Chapter 10

Chapter 10: Differential forms PhL 2.27.16 Chapter 10: Differential forms PhL 2.27.16 1 1. Differential Forms Defined 1 2. Differential Forms on Manifolds 3 3. The exterior derivative of a differential form 4 4. Commutation properties of differential forms 6 5. Closed and Exact, Poincaré and the Angle Form 7 6. The pullback of a differential form 9 10.1. Differential Forms Defined A differential form is in fact just an element of the space Λk described in Chapter 8. The reason for the word "differential" will become apparent later when we look at integration on surfaces. Recall that our most general element of Λk was written in symmetric sum notation as T^ = Σii....i Tii....i (λi ^ λi .....^ λi) . T^ = ΣITIλ^I (8.4.4) (10.1.1) This sum is redundant since each basis vector appears k! times. In the ordered sum form, each basis vector of Λk appears only once, T^ = Σ1≤i<i<....<i≤n Aii...i (λi ^ λi .....^ λi) T^ = Σ'IAIλ^I (8.4.7) (10.1.2) where Σ'I indicates the ordered summation. The tensor A is related to tensor T by A = k!Alt(T) (8.4.16) (10.1.3) so that in components Aii...i = k! Alt(Tii...i) or AI = k! Alt(TI) . (10.1.4) Written out this says, Aii...i = k! Alt(Tii...i) = Tii....i - Tii....i + all other signed permutations . (10.1.5) Whereas T is an arbitrary rank-k tensor, A is an arbitrary totally antisymmetric rank-k tensor. Notice that if one is given T one can find the corresponding A from (10.1.4). If instead one is given A, one can take T = (1/k!) A as a viable T since then k!Alt(T) = k![(1/k!)Alt(A)] = Alt(A) = A. As shown in the generic language of (A.4), T can be any tensor whose decomposition has (1/k!)A for its totally antisymmetric component. Below we shall treat the objects T and A as rank-k tensor fields with an argument in x-space, so we will then have for example, Aii...i(x) x ϵ Rn ( "x-space") . (10.1.6) In the usual presentation of the theory of differential forms, the dual-space basis vector λi is given the purely cosmetic name dxi dxi ≡ λi = <ei| . (10.1.7) This object dxi is different from the normal calculus differential dxi, and for that reason we write dxi in a red italic font. For example, one can then write, dxi(v) = λi(v) = <ei| v> = vi In contrast, there is no calculus differential object called dxi(v). The differential forms (elements of Λk) shown above in (1.1) and (1.2) are now written in cosmetic notation as T^ = Σii....i Tii....i ( dxi ^ dxi .....^ dxi) T^ = ΣITI dx^I (10.1.7) T^ = Σ1≤i<i<....<i≤n Aii...i ( dxi ^ dxi .....^ dxi) T^ = Σ'IAI dx^I . (10.1.8) We have used the hat subscript notation to distinguish dual tensors in V*k from those in Λk , λI = λi λi ..... λi // basis vector in dual space V*k λ^I = λi ^ λi .....^ λi // basis vector in dual space Λk . (10.1.9) Thus, dx^I ≡ dxi ^ dxi .....^ dxi . (10.1.10) The traditional names for differential forms are α, β and so on, so we take T^ → α and write our arbitrary differential form (10.1.8) now as α = Σ'I fI(x) λ^I = Σ'I fI(x) dx^I α ϵ Λk(V) V = x-space = Rn (10.1.11) where fI is the more traditional name for AI. We have here taken V = Rn, Euclidean space, where the basis vectors ei = |ei> are independent of x, and so the λi = <ei| are also independent of x. 10.2. Differential Forms on Manifolds Now suppose x is a point lying on some surface embedded within Rn . We shall assume this surface is a "manifold" M Rn . We can define a differential form α at a point x ϵ M in this way (αx = xα ), αx = Σ'I fI(x) xλ^I = Σ'I fI(x) dx^I (10.2.1) Assume that the manifold M is a "surface" of dimension m within Rn. The manifold M could be some open set within Rn (or all of Rn), in which case it has dimension m = n. If the manifold is a "hypersurface" in Rn it then has dimension m = n-1. In general M is some "surface" where 1 ≤ m ≤ n. At a point x on M one usually constructs an x-dependent set of n basis vectors xei . The first m of these basis vectors are all tangent to the "surface" M, while the last n-m are normal to M. For example, for a manifold that is a smooth non-self-intersecting 3D curve embedded in R3, one would take xe1 to be tangent to the curve at x, and then xe2 and xe3 are both normal to the curve at x. If the curve is described by x = φ(t) where t is a parameter, then dx = ∂tφ(t)dt and one normally takes xe1 = ∂tφ(t). On the other hand, if M is a 2D surface in R3, xe1 and xe2 are taken to be two linearly independent tangent vectors at x, and xe3 would be normal to M at x. The set of m linearly independent basis vectors {xe1...xem} which are tangent to M at x are first thought of as having their tails right at the point x on M. When these vectors are translated so their tails are all at the origin, the {xe1...xem} then span an m-dimensional vector space. This vector space is usually written TxM and is called the tangent space to M at point x on M, dimension m. As with any vector space, there is a corresponding dual space. The dual space to the tangent space is called the cotangent space and it is the set of all rank-m linear functionals of vectors in TxM. The name cotangent is like the name covector mentioned below (2.11.a.3) and has nothing to do with the cotangent of any angle. As one moves from x to a nearby point x + dx on M, the basis vectors in general will move slightly (M is "smooth"). The dual basis vectors xei of course also move to maintain xei xej = δij. Thus we have xλi = <xei| also depending on x. We don't want to write this xλi as λi(x) because then we have to write <xei|v> = (λi(x))(v) which is rather messy (although Spivak uses this kind of notation with x = p in various places). We hesitate to write the left side xα as α(x) because this makes α look like a function, but it is in fact a differential form. Notice another benefit of the cosmetic notation xλi = dxi . The dependence on x can be regarded as being implied by us writing dxi instead of say dyi. So we don't need to write xdxi . It is customary to abbreviate the left side xα as just α with the understanding that it is at some point x on M. Similarly one writes ei understanding that it is xei . As noted, a simple example of a manifold is a non-self-intersecting and "smooth" finite piece of 3D curve hanging in R3 which is defined by some function x = φ(t) where t is a scalar parameter which marks points on the curve. In this case αx is a differential 1-form defined at every point x along that curve, and the tangent space as noted is one dimensional and contains the tangent vector to the curve at some x. Our second example of a manifold is a non-self-intersecting and "smooth" finite piece of 2D surface embedded in R3 which is defined by some function x = φ(t) with t = (t1,t2) where every point on the surface is marked by a unique value of t. In this case αx is a differential form defined at every point x on that surface. The tangent space at any point x on M is 2 dimensional. See Sjamaar Chapter 6 or elsewhere for a formal definition of a manifold and smoothness. A manifold is roughly a smooth "surface" which can be cobbled together from a set of smooth mappings x = φi(t) which are said to cover the manifold, the way an atlas of flat maps can cover the entire globe of the Earth. A manifold is a "surface" which is locally smooth in the region of any point x on the manifold. Since each x = φi(t) must be 1-to-1 between the parameter t-space and x-space, the manifold cannot be self-intersecting, since a mapping which included a self-intersecting point would not be 1-to-1. Each mapping has some open domain Ui in Rm and one writes φi: Ui → M and φ must be 1-to-1 as noted. But (∂φi/∂tj) : Ui→M must also be 1-to-1 to provide clean differentiability at all points on M. This is often stated as (Dφi) must be 1-to-1. The conglomeration of all the tangent spaces TxM on M has the structure of a fiber bundle and is often called the tangent bundle. There is a corresponding dual cotangent bundle. See Spivak [1999] Chapter 3 or wiki on tangent bundles. 10.3. The exterior derivative of a differential form In Section 1 we noted that TI = Tii....i(x) and AI = Aii...i(x) were rank-k tensor fields with respect to some unspecified Chapter 2 transformation x' = F(x) and dx' = Rdx. We now regard these objects as being just scalar-valued functions which happen to have label I. They and fI = AI are then just scalar coefficient functions in the expansions (10.1.1) and (10.2.2). Such a function by itself is a 0-form because it has no λi factors. That is, the object f , f = fI(x) ϵ Λ0, (10.3.1) is a differential 0-form (abbreviated 0-form) having a label I. The exterior derivative of such a 0-form is written df and is defined as df ≡ Σj=1n [∂fI(x)/∂xj] λj = Σj=1n [∂jfI(x)] λj . (10.3.2) Here we put df in red italic so it won't be confused with a calculus differential df of a function f(x) = fI(x). We could have written the 0-form f as f , but since then f = f there is no reason to do so. The first thing we can see is that, since f is a 0-form, df is a 1-form since it has a single λk dual basis vector. Using the cosmetic notation defined above, we then write (3.2) as, df = Σj=1n [∂jfI(x)] dxj . (10.3.3) Now we begin to see the motivation for the cosmetic notation dxj . The above equation looks just like the corresponding calculus equation df = Σj=1n [∂jfI(x)] dxj f(x) = fI(x) . (10.3.4) In this last equation df(v) would make no sense, but in (10.3.3) we could write df (v) = Σj=1n [∂jfI(x)] λj(v) = Σj=1n [∂jfI(x)] vj . // (2.11.c.5) (10.3.5) The exterior derivative of a general differential form α has an extremely simple definition. Reverting from fI back to AI and its corresponding TI, one has α = Σ'I fI(x) λI general k-form α ϵ Λk dα ≡ Σ'I (dfI(x)) ^ λ^I = Σ'I ( Σj=1n [∂jfI(x)] λj) ^ λ^I // from (10.3.2) = Σ1≤i<i<...<i≤n Σj=1n [∂jfii...i(x)] λj ^ λi ^ λi ...^ λi . (10.3.6) Since there are now k+1 wedged dual basis vectors λr, this dα must be a (k+1)-form. Fact: If one defines dα using the symmetric sum, the resulting dα is the same as that shown above. Proof: Let α = Σ'IAI(x)λ^I = ΣITI(x)λ^I (10.3.7) and recall from (10.3.6) that dα ≡ Σ'I(dAI(x))λ^I = Σ'I Σj=1n [∂jAI(x)] λj ^ λ^I = Σj=1n λj ^ ( Σ'I [∂jAI(x)] λ^I) . (10.3.8) Now define dα" as follows, involving the symmetric sum ΣI, dα" ≡ ΣI(dTI(x))λ^I = Σj=1n λj ^ ( ΣI [∂jTI(x)] λ^I) . (10.3.9) We will have shown that dα = dα" if we can show that Σ'I [∂jAI(x)] λ^I = ΣI [∂jTI(x)] λ^I ? (10.3.10) where AI and TI are related by (10.1.4), AI(x) = k!Alt(TI(x)). But AI(x) = k!AltI(TI(x)) [∂jAI(x)] = k! AltI [∂jTI(x)] CI(j) = k!AltI(DI(j)) (10.3.11) where CI(j) ≡ ∂jAI(x) and DI(j) ≡ ∂jTI(x) and j is regarded as a passive label. But (10.3.11) is the relationship which says that a form β = Σ'I CI(j)λ^I can also be written as β = ΣI DI(j)λ^I. Therefore we conclude that Σ'I CI(j)λ^I = ΣI DI(j)λ^I so (10.3.10) is true. QED So far we have shown that if α is a k-form, then dα is a (k+1)-form. What can be said about d2α ≡ d(dα) ? One might reasonably think this would be a (k+2)-form, but that is not correct. In fact: Fact: d2α = 0 for any k-form α (differential forms have zero "curvature") . (10.3.12) Proof: The proof is quite simple if we use the redundant symmetric sum (10.1.1) to express α. Then α = ΣI TI(x) λ^I dα = ΣI(dTI(x)) ^ λ^I = ΣI ( Σr=1n [∂rTI(x)] λr) ^ λ^I = ΣIΣr=1n [∂rTI(x)] (λr ^ λ^I) d(dα) = ΣI Σr=1n d[∂rTI(x)] (λr ^ λ^I) = ΣI Σr=1n (Σs=1n∂s[∂rTI(x)] λs ) ^ (λr ^ λ^I) = ΣI Σr=1n Σs=1n[ ∂s∂rTI(x)] (λs ^ λr ^ λ^I) = 0 . QED The result is 0 because in the symmetric sum Σrs the object ∂s∂rFI(x) is symmetric under r↔ s while the object (λs ^ λr ^ λI) is antisymmetric under r↔s. That is to say, if S is symmetric and A antisymmetric, swap names r↔s use symmetries sum = Σrs SrsArs = Σsr SsrAsr = Σrs (+Srs)(-Ars) = - Σrs SrsArs = - sum = 0 (10.3.13) 10.4. Commutation properties of differential forms Recall these results from Chapter 8 concerning elements of Λ(V), S^^ T^ = (-1)kk'T^^ S^ ranks of the two dual tensors are k and k' . (8.9.c.6) Fact: In a product of tensors T1^T2^T3.... of rank k1, k2, k3 ... , if two tensors are swapped Tr ↔ Ts (with r < s), the resulting tensor incurs the following sign relative to the starting tensor, sign = (-1)m where m = (kr+1+kr+2 ...+ks-1)(kr+ks) + krks . (8.9.e.6) T^N = 0 for any N ≥ n+1 assuming k ≠ 0. (8.9.d.9) In the language of differential forms these become α ^ β = (-1)kk'β ^ α α = k-form, β = k'-form (10.4.1) α1 ^ α2 ^ ... αr ... αs ... ^ αk = (-1)m α1 ^ α2 ^ ... αs ... αr ... ^ αk where m = (kr+1+kr+2 ...+ks-1)(kr+ks) + krks (10.4.2) αN = 0 for N ≥ n+1 dim(V) = n α = any k-form with k ≥ 1 where αN ≡ α ^ α ... ^ α . (10.4.3) Equations (10.4.1) and (10.4.3) appear in Sjamaar as "2.1 Proposition" on page 19 and the preceding equation. In Sjamaar, Buck and many other source all ^ symbols are suppressed so (4.1) is written αβ = (-1)kk'βα and one must understand that these are wedge products in Λ(V). 10.5. Closed and Exact, Poincaré and the Angle Form Closed: If dα = 0 for a k-form α, α is said to be closed. The analogous fact for a function f(x) with df = 0 would be that f(x) = constant. (10.5.1) Exact: Sometimes one finds that a form α can be written α = dβ where β is some other form. If α is a k-form, we know from below (10.3.6) that β must be a (k-1)-form. When α = dβ for some form β, α is said to be exact. (10.5.2) We showed in (10.3.12) that d2β = 0 for any form β, so it follows that if α = dβ, then dα = 0 and α is closed. Thus we have shown that : Fact: If α is exact, then α is closed. (10.5.3) In 1D calculus if f = dh/dx one says that f is a "perfect differential" and one then writes !Syntax Error, If(x) dx = !Syntax Error, I() dx = !Syntax Error, Idh = h(a) - h(b) dh = () dx (10.5.4) and this is the analogy to exactness for a function. In Rn being a perfect differential means that one can write a vector function as f = h. The above integral then becomes !Syntax Error, If(x) dx =!Syntax Error, Ih dx = ∫c dh = h(a) - h(b) where dh = h dx = Σi=1n (∂ih(x))dxi = Σi=1n fi(x) dxi = f dx . (10.5.5) The line integral depends only on the line endpoints a and b, and not on the particular shape of the curve c joining a and b. For a closed curve a = b and we find ∫C dh = h(a) - h(b) dh = h(a) - h(a) = 0 . (10.5.6) We shall see below a similar theorem for α = dβ where β is a 0-form (a function) and α is 1-form: ∫c α = ∫c dβ = g(φ(a)) - g(φ(b)) if α = dβ (α is exact) (10.5.7) where g is a certain function related to α, and where x = φ(t) describes the curve c for t in [a,b]. Thus for a closed curve we get α = dβ = g(φ(a)) - g(φ(a)) = 0 if α = dβ (α is exact) (10.5.8) In some sense, a 1-form α being exact is like a vector function being a perfect differential. In physics if the function f(x) in (10.5.5) is a force field and f(x) dx = dW (differential work done by the force moving a particle distance dx), and if the force can be written f = h (h is called a potential), then f is a conservative force field. Fact (10.5.3) above says α exact α closed. Is it possibly also true that α closed α exact and so then the two descriptions are one in the same? The answer is "not quite" as expressed in this claim: Poincaré Lemma: If any differential form α on Rn is closed for x in some open star-shaped domain in Rn which includes the origin, then α is exact. (10.5.9) This Lemma appears on p 38 of Spivak from which we quote, and Spivak proceeds to give a detailed proof. In topological language, the star-shaped domain is any domain that is "contractible to a point". Certainly the Lemma is valid for a domain which is an open "cube" or "sphere" (n dimensions) about the origin. The domain need not be convex. The classic example of this theorem involves the so-called angle form defined on R2 with coordinates (x1,x2), α = Σi=12 fi(x)λi where f1(x) = - (x2/r2) r2 = x12 + x22 f2(x) = (x1/r2) Then dα = Σi dfi(x)λi = Σij (∂jfi) λj ^ λi . Notice that, using the fact that ∂ir = xi/r, (∂1f2) = ∂1(x1/r2) = [ r2 * 1 - x1 (∂1r2)] / r4 = [ r2 - x12r (∂1r)] / r4 = - [r2 - x12r(x1/r)] / r4 = [r2 - 2x12] / r4 = [x12 + x22 - 2x12] / r4 = (x22 - x12) / r4 and (∂2f1) = - ∂2 (x2/r2) = - [ r2 * 1 - x2 (∂2r2)] / r4 = - [ r2 - x22r (∂2r)] / r4 = - [r2 - x22r(x2/r)] / r4 = - [r2 - 2x22] / r4 = - [x12 + x22 - 2x22] / r4 = (x22 - x12) / r4 = (∂1f2) . Thus it turns out that the quantity (∂jfi) is symmetric under i ↔ j. Then by the argument (10.3.13) we get dα = Σij (∂jfi) λj ^ λi = Σij (Sij)(Aji) = 0 so α is a closed 2-form. As we shall show below, the line integral of α around a circle centered at the origin gives α = 2π. Thus the angle form is not exact because if it were one would have α = 0. So here is a form α which is closed, but which is not exact. The condition of the Poincaré Lemma must therefore be violated, and that is indeed the case since the form α is undefined for r=0 where f1 and f2 blow up, so α is then defined on R2 punctured at the origin, sometimes written R2/ {0} or R2 - {0}. Thus we can't have any open star-shaped set including the origin for α, so Poincaré's Lemma does not apply. Our plan next is first to define the pullback of a differential form, and then in later sections to use the pullback to define the meaning of integration of a differential form over a manifold. ok to here 10.6. The pullback of a differential form : Method 1 Recall from Section 2 above that a smooth piece V of a manifold M (surface of dimension m in Rn) can be generated by applying the 1-to-1 transformation x = φ(t) to a simple open region U of t-space. We then have a mapping: φ : U Rm → V M Rn x = (x1,x2.....xn) = φ(t1,t2...tm) = φ(t) (10.6.1) t-space x-space One might say that in this mapping, a point t in U is "pushed forward" to a point x on manifold M. A pullback takes a differential form α defined at point x on M (x-space), and "pulls it back" to a different differential form β = φ*(α) defined at point t in U (t-space). Here φ* is a pullback operator which is associated with the transformation x = φ(t) . The operator φ* acts on a k-form α to produce (it turns out) another k-form β, where α exists in x-space and β exists in t-space. One sometimes says that β is the pullback of α along φ. How then does one compute the pulled back form β from the original form α ? Recall from (2.1) how xλi = <xei| refers to a dual basis vector λi at some point x in x-space. In our current context, x would be a point in V on manifold M within Rn as shown in (6.1). Similarly we can refer to the dual basis in t-space as tλj where t lies in U with M. There are n dual vectors xλi in x-space, while there are m dual vectors tλj in t-space. (a) Definition of a pullback acting dual basis vectors λi What is the distinction between m and k ???? k = 1 We start by defining the action of pullback φ* on a the simplest 1-form which is a basis vector xλi in x-space, φ*(xλi) ≡ Σj=1m (Dφ)ij tλj = Σj=1m Rij tλj i = 1,2...n . (10.6.2) Some comments on notations (Dφ) and R are certainly due at this point. First, (Dφ)ij ≡ (∂φi(t)/∂tj) = ≡ Rij i = 1,2...n j = 1,2..,m (10.6.3) The object (Dφ) is thus a matrix of derivatives having n rows (first i index) and m columns (second j index). In Chapter 2 where we considered the general transformation x' = F(x) we referred to the differential of the transformation as the R-matrix where Rab = (∂x'a/∂xb) as in (2.1.2). Here we have x = φ(t) as our general transformation and the R-matrix is Rab = (∂xa/∂tb). In the discussion of Chapter 2, taken from our document Tensor, we had in mind that the transformation x' = F(x) was a mapping between x-space and x'-space where both spaces were of dimension n, so R was an nxn matrix. But now for x = φ(t) the transformation maps from t-space of dimension m to x-space of dimension n, so the differential R matrix is only square nxn if both t-space and x-space have dimension n, which means that the manifold M is some full chunk of Rn . For a general m-dimensional "surface" or manifold in Rn, the differential matrix R = (Dφ) is an n x m matrix. The notation (Dφ) is favored by Spivak and other authors, but we like R in dense equations because it is 1 symbol instead of 4 symbols. We now continue to study (10.6.2) above, φ*(xλi) ≡ Σj=1m (Dφ)ij tλj = Σj=1m Rij tλj ≡ βt i = 1,2...n . (10.6.2) The object xλi is a 1-form in x-space, xλi ϵ Λ1(V M Rn). The object on the right is a linear combination of the dual basis vectors tλj ϵ Λ1(U Rm) and so is a 1-form in t-space. We call this 1-form βt just to give it a name. We can now compare the transformations φ and φ* : φ : U Rm → V M Rn x = φ(t) t-space x-space x = (x1,x2.....xn) t = (t1,t2...tm) φ* : Λ1(V M Rn) → Λ1(U Rm) βt = φ*(xλi) pullback (10.6.4) 1-form in x-space 1-form in t-space where the notation Λ1(V) means the space of rank-1 linear functionals defined on vectors in V. The first mapping pushes forward from t-space to x-space, the second mapping pulls back from x-space to t-space. The following drawing illustrates the discussion above. (10.6.5) The mapping φ takes the unit interval [0,1] in t-space to the thick part of the red curve in x-space. This kind of mapping is called a 1-cube. The thick red curve on the right is the image of this 1-cube mapping, though one can think of it as being the 1-cube (multiple mappings can yield the same image, however). We also think of the domain as being a 1-cube, since it is a unit one-dimensional cube. Thus we in fact have three distinct meanings for the term 1-cube. The same is true for 2-cubes and k-cubes. Comments 1. We are free to add more dimensions to t-space on the left of the above picture, to bring it up to Rm, and the figure would still describe the pullback of a 1-form (k = 1). As we know, the Rm space will support k-forms for any k ≤ m. Once cannot have k > m because then there would need to be k basis vectors λi wedged together for such a k-form, but such a wedge product vanishes if k > m because one can only produce m linearly independent basis vectors in Rm, see (8.2.6). 2. If one has Rm on the left for t-space, the mapping x = φ(t) cannot create a surface in x-space on the right that has dimension more than m. The surface on the right is supposed to be embedded in Rn so we always have in mind that n ≥ m. 3. Officially, the set U in t-space on the left can be any open set in t-space, not just the 1-cube we have shown (or the c-cube shown next). STOP. Then you cannot do what I have shown below because a curve in t-space is NOT an open set in t-space if t-space is dimension 2. I am confused. Does this really mean that we must have m = k ? I am confused indeed, manana! See page 37 Sjamaar for help. k = 2 The pullback on a basis vector in Λ2(V M Rn) is defined as φ*(xλi ^ xλi) ≡ Σjj=1m (Dφ)ij (Dφ)ij (tλj ^ tλj) = Σjj=1m RijRij (tλj ^ tλj) ≡ βt . i1, i2 ϵ {1,2...n} (10.6.6) This mapping takes a basic 2-form at a point x in V on M in x-space and maps it to (pulls it back to) some 2-form at point t in U in t-space. We call this 2-form βt just to give it a name (it is of course a different βt from that appearing earlier). Notice the symmetric sums on j1 and j2. Here is a drawing illustrating the k=2 case. As in the previous drawing, we think of x-space as being Rn but we draw it as if it were R3 just to be able to draw something. For k = 1 the manifold M on the right was the thin red curve embedded in R3, while V was the fat piece of this curve. For k = 2 the manifold of interest is a torus and V is a patch on that torus which is mapped into from a 2-cube in t-space by the mapping x = φ(t). The 2-cube (as the mapping domain) is often written [0,1]2. (10.6.7) Even in R3 there are three distinct basis vectors xλi ^ xλi one of which is xλ1 ^ xλ3 . Recall that the dimension of Λ2 in R3 is binomial (3,2) = 3. When t-space is Rm, it is possible to have k-forms defined on Rm for any k ≤ m. How can equation (10.6.6) concerning rank-2 dual tensors be stated in terms of rank-2 tensor functions? φ*(xλi ^ xλi) = Σjj=1m RijRij (tλj ^ tλj) . (10.6.6) To find out, we close both sides of bra (10.6.6) onto the ket |v1> |v2> = |v1,v2> where vi are both in Rm = t-space so these vectors each have m components, see (2.11.e.7). Doing this gives [φ*(xλi ^ xλi)] (v1,v2) = Σjj=1m RijRij (tλj ^ tλj)(v1,v2) (10.6.8) where both sides of this equation are rank-2 tensor functions in Λ2f(V) [defined above (4.4.34)]. We shall now process the right hand side of (10.6.8). First use (4.4.20b) to write (tλj ^ tλj)(v1,v2) = [ (v1)j(v2)j - (v1)j(v2)j] (10.6.9) so then RHS(10.6.8) = (1/2) Σjj=1m RijRij [ (v1)j(v2)j - (v2)j(v1)j] // (4.4.20b) = (1/2) { Σjj=1m RijRij(v1)j(v2)j - (v1↔ v2) } // do ↔ to make 2nd term = (1/2) { [Σj=1mRij(v1)j] [Σj=1mRij(v2)j] - (v1↔ v2) } // reorder = (1/2) { (Rv1)i(Rv2)i - (v1↔ v2) } // matrix multiplication = (1/2) [ xλi(Rv1)xλi(Rv2) - xλi(Rv1)xλi(Rv2)] // (2.11.c.5) λi(v) = vi = (xλj ^ xλj)(Rv1, Rv2) . // (4.4.20b) (10.6.10) We have thus shown that the tensor function representation of the pullback definition is this: [φ*(xλi ^ xλi)](v1,v2) = (xλj ^ xλj)(Rv1, Rv2) . (10.6.11) k = k Finally we define the pullback on a basis vector in Λk : φ*(xλi ^ xλi....^ xλi) = Σjj...j=1m RijRij ....Rij (tλj ^ tλj ....^ tλj) or φ*(xλI) = ΣJ RIJ tλJ . (10.6.12) The pullback mapping is then, φ* : Λk(V M Rn) → Λk(U Rm) . (10.6.13) k-form in x-space k-form in t-space As we did with k=2, we again wish to evaluate the tensor function statement of (10.6.11) (λt→ λ for now). As before we close with an appropriate ket to get, [φ*(λxI)](vn,vn...vn) = ΣJ RIJ (λj ^ λj ....^ λj)(vn,vn...vn) . (10.6.14) We now process the right hand side of (10.6.11). There are quite a few steps, each explained: RHS (6.9) = ΣJ RIJ (λj ^ λj ....^ λj)(vn,vn...vn) = ΣJ RIJ AltJ[(λj λj .... λj)(vn,vn...vn)] // (8.3.8) = ΣJ RIJ AltJ[(λj(vn)λj(vn) ...λj(vn)] // (6.1.3) = ΣJ RIJ AltN[(λj(vn)λj(vn) ...λj(vn)] // (A.8.29) = AltN [ ΣJ RIJ (λj(vn)λj(vn) ...λj(vn) ] // (A.8.10) = AltN [ ΣJ RIJ (vn)j(vn)j ... λ(vn)j ] // (2.11.c.5) = AltN [ Σjj...j=1m RijRij ....Rij (vn)j(vn)j ... λ(vn)j ] = AltN { [ΣjRij(vn)j] [ΣjRij(vn)j] ... [ΣjRij(vn)j] } = AltN { (Rvn)i(Rvn)i... (Rvn)i } // (Rv)i = ΣjRijvj = AltN {λi(Rvn)λi(Rvn)... λi(Rvn) } // (2.11.c.5) = AltI {λi(Rvn) λi(Rvn)... λi(Rvn) } // (A.8.29) = AltI {(λi λi .... λj) (Rvn, Rvn ...Rvn)} // (6.1.3) = ( λi ^ λi... ^ λi )(Rvn,Rvn...Rvn) . // (8.3.8) (10.6.15) We have thus shown that the tensor function representation of the pullback definition is this: [φ*(xλi ^ xλi....^ xλi)](vn,vn...vn) = ( tλi ^ tλi... ^ tλi )(Rvn,Rvn...Rvn) or [φ*(xλI)] (vn,vn...vn) = tλI(Rvn,Rvn...Rvn) . (10.6.16) Finally, having held off a long time, we can rewrite equations above using the cosmetic notations introduced in (10.1.7). dxi ≡ xλi dtj ≡ tλj (10.6.17) to get these Cosmetic equations for pulling back basis vectors and their tensor functions, // R = (Dφ). φ*(dxi ) ≡ Σj=1m Rij dtj R ≡ (Dφ) i = 1,2...n . (10.6.2)C φ*(dxi ^ dxi) ≡ Σjj=1m RijRij dtj ^ dtj i1, i2 ϵ {1,2...n} (10.6.6)C [φ*(dxi ^ dxi)](v1,v2) = (dxi ^ dxi)(Rv1, Rv2) // tensor function (10.6.11)C φ*(dxi ^ dxi....^ dxi.) ≡ Σjj...j=1m RijRij .... Rij (dtj ^ dtj ....^ dti) or (10.6.12)C φ*(dxI) = ΣJ RIJ dtJ . [φ*(dxi ^ dxi....^ dxi)](vn,vn...vn) = ( dxi ^ dxi.... ^ dxi )(Rvn,Rvn...Rvn) or (10.6.16)C [φ*(dxI)](vn,vn...vn) = (dxI)(Rvn,Rvn...Rvn) vi ϵ Rm , Rvi ϵ Rn k = 0 A 0-form in x-space is just a function f(x) ϵ Λ0(V M Rn). The pullback of this 0-form in x-space to a 0-form in t-space is defined by φ*(f(x)) ≡ (f o φ)(t) = f(φ(t)) ≡ F(t) ≡ βt ϵ Λ0(U Rm) x ϵ V, t ϵ U (10.6.18) That is to say, we just replace the x in f(x) by x = φ(t) to get the pullback βt . ok to here 3 PM 2.29.16 can I do the properties at this point? No! (b) Action of a pullback on a differential form α The hard part is done! The pullback of a differential k-form α in x-space, αx = Σ'I fI(x) xλI (10.2.1) (10.6.19) is defined to be the following k-form in t-space, φ*(αx) ≡ Σ'I φ*(fI(x)) φ*(xλI) = Σ'I fI(φ(t)) ΣJ RIJ tλJ // (10.6.18) and (10.6.12) = ΣJ [ Σ'I fI(φ(t)) RIJ ] (tλJ) // reorder = ΣJ GJ(t) (tλJ) GJ(t) ≡ Σ'I fI(φ(t)) RIJ = Σ'J gJ(t) (tλJ) gJ(t) ≡ k! AltJ[GJ(t)] , (10.6.20) where the last line comes from (10.1.1) through (10.1.3). We have stated the result both in terms of symmetric ΣJ and ordered Σ'J summations. The coefficient functions gJ(t) can be re-expressed as, gJ(t) ≡ k!AltJ[GJ(t)] = k!AltJ[ Σ'I fI(φ(t)) RIJ] = k! Σ'I fI(φ(t)) AltJ[RIJ] // Alt is linear = k! Σ'I fI(φ(t)) [ (1/k!) det(RIJ) ] // (A.8.30) = Σ'I fI(φ(t)) det(RIJ) . (10.6.21) The pullback of αx along φ is then φ*(αx) = Σ'J Σ'I fI(φ(t)) det(RIJ(t)) (tλJ) = Σ'J Σ'I fI(φ(t)) det[(Dφ)IJ(t)] (tλJ) ≡ βt (10.6.22) Comment: This equation appears in Sjamaar page 42 old A. We can redraw our figure one more time, showing how a general m-form αx in x-space is pulled back along φ into the m-form βt = φ*(αx) in t-space : (10.6.23) Recall that (Dφ)ij is a function of t, so [Dφ(t)]ij ≡ (∂φi(t)/∂tj) = ≡ R(t)ij i = 1,2...n j = 1,2..,m . (10.6.3) We pause again to write equations in Cosmetic notation : αx = Σ'I fI(x) dxI // k-form in x-space (10.6.19)C φ*(xα) = ΣJ GJ(t) dtJ GJ(t) ≡ Σ'I fI(φ(t)) RIJ R = (Dφ) (10.6.20)C φ*(xα) = Σ'J gJ(t) dtJ gJ(t) ≡ Σ'I fI(φ(t)) det(RIJ) . (10.6.22)C Next, we want to obtain a tensor function definition of the pullback of a k-form. Start with [φ*(xα)](vn,vn...vn) ≡ Σ'I φ*(fI(x)) { [φ*(xλI)](vn,vn...vn) } = Σ'I fI(φ(t)) tλI(Rvn,Rvn...Rvn) // (10.6.16) = xα (Rvn,Rvn...Rvn) . (10.6.23a) In more conventional form [φ*(αx)](v1,v2...vk) = αφ(t) (Rv1,Rv2...Rvk) . (10.6.23b) = αφ(t)((Dφ)v1,(Dφ)v2...(Dφ)vk) . (10.6.23d) Everything on the right depends on t, not x. For example, one should write [R(t)]vi and [(Dφ)(t)]v1 Since the k-form on the right is to be expressed entirely in t-space terms, we have replaced x = φ(t) on the αx label to make it αφ(t). Showing all t dependence, we get [φ*(αx)](v1,v2...vk) = αφ(t)(R(t)v1,R(t)v2...R(t)vk) = αφ(t)([(Dφ)(t)]v1,[(Dφ)(t)]v2...[(Dφ)(t)]vk) . (10.6.23c) Suppressing this t dependence gives a less cluttered result, [φ*(αx)](v1,v2...vk) = αφ(t)(Rv1,Rv2...Rvk) = αφ(t)((Dφ)v1,(Dφ)v2...(Dφ)vk) . (10.6.23d) The tensor function αx(w1,w2...wk) ϵ Λkf(Rn) is of course k-multilinear and alternating in its vector arguments wi, each of which has n components. In (6.21) we have for example w1 = (Dφ)v1 where vi is a vector having m components. Since we know from above that R = (Dφ) is a matrix with n rows and m columns, things conform properly. The right side of (10.6.23d) is seen to be linear in the vi due to the matrix form Rvi for each argument. It alternates if we swap Rvi ↔ Rvj so it then alternates under vi ↔ vj. Therefore, since the right side is a k-multilinear alternating function of the vectors v1,v2...vk, it is in fact an rank-k tensor function in Λkf(Rm). The name of this rank-k tensor function is [φ*(αx)](v1,v2...vk) . Equation (10.6.23d) says: αx is a k-form in x-space, and φ*(αx) is the pulled back k-form in t-space. We make this pulled back form be a rank-k tensor function associated with t-space by adding the k vector arguments (v1,v2...vk) where the vi are vectors in t-space. This is the same as closing the bra t<φ*(αx)| with the ket |v1,v2...vk>t in the tensor product space, where the scalar product t< | >t is for V*k on the left and Vk on the right, where V = Rm = t-space. Note that t<φ*(αx)| ϵ Λk(Rm) V*k = (Rm)*k . In a normal treatment of pullbacks, the tensor function equation (10.6.23d) is taken to be the definition of the pullback of a k-form from x-space to t-space. This definition is equivalent to our definition in terms of the action on dual basis vectors and on functions f(x). Our definition work took place directly in the space of dual tensor functionals, not in the space of tensor functions. Notice that (10.6.23d) does not involve any dual basis vectors like dxi ≡ xλi . One fringe benefit of (10.6.23d) is that it also gives the φ* rule for rank-0 forms. If there are no arguments it just says φ*(αx) = αφ(t) so if αx = f(x), then φ*(f(x)) = φ*(αx) = αφ(t) = f(φ(t)) = (f o φ)(t) as in (10.6.18) above. Equation (10.6.23d) appears in Spivak but not quite as we have written it. Spivak says on the bottom of page 89 and the top of page 90, more or less, f*ω(p)(v1,v2...vk) = ω(f(p))(f*(v1), f*(v2), ... f*(vk) ) where f*(v) = (Df)v . His actual notation includes some subtleties about where the tails of vectors are located. To translate to our notation, the first step to replace ω(s) by ωs in two places, [f*ωp](v1,v2...vk) = ωf(p)((Df)v1, (Df)v2, ... (Df)vk ) . We then replace f→ φ, p→ x and ω → α to get [φ*αx](v1,v2...vk) = αφ(x)((Dφ)v1, (Dφ)v2, ... (Dφ)vk ) and we arrive at (10.6.23d). (c) Properties of the pullback operator φ*