Ivan G. Avramidi
New Mexico Institute of Mining and Technology
October, 2011
Author: IGA File: su2.tex; Date: November 29, 2011; Time: 10:07
1 Group S O(3)
1.1 Algebra so(3)
The generators Xi = Xi, i= 1, 2, 3, of the algebra so(3) are 3×3 real antisymmetric matrices defined by
X1 =
0 0 0
0 0 1
0 −1 0
, X2=
0 0 −1
0 0 0
1 0 0
, X3 =
0 1 0
−1 0 0
0 0 0
, (1.1)
or
(Xi)kj = −εki j = εik j (1.2) 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
XiXj = Eji− Iδi j, (1.3) and
εi jkXk = Ei j − Eji (1.4) where
(Eji)kl = δkjδli (1.5) Therefore, they satisfy the algebra
[Xi, Xj]= −εki jXk (1.6) The Casimir operator is defined by
X2 = XiXi (1.7)
It is equal to
X2 = −2I (1.8)
and commutes with all generators
[Xi, X2]= 0 . (1.9)
A general element of the group S O(3) in canonical coordinates yihas 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)
where
r= p
yiyi, θi = yi
r (1.12)
X(y)= Xiyi, and the matricesΠ and P are defined by
Pij = θiθj (1.13)
Πij = δij−θiθj (1.14)
In particular, for r= π we have
exp (X(y)) = P − Π (1.15)
Notice that
tr P= 1, trΠ = 2 (1.16)
and, therefore,
tr exp (X(y))= 1 + 2 cos r . (1.17) Also,
detsinh X(y)
X(y) = sin r r
!2
(1.18)
1.2 Representations of S O(3)
Let Xibe the generators of S O(3) in a general irreducible representation satisfying the algebra
[Xi, Xj]= −εi jkXk (1.19) They are determined by symmetric traceless tensors of type ( j, j) (with j a positive integer)
(Xi)l1...ljk1...kj = − jε(l1i(k1δlk2
2· · ·δlkj)
j) (1.20)
Then
X2= − j( j + 1)I (1.21)
where
Il1...ljk1...kj = δ(l(k11· · ·δlkj)
j) (1.22)
Another basis of generators is (J+, J−, J3), where
J+ = −iX1+ X2 (1.23)
J− = −iX1− X2 (1.24)
J3 = −iX3 (1.25)
so that
J†+= J−, J3†= J3. (1.26) They satisfy the commutation relations
[J+, J−]= 2J3, [J3, J+]= J+, [J3, J−]= −J−. (1.27) The Casimir operator in this basis is
X2 = −J+J−− J23+ J3 = −J−J+− J23− J3 (1.28) From the commutation relations we have
J3J+ = J+(J3+ 1) (1.29)
J3J− = J−(J3− 1) (1.30)
J−J+ = −X2− J23− J3 (1.31) J+J− = −X2− J23+ J3 (1.32) Since J3 commutes with X2 they can be diagonalized simultaneously. Since they are both Hermitian, they have real eigenvalues. Notice also that since Xi are anti-Hermitian, then
X2≤ 0 . (1.33)
Let |λ, mi be the basis in which both operators are diagonal, that is,
X2|λ, mi = −λ|λ, mi (1.34)
J3|λ, mi = m|λ, mi (1.35)
There are many ways to show that the eigenvalues m of the operator J3are integers.
Now, since
X12+ X22 = X2− X32= X2+ J32≤ 0 (1.36)
then the eigenvalues of J3 are bounded by a maximal integer j,
|m| ≤ j (1.37)
that is, m takes (2 j+ 1) values
− j, − j + 1, · · · , −1, 0, 1, · · · , j − 1, j (1.38)
¿From the commutation relations above we have
J+|λ, mi = const |λ, m + 1i (1.39) J−|λ, mi = const |λ, m − 1i (1.40) Let |λ, 0i be the eigenvector corresponding to the eigenvalue m= 0, that is
J3|λ, 0i = 0 . (1.41)
Then for positive m > 0
|λ, mi = aλmJ+m|λ, 0i (1.42)
|λ, −mi = aλ,−mJ−m|λ, −mi (1.43) Then
J+|λ, ji = 0 (1.44)
J−|λ, − ji = 0 (1.45)
Therefore,
0= J−J+|λ, ji =
−X2− J32− J3
|λ, ji (1.46)
0= J+J−|λ, − ji =
−X2− J32+ J3
|λ, − ji (1.47) Thus,
λ2− j2− j
|λ, ji = 0 (1.48)
So, the eigenvalues of the Casimir operator X2are labeled by a non-negative inte- ger j ≥ 0,
λj = j( j + 1) . (1.49)
These eigenvalues have the multiplicity
dj = 2 j + 1 . (1.50)
In particular, for j = 1 we get λ1 = 2 as expected before.
From now on we will denote the basis in which the operators X2 and J3 are diagonal by
|λj, mi =
j m
. (1.51)
The matrix elements of the generators J±can be computed as follows. First of all,
j m
J3
j m0
= δmm0m (1.52)
We have
j( j+ 1) = j m
J+
j m −1
j m −1
J−
j m
+ m2− m (1.53) and
j m
J+
j m −1
= j m −1
J−
j m
(1.54) This gives
j m
J+
j m0
= δm0,m−1p
( j+ m)( j − m + 1) (1.55)
j m
J−
j m0
= δm0,m+1p
( j − m)( j+ m + 1) (1.56) The matrix elements of the generators Xiare
j m
X1
j m0
= δm0,m−1i 2
p( j+ m)( j − m + 1) +δm0,m+1i
2
p( j − m)( j+ m + 1) (1.57)
j m
X2
j m0
= δm0,m−11 2
p( j+ m)( j − m + 1)
−δm0,m+11 2
p( j − m)( j+ m + 1) (1.58)
j m
X3
j m0
= δmm0im (1.59)
Thus, irreducible representations of S O(3) are labeled by a non-negative inte- ger j ≥ 0. The matrices of the representation j in canonical coordinates yi will be denoted by
Dmmj 0(y)= j m
expX(y)
j m0
(1.60) where X(y) = Xiyi.
The characters of the irreducible representations are χj(y)=
j
X
m=− j
Dmmj (y)=
j
X
m=− j
eimr = 1 + 2
j
X
m=1
cos(mr) (1.61)
2 Group S U(2)
2.1 Algebra su(2)
Pauli matrices σi = σi, i= 1, 2, 3, are defined by
σ1 =
0 1 1 0
, σ2 =
0 −i
i 0
, σ3 =
1 0
0 −1
. (2.1)
They have the following properties
σ†i = σi (2.2)
σ2i = I (2.3)
det σi = −1 (2.4)
tr σi = 0 , (2.5)
σ1σ2 = iσ3 (2.6)
σ2σ3 = iσ1 (2.7)
σ3σ1 = iσ2 (2.8)
σ1σ2σ3 = iI (2.9)
(2.10) which can be written in the form
σiσj = δi jI+ iεi jkσk. (2.11)
They satisfy the following commutation relations
[σi, σj]= 2iεi jkσk (2.12) The traces of products are
tr σiσj = 2δi j (2.13)
tr σiσjσk = 2iεi jk (2.14) Also, there holds
σiσi = 3I . (2.15)
The Pauli matrices satisfy the anti-commutation relations
σiσj+ σjσi = 2δi jI (2.16) Therefore, Pauli matrices form a representation of Clifford algebra in 3 dimen- sions and are, in fact, Dirac matrices σi = (σiA
B), where A, B= 1, 2, in 3 dimen- sions. The spinor metric E = (EAB) is defined by
E =
0 1
−1 0
. (2.17)
Notice that
E = iσ2 (2.18)
and the inverse metric E−1= (EAB) is
E−1= −E (2.19)
Then
σTi = −EσiE−1 (2.20)
This allows to define the matrices Eσi = (σi AB) with covariant indices Eσ1 = σ3=
1 0
0 −1
, (2.21)
Eσ2 = iI =
i 0 0 i
, (2.22)
Eσ3 = −σ1 =
0 −1
−1 0
. (2.23)
Notice that the matrices Eσi are all symmetric
(Eσi)T = Eσi (2.24)
Pauli matrices satisfy the following completeness relation σi
A Bσi
C
D = 2δADδCB−δABδCD (2.25) which implies
σi(A (BσiC)
D) = δ(A(BδC)D) (2.26) This gives
σi ABσi CD = 2EADECB− EABECD (2.27) In a more symmetric form
σi ABσi CD = EADECB+ EBDECA (2.28) A spinor is a two dimensional complex vector ψ= (ψA), A= 1, 2 in the spinor space C2. We use the metric E to lower the index to get the covariant components
ψA = EABψB (2.29)
which defines the cospinor ˜ψ = (ψA)
ψ = Eψ˜ (2.30)
We also define the conjugated spinor ¯ψ = ( ¯ψA) by
ψ¯A = ψA∗ (2.31)
On the complex spinor space there are two invariant bilinear forms
hψ, ϕi = ¯ψϕ = ψA∗ϕA (2.32)
and
ψϕ = ψ˜ AϕA = EABψBϕA = ψ1ϕ2−ψ2ϕ1 (2.33) Note that
hψ, ψi = ¯ψψ = ψA∗ψA = |ψ1|2+ |ψ2|2 (2.34) and
ψψ = ψ˜ AψA = 0 (2.35)
The matrices
Xi = i
2σi (2.36)
are the generators of the Lie algebra su(2) with the commutation relations
[Xi, Xj]= −εi jkXk (2.37) The Casimir operator is
X2 = XiXi (2.38)
It commutes with all generators
[Xi, X2]= 0 . (2.39)
There holds
X2 = −3
4I (2.40)
tr XiXj = −1
2δi j (2.41)
Notice that
X2= − j( j + 1)I (2.42)
with j= 12, which determines the fundamental representation of S U(2).
An arbitrary element of the group S U(2) in canonical coordinates yi has the form
exp
i 2σjyj
= I cosr
2 + iσjθjsinr
2, (2.43)
This can be written in the form
exp (X(y)) = I cos(r/2) + X(y)sin(r/2)
r/2 . (2.44)
where X(y) = Xiyi. In particular, for r= 2π
exp (X(y)) = −I. (2.45)
Obviously,
tr exp (X(y))= 2 cos(r/2) (2.46)
2.2 Representations of S U(2)
We define symmetric contravariant spinors ψA1...A2 j of rank 2 j. Each index here can be lowered by the metric EAB. The number of independent components of such spinor is precisely
2 j
X
i=0
1= (2 j + 1) (2.47)
Let Xi be the generators of S U(2) in a general irreducible representation sat- isfying the algebra
[Xi, Xj]= −εi jkXk (2.48) They are determined by symmetric spinors of type (2 j, 2 j)
(Xk)A1...A2 jB1...B2 j = i jσk(A1
(B1δAB2
2· · ·δAB2 j)
2 j) (2.49)
Then
X2= − j( j + 1)I (2.50)
where
IBA1...A2 j
1...B2 j = δ(A(B11· · ·δAB2 j)
2 j) (2.51)
Another basis of generators is
J+ = −iX1+ X2 (2.52)
J− = −iX1− X2 (2.53)
J3 = −iX3 (2.54)
They have all the properties of the generators of S O(3) discussed above except that the operator J3 has integer eigenvalues m. One can show that they can be either integers or half-integers taking (2 j+ 1) values
− j, − j + 1, . . . , j − 1, j , |m| ≤ j (2.55) with j being a positive integer or half-integer. If j is integer, then all eigenvalues mof J3 are integers including 0. If j is a half-integer, then all eigenvalues m are half-integers and 0 is excluded.
We choose a basis in which the operators X2 and J3 are diagonal. Let ψ = ψA1...A2 j be a symmetric spinor of rank 2 j. Let e1 = (e1A
) and e2 = (e2A
) be the basis cospinors defined by
e1A = δA1, e2A = δ2A. (2.56)
Let
ej,mA1...Aj+mB1...Bj−m = e1(A1· · · e1Aj+me2B1· · · e2Bj−m)
= δ(A1 1· · ·δ1Aj+mδB21· · ·δ2Bj−m) (2.57) Then this is an eigenspinor of the operator J3with the eigenvalue m
J3ej,m= mej,m (2.58)
Then the spinors
j m
= (2 j)!
( j+ m)!( j − m)!
!1/2
ej,m (2.59)
form an orthonormal basis in the space of symmetric spinors of rank 2 j.
The matrix elements of the generators Xiare given by exactly the same formu- las
j m
X1
j m0
= δm0,m−1i 2
p( j+ m)( j − m + 1) +δm0,m+1i
2
p( j − m)( j+ m + 1) (2.60)
j m
X2
j m0
= δm0,m−11 2
p( j+ m)( j − m + 1)
−δm0,m+11 2
p( j − m)( j+ m + 1) (2.61)
j m
X3
j m0
= δmm0im (2.62)
but now j and m are either both integers or half-integers.
Thus, irreducible representations of S U(2) are labeled by a non-negative half- integer j ≥ 0. The matrices of the representation j in canonical coordinates yi are
Dmmj 0(y)= j m
expX(y)
j m0
(2.63) where X(y) = Xiyi. The representations of S U(2) with integer j are at the same time representations of S O(3). However, representations with half-integer j do not give representations of S O(3), since for r = 2π
Dmmj 0(y)= −δmm0. (2.64)
The characters of the irreducible representations are: for integer j χj(y)=
j
X
m=− j
Dmmj (y)=
j
X
m=− j
eimr = 1 + 2
j
X
m=1
cos(mr) (2.65)
and for half-integer j
χj(y)=
−1
2+ j
X
k=−12− j
Dj1
2+k,12+k(y)=
−1
2+ j
X
k=−12− j
exp
"
i 1 2 + k
! r
#
= 2
j
X
k=1
cos" 1 2 + k
! r
#
(2.66) Now, let Xiand Yj be the generators of two representations and let
Gi = Xi ⊗ IY + IX⊗ Yi (2.67) Then Gigenerate the product representation X ⊗ Y and
exp[G(y)]= exp[X(y)] exp[Y(y)] (2.68) Therefore, the character of the product representation factorizes
χG(y)= χX(y)χY(y) (2.69)
2.3 Double Covering Homomorphism S U(2) → S O(3)
Recall that the matrices Eσi = (σi AB) are symmetric and the matrix E = (EAB) is anti-symmetric. Let ψABbe a symmetric spinor. Then one can form a vector
Ai = σi BAψAB = tr (Eσiψ) . (2.70) Conversely, given a vector Ai we get a symmetric spinor of rank 2 by
ψAB = 1
2Aiσi AB (2.71)
or
ψ = 1
2AiσiE−1 (2.72)
More generally, given a symmetric spinor of rank 2 j we can associate to it a symmetric tensor of rank j by
Ai1...ij = σi1A1B1· · ·σijAjBjψA1...AjB1...Bj (2.73)
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
ψA1...AjB1...Bj = 1
2jAi1...ijσi1A1B1· · ·σijAjBj (2.74) The double covering homomorphism
Λ : S U(2) → S O(3) (2.75)
is defined as follows. Let U ∈ S U(2). Then the matrices UσiU−1 satisfy all the properties of the Pauli matrices and are therefore in the algebra su(2). Therefore,
UσiU−1= Λji(U)σj. (2.76) That is,
Λij(U)= 1
2tr σiUσjU−1 (2.77)
The matrixΛ(U) depends on the matrix U and is, in fact, in S O(3). For infinites- imal transformations
U = exp(Xkyk)= I + i
2σkyk+ · · · (2.78) the matrixΛ is
Λi j = δi j+ εi jkyk+ · · · =h
exp(Xkyk)i
i j (2.79)
It is easy to see that
Λ(I) = Λ(−I) = I . (2.80)
So, in short
Λ(exp(Xiyi))= exp(Λ(Xi)yi) (2.81) where
[Λ(σk)]i j = 2iεik j (2.82)
3 Invariant Vector Fields
Let yibe the canonical coordinates on the group S U(2) ranging over (−π, π). The position of the indices on the coordinates will be irrelevant, that is, yi = yi. We introduce the radial coordinate r = |y| = pyiyi, so that the geodesic distance
from the origin is equal to r, and the angular coordinates (the coordinates on S2) θi = ˆyi = yi/r .
Let us consider the fundamental representation of the group S U(2) with gen- erators Ta = iσa, where σaare the Pauli matrices satisfying the algebra
[Ta, Tb]= −2εcabTc. (3.83) Let T (y) = Tiyi. Then
T(y)T (p)= −(y, p) − T(y ∧ p) (3.84) where (y, p)= yipi, (y ∧ p)k = εi jkyipj and
[T (y)]2 = −|y|2I (3.85)
Therefore,
exp[rT (ˆy)]= cos r + T(ˆy) sin r (3.86) where ˆy is a unit vector. Next, we compute
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
F(ω, F(p, q))= F(F(ω, p), q) (3.88) and the inverse property
F(−p, −ω)= −F(ω, p). (3.89)
This map defines all the properties of the group. We introduce the matrices Lαb(x) = ∂
∂ybFα(y, x)
y=0, (3.90)
Rαb(x) = ∂
∂ybFα(x, y)
y=0, (3.91)
Let X and Y be the inverses of the matrices L and R, that is,
LX = XL = RY = YR = I . (3.92)
Also let
D= XR , D−1= YL (3.93)
so that
L= RD−1, X = DY (3.94)
Let Cabe the matrices of the adjoint representation defined by
(Ca)bc = −2εbac (3.95)
and C = C(x) be the matrix defined by
C = Caxa. (3.96)
Then one can show that
X= exp(C) − I
C , Y = I − exp(−C)
C , (3.97)
L= C
exp(C) − I, R= C
I − exp(−C). (3.98)
It is easy to see also that
xa= Labxb= Rabxb = Xabxb= Yabxb. (3.99) and
X = YT, L= RT (3.100)
and
XY = YX = sinh(C/2) C/2
!2
, (3.101)
LR= RL = C/2 sinh(C/2)
!2
, (3.102)
as well as
D= XR = RX = exp(C) , DT = YL = LY = exp(−C) . (3.103)
Note that
C2 = −4r2Π , (3.104)
where r = |x| and
Πij = δij−θiθj (3.105) with θi = xi/r. Therefore,
C2n = (−1)n(2r)2nΠ , C2n+1= (−1)n(2r)2nC (3.106) Therefore, the eigenvalues of the matrix C are (2ir, −2ir, 0), and for any analytic function
f(C) = f (0)(I − Π) + Π1
2 f (2ir)+ f (−2ir) + 1
4irC f (2ir) − f (−2ir
and, therefore,
tr f (C) = f (0) + f (2ir) + f (−2ir) (3.107) det f (C) = f (0) f (2ir) f (−2ir) (3.108) This enables one to compute
Y = I − Π + sin r
r cos rΠ − 1 2
sin2r
r2 C (3.109)
X = I − Π + sin r
r cos rΠ + 1 2
sin2r
r2 C (3.110)
R = I − Π + r cot rΠ + 1
2C (3.111)
L = I − Π + r cot rΠ − 1
2C (3.112)
D = I − Π + Π cos(2r) +sin(2r)
2r C (3.113)
Then one can show that
∂
∂xµFα(x, y) = Lαγ(F(x, y))Xγµ(x) , (3.114)
∂
∂xµFα(y, x) = Rαγ(F(y, x))Yγµ(x) . (3.115) These matrices satisfy important identities
∂µXaν−∂νXaµ = −2εabcXbµXcν, (3.116)
∂µYaν−∂νYaµ = 2εabcYbµYcν. (3.117)
and
Lµa∂µLνb− Lµb∂µLνa = 2εcabLγc, (3.118) Rµa∂µRνb− Rµb∂µRνa = −2εcabRνc. (3.119) Let us define the one-forms
σa−= Xaµ(x)dxµ, σa+= Yaµ(x)dxµ. They satisfy important identities
dσa− = −εabcσb−∧σc−, (3.120) dσa+ = εabcσb+∧σc+. (3.121) Next, we define the vector fields.
Ka− = Lµa(x)∂µ, Ka+ = Rµa(x)∂µ.
These are the right-invariant and the left-invariant vector fields on the group. They are related by
Ka− = Dab
Kb+ (3.122)
They form two representations of the group S U(2) and satisfy the algebra
[Ka+, Kb+] = −2εcabKc+ (3.123) [Ka−, Kb−] = 2εcabKc− (3.124)
[Ka+, Kb−] = 0 (3.125)
That is, the left-invariant vector fields commute with the right-invariant ones.
Now, let Ta be the generators of some representation of the group S U(2) sat- isfying the same algebra. Then, of course,
exp[T (ω)] exp[T (p)]= exp[T(F(ω, p))] (3.126) First, one can derive a useful commutation formula
exp[T (ω)]Tbexp[−T (ω)]= DabTa, (3.127) where the matrix D = (Dab) is defined by D= exp C .
More generally,
exp[T (ω)] exp[T (p)] exp[−T (ω)]= exp[T(D(ω)p)] . (3.128)
Next, one can show that
exp[−T (ω)] ∂
∂ωi exp[T (ω)]= YaiTa, (3.129) This means that
∂
∂ωi exp[T (ω)]= Yaiexp[T (ω)]Ta = YiaTaexp[T (ω)] , (3.130) The Killing vectors have important properties
Ka+exp[T (ω)]= (exp[T(ω)])Ta (3.131) Ka−exp[T (ω)]= Ta(exp[T (ω)]) (3.132) This immediately leads to further important equations
exp[K+(p)] exp[T (ω)]= exp[T(ω)] exp[T(p)] (3.133) and
exp[K−(q)] exp[T (ω)]= exp[T(q)] exp[T(ω)] (3.134) where K±(p) = paKa±. 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)]
ω=0 = exp[T(q)] exp[T(p)] = exp[T(F(q, p))]
(3.136) In particular, for the adjoint representation this formulas take the form
DCb = DabCaD (3.137)
Ka+D= DCa (3.138)
Ka−D= CaD (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) (3.142)
Then for a scalar function f (ω) the action of the right-invariant and left- invariant 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
exp[K+(p)+ K−(q)] f (ω)= f (F(q, F(ω, p))). (3.144) Therefore,
exp[K+(p)+ K−(q)] f (ω)
ω=0 = f (F(q, p)). (3.145) The bi-invariant metric is defined by
gαβ = δabYaαYbβ = δabXaαXbβ (3.146) gαβ = δabLαaLβb = δabRαaRβb, (3.147) that is,
g= sinh[C/2]
C/2
!2
= I − Π + sin2r
r2 Π . (3.148)
The Riemannian volume elements of the metric g is defined as usual
dvol = g1/2(x)dx1∧ dx2∧ dx3 (3.149) where
g1/2 = (det gµν)1/2 = sin2r
r2 . (3.150)
Thus the bi-invariant volume element of the group is dvol (x)= sin2r
r2 dx= sin2r dr dθ (3.151) where dx = dx1dx2dx3, and dθ is the volume element on S2. The invariance of the volume element means that for any fixed p
dvol (x)= dvol (F(x, p)) = dvol (F(p, x)) (3.152) Notice also, that |F(p, q)| is nothing but the geodesic distance between the points
pand q.
We will always use the orthonormal basis of the right-invariant vector fields Ka+and one-forms σa+and denote the covariant derivative with respect to Ka+sim- ply by ∇a. The Levi-Civita connection ∇ of the bi-invariant metric in the right- invariant basis is defined by
∇aKb+ = εcbaKc+ (3.153) so that the coefficients of the affine connection are
ωabc = σa+(∇cKb+)= εabc. (3.154) so that for an arbitrary vector
∇aV =h
(Ka+Vb)+ εbcaVci
Kb+ (3.155)
Then
∇aKd−= MbdaKb+ (3.156) where
Mbda = Ka+Ddb+ εbcaDdc
= −2εbcaDdc+ εbcaDdc = −εbcaDdc
(3.157) so that
∇aKd− = −εbcaDdc
Kb+ (3.158)
Now, the curvature tensor is
Rabcd = −εfcdεaf b. (3.159) The Ricci curvature tensor and the scalar curvature are
Rab= 2δab, R= 6 . (3.160)
The scalar Laplacian is
∆0 = δabKa−Kb−= δabKa+Kb+. (3.161) One can show that
∆0
|ω|
sin |ω| = |ω|
sin |ω| (3.162)
Further,
∆0exp[T (ω)]= (exp[T(ω)])T2 = T2exp[T (ω)] (3.163)
where T2= TaTa. This means that
exp[t∆20] exp[T (ω)]= exp(tT2) exp[T (ω)] (3.164) The Killing vector fields Ka± are divergence free, which means that they are anti-self-adjoint with respect to the invariant measure dvol (ω). that is, Thus the Laplacian∆0is self-adjoint 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 one-forms
A= σa+Ga = GaYaµdxµ. (3.165) and F = dA + A ∧ A. Then, by using the above properties of the right-invariant one-forms we compute
F = 1
2Fabσa+∧σb+= 1
2FabYaµYbνdxµ∧ dxν (3.166) where
Fab = εcabGc (3.167)
Thus A is the Yang-Mills curvature on the vector bundle V with covariantly con- stant curvature.
The covariant derivative of a section of the bundle V is then (recall that ∇a =
∇K+ a)
∇aϕ = (Ka++ Ga)ϕ (3.168)
and the covariant derivative of a section of the endomorphism bundle End(V) along the right-invariant basis is then
∇aQ= Ka+Q+ [Ga, Q] (3.169) Therefore, we find
∇aGb = εcbaGc+ [Ga, Gb]= 0 . (3.170) Then the derivatives along the left-invariant vector fields are
∇K−
aϕ = (Ka−+ Ba)ϕ (3.171)
∇K−aQ= Ka−Q+ [Ba, Q] (3.172) where
Ba = Dab
Gb (3.173)
The Laplacian takes the form
∆ = ∇a∇a = (Ka++ Ga)(Ka++ Ga)= ∆0+ 2GaKa++ G2 (3.174) where G2= GiGi.
We want to rewrite the Laplacian in terms of Casimir operators of some rep- resentations of the group S U(2). The covariant derivatives ∇ado not form a rep- resentation of the algebra S U(2). 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
Lξ = ∇ξ− 1
2σa+(∇bξ)εbcaGc (3.175) By denoting the Lie derivatives along the Killing vectors Ka±by Ka±this gives for the right-invariant and the left-invariant bases
Ka+ = LK+a = ∇a+ Ga = Ka++ 2Ga. (3.176) Ka− = LK−a = ∇Ka−− Ba= Ka−. (3.177) It is easy to see that these operators form the same algebra S U(2) × S U(2)
[Ka+, Kb+] = −2εcabKc+ (3.178) [Ka−, Kb−] = 2εcabKc− (3.179)
[Ka+, Kb−] = 0 (3.180)
Now, the Laplacian is given now by the sums of the Casimir operators
∆ = K2− G2 (3.181)
where
K2= 1
2K+2+ 1
2K−2 (3.182)
and K±2= Ka±Ka±.
4 Heat Kernel on S
3Our goal is to evaluate the heat kernel diagonal. Since it is constant we can eval- uate it at any point, say, at the origin. We use the geodesic coordinates yi defined
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
−tG2 exp
tK2 . (4.1)
Therefore, the heat kernel is equal to U(t, y)= exp
−tG2 Ψt 2, y
(4.2) whereΨ(t, y) = exp(2tK2)δ(y) is the heat kernel of the operator 2K2.
4.1 Scalar Heat Kernel
Let
Φ(t, ω) = (4πt)−3/2et
∞
X
n=−∞
|ω| + 2πn
sin |ω| exp −(|ω|+ 2πn)2 4t
!
(4.3) One can show by direct calculation that this function satisfies the equation
∂tΦ = ∆20Φ (4.4)
with the initial condition
Φ(0, ω) = δS3(ω) (4.5)
Therefore, this is the scalar heat kernel on the group S U(2) and, therefore, on S3. Now, let us compute the integral
Ψ(t) = Z
S U(2)
dvol (ω)Φ(t, ω) exp[T(ω)] (4.6) Obviously,Ψ(0) = 1. Next, we have
∂tΨ = Z
S U(2)
dvol (ω) exp[T (ω)]∆0Φ(t, ω) (4.7) Now, by integrating by parts we get
∂tΨ = Z
S U(2)
DωΦ(t, ω)∆0exp[T (ω)] (4.8) which gives
∂tΨ = ΨT2 (4.9)
Thus, we obtain a very important equation exp(tT2)=Z
S U(2)
dvol (ω)Φ(t, ω) exp[T(ω)] (4.10) This is true for any representation of S U(2).
We derive two corollaries. First, we get exp(tT2)= Z
S U(2)
dvol (ω)Φ(t, ω) exp[T(p)] exp[T(ω)] exp[−T(p)] (4.11) which means that for any p
Φ(t, ω) = Φ(t, F(−p, F(ω, p)). (4.12) or
Φ(t, F(p, q)) = Φ(t, F(q, p)). (4.13) Also, we see that
exp[(t+ s)T2]= Z
S3×S3
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 7→ −p we obtain exp[(t+ s)T2]= Z
S U(2)
dvol (z) Z
S U(2)
dvol (p)Φ(t, F(p, z))Φ(s, p) exp[T(z)]
(4.15) which means that
Z
S U(2)
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
Z
S U(2)
dvol (p)Φ(t, p)Φ(t, p) = Φ(t, 0) (4.17)