Let U be an open subset of R n . Part I. Differential forms. The - PDF document

Let U be an open subset of R n . Part I. Differential forms. The differential forms are an algebra over C ∞ functions f : U → R . The algebra contains all the functions f (of degree 0), together with abstract elements dx 1 , dx 2 , . . . , dx n and their linear combinations f 1 dx 1 + f 2 dx 2 + · · · + f n dx n . These are the elements of degree 1. The rest of the algebra is generated by products of elements of this form. In general, a differential form of homogenous degree d looks like this. a i 1 ,i 2 ,...i d dx i 1 ∧ dx i 2 ∧ · · · ∧ dx i d � ω = 1 ≤ i 1 <i 2 < ··· <i d ≤ n The elements dx i in this sum are stuck together by an associative product, the wedge product , which acts on differential forms as follows. Let ω and η be two differential forms, and let f be a smooth function from U to R . Then f ∧ ω = ω ∧ f = fω. Here fω is the differential form obtained from ω by multiplying all the coefficients by f . In particular, if g is another smooth function, f ∧ g = g ∧ f is the same as fg . Also, functions pass through the wedge product. fω ∧ η = ω ∧ fη. For all elements dx i , dx j , the product is antisymmetric. dx i ∧ dx j = − dx j ∧ dx i . In particular, dx i ∧ dx i = 0 . So, if we have two forms α = � a j dx j and β = � b j dx j , where a j and b j are smooth functions U → R , then ( a j b l − a l b j ) dx j ∧ dx l = − β ∧ α. � α ∧ β = j<l In general, we can permute the elements of a monomial form like dx i 1 ∧ dx i 2 ∧ · · · ∧ dx i d in any way we like by successive transpositions of adjacent elements, pro- vided that we reverse the sign if the permutation is odd. Convince yourself that if α and β are differential forms of homogenous degree a and b in the generating elements dx j , then α ∧ β = ( − 1) ab β ∧ α. A general form is the sum of forms of degree 0, 1, . . . , n . No form can have degree more than n , since any monomial in a wedge of n + 1 terms has to look like ( · · · ∧ dx i ∧ · · · ∧ dx i ∧ · · · ), which is zero by the antisymmetry rules. 1

2 For example, in R 3 , the differential forms are linear combinations of these forms. degree 0 f f dx 1 + g dx 2 + h dx 3 degree 1 f dx 2 ∧ dx 3 + g dx 3 ∧ dx 1 + h dx 1 ∧ dx 2 degree 2 f dx 1 ∧ dx 2 ∧ dx 3 , degree 3 where f, g, h ∈ C ∞ ( U ). In dimension n , the C ∞ ( U )-vector space of all the differential forms of � n � degree d has dimension . d A digression about the Maxwell equations. For another example, in R 4 , the forms of degree 2 all look like − E x dt ∧ dx − E y dt ∧ dy − E z dt ∧ dz + B x dy ∧ dz + B y dz ∧ dx + B z dx ∧ dy. Here E x , E y , E z , B x , B y , B z ∈ C ∞ ( U ). Then Maxwell’s equations look like � ∂ B � ∗ dω = ic ∇ · B dt + 1 ∂t + ∇ × E = 0 ic � − 1 ∂ E � = q j ∗ d ∗ ω = ∇ · E dt + ∂t + ∇ × B dt + c 2 ε 0 c 2 ε 0 which is pretty cool. Here the ‘ ∗ ’ operator is defined in terms of the Riemannian metric, which is where ‘ c ’ comes in. It works as follows: Pick an orientation on R n , or just pick some form ω of degree n . Let ω 1 , ω 2 , . . . , ω n be any orthonormal basis in R n that has that orientation, or any orthonormal basis for which the wedge ω 1 ∧ ω 2 ∧· · ·∧ ω n , which is always equal to Jω for some constant J � = 0, is set up so that J > 0. Then ∗ ( ω j ∧ ω j +1 ∧ · · · ∧ ω n ) = ω 1 ∧ ω 2 ∧ · · · ∧ ω j − 1 and for any permutation σ = σ 1 · · · σ j − 1 σ j · · · σ n , ∗ ( ω σ ( j ) ∧ ω σ ( j +1) ∧ · · · ∧ ω σ ( n ) ) = sign( σ ) ω σ (1) ∧ ω σ (2) ∧ · · · ∧ ω σ ( j − 1) and we can extend by linearity. We made a choice of basis. The definition only depends on the orientation and not on the basis, but this isn’t obvious (well, it may be obvious to you). For example, an orthonormal basis in Minkowski R 4 is dx, dy, dz, ic dt . Ordinarily our differential forms have only real components, but since we are hanging out in Minkowski space we have to make some concessions. We have ∗ ( dz ∧ dt ) = (1 /ic ) ∗ ( dz ∧ ic dt ) = (1 /ic ) dx ∧ dy , for example.

3 The operator ‘d’. Let f be a function. Then we say n ∂f � ∂x j dx j . d f = j =1 Notice especially that d ( x j ) = dx j . Let ω = f dx I be a monomial differential form. Here I is supposed to be a sequence i 1 , · · · , i d , so dx I means dx i 1 ∧ dx i 2 ∧ · · · ∧ dx i d . Then we define d by f ∧ dx I , dω = d extending by linearity to all differential forms. This operation increases the degree of a differential form by one. Forms of degree n vanish under the operation d , since any form of degree n + 1 has to be zero. It also has the property that d 2 ω = 0 . This is because � ∂f ∂ 2 f � � � d 2 f dx I = d ∂x j dx j ∧ dx I ∂x i ∂x j dx i ∧ dx j ∧ dx I = = 0 by antisymmetry and the fact that partial derivatives commute. Also, if ω and η are two differential forms of degree d and e , then d ( ω ∧ η ) = d ( ω ) ∧ η + ( − 1) d ω ∧ d ( η ) . First, make the observation that d ( fg ) = g d f + f dg. If we consider monomials ω = f dx I and η = f dx J , then d ( ω ∧ η ) = d ( fg dx I ∧ dx J ) f + f dg ) ∧ dx I ∧ dx J = ( g d f ∧ dx I ) ∧ ( g dx J ) + ( − 1) d ( f dx I ) ∧ ( dg ∧ dx J ) = ( d = d ( ω ) ∧ η + ( − 1) d ω ∧ d ( η ) . The factor of ( − 1) d comes from exchanging dg with dx I . The pullback map. Let F : U → U ′ be a smooth map between two open sets, U ⊂ R n , and U ′ ⊂ R m ; then we define the pullback map F ∗ as follows. If ω = f dx ′ i 1 ∧ dx ′ i 2 ∧ · · · ∧ dx ′ i d is a differential form on U ′ , then F ∗ ( ω ) is F ∗ ( ω ) = ( f ◦ F ) d ( x ′ i 1 ) ∧ d ( x ′ i 2 ) ∧ · · · ∧ d ( x ′ i d ) We are thinking of x ′ j as functions on U , so d ( x ′ j ) = ∂F j ∂x i dx i .

4 We now have the really nice property that dF ∗ = F ∗ d . Let’s see why. Let ω = f dx ′ I be a monomial differential form. We compare the value of F ∗ dω and dF ∗ ω . � ∂f ∂x ′ j � = ∂f ∂x ′ j dx ′ j ∧ dx ′ I ∂x i dx i ∧ dx ′ I = d ( f ◦ F ) ∧ dx ′ I F ∗ dω = F ∗ ∂x ′ j and dF ∗ ω = d [( f ◦ F ) dx ′ I ] = d ( f ◦ F ) ∧ dx ′ I since d ( dx ′ j ) = d 2 x ′ j = 0. The general case follows by linearity. Another important property is that, when we have two maps F G → U ′ → U ′′ , U − − then ( G ◦ F ) ∗ = F ∗ ◦ G ∗ . This map takes differential forms on U ′′ to differential forms on U . This is easy to check for functions, and for elements dx ′′ i we write � ∂x ′′ i ( G ◦ F ) ∗ ( dx ′′ i ) = ∂x ′′ i ∂x j dx j = ∂x ′′ i ∂x ′ m � ∂x j dx j = F ∗ ∂x ′ m dx m = F ∗ ( G ∗ ( dx ′′ i )) . ∂x ′ m A similar computation works for all monomial differential forms, and so it’s true for all forms by linearity. The definition of a differentiable manifold. A topological manifold of dimension m is a topological space M where each point x ∈ M has a neighbourhood U x which is homeomorphic to an open set in R m . These homeomorphisms ξ x : U ⊂ R m → U x are called coordinate charts. A differentiable manifold of dimension m is a manifold where we have fixed a collection of coordinate charts, and where they match up smoothly. x = ξ − 1 This means that if U x ∩ U y isn’t empty, and U ′ x ( U x ), then ξ − 1 x ∩ ξ − 1 y ∩ ξ − 1 ◦ ξ x : U ′ x ( ξ y ( U ′ y )) → U ′ y ( ξ x ( U ′ x )) y is a diffeomorphism between open subsets of R n . We say that a map F : M → N between two manifolds is smooth if for any two points x ∈ M and y ∈ N , the composite map ξ − 1 ◦ F ◦ ξ x , y a map from a subset of R m to a subset of R n , is smooth on the open set x ∩ ( ξ − 1 ◦ F ◦ ξ x ) − 1 ( U ′ U ′ y ) where it’s defined. y The important thing that the coordinate charts give us is the concept of a differentiable function, a smooth function, and so on, on this topological space. We now need to give some examples of manifolds, and say what it means to have a differential form on a manifold. Examples of manifolds. We can consider R n or any open subset of R n to be a differentiable manifold with one coordinate chart, the identity map. There’s only one other differentiable manifold that we use in the proof of the Brouwer fixed point theorem, namely the n -dimensional sphere, S n .