TA8 - More Symmetries and Elements of Lie Theory

Erez Urbach

Contents

1 Lie algebra of a Lie group ...... 1 2 Representations ...... 3 2.1 Reminder: representations of the group ...... 3 2.2 Representations of the algebra ...... 4 2.3 Examples for algebra representations ...... 5 2.3.1 The trivial representation ...... 5 2.3.2 The fundamental representation ...... 6 2.3.3 The adjoint representation ...... 6 3 From the algebra back to the group ...... 7 4 Spoiler to the next topic ...... 8

1 Lie algebra of a Lie group

Given a continuous (Lie) group G (the examples we saw were SO (n) \SU (n) etc.), the Lie algebra g (or Lie (G)) is “tangent space of the group at the identity”. By that we meant that we take all the possible paths in the Group Mˆ (α) ∈ G that start at the identity Mˆ (0) = 1n×n and expand to linear order

2 Mˆ (α) = 1n×n + α · t + O α (1)

All the possible ts form the the Lie algebra. In the last exercise you saw that the algebra of the group SU (2) is

su (2) = tˆ| tˆ† = tˆ and tr tˆ = 0 (2)

Question: is the algebra g the same as the group G? Answer: NO. These are two diﬀerent objects. For SU (2) the group is made of unitary matrices and the algebra of hermitian matrices. The two are not equal. More algebrically: the group has only a group (multiplication) structure, whereas the algebra is also a vector space. But as we just said, the algebra of a group is built using the group.

1 1 Lie algebra of a Lie group 2

Starting from a matrix group, one can show that the algebra is closed to commutation. For example, given t1, t2 ∈ su (2), also

[t1, t2] = t1t2 − t2t1 ∈ su (2) . (3)

Starting from the Group G, one can build a bracket operator on the corresponding algebra [., .]: g × g → g that is 1. Bi-linear (remember that g is a vector space!). 2. Anti-Symmetric [t1, t2] = − [t2, t1] . (4)

3. Satisfy the Jacobi identity

[t1, [t2, t3]] + [t2, [t3, t1]] + [t3, [t1, t2]] = 0. (5)

In the cases we described here (n × n matrix groups and algebras) we deﬁned the bracket as the matrix commutator

[t1, t2] = t1t2 − t2t1 (6) and these three properties are immediate. The general case is not as easy and one has only these three properties to work with. As the Lie algebra g is a vector space, we would like to work with a basis for this space. We denote the basis elements as

X1, ..., Xm ∈ g (7)

We call the Xas the generators of the algebra (but its just a basis really). Example: for su (2) algebra you saw in the exercise that the Pauli matrices

0 1 0 −i 1 0 σ = , σ = , σ = (8) 1 1 0 2 i 0 3 0 −1 form a basis. Thus in this case we choose 1 X = σ a = 1, 2, 3. (9) a 2 a Because the bracket is linear, its action is encoded entirely in its action on basis elements. For every two elements Xa,Xb we can expand the bracket in the basis: m X c [Xa,Xb] = fab Xc (10) c=1 c where fab are simply the vector coeﬃcients of the RHS in the Xs basis. We c call the 3-index “tensor” fab the structure constants of the algebra. They fully encode the bracket and thus the entire algebra (at a given basis). 2 Representations 3

So far we only used the bi-linearity of the bracket. The anti-symmetry of the bracket is equivalent to c c fab = −fba (11) and Jacobi identity to

[[Xa,Xb] ,Xc] + [[Xb,Xc] ,Xa] + [[Xc,Xa] ,Xb] = 0 ↓ d e d e d e fab fdc + fbc fda + fca fdb = 0 (12) Example: for su (2) we chose Pauli as our basis. Because

[σ1, σ2] = 2iσ3 (13)

[σ2, σ3] = 2iσ1 (14)

[σ3, σ1] = 2iσ2 (15) we can write

[X1,X2] = iX3 (16)

[X2,X3] = iX1 (17)

[X3,X1] = iX2 (18) or k fij = iijk (19) where ijk is the Levi-Civita symbol.

2 Representations

The main idea: for every representation of the group there exist a representation of the corresponding algebra. Representations of the algebra are much easier to ﬁnd.

2.1 Reminder: representations of the group Representation of a group is simply the way the group acts on a vector space V with some dim V = d. In the physical cases we considered so far, the vector space was ﬁelds in the action. The examples we saw were: for every M ∈ SU (2) 1. Fundamental: ~ ~ TM φ = Mφ (20) here the vector space was of dimension d = 2. 2. Adjoint: −1 TM Aˆ = MAMˆ (21) ˆ P3 where A = i=1 Aiσi. Here the vector space (A1,A2,A3) is of dimension d = 3. 2 Representations 4

Deﬁnition: Given a vector space V and group G, a representation T : G → GL (V ) gives an invertible linear transformation (d × d matrix) TM for every group element M ∈ G, in a way that respect the group structure: For every M1,M2 ∈ G, the matrix multiplication gives

TM1 · TM2 = TM1·M2 . (22)

2.2 Representations of the algebra Deﬁnition: an algebra representation is a linear map T : g → gl (V ) that gives a matrix Tt for every algebra vector t ∈ g, and respect the algebra’s bracket: for every t1, t2 ∈ g

T[t1,t2] = [Tt1 ,Tt2 ] = Tt1 Tt2 − Tt2 Tt1 (23)

Notice that on the LHS [t1, t2] is an algebra bracket, and on the RHS [Tt1 ,Tt2 ] is a matrix commutator between two matrices of size d × d (d = dim V ). Important point: because Tt is linear in t ∈ g, it is enough to know the representation of the algerbra’s basis elements Xa. We deﬁne:

Ta ≡ TXa (24)

Question: Is Ta just another name for Xa? Answer: No. Xa is a speciﬁc element of the algebra, whereas Ta depend on the representation that we choose. The dimensions of Ta are d × d. As we said to know the Ta deﬁned the algebra’s representation. What the Ta has to satisfy? From (23) and (10) we have

m X c [Ta,Tb] = fab Tc (25) c=1

Example: For su (2), every algebra representation is consist of some three matrices T1,T2,T3 with (using (19))

3 X [Ti,Tj] = iijkTj (26) k=1 That’s it. To find a representation of an algebra g all we need to do is to find m (the dimension of the algebra) matrices of the same size d × d that satisfy the commutation relation (25)! Compare this task to the mission of finding a representation of the group, which involve finding infinitely many matrices Tg (for every g ∈ G) with infinitely many constraints (22). This is why we prefer looking at algebra representations... 2 Representations 5

Given a group representation we would like to build its corresponding algebra representation. Given a group representation, we have a matrix Tg for every group element g ∈ G. Now for every algebra element t ∈ g we have a path M (α) ∈ G with M (α) = id + α t + O α2 . (27)

As function of α, we can expand the matrix representation of the path TM(α). The linear order will be the algebra representation of t:

2 TM(α) = id + α Tt + O α (28) this way we can deﬁne the matrix Tt for every t ∈ g. It can be shown that if Tg is a group representation (22), the corresponding Tt will be an algebra representation (23).

2.3 Examples for algebra representations 2.3.1 The trivial representation Somewhat stupidly, we can always take the group representation that does noth- ing: TM = 1 (29) it will trivially satisfy (22). Note that this representation is of dimension d = 1 as TM is a 1 × 1 matrix. Expanding TM(α) to linear order (28) will give zero as it is a constant function. Hence the corresponding algebra representation is

Tt = 0 (30) for every t ∈ g, which trivially satisfy (25). Taking t = Xa gives the generators representation Ta = 0 (31) Example: for su (2), the trivial representation generators are

Ta = 0 (32) for a = 1, ..., 3, which indeed satisfy

3 X [Ti,Tj] = iijkTk. (33) k=1 This 1-dimensional representation is also called the spin-0 representation and denoted by “0”. 2 Representations 6

2.3.2 The fundamental representation Here we are dealing with matrix groups only (SO (n), SU (n) etc). The group fundamental representation is a dimension d = n with

TM = M (34)

Which is another way of saying in this representation the group element M ∈ G simply act on the vector “as a vector”. Expanding in α following (28) gives the fundamental representation of the algebra (see the TA7) Tt = t (35) and so also Ta = Xa. (36)

Note: the identity between Ta and Xa is a property of the fundamental representation, and not general. Also note that (25) here is automatically correct from the very deﬁnition of the structure constants (10) of a matrix group. Example: For t ∈ su (2) we would also have

Tt = t (37) and so the representation of the generators in this case would be Ta = Xa: 1 1 0 2 T1 = σ1 = 1 (38) 2 2 0 1 i 0 − 2 T2 = σ2 = i (39) 2 2 0 1 1 2 0 T3 = σ3 = 1 (40) 2 0 − 2

This is a 2-dimensional representation (as the Ta are 2×2), which indeed satisfy:

3 X [Ti,Tj] = iijkTk (41) k=1

1 1 This representation is called the spin- 2 representation and denoted by “ 2 ”.

2.3.3 The adjoint representation As we discussed last time, here the vector space itself is the algebra: V = g. Therefore the dimension of the representation is d = m. We deﬁned the group adjoint representation as −1 TM A = MAM (42) 3 From the algebra back to the group 7

for M ∈ G and vector A ∈ g. Expanding TM(α) to linear order gives

2 TM(α)A = (1 + α t) A (1 − α t) + O α (43) = 1 + α [t, A] + O α2 (44) ↓ (45)

TtA = [t, A] (46) Note that the brackets are the Lie algebra brackets! Now we would like to find the matrices Ta. As a linear transformation (substituting t = Xa) acting on a vector A ∈ g: Ta (A) = [Xa,A] . (47) To find its matrix elements we need to choose a basis for V = g. But we already have a convenient basis for g: the Xas! The matrix elements in this basis (Ta)b,c are defined by 3 X Ta (Xb) = [Xa,Xb] = Xc · (Ta)c,b (48) c=1 comparing with (10), these are exactly the structure constant:

c (Ta)c,b = fab (49) You can show that the fact that these matrices satisfy (25) because of the Jacobi identity (12). Example: For su (2) the generator adjoint representation

j (Ti)j,k = fi,k = iikj (50) or explicitly: 0 0 0   0 0 1 0 −1 0 T1 = i 0 0 −1 ,T2 = i  0 0 0 ,T3 = i 1 0 0 (51) 0 1 0 −1 0 0 0 0 0

You can check explicitly that these matrices satisfy (26). This is a 3- dimensional representation, and it is also called the spin-1 representation and denoted by “1”. In the exercise you will construct all the spin-j representations of SU (2).

3 From the algebra back to the group

As we said, ﬁnding new algebra representations is a doable task. Say that we found a new algebra representation {T1, ..., Tm} with (25), how do we “lift it” to a group representation? For that we ﬁrst need to ask: how to go from the algebra g itself back to the group G? Observation: for every α1 ∈ R

M (α1) ≡ exp (iα1σ1) ∈ SU (2) . (52) 4 Spoiler to the next topic 8

To show it we ﬁrst show M (α1) is unitary

† † MM = exp (iα1σ1) · (exp (iα1σ1)) (53)

= exp (iα1σ1) · exp (−iα1σ1) (54)

= 12×2. (55)

In the second line we used the fact σ1 is hermitian. We leave the proof for P3 det M = 1 as an exercise. Similarly, every t ∈ su (2) can be written as i=1 αiσi satisfy 3 ! X exp (it) = exp i αiσi ∈ SU (2) . (56) i=1 Pm Similarly, the exponentiation of every algebra element a=1 αaXa ∈ g gives an element of the group:

m ! X exp i αaXa ∈ G (57) a=1 For matrix group by exponent we mean explicitly exponentiating the matrix Pm 1 a=1 αaXa. Now given a representation of the algebra {Ta}. What would be the corresponding group matrix Tg for g ∈ G? If we can ﬁnd αas with

m ! X g = exp i αaXa , (58) a=1 we can guess m ! X Tg = exp i αaTa . (59) a=1 where here we exponentiate the representation matrix instead! Possible problems:

• What if no αas exist to satisfy (58)? (happens when G is not connected)

• What if two diﬀerent αas exist with (58) but gives two diﬀerent matrices Tg? (happens when G is not simply-connected)

4 Spoiler to the next topic

Let’s consider the following physical scenario. We have a theory described by an action: Z 3 S = dt d xL (φi (~x,t)) . (60)

1 P∞ 1 n The exponent of a matrix can be deﬁned as the series exp (A) = n=1 n! A . 4 Spoiler to the next topic 9

Lets say we know the theory is symmetric to rotations. The most general statement is that the ﬁelds posses an SO (3) symmetry. In other words there exist a representation T of SO (3), such that for R ∈ SO (3):

S [φ0] = S [φ] , (61)

0 i −1 (φ )i (~x) = TRφ R ~x (62) d X −1 = (TR)i,j φj R ~x . (63) j=1

In general TR can be any representation of SO (3). But we can focus only on the irreducible representation. This leads us to ask: what are the possible irreducible representations of the group SO (3)? As we discussed today, one way to do it is: 1. Find the corresponding algebra so (3). 2. Find the irreducible representations of the algebra so (3) (easier task). 3. Exponentiate them to get back group representations of SO (3).

The ﬁrst step is easy: by expanding elements of the group SO (3) around the identity we can show (similar to the previous exercise) that so (3) is the space of all the real anti-symmetric matrices. The most general one is

 0 −c b  0 0 0   0 0 1 0 −1 0  c 0 −a = a 0 0 −1 + b  0 0 0 + c 1 0 0 (64) −b a 0 0 1 0 −1 0 0 0 0 0

We can therefore choose as a basis (=generators):

0 0 0   0 0 1 0 −1 0 X1 = 0 0 −1 ,X2 =  0 0 0 ,X3 = 1 0 0 (65) 0 1 0 −1 0 0 0 0 0

You will see in the exercise that these are exactly the inﬁnitesimal rotations around the x, y, z axis correspondingly. The commutation relations are

3 X [Xi,Xj] = ijkXk (66) k=1

Up to minuses and is (that we can ignore), the so (3) algebra has the same structure constants as su (2)! This means that the algebra representations of both are the same, and we can copy the matrices we found for su (2)! 4 Spoiler to the next topic 10

• Spin 0: Ti = 0 (67) Exponentiating, we would still get the group’s trivial representation (see above), which in our case gives

φ0 (~x) = φ R−1~x . (68)

This is the way the mass density transforms under rotations for example.

1 • Spin 2 : 1 T = σ (69) i 2 i This representation acts on two complex fields. No object you heard off in classical field theory transforms like that under rotations. In fact, in this case the exponentiation of this algebra representation will not give a well defined a representation of the group SO (3). Take X3 =  0 1 0 −1 0 0 for example: explicit calculation gives 0 0 0

cos θ − sin θ 0 g (θ) = exp (θX3) = sin θ cos θ 0 , (70) 0 0 1

and so the identity is g (0) = g (2π) = 13×3. On the other hand the exponentiation of the representation matrix

T (θ) = exp (iθT3) (71) θ = exp i σ (72) 2 3 " i θ # e 2 0 = −i θ (73) 0 e 2

gives T (2π) = −12×2 6= T (0)! We got an inconsistency here that can’t be resolved! This is an example of an algebra representation that doesn’t correspond to any group representation. Put differently, when we will try to rotate the field by 360◦ we will get minus the field instead of the original one! • Spin 1: Ti = Xi (74)

where here we mean the Xi deﬁned for so (3) (65). Notice that this representation of SO (3) is both the fundamental and the adjoint. As this is a 3-dimensional representation, maybe it won’t surprise you 4 Spoiler to the next topic 11

that (after exponentiating) this is the way vectors are transformed under rotations. For example the velocity ﬁeld

~v0 (~x) = R · ~v R−1~x (75)

transforms in this representation of SO (3).

In class you will face a more complicated question. Instead of the rotation group, relativistic field theories have the full Lorentz group as a symmetry. In the same way, studying possible field theories lead us to the question: what are the irreducible representations of the Lorentz group? (as the different ways fields can transform under Lorentz transformation) As you will see, the first step is to consider the algebra of the Lorentz group. The second step would be to study irreducible representations of the Lorentz algebra. It will turn out to be a somewhat similar story to what you saw above for SO (3).