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Ivan G. Avramidi

New Mexico Institute of Mining and Technology

October, 2011

Author: IGA File: su2.tex; Date: November 29, 2011; Time: 10:07

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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)

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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)

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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 = −JJ+− J23− J3 (1.28) From the commutation relations we have

J3J+ = J+(J3+ 1) (1.29)

J3J = J(J3− 1) (1.30)

JJ+ = −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)

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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,−mJm|λ, −mi (1.43) Then

J+|λ, ji = 0 (1.44)

J|λ, − ji = 0 (1.45)

Therefore,

0= JJ+|λ, ji = 

−X2− J32− J3

|λ, ji (1.46)

0= J+J|λ, − ji = 

−X2− J32+ J3

|λ, − ji (1.47) Thus,

2− 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)

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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

j, 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)

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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

σi = 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)

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They satisfy the following commutation relations

i, 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)

2 = iI =









 i 0 0 i









, (2.22)

3 = −σ1 =









0 −1

−1 0









. (2.23)

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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)

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The matrices

Xi = i

i (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

i 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)

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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)

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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)

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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)

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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,

iU−1= Λji(U)σj. (2.76) That is,

Λij(U)= 1

2tr σijU−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

kyk+ · · · (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

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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)

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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)

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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)

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and

LµaµLνb− LµbµLνa = 2εcabLγc, (3.118) RµaµRνb− RµbµRνa = −2εcabRνc. (3.119) Let us define the one-forms

σa= Xaµ(x)dxµ, σa+= Yaµ(x)dxµ. They satisfy important identities

a = −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)

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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) Kaexp[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)

KaD= 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)

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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.

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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 = δabKaKb= 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)

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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)

KaQ= KaQ+ [Ba, Q] (3.172) where

Ba = Dab

Gb (3.173)

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The Laplacian takes the form

∆ = ∇aa = (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

a+(∇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 = LKa = ∇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

2K2 (3.182)

and K±2= Ka±Ka±.

4 Heat Kernel on S

3

Our 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

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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)

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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)

References

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