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

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 antisymmetricmatrices defined by

0 −1 0

, X2 =

0 0 −1

, X3 =

−1 0 0

, (1.1)

or(Xi)k

j = −εki j = εik j (1.2)

Raising and lowering a vector index does not change anything; it is done just forconvenience of notation. These matrices have a number of properties

XiX j = E ji − Iδi j , (1.3)

andεi jkXk = Ei j − E ji (1.4)

where(E ji)kl = δk

jδli (1.5)

Therefore, they satisfy the algebra

[Xi, X j] = −εki jXk (1.6)

The Casimir operator is defined by

X2 = XiXi (1.7)

It is equal toX2 = −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 yi 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 =√

yiyi, θi =yi

r(1.12)

X(y) = Xiyi, and the matrices Π and P are defined by

Pij = θiθ j (1.13)

Πij = δi

j − θiθ j (1.14)

In particular, for r = π we have

exp (X(y)) = P − Π (1.15)

Notice thattr P = 1, tr Π = 2 (1.16)

and, therefore,

tr exp (X(y)) = 1 + 2 cos r . (1.17)

detsinh X(y)

(sin r

(1.18)

1.2 Representations of S O(3)

Let Xi be the generators of S O(3) in a general irreducible representation satisfyingthe algebra

[Xi, X j] = −εi jkXk (1.19)

They are determined by symmetric traceless tensors of type ( j, j) (with j a positiveinteger)

(Xi)l1...l jk1...k j = − jε(l1

i(k1δl2k2· · · δ

l j)k j)

(1.20)

ThenX2 = − j( j + 1)I (1.21)

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whereIl1...l j

k1...k j = δ(l1(k1· · · δ

l j)k 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 thatJ†+ = J− , J†3 = 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+ − J2

3 − J3 (1.28)

From the commutation relations we have

J3J+ = J+(J3 + 1) (1.29)J3J− = J−(J3 − 1) (1.30)J−J+ = −X2 − J2

3 − J3 (1.31)J+J− = −X2 − J2

3 + J3 (1.32)

Since J3 commutes with X2 they can be diagonalized simultaneously. Sincethey are both Hermitian, they have real eigenvalues. Notice also that since Xi areanti-Hermitian, then

X2 ≤ 0 . (1.33)

Let |λ,m〉 be the basis in which both operators are diagonal, that is,

X2|λ,m〉 = −λ|λ,m〉 (1.34)J3|λ,m〉 = m|λ,m〉 (1.35)

There are many ways to show that the eigenvalues m of the operator J3 are integers.Now, since

X21 + X2

2 = X2 − X23 = X2 + J2

3 ≤ 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,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

J3|λ, 0〉 = 0 . (1.41)

Then for positive m > 0

|λ,m〉 = aλmJm+ |λ, 0〉 (1.42)

|λ,−m〉 = aλ,−mJm− |λ,−m〉 (1.43)

J+|λ, j〉 = 0 (1.44)J−|λ,− j〉 = 0 (1.45)

Therefore,

0 = J−J+|λ, j〉 =(−X2 − J2

3 − J3

)|λ, j〉 (1.46)

0 = J+J−|λ,− j〉 =(−X2 − J2

3 + J3

)|λ,− j〉 (1.47)

Thus, (λ2 − j2 − j

)|λ, j〉 = 0 (1.48)

So, the eigenvalues of the Casimir operator X2 are labeled by a non-negative inte-ger j ≥ 0,

λ j = j( j + 1) . (1.49)

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These eigenvalues have the multiplicity

d j = 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

|λ j,m〉 =

∣∣∣∣∣ jm

⟩. (1.51)

The matrix elements of the generators J± can be computed as follows. First of all,⟨ jm

∣∣∣∣∣ J3

∣∣∣∣∣ jm′

⟩= δmm′m (1.52)

We havej( j + 1) =

⟨ jm

∣∣∣∣∣ J+

∣∣∣∣∣ jm − 1

⟩ ⟨ jm − 1

∣∣∣∣∣ J−∣∣∣∣∣ jm

⟩+ m2 − m (1.53)

and ⟨ jm

∣∣∣∣∣ J+

∣∣∣∣∣ jm − 1

⟨ jm − 1

∣∣∣∣∣ J−∣∣∣∣∣ jm

⟩(1.54)

This gives ⟨ jm

∣∣∣∣∣ J+

∣∣∣∣∣ jm′

⟩= δm′,m−1

√( j + m)( j − m + 1) (1.55)

⟨ jm

∣∣∣∣∣ J−∣∣∣∣∣ jm′

⟩= δm′,m+1

√( j − m)( j + m + 1) (1.56)

The matrix elements of the generators Xi are⟨ jm

∣∣∣∣∣ X1

∣∣∣∣∣ jm′

⟩= δm′,m−1

√( j + m)( j − m + 1)

+δm′,m+1i2

√( j − m)( j + m + 1) (1.57)⟨ j

∣∣∣∣∣ X2

∣∣∣∣∣ jm′

⟩= δm′,m−1

√( j + m)( j − m + 1)

−δm′,m+112

√( j − m)( j + m + 1) (1.58)⟨ j

∣∣∣∣∣ X3

∣∣∣∣∣ jm′

⟩= δmm′im (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 bedenoted by

D jmm′(y) =

⟨ jm

∣∣∣∣∣ exp[X(y)

] ∣∣∣∣∣ jm′

⟩(1.60)

where X(y) = Xiyi.The characters of the irreducible representations are

χ j(y) =

j∑m=− j

D jmm(y) =

j∑m=− j

eimr = 1 + 2j∑

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

, σ2 =

0 −i

, σ3 =

0 −1

. (2.1)

They have the following properties

σ†i = σi (2.2)σ2

i = I (2.3)detσi = −1 (2.4)

trσi = 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

σiσ j = δi jI + iεi jkσk . (2.11)

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They satisfy the following commutation relations

[σi, σ j] = 2iεi jkσk (2.12)

The traces of products are

trσiσ j = 2δi j (2.13)trσiσ jσk = 2iεi jk (2.14)

Also, there holdsσiσi = 3I . (2.15)

The Pauli matrices satisfy the anti-commutation relations

σiσ j + σ jσi = 2δi jI (2.16)

Therefore, Pauli matrices form a representation of Clifford algebra in 3 dimen-sions and are, in fact, Dirac matrices σi = (σi

AB), where A, B = 1, 2, in 3 dimen-

sions. The spinor metric E = (EAB) is defined by

−1 0

. (2.17)

Notice thatE = iσ2 (2.18)

and the inverse metric E−1 = (EAB) is

E−1 = −E (2.19)

ThenσT

i = −EσiE−1 (2.20)

This allows to define the matrices Eσi = (σi AB) with covariant indices

Eσ1 = σ3 =

0 −1

, (2.21)

Eσ2 = iI =

, (2.22)

Eσ3 = −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

D = 2δADδ

CB − δ

CD (2.25)

which impliesσi

(A(Bσi

C)D) = δ(A

(BδC)

D) (2.26)

This givesσi 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 spinorspace 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

ψA = ψA∗ (2.31)

On the complex spinor space there are two invariant bilinear forms

〈ψ, ϕ〉 = ψϕ = ψA∗ϕA (2.32)

andψϕ = ψAϕ

A = EABψBϕA = ψ1ϕ2 − ψ2ϕ1 (2.33)

Note that〈ψ, ψ〉 = ψψ = ψA∗ψA = |ψ1|2 + |ψ2|2 (2.34)

andψψ = ψAψ

A = 0 (2.35)

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The matricesXi =

i2σi (2.36)

are the generators of the Lie algebra su(2) with the commutation relations

[Xi, X j] = −εi jkXk (2.37)

The Casimir operator isX2 = XiXi (2.38)

It commutes with all generators

[Xi, X2] = 0 . (2.39)

There holdsX2 = −

I (2.40)

tr XiX j = −12δi j (2.41)

Notice thatX2 = − 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 theform

exp( i2σ jy j

)= I cos

+ iσ jθj sin

r2, (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 herecan be lowered by the metric EAB. The number of independent components ofsuch spinor is precisely

2 j∑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, X j] = −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(B1δ

A2B2· · · δ

A2 j)B2 j)

(2.49)

ThenX2 = − j( j + 1)I (2.50)

whereIA1...A2 j

B1...B2 j= δ(A1

(B1· · · δ

A2 j)B2 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 exceptthat the operator J3 has integer eigenvalues m. One can show that they can beeither 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 eigenvaluesm of J3 are integers including 0. If j is a half-integer, then all eigenvalues m arehalf-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 = (e2

A) be thebasis cospinors defined by

e1A = δA

1 , e2A = δA

2 . (2.56)

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e j,mA1...A j+mB1...B j−m = e1

(A1 · · · e1A j+me2

B1 · · · e2B j−m)

= δ(A11 · · · δ

1 δB12 · · · δ

B j−m)2 (2.57)

Then this is an eigenspinor of the operator J3 with the eigenvalue m

J3e j,m = me j,m (2.58)

Then the spinors ∣∣∣∣∣ jm

((2 j)!

( j + m)!( j − m)!

e j,m (2.59)

form an orthonormal basis in the space of symmetric spinors of rank 2 j.The matrix elements of the generators Xi are given by exactly the same formu-

las ⟨ jm

∣∣∣∣∣ X1

∣∣∣∣∣ jm′

⟩= δm′,m−1

√( j + m)( j − m + 1)

+δm′,m+1i2

√( j − m)( j + m + 1) (2.60)⟨ j

∣∣∣∣∣ X2

∣∣∣∣∣ jm′

⟩= δm′,m−1

√( j + m)( j − m + 1)

−δm′,m+112

√( j − m)( j + m + 1) (2.61)⟨ j

∣∣∣∣∣ X3

∣∣∣∣∣ jm′

⟩= δmm′im (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

areD j

mm′(y) =

⟨ jm

∣∣∣∣∣ exp[X(y)

] ∣∣∣∣∣ jm′

⟩(2.63)

where X(y) = Xiyi. The representations of S U(2) with integer j are at the sametime representations of S O(3). However, representations with half-integer j donot give representations of S O(3), since for r = 2π

D jmm′(y) = −δmm′ . (2.64)

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The characters of the irreducible representations are: for integer j

χ j(y) =

j∑m=− j

D jmm(y) =

j∑m=− j

eimr = 1 + 2j∑

cos(mr) (2.65)

and for half-integer j

χ j(y) =

− 12 + j∑

k=− 12− j

D j12 +k, 1

2 +k(y) =

− 12 + j∑

k=− 12− j

exp[i(12

= 2j∑

(2.66)Now, let Xi and Y j be the generators of two representations and let

Gi = Xi ⊗ IY + IX ⊗ Yi (2.67)

Then Gi generate 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) isanti-symmetric. Let ψAB be 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 =12

Aiσi AB (2.71)

orψ =

AiσiE−1 (2.72)

More generally, given a symmetric spinor of rank 2 j we can associate to it asymmetric tensor of rank j by

Ai1...i j = σi1 A1B1 · · ·σi j A jB jψA1...A jB1...B j (2.73)

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One can show that this tensor is traceless. Indeed by using the eq. (??) and thefact that the spinor is symmetric we see that the tensor A is traceless. The inversetransformation is defined by

ψA1...A jB1...B j =12 j A

i1...i jσi1 A1B1 · · ·σi j A jB j (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 theproperties of the Pauli matrices and are therefore in the algebra su(2). Therefore,

UσiU−1 = Λ ji(U)σ j . (2.76)

That is,

Λij(U) =

trσiUσ jU−1 (2.77)

The matrix Λ(U) depends on the matrix U and is, in fact, in S O(3). For infinites-imal transformations

U = exp(Xkyk) = I +i2σkyk + · · · (2.78)

the matrix Λ isΛi j = δi j + εi jkyk + · · · =

[exp(Xkyk)

(2.79)

It is easy to see thatΛ(I) = Λ(−I) = I . (2.80)

So, in shortΛ(exp(Xiyi)) = exp(Λ(Xi)yi) (2.81)

where[Λ(σk)]i

j = 2iεik j (2.82)

3 Invariant Vector FieldsLet yi be the canonical coordinates on the group S U(2) ranging over (−π, π). Theposition of the indices on the coordinates will be irrelevant, that is, yi = yi . Weintroduce the radial coordinate r = |y| =

√yiyi , so that the geodesic distance

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from the origin is equal to r, and the angular coordinates (the coordinates on S 2)θi = yi = yi/r .

Let us consider the fundamental representation of the group S U(2) with gen-erators Ta = iσa, where σa are 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) = yi pi, (y ∧ p)k = εi jkyi p j and

[T (y)]2 = −|y|2I (3.85)

Therefore,exp[rT (y)] = cos r + T (y) sin r (3.86)

where y is a unit vector. Next, we compute

exp[T (F(rq, ρp))] = cos r cos ρ − (q, p) sin r sin ρ+ cos r sin ρT ( p) + cos ρ sin rT (q)− sin r sin ρT (q ∧ p) (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 ofthe 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) =∂

∂yb Fα(y, x)∣∣∣∣y=0

, (3.90)

Rαb(x) =

∂yb Fα(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 letD = XR , D−1 = YL (3.93)

so thatL = RD−1, X = DY (3.94)

Let Ca be the matrices of the adjoint representation defined by

(Ca)bc = −2εb

ac (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)

exp(C) − I, R =

CI − exp(−C)

. (3.98)

It is easy to see also that

xa = Labxb = Ra

bxb = Xabxb = Ya

bxb . (3.99)

andX = YT , L = RT (3.100)

XY = YX =

(sinh(C/2)

, (3.101)

LR = RL =

sinh(C/2)

, (3.102)

as well as

D = XR = RX = exp(C) , DT = YL = LY = exp(−C) . (3.103)

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Note thatC2 = −4r2Π , (3.104)

where r = |x| andΠi

j = δij − θ

iθ j (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 analyticfunction

f (C) = f (0)(I − Π) + Π12

[f (2ir) + f (−2ir)

C[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

rcos rΠ −

sin2 rr2 C (3.109)

X = I − Π +sin r

rcos rΠ +

sin2 rr2 C (3.110)

R = I − Π + r cot rΠ +12

C (3.111)

L = I − Π + r cot rΠ −12

C (3.112)

D = I − Π + Π cos(2r) +sin(2r)

2rC (3.113)

Then one can show that∂

∂xµFα(x, y) = Lαγ(F(x, y))Xγ

µ(x) , (3.114)

∂xµFα(y, x) = Rα

γ(F(y, x))Yγµ(x) . (3.115)

These matrices satisfy important identities

∂µXaν − ∂νXa

µ = −2εabcXb

µXcν , (3.116)

∂µYaν − ∂νYa

µ = 2εabcYb

µYcν . (3.117)

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Lµa∂µLνb − Lµb∂µLνa = 2εcabLγc , (3.118)

Rµa∂µRν

b − Rµb∂µRν

a = −2εcabRν

c . (3.119)

Let us define the one-forms

σa− = Xa

µ(x)dxµ , σa+ = Ya

µ(x)dxµ .

They satisfy important identities

dσa− = −εa

bcσb− ∧ σ

c− , (3.120)

dσa+ = εa

bcσb+ ∧ σ

c+ . (3.121)

Next, we define the vector fields.

K−a = Lµa(x)∂µ , K+a = Rµ

a(x)∂µ .

These are the right-invariant and the left-invariant vector fields on the group. Theyare related by

K−a = DabK+

b (3.122)

They form two representations of the group S U(2) and satisfy the algebra

[K+a ,K

+b ] = −2εc

abK+c (3.123)

[K−a ,K−b ] = 2εc

abK−c (3.124)[K+

a ,K−b ] = 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)]Tb exp[−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)] = Yai exp[T (ω)]Ta = Yi

aTa exp[T (ω)] , (3.130)

The Killing vectors have important properties

K+a exp[T (ω)] = (exp[T (ω)])Ta (3.131)

K−a 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)

andexp[K−(q)] exp[T (ω)] = exp[T (q)] exp[T (ω)] (3.134)

where K±(p) = paK±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)]∣∣∣∣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)

K+a D = DCa (3.138)

K−a 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)

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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 = δabYaαYb

β = δabXaαXb

β (3.146)gαβ = δabLαaLβb = δabRα

aRβb , (3.147)

that is,

(sinh[C/2]

= I − Π +sin2 r

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)

g1/2 = (det gµν)1/2 =sin2 r

r2 . (3.150)

Thus the bi-invariant volume element of the group is

dvol (x) =sin2 r

r2 dx = sin2 r dr dθ (3.151)

where dx = dx1dx2dx3, and dθ is the volume element on S 2. The invariance ofthe 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 pointsp and q.

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We will always use the orthonormal basis of the right-invariant vector fieldsK+

a and one-forms σa+ and denote the covariant derivative with respect to K+

a sim-ply by ∇a. The Levi-Civita connection ∇ of the bi-invariant metric in the right-invariant basis is defined by

∇aK+b = εc

baK+c (3.153)

so that the coefficients of the affine connection are

ωabc = σa

+(∇cK+b ) = εa

bc . (3.154)

so that for an arbitrary vector

∇aV =[(K+

a Vb) + εbcaVc

b (3.155)

Then∇aK−d = Mb

daK+b (3.156)

Mbda = K+

a Ddb + εb

= −2εbcaDd

c + εbcaDd

c = −εbcaDd

c (3.157)

so that∇aK−d = −εb

caDdcK+

b (3.158)

Now, the curvature tensor is

Rabcd = −ε f

f b . (3.159)

The Ricci curvature tensor and the scalar curvature are

Rab = 2δab , R = 6 . (3.160)

The scalar Laplacian is

∆0 = δabK−a K−b = δabK+a K+

b . (3.161)

One can show that∆0|ω|

sin |ω|=|ω|

sin |ω|(3.162)

Further,∆0 exp[T (ω)] = (exp[T (ω)])T 2 = T 2 exp[T (ω)] (3.163)

su2.tex; November 29, 2011; 10:07; p. 20

where T 2 = TaT a. This means that

exp[t∆20] exp[T (ω)] = exp(tT 2) exp[T (ω)] (3.164)

The Killing vector fields K±a are divergence free, which means that they areanti-self-adjoint with respect to the invariant measure dvol (ω). that is, Thus theLaplacian ∆0 is self-adjoint with respect to the same measure as well.

Let ∇ be the total connection on a vector bundle V realizing the representationG. Let us define the one-forms

A = σa+Ga = GaYa

µdxµ. (3.165)

and F = dA +A∧A. Then, by using the above properties of the right-invariantone-forms we compute

F =12Fabσ

a+ ∧ σ

12FabYa

µYbνdxµ ∧ dxν (3.166)

whereFab = εc

abGc (3.167)

ThusA 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 = (K+a + 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 = K+a Q + [Ga,Q] (3.169)

Therefore, we find∇aGb = εc

baGc + [Ga,Gb] = 0 . (3.170)

Then the derivatives along the left-invariant vector fields are

∇K−a ϕ = (K−a + Ba)ϕ (3.171)

∇K−a Q = K−a Q + [Ba,Q] (3.172)

whereBa = Da

bGb (3.173)

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The Laplacian takes the form

∆ = ∇a∇a = (K+

a + Ga)(K+a + Ga) = ∆0 + 2GaK+

a + 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 ∇a do not form a rep-resentation of the algebra S U(2). The operators that do are the covariant Liederivatives. The covariant Lie derivatives along a Killing vector ξ of sections ofthis vector bundle are defined by

Lξ = ∇ξ −12σa

+(∇bξ)εbcaGc (3.175)

By denoting the Lie derivatives along the Killing vectors K±a by K±a this gives forthe right-invariant and the left-invariant bases

K+a = LK+

a = ∇a + Ga = K+a + 2Ga . (3.176)

K−a = LK−a = ∇K−a − Ba = K−a . (3.177)

It is easy to see that these operators form the same algebra S U(2) × S U(2)

[K+a ,K

+b ] = −2εc

abK+c (3.178)

[K−a ,K−b ] = 2εc

abK−c (3.179)

[K+a ,K

−b ] = 0 (3.180)

Now, the Laplacian is given now by the sums of the Casimir operators

∆ = K2 −G2 (3.181)

whereK2 =

+ +12K2− (3.182)

and K2± = K±aK

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

su2.tex; November 29, 2011; 10:07; p. 22

above. That is why, we will evaluate the heat kernel when one point is fixed at theorigin. We will denote it simply by U(t; y). We obviously have

exp(t∆) = exp(−tG2

). (4.1)

Therefore, the heat kernel is equal to

U(t, y) = exp(−tG2

( t2, y

)(4.2)

where Ψ(t, y) = exp(2tK2)δ(y) is the heat kernel of the operator 2K2.

4.1 Scalar Heat KernelLet

Φ(t, ω) = (4πt)−3/2et∞∑

n=−∞

|ω| + 2πnsin |ω|

exp(−

(|ω| + 2πn)2

)(4.3)

One can show by direct calculation that this function satisfies the equation

∂tΦ = ∆20Φ (4.4)

with the initial conditionΦ(0, ω) = δS 3(ω) (4.5)

Therefore, this is the scalar heat kernel on the group S U(2) and, therefore, on S 3.Now, let us compute the integral

Ψ(t) =

∫S U(2)

dvol (ω)Φ(t, ω) exp[T (ω)] (4.6)

Obviously, Ψ(0) = 1. Next, we have

∂tΨ =

∫S U(2)

dvol (ω) exp[T (ω)]∆0Φ(t, ω) (4.7)

Now, by integrating by parts we get

∂tΨ =

∫S U(2)

DωΦ(t, ω)∆0 exp[T (ω)] (4.8)

which gives∂tΨ = ΨT 2 (4.9)

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Thus, we obtain a very important equation

exp(tT 2) =

∫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(tT 2) =

∫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)

orΦ(t, F(p, q)) = Φ(t, F(q, p)). (4.13)

Also, we see that

exp[(t + s)T 2] =

∫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 7→ −p we obtain

exp[(t + s)T 2] =

∫S U(2)

dvol (z)∫

S U(2)dvol (p)Φ(t, F(p, z))Φ(s, p) exp[T (z)]

(4.15)which means that∫

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∫S U(2)

dvol (p)Φ(t, p)Φ(t, p) = Φ(t, 0) (4.17)

su2.tex; November 29, 2011; 10:07; p. 24

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