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TA8 - More Symmetries and Elements of Lie Theory

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TA8 - More Symmetries and Elements of Lie Theory 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) nSU (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 M^ (α) 2 G that start at the identity M^ (0) = 1n×n and expand to linear order 2 M^ (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) = t^j t^y = t^ and tr t^ = 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 2 su (2), also [t1; t2] = t1t2 − t2t1 2 su (2) : (3) Starting from the Group G, one can build a bracket operator on the corre- sponding 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 2 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 2 SU (2) 1. Fundamental: ~ ~ TM φ = Mφ (20) here the vector space was of dimension d = 2. 2. Adjoint: −1 TM A^ = MAM^ (21) ^ 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 2 G, in a way that respect the group structure: For every M1;M2 2 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 2 g, and respect the algebra’s bracket: for every t1; t2 2 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 2 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 2 G) with in- finitely many constraints (22). This is why we prefer looking at algebra repre- sentations... 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 2 G. Now for every algebra element t 2 g we have a path M (α) 2 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 2 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 2 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 2 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 rep- resentation, 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 2 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 ”.
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