Notes on Groups S O(3) and S U(2)


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1 Notes on Groups S O(3) and S U() Ivan G. Avramidi New Mexico Institute of Mining and Technology October, 011 Author: IGA File: su.tex; Date: November 9, 011; Time: 10:07
2 1 1 Group S O(3) 1.1 Algebra so(3) The generators X i = X i, i = 1,, 3, of the algebra so(3) are 3 3 real antisymmetric matrices defined by or X 1 = , X = , X 3 = , (1.1) (X i ) k j = ε k i j = ε ik j (1.) Raising and lowering a vector index does not change anything; it is done just for convenience of notation. These matrices have a number of properties and where Therefore, they satisfy the algebra The Casimir operator is defined by X i X j = E ji Iδ i j, (1.3) ε i jk X k = E i j E ji (1.4) (E ji ) kl = δ k j δl i (1.5) [X i, X j ] = ε k i jx k (1.6) X = X i X i (1.7) It is equal to and commutes with all generators X = I (1.8) [X i, X ] = 0. (1.9) su.tex; November 9, 011; 10:07; p. 1
3 where A general element of the group S O(3) in canonical coordinates y i has the form exp (X(y)) = I cos r + P(1 cos r) + X(y) sin r r (1.10) = P + Π cos r + X(y) sin r, r (1.11) r = y i y i, θ i = yi r X(y) = X i y i, and the matrices Π and P are defined by In particular, for r = π we have (1.1) P i j = θ i θ j (1.13) Π i j = δ i j θ i θ j (1.14) exp (X(y)) = P Π (1.15) Notice that and, therefore, tr P = 1, tr Π = (1.16) tr exp (X(y)) = 1 + cos r. (1.17) Also, det sinh X(y) X(y) ( ) sin r = (1.18) r 1. Representations of S O(3) Let X i be the generators of S O(3) in a general irreducible representation satisfying the algebra [X i, X j ] = ε i jk X k (1.19) They are determined by symmetric traceless tensors of type ( j, j) (with j a positive integer) (X i ) l 1...l j k1...k j = jε (l 1 i(k1 δ l k δ l j) k j ) (1.0) Then X = j( j + 1)I (1.1) su.tex; November 9, 011; 10:07; p.
4 3 where Another basis of generators is (J +, J, J 3 ), where I l 1...l jk1...k j = δ (l 1 (k 1 δ l j) k j ) (1.) J + = ix 1 + X (1.3) J = ix 1 X (1.4) J 3 = ix 3 (1.5) so that J + = J, J 3 = J 3. (1.6) They satisfy the commutation relations [J +, J ] = J 3, [J 3, J + ] = J +, [J 3, J ] = J. (1.7) The Casimir operator in this basis is X = J + J J 3 + J 3 = J J + J 3 J 3 (1.8) From the commutation relations we have J 3 J + = J + (J 3 + 1) (1.9) J 3 J = J (J 3 1) (1.30) J J + = X J 3 J 3 (1.31) J + J = X J 3 + J 3 (1.3) Since J 3 commutes with X they can be diagonalized simultaneously. Since they are both Hermitian, they have real eigenvalues. Notice also that since X i are antihermitian, then X 0. (1.33) Let λ, m be the basis in which both operators are diagonal, that is, X λ, m = λ λ, m (1.34) J 3 λ, m = m λ, m (1.35) There are many ways to show that the eigenvalues m of the operator J 3 are integers. Now, since X 1 + X = X X 3 = X + J 3 0 (1.36) su.tex; November 9, 011; 10:07; p. 3
5 4 then the eigenvalues of J 3 are bounded by a maximal integer j, m j (1.37) that is, m takes ( j + 1) values j, j + 1,, 1, 0, 1,, j 1, j (1.38) From the commutation relations above we have J + λ, m = const λ, m + 1 (1.39) J λ, m = const λ, m 1 (1.40) Let λ, 0 be the eigenvector corresponding to the eigenvalue m = 0, that is J 3 λ, 0 = 0. (1.41) Then for positive m > 0 λ, m = a λm J+ m λ, 0 (1.4) λ, m = a λ, m J m λ, m (1.43) Then J + λ, j = 0 (1.44) J λ, j = 0 (1.45) Therefore, 0 = J J + λ, j = ( X J 3 J 3) λ, j (1.46) 0 = J + J λ, j = ( X J 3 + J 3) λ, j (1.47) Thus, ( λ j j ) λ, j = 0 (1.48) So, the eigenvalues of the Casimir operator X are labeled by a nonnegative integer j 0, λ j = j( j + 1). (1.49) su.tex; November 9, 011; 10:07; p. 4
6 5 These eigenvalues have the multiplicity d j = j + 1. (1.50) In particular, for j = 1 we get λ 1 = as expected before. From now on we will denote the basis in which the operators X and J 3 are diagonal by λ j, m = j. (1.51) m The matrix elements of the generators J ± can be computed as follows. First of all, j m J j 3 = δ m mm m (1.5) We have and This gives j j( j + 1) = m J j j + m 1 m 1 J j + m m (1.53) m j m J j j + = m 1 m 1 J j m j m J j + m j m J j m (1.54) = δ m,m 1 ( j + m)( j m + 1) (1.55) = δ m,m+1 ( j m)( j + m + 1) (1.56) The matrix elements of the generators X i are j m X j i 1 = δ m m,m 1 ( j + m)( j m + 1) i +δ m,m+1 ( j m)( j + m + 1) (1.57) j m X j 1 = δ m m,m 1 ( j + m)( j m + 1) 1 δ m,m+1 ( j m)( j + m + 1) (1.58) j m X j 3 = δ m mm im (1.59) su.tex; November 9, 011; 10:07; p. 5
7 6 Thus, irreducible representations of S O(3) are labeled by a nonnegative integer j 0. The matrices of the representation j in canonical coordinates y i will be denoted by j D j mm (y) = m exp [ X(y) ] j (1.60) m where X(y) = X i y i. The characters of the irreducible representations are χ j (y) = j Dmm(y) j = m= j Group S U() j e imr = 1 + m= j j cos(mr) (1.61) m=1.1 Algebra su() Pauli matrices σ i = σ i, i = 1,, 3, are defined by 0 1 σ 1 = 1 0, σ 0 i = i 0, σ 3 = They have the following properties (.1) σ i = σ i (.) σ i = I (.3) det σ i = 1 (.4) tr σ i = 0, (.5) σ 1 σ = iσ 3 (.6) σ σ 3 = iσ 1 (.7) σ 3 σ 1 = iσ (.8) σ 1 σ σ 3 = ii (.9) (.10) which can be written in the form σ i σ j = δ i j I + iε i jk σ k. (.11) su.tex; November 9, 011; 10:07; p. 6
8 7 They satisfy the following commutation relations The traces of products are Also, there holds [σ i, σ j ] = iε i jk σ k (.1) tr σ i σ j = δ i j (.13) tr σ i σ j σ k = iε i jk (.14) σ i σ i = 3I. (.15) The Pauli matrices satisfy the anticommutation relations σ i σ j + σ j σ i = δ i j I (.16) Therefore, Pauli matrices form a representation of Clifford algebra in 3 dimensions and are, in fact, Dirac matrices σ i = (σ i B ), where A, B = 1,, in 3 dimen A sions. The spinor metric E = (E AB ) is defined by 0 1 E = 1 0. (.17) Notice that and the inverse metric E 1 = (E AB ) is Then E = iσ (.18) E 1 = E (.19) σ T i = Eσ i E 1 (.0) This allows to define the matrices Eσ i = (σ i AB ) with covariant indices Eσ 1 = 1 0 σ 3 = 0 1, (.1) Eσ = i 0 ii = 0 i, (.) Eσ 3 = 0 1 σ 1 = 1 0. (.3) su.tex; November 9, 011; 10:07; p. 7
9 8 Notice that the matrices Eσ i are all symmetric (Eσ i ) T = Eσ i (.4) Pauli matrices satisfy the following completeness relation which implies This gives In a more symmetric form σ i A B σ i C D = δ A Dδ C B δ A Bδ C D (.5) σ i (A (B σ i C) D) = δ (A (Bδ C) D) (.6) σ i AB σ i CD = E AD E CB E AB E CD (.7) σ i AB σ i CD = E AD E CB + E BD E CA (.8) A spinor is a two dimensional complex vector ψ = (ψ A ), A = 1, in the spinor space C. We use the metric E to lower the index to get the covariant components which defines the cospinor ψ = (ψ A ) We also define the conjugated spinor ψ = ( ψ A ) by ψ A = E AB ψ B (.9) ψ = Eψ (.30) ψ A = ψ A (.31) On the complex spinor space there are two invariant bilinear forms ψ, ϕ = ψϕ = ψ A ϕ A (.3) and Note that and ψϕ = ψ A ϕ A = E AB ψ B ϕ A = ψ 1 ϕ ψ ϕ 1 (.33) ψ, ψ = ψψ = ψ A ψ A = ψ 1 + ψ (.34) ψψ = ψ A ψ A = 0 (.35) su.tex; November 9, 011; 10:07; p. 8
10 9 The matrices X i = i σ i (.36) are the generators of the Lie algebra su() with the commutation relations [X i, X j ] = ε i jk X k (.37) The Casimir operator is It commutes with all generators X = X i X i (.38) [X i, X ] = 0. (.39) There holds X = 3 4 I (.40) Notice that tr X i X j = 1 δ i j (.41) X = j( j + 1)I (.4) with j = 1, which determines the fundamental representation of S U(). An arbitrary element of the group S U() in canonical coordinates y i has the form ( i ) exp σ jy j = I cos r + iσ jθ j sin r, (.43) This can be written in the form exp (X(y)) = I cos(r/) + X(y) sin(r/). (.44) r/ where X(y) = X i y i. In particular, for r = π exp (X(y)) = I. (.45) Obviously, tr exp (X(y)) = cos(r/) (.46) su.tex; November 9, 011; 10:07; p. 9
11 10. Representations of S U() We define symmetric contravariant spinors ψ A 1...A j of rank j. Each index here can be lowered by the metric E AB. The number of independent components of such spinor is precisely j 1 = ( j + 1) (.47) i=0 Let X i be the generators of S U() in a general irreducible representation satisfying the algebra [X i, X j ] = ε i jk X k (.48) They are determined by symmetric spinors of type ( j, j) Then (X k ) A 1...A j B1...B j = i jσ k (A 1(B1 δ A B δ A j) B j ) (.49) X = j( j + 1)I (.50) where I A 1...A j B 1...B j = δ (A 1 (B 1 δ A j) B j ) (.51) Another basis of generators is J + = ix 1 + X (.5) J = ix 1 X (.53) J 3 = ix 3 (.54) They have all the properties of the generators of S O(3) discussed above except that the operator J 3 has integer eigenvalues m. One can show that they can be either integers or halfintegers taking ( j + 1) values j, j + 1,..., j 1, j, m j (.55) with j being a positive integer or halfinteger. If j is integer, then all eigenvalues m of J 3 are integers including 0. If j is a halfinteger, then all eigenvalues m are halfintegers and 0 is excluded. We choose a basis in which the operators X and J 3 are diagonal. Let ψ = ψ A 1...A j be a symmetric spinor of rank j. Let e 1 = (e A 1 ) and e = (e A ) be the basis cospinors defined by e 1 A = δ A 1, e A = δ A. (.56) su.tex; November 9, 011; 10:07; p. 10
12 11 Let e j,m A 1...A j+m B 1...B j m = e 1 (A1 e 1 A j+m e B1 e B j m ) = δ (A 1 1 δ A j+m 1 δ B 1 δb j m) (.57) Then this is an eigenspinor of the operator J 3 with the eigenvalue m Then the spinors J 3 e j,m = me j,m (.58) j ( ) 1/ ( j)! = e j,m (.59) m ( j + m)!( j m)! form an orthonormal basis in the space of symmetric spinors of rank j. The matrix elements of the generators X i are given by exactly the same formulas j m X j i 1 = δ m m,m 1 ( j + m)( j m + 1) i +δ m,m+1 ( j m)( j + m + 1) (.60) j m X j 1 = δ m m,m 1 ( j + m)( j m + 1) 1 δ m,m+1 ( j m)( j + m + 1) (.61) j m X j 3 = δ m mm im (.6) but now j and m are either both integers or halfintegers. Thus, irreducible representations of S U() are labeled by a nonnegative halfinteger j 0. The matrices of the representation j in canonical coordinates y i are j D j mm (y) = m exp [ X(y) ] j (.63) m where X(y) = X i y i. The representations of S U() with integer j are at the same time representations of S O(3). However, representations with halfinteger j do not give representations of S O(3), since for r = π D j mm (y) = δ mm. (.64) su.tex; November 9, 011; 10:07; p. 11
13 1 The characters of the irreducible representations are: for integer j χ j (y) = and for halfinteger j χ j (y) = 1 + j j Dmm(y) j = m= j k= 1 j D j 1 +k, 1 +k(y) = 1 + j j e imr = 1 + m= j k= 1 j exp j cos(mr) (.65) m=1 [ ( ) ] 1 i + k r = Now, let X i and Y j be the generators of two representations and let Then G i generate the product representation X Y and j k=1 [( ) ] 1 cos + k r (.66) G i = X i I Y + I X Y i (.67) exp[g(y)] = exp[x(y)] exp[y(y)] (.68) Therefore, the character of the product representation factorizes χ G (y) = χ X (y)χ Y (y) (.69).3 Double Covering Homomorphism S U() S O(3) Recall that the matrices Eσ i = (σ i AB ) are symmetric and the matrix E = (E AB ) is antisymmetric. Let ψ AB be a symmetric spinor. Then one can form a vector A i = σ i BA ψ AB = tr (Eσ i ψ). (.70) Conversely, given a vector A i we get a symmetric spinor of rank by ψ AB = 1 Ai σ i AB (.71) or ψ = 1 Ai σ i E 1 (.7) More generally, given a symmetric spinor of rank j we can associate to it a symmetric tensor of rank j by A i1...i j = σ i1 A 1 B 1 σ i j A j B j ψ A 1...A j B 1...B j (.73) su.tex; November 9, 011; 10:07; p. 1
14 13 One can show that this tensor is traceless. Indeed by using the eq. (??) and the fact that the spinor is symmetric we see that the tensor A is traceless. The inverse transformation is defined by ψ A1...A j B 1...B j = 1 j Ai 1...i j σ i1 A 1 B 1 σ i j A j B j (.74) The double covering homomorphism Λ : S U() S O(3) (.75) is defined as follows. Let U S U(). Then the matrices Uσ i U 1 satisfy all the properties of the Pauli matrices and are therefore in the algebra su(). Therefore, Uσ i U 1 = Λ j i(u)σ j. (.76) That is, Λ i j(u) = 1 tr σi Uσ j U 1 (.77) The matrix Λ(U) depends on the matrix U and is, in fact, in S O(3). For infinitesimal transformations U = exp(x k y k ) = I + i σ ky k + (.78) the matrix Λ is Λ i j = δ i j + ε i jk y k + = [ exp(x k y k ) ] i j (.79) It is easy to see that So, in short where Λ(I) = Λ( I) = I. (.80) Λ(exp(X i y i )) = exp(λ(x i )y i ) (.81) [Λ(σ k )] i j = iε i k j (.8) 3 Invariant Vector Fields Let y i be the canonical coordinates on the group S U() ranging over ( π, π). The position of the indices on the coordinates will be irrelevant, that is, y i = y i. We introduce the radial coordinate r = y = y i y i, so that the geodesic distance su.tex; November 9, 011; 10:07; p. 13
15 14 from the origin is equal to r, and the angular coordinates (the coordinates on S ) θ i = ŷ i = y i /r. Let us consider the fundamental representation of the group S U() with generators T a = iσ a, where σ a are the Pauli matrices satisfying the algebra Let T(y) = T i y i. Then where (y, p) = y i p i, (y p) k = ε i jk y i p j and Therefore, where ŷ is a unit vector. Next, we compute [T a, T b ] = ε c abt c. (3.83) T(y)T(p) = (y, p) T(y p) (3.84) [T(y)] = y I (3.85) exp[rt(ŷ)] = cos r + T(ŷ) sin r (3.86) exp[t(f(r ˆq, ρ ˆp))] = cos r cos ρ (ˆq, ˆp) sin r sin ρ + cos r sin ρt( ˆp) + cos ρ sin rt(ˆq) sin r sin ρt(ˆq ˆp) (3.87) This defines the group multiplication map F(q, p) in canonical coordinates. This map has a number of important properties. In particular, F(0, p) = F(p, 0) = p and F(p, p) = 0. Another obvious but very useful property of the group map is that if q = F(ω, p), then ω = F(q, p) and p = F( ω, q). Also, there is the associativity property and the inverse property F(ω, F(p, q)) = F(F(ω, p), q) (3.88) F( p, ω) = F(ω, p). (3.89) This map defines all the properties of the group. We introduce the matrices L α b(x) = y b Fα (y, x), (3.90) y=0 R α b(x) = y b Fα (x, y), (3.91) y=0 su.tex; November 9, 011; 10:07; p. 14
16 15 Let X and Y be the inverses of the matrices L and R, that is, Also let so that LX = XL = RY = YR = I. (3.9) D = XR, D 1 = YL (3.93) L = RD 1, X = DY (3.94) Let C a be the matrices of the adjoint representation defined by and C = C(x) be the matrix defined by Then one can show that X = exp(c) I C L = It is easy to see also that (C a ) b c = ε b ac (3.95) C = C a x a. (3.96), Y = I exp( C) C, (3.97) C exp(c) I, R = C I exp( C). (3.98) x a = L a bx b = R a bx b = X a bx b = Y a bx b. (3.99) and and as well as X = Y T, L = R T (3.100) ( ) sinh(c/) XY = YX =, C/ (3.101) ( ) C/ LR = RL =, sinh(c/) (3.10) D = XR = RX = exp(c), D T = YL = LY = exp( C). (3.103) su.tex; November 9, 011; 10:07; p. 15
17 16 Note that where r = x and with θ i = x i /r. Therefore, C = 4r Π, (3.104) Π i j = δ i j θ i θ j (3.105) C n = ( 1) n (r) n Π, C n+1 = ( 1) n (r) n C (3.106) Therefore, the eigenvalues of the matrix C are (ir, ir, 0), and for any analytic function f (C) = f (0)(I Π) + Π 1 and, therefore, This enables one to compute Then one can show that [ f (ir) + f ( ir) ] + 1 4ir C [ f (ir) f ( ir ] tr f (C) = f (0) + f (ir) + f ( ir) (3.107) det f (C) = f (0) f (ir) f ( ir) (3.108) Y = I Π + sin r r X = I Π + sin r r cos rπ 1 sin r cos rπ + 1 C (3.109) r sin r C (3.110) r R = I Π + r cot rπ + 1 C (3.111) L = I Π + r cot rπ 1 C (3.11) D = I Π + Π cos(r) + sin(r) C (3.113) r x µ Fα (x, y) = L α γ(f(x, y))x γ µ(x), (3.114) x µ Fα (y, x) = R α γ(f(y, x))y γ µ(x). (3.115) These matrices satisfy important identities µ X a ν ν X a µ = ε a bcx b µx c ν, (3.116) µ Y a ν ν Y a µ = ε a bcy b µy c ν. (3.117) su.tex; November 9, 011; 10:07; p. 16
18 17 and Let us define the oneforms They satisfy important identities Next, we define the vector fields. L µ a µ L ν b L µ b µ L ν a = ε c abl γ c, (3.118) R µ a µ R ν b R µ b µ R ν a = ε c abr ν c. (3.119) σ a = X a µ(x)dx µ, σ a + = Y a µ(x)dx µ. dσ a = ε a bcσ b σ c, (3.10) dσ a + = ε a bcσ b + σ c +. (3.11) K a = L µ a(x) µ, K + a = R µ a(x) µ. These are the rightinvariant and the leftinvariant vector fields on the group. They are related by K a = D a b K + b (3.1) They form two representations of the group S U() and satisfy the algebra [K + a, K + b ] = εc abk + c (3.13) [K a, K b ] = εc abk c (3.14) [K + a, K b ] = 0 (3.15) That is, the leftinvariant vector fields commute with the rightinvariant ones. Now, let T a be the generators of some representation of the group S U() satisfying the same algebra. Then, of course, exp[t(ω)] exp[t(p)] = exp[t(f(ω, p))] (3.16) First, one can derive a useful commutation formula exp[t(ω)]t b exp[ T(ω)] = D a bt a, (3.17) where the matrix D = (D a b) is defined by D = exp C. More generally, exp[t(ω)] exp[t(p)] exp[ T(ω)] = exp[t(d(ω)p)]. (3.18) su.tex; November 9, 011; 10:07; p. 17
19 18 Next, one can show that This means that exp[ T(ω)] ω i exp[t(ω)] = Ya it a, (3.19) ω i exp[t(ω)] = Ya i exp[t(ω)]t a = Y i a T a exp[t(ω)], (3.130) The Killing vectors have important properties K + a exp[t(ω)] = (exp[t(ω)])t a (3.131) K a exp[t(ω)] = T a (exp[t(ω)]) (3.13) This immediately leads to further important equations and exp[k + (p)] exp[t(ω)] = exp[t(ω)] exp[t(p)] (3.133) exp[k (q)] exp[t(ω)] = exp[t(q)] exp[t(ω)] (3.134) where K ± (p) = p a K ± a. More generally, exp[k + (p) + K (q)] exp[t(ω)] = exp[t(q)] exp[t(ω)] exp[t(p)] (3.135) and, therefore, exp[k + (p) + K (q)] exp[t(ω)] = exp[t(q)] exp[t(p)] = exp[t(f(q, p))] ω=0 (3.136) In particular, for the adjoint representation this formulas take the form DC b = D a bc a D (3.137) K + a D = DC a (3.138) K a D = C a D (3.139) exp[k + (p)]d(x) = D(x)D(p) (3.140) exp[k (q)]d(x) = D(q)D(x) (3.141) exp[k + (p) + K (q)]d(x) = D(q)D(x)D(p) (3.14) su.tex; November 9, 011; 10:07; p. 18
20 19 Then for a scalar function f (ω) the action of the rightinvariant and leftinvariant vector fields is simply exp[k + (p)] f (ω) = f (F(ω, p)), exp[k (q)] f (ω) = f (F(q, ω)). (3.143) Since the action of these vector fields commutes, we have more generally Therefore, that is, exp[k + (p) + K (q)] f (ω) = f (F(q, F(ω, p))). (3.144) exp[k + (p) + K (q)] f (ω) = f (F(q, p)). (3.145) ω=0 The biinvariant metric is defined by g αβ = δ ab Y a αy b β = δ ab X a αx b β (3.146) g αβ = δ ab L α al β b = δ ab R α ar β b, (3.147) ( ) sinh[c/] g = = I Π + sin r Π. (3.148) C/ r The Riemannian volume elements of the metric g is defined as usual where dvol = g 1/ (x)dx 1 dx dx 3 (3.149) g 1/ = (det g µν ) 1/ = sin r r. (3.150) Thus the biinvariant volume element of the group is dvol (x) = sin r r dx = sin r dr dθ (3.151) where dx = dx 1 dx dx 3, and dθ is the volume element on S. The invariance of the volume element means that for any fixed p dvol (x) = dvol (F(x, p)) = dvol (F(p, x)) (3.15) Notice also, that F(p, q) is nothing but the geodesic distance between the points p and q. su.tex; November 9, 011; 10:07; p. 19
21 0 We will always use the orthonormal basis of the rightinvariant vector fields K + a and oneforms σ a + and denote the covariant derivative with respect to K + a simply by a. The LeviCivita connection of the biinvariant metric in the rightinvariant basis is defined by a K + b = εc bak + c (3.153) so that the coefficients of the affine connection are so that for an arbitrary vector ω a bc = σ a +( c K + b ) = εa bc. (3.154) a V = [ (K + a V b ) + ε b cav c] K + b (3.155) Then where a K d = Mb dak + b (3.156) M b da = K + a D d b + ε b cad d c = ε b cad d c + ε b cad d c = ε b cad d c (3.157) so that Now, the curvature tensor is The Ricci curvature tensor and the scalar curvature are The scalar Laplacian is One can show that Further, a K d = εb cad d c K + b (3.158) R a bcd = ε f cdε a f b. (3.159) R ab = δ ab, R = 6. (3.160) 0 = δ ab K a K b = δab K + a K + b. (3.161) 0 ω sin ω = ω sin ω (3.16) 0 exp[t(ω)] = (exp[t(ω)])t = T exp[t(ω)] (3.163) su.tex; November 9, 011; 10:07; p. 0
22 1 where T = T a T a. This means that exp[t 0 ] exp[t(ω)] = exp(tt ) exp[t(ω)] (3.164) The Killing vector fields K a ± are divergence free, which means that they are antiselfadjoint with respect to the invariant measure dvol (ω). that is, Thus the Laplacian 0 is selfadjoint with respect to the same measure as well. Let be the total connection on a vector bundle V realizing the representation G. Let us define the oneforms A = σ a +G a = G a Y a µdx µ. (3.165) and F = da + A A. Then, by using the above properties of the rightinvariant oneforms we compute where F = 1 F abσ a + σ b + = 1 F aby a µy b νdx µ dx ν (3.166) F ab = ε c abg c (3.167) Thus A is the YangMills curvature on the vector bundle V with covariantly constant curvature. The covariant derivative of a section of the bundle V is then (recall that a = K + a ) a ϕ = (K + a + G a )ϕ (3.168) and the covariant derivative of a section of the endomorphism bundle End(V) along the rightinvariant basis is then Therefore, we find a Q = K + a Q + [G a, Q] (3.169) a G b = ε c bag c + [G a, G b ] = 0. (3.170) Then the derivatives along the leftinvariant vector fields are K a ϕ = (K a + B a )ϕ (3.171) where K a Q = K a Q + [B a, Q] (3.17) B a = D a b G b (3.173) su.tex; November 9, 011; 10:07; p. 1
23 The Laplacian takes the form = a a = (K + a + G a )(K + a + G a ) = 0 + G a K + a + G (3.174) where G = G i G i. We want to rewrite the Laplacian in terms of Casimir operators of some representations of the group S U(). The covariant derivatives a do not form a representation of the algebra S U(). The operators that do are the covariant Lie derivatives. The covariant Lie derivatives along a Killing vector ξ of sections of this vector bundle are defined by L ξ = ξ 1 σa +( b ξ)ε bc ag c (3.175) By denoting the Lie derivatives along the Killing vectors K ± a by K ± a this gives for the rightinvariant and the leftinvariant bases K + a = L K + a = a + G a = K + a + G a. (3.176) K a = L K a = K a B a = K a. (3.177) It is easy to see that these operators form the same algebra S U() S U() [K + a, K + b ] = εc abk + c (3.178) [K a, K b ] = εc abk c (3.179) [K + a, K b ] = 0 (3.180) Now, the Laplacian is given now by the sums of the Casimir operators = K G (3.181) where K = 1 K K (3.18) and K ± = K ± a K ± a. 4 Heat Kernel on S 3 Our goal is to evaluate the heat kernel diagonal. Since it is constant we can evaluate it at any point, say, at the origin. We use the geodesic coordinates y i defined su.tex; November 9, 011; 10:07; p.
24 3 above. That is why, we will evaluate the heat kernel when one point is fixed at the origin. We will denote it simply by U(t; y). We obviously have exp(t ) = exp ( tg ) exp ( tk ). (4.1) Therefore, the heat kernel is equal to U(t, y) = exp ( tg ) ( t ) Ψ, y (4.) where Ψ(t, y) = exp(tk )δ(y) is the heat kernel of the operator K. 4.1 Scalar Heat Kernel Let Φ(t, ω) = (4πt) 3/ e t n= ω + πn sin ω ) ( ω + πn) exp ( 4t One can show by direct calculation that this function satisfies the equation (4.3) t Φ = 0 Φ (4.4) with the initial condition Φ(0, ω) = δ S 3(ω) (4.5) Therefore, this is the scalar heat kernel on the group S U() and, therefore, on S 3. Now, let us compute the integral Ψ(t) = dvol (ω)φ(t, ω) exp[t(ω)] (4.6) S U() Obviously, Ψ(0) = 1. Next, we have t Ψ = dvol (ω) exp[t(ω)] 0 Φ(t, ω) (4.7) S U() Now, by integrating by parts we get t Ψ = DωΦ(t, ω) 0 exp[t(ω)] (4.8) S U() which gives t Ψ = ΨT (4.9) su.tex; November 9, 011; 10:07; p. 3
25 4 Thus, we obtain a very important equation exp(tt ) = dvol (ω) Φ(t, ω) exp[t(ω)] (4.10) S U() This is true for any representation of S U(). We derive two corollaries. First, we get exp(tt ) = dvol (ω) Φ(t, ω) exp[t(p)] exp[t(ω)] exp[ T(p)] (4.11) S U() which means that for any p or Also, we see that exp[(t + s)t ] = Φ(t, ω) = Φ(t, F( p, F(ω, p)). (4.1) Φ(t, F(p, q)) = Φ(t, F(q, p)). (4.13) S 3 S 3 dvol (q)dvol (p)φ(t, q)φ(s, p) exp[t(q)] exp[t(p)] (4.14) By changing the variables z = F(q, p) and p p we obtain exp[(t + s)t ] = dvol (z) dvol (p)φ(t, F(p, z))φ(s, p) exp[t(z)] which means that S U() S U() S U() (4.15) dvol (p)φ(t, F(p, z))φ(s, p) = Φ(t + s, z) (4.16) In particular, for z = 0 and s = t this becomes dvol (p)φ(t, p)φ(t, p) = Φ(t, 0) (4.17) S U() su.tex; November 9, 011; 10:07; p. 4
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