Nakahara M. Geometry, Topology and Physics

Подождите немного. Документ загружается.

The exterior product ω ∧∗η is an m-form:

ω ∧∗η =

(r!)

...µ

...ν

√

|g|

(m −r )!

...ν

r+1

...µ

× dx

∧ ...∧ dx

∧ dx

r+1

∧ ...∧ dx



µν

...µ

...ν

r!(m −r)!

...ν

r+1

...µ

× ε

...µ

r+1

...µ



|g|dx

∧ ...∧ dx

...µ



|g|dx

∧ ...∧ dx

. (7.178)

This expression shows that the product is symmetric:

ω ∧∗η = η ∧∗ω. (7.179)

Let {

} be the non-coordinate basis and

ω =

...α

∧ ...∧

η =

...α

∧ ...∧

Equation (7.178) is rewritten as

ω ∧∗η =

...α

∧ ...∧

. (7.180)

Since α ∧∗β is an m-form, its integral over M is well deﬁned. Deﬁne the

inner product (ω, η) of two r-forms by

(ω, η) ≡



ω ∧∗η



...µ



|g|dx

...dx

. (7.181)

Since ω ∧∗η = η ∧∗ω, the inner product is symmetric,

(ω, η) = (η, ω). (7.182)

If (M, g) is Riemannian, the inner product is positive deﬁnite,

(α, α) ≥ 0. (7.183)

where the equality holds only when α = 0. This is not true if (M, g) is Lorentzian.

7.9.4 Adjoints of exterior derivatives

Deﬁnition 7.6. Let d : 

r−1

(M) → 

(M) be the exterior derivative operator.

The adjoint exterior derivative operator d

†

: 

(M) → 

r−1

(M) is deﬁned by

†

= (−1)

mr+m+1

∗ d∗ (7.184a)

if (M, g) is Riemannian and

†

= (−1)

mr+m

∗ d∗ (7.184b)

if Lorentzian, where m = dim M.

In summary, we have the following diagram (for a Riemannian manifold),



m−r

(M)

(−1)

mr+m+1

−−−−−−−−−−→ 

m−r+1

(M)



∗



∗



(M)

†

−−−−−−−−−−→ 

r−1

(M).

(7.185)

The operator d

†

is nilpotent since d is: d

†2

=∗d ∗∗d∗∝∗d

∗=0.

Theorem 7.5. Let (M, g) be a compact orientable manifold without a boundary

and α ∈ 

(M), β ∈ 

r−1

(M).Then

(dβ, α) = (β, d

†

α). (7.186)

Proof. Since both dβ ∧∗α and β ∧∗d

†

α are m-forms, their integrals over M are

well deﬁned. Let d act on β ∧∗α,

d(β ∧∗α) = dβ ∧∗α − (−1)

β ∧ d ∗ α.

Suppose (M, g) is Riemannian. Noting that d ∗ α is an (m − r + 1)-form and

inserting the identity map (−1)

(m−r+1)[m−(m−r+1)]

∗∗=(−1)

mr+m+r+1

∗∗in

front of d ∗ α in the second term, we have

d(β ∧∗α) = dβ ∧∗α − (−1)

mr+m+1

β ∧∗(∗d ∗α).

Integrating this equation over M,wehave



dβ ∧∗α −



β ∧∗[(−1)

mr+m+1

∗ d ∗ α]=



d(β ∧∗α)



∂ M

β ∧∗α = 0

where the last equality follows by assumption. This shows that (dβ, α) =

(β, d

†

α). The reader should check how the proof is modiﬁed when (M, g) is

Lorentzian.

7.9.5 The Laplacian, harmonic forms and the Hodge decomposition

theorem

Deﬁnition 7.7. The Laplacian  : 

(M) → 

(M) is deﬁned by

 = (d + d

†

)

= dd

†

+ d

†

d. (7.187)

As an example, we obtain the explicit form of  : 

(M) → 

(M).Let

f ∈

(M).Sinced

†

f = 0, we have

 f = d

†

d f =−∗d ∗(∂

f dx

)

=−∗d



√

|g|

(m − 1)!

∂

µλ

λν

...ν

∧ ...∧ dx



=−∗

(m − 1)!

∂

[



|g|g

λµ

∂

f ]ε

λν

...ν

∧ dx

∧ ...∧ dx

=−∗∂

[



|g|g

νµ

∂

f ]g

−1

∧ ...∧ dx

=−

√

|g|

∂

[



|g|g

νµ

∂

f ]. (7.188)

Exercise 7.21. Take a one-form ω = ω

in the Euclidean space (

,δ).

Show that

ω =−



µ=1

∂

∂x

Example 7.16. In example 5.11, it was shown that half of the Maxwell equations

are reduced to the identity, dF = d

A = 0, where A = A

is the vector

potential one-form and F = dA is the electromagnetic two-form. Let ρ be the

electric charge density and j the electric current density and form the current one-

form j = η

µν

=−ρ dt + j · dx. Then the remaining Maxwell equations

become

†

F = d

†

dA = j. (7.189a)

The component expression is

∇·E = ρ ∇×B −

∂ E

∂t

= j. (7.189b)

The vector potential A has a large number of degrees of freedom and we can

always choose an A which satisﬁes the Lorentz condition d

†

A = 0. Then

(7.189a) becomes (dd

†

+ d

†

d)A = A = j .

Let (M, g) be a compact Riemannian manifold. The Laplacian  is a

positive operator on M in the sense that

(ω, ω) = (ω, (d

†

d + dd

†

)ω) = (dω, dω) + (d

†

ω, d

†

ω) ≥ 0 (7.190)

where (7.183) has been used. An r -form ω is called harmonic if ω = 0and

closed (coclosed)ifdω = 0 (d

†

ω = 0). The following theorem is a direct

consequence of (7.190).

Theorem 7.6. An r -form ω is harmonic if and only if ω is closed and coclosed.

An r -form ω is called coexact if it is written globally as

= d

†

r+1

(7.191)

where β

r+1

∈ 

r+1

(M) [cf a form ω

∈ 

(M) is exact if ω

= dα

r−1

∈ 

r−1

(M)]. We denote the set of harmonic r-forms on M by Harm

(M)

and the set of exact r-forms (coexact r -forms) by d

r−1

(M)(d

†



r+1

(M)).

[Note: The set of exact r-forms has been denoted by B

(M) so far.]

Theorem 7.7. (Hodge decomposition theorem)Let(M, g) be a compact

orientable Riemannian manifold without a boundary. Then 

(M) is uniquely

decomposed as



(M) = d

r−1

(M) ⊕d

†



r+1

(M) ⊕ Harm

(M). (7.192a)

[That is, any r-form ω

is written globally as

= dα

r−1

+ d

†

r+1

+ γ

(7.192b)

where α

r−1

∈ 

r−1

(M), β

r+1

∈ 

r+1

(M) and γ

∈ Harm

(M).]

If r = 0, we deﬁne 

−1

(M) ={0}. The proof of this theorem requires the

results of the following two easy exercises.

Exercise 7.22. Let (M, g) be as given in theorem 7.7. Show that

(dα

r−1

, d

†

r+1

) = (dα

r−1

,γ

) = (d

†

r+1

,γ

) = 0. (7.193)

Show also that if ω

∈ 

(M) satisﬁes

(dα

r−1

,ω

) = (d

†

r+1

,ω

) = (γ

,ω

) = 0 (7.194)

for any dα

r−1

∈ d

r−1

(M),d

†

r+1

∈ d

†



r+1

(M) and γ

∈ Harm

(M),then

= 0.

Exercise 7.23. Suppose ω

∈ 

(M) is written as ω

= ψ

for some ψ

∈



(M). Show that (ω

,γ

) = 0foranyγ

∈ Harm

(M). The proof of the

converse ‘if ω

is orthogonal to any harmonic r-form, then ω

is written as ψ

for some ψ

∈ 

(M)’ is highly technical and we just state that the operator 

−1

(the Green function) is well deﬁned in the present problem and ψ

is given by



−1

Let P : 

(M) → Harm

(M) be a projection operator to the space of

harmonic r-forms. Take an element ω

∈ 

(M).Sinceω

− Pω

is orthogonal

to Harm

(M), it can be written as ψ

for some ψ

∈ 

(M).Thenwehave

= d(d

†

) + d

†

(dψ

) + Pω

. (7.195)

This realizes the decomposition of theorem 7.7.

7.9.6 Harmonic forms and de Rham cohomology groups

We show that any element of the de Rham cohomology group has a unique

harmonic representative. Let [ω

]∈H

(M). We ﬁrst show that ω

∈

Harm

(M) ⊕ d

r−1

(M). According to (7.192b), ω

is decomposed as ω

+ dα

r−1

+ d

†

r+1

.Sincedω

= 0, we have

0 = (dω

,β

r+1

) = (dd

†

r+1

,β

r+1

) = (d

†

r+1

, d

†

r+1

This is satisﬁed if and only if d

†

r+1

= 0. Hence, ω

= γ

+dα

r−1

. From (7.195)

we have

= Pω

+ d(d

†

ψ) = Pω

+ dd

†



−1

. (7.196a)

≡ Pω

is the harmonic representative of [ω

].Let(ω

be another representative

of [ω

]: (ω

−ω

= dη

r−1

, η

r−1

∈ 

r−1

(M). Corresponding to (7.196a), we have

(ω

= P(ω

+ d(d

†



−1

(ω

) = Pω

+ d(...) (7.196b)

where the last equality follows since dη

r−1

is orthogonal to Harm

(M) and hence

its projection to Harm

(M) vanishes. (7.196a) and (7.196b) show that [ω

] has a

unique harmonic representative Pω

This proof shows that H

(M) ⊂ Harm

(M). Now we prove that H

(M) ⊃

Harm

(M).Sincedγ

= 0foranyγ

∈ Harm

(M),weﬁndthatZ

(M) ⊃

Harm

(M).WealsohaveB

(M) ∩Harm

(M) =∅since B

(M) = d

r−1

(M),

see (7.192a). Thus, every element of Harm

(M) is a non-trivial member of

(M) and we ﬁnd that Harm

(M) is a vector subspace of H

(M) and hence

Harm

(M) ⊂ H

(M). We have proved:

Theorem 7.8. (Hodge’s theorem) On a compact orientable Riemannian manifold

(M, g), H

(M) is isomorphic to Harm

(M):

(M)

∼

Harm

(M). (7.197)

The isomorphism is provided by identifying [ω]∈H

(M) with Pω ∈

Harm

(M).

In particular, we have

dim Harm

(M) = dim H

(M) = b

(7.198)

being the Betti number. The Euler characteristic is given by

χ(M) =



(−1)



(−1)

dim Harm

(M) (7.199)

see theorem 3.7. We note that the LHS is a topological quantity while the RHS is

an analytical quantity given by the eigenvalue problem of the Laplacian .

7.10 Aspects of general relativity

7.10.1 Introduction to general relativity

The general theory of relativity is one of the most beautiful and successful

theories in classical physics. There is no disagreement between the theory

and astrophysical and cosmological observations such as solar system tests,

gravitational radiation from pulsars, gravitational red shifts, the recently

discovered gravitational lens effect and so on. Readers not very familiar with

general relativity may consult Berry (1989) or the primer by Price (1982).

Einstein proposed the following principles to construct the general theory of

relativity

(I) Principle of General Relativity: All laws in physics take the same forms in

any coordinate system.

(II) Principle of Equivalence: There exists a coordinate system in which the

effect of a gravitational ﬁeld vanishes locally. (An observer in a freely falling

lift does not feel gravity until it crashes.)

Any theory of gravity must reduce to Newton’s theory of gravity in the weak-

ﬁeld limit. In Newton’s theory, the gravitational potential  satisﬁes the Poisson

equation

 = 4π Gρ (7.200)

where ρ is the mass density. The Einstein equation generalizes this classical result

so that the principle of general relativity is satisﬁed.

In general relativity, the gravitational potential is replaced by the components

of the metric tensor. Then, instead of the LHS of (7.200), we have the Einstein

tensor deﬁned by

µν

≡ Ric

µν

−

µν

. (7.201)

Similarly, the mass density is replaced by a more general object called the

energy–momentum tensor T

µν

.TheEinstein equation takes a very similar

form to (7.200):

µν

= 8π GT

µν

. (7.202)

The constant 8π G is chosen so that (7.202) reproduces the Newtonian result in

the weak-ﬁeld limit. The tensor T

µν

is obtained from the matter action by the

variational principle. From Noether’s theorem, T

µν

must satisfy a conservation

equation of the form ∇

µν

= 0. A similar conservation law holds for G

µν

(but

not for Ric

µν

). We shall see in the next subsection that the LHS of (7.202) is also

obtained from the variational principle.

Exercise 7.24. Consider a metric

=−1 −

2

= 0 g

= δ

1 ≤ i, j ≤ 3

and T

µν

given by T

= ρc

, T

= T

= 0 which corresponds to dust at

rest. Show that (7.202) reduces to the Poisson equation in the weak-ﬁeld limit

(/c

' 1).

7.10.2 Einstein–Hilbert action

This and the next example are taken from Weinberg (1972). The general theory

of relativity describes the dynamics of the geometry, that is, the dynamics of

µν

. What is the action principle for this theory? As usual, we require that the

relevant action should be a scalar. Moreover, it should contain the derivatives of

µν



√

|g|d

x cannot describe the dynamics of the metric. The simplest guess

will be S

∝



√

|g|d

x.Since is a scalar and

√

|g|dx

...dx

the invariant volume element, S

is a scalar. In the following, we show that

indeed yields the Einstein equation under the variation with respect to the

metric. Our connection is restricted to the Levi-Civita connection. We ﬁrst prove

a technical proposition.

Proposition 7.2. Let (M, g) be a (pseudo-)Riemannian manifold. Under the

variation g

µν

→ g

µν

+ δg

µν

, g

µν

, g and Ric

µν

change as

(a)δg

µν

=−g

µκ

λν

δg

κλ

(7.203)

(b)δg = gg

µν

δg

µν

,δ



|g|=



|g|g

µν

δg

µν

(7.204)

(c)δRic

µν

=∇

δ

νµ

−∇

δ

κµ

(Palatini identity). (7.205)

Proof.(a)Fromg

κλ

λν

= δ

, it follows that

0 = δ(g

κλ

λν

) = δg

κλ

λν

+ g

κλ

δg

λν

Multiplying by g

µκ

we ﬁnd that δg

µν

=−g

µκ

λν

δg

κλ

(b) We ﬁrst note the matrix identity ln(det g

µν

) = tr(ln g

µν

). This can be

proved by diagonalizing g

µν

. Under the variation δg

µν

, the LHS becomes δg ·g

−1

while the RHS yields g

µν

· δg

µν

, hence δg = gg

µν

δg

µν

. The rest of (7.204) is

easily derived from this.

(

 be two connections. The difference δ ≡

(

 −  is a tensor

of type (1, 2), see exercise 7.5. In the present case, we take

(

 to be a connection

associated with g + δg and  with g. We will work in the normal coordinate

system in which  ≡ 0 (of course ∂ = 0 in general); see section 7.4. We ﬁnd

δ Ric

µν

= ∂

δ

νµ

− ∂

δ

κµ

=∇

δ

νµ

−∇



κµ

[The reader should verify the second equality.] Since both sides are tensors, this

is valid in any coordinate system.

We deﬁne the Einstein–Hilbert action by

≡

16π G



√

−g d

x. (7.206)

The constant factor 1/16π G is introduced to reproduce the Newtonian limit when

matter is added; see (7.214). We prove that δS

= 0 leads to the vacuum Einstein

equation. Under the variation g → g + δg such that δg → 0as|x|→0, the

integrand changes as

δ(

√

−g) = δ(g

µν

Ric

µν

√

−g)

= δg

µν

Ric

µν

√

−g + g

µν

δ Ric

µν

√

−g + δ(

√

−g)

=−g

µκ

λν

δg

κλ

Ric

µν

√

−g

+ g

µν

(∇

δ

νµ

−∇



κµ

)

√

−g +

√

−gg

µν

δg

µν

We note that the second term is a total divergence,

∇

µν

δ

νµ

√

−g) −∇

µν

δ

κµ

√

−g)

= ∂

µν

δ

µν

√

−g) − ∂

µν

δ

κµ

√

−g)

and hence does not contribute to the variation. From the remaining terms we have

δS

16π G





−Ric

µν



δg

µν

√

−g d

x. (7.207)

If we require that δS

= 0 under any variation δg, we obtain the vacuum Einstein

equation,

µν

= Ric

µν

−

µν

= 0 (7.208)

where the symmetric tensor G is called the Einstein tensor.

So far we have considered the gravitational ﬁeld only. Suppose there exists

matter described by an action

≡



(φ)

√

−g d

x (7.209)

where

(φ) is the Lagrangian density of the theory. Typical examples are the real

scalar ﬁeld and the Maxwell ﬁelds,

≡−



µν

∂

φ∂

φ +m

]

√

−g d

x (7.210a)

≡−



µν

√

−g d

x (7.210b)

where F

µν

= ∂

− ∂

=∇

−∇

. If the matter action changes by

δS

under δg,theenergy–momentum tensor T

µν

is deﬁned by

δS



µν

δg

µν

√

−g d

x. (7.211)

Since δg

µν

is symmetric, T

µν

is also taken to be so. For example, T

µν

of a real

scalar ﬁeld is given by

µν

(x ) = 2

√

−g

δg

µν

(x )

= ∂

φ∂

φ −

µν

κλ

∂

φ∂

φ +m

). (7.212)

Suppose we have a gravitational ﬁeld coupled with a matter ﬁeld whose

action is S

. Now our action principle is

δ(S

+ S

) = 0 (7.213)

under g → g + δg. From (7.207) and (7.211), we obtain the Einstein equation

µν

= 8π GT

µν

. (7.214)

Exercise 7.25. We may add an extra scalar to the scalar curvature without spoiling

the invariance of the action. For example, we can add a constant called the

cosmological constant ,

(

16π G



( + )

√

−g d

x. (7.215)

Write down the vacuum Einstein equation. Other possible scalars may be such

terms as

, Ric

µν

Ric

µν

or R

κλµν

7.10.3 Spinors in curved spacetime

For concreteness, we consider a Dirac spinor ψ in a four-dimensional Lorentz

manifold M. The vierbein e

deﬁned by

µν

= e

αβ

(7.216)

deﬁnes an orthonormal frame {

= e

} at each point p ∈ M. As noted

before, α,β,γ,... are the local orthonormal indices while µ,ν,λ,... are the

coordinate indices. With respect to this frame, the Dirac matrices γ

= e

satisfy {γ

,γ

}=2η

αβ

. Under a local Lorentz transformation 

( p),the

Dirac spinor transforms as

ψ(p) → ρ()ψ(p)

ψ(p) →

ψ(p)ρ()

−1

(7.217)

where

ψ ≡ ψ

†

and ρ() is the spinor representation of . To construct an

invariant action, we seek a covariant derivative ∇

ψ which is a local Lorentz

vector and transforms as a spinor,

∇

ψ → ρ()

∇

ψ. (7.218)

Ifweﬁndsucha∇

ψ, an invariant Lagrangian may be given by



iγ

∇

+ m

ψ (7.219)

m being the mass of ψ. We note that e

∂

ψ transforms under (p) as

∂

ψ → 

∂

ρ()ψ = 

[ρ()∂

ψ + ∂

ρ()ψ]. (7.220)

Suppose ∇

is of the form

∇

ψ = e

[∂

+ 

]ψ. (7.221)

From (7.218) and (7.220), we ﬁnd that 

satisﬁes



→ ρ()

ρ()

−1

− ∂

ρ()ρ()

−1

. (7.222)

To ﬁnd the explicit form of 

, we consider an inﬁnitesimal local Lorentz

transformation 

( p) = δ

+ ε

( p). The Dirac spinor transforms as

ψ → exp[

iε

αβ



αβ

]ψ [1 +

iε

αβ



αβ

]ψ (7.223)

where 

αβ

≡

,γ

is the spinor representation of the generators of the

Lorentz transformation. 

αβ

satisﬁes the (1, 3) Lie algebra

i[

αβ

,

γδ

]=η

γβ



αδ

− η

γα



βδ

+ η

δβ



γα

− η

δα



γβ

. (7.224)

Under the same Lorentz transformation, 

transforms as



→ (1 +

iε

αβ



αβ

)

(1 −

iε

γδ



γδ

) −

i∂

αβ



αβ

(1 −

iε

γδ



γδ

)

= 

iε

αβ

[

αβ

,

]−

i∂

αβ



αβ

. (7.225)

We recall that the connection one-form ω

transforms under an inﬁnitesimal

Lorentz transformation as (see (7.152))

→ ω

+ ε

− ω

− dε

(7.226a)

or in components,



µβ

→ 

µβ

+ ε



µβ

− 

µγ

− ∂

. (7.226b)

From (7.224), (7.225) and (7.226b), we ﬁnd that the combination



≡

i



αβ

∇

βν



αβ

(7.227)

satisﬁes the transformation property (7.222). In fact,

i



αβ

→

i(

+ ε



− 

µγ

γβ

− ∂

αβ

)

αβ

i



αβ

i(ε





αβ

− 

µγ

γβ



αβ

)

−

i∂

αβ



αβ

i



αβ

iε

αβ

[

αβ

i



γδ

]−

i∂

αβ



αβ