GROUP THEORY PRIMER. New terms: so(2n), so(2n+1), symplectic algebra sp(2n)

GROUP THEORY PRIMER New terms: so(2n), so(2n+1), symplectic algebra sp(2n) 1. Some examples of semi-simple Lie algebras In the previous chapter, we developed the idea of understanding semi-simple Lie algebras by their simple roots. It turned out that the relative lengths of simple roots and the angles between them are highly constrained. In fact, they are so constrained that the simple Lie algebras turn out to be easy to classify. They fall into four infinite families called A n, B n, C n, D n by Cartan, where n = 1, 2, 3, 4,... and then five exceptional algebras which do not fit into these families G 2, F 3, E 6, E 7, E 8. The Dynkin diagrams for these algebras are shown in figure 1. Note that the Dynkin diagrams for these algebras are rather un-complicated. As we shall see, this is a result of geometry and the constraints on relative lengths and angles of root systems. Before we get on with this, in this chapter, we will pause to discuss the examples A n, B n, C n, D n. We have already discussed A n in the assignments. It is the simple Lie algebra su(n + 1). Its simple roots all have the same length. Once they all have the same length, the angle between them can only be π or 3π. This is indeed the case for the 2 2 simple root system A n. In this chapter, we will study the other infinite families B n, C n, D n in some detail. Figure 1. The Dynkin diagrams for the semi-simple Lie algebras. They consist of four infinite families of algebras, A n, B n, C n, D n in Cartan s classification and five exceptional algebras, G 2, F 4, E 6, E 7, E 8. 1

2 GROUP THEORY PRIMER 1.1. so(2n). The semi-simple Lie algebra so(n) is associated with the group SO(N) of proper rotations in N-dimensional euclidean space. The elements of this group are real orthogonal N N matrices R which obey the equation RR t = I It is easy to see that the set of all such matrices from a Lie group. We can think of these matrices as comprising a defining representation of the group. The Lie algebra that is associated with this Lie group is found by studying the matrices in an infinitesimal neighbourhood of the identity, R is a real orthogonal matrix, R I + iω RR t I + iω + iω t = I if ω anti-symmetric matrix, ω = ω t We also remember that the elements of the matrix R are real numbers. With our definition, the elements of ω must be imaginary ω ij = ω ij The diagonal elements of an anti-symmetric matrix must always be zero. In such a matrix, once we fix the off-diagonal components which are above the diagonal, in the upper triangle, the anti-symmetry automatically fixes the components which are below the diagonal. Thus, the number of parameters that are needed to specify an imaginary, anti-symmetric matrix is equal to the number of entries in the upper triangle. Thus the dimension of the Lie algebra so(n) is n(n 1)/2. We can choose a basis for the space of imaginary anti-symmetric matrices as (1) [M jk ] lm = i [δ jm δ kl δ jl δ km ] In other words, the generator M jk has i in its j th row and k th column entry, an i in the k th row and j th column entry and zeros everywhere else. In order to proceed, we need to identify a Cartan sub algebra, that is, a maximal commuting set of generators. We begin with the generator H 1 = M 12 It is easy to see that the remaining matrices which commute with this one are all of the M ij with i > 2 and j > 2. The next Cartan generator can then be chosen to be and, in general H 2 = M 34 H k = M 2k 2k+1, k = 1,..., n

GROUP THEORY PRIMER 3 Or, more explicitly, (2) 0 i 0 0 0 0... 0 0 0 i 0 0 0 0 0... 0 0 0 H 1 = 0 0 0 i 0 0... 0 0 0 0 0 i 0 0 0... 0 0 0 H 2 =... H n = 0 0 0 0 0 0... 0 0 i 0 0 0 0 0 0... 0 i 0 This is a maximal set of commuting matrices and it spans the Cartan sub-algebra. There are n such matrices and the Lie algebra has rank n. There is a basis of the [ vector ] space on which the H 1,..., H n act where they are 0 i diagonal. The matrix has eigenvalues 1 and 1. In the basis where they i 0

4 GROUP THEORY PRIMER are diagonal, the Catran generators have the form 1 0 0 0 0 0... 0 0 0 0 1 0 0 0 0... 0 0 0 H 1 = 0 0 1 0 0 0... 0 0 0 0 0 0 1 0 0... 0 0 0 H 2 =... H n = 0 0 0 0 0 0... 0 1 0 0 0 0 0 0 0... 0 0 1 The weights of this 2n-dimensional defining representation are simply read off from the eigenvalues. They are the 2n vectors (3) ±ê 1, ±ê 2,..., ±ê n where ê i is the unit vector aligned with the positive i-axis in n-dimensional space. In other words, (4) [ê i ] j = δ ij These are the weights of the defining representation. The vectors that connect the weights must be roots. The total number of roots must be equal to the dimension of the algebra, 2n(2n 1)/2 and the number of zero roots is equal to the number of Cartan generators, the rank, which is n in this case.

GROUP THEORY PRIMER 5 There should thus be 2n(2n 1) 2 n = 2n(n 1) roots. A candidate for these is the set of all possible the differences of the weights, (5) (6) (7) ± (ê i ê j ), i < j ± (ê i + ê j ), i < j for i, j = 1,..., n. The number of these is equal to four sign possibilities times the number of pairs of indices, n(n 1)/2 which totals the correct number 2n(n 1). This, together with the n zero roots, turns out to be a consistent root system. It is possible to see that this is the correct root system for so(2n) by an explicit construction. In the representation where H 1,..., H n are the anti-symmetric matrices given in equation (2), for example, (8) Eê1 ê 2 = 1 8 Eê1 +ê 2 = 1 8 0 0 1 i 0 0... 0 0 0 0 0 i 1 0 0... 0 0 0 1 i 0 0 0 0... 0 0 0 i 1 0 0 0 0... 0 0 0 0 0 i 1 0 0... 0 0 0 0 0 1 i 0 0... 0 0 0 i 1 0 0 0 0... 0 0 0 1 i 0 0 0 0... 0 0 0 Note that these are linear combinations of our previous basis for the anti-symmetric matrices, M ij in equation (1), with complex coefficients. They have the correct properties to be raising and lowering operators, [ H, Eê1 ê2] = (ê 1 ê 2 )Eê1 ê 2, [ H, Eê1+ê2] = (ê 1 + ê 2 )Eê1 +ê 2 One can easily proceed to construct the full set of eigen-operators E α and E β for all of the roots and to demonstrate that, together with the Cartan sub- algebra, they indeed span the space of 2n 2n anti-symmetric matrices.

6 GROUP THEORY PRIMER The positive roots are (9) (10) The simple roots are ê i ê j, i < j ê i + ê j, i < j α 1 = ê 1 ê 2 α 2 = ê 2 ê 3... α n 1 = ê n 1 ê n α n = ê n 1 + ê n The roots have equal lengths ê i ê j = 2 when i j. The allowed angles between the simple roots are either π/2 or 2π/3. We can find the angles directly by considering, for example, for successive simple roots, (ê i ê i+1 ) (ê i+1 ê i+2 )/ ê i ê i+1 ê i+1 ê i+2 = 1/2 = cos θ i,i+1, which tells us that the angles between the first successive n 1 simple roots are 2π/3 radians. Also the last simple root ê n 1 + ê n is orthogonal to the second-last simple root, ê n 1 ê n, and is at 2π/3 radians to ê n 2 ê n 1. The Dynkin diagram of this root system is depicted in figure 2. Recall that the fundamental weights, which are the highest weights of the fundamental representations are r = n weight vectors µ i which obey the formula 2 α i µ j α i 2 The fundamental weights of so(2n) are = δ ij µ 1 = ê 1 µ 2 = ê 1 + ê 2 µ 3 = ê 1 + ê 2 + ê 3... µ n 2 = ê 1 + ê 2 +... + ê n 2 µ n 1 = 1 2 (ê 1 + ê 2 +... + ê n 1 ê n ) µ n 2 = 1 2 (ê 1 + ê 2 +... + ê n 1 + ê n ) 1.2. so(4). The Dynkin diagram of so(4) is given by two circles which are not connected by lines, as depicted in figure 3. If a Dynkin diagram is disconnected, the Lie algebra is not simple. It decomposes into sub-algebras which are simple Lie algebras. In this case the two sub-algebras are those whose Dynkin diagram is just a circle,

GROUP THEORY PRIMER 7 Figure 2. The Dynkin diagram for the semi-simple Lie algebra so(2n). This algebra has rank n. It was originally named D n by Cartan. Figure 3. The Dynkin diagram for the semi-simple Lie algebra so(4). This algebra is has rank 2 and it is equivalent to two commuting su(2) algebras. that is su(2). This means that, at the level of the Lie algebra, so(4) is identical to two copies of su(2). The representations of the so(4) Lie algebra must then be determined by the representations of the subalgebras, (2j 1, 2j 2 ) where j 1 and j 2 are the highest weights of the su(2) representations. As a very simple example of how we can use the fundamental weights to construct representations, let us derive this using the formal approach. The simple roots of so(4) are The fundamental weights are α 1 = ê 1 ê 2 α 1 = ê 1 + ê 2 µ 1 = 1 2 (ê 1 ê 2 ) = 1 2 α 1 µ 2 = 1 2 (ê 1 + ê 2 ) = 1 2 α 2

8 GROUP THEORY PRIMER Figure 4. The Dynkin diagram for the semi-simple Lie algebra so(6). This algebra is has rank 3 and it is identical to the Dynkin diagram of the su(4) Lie algebra. We take µ 1 as the highest weight of a fundamental representation. This means that it can be lowered once by α 1 and it cannot be raised or lowered by α 2. Once we lower it by α 1 to get the weight µ 1 α 1 = 1 2 α 1 we find that 2 α 2 ( 1 2 α 1) α 2 2 = 0 and it is consistent to assume that the weight 1 2 α 1 cannot be raised or lowered by α 2. Thus the first fundamental representation is two-dimensional with the weights + 1 2 α 1 1 2 α 1 The second fundamental representation will also have two weights, + 1 2 α 2 1 2 α 2 In terms of su(2) representations, these are the (2j 1, 2j 2 ) = (1, 0) and the (2j 1, 2j 2 ) = (0, 1) representations. It is easy to see that the highest weight of the (2j 1, 2j 2 ) representation is ν = 2j 1 µ 1 + 2j 2 µ 2. 1.3. so(6). The so(6) Lie algebra has a Dynkin diagram with three simple roots in a chain which are connected by a single line, as depicted in figure 4. This is the same Dynkin diagram that represents the Lie algebra su(4) and at the level of the

GROUP THEORY PRIMER 9 Lie algebras, so(6) should be identical to the su(4). The simple roots of so(6) are α 1 = ê 1 ê 2 α 2 = ê 2 ê 3 The fundamental weights are α 3 = ê 2 + ê 3 µ 1 = ê 1 µ 2 = 1 2 (ê 1 + ê 2 ê 3 ) µ 3 = 1 2 (ê 1 + ê 2 + ê 3 ) Now, consider the representation whose highest weight is ê 1. By definition, this highest weight cannot be raised at all and it can only be lowered by α 1 = ê 1 ê 2 to obtain another weight ê 1 α 1 = ê 2. We must test this new weight to see whether it can be raised and lowered by α 2 or α 3. (We already know that it can be raised by α 1 once and it cannot be lowered by α 1. ) For this, we must compute 2 α 2 ê 2 α 2 2 = 1, 2 α 3 ê 2 α 2 3 From this we conclude that ê 2 can be lowered once by α 2 and it can be lowered once by α 3. The additional weights that are generated are ê 2 α 2 = ê 3 and ê 2 α 3 = ê 3. We must now test these weights to see whether they can be raised or lowered. In the case of ê 3 what is needed is 2 α 1 ê 3 α 2 1 = 0, 2 α 3 ê 3 α 2 3 which we can take to mean that ê 3 cannot be raised or lowered by α 1 and that ê 3 α 3 = ê 2 is a weight. In the case of ê 3 what is needed is 2 α 1 ( ê 3 ) α 2 1 = 0, = 1 = 1 2 α 2 ( ê 3 ) α 2 2 which we can take to mean that ê 3 cannot be raised or lowered by α 1 and that ê 3 α 2 = ê 2 is a weight (which we already know about). To summarize, so far we have found weights ê 1, ê 3, ê 3, ê 2, ê 2. The weight which we have not tested yet is ê 2. We know that we found ê 2 by lowering ê 3 by α 3. So, we need to test whether it can be raised or lowered by α 1 or α 2, 2 α 1 ( ê 2 ) α 2 1 = 1, 2 α 2 ( ê 2 ) α 2 2 = 1 = 1

10 GROUP THEORY PRIMER These tell us that ê 2 α 1 = ê 1 and ê 2 + α 3 = ê 3 are roots. We already know about ê 3. Testing ê 1 gives 2 α 2 ( ê 1 ) α 2 2 = 0, 2 α 3 ( ê 1 ) α 2 3 we come to the conclusion that ê 1 can only be raised by α 1 and cannot be lowered at all (it is the lowest weight ). We have found a consistent root system of the first fundamental representation (1, 0, 0) of so(6), ±ê 1, ±ê 2, ±ê 3 which are the weights of the six-dimensional defining representation. Now, consider the representation (0, 0, 1), that is, the representation whose highest weight is µ 3 = 1 2 (ê 1 + ê 2 + ê 3 ). By definition, this weight cannot be raised at all and it can only be lowered by α 3 to get 1 2 (ê 1 + ê 2 + ê 3 ) α 3 = 1 2 (ê 1 ê 2 ê 3 ). Now, we must test this weight to see if it can be raised or lowered in the α 1 and α 2 directions. 2 α 1 1 2 (ê 1 ê 2 ê 3 ) α 2 1 = 1, = 0 2 α 2 1 2 (ê 1 ê 2 ê 3 ) α 2 2 This tells us that 1 2 (ê 1 ê 2 ê 3 ) α 1 = 1 2 ( ê 1 + ê 2 ê 3 ) is a weight. We must test whether this weight can be raised or lowered in the α 2 or α 3 directions, 2 α 2 1( ê 2 1 + ê 2 ê 3 ) 2 α 3 1 = 1, ( ê 2 1 + ê 2 ê 3 ) = 0 α 2 2 α 3 2 which tells us that 1( ê 2 1 + ê 2 ê 3 ) α 2 = 1( ê 2 1 ê 2 + ê 3 ) is a weight. Now, we must find 2 α 1 1( ê 2 1 ê 2 + ê 3 ) 2 α 3 1 = 0, ( ê 2 1 ê 2 + ê 3 ) = 0 α 1 2 α 3 2 which we take to mean that it cannot be raised or lowered in either the α 1 or the α 3 directions. We have now found a consistent system with four weights 1 2 (ê 1 1 + ê 2 + ê 3 ), 2 (ê 1 1 ê 2 ê 3 ), 2 ( ê 1 1 + ê 2 ê 3 ), 2 ( ê 1 ê 2 + ê 3 ) These are the weights of the fundamental representation (0, 0, 1) of so(6) and, at the same time, the four-dimensional defining representation of su(4). In fact the other fundamental representation (0, 1, 0) of so(6) has weights which are the negatives of weights of (0, 0, 1) which we have constructed. It is just the four-dimensional representation which is the complex conjugate of the defining representation of su(4). That it should be a representation follows from the simple observation that, if T a satisfy the Lie algebra [T a, T b ] = if abc T c with real structure constants f abc, then Ta is also a representation. Here, we take a simple complex conjugate, without the transposition which one would do to take a hermitian conjugate. For the Cartan = 0

GROUP THEORY PRIMER 11 generators, which are real, this tell us that, if there is a representation with weights ν 1, ν 2,..., there must also be a representation with weights ν 1, ν 2,.... 1.4. so(2n + 1). For the Lie algebra so(2n + 1), we use the same basis for the (2n + 2) (2n + 1)-dimensional matrices of the defining representation, [M jk ] lm = i [δ jm δ kl δ jl δ km ] where, now, there are d = 2n(2n + 1)/2 = n(2n + 1) generators, so the dimension of so(2n + 1) is d = n(2n + 1). The rank is n. The Cartan generators differ from those of so(2n) only by an extra row and column all of whose entries are zero, (11) or, explicitly, H k = M 2k 2k+1 0 i 0 0 0 0... 0 0 0 i 0 0 0 0 0... 0 0 0 H 1 = 0 0 0 i 0 0... 0 0 0 0 0 i 0 0 0... 0 0 0 H 2 =... H n = 0 0 0 0 0 0... 0 i 0 0 0 0 0 0 0... i 0 0

12 GROUP THEORY PRIMER Again, they can be diagonalized as 1 0 0 0 0 0... 0 0 0 0 1 0 0 0 0... 0 0 0 H 1 = 0 0 1 0 0 0... 0 0 0 0 0 0 1 0 0... 0 0 0 H 2 =... H n = 0 0 0 0 0 0... 1 0 0 0 0 0 0 0 0... 0 1 0 The weights of this representation are the vectors ±ê 1, ±ê 2,..., ±ê n, 0 where the zero vector appears in addition to all of the positive and negative unit vectors, which we had for so(2n). There are 2n + 1 weight vectors for this (2n + 1)- dimensional, defining representation of so(2n + 1).

GROUP THEORY PRIMER 13 The roots are (ê i ê j ), i < j (ê i ê j ), i < j (e i + e j ), i < j (e i + e j ), i < j ê i ê i There are 4 n(n 1) 2 + 2n = 2n 2 roots. The dimension is then the rank, n, plus the number of roots, n+2n 2 = n(2n+1) = (2n+1)(2n)/2 which is the correct dimension for the space of (2n + 1)-dimensional imaginary anti-symmetric matrices. The positive roots are (12) The simple roots are (ê i ê j ), i < j (e i + e j ), i < j ê i, i = 1,..., n α 1 = ê 1 ê 2 α 2 = ê 2 ê 3... α n 1 = e n 1 ê n α n = ê n and the Dynkin diagram is depicted in figure 5. 1.4.1. so(5). The Lie algebra so(5) is of rank two. Its simple roots are α 12 = ê 1 ê 2 and the fundamental weights are γ 2 = ê 2 µ 1 = ê 1 (13) µ 2 = 1 2 (ê 1 + ê 2 )

14 GROUP THEORY PRIMER From the general argument of the previous subsection, we already know that the fundamental representation with highest weight ê 1 will the the five-dimensional defining representation. What about the representation whose highest weight is 1(ê 2 1+ê 2 )? By definition, this weight can be lowered once by α 2, that is, 1(ê 2 1 + ê 2 ) α 2 = 1(ê 2 1 ê 2 ) is a weight. Now 2 α 12 [ 1(ê 2 1 ê 2 )] = 1 α 12 2 tells us that we can lower 1 2 (ê 1 ê 2 ) once by α 12, that is, that 1 2 (ê 1 ê 2 ) α 12 = 1 2 (ê 1 ê 2 ) is a weight. Then, 2 α 2 [ 1 2 (ê 1 ê 2 )] α 2 2 = 1 tells us that 1 2 (ê 1 ê 2 ) α 2 = 1 2 (ê 1 + ê 2 ) is a root. Then, this weigh cannot be raised or lowered by α 12 and we are done. The weights are 1 2 (ê 1 + ê 2 ) 1 2 (ê 1 + ê 2 ) 1 2 (ê 1 ê 2 ) 1 2 (ê 1 ê 2 ) and the representation is four dimensional. This is a spinor representation of so(5). 1.4.2. The defining representation is a fundamental representation. The fundamental weights of so(2n + 1) are µ 1 = ê 1 µ 2 = ê 1 + ê 2 µ 3 = ê 1 + ê 2 + ê 3... µ n 1 = ê 1 + ê 2 +... + ê n 1 µ n = ê 1 + ê 2 +... + ê n Consider the representation whose highest weight is ê 1. This weight cannot be raised and it can be lowered only by α 1 which means that ê 1 α 1 = ê 2 is also a weight. Now, we must understand whether ê 2 can be raised or lowered. That is, we

GROUP THEORY PRIMER 15 must find 2 α i ê 2 α 2 i for each simple root. The only non-zero results are 2 α 1 ê 2 α 2 1 = 1, 2 α 2 ê 2 α 2 2 The first of these we already know, since ê 2 was gotten by lowering ê 1 by α 1. The second of these tell us that ê 2 α 2 = ê 3 is a weight. This process continues and we find weights ê 1, ê 2,..., ê n. Then, we ask our question as to whether ê n can be raised or lowered, 2 α n 1 ê n α 2 n 1 = 1, 2 α n ê n α 2 n The second of these tells us that ê n α n = 0 and ê n 2 α n = ê n is a weight. Now that we know that ê n is a weight, 2 α n 1 ( ê n ) α 2 n 1 = 1, = 1 2 α n ( ê n ) α 2 n = 2 = 2 and we see that ê n can be lowered by α n 1 and ê n α n 1 = ê n 1 is a weight. Now, we simply climb up the chain showing that ê n 2, ê n 3,..., ê 1 are weights. This produces weights, 0, ±e 1, ±e 2,..., ±e n These are the weights of the (2n + 1)-dimensional defining representation which we found at the beginning when we constructed the Cartan sub-algebra explicitly in that representation. 1.5. sp(2n). The remaining series of classical semi-simple Lie algebras are the symplectic algebras. An element of the syplectic group is a 2n 2n unitary matrix, U, which, as well as being unitary, UU = I, preserves the skew-diagonal matrix [ ] 0 I UΓU t = Γ, Γ = I 0 where, in Γ, the I is the n n unit matrix. If we look near the identity, U I + iω this is an element of the symplectic group if (14) (15) (16) Ω = Ω ΩΓ ΓΩ = 0 [ ] A B Ω = B D

16 GROUP THEORY PRIMER Unitarity requires that A = A and D = D are hermitian matrices. B is a complex matrix. The symplectic requirement is [ ] [ ] [ ] [ ] [ ] A B 0 I 0 I A t B B + t B + B t A + D D I 0 I 0 B t D t = t D A t B B t = 0 which requires that B is a symmetric matrix and that D = A t. The number of real parameters that are needed give the dimension of the algebra. The complex symmetric matrix B requires 2 n(n+1) real parameters. The n n hermitain matrix 2 A requires n 2 parameters. The dimension of the algebra is therefore 2 n(n+1) + n 2 = 2 n(2n + 1). We can see that the matrices form a Lie algebra, [[ ] A B i B A t, [ ]] à B B Ãt [ ] [ ] [ ] [ ] A B à B à B A B = i B A t B + i Ãt B Ãt B A t = [ A à + B = i B ÃA BB A B ] BÃt t ÃB + BA B à A t B B A + Ãt B B B + A t à t B B Ãt A t which, we shall see, has the properties of a symplectic matrix. For this to be so, we need and As well, we need that Aà + B B ÃA BB = (Aà + B B ÃA BB ) Aà + B B ÃA BB = (B B + A t à t B B Ãt A t ) t is a symmetric matrix and A B BÃt t ÃB + BA A B BÃt ÃB + BA t = (B à A t B B A + Ãt B ) These indeed hold and the commutator of two symplectic matrices produces i times a symplectic matrix. The set of all symplectic matrices, that is all 2n 2n matrices Ω which satisfy equations (14) and (15), thus forms a Lie algebra. To study this simple Lie algebra, we must begin by identifying the Cartan subalgebra. The Cartan subalgebra can be chosen as the matrices (in the example of

GROUP THEORY PRIMER 17 sp(6)) Figure 5. The Dynkin diagram for the semi-simple Lie algebra so(2n + 1). This algebra is has rank n and the last root is the short root, thus the arrow drawn on the double line. This infinite series of simple Lie algebras was originally named B n by Cartan and, in general, (17) 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 H 1 = 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 H 2 = 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 H 3 = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 [H k ] ij = δ ij [δ ik δ i,k+n ], k = 1,..., n, i, j = 1,..., 2n This semi-simple Lie algebra has rank n. The weights of this representation are the 2n unit vectors ±ê 1, ±ê 2,..., ±ê n This is precisely the same set of weights that we found for the defining representation of so(2n). There, we chose the roots as certain differences between the weight vectors. Here, we shall also make a choice but it will be different from the choice made for so(2n). The implication is that, for this system of weights, more than one consistent root system can be found with roots given by the differences of the weights. A useful

18 GROUP THEORY PRIMER guideline is one that we have employed, the number of roots must be equal to the dimension of the algebra minus the rank. In addition to this, in the case of so(2n) we argued that the generators E α corresponding to the roots could be constructed explicitly. Then, the roots are found from their commutators with the Cartan subalgebra generators. In principle, although we will not do it here, this could be done for the present case of sp(2n). The roots are ± (ê i ê j ), i < j ± (ê i + ê j ), i < j ± 2ê i, i, j = 1,..., n We can check that we have found the correct number of roots. We recall that the dimension of the algebra is 2n 2 + n and this should be equal to the total number of roots, including the n zero roots. Here, there are 4 n(n 1) 2 + 2n = 2n 2 non-zero roots and there are n zero roots, which confirms that we have identified the correct number of roots. The positive roots are The simple roots are (ê i ê j ), i < j (ê i + ê j ), i < j 2ê i, i, j = 1,..., n α 1 = ê 1 ê 2 α 2 = ê 2 ê 3... α n 1 = ê n 1 ê n α n = 2ê n and the Dynkin diagram is the one shown in figure 6. The diagram differs from that of the so(2n + 1) algebra in that the last root is the long root, rather than the short root.

GROUP THEORY PRIMER 19 Figure 6. The Dynkin diagram for the semi-simple Lie algebra sp(2n). This algebra is has rank n and the last root is the long root, thus the arrow drawn on the double line pointing toward the shorter roots. This infinite series of simple Lie algebras was originally named C n by Cartan. Figure 7. The Dynkin diagram of the allowed connected systems of three simple roots. The fundamental weights are given by µ 1 = ê 1 µ 2 = ê 1 + ê 2 µ n 1 = ê 1 + ê 2 +... + ê n 1 µ n = ê 1 + ê 2 +... + ê n The fundamental weight µ 1 corresponds to the highest weight of the defining representation, which is therefore one of the fundamental representations. 2. Classification of the simple Lie algebras So far, we have developed some intuition about weights and roots and we have identified the four infinite series of consistent simple root systems, corresponding to the classical compact semisimple Lie algebras su(n + 1), so(2n), so(2n + 1) and sp(2n). Now, we can ask the question as to whether we can use the constraints on simple roots that we have found to look for all possible root systems which are compatible with those constraints. The consistent connected simple root systems correspond to distinct simple compact Lie algebras. A connected simple root system is one whose Dynkin diagram is connected. We will look for connected simple root systems. One observation which will help us a lot

20 GROUP THEORY PRIMER Figure 8. A Dynkin diagram for a simple root system which has a subsystem with two roots, α and β, connected a single line. Figure 9. If the Dynkin diagram in figure 8 represents a consistent simple root system, then the simple root system with Dynkin diagram depicted here is also a consistent simple root system. The two simple roots which were connected by a single line in figure 8 have been merged to form a single root. If the two simple roots in figure 8 were α and β, the root to which they were merged is α + β. is that any connected sub-system of a simple root system must itself be an allowed simple root system. The only connected systems with three simple roots are the ones whose Dynkin diagrams are sketched in figure 7. This is a consequence of the fact that the sum of the angles between three linearly independent vectors must be

GROUP THEORY PRIMER 21 Figure 10. The only consistent simple root system with a triple line has rank 2 and corresponds to the exceptional simple Lie algebra G 2.. less than 2π radians. To see this, begin with two non-parallel vectors placed tail-to-tail so that they lie in a two-dimensional subspace of any three dimensional space. They are linearly independent if they are not parallel. Then, the third vector either lies in the same two-dimensional space, in which case the three vectors are linearly dependent, or it does not, in which case they are linearly independent. If the third vector lies in the same plane, the angles between the vectors ad up to 2π radians. If we lift the third vector out of the plane, into a third dimension, which it then defines, and whence it becomes linearly independent, the angles between the third vector and the first two vectors decrease, and the three angles must now sum to less than 2π radians. Thus, for three linearly independent vectors, the angles between them add to less than 2π radians. No other system of three connected simple roots is allowed. An implication of this and the fact that any connected subsystem of a consistent connected root system is that the only consistent simple root system with a triple line is the rank 2 system depicted on figure 10. This is the Dynkin diagram of the exceptional simple Lie algebra g 2.. If a root system has two subsystems which are connected by a single line as depicted in 8, the root system where the line is removed and the roots, α and β are replaced by α + β, as depicted in figure 9, must also be a consistent simple root system. This implies that any consistent simple root system leads to a family of consistent systems obtained by shrinking away the single lines and the smaller systems that are so obtained must also be consistent. The implication of this point is that a consistent connected simple root system cannot contain more than one double line and it cannot contain a closed loop. If a consistent simple root system has a root, α, which is connected by a single line to each of two roots, β and γ as depicted in figure 11, there is another consistent simple root system which has a single root α connected by a double line to another single root β + γ, as depicted in figure 12. This implies that a consistent simple root system can contain only a single three-fold branch and it cannot contain both a branch and a double line.

22 GROUP THEORY PRIMER Figure 11. The Dynkin diagram of a simple root system which has a subdiagram with containing a single root γ connected to two other roots, α and β by single lines. The points above restrict consistent Dynkin diagrams to those with a tree-like structure with a single three-fold branch or a single occurrence of a double line. Now, we shall find further constraints on these structures. This is encapsulated in the observation that the vectors forming the root systems in figures 13 and 14 are linearly dependent. The nonzero coefficients superposition of the vectors which gives the zero vector are drawn inside the circles which denote the simple roots. The fact that these cannot be consistent simple root systems limits the position of the branch and the double line to just a few possibilities. It leaves us with the four infinite series of algebras a n, b n, c n, d n associated with the Lie groups A n, B n, C n, D n and the five exceptional Lie algebras g 2, f 4, e 6, e 7, e 8 associated with the Lie groups G 2, F 4, E 6, E 7, E 8. displayed in figure 1.

GROUP THEORY PRIMER 23 Figure 12. If the root system that is displayed in figure 11 is consistent then the root system depicted in this figure is also consistent. The simple roots α and β in in figure 11 are merged to a root α + β which is joined to the root γ by a double line. Figure 13. The simple roots in this system are not linearly independent. α 1 +2 α 2 +3 α 3 +2 α 4 + α 5 = 0 if α 2 > α 3, that is α 2 = 2 α 3, and 2 α 1 + 4 α 2 + 3 α 3 + 2 α 4 + α 5 = 0 if α 2 < α 3, that is α 2 = 1 2 α 3. To see this, it is easy to show that ( α 1 + 2 α 2 + 3 α 3 + 2 α 4 + α 5 ) 2 = 0 if α 2 > α 3 and (2 α 1 + 4 α 2 + 3 α 3 + 2 α 4 + α 5 ) 2 = 0, respectively.

24 GROUP THEORY PRIMER Figure 14. The simple roots in each of these three systems are not linearly independent. In the first, ( α 3 + α 5 + α 7 +2( α 2 + α 4 + α 6 )+3 α 1 ) 2 = 0, in the second ( α 5 + α 8 + 2( α 4 + α 7 ) + 3( α 3 + α 6 ) + 4 α 2 + 2 α 2 ) 2 = 0, and in the third, ( α 9 +2 α 8 +3 α 7 +4 α 6 +5 α 5 +6 α 1 +2 α 4 +4 α 3 +3 α 2 ) 2 = 0.