Group Theory - QMII 017 Reminder Last time we said that a group element of a matrix lie group can be written as an exponent: U = e iαaxa, a = 1,..., N. We called X a the generators, we have N of them, they span a basis for the Lie algebra, and they can be found by taking the derivative with respect to α a at α = 0. The generators are closed under the Lie product ( [, ]), and related by the structure constants [X a, X b ] = if abc X c. (1) We end with the Jacobi identity [X a, [X b, X c ]] + [X b, [X c, X a ]] + [X c, [X a, X b ]] = 0. () 1 Representations of Lie Algebra 1.1 The adjoint rep. part- I One of the most important representations is the adjoint. There are two equivalent ways to defined it. Here we follow the definition by Georgi. By plugging Eq. (1) into the Jacobi identity Eq. () we get the following: f bcd f ade + f abd f cde + f cad f bde = 0 (3) Proof: 0 = [X a, [X b, X c ]] + [X b, [X c, X a ]] + [X c, [X a, X b ]] = if bcd [X a, X d ] + if cad [X b, X d ] + if abd [X c, X d ] = (f bcd f ade + f cad f bde + f abd f cde ) X e (4) 1
Now we can define a set of matrices T a s.t. [T a ] bc = if abc. (5) Then by using the relation of the structure constants Eq. (3) we get [T a, T b ] = if abc T c. (6) Proof: [T a, T b ] ce = [T a ] cd [T b ] de [T b ] cd [T a ] de = f acd f bde + f bcd f ade which means that: = f cad f bde + f bcd f ade = f abd f cde = f abd f dce = if abd [T d ] ce (7) [T a, T b ] = if abd T d. (8) Therefore the structure constants themselves generate a representation. This representation is called the adjoint representation. The adjoint of su() We already found that the structure constants of su() are given by the Levi-Civita tensor. Therefore we get: 0 0 0 0 0 i 0 i 0 T 1 = 0 0 i, T = 0 0 0, T 3 = i 0 0 (9) 0 i 0 i 0 0 0 0 0 Back to the adjoint We can define a scalar product on the linear space of the generators in the adjoint representation as tr(t a T b ). This is a real symmetric matrix, and hence can be diagonalized with orthogonal matrix. By performing a linear
transformation on the generators: X a X a = S ab X b (10) with some orthogonal matrix S, Eq. (10) induces the following transformation of the structure constants f abc f abc = S ad S be f deg S 1 gc. (11) The matrices T a of the adjoint representation transform by T a T a = S ab ( STb S 1) (1) which is a combination of a similarity transformation and simultaneously a linear transformation of the form (10). In the trace, however, only the latter transformation matters and we obtain tr(t a T b ) tr(t at b) = S ac S bd tr(t c T d ) = S ac tr(t c T d )(S T ) db (13) tr(t T ) S tr(t T ) S T Selecting a diagonal basis, we write tr(t a T b ) = κ a δ ab (no sum!) (14) However, we still have freedom to re-scale our generators. We shall assume that all of the κ a are positive. This will be the case if the algebra is compact 1 and semisimple. Then we can just go ahead and write tr(t a T b ) = λδ ab (15) with some positive λ. 1 Non-compact Lie algebras, in which some of the κ a are negative, do not have nontrivial, finite dimensional unitary representations. They still, however, play an important role in physics: The lorentz algebra is an example. A semisimple group contains no nontrivial abelian invariant subgroup. 3
A nice feature of this basis comes from the following: tr([t a, T b ]T c ) = if abd tr(t d T c ) = iλf abc f abc = iλ 1 tr([t a, T b ]T c ). (16) Thanks to the trace we find that the structure constants in this basis are completely anti-symmetric! 1. Adjoint, states, and why we call it a rep. So far we didn t define a representation of a Lie algebra. In fact the definition is straight forward once we defined it for a group: Representation - A rep. of a Lie algebra g is a linear map π : g gl(n, C) (17) that preserves the product, X, Y g π([x, Y ]) = [π(x), π(y )]. (18) A linear map between two Lie algebras that preserves the product is called a Lie algebra homomorphism. If this map is 1-to-1 and onto it is called an isomorphism. 1..1 States and the adjoint Here we would like to generalize our construction of the regular representation to semi-simple Lie groups. 1. For each generator T a g associate a vector T a s.t. they form an orthonormal basis, T a T b = λ 1 tr(t at b ) = δ 3 ab.. Define the adjoint representation on this vector space as ad(t a ) T b = [T a, T b ] (19) 3 The dagger will become important in the following sections. 4
exercise check that this is a rep, i.e. [ad(x), ad(y )] = ad([x, Y ]). 3. The components of the matrices are given by [ad(t a )] bc T b ad(t a ) T c = if acd T b T d = if abc = [T a ] bc (0) So the remarkable thing is that ad(t a ) = T a. Therefore it sometimes said that the adjoint representation is the action of the algebra on itself. Note that there are few different (but equivalent) definitions for the adjoint representation. We will make use of the construction given above in For completeness we define the following: Irrep. - A Lie algebra rep π : g gl(v ) is irreducible if there is no nontrivial subspace W V such that X g π(x)w W. (1) Subalgebra and the Cartan subalgebra Lie Subalgebra - For a given Lie algebra g, if a subspace h g is a Lie subalgebra if it is closed under the Lie product, X, Y h, i[x, Y ] h. () Example -su(3) and su() You saw in class that the Gell-Mann matrices are a good basis for su(3): σ λ i = 1 0 i 0, i = 1,, 3 λ 4 = 1 0 0 1 0 0 0 λ 5 = 1 0 0 i 0 0 0 0 0 0 1 0 0 i 0 0 λ 6 = 1 0 0 0 0 0 1 λ 7 = 1 0 0 0 0 0 i λ 8 = 1 1 0 0 0 1 0 3 0 1 0 0 i 0 0 0 5
It s easy to see that [λ i, λ j ] = iɛ ijk λ k for i, j, k = 1,, 3. This is also the algebra of su(). Therefore su() is a subalgebra of su(3)..1 Cartan subalgebra Cartan Subalgebra - A Cartan Subalgebra h, is the maximal commuting subalgebra of a Lie algebra g. Namely, 1. H 1, H h, [H 1, H ] = 0.. If X g and [X, H] = 0 for all H h, then X h. Example -su(3) The Cartan subalgebra of su(3) can be chosen as λ 3 = 1 1 0 0 0 1 0 λ 8 = 1 1 0 0 0 1 0 3 0 0 0 0 0 All Cartan subalgebras of g have the same dimension. We define the rank of an algebra to be the dimension of its Cartan subalgebra..1.1 Casimir operator A Casimir operator is an operator (in the enveloping algebra) that commutes with all members of the algebra. For example, J = Jx +Jy +Jz is a Casimir of su(). The number of Casimirs is equal to the rank of the algebra. 6
3 Roots and Weights The philosophy of Roots and weights is to give a geometrical interpretation to a minimal (to be defined later) number of generators, and then by using this geometrical interpretation to construct the whole algebra, and to construct all irreps. of the group. 3.1 Weights Recall that any number of commuting linear operators is simultaneously diagonalizable, thus we can choose a basis in which all of the H i are diagonalized. In this basis their action on the states of some irrep of dim D is H i µ, D = µ i µ, D. (3) The entries µ i are the weights. Since H i are hermitian µ i R. The vector µ is called the weight vector. It s a real vector of length M. We have D of those. In different irreps we will get different weight vectors. 3. Roots The weights of the adjoint representation are called Roots. 1. We choose an orthonormal basis for the Cartan states H i H j = λ 1 tr(h i H j ) = δ ij. (4). The Cartan generators have zero roots since H i H j = [H i, H j ] = 0. (5) 7
3. The other states of the adjoint, those not corresponding to the Cartan, have non-zero roots. H i E α = α i E α [H i, E α ] = α i E α. (6) α is called the root vector. 4. Since α is real [ ] Hi, E α = αi E α E α = E α. (7) It means that in the basis where the Cartan generators are diagonal, the rest of the generators are no longer hermitian! As we will see, E ±α will play the role of J ± in SU(). 5. Since E α are not hermitian we defined their inner product as E α E β = λ 1 tr(e αe β ) = δ αβ. (8) 6. The state E α E α has zero root. Therefore [E α, E α ] is in the Cartan [E α, E α ] = β i H i (9) By using orthonormality of the Cartan generators we can calculate β i β i = H i E α E α = λ 1 tr (H i [E α, E α ]) = λ 1 tr (E α [H i, E α ]) = α i So we conclude that [E α, E α ] = α i H i. (30) 8