Lecture 13. Differential forms

Transcription

1 Lecture 13. Differential forms In the last few lectures we have seen how a connection can be used to differentiate tensors, and how the introduction of a Riemannian metric gives a canonical choice of connection. Before exploring the properties of Riemannian spaces more thoroughly, we will first loo at a special class of tensors for which there is a notion of differentiation that maes sense even without a connection or a metric. These are called differential forms, and they play an extremely important role in differential geometry Alternating tensors We will first loo a little more at the linear algebra of tensors at a point. We will consider a natural subspace of the space of -tensors, namely the alternating tensors. Definition A -tensor ω T x M is alternating if it is antisymmetric under interchange of any two of its arguments. Equivalently, for any vectors v 1,...,v T x M, and any permutation σ S, ω(v σ(1),...,v σ() )=sgnσω(v 1,...,v ), where sgnσ = 1 if σ is an even permutation, sgnσ = 1 is σ is an odd permutation. The space of alternating -tensors at x is denoted Λ T M. Note that Λ 1 Tx M = Tx M, soalternating 1-tensors are just covectors. There is a natural projection A : Tx M Λ Tx M defined as follows: AT (v 1,...,v )= 1 sgnσt(v σ(1),...,v σ() ).! σ S Then T is alternating if and only if AT = T. Example The geometric meaning of this definition is probably not clear at this stage. An illustrative example is the following: Choose an orthonormal

2 120 Lecture 13. Differential forms basis {φ 1,...,φ n } for Tx M. Then we can define an alternating n-tensor A by taing A(v 1,...,v n )=det[φ i (v j )]. This is antisymmetric since the determinant is antisymmetric under interchange of any pair of columns. Geometrically, the result A(v 1,...,v n )isthe (signed) n-dimensional volume of the parallellopiped generated by v 1,...,v n. Given a basis { 1,..., n } for T x M,wecan define a natural basis for Λ T x M: Foreach -tuple i 1,...,i,wedefine dx i1 dx i2...dx i =!A ( dx i1... dx i ). Note that this is zero if the -tuple is not distinct, and that if we change the order of the -tuple then the result merely changes sign (depending whether the -tuple is rearranged by an even or an odd permutation). The factor! isincluded in our definition for the following reason: If we apply the alternating -tensor dx i1... dx i (where i 1,...,i are distinct) to the vectors i1,..., i, the result is 1. If we apply it to the same vectors in a different order (say, rearranged by some permutation σ), then the result is just the sign of σ. Any other vectors yield zero. These elementary alternating -tensors have a geometric interpretation similar to that in Example : The value of dx I (v 1,...,v )isthe determinant of the matrix with (m, n) coefficient dx im (v n ), and this gives the signed -dimensional volume of the projection of the parallelopiped generated by v 1,...,v onto the subspace generated by i1,..., i. This relationship between alternating forms and volumes will be central in the next lecture when we define integration of differential forms and prove Stoes theorem. Proposition (1). dx i1... dx i = σ S sgnσdx i σ(1)... dx i σ() =!A(dx i 1... dx i ). (2). For each, {dx i1... dx i : 1 i 1 <...<i n} is a basis for Λ T x M. Inparticular the space of alternating -tensors at x has dimension ( n )= n!!(n )!. Proof. (1) is immediate, since the value on any -tuple of coordinate basis vectors agrees. To prove (2), we note that by Proposition , any alternating tensor can be written as a linear combination of the basis elements dx i1... dx i.invariance under A shows that this is the same as a linear combination of -forms of the form A(dx i1... dx i ), and these are all of the form given. It remains to show the supposed basis is linearly independent, but this is also immediate since if I =(i 1,...,i ) then dx I ( i1,..., i )=1, but dx J ( i1,..., i )=0for any increasing -tuple J I. It follows that any alternating -tensor T can be written in the form

3 T = 1 i 1<...<i n 13.2 The wedge product 121 T i1...i dx i1... dx i for some coefficients T i1...i. Some caution is required here, because T can also be written in the form T = 1 T i1...i! dx i1... dx i i 1,...,i where the coefficients are the same as above for increasing -tuples, and to be given from these by antisymmetry in other cases. Thus the coefficients in this expression differ by a factor! from those given in the expression after Proposition The wedge product The projection A immediately gives us a notion of product on alternating tensors, which we have already implicity built into our notation for the basis elements for Λ T x M: Definition Let S Λ Tx M and T Λ l Tx M be alternating tensors. Then the wedge product S T of S and T is the alternating + l-tensor given by ( + l)! S T = A(S T ).!l! This may not seem the obvious definition, because of the factor on the right. This is chosen to mae our notation consistent with that in our definition of the basis elements: Tae an increasing -tuple i 1,...,i and an increasing l-tuple j 1,...,j l, and assume for simplicity that i <j 1. Then we can form the alternating tensors dx i1... dx i, dx j1... dx j l and dx j1... dx i dx j1... dx j l, and we would lie to now that the third of these is the wedge product of the first two. Proposition The wedge product is characterized by the following properties: (i). Associativity: f (g h) =(f g) h; (ii). Homogeneity: (cf) g = c(f g) =f (cg); (iii). Distributivity: If f and g are in Λ Tx M then (f + g) h =(f h)+(g h); (iv). Anticommutativity: If f Λ T x M and g Λ l T x M, then g f =( 1) l f g;

4 122 Lecture 13. Differential forms (v). In any chart, (dx i1... dx i ) (dx j1... dx j l )=dx i1... dx i dx j1... dx j l. Proof. We start by proving (v). Choose a chart about x. Then (dx i1... dx i ) (dx j1... dx j l ) = = ( + l)!!l! ( + l)!!l! = 1!l! A ( (dx i1... dx i ) (dx j1... dx j l ) ) A sgnσ sgnτdx i σ(1)... dx i σ() dx j τ(1)... dx j τ(l) σ S,τ S l σ S,τ S l sgnσ sgnτdx iσ(1)... dx iσ() dx jτ(1)... dx jτ(l) = dx i1... dx i dx j1... dx j l. The homogeneity and distributivity properties of the wedge product are immediate from the definition. From this we can deduce the following expression for the wedge product in local coordinates: For S = S i1...i dx i1... dx i and T = T j1...j l dx j1... dx j l (summing over increasing -tuples and l-tuples respectively) S T = 1 S i1...i!l! T j1...j l dx i1... dx i dx j1... dx jl. i 1,...,i,j 1,...,j l The associativity property can now be checed straightforwardly. Finally, we derive the anticommutativity property (iv): If g = g i1...i dx i1... dx i and f = f j1...j l dx j1... dx j l, then g f = 1 g i1...i!l! f j1...j l dx i1... dx i dx j1... dx jl i 1,...,i,j 1,...,j l = ( 1) g i1...i!l! f j1...j l dx j1 dx i1... dx i dx j2... dx j l... = ( 1)l!l! i 1,...,i,j 1,...,j l =( 1) l f g. i 1,...,i,j 1,...,j l g i1...i f j1...j l dx j1... dx jl dx i1... dx i

5 13.4 The exterior derivative Differential forms Definition A -form ω on a differentiable manifold M is a smooth section of the bundle of alternating -tensors on M. Equivalently, ω associates to each x M an alternating -tensor ω x,insuch a way that in any chart for M, the coefficients ω i1...i are smooth functions. The space of -forms on M is denoted Ω (M). In particular, a 1-form is a covector field. We will also interpret a 0-form as being a smooth function on M, soω 0 (M) =C (M). By using the local definition in section 13.2, we can mae sense of the wedge product as an operator which taes a -form and an l-form to a + l-form, which is associative, C -linear in each argument, distributive and anticommutative The exterior derivative Now we will define a differential operator on differential -forms. Proposition There exists a unique linear operator d:ω (M) Ω +1 (M) such that (i). If f Ω 0 (M) =C (M), then df agrees with the differential of f (Definition 4.2.1); (ii). If ω Ω (M) and η Ω l (M), then (iii). d 2 =0. d(ω η) =(dω) η +( 1) ω (dη); Proof. Choose a chart ϕ : U V with coordinate tangent vector fields 1,..., n. We will first produce the operator d acting on differential forms on U M. On this region we have the smooth functions x 1,...,x n given by the components of the map ϕ. The differentials of these are the one-forms dx 1,...,dx n, and in agreement with condition (iii) we assume that d(dx i )=0for each i. By induction and condition (ii), we deduce that d(dx i1... dx i )=0 for any -tuple i 1,...,i. Now let f = 1! i 1,...,i f i1...i dx i1... dx i. The linearity of d, together with condition (ii) and condition (i), imply df = 1 i0 f i1...i! dx i0 dx i1... dx i. i 0,i 1,...,i

6 124 Lecture 13. Differential forms One can easily chec that this formula defines an operator which satisfies the required conditions. In particular we can compute d 2 to chec that it vanishes: d 2 ( ω i1...i dx i1... dx i ) =d ( i ω i1...i dx i dx i1... dx i ) = j i ω i1...i dx j dx i dx i1... dx i = 1 ( 2 ) ω i1...i 2 x j x i 2 ω i1...i x i x j dx j dx i dx i1... dx i =0. To extend this definition to all of M we need to chec that it does not depend on the choice of coordinate chart. Let η be any other chart, with components y 1,...,y n.onthe common domain of η and ϕ, wehave x i = F i (y), where F = ϕ η 1, and This implies that dx i = Fi y j dyj. dx i1... dx i = Fi1 y j1 i 1,...,i,j 1,...,j I,J... Fi y j dy j1... dy j. Now we can chec that if we define the operator d in the y coordinates, then d(dx i1... dx i )=0: d ( dx i1... dx i ) ( F i 1 ) = d y j1... Fi y j dy j1... dy j = = 1 2 =0 I,J,j 0 m=1 I,J,j 0 m=1 2 F im y jm y j0 p m p m F ip y jp dyj0... dy j ( 2 ) F im y jm y 2 F im j0 y j0 y jm F ip dy j0... dy j y jp It follows (by linearity and distributivity) that the differential operators defined in the two charts agree. The differential operator may seem somewhat mysterious. The following example may help:

7 13.5 Pull-bac invariance 125 Example (The exterior derivative on R 3 ) The exterior derivative in R 3 captures the differential operators which are normally defined as part of vector calculus: First, the differential operator of a 0-form (i.e. a function f) is just the differential of the function, which we can identify with the gradient vector field f. Next, consider d applied to a 1-form: For purposes of visualisation, we can identify a 1-form with a vector field by duality: The 1-form ω = ω 1 dx 1 ω 2 dx 2 + ω 3 dx 3 is identified with the vector field (ω 1,ω 2,ω 3 ). Applying d to ω, weobtain dω =d(ω i dx i ) = j ω i dx j dx i =( 1 ω 2 2 ω 1 )dx 1 dx 2 +( 2 ω 3 3 dx 2 )dx 2 dx 3 +( 3 ω 1 1 ω 3 )dx 3 dx 1. The result is a 2-form. We identify 2-forms with vector fields again, by sending adx 1 dx 2 + bdx 2 dx 3 + cdx 3 dx 1 to the vector field (b, c, a). With this identification, the exterior derivative on 1-forms is equivalent to the curl operator on vector fields. Finally, consider d applied to a 2-form (which we again associate to a vector field V =(V 1,V 2,V 3 )). We find d ( V 3 dx 1 dx 2 + V 1 dx 2 dx 3 + V 2 dx 3 dx 1) = ( 3 V V V 2 )dx 1 dx 2 dx 3) = (divv ) dx 1 dx 2 dx 3. Thus the exterior derivative acting on 2-forms is equivalent to the divergence operator acting on vector fields. The familiar identities from vector calculus that the curl of a gradient is zero and that the divergence of a curl is zero are therefore special cases of the identity d 2 = Pull-bac invariance Now we will prove a remarable result which really maes the theory of differential forms wor: Proposition Suppose M and N are differentiable manifolds, and F : M N is a smooth map. Then for any ω Ω (N) and η Ω l (N), F (ω η) =F (ω) F η and d(f ω)=f (dω).

8 126 Lecture 13. Differential forms Proof. The proof of the first statement is immediate from the definition. The second statement is proved by an argument identical to that used to prove that the definition of the exterior derivative does not depend on the chart in Proposition , except that the map F may be a smooth map between Euclidean spaces of different dimension Differential forms and orientability There is a useful relationship between orientability of a differentiable manifold M n and the space of n-forms Ω n (M): Proposition A differentiable manifold M is orientable if and only if there exists an n-form ω Ω n (M) which is nowhere vanishing on M. Proof. Suppose there exists such an n-form ω. Let A be the set of charts ϕ : U V for M for which ω( 1,..., n ) > 0. Then A is an atlas for M, since any chart for M is either in A or has its composition with a reflection in A inparticular charts in A cover M. Furthermore A is an oriented atlas: For any pair of charts ϕ and η in A with non-trivial common domain of definition in M, wehave (η) i = ( D i η ϕ 1) j i (ϕ) j, and therefore by linearity and antisymmetry of ω, ω( (η) 1,..., (η) n )=det D(η ϕ 1 )ω( (ϕ) 1,..., n (ϕ) ). By assumption, ω( (η) 1,..., (η) n ) and ω( (ϕ) 1,..., n (ϕ) ) are positive and nonzero. It follows that det D(η ϕ 1 ) > 0. Conversely, suppose M has an oriented atlas A = {ϕ α : U α V α } α A. Let {ρ β } β J beapartition of unity subordinate to the cover {U α : α I}, so that for each β J there exists α(β) Isuch that suppρ β U α(β). Define ω = ρ β dx 1 ϕ α(β)... dx n ϕ α(β). β J Then ω is everywhere non-zero, since dx 1 ϕ... dxn α(β1 ) ϕ α(β1 ) is a positive multiple of dx 1 ϕ... dxn α(β2 ) ϕ for β α(β2 ) 1 β 2. We can interpret this in a slightly different way: For each x M, let Or x M be the set of equivalence classes of non-zero alternating n-tensors at x, where ω η if ω is a positive multiple of η. Or x M has exactly two elements for each x M. Then we tae OrM = x M Or x M, which is the orientation bundle of M. Onany chart ϕ : U V for M, the restriction of this bundle to U is diffeomorphic to U Z 2, but this is not necessarily true globally.

9 13.7 Frobenius Theorem revisited 127 A slight modification of the proof of Proposition gives the result that M is orientable if and only if the orientation bundle of M is trivial (that is, diffeomorphic to M Z 2 ) Frobenius Theorem revisited Differential forms allow an alternative formulation of the Theorem of Frobenius that we proved in Lecture 7 (Proposition 7.3.5). In order to formulate this, let D be a -dimensional distribution on M. We relate this distribution to differential forms by considering the subspace Ω 0 (D) ofω(m) consisting of differential forms which yield zero when applied to vectors in the distribution D. This subspace is closed under C scalar multiplication and under wedge products. Proposition The distribution D is integrable if and only if the subspace Ω 0 (D) is closed under exterior differentiation. Proof. First suppose D is integrable. Then locally we can choose charts ϕ : U V R R n where the first directions are tangent to D. In such a chart, forms in Ω l 0(D) have the form ω i1...i l dx i1... dx i l where ω i1...i l =0whenever i 1,...,i l. This implies that m ω i1...i l =0for i 1,...,i l and arbitrary m. Applying the exterior derivative, we find dω = m ω i1...i dx m dx i1... dx i which is clearly in Ω l+1 0 (D). Next suppose D is not integrable. Then by Frobenius s theorem we can find vector fields X and Y in X (D) such that [X, Y ] / X(D), say in particular [X, Y ] x / D x for some x M. Choose a 1-form ω Ω 1 0(D) such that ω x ([X, Y ] x ) 0(How would you construct such a 1-form?) Then we compute: dω(x, Y )=X i Y j ( i ω j j ω i ) = X i i (Y j ω j ) Y j j (X i ω i ) X i ( i Y j )ω j + Y j ( j X i )ω i = Xω(Y ) Yω(X) ω([x, Y ]) 0 atx, since ω(y ) = 0andω(X) = 0everywhere, and ω([x, Y ]) 0atxby assumption. Therefore Ω 0 (D) isnot closed under exterior differentiation.

10 128 Lecture 13. Differential forms Exercise The identity dω(x, Y )=Xω(Y ) Yω(X) ω([x, Y ]) for the exterior derivative of a one-form generalises to an expression for exterior derivatives of -forms: If ω Ω (M), then dω(x 0,...,X )= ( 1) i X i ω(x 0,..., ˆX i,...,x ) i=0 + ( 1) i+j ω([x i,x j ],X 0,..., ˆX i,..., ˆX j,...,x ). Prove this identity. 0 i<j