Group Theory (PHYS 507) Solution Set #8 5/26/17 1. Lie Algebra of so(5) from root diagram. In HW7 #3 you constructed the Cartan matrix and positive roots for the Lie algebra so(5) (two simple roots at an angle of 135◦). Use the resulting root diagram to determine all the nonvanishing commutators involving E±α and Hi, as we did in lecture for su(3). In this way you are determining the structure constants of the algebra. We recall from HW7 #3 that there are four positive roots: α(1) = (0; 1) ; α(2) = (1; −1) ; β = α(1) + α(2) = (1; 0) ; γ = 2α(1) + α(2) = (1; 1) ; and so 10 roots in all (four positive, four negative and two Cartan generators). The root diagram is (2) 1 (1) -α α γ -1 -β 1 β -γ -α(1) α(2) -1 Next we need to determine all the commutators, i.e. the Lie algebra. We know the commutators of the (two) generators of the Cartan subalgebra: [Hi;Hj] = 0 ; [Hi;E±φ] = ±φiE±φ ; with φ any positive root. We also know that [Eφ;E−φ] = φ · H: What remains to be determined are [Eφ;Eφ0 ] and [E−φ;Eφ0 ]. All other commu- tators can then be obtained by hermitian conjugation. 1 From the root diagram we know that certain commutators must vanish: [Eα(1) ;Eγ] = [Eα(2) ;Eβ] = [Eα(2) ;Eγ] = [E−α(1) ;Eα(2) ] = [E−α(2) ;Eγ] = 0 : The remaining commutators are non-zero. These are [Eα(1) ;Eα(2) ], [Eα(1) ;Eβ], [E−α(1) ;Eβ], [E−α(1) ;Eγ], [E−α(2) ;Eβ], and [E−β;Eγ]. These can be calculated using the properties of the raising and lowering operators in the SU(2) sub- groups, as well as the Jacobi identity. (1) (2) First note that Eα(1) = E+ for its SU(2), since jα j = 1. Roots α , β and γ form a triplet under this SU(2), and we can choose the phases of the states in the adjoint irrep so that jEα(2) i = j1; −1i ; jEβi = j1; 0i ; jEγi = j1; 1i : We also need the matrix elements of E± for the triplet irrep (which we wrote down earlier in lecture): E+j1; −1i = j1; 0i ;E+j1; 0i = j1; 1i ;E−j1; 1i = j1; 0i ;E−j1; 0i = j1; −1i : Now we are ready to proceed. • Raising α(2) with α(1) we obtain j[Eα(1) ;Eα(2) ]i = Eα(1) jEα(2) i = E+j1; −1i = j1; 0i = jEβi ; so that [Eα(1) ;Eα(2) ] = Eβ. • Similarly, raising β with α(1) we obtain j[Eα(1) ;Eβ]i = Eα(1) jEβi = E+j1; 0i = j1; 1i = jEγi ; so that [Eα(1) ;Eβ] = Eγ. • Next, lowering β with −α(1) we find j[E−α(1) ;Eβ]i = E−α(1) jEβi = E−j1; 0i = j1; −1i = jEα(2) i ; so that [E−α(1) ;Eβ] = Eα(2) . This can also be obtained using the Jacobi identity after substituting Eβ = [Eα(1) ;Eα(2) ]. • Lowering γ with −α(1) we find j[E−α(1) ;Eγ]i = E−α(1) jEγi = E−j1; 1i = j1; 0i = jEβi ; so that [E−α(1) ;Eγ] = Eβ. This can also be obtained using the Jacobi identity. 2 • Next we consider [E−α(2) ;Eβ]. This must equal c × Eα(1) , and using low- ering operators as above one finds that jcj = 1. We do not have the freedom, however, to choose the phase here|all phases have been fixed. To determine c we must use the Jacobi identity [E−α(2) ;Eβ] = [E−α(2) ; [Eα(1) ;Eα(2) ]] = −[Eα(1) ; [Eα(2) ;E−α(2) ] (1) (2) = α · α Eα(1) = −Eα(1) : • Finally, by a similar calculation, we obtain [E−β;Eγ] = [E−β; [Eα(1) ;Eβ]] = −[Eα(1) ; [Eβ;E−β]] − [Eβ; [E−β;Eα(1) ]] (1) = β · α Eα(1) − [Eβ;E−α(2) ] = −Eα(1) where I have used β · α(1) = α(2) · α(1) + α(1) · α(1) = 0. 2. Fun with sO(2N). In class we determined the roots of the Lie algebra sO(2N) by taking differences between weights of the fundamental irrep. We found that they were ±^j ± k^ ≡ ±ej ± ek ; j < k ; where ^j ≡ ej are unit vectors lying along one of the N axes of a Cartesian basis for the Cartan subalgebra (ej is Georgi's notation). Note that the two \±" symbols are independent, so that, for example, for so(4) there are four roots. What we did not do in class, however, was determine the generators corresponding to these roots. (a) For the case of so(4), determine which linear combinations of generators correspond to the four roots. (Recall that SO(4) generators are imaginary and antisymmetric, and that a standard basis for the Cartan subalgebra is σ2 0 0 0 H1 = ;H2 = : 0 0 0 σ2 The roots can in general be complex linear combinations of the generators. The four remaining generators are 0 0 0 −i 0 1 0 0 0 0 −i 1 0 0 0 0 0 1 0 0 0 0 0 1 B 0 0 0 0 C B 0 0 0 0 C B 0 0 −i 0 C B 0 0 0 −i C B C ; B C ; B C ; B C : @ 0 i 0 0 A @ 0 0 0 0 A @ 0 i 0 0 A @ 0 0 0 0 A 0 0 0 0 i 0 0 0 0 0 0 0 0 i 0 0 3 Taking (complex) linear combinations one can write them in the form 0 B E = ; B −BTr 0 in which B is a linear combination of the four matrices 1 and σi. Noting that 0 σ2B 0 Bσ2 [H1;EB] = Tr ; [H2;EB] = − Tr : −(σ2B) 0 (−Bσ2) 0 we see we must choose B's that are both left and right eigenvectors of σ2, i.e. σ2B = ±B and Bσ2 = ±B (with the two ± not necessarily the same). We know that the eigenvalues must be ±1 from the given values for the roots. It is easy to see that the appropriate B's are (1 ± σ2) (with left and right eigenvalues ±1) and (σ3 ± iσ1) (with left eigenvalue ±1 and right eigenvalue ∓1). Putting this together, one finds the generators corresponding to the four roots have B's as follows: root = (+1; +1) : B = σ3 + iσ1 ; root = (+1; −1) : B = 1 + σ2 ; root = (−1; +1) : B = 1 − σ2 ; root = (−1; −1) : B = σ3 − iσ1 ; (b) Sketch briefly how your result generalizes to so(2N), with N > 2. 2 Divide the matrices into N 2 × 2 blocks. The Hj, j = 1;N then have a σ2 in the j'th diagonal block. The Eα have one of the same four B matrices in the off-diagonal jk'th block (j < k), and −BTr in the kj'th block. The root vectors corresponding to the four choices of B will be as in SO(4), but with the non-zero components being the j'th and k'th ones. 3. More fun with so(4). The Dynkin diagram for so(4) is decomposable: it consists of two unattached circles. Thus it is the same algebra as that for SU(2) × SU(2). This decomposition is clear from the root diagram of so(4). The two su(2)'s are generated by the two simple roots, i.e. α(1) = 1^ + 2^ and α(2) = 1^ − 2^. Since these are orthogonal, it follows that the corresponding gen- erators commute: [E±α(1) ;E±α(2) ] = 0 (where the two ± are independent). The Cartan subalgebra decomposes into α(1) · H and α(2) · H. This decomposition of the Lie algebra of SO(4) is very useful for determining its irreps, since we know those of SU(2). In this problem we see how rotations of 4-d vectors can be parametrized by two SU(2) transformations. An SO(4) 0 tranformation rotates vµ (a real Euclidean four-vector) into another one vµ by 4 an orthogonal matrix with unit determinant. We can package four-vectors into 2 × 2 matrices by contracting them with a vector of matrices: 4 X V ≡ vµσµ ; σµ = (iσ1; iσ2; iσ3; 1) : µ=1 This is nothing other than a (matrix representation of a) quarternion. (a) Since V has only four real parameters, whereas a general 2 × 2 matrix has eight, V must be constrained in some way. In fact V = jvjU, where qP 2 U 2 SU(2), and jvj = µ vµ. Show this result. The result V = jvjU is equivalent to V=jvj being unitary and having unit determinant. First we calculate det V , which can be done directly using the expressions for the Pauli matrices: v0 + iv3 iv1 + v2 X 2 2 det V = det = vµ ≡ v iv1 − v2 v0 − iv3 µ Thus det(V=jvj) = (det V )=v2 = 1, as desired. Note that this is true irrespective of whether vµ are real or complex. Next we check unitarity by direct computation, using σaσb = iabcσc (which y implies σaσb + σbσa = 2δab1) and σa = σa: V yV 1 v2 = (v − i~v · ~σ)(v + i~v · ~σ) = 1 = 1 : det V v2 0 0 v2 This does rely on vµ being real. Thus V=jvj is unitary. To summarize, the propertiesp of V are that det V is real and positive semi- definite and that V= det V 2 SU(2). 0 y (b) Now we perform the transformation V ! V = LV R , with L 2 SU(2)L 0 P 0 0 0 2 2 and R 2 SU(2)R.