De Rham Cohomology and relation toPhysics

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August 5, 2017

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De Rham Cohomology

Definition 1.1

A differential form ω on a manifold M is said to be closed if dω = 0, andexact if ω = dτ for some τ of degree one less.

Corollary 1.2

Since d2 = 0, every exact form is closed.

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Remark 1.3

Whether every closed form on a manifold is exact depends on thetopology of the manifold.

For instance, on R2 every closed k-form is exact for k > 0.

But on the punctured plane R2 − {(0, 0)} there are closed 1-formsthat are not exact. (e.g. ω = x dy−y dx

x2+y2 )

The extent to which closed forms are not exact is measured by the deRham cohomology.

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Definition 1.4

Let Z k(M) be the vector space of all closed k-forms on M.Let Bk(M) be the vector space of all exact k-forms on M.Since every exact form is closed, hence Bk(M) ⊆ Z k(M).The de Rham cohomology of M in degree k is defined as the quotientvector space

Hk(M) := Z k(M)/Bk(M).

The quotient vector space construction induces an equivalence relation onZ k(M):w ′ ∼ w in Z k(M) iff w ′ − w ∈ Bk(M) iff w ′ = w + dτ for some exactform dτ .The equivalence class of a closed form ω is called its cohomology classand denoted by [ω].

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Proposition 1.5

If a manifold M has k connected components, then its de Rhamcohomology in degree 0 is H0(M) ∼= Rk .

Proof.

Since there are no nonzero exact 0-forms,

H0(M) ∼= Z 0(M) = {closed 0-forms}.

Let f be a closed 0-form on M; i.e. f is a smooth function on M suchthat df = 0. On any chart (U, x1, . . . , xn), we have

df =n∑

i=1

∂f

∂x idx i .

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Proof.

Hence

df = 0 on U ⇐⇒ ∂f

∂x i≡ 0 (for all i) on U

⇐⇒ f is locally constant on U.

∴ {closed 0-forms on M} = {locally constant functions on M}.A locally constant function is constant on each connected componentof M. If M has k connected components, then a locally constantfunction on M can be specified by an ordered k-tuple of real numbers.∴ Z 0(M) ∼= Rk .

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Proposition 1.6

Let M be a manifold of dimension n. Then Hk(M) = 0 for k > n.

Proof.

For any point p ∈ M, the tangent space TpM is a vector space ofdimension n.

Let ω be a k-form on M, then the k-covector ωp ∈ Ak(TpM), thespace of alternating k-linear functions on TpM.

Let {dx1, . . . , dxn} be the dual basis for T ∗pM, then

{dx i1 ∧ · · · ∧ dx ik} form a basis for Ak(TpM). In dx i1 ∧ · · · ∧ dx ik , atleast two of the factors must be the same, say dx j = dx l . Sincedx j ∧ dx l = 0, so dx i1 ∧ · · · ∧ dx ik = 0. Thus Ak(TpM) = 0.

Hence, for k > n, the only k-form on M is the zero k-form.

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De Rham Cohomology of the Real Line

Since the real line R1 is connected, H0(R1) = R.

For dimensional reasons, on R1 there are no nonzero 2-forms. Henceevery 1-form on R1 is closed.

A 1-form f (x) dx on R1 is exact iff ∃ a C∞ function g(x) on R1 s.t.

f (x) dx = dg = g ′(x) dx .

By FTC, we may take g(x) =∫ x0 f (t) dt. Thus every 1-form on R is

exact. Therefore, H1(R1) = Z 1(R1)/B1(R1) = 0.

Summary

Hk(R1) =

{R for k = 0,

0 for k ≥ 1.

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Mayer-Vietoris sequence

Definition 1.7

The Mayer-Vietoris sequence is the following long exact sequence

. . .j∗−→ Hk−1(U ∩ V )

d∗−→ Hk(M)

i∗−→ Hk(U)⊕ Hk(V )j∗−→ Hk(U ∩ V )

d∗−→ Hk+1(M)

i∗−→ . . . .

where

i∗[σ] = [i(σ)] = ([σ|U ], [σ|V ]) ∈ Hk(U)⊕ Hk(V ),

j∗([ω], [τ ]) = [j(ω, τ)] = [τ |U∩V − ω|U∩V ] ∈ Hk(U ∩ V ).

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De Rham Cohomology of the Circle

Cover the circle S1 with two open arcs U and V as shown.

The intersection U ∩ V is the disjoint union of two open arcs, labelledA and B.

Figure: {U,V } is an open cover of S1.

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Since an open arc is diffeomorphic to an open interval and hence toR1,

H∗(U) ∼= H∗(V ) ∼= H∗(R1),

andH∗(U ∩ V ) ∼= H∗(R1 t R1) ∼= H∗(R1)⊕ H∗(R1).

From the Mayer-Vietoris sequence

0→ H0(S1)i∗−→ H0(U)⊕ H0(V )

j∗−→ H0(U ∩ V )

d∗−→ H1(S1)

i∗−→ H1(U)⊕ H1(V )→ . . .

we get the exact sequence

0→ R i∗−→ R⊕ R j∗−→ R⊕ R d∗−→ H1(S1)→ 0.

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Lemma: Let 0→ A0 d0−→ A1 d1−→ A2 d2−→ · · · → Am → 0 be an exactsequence of finite-dimensional vector spaces. Then

m∑k=0

(−1)k dimAk = 0.

Conclude that 1− 2 + 2− dimH1(S1) = 0, so dimH1(S1) = 1. Hence

Hk(S1) =

{R for k = 0, 1

0 for k ≥ 2.

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Relation to Electromagnetic Field and Potentials

De Rham cohomology can be viewed as a generalization of Maxwell’stheory of electromagnetism.

Definition 1.8

The Maxwell equations for the electric vector field E and the magneticvector field B in a vacuum are:

∇ · E =ρ

ε0∇ · B = 0

∇× E = −∂B∂t

∇× B = µ0(J + ε0∂E

∂t).

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Definition 1.9 (Stationary Maxwell equations)

In the special stationary case, all the functions do not depend on time t.This gives us the stationary Maxwell equations:

∇ · E =ρ

ε0,

∇ · B = 0,

∇× E = 0,

∇× B = µ0J on O,

where O is an open subset of R3.

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Stationary Electric and Magnetic Potential

The stationary Maxwell equations motivate the study of the following fourequations:

∇× E = 0 on O, (1)

∇ · B = 0 on O, (2)

E = −∇U on O, (3)

B = ∇× A on O. (4)

The solutions of these equations depend critically on the topology(i.e. the Betti numbers βk(M) := dimHk(M)) of the open subsetO ⊆ R3.

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Due to the identities

curl gradU = ∇× (∇U) = 0

div curl v = ∇ · (∇× v) = 0

on O, there exists trivial solutions and trivial constraints.

The field E = ∇U (resp. B = ∇× A) is a trivial solution to∇× E = 0 (resp. ∇ · B = 0).

If E = −∇U (resp. B = ∇× A) has a solution, then ∇× E = 0(resp. ∇ · B = 0) are automatically satisfied on O.

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The electric (resp. magnetic) field sees the Betti number β1 (resp.β2).

In other words, the electric (resp. magnetic) field sees the nontrivial1-cycles (resp. 2-cycles) of the set O.

The reason for that is because the electric field corresponds to adifferential 1-form, whereas the magnetic field corresponds to adifferential 2-form.

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Let E = E 1i + E 2j + E 3k, B = B1i + B2j + B3k.

The corresponding differential forms are

ωE = E 1 dx + E 2 dy + E 3 dz ,

ωB = B1 dy ∧ dz + B2dz ∧ dx + B3 dx ∧ dy .

The corresponding exterior derivatives are

dωE = dE 1 ∧ dx + dE 2 ∧ dy + dE 3 ∧ dz

= (E 3y − E 2

z )dy ∧ dz + (E 1z − E 3

x )dz ∧ dx + (E 2x − E 1

y )dx ∧ dy

= (curl E)1dy ∧ dz + (curl E)2dz ∧ dx + (curl E)3dx ∧ dy

dωB = dB1 ∧ dy ∧ dz + dB2 ∧ dz ∧ dx + dB3 ∧ dx ∧ dy

= (B1x + B2

y + B3z )dx ∧ dy ∧ dz

= divB · dx ∧ dy ∧ dz .

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Note that

d(dωE) = 0 is equivalent to div curl E = 0.

d(dωB) = 0 is always satisfied.

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Summary

Remark 1.10

In

E = −∇U,B = ∇× A,

U is called a scalar potential, A is called a vector potential.

De Rham cohomology helps to determine necessary and sufficientconditions for the existence of the potentials of stationary electric andmagnetic fields.

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References I

John M. Lee.Introduction to Smooth Manifolds.Springer, 2013.

Loring W. Tu.An Introduction to Manifolds.Springer, 2010.

Eberhard Zeidler.Quantum Field Theory III: Gauge Theory.Springer, 2011.

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The End

Thank you very much for your attention.

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