Calculus of differential forms

A P P E N D I X

B

Calculus of Differential Forms

Contents B.1

B.2

Exterior Differentiation of the Forms B.1.1 Exterior differential operator in multidimensional space En B.1.2 Exterior differential operator in four-dimensional space M4 Integration of the Forms B.2.1 Three-dimensional space E3 B.2.2 Beyond three-dimensional space

799 800 802 806 806 808

B.1 Exterior Differentiation of the Forms In the beginning of Chapter 1 we introduced a physical field as any physical value which is a function of the observation point. We have also noticed that physical fields can be characterized by different mathematical objects. For example, we can consider a field formed by a spatial distribution of the p-forms. Definition 44 We say that we have a p-form field Ω(r) in the Euclidean space En , if, for any point r = (x1 , x2 , . . . . xn ) from En , one specific p-form Ω(r) is determined by the following expression: Ω (r) = Ω (r) = (p)

n 

ai1 ,...ip (r) dxi1 ∧ .... ∧ dxip ,

(B.1)

i1 ,...ip =1

where all components ai1 ,...ip are the differentiable functions of the coordinates: ai1 ,...ip (r) = ai1 ,...ip (x1 , x2 , . . . . xn ). In the case of the vector field theory. we have introduced the rules of the vector differential operations on these fields. However, those rules were strictly defined only for three-dimensional space. The advantage of the differential form theory is that we can introduce the differential operations for a p-form field defined in a multidimensional space. The remarkable property of this operation is that, in the 799

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case of three-dimensional space, it reduces to the conventional vector differential operations.

B.1.1 Exterior differential operator in multidimensional space En Let us consider a 0-form Ω = f (r). The differential operator d operating on a (0)

scalar function f (r) is defined as the following 1-form: n  ∂f dxi . df = ∂xi i=1

(B.2)

Thus, operator d is similar to the conventional differential operator applied to the function of many variables. However, we treat the result of this operator application as a 1-form, Ω = (0)

df, with the components given by the partial derivatives, ∂f/∂xi . Obviously, this operation satisfies the Leibniz rule (or chain rule), if we apply it to the product of two functions: d( fg) = fdg + gdf. We can now determine the differential operation for a p-form according to the formula: n 

dΩ = (p)

dai1 ,...ip ∧ dxi1 ∧ . . . . ∧ dxip ,

(B.3)

i1 ,...ip =1

where the differential of the scalar components ai1 ,...ip of the p-form is calculated according to the rule (B.2). The operation described by equation (B.3) is called exterior differentiation. The exterior differential has the following properties. (i) Differential operator d is a linear operator: 

 d c1 Ω +c1 Ψ ( p)

( p)

= c1 d Ω +c1 d Ψ , ( p)

(B.4)

( p)

where c1 and c2 are some scalar coefficients. (ii) Differential operator d satisfies a generalized Leibniz rule:  d

 Ω ∧Θ

( p)

(q)

p

= d Ω ∧ Θ + (−1) Ω ∧ d Θ . ( p)

(q)

( p)

(q)

(B.5)

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B.1 Exterior Differentiation of the Forms

(iii) Double differential operator d is always equal to zero for any p-form:   d d Ω = 0. ( p)

(B.6)

The proof of properties (i) and (ii) is straightforward, and we leave this to the reader. The property (iii) of the exterior differentiation operator seems to be very unusual from the first look. However, it can be easily proved as well. Proof of property (iii): According to definition (B.3), we have ⎞ ⎛   n  dai1 ,...ip ∧ dxi1 ∧ .... ∧ dxip ⎠, d d Ω = d⎝ (p)

(B.7)

i1 ,...ip =1

where the differential dai1 ,...ip is determined by expression (B.2): dai1 ,...ip =

n  ∂ai1 ,...ip

∂xi

i=1

dxi .

(B.8)

Substituting (B.8) back into (B.7), and applying the rule (B.2) again, we find: ⎡ ⎤   n n  n 2   ∂ ai1 ,...ip ⎣ dxj ∧ dxi ⎦ ∧ dxi1 ∧ .... ∧ dxip . (B.9) d dΩ = ∂x ∂x (p) i j i ,...i =1 i=1 j=1 1

p

Note that the term in the square brackets is equal to zero, n  n  ∂ 2 ai1, ...ip i=1 j=1

∂xi ∂xj

dxj ∧ dxi = 0,

because for the differentiable functions the second order partial derivatives are equal, ∂ 2 ai1, ...ip ∂ 2 ai1, ...ip = , ∂xi ∂xj ∂xj ∂xi while the exterior product of two differentials is anticommutative, dxj ∧ dxi = −dxi ∧ dxj . Hence,

  d d Ω = 0. (p)

We will give the following definitions.



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Definition 45 We say that the differential p-form Ω (r) is closed if dΩ (r) = 0. Definition 46 A differential p-form Ψ (r) is exact if there exists a (p − 1)-form Φ (r) such that Ψ (r) = dΦ (r) . Note that, from property (iii) of the exterior differential operator, it follows immediately that every exact form is closed. This simple fact is called Poincare’s lemma. The inverse statement of Poincare’s lemma is given by the following theorem. Theorem 47 de Rham’s theorem: if Ω is a p-form (1 ≤ p < n) in multidimen( p)

sional space En and d Ω = 0, there exist ( p − 1)-form Ω =d Ω .

( p)

( p)

Ω

( p−1)

with the property that

( p−1)

B.1.2 Exterior differential operator in four-dimensional space M4 In four-dimensional Minkowskian space the exterior differential operator d = dM4 can be formally written as a 1-form: dM4 = dx

∂ ∂ ∂ ∂ ∂ + dy + dz + dτ = dE3 + dτ . ∂x ∂y ∂z ∂τ ∂τ

(B.10)

Applying this operator to the 0-form field, f (r, t) , which is a scalar field in the space M4 , we have: df = dM4 f = dE3 f + dτ = gradf · dr + dτ

∂ f ∂τ

∂ f. ∂τ

(B.11)

The exterior differentiation of a 1-form field, Ω = ϕ · dr + fdτ , is calculated as: (1)



 ∂ d Ω = dM4 ∧ Ω = dE3 + dτ ∧ (ϕ · dr + fdτ ) ∂τ (1) (1)   ∂ϕ = dE3 ∧ (ϕ · dr) + dτ ∧ · dr + dE3 f ∧ dτ , ∂t where we use the linear property (i) of the d operator. Note that, according to equation (1.81) we have: dE3 ∧ (ϕ · dr) = curlϕ · dΣ

(B.12)

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B.1 Exterior Differentiation of the Forms

and d Ω = curlϕ · dΣ + dτ ∧ (1)

= curlϕ · dΣ + dτ ∧

 

∂ϕ ∂t



· dr

+ (gradf · dr) ∧ dτ

 ∂ϕ −gradf · dr, ∂t

(B.13)

where we took into account the anticommutative law (A.8), according to which (gradf · dr) ∧ dτ = −dτ ∧ (gradf · dr),

because (gradf · dr) and dτ are 1-forms. In a similar way we can calculate the exterior derivative of the 2-form field:  d Ω = dM4 ∧ Ω = (2)

(2)

dE3 + dτ

∂ ∂τ



∧ [ψ · dΣ + (η · dr) ∧ dτ ]

∂ ∧ [ψ · dΣ + (η · dr) ∧ dτ ] ∂τ ∂ ∧ (ψ · dΣ) . ∧ (η · dr)] ∧ dτ + dτ ∂τ

= dE3 ∧ [ψ · dΣ + (η · dr) ∧ dτ ] + dτ = dE3 ∧ ψ · dΣ + [dE3

According to equations (1.81) and (1.82), dE3 ∧ ψ · dΣ = (divψ ) dv and dE3 ∧ (η · dr) = curlη · dΣ. Therefore, we finally have d Ω = dM4 ∧ Ω = (2)

(2)

= (divψ ) dv + (curlη · dΣ) ∧ dτ + dτ ∧

 = (divψ ) dv + dτ ∧

∂ ψ + curlη ∂τ





 ∂ ψ · dΣ ∂τ

· dΣ,

(B.14)

where we took into account the graded commutativity law (A.47), according to which (curlη · dΣ) ∧ dτ = (−1)

2×1

dτ ∧ (curlη · dΣ) = dτ ∧ (curlη · dΣ) ,

because (curlη · dΣ) is a 2-form, and dτ is a 1-form.

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Calculus of Differential Forms

For an exterior differential of the 3-form in the Minkowskian space, taking into account equation (1.82), we find:   ∂ ∧ [hdv + (χ · dΣ) ∧ dτ ] d Ω = dM4 ∧ Ω = dE3 + dτ ∂τ (3) (3) = dE3 ∧ (χ · dΣ) ∧ dτ + dτ

∂ ∧ hdv ∂τ

 ∂ ∂ = (divψ ) dv ∧ dτ + dτ ∧ hdv = dτ ∧ h − divψ dv. ∂τ ∂τ

(B.15)

In the last formula we took into account the graded commutativity law (A.47), according to which 3×1

(divψ ) dv ∧ dτ = (−1)

dτ ∧ (divψ ) dv = −dτ ∧ (divψ ) dv,

because (divψ ) dv is a 3-form, and dτ is a 1-form. Finally, it is evident based on the definition that the exterior derivative of the 4-form in a four-dimensional space is just zero:   ∂ ∧ (θdv ∧ dτ ) = 0. d Ω = dM4 ∧ Ω = dE3 + dτ ∂τ (4) (4) For convenience, we summarize these results in the following table: 0-forms: dM4 Ω = gradf · dr + dτ (0)

∂ f ∂τ

(B.16) 

 ∂ϕ − gradf · dr, 1-forms: dM4 Ω = curlϕ · dΣ + dτ ∧ ∂t (1)   ∂ ψ + curlη · dΣ, 2-forms: dM4 Ω = (divψ ) dv + dτ ∧ ∂τ (2)

 ∂ h − divψ dv, 3-forms: dM4 Ω = dτ ∧ ∂τ (3) 4-forms: dM4 Ω = 0. (4)

(B.17) (B.18) (B.19) (B.20)

Finally, we will present the same equations, (B.16) through (B.20), in a way, which is more convenient for application of these formulas in field theory. We introduce the arbitrary differentiable scalar functions U (r, τ ) and q (r, τ ) and the

B.1 Exterior Differentiation of the Forms

805

vector functions A (r, τ ), H (r, τ ) , D (r, τ ), and j (r, τ ), in the Minkowskian space, M4 . Using these functions, we can define differential forms of five different orders in the four-dimensional space M4 . According to equation (A.56), these forms can be expressed using the three-dimensional vector notations as: 0-forms: 1-forms: 2-forms: 3-forms: 4-forms:

Ω = U,

(B.21)

Ω = A · dr − Udτ ,

(B.22)

Ω = D · dΣ − (H · dr) ∧ dτ ,

(B.23)

Ω = qdv − (j · dΣ) ∧ dτ ,

(B.24)

Ω = qdv ∧ dτ .

(B.25)

(0)

(1)

(2)

(3)

(4)

Note that we use the negative sign in the second term of equations (B.23) and (B.24) in order to have a structure of the differential form equations similar to the Maxwell’s equations (see Chapters 2 and 3). Using these notations, we can rewrite expressions (B.16) through (B.20) for the exterior derivatives of these forms as ∂ Udτ , ∂t (0) (1)   ∂A 1-forms: dM4 Ω = Φ = curlA · dΣ − gradU + · dr ∧ dτ , ∂t (1) (2)   ∂ D − curlH · dΣ ∧ dτ , 2-forms: dM4 Ω = Φ = (divD) dv + ∂t (2) (3) 

∂ q dv ∧ dτ 3-forms: dM4 Ω = Φ = − divj + ∂τ (3) (4) 0-forms: dM4 Ω = Φ = gradU · dr +

4-forms: dM4 Ω = 0. (4)

(B.26) (B.27) (B.28) (B.29) (B.30)

Equations (B.26) through (B.30) constitute the fundamental system of differential form equations in the Minkowskian space, M4 .

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Calculus of Differential Forms

B.2 Integration of the Forms B.2.1 Three-dimensional space E3 We begin our analysis of the basic rules of integration with the differential forms in three-dimensional Euclidean space, E3 . We can recall that 1-, 2-, and 3-forms were introduced as differential expressions, which were integrated over a curve L, surface S, or volume D, respectively. Therefore, according to equations (1.69), (1.70), and (1.71), we can introduce a linear integral of a 1-form, a surface integral of a 2-form, and a volume integral of a 3-form as:      ϕ = ϕx dx + ϕy dy + ϕz dz = ϕ · dr, (B.31) (1)

 L

L 

 S

S 

 θdx ∧ dy ∧ dz =

θ =

(3) D





ψyz dy ∧ dz + ψzx dz ∧ dx + ψxy dx ∧ dy =

ψ =

(2)

L



D

ψ · dΣ, (B.32)

S

θdv.

(B.33)

D

Let us assume that we are given an exact 1-form, Ω = dE3 f,

(B.34)

(1)

where f = f (r) is a scalar field. Substituting equation (1.80) into (B.34) and integrating over a curve L, we have: 

 Ω=

 dE3 f =

(1)

L

L

gradf(r) · dr = f (rB ) − f (rA ) ,

(B.35)

L

where rA and rB are two end points of the curve L. We assume also that the exact 2-form is equal to Ω = dE3 ∧ ϕ = dE3 ∧ (ϕ · dr) . (1)

(2)

Integrating this exact 2-form over the surface S according to formula (B.32), we find    curlϕ · dΣ. (B.36) dE3 ∧ ϕ = Ω= (1)

(2)

S

S

S

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B.2 Integration of the Forms

According to Stoke’s theorem of vector calculus (1.44), the scalar flux of the curl of the vector field ϕ through an open surface S is equal to the circulation of this field along the close boundary L of the surface, where the orientations of the curve L and surface S are selected based on the standard convention illustrated in Figure 1.4:    ϕ. (B.37) curlϕ · dΣ = ϕ · dr = (1)

S

L

L

Substituting equation (B.37) into (B.36), we have:   ϕ. dE3 ∧ ϕ =

(B.38)

(1)

(1)

L

S

Let us consider now the exact 3-form equal to: Ω = dE3 ∧ ψ = dE3 ∧ (ψ · dΣ).

(3)

(B.39)

(2)

The volume integral of this 3-form can be calculated, using the Gauss’s theorem (1.38), as follows:    = d ∧ ψ = dE3 ∧ (ψ · dΣ) Ω E3 (3)

(2)

D

D

D



 (divψ ) dv =

=

S

D

ψ · dΣ =

 ψ,

S

(2)

(B.40)

where volume D is enclosed by the surface S, and the positive normal to S according to a standard rule is oriented outside the volume. We can write all three equations, (B.35), (B.38), and (B.40), in a unified symbolic notations, using the language of the differential forms:   (B.41) dE3 ∧ Ω = Ω , k = 1, 2, 3; (k−1)

Ck

∂Ck

(k−1)

where C3 denotes volume D, C2 denotes Surface S, and C1 denotes curve L, while Ck stands for the volume integral (k = 3), surface integral (k = 2), or linear integral (k = 1), respectively. We use a standard notation of multidimensional calculus, ∂Ck , for a boundary of the geometrical element Ck . For example, ∂C3 denotes an oriented (according to the standard rule discussed above) surface, S, of the volume, D. The element ∂C2 denotes an oriented (according to the standard rule discussed above) boundary line, L, of the surface, S. The element ∂C1 is formed by two end points, rA and rB , of the curve L.

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Calculus of Differential Forms

Thus, formula (B.41) combines in one equation all three fundamental theorems of vector field integration: the work theorem 10, and the Stokes’s and Gauss’s theorems, equations (1.44) and (1.38). In other words, equation (B.41) combines together the properties of the vector field work, and scalar and vector fluxes. This formula is often referred to as a Stokes’s theorem for the differential forms. The most important fact is that Stokes’s theorem (B.41) can be naturally extended to multidimensional spaces, thus allowing us to develop a multidimensional analogue of the fundamental integral theorems of vector calculus.

B.2.2 Beyond three-dimensional space Consider now the Euclidean space En . A general differential (p − 1)-form in space En is defined by equation (B.1): n 

Ω (r) = Ω (r) = (p−1)

ai1, ...ip dxi1 ∧ .... ∧ dxip−1 .

i1, ...ip−1 =1

Let us introduce the exact p-form equal to: Ω (r) = d Ω =

(p)

(p−1)

n 

dai1, ...ip−1 ∧ dxi1 ∧ .... ∧ dxip−1 .

(B.42)

i1, ...ip−1 =1

The general Stokes’s theorem states that 

 Ω=

d Ω =

(p) Cp

 (p−1)

Cp

Ω , p = 1, 2, ...n;

∂Cp

(p−1)

(B.43)

where ∂Cp stands for the boundary of the geometrical element Cp from the space En . Note that geometrical element Cp can be treated as a p-dimensional domain in a multidimensional space. The interested reader can find a rigorous definition of multidimensional domains and the proof of the general Stokes’s theorem in textbooks on the mathematical theory of differential forms. In the present text we restrict our discussion to the intuitive explanation of this important mathematical theory. For example, the following formal derivation should give some idea of how we can arrive at the Stokes’s theorem. Indeed, we can substitute representation (B.42) into the left-hand part of equation (B.43),

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B.2 Integration of the Forms

and, after some algebra, we obtain the right-hand part of this formula: 

 Ω=

d Ω =

(p) Cp

  (p−1)

Cp



n 

... dai1, ...ip−1 ∧ dxi1 ∧ .... ∧ dxip−1   i1, ...ip−1 =1 p times

  =







n 

...   i1, ...ip−1 =1

 dai1, ...ip−1 ∧ dxi1 ∧ .... ∧ dxip−1

p−1 times

  =



  n  ... ai1, ...ip−1 ∧ dxi1 ∧ .... ∧ dxip−1 =   i1, ...ip−1 =1 ∂Cp

Ω .

(p−1)

p−1 times

In conclusion, we note that the three-dimensional form of the Stokes’s theorem (1.44) comes as a special case of the general formula (B.43).