 so that w is the composition of those reflections. We have labeled the elements of W in such a way that ( 1, if i is odd; ε(wi ) = −1, if i is even. (17) Now we are ready to introduce the Weyl character formula. Given a compact, connected Lie group G whose Lie algebra has root system Φ, the character of an element γ ∈ G in an irreducible representation Vλ is given by: cλ (γ) = Nλ (γ) , ∆(γ) (18) where the numerator is Nλ (γ) = ∑ ε(w)ew(λ +ρ) (γ), (19) w∈W and the denominator is ∆(γ) = e−ρ (γ) ∏+ (eα (γ) − 1) . (20) α∈Φ Note that (18) is defined only for elements γ such that ∆(γ) 6= 0. Also note that ∆γ = 0 if and only if eα (γ) = 1 for at least one α ∈ Φ+ . This leads to the following definition. Definition 3.1. An element γ of a Lie group G is called irregular if eα (γ) = 1 for at least one α ∈ Φ+ , and γ is called regular if eα (γ) 6= 1 for all α ∈ Φ+ . At times we refer to the conjugacy classes of irregular elements as irregular classes and classes of regular elements as regular classes. Now we address the question: when does eα (γ) = 1? Given a representative element 35 γ = (a, b, c) in the torus T j , exponentiation by the roots in Φ+ gives the six values a , b a , c b , c a, b, c. Observation 3.2. Note that an element γ = (a, b, c) ∈ T j satisfies eα (γ) = 1 for some α ∈ Φ+ if and only if at least one of the following two conditions occur: 1. At least two of a, b, and c are equal. 2. At least one of a, b, and c is 1. In Table 5 we classify each of the conjugacy classes of G2 (2) as either regular or irregular. Since G2 has some irregular conjugacy classes, the Weyl character formula is not directly applicable to them. In Section 3.3, we give a way to calculate the traces on the irregular classes. However, first we use the Weyl character formula directly to calculate the trace formulas for the regular classes. Representative 1a 2a and 2b 3a 3b 4a and 4c 4b 6a 7a 8a 12a, 12b, and 12c 6b 8b (1, 1, 1) (−1, −1, 1) (ζ3 , ζ3 , ζ3 ) (ζ3 , 1, ζ3−1 ) (i, 1, −i) (i, i, −1) (ζ6 , ζ6 , ζ6−2 ) (ζ7 , ζ72 , ζ74 ) (ζ8 , ζ83 , ζ84 ) 4 ,ζ7 ) (ζ12 , ζ12 12 2 (ζ6 , ζ6 , ζ63 ) (ζ8 , ζ82 , ζ85 ) Regularity Irregular Irregular Irregular Irregular Irregular Irregular Irregular Regular Regular Regular Regular Regular Table 5: Table of Representatives for G2 (2) in SU(3) 36 3.2. A TRACE FORMULA FOR REGULAR ELEMENTS Since the Weyl character formula is quite complicated for elements in G2 , we attempt to write it in a way that reduces calculation. First, we make the following observation: Observation 3.3. As we have shown, each conjugacy class of G2 of order n has a primitive nth root of unity in its representative element. Let Gn be the Galois group of the field extension Q(ζn )/Q. Then for each regular conjugacy class A of order n there is an injective homomorphism ϕA : Gn → W where W is the Weyl group of G2 . We provide the precise homomorphism for each regular class in the following subsections. Recall that a field extension has a trace map with the following definition: Definition 3.4. Let E/F be a Galois extension of finite degree k with Galois group G = {σ1 , . . . , σk }. Then for all x ∈ E, we define the field trace to be k Tr(x) = ∑ σi (x). i=1 For each regular conjugacy class A we use the injective homomorphisms from Observation 3.3 and Definition 3.4 to define another map TrA . Let γ ∈ T j be the representative element for A. Let WA be the image of ϕA in W . We define TrA (x) = ∑ ε(w)σw (x). (21) w∈WA Where σw = ϕA−1 (w) for each w ∈ WA . We choose another subset WA∗ of W such that for all wi , w j ∈ WA∗ with wi 6= w j , we have wiWA ∩ w jWA = 0, / (22) 37 and [ wWA = W, (23) w∈WA∗ where wiW is the subset of W obtained by multiplying each element of WA on the left by wi . Then the following theorem allows us to use the generalized trace map (21) to simplify calculation of the numerator of the Weyl character formula for the regular classes of G2 (2). Theorem 3.5. The numerator of the Weyl character formula for the representative element γ ∈ T j of a regular conjugacy class A of G2 (2) in G2 is given by Nλ (γ) = ∗ TrA ew (λ +ρ) (γ) . ∑ (24) w∗ ∈WA∗ where WA∗ is a subset of the Weyl group satisfying (22) and (23). Proof. We start with the numerator of the Weyl character formula as in (19): Nλ (γ) = ∑ ε(w)ew(λ +ρ) (γ). w∈W Now we split up the sum according to the subsets of W defined by left-multiplication by elements of WA∗ : # " Nλ (γ) = ∑∗ ∗ w ∈WA ∑∗ w∈w WA " = ∑ ε(w)ew(λ +ρ) (γ) ∑ # ∗ w w(λ +ρ) ε(w w)σw e (γ) ∗ w∗ ∈WA∗ w∈WA = ∑ ∗ ε(w∗ )TrA ew (λ +ρ) (γ) . w∗ ∈WA∗ This theorem decreases the amount of calculation needed to find the trace formula for each of the regular conjugacy classes. The usefulness of this reformulation of the Weyl character formula is most apparent in the calculation of the trace formula for the class of elements of order 7. 38 3.2.1. Conjugacy Class 6b: Elements of Type (ζ6 , ζ62 , ζ63 ) For this class, we have G6 = Gal(Q(ζ6 )/Q) ∼ = Z/2. And G6 is defined by the action of the elements σ1 and σ2 on ζ6 : σ1 (ζ6 ) = ζ6 , σ2 (ζ6 ) = ζ65 , σ1 7→ w1 , σ2 7→ w7 . Define ϕ6b by ∗ = {w , . . . , w } satisfies the conditions (22) and (23). Then W6b = {w1 , w7 }. The set W6b 1 6 So by Theorem 3.5 the numerator of the character formula is 6 Nλ (γ6b ) = ∑ Tr6b ewi (λ +ρ) (γ6b ) . i=1 We calculate the six exponentials in this sum: ew1 (λ +ρ) (γ6b ) = ζ63m+4n+1 , ew2 (λ +ρ) (γ6b ) = ζ63m−n+2 , ew3 (λ +ρ) (γ6b ) = ζ62m−n+1 , ew4 (λ +ρ) (γ6b ) = ζ62m+n+3 , ew5 (λ +ρ) (γ6b ) = ζ6n−m , ew6 (λ +ρ) (γ6b ) = ζ62n−m+1 . Now we calculate the traces. By (17) we have   Tr (ewi (λ +ρ) (γ )), if i is odd; 6 6b wi (λ +ρ) Tr6b (e (γ6b )) = w (λ +ρ) −Tr6 (e i (γ6b )), if i is even. Where Tr6 is the field trace of the extension Q(ζ6 )/Q. So the numerator is Nλ (γ6b ) =Tr6 ζ63m+4n+1 − Tr6 ζ63m−n+2 + Tr6 ζ62m−n+1 − Tr6 ζ62m+n+3 + Tr6 ζ6n−m − Tr6 ζ62n−m+1 , We set m = n = 0 in Nλ (γ6b ) to obtain ∆(γ6b ) = Tr6 (ζ6 ) − Tr6 (ζ62 ) + Tr6 (ζ6 ) − Tr6 (ζ63 ) + Tr6 (1) − Tr6 (ζ6 ) = 6. 39 So we have cλ (γ6b ) = Nλ (γ6b ) . 6 (25) 3.2.2. Conjugacy Class 7a: Elements of Type (ζ7 , ζ72 , ζ74 ) For this class, Theorem 3.5 is much more useful than in the previous class. We have G = Gal(Q(ζ7 )/Q) ∼ = Z/6, and G is defined by the action of the elements σ1 , . . . , σ6 , where σi (ζ6 ) = (ζ6 )i . The homomorphism ϕ7a is defined by σ1 7→ w1 , σ2 7→ w9 , σ3 7→ w11 , σ4 7→ w5 , σ5 7→ w3 , σ6 7→ w7 . ∗ = {w , w } satisfies the conditions (22) Then W7a = {w1 , w3 , w5 , w7 , w9 , w11 }. The set W7a 1 2 and (23). Thus by Theorem 3.5 the numerator of the Weyl character formula is 2 Nλ (γ7a ) = ∑ Tr7a ewi (λ +ρ) (γ7a ) . i=1 This time, we only have two exponentials in the sum to calculate: 3(m−n) ew1 (λ +ρ) (γ7a ) = ζ7 , ew2 (λ +ρ) (γ7a ) = ζ73m−2n+1 . Now we calculate the traces. As before, we have    Tr7 ewi (λ +ρ) (γ7a ) , if i is odd;  Tr7a ewi (λ +ρ) (γ7a ) =   −Tr7 ewi (λ +ρ) (γ7a ) , if i is even. The map Tr7 is the field trace of the extension Q(ζ7 )/Q. So we calculate the numerator to be 3(m−n) Nλ (γ7a ) = Tr7 ζ7 − Tr7 ζ73m−2n+1 . 40 By setting m = n = 0 in Nλ (γ7a ), we obtain ∆(γ7a ) = Tr7 (1) − Tr7 (ζ7 ) = 6 − (−1) = 7. So we have cλ (γ7a )) = Nλ (γ7a ) . 7 (26) 3.2.3. Conjugacy Class 8a: Elements of Type (ζ8 , ζ83 , ζ84 ) For this class we have G = Gal(Q(ζ8 )/Q) ∼ = Z/2 × Z/2. The Galois group G is defined by the action of the elements σ1 , σ3 , σ5 , and σ7 where σ1 (ζ8 ) = ζ8 , σ2 (ζ8 ) = ζ83 , σ3 (ζ8 ) = ζ85 , σ4 (ζ8 ) = ζ87 . σ2 7→ w12 , σ3 7→ w10 , σ4 7→ w7 . Then ϕ8a is defined by σ1 7→ w1 , ∗ = {w , w , w } satisfies the conditions Therefore W8a = {w1 , w7 , w10 , w12 }. The set W8a 1 3 5 (22) and (23). So by Theorem 3.5 the numerator of the character formula is Nλ (γ8a ) = Tr8a ew1 (λ +ρ) (γ8a ) + Tr8a ew3 (λ +ρ) (γ8a ) + Tr8a ew5 (λ +ρ) (γ8a ) . We calculate the three exponentials in the sum to be ew1 (λ +ρ) (γ8a ) = ζ84m−3n+1 , ew3 (λ +ρ) (γ8a ) = ζ83m−n+2 , ew5 (λ +ρ) (γ8a ) = ζ82n−m+1 . This class is a little bit different than the two we have previously calculated, the reason being that in the previous two cases ε was constant on the sets w∗WA in (24). This allowed us to write the numerator directly in terms of the field trace of the Galois extension. Here we do not have that luxury, since ε(w1 ) = ε(w7 ) = 1, whereas ε(w2 ) = ε(w8 ) = −1. So we calculate the numerator to be Nλ (γ8a ) = Tr8a ζ84m−3n+1 + Tr8a ζ83m−n+2 + Tr8a ζ82n−m+1 . 41 By setting m = n = 0 in Nλ (γ8a ), we obtain √ ∆(γ8a ) = Tr8a (ζ8 ) + Tr8a (ζ82 ) + Tr8a (ζ8 ) = 4 2. So we have cλ (γ8a ) = Nλ (γ8a ) √ . 4 2 (27) 3.2.4. Conjugacy Class 8b: Elements of Type (ζ8 , ζ82 , ζ85 ) Again we have G = Gal(Q(ζ8 )/Q) ∼ = Z/2 × Z/2. Though the Galois group G is defined in the same way as in the class 8a, the map ϕ8b is defined by σ1 7→ w1 , σ2 7→ w2 , σ3 7→ w8 , σ4 7→ w7 . ∗ = {w , w , w } satisfies the conditions (22) and Thus W8b = {w1 , w2 , w7 , w8 }. The set W8b 1 3 5 (23). So by Theorem 3.5 the numerator of the character formula is Nλ (γ8b ) = Tr8b ew1 (λ +ρ) (γ8b ) + Tr8b ew3 (λ +ρ) (γ8b ) + Tr8b ew5 (λ +ρ) (γ8b ) . We calculate the three exponentials in this sum to be ew1 (λ +ρ) (γ8b ) = ζ83m+4n−1 , ew3 (λ +ρ) (γ8b ) = ζ82m−3n−1 , ew5 (λ +ρ) (γ8b ) = ζ8n−m . So we calculate the numerator to be Nλ (γ8b ) = Tr8b ζ83m+4n−1 + Tr8b ζ82m−3n−1 + Tr8b ζ8n−m . Setting m = n = 0 in in Nλ (γ8b ) gives √ ∆(γ8b ) = Tr8b (ζ87 ) + Tr8b (ζ87 ) + Tr8b (1) = 4 2. So we have cλ (γ8b ) = Nλ (γ8b ) √ . 4 2 (28) 42 4 ,ζ7 ) 3.2.5. Conjugacy Classes 12a, 12b, and 12c: Elements of Type (ζ12 , ζ12 12 For these three classes we have G = Gal(Q(ζ12 )/Q) ∼ = Z/2 × Z/2. The Galois group G is defined by the action of the elements σ1 , σ2 , σ3 , and σ4 where σ1 (ζ12 ) = ζ12 , 5 σ2 (ζ12 ) = ζ12 , 7 , σ3 (ζ12 ) = ζ12 11 σ4 (ζ12 ) = ζ12 . We define ϕ12a as the homomorphism σ1 7→ w1 , σ2 7→ w2 , σ3 7→ w8 , σ4 7→ w7 . ∗ = {w , w , w } satisfies the conditions (22) Then W12a = {w1 , w2 , w7 , w8 }. The set W12a 1 3 5 and (23). So by Theorem 3.5 the numerator of the character formula is Nλ (γ12a ) = Tr12a ew1 (λ +ρ) (γ12a ) + Tr12a ew3 (λ +ρ) (γ12a ) + Tr12a ew5 (λ +ρ) (γ12a ) . We calculate the three exponentials in this sum to be 5m+6n−1 4m−3n+1 3n−m+2 ew1 (λ +ρ) (γ12a ) = ζ12 , ew3 (λ +ρ) (γ12a ) = ζ12 , ew5 (λ +ρ) (γ12a ) = ζ12 . So we calculate the numerator to be 4m−3n+1 3n−m+2 5m+6n−1 Nλ (γ12a ) = Tr12a ζ12 + Tr12a ζ12 + Tr12a ζ12 . By setting m = n = 0 in Nλ (γ12a ) we obtain √ 2 11 ∆(γ12a ) = Tr12a (ζ12 ) + Tr12a (ζ12 ) + Tr12a (ζ12 ) = 4 3. So we have cλ (γ12a ) = Nλ (γ12a ) √ . 4 3 (29) 43 3.3. A TRACE FORMULA FOR IRREGULAR ELEMENTS Now that we have given a character formula for each of the regular classes, we consider the case of irregular classes. First we approach the problem of irregular elements in some generality, obtaining a general formula which we then apply to the case of irregular elements of G2 (2) in irreducible representations of G2 . Let G be a compact, simply connected Lie group, T a maximal Cartan subgroup, and Φ the corresponding root system. Let Vλ be an irreducible representation of G with highest weight λ . Let W be the Weyl group, and denote by C + the positive Weyl chamber. Let Φ+ be the subset of Φ of positive roots. Let Γ be a finite subgroup of G. Let γ ∈ Γ. If γ were regular, the Weyl character formula would be directly applicable as we have shown. We consider the case where γ is irregular. Recall that this implies the denominator ∆(γ) = 0, which means that the Weyl character formula is undefined for this γ. We know that for at least one α we have eα = 1. We let ΦK = {α ∈ Φ|eα (γ) = 1}. Then ΦK is the root system corresponding to a Lie subgroup K of G. We define the positive Weyl chamber CK+ to be the Weyl chamber which contains C + . We define δ= 1 ∑+ α. 2 α∈Φ k We further define WK ≤ W to be the Weyl group for K. Let m = [W : WK ], and let w1 = 1 and w2 , . . . , wm be the elements of W such that wi (λ + ρ) lands in CK+ . Then it is easy to see that w1WK , . . . , wmWK are the m left cosets of WK in W . Then for i = 1, . . . , m µi = wi (λ + ρ) − δ is the highest weight for a representation Uµi of K. We may well ask: what is the trace of γ in Uµi ? Recall that we defined K so that its root system consists precisely of the roots 44 α ∈ Φ such that eα (γ) = 1. This implies that 0 ew(µi +δ ) (γ) = ew (µi +δ ) (γ) (30) for all w, w0 ∈ WK . So the Weyl character formula gives the trace of γ in Uµi : cµi (γ) = ∑w∈WK ε(w)ew(µi +δ ) (γ) . e−δ (γ) ∏α∈Φ+ (eα (γ) − 1) K Equation (30) along with the fact that w1 = 1 ∈ WK allow us to rewrite the formula as ! ε(w) ∑ w∈W K cµi (γ) = eµi +δ (γ) −δ . (31) e (γ) ∏α∈Φ+ (eα (γ) − 1) K We note that replacing γ with 1 inside the large parentheses of (31) does not change the value of any of the exponentials, yet doing so yields the Weyl character formula for the identity in the representation of K with highest weight µi . Technically, the Weyl character formula is undefined in this case since the denominator is 0. However, one may work around this by using formal differentiation, defined as the operator ∂α eλ = hα, λ ieλ where h·, ·i is the Killing form (we do not explain the Killing form except to say that for our purposes we may treat it like a dot product). Then one formally applies L’Hospital’s rule, operating by ∂α in the numerator and denominator for every root in Φ+ K. This type of formal differentiation may be used to derive the Weyl dimension formula. The details are readily available. For instance, one may consult [6]. The trace of the identity in an irreducible representation of G with highest weight λ is found to be: dim(Vλ ) = ∏α∈Φ+ hα, λ + ρi . ∏α∈Φ+ hα, ρi By applying this formal L’Hospital’s rule to (31), we obtain ! + hα, µi + δ i ∏ α∈ΦK = eµi +δ (γ) dim Uµi . cµi (γ) = eµi +δ (γ) ∏α∈Φ+ hα, δ i K (32) (33) 45 In the following theorem, we show that the trace formula for γ in irreducible representations of G2 may be written in terms of the trace formulas for γ in the representations Uµ1 , . . . ,Uµm . Theorem 3.6. The trace of γ in an irreducible representation of G with highest weight λ is given by Nλ (γ) , ∆G/K cλ (γ) = (34) where m Nλ (γ) = ∑ ε(wi )dim(Uµi )ewi (λ +ρ) (γ), i=1 and ∆G/K = e−ρ (γ) ∏ (eα (γ) − 1). α∈Φ+ \Φ+ K Proof. Note that WK consists precisely of the elements of W such that given any root α ∈ Φ, the action of all elements of WK on α forms a root space of the same type as ΦK . We begin by rewriting the numerator and denominator of the Weyl character formula as follows: h i w◦w (λ +ρ) i ε(wi ) ∑w∈Wk ε(w)e . e−ρ+δ ∏α∈Φ+ \Φ+ (eα − 1) e−δ ∏α∈Φ+ (eα − 1) K K ∑m i=1 (35) The numerator we have simply written as a sum of left cosets of WK in W . Next, we consider the inner sum in the numerator divided by the second product in the denominator of (35): ∑w∈Wk ε(w)ew◦wi (λ +ρ) . e−δ ∏α∈Φ+ (eα − 1) (36) K This is precisely the Weyl character formula for the element γ in the irreducible representation of K with highest weight µi = wi (λ + ρ) − δ . So by (33), we may rewrite (36) as dim(Uµi )ewi (λ +ρ) . By substituting this result back into (35), we obtain wi (λ +ρ) ∑m i=1 ε(wi ) dim(Uµi )e . e−ρ+δ ∏α∈Φ+ \Φ+ (eα − 1) K 46 α Note that since δ is defined as half the sum of the positive roots in Φ+ K , and since e (γ) = 1 2δ for all α ∈ Φ+ K , we have e (γ) = 1. Therefore 1/2 e−ρ+δ (γ) = e−ρ (γ)eδ (γ) = e−ρ (γ) e2δ (γ) = e−ρ . This completes the proof. 3.3.1. Trace Formulas for the Irregular Elements of T j of Finite Order On the way to finding the formula for G2 (2). We may as well write the formula for all elements of finite order in the torus T j . For regular elements of T j , one applies the Weyl character formula directly as we have done for the regular representatives. For irregular elements of finite order we apply (34). But what do elements of finite order in T j look like? The following proposition gives the answer. Proposition 3.7. Let ω3 be a primitive third root of unity. Let T be the subset of norm 1 elements in C which are roots of 1. Let s ∈ T \ {±1, ω3 , ω32 }, and t ∈ T \ {±1}. Then any element of T j of finite order looks like one of the following: (1, 1, 1), (−1, −1, 1), (ω3 , ω3 , ω3 ), (s, s, s−2 ), or (t, 1,t −1 ). Proof. We know that any element of T j of finite order n looks like (a, b, c) for some a, b, c ∈ C such that abc = 1. We also know that a, b, and c are roots of 1. By Observation 3.2, the the fact that the element (a, b, c) is irregular implies one of two cases. Either at least one of a, b, or c is 1; or at least two of a, b, or c are the same. We treat the cases separately. 1. We may as well assume b = 1. Then a = c−1 . So we may have (1, 1, 1), (−1, 1, −1) which is the same as (−1, 1, 1), or (t, 1,t −1 ) for some t ∈ T \ {±1}. 2. We may assume that a = b. Then either a = c, in which case a = b = c = ω3 , where ω3 is a primitive third root of unity, or a 6= c, in which case c = (ab)−1 = a−2 . So we have the form (s, s, s−2 ) where s ∈ T \ {±1, ω3 , ω32 }. 47 Now we find the trace formulas for the irregular elements described in Proposition 3.7. Since the representative elements from each of the remaining conjugacy classes of G2 (2) fit one of these forms, these formulas allow us to compute the traces of the remaining irregular conjugacy classes of G2 (2) in G2 . 3.3.2. Trace Formula for (1, 1, 1) The trace formula for the identity, or in other words the dimension formula for irreducible representations of G2 may be calculated using (32). We calculate the products in (32) for G2 with λ + ρ = (m + 1)ω1 + (n + 1)ω2 : hα1 , λ + ρi = m + 1, hα1 , ρi hβ1 , λ + ρi m + 3n + 4 = , hβ1 , ρi 4 hω1 , λ + ρi 2m + 3n + 5 = , hω1 , ρi 5 hα2 , λ + ρi = n + 1, hα2 , ρi hβ2 , λ + ρi m + n + 2 = , hβ2 , ρi 2 hω2 , λ + ρi m + 2n + 3 = . hω2 , ρi 3 Then the character of the identity in an irreducible representation of G2 with highest weight λ is given by cλ (1) = (m + 1)(n + 1)(m + 3n + 4)(m + n + 2)(2m + 3n + 5)(m + 2n + 3) . 120 (37) 3.3.3. Trace Formula for (−1, −1, 1) We use formula (34), which we established in Section 3.3. First, we need to know for which positive roots α we have eα (γ) = 1. We exponentiate the roots as follows: eα1 (γ) = −1, eα2 (γ) = 1, eβ1 (γ) = −1, eβ2 (γ) = −1, eω1 (γ) = 1, eω2 (γ) = −1. Therefore ΦK = {±α2 , 0, ±ω1 }, which is an A1 × A1 root system. Thus K ∼ = SO(4). Figure 3 gives a picture of how ΦK looks in the root system of G2 . 48 CK+ ω2 C+ β1 β2 ω1 α2 −α1 α1 −β2 −ω1 −β1 −α2 −ω2 Figure 3: Positive Weyl chambers C + and CK+ in the case where γ = (−1, −1, 1). The A1 × A1 root system ΦK is shown in bold. The elements of the Weyl group which send λ + ρ to other weights in CK+ are w1 , w11 , and w12 . Since µi + δ = wi (λ + ρ), we apply (32) to get the dimension formula for Uµi as follows: dim(Uµi ) = ∏α∈Φ+K hα, wi (λ + ρ)i ∏α∈Φ+K hα, δ i . We use this formula to calculate: dim(Uµ1 ) = (n + 1)(2m + 3n + 5), dim(Uµ11 ) = (m + 2n + 3)(m + 1), dim(Uµ12 ) = (m + n + 2)(m + 3n + 4). Finally, we exponentiate wi (λ + ρ) for i = 1, 11, 12: ew1 (λ +ρ) = (−1)n+1 , ew11 (λ +ρ) = (−1)m+1 , ew12 (λ +ρ) = (−1)m+n+2 . (38) 49 So we write the numerator of the Weyl character formula as Nλ (−1, −1, 1) = (n + 1)(2m + 3n + 5)(−1)n+1 − (m + n + 2)(m + 3n + 4)(−1)m+n+2 + (m + 2n + 3)(m + 1)(−1)m+1 . We calculate e−ρ = −1. Then we calculate the denominator to be −1(−2)4 = −16. So the character formula for elements of the type (−1, −1, 1) is cλ (−1, −1, 1) = Nλ (−1, −1, 1) . −16 (39) 3.3.4. Trace Formula for (ω3 , ω3 , ω3 ) We follow the same process that we used for the element (−1, −1, 1). First, we need to know for which positive roots α we have eα (γ) = 1. We exponentiate the roots as follows: eα1 (γ) = t, eα2 (γ) = 1, eβ1 (γ) = t, eβ2 (γ) = 1, eω1 (γ) = t −1 , eω2 (γ) = 1. Therefore ΦK = {±α2 , ±β2 , ±γ2 , 0} which is an A2 root system. Thus K ∼ = SU(3). The elements of the Weyl group which send λ + ρ to the other weights in CK+ are w1 and w12 . We use (38) to calculate the dimensions of the irreducible representations of K as follows: 1 dim(Uµ1 ) = (n + 1)(m + 2n + 3)(m + n + 2), 2 1 dim(Uµ2 ) = (n + 1)(m + 2n + 3)(m + n + 2). 2 Finally, we exponentiate w1 (λ + ρ) and w12 (λ + ρ) to obtain −(m+1) ew1 (λ +ρ) (γ) = ω3 , ew12 (λ +ρ) (γ) = ω3m+1 . Therefore we calculate the numerator of the formula to be 1 −(m+1) Nλ (ω3 , ω3 , ω3 ) = (n + 1)(m + 2n + 3)(m + n + 2) ω3 − ω3m+1 . 2 50 We calculate e−ρ = ω3 . So the denominator of the Weyl character formula is ω3 (ω3 − 1)(ω3 − 1)(ω3 − 1)(ω3−1 − 1) = 6(ω3 − ω3−1 ). Thus we can write the complete formula for elements of type (ω3 , ω3 , ω3 ) as −(m+1) (n + 1)(m + 2n + 3)(m + n + 2) ω3m+1 − ω3 . cλ (ω3 , ω3 , ω3 ) = 6(ω3 − ω3−1 ) (40) 3.3.5. Trace Formula for (t, 1,t −1 ) First, we need to know for which positive roots α we have eα (γ) = 1. We exponentiate the roots as follows: eα1 (γ) = 1, eα2 (γ) = ω3 , eβ1 (γ) = ω3 , eβ2 (γ) = ω3 , eω1 (γ) = ω3 , eω2 (γ) = ω32 . Therefore ΦK = {α1 , 0, −α1 }. Thus K ∼ = SU(2). The elements of the Weyl group which send λ + ρ to the other weights in CK+ are w1 , . . . , w6 . We use (38) to calculate the dimensions of the representations Uµi as follows: dim(Uµ1 ) = (m + 1), dim(Uµ2 ) = (m + 3n + 4), dim(Uµ3 ) = (2m + 3n + 5), dim(Uµ4 ) = (2m + 3n + 5), dim(Uµ5 ) = (m + 3n + 4), dim(Uµ6 ) = (m + 1). Finally, we exponentiate wi (λ + ρ) for i = 1, . . . , 6 and obtain: ew1 (λ +ρ) (γ) = t m+2n+3 , ew2 (λ +ρ) (γ) = t m+n+2 , ew3 (λ +ρ) (γ) = t n+1 , ew4 (λ +ρ) (γ) = t −(n+1) , ew5 (λ +ρ) (γ) = t −(m+n+2) , ew6 (λ +ρ) (γ) = t −(m+2n+3) . 51 Then the numerator of the formula is Nλ (t, 1,t −1 ) = (m + 1) t m+2n+3 − t −(m+2n+3) − (m + 3n + 4) t m+n+2 − t −(m+n+2) + (2m + 3n + 5) t n+1 − t −(n+1) . We calculate e−ρ = t −3 . So the denominator is t −3 (t − 1)4 (t 2 − 1). Thus the trace formula for elements of type (t, 1,t −1 ) is cλ (t, 1,t −1 ) = Nλ (t, 1,t −1 ) . t −3 (t − 1)4 (t 2 − 1) (41) 3.3.6. Trace Formula for (s, s, s−2 ) First, we need to know for which positive roots α we have eα (γ) = 1. We exponentiate the roots as follows: eα1 (γ) = s, eα2 (γ) = 1, eβ1 (γ) = s, eβ2 (γ) = s3 , eω1 (γ) = s2 , eω2 (γ) = s3 . Therefore ΦK = {α2 , 0, −α2 }. Thus K ∼ = SU(2). The elements of the Weyl group which send λ + ρ to to the other weights in CK+ are w1 and w8 , . . . , w12 . We use (38) to calculate the dimensions of the irreducible representations Uµi as follows: dim(Uµ1 ) = n + 1, dim(Uµ8 ) = n + 1, dim(Uµ9 ) = m + n + 2, dim(Uµ10 ) = m + 2n + 3, dim(Uµ11 ) = m + 2n + 3, dim(Uµ12 ) = m + n + 2. Therefore the numerator of the formula to is 2m+3n+5 −(2m+3n+5) m+3n+4 −(m+3n+4) Nλ (s, s, s ) = (n + 1) s −s − (m + n + 2) s −s + (m + 2n + 3) sm+1 − s−(m+1) . −2 52 We calculate e−ρ = s−5 . Therefore the denominator is s−5 (s − 1)2 (s2 − 1)(s3 − 1)2 . Thus the formula for elements of the type (s, s, s−2 ) is cλ (s, s, s−2 ) = Nλ (s, s, s−2 ) . s−5 (s − 1)2 (s2 − 1)(s3 − 1)2 (42) 3.3.7. Application to G2 (2) The formula for (1, 1, 1) already applies directly for the identity element of G2 (2) in irreducible representations of G2 . Also, the formula for the element (−1, −1, 1) already applies directly to the conjugacy classes 2a and 2b of G2 (2). By substituting ζ3 for ω3 in (40) we obtain the character formula for the conjugacy class 3a: cλ (γ3a ) = −(m+1) (n + 1)(m + 2n + 3)(m + n + 2) ζ3m+1 − ζ3 6(ζ3 − ζ3−1 ) . (43) By again substituting ζ3 , this time for t in (41), we obtain the character formula for the conjugacy class 3b: cλ (γ3b ) = Nλ (γ3b ) , 9(ζ3 − ζ3−1 ) (44) where the numerator is given by −(m−n) −(m+n+2) − (m + 3n + 4) ζ3m+n+2 − ζ3 Nλ (γ3b ) = (m + 1) ζ3m−n − ζ3 −(n+1) + (2m + 3n + 5) ζ3n+1 − ζ3 . Next, we substitute i for t in (41) to obtain the trace formula for the conjugacy classes 4a and 4c cλ (γ4a ) = cλ (γ4c ) = Nλ (γ4a ) , 8i (45) where the numerator is given by Nλ (i, 1, −i) = (m + 1) im+2n+3 − i−(m+2n+3) − (m + 3n + 4) im+n+2 − i−(m+n+2) + (2m + 3n + 5) in+1 − i−(n+1) . 53 By replacing s with i in (42) we obtain the formula for the conjugacy class 4b: cλ (γ4b ) = Nλ (γ4b ) , 8i (46) where the numerator is given by Nλ (i, i, −1) = (n + 1) i2m+3n+5 − i−(2m+3n+5) − (m + n + 2) im+3n+4 − i−(m+3n+4) + (m + 2n + 3) im+1 − i−(m+1) . Finally, by replacing s with ζ6 in (42) we obtain the formula for the conjugacy class 6a: cλ (γ6a ) = Nλ (γ6a ) , 4(ζ62 − ζ6−2 ) (47) where the numerator is given by −(2m+3n+5) −(m+3n+4) Nλ (γ6a ) = (n + 1) ζ62m+3n+5 − ζ6 − (m + n + 2) ζ6m+3n+4 − ζ6 −(m+1) + (m + 2n + 3) ζ6m+1 − ζ6 . 54 4. BRANCHING FROM G2 TO G2 (2) Now that we have a complete trace formula for G2 (2) in irreducible representations of G2 , we have all the tools we need to calculate the branching from any irreducible representation of G2 to G2 (2). Recall that in order to calculate branching, we need the traces of all 16 conjugacy classes of G2 (2) in irreducible representations of G2 . Then given any irreducible representation Vλ of G2 we need only calculate di = hχi , χλ i for i = 1, . . . 16, where χλ is the character of the restriction of the irreducible representation Vλ to G2 (2). Then the restriction of Vλ decomposes over G2 (2) as Vλ |G2 (2) = d1V1 ⊕ · · · ⊕ d16V16 , where V1 , . . .V16 are the irreducible representations of G2 (2) corresponding to the irreducible characters χ1 , . . . , χ16 . In order to actually calculate this branching for irreducible representations of G2 , we use Matlab. The next section gives the code which we implemented to calculate this branching. 4.1. IMPLEMENTATION Our first Matlab function uses the trace formulas we have calculated to give the traces on all sixteen conjugacy classes of G2 (2) in irreducible representations of G2 . 4.1.1. Matlab Function for Traces function [trace] = TraceG2of2(m,n) %This function receives m and n and calculates the traces on the %16 conjugacy classes of G_2(2) in the irreducible representation %of G_2 with highest weight lambda = m*omega_1+n*omega_2, where 55 %omega_1 and omega_2 are the short and long roots corresponding to %7 and 14-dimensional representations respectively. %Build the matrix for Traces trace = zeros(16,2); trace(:,1) = [1:16]; %Prepare the primitive roots of unity z3 = e^(2*pi*i/3); z6 = e^(2*pi*i/6); z7 = e^(2*pi*i/7); z8 = e^(2*pi*i/8); z12 = e^(2*pi*i/12); %Trace of 1a: (identity) Num1a = (m+1)*(n+1)*(m+3*n+4)*(m+n+2)*(2*m+3*n+5)*(m+2*n+3); Den1a = 120; T1a = Num1a/Den1a; %Trace of 2a and 2b: (-1,-1,1) Num2a1=(n+1)*(2*m+3*n+5)*(-1)^(n+1); Num2a2=(m+n+2)*(m+3*n+4)*(-1)^(m+n+2); Num2a3=(m+2*n+3)*(m+1)*(-1)^(m+1); Num2a = Num2a1-Num2a2+Num2a3; Den2a = -16; T2a = Num2a/Den2a; T2b = T2a; %Trace of 3a: (z3,z3,z3) Num3a = (n+1)*(m+2*n+3)*(m+n+2)*(z3^(m+1)-z3^(-(m+1))); Den3a = 6*(z6-z6^(-1)); T3a = Num3a/Den3a; %Trace of 3b: (z3,1,z3^(-1)) Num3b1=(m+1)*(z3^(m-n)-z3^(-(m-n))); Num3b2=(m+3*n+4)*(z3^(m+n+2)-z3^(-(m+n+2))); Num3b3=(2*m+3*n+5)*(z3^(n+1)-z3^(-(n+1))); Num3b = Num3b1-Num3b2+Num3b3; Den3b = 9*(z3-z3^(-1)); T3b = Num3b/Den3b; %Trace of 4a and 4c: (i,1,-i) Num4a1=(m+1)*(i^(m+2*n+3)-i^(-(m+2*n+3))); Num4a2=(m+3*n+4)*(i^(m+n+2)-i^(-(m+n+2))); Num4a3=(2*m+3*n+5)*(i^(n+1)-i^(-(n+1))); 56 Num4a Den4a T4a = T4c = =Num4a1-Num4a2+Num4a3; = 8*i; Num4a/Den4a; T4a; %Trace of 4b: (i,i,-1) Num4b1=(n+1)*(i^(2*m+3*n+5)-i^(-(2*m+3*n+5))); Num4b2=(m+n+2)*(i^(m+3*n+4)-i^(-(m+3*n+4))); Num4b3=(m+2*n+3)*(i^(m+1)-i^(-(m+1))); Num4b = Num4b1-Num4b2+Num4b3; Den4b = 8*i; T4b = Num4b/Den4b; %Trace of 6a: (z6,z6,z6^(-2)) Num6a1 = (n+1)*(z6^(2*m+3*n+5)-z6^(-(2*m+3*n+5))); Num6a2 = (m+n+2)*(z6^(m+3*n+4)-z6^(-(m+3*n+4))); Num6a3 = (m+2*n+3)*(z6^(m+1)-z6^(-(m+1))); Num6a = Num6a1-Num6a2+Num6a3; Den6a = 4*(z6^2-z6^(-2)); T6a = Num6a/Den6a; %Trace of 7a: (z7,z7^2,z7^4) T1 = z7^(3*(m-n)); T2 = z7^(3*m-2*n+1); Num7a1 = T1+T1^2+T1^3+T1^4+T1^5+T1^6; Num7a2 = T2+T2^2+T2^3+T2^4+T2^5+T2^6; Num7a = Num7a1-Num7a2; Den7a = 7; T7a = Num7a/Den7a; %Trace of 8a: (z8,z8^3,z8^4) T1 = z8^(4*m-3*n+1); T2 = z8^(3*m-n+2); T3 = z8^(2*n-m+1); Num8a1 = T1-T1^3-T1^5+T1^7; Num8a2 = T2-T2^3-T2^5+T2^7; Num8a3 = T3-T3^3-T3^5+T3^7; Num8a = Num8a1+Num8a2+Num8a3; Den8a = 4*sqrt(2); T8a = Num8a/Den8a; %Trace of 12a, 12b, and 12c: (z12,z12^4,z12^7) T1 = z12^(5*m+6*n-1); T2 = z12^(4*m-3*n+1); T3 = z12^(3*n-m+2); 57 Num12a1 = T1-T1^5-T1^7+T1^11; Num12a2 = T2-T2^5-T2^7+T2^11; Num12a3 = T3-T3^5-T3^7+T3^11; Num12a = Num12a1+Num12a2+Num12a3; Den12a = 4*sqrt(3); T12a = Num12a/Den12a; T12b = T12a; T12c = T12a; %Trace of 6b: (z6,z6^2,z6^3) Num6b1 = z6^(3*m-2*n+1)+(z6^(3*m-2*n+1))^5; Num6b2 = z6^(3*m-n+2)+(z6^(3*m-n+2))^5; Num6b3 = z6^(2*m-n+1)+(z6^(2*m-n+1))^5; Num6b4 = z6^(2*m+n+3)+(z6^(2*m+n+3))^5; Num6b5 = z6^(n-m)+(z6^(n-m))^5; Num6b6 = z6^(2*n-m+1)+(z6^(2*n-m+1))^5; Num6b = Num6b1-Num6b2+Num6b3-Num6b4+Num6b5-Num6b6; Den6b = 6; T6b = Num6b/Den6b; %Trace of 8b: (z_8,z8^2,z8^5) T1 = z8^(3*m+4*n-1); T2 = z8^(2*m-3*n-1); T3 = z8^(n-m); Num8b1 = T1-T1^3-T1^5+T1^7; Num8b2 = T2-T2^3-T2^5+T2^7; Num8b3 = T3-T3^3-T3^5+T3^7; Num8b = Num8b1+Num8b2+Num8b3; Den8b = 4*sqrt(2); T8b = Num8b/Den8b; %Return the matrix of traces trace(:,2) = round(real([T1a;T2a;T3a;T3b;T4a;T4b; T6a;T7a;T8a;T12a;T2b;T4c;T6b;T8b;T12b;T12c])); end This function outputs the traces of the 16 conjugacy classes of G2 (2) in the irreducible representation of G2 with highest weight λ = mω1 + nω2 . An example output follows: --> TraceG2of2(2,1) ans = 1 189 58 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 --> -3 0 0 -3 -3 0 0 1 0 -3 -3 0 1 0 0 4.1.2. Matlab Function for Multiplicities The following Matlab function gives the multiplicities of irreducible representations of G2 (2) in irreducible representations of G2 . function [mult] = MultG2of2(m,n) %This function receives m and n and calculates the multiplicity %of the irreducible representations V_k of G_2(2) corresponding %to the irreducible characters chi_k in the irreducible %representation of G_2 with highest weight lambda = m*omega_1 %+n*omega_2, where omega_1 and omega_2 are the short and long %roots corresponding to 7 and 14-dimensional representations %respectively. %Build a vector of the orders of the conjugacy classes ClassNum = [1;63;56;672;126;378;504;1728;1512;1008; 252;252;2016;1512;1008;1008]; %Build a matrix for the multiplicities mult = zeros(16,2); mult(:,1) = [1:16]; %Build the character table for G_2(2) A=i*sqrt(3); table = zeros(16,16); 59 table(1,:)=ones(16,1); table(2,:)=[1,1,1,1,1,1,1,1,1,1,-1,-1,-1,-1,-1,-1]; table(3,:)=[6,-2,-3,0,-2,2,1,-1,0,1,0,0,0,0,A,-A]; table(4,:)=[6,-2,-3,0,-2,2,1,-1,0,1,0,0,0,0,-A,A]; table(5,:)=[7,-1,-2,1,3,-1,2,0,-1,0,1,-3,1,-1,0,0]; table(6,:)=[7,-1,-2,1,3,-1,2,0,-1,0,-1,3,-1,1,0,0]; table(7,:)=[14,6,-4,2,-2,2,0,0,0,-2,0,0,0,0,0,0]; table(8,:)=[14,-2,5,-1,2,2,1,0,0,-1,2,-2,-1,0,1,1]; table(9,:)=[14,-2,5,-1,2,2,1,0,0,-1,-2,2,1,0,-1,-1]; table(10,:)=[21,5,3,0,1,1,-1,0,-1,1,3,-1,0,1,-1,-1]; table(11,:)=[21,5,3,0,1,1,-1,0,-1,1,-3,1,0,-1,1,1]; table(12,:)=[42,2,6,0,-6,-2,2,0,0,0,0,0,0,0,0,0]; table(13,:)=[27,3,0,0,3,-1,0,-1,1,0,3,3,0,-1,0,0]; table(14,:)=[27,3,0,0,3,-1,0,-1,1,0,-3,-3,0,1,0,0]; table(15,:)=[56,-8,2,2,0,0,-2,0,0,0,0,0,0,0,0,0]; table(16,:)=[64,0,-8,-2,0,0,0,1,0,0,0,0,0,0,0,0]; %Calculate traces in the irreducible representation of G_2 trace = TraceG2of2(m,n); %Return a matrix of multiplicities mult(:,2) =real(table*(ClassNum.*trace(:,2))/12096); end An example output follows: --> MultG2of2(2,1) ans = 1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 0 10 0 11 0 12 1 13 0 14 1 60 15 16 --> 1 1 The information we receive from the outputs of the two functions in our examples allows us to deduce that the irreducible representation of G2 with highest weight λ = 2ω1 + ω2 is a 189-dimensional representation of G2 which decomposes as a representation of G2 (2) into the direct sum V(2,1) |G2 (2) ∼ = V12 ⊕V14 ⊕V15 ⊕V16 . 4.2. AGREEMENT WITH FORMER RESULTS In conclusion, we give evidence that our formula calculates the correct results. The following table gives information about the branching from the smallest ten nontrivial irreducible representations of G2 to G2 (2). The justification for which is given in [7] using entirely different methods. We write λ = (m, n) as shorthand for λ = mω1 + nω2 . λ dim(Vλ ) χλ |G (1, 0) (0, 1) (2, 0) (1, 1) (0, 2) (3, 0) (4, 0) (2, 1) (0, 3) (1, 2) 7 14 27 64 77 77 182 189 273 286 χ6 χ9 χ13 χ16 χ8 + χ10 + χ12 χ11 + χ15 χ7 + χ10 + χ13 + χ15 + χ16 χ12 + χ14 + χ15 + χ16 χ2 + χ8 + χ9 + χ11 + 2χ12 + χ14 + 2χ15 χ3 + χ4 + χ5 + χ7 + χ12 + χ14 + χ15 + χ16 2 (2) Table 6: Findings of Magaard and Savin in [7] We have found that in every case our formulas agree with Table 6. We demonstrate the usefulness of our formulas in Table 7 by using them to find the next five irreducible representations of G2 restricted to G2 (2). 61 λ dim(Vλ ) χλ |G (3, 1) (2, 2) (0, 4) (1, 3) (4, 1) 448 729 748 896 924 χ8 + χ9 + χ10 + χ11 + 2χ12 + χ13 + χ14 + 2χ15 + 2χ16 χ5 + χ7 + χ8 + 2χ10 + χ11 + 3χ12 + 2χ13 + χ14 + 3χ15 + 4χ16 χ1 + 3χ8 + 2χ9 + 3χ10 + 2χ11 + 4χ12 + 3χ13 + χ14 + 3χ15 + 2χ16 χ3 + χ4 + χ5 + χ6 + 2χ7 + χ10 + χ11 + 2χ12 + 2χ13 + 2χ14 + 4χ15 + 6χ16 χ3 + χ4 + χ6 + χ7 + χ8 + χ9 + χ10 + 2χ11 + 2χ12 + 2χ13 + 2χ14 + 4χ15 + 6χ16 2 (2) Table 7: Our findings continuing Table 6 62 5. REFERENCES [1] J. Adams, Lectures on Exceptional Lie Groups, The University of Chicago Press, Chicago and London, 1996. [2] M. Artin, Algebra, Prentice Hall, Boston, MA, 2011. [3] C. Chevalley and S. Chern, Élie Cartan and His Mathematical Work, Bulletin of the American Mathematical Society 58 (1952). [4] J. Conway and D. Smith, On Quaternions and Octonions: Their Geometry, Arithmetic, and Symmetry, A K Peters, Wellesley, MA, 2004. [5] H. S. M. Coxeter, Integral Cayley Numbers, Duke Mathematics Journal 13 (1946). [6] W. Fulton and J. Harris, Representation Theory: A First Course, Springer, New York, NY, 2004. [7] K. Magaard and G. Savin, Invariants of G2 (2) on G2 (C)-modules, preprint (2011). [8] M. Reeder, Notes on Lie Groups, preprint (2010). [9] G. Savin, Dual Pair PGL(3) × G2 and (g2 , SL(3))-Modules, International Mathematics Research Notices 4 (1994). [10] H. Weyl, Theorie der Darstellung kontinuierlicher halb-einfacher Gruppen durch lineare Transformationen. I, Mathematische Zeitschrift 23 (1925). [11] , Theorie der Darstellung kontinuierlicher halb-einfacher Gruppen durch lineare Transformationen. II, Mathematische Zeitschrift 24 (1926). [12] , Theorie der Darstellung kontinuierlicher halb-einfacher Gruppen durch lineare Transformationen. III, Mathematische Zeitschrift 24 (1926).