Blas de Otero

Lie algebras: representation theory

Now that we have determined all simple and semi-simple Lie algebras, the next step is to represent them in Hilbert spaces, where connections with physics may be established. The technique for building irreducible representations is rather simple to state: identify a highest weight state and then generate its other states by the repeated application of lowering operators until the process terminates, and you run out of steam. We have seen that this process is very efficient for SU(2) which has only one lowering operator. However, a rank-r Lie algebra has r independent non-commuting annihilation operators, and the process becomes rather unwieldy, and a systematic approach must be adopted.

8.1 Representation basics

We work in the Dynkin basis where all states are labeled by r integers (positive, negative or zero), denoted by the ket

| λ ,

whereλ is a vector whose components are the Dynkin labels of the state

λ = ( a1a₂ . . . an). (8.1) To each simple root, associate a lowering operator T₋⁽ⁱ⁾ which, together with its hermitian conjugate, generates an SU(2) algebra

[ T₊⁽ⁱ⁾, T₋⁽ⁱ⁾] = T3⁽ⁱ⁾, i = 1, 2 . . . , r. (8.2) We have

T₃⁽ⁱ⁾| λ  = aⁱ| λ . (8.3) 143

The action of the lowering operators on this state yields another state with labelλ − αi

T₋⁽ⁱ⁾| λ  ∼ | λ − αi. (8.4) When all ai are positive, the state is said to be inside the Weyl chamber of the algebra’s lattice, or on its wall if any of the entries are zero. A state inside or on the walls of the Weyl chamber is said to be dominant; denote it a capitalized vector

= ( a1a₂ . . . an), a_i ≥ 0.

It is annihilated by the r raising operators

T₊⁽ⁱ⁾|  = 0, i = 1, 2 . . . , r, (8.5) since the commutator of two raising operators is also a raising operator. Any domi-nant weight generates a representation of the algebra by repeated application of the lowering operators. For this reason, it is called the highest weight state of the rep-resentation. Representations of the Lie algebra are in one-to-one correspondence with the states inside or on the walls of the Weyl chamber. This is in contrast with the physics way to label representations in terms of the Casimir operators of the algebra.

Consider a highest weight state with entry ai. We can subtract the simple root αi ai times, by applying T₋⁽ⁱ⁾ = E(−αi) as many times. The simple roots are expressed in Dynkinese by the rows of the Cartan matrix.

The number of applications of the annihilation operators to generate all states of a representations is called the level or height of the representation. It is given by the dot product of the Dynkin indices of the highest weight state with the level vector R= (R1, R2, . . . , Rr), whose components are given by twice the sum of the rows of the inverse Cartan matrix

R_i = 2 

j

A⁻¹_{i j}. (8.6)

The fundamental representations are those which are labeled by the Dynkin labels ( 0^p1 0^n−p−1), where p = 1, 2, . . . n, and 0^pis shorthand for p zero entries.

8.2 A₃fundamentals

In this section, we use this technique to build the fundamental representations of A3. There are three fundamental representations, since its Dynkin diagram is a linear chain with three nodes. Its three simple roots, expressed in the Dynkin basis, are the rows of the Cartan matrix (1 stands for−1)

α1= ( 2 1 0 ), α2= ( 1 2 1 ), α3= ( 0 1 2 ).

8.2 A₃fundamentals 145 From its Cartan matrix, we find the level vector

R_SU(4)= (3, 4, 3), (8.7)

whose entries give the levels of the fundamental representations.

The highest weight state of the first fundamental representation is( 1 0 0 ). With only one positive entry in the first spot, the only way to lower it is by appli-cation of the operator associated with α1. This produces the state labeled by ( 1 1 0 ) = ( 1 0 0 ) − ( 2 1 0 ). This new state has a positive entry in the second spot: we apply T₋⁽²⁾ which subtracts the second rootα2. The result is a new state ( 0 1 1 ) = ( 1 1 0 ) − ( 1 2 1 ), with positive unit entry in the third spot. Then we apply T₋⁽³⁾which subtracts the third simple root to get( 0 0 1 ) = ( 0 1 1 ) − ( 0 1 2 ).

We are done because this state has no positive entries: application of any lowering operator gives zero. We have generated the four-dimensional representation of A3. We observe that it has three levels, as expected.

Let us apply the same procedure to( 0 1 0 ). We first subtract α2to get( 1 1 1 ).

We now have positive entries in the first and third spots, so we need to subtract with α1 andα3, to produce two states. Then, as shown in the figure, the next level of subtractions produces the same state, so that the diagram is spindle-shaped. This is a general feature of Lie algebra representations. At the fourth level we finally arrive at a state without positive entries and the representation terminates. The height of this representation is four and it contains six states.

The highest weight state of the third fundamental is( 0 0 1 ). The first level is obtained by subtractingα3, yielding( 0 1 1 ). Subtraction by α2produces( 1 1 0 );

finally subtraction byα1yields the lowest weight state( 1 0 0 ); as expected it has three levels.

These constructions are summarized by the following diagrams.

SU(4)fundamental representations

We see that the states in(0 0 1) have opposite Dynkin labels from those in (1 0 0).

These representations are complex conjugates of one another, a consequence of the Dynkin diagram’s reflection symmetry. Also the weights in the(0 1 0) are all accompanied by their negatives: this representation is self-conjugate, as expected from the Dynkin diagram.

Another method, more familiar to some physicists, is to represent the quadru-plet(1 0 0) by a complex vector with upper index, the sextet (0 1 0) by a complex antisymmetric tensor with two upper indices, and the antiquadruplet(0 0 1) with a complex antisymmetric tensor with three upper indices

(1 0 0) ∼ T^a, (0 1 0) ∼ T^ab, (0 0 1) ∼ T^abc.

We could have gone the other way, associating to the antiquadruplet a complex vector with one lower index, with the sextet represented by a tensor with two lower indices, and the quadruplet with a tensor with three lower indices

(0 0 1) ∼ Ta, (0 1 0) ∼ Tab, (1 0 0) ∼ Tabc.

This dual tensor description of the same representation is resolved by using the Levi–Civita tensor; for SU(4), its four upper(lower) indices enable to trade one lower(upper) index for three upper(lower) indices and vice-versa. It allows us to write

T_a= 1

3!abcdT^bcd. (8.8)

This is called Poincaré duality. The sextet which sits in the middle of the Dynkin diagram is represented in two ways by a complex two-form (twelve real parameters), with upper or lower indices related by complex conjugation

T_ab = Tab. (8.9)

The same duality relation

T^ab = 1

4!^abcdT_cd, (8.10)

cuts the number of real parameters in half to six, the required number needed to describe the real sextet.

8.2 A₃fundamentals 147 This pattern readily generalizes to all SU(n + 1) algebras, for which the basic representations correspond to the antisymmetric tensor products of its fundamen-tal (n + 1)-plet. One moves along the Dynkin diagram by adding an index and antisymmetrizing. Half way through the Dynkin diagram, one can use the totally antisymmetric Levi–Civita tensor with(n + 1) indices

^a1a2...an+1, to lower the indices.

The dimensions of the fundamental representations of Anare summarized in the following table.

In this table we have used duality in the last two entries. Each of these representa-tions corresponds to one node in the Dynkin diagram.

The adjoint representation(1 0 1) is real but is not a fundamental representation.

Since the highest weight has two positive entries, there are two states at the next level. These produce two weights each of which have two positive entries, resulting again in two paths to the next level. There, redundancy occurs and there are only three states at this level, each of which has only one positive entry. The next level has three states, each of which has zero Dynkin labels; they correspond to the rank of the algebra. Then the process continues in reverse, ending with the lowest state ( 1 0 1 ). This produces fifteen states, arranged in a self-conjugate spindle-shaped arrangements, with the conjugate weights appearing in the bottom half of the spindle.

SU(4)adjoint representation

( 1 0 1 )

@@R −α³

−α₁

( 1 1 1 ) ( 1 1 1 )

@@R @@R −α₂

−α₂ −α₃ −α₁

( 0 1 2 ) ( 1 2 1 ) ( 2 1 0 )

−α₂ −α₁

−α₃ ? ? ?

( 0 0 0 ) ( 0 0 0 ) ( 0 0 0 )

−α₃ ? −α₂ ? −α₁ ?

( 0 1 2 ) ( 1 2 1 ) ( 2 1 0 )

@@R @@R −α₂

−α₂ −α₃ −α₁

( 1 1 1 ) ( 1 1 1 )

( 1 0 1 )

@@R −α³

−α₁

The positive roots lie in the upper half of the spindle; they are easily identified as(α1+ α2+ α3), (α2+ α3), (α1+ α2), and the three simple roots. Half the sum of all positive roots, the Weyl vector, is then

ρ = 1

2(3α1+ 4α2+ 3α3).

Does this ring a bell? Yes you noticed it, the numerical factors are the same as in the level vector. This turns out to be a feature of the Lie algebras An, Dn, and E6,7,8 which have a symmetric Cartan matrix, with roots of the same length. For the others, it is not true. The Weyl vector looks particularly simple in the Dynkin basis

ρ = ω1+ ω2+ ω3= ( 1 1 1 ). (8.11) This is a general feature: the components of the Weyl vector of any simple Lie algebra are all equal to one in the Dynkin basis.

The height of the adjoint representation, ladj (six for SU(4)), is related to the Coxeter number of the algebra

h= 1

2l_adj+ 1, (8.12)

8.3 The Weyl group 149 which relates D, the dimension of the algebra, to its rank r ,

D= r (h + 1). (8.13)

8.3 The Weyl group