Differential Forms in R n

Transcription

1 Wintersemester 2014/2015 University of Heidelberg Differential Forms seminar Differential Forms in R n by Stephan Schmitt Sven Grützmacher

2 Contents Differential Forms in R 3 1 Differential Forms in R n 2 Exterior Product Pullback exterior derivative divergence and gradient laplacian

3 Stephan Schmitt DFFERENTAL FORMS N R 3 Differential Forms in R 3 We will start with an introduction to Differential Forms in R 3, after that we will naturally generalize the concept to R n and proof important properties about them. Definition 1.1. Let p R 3, denote the set {q p q R 3 } by R 3 p. This space is called the tangent space(of R 3 in p). The vectors e i ; i = 1, 2, 3 form a canonical basis for R 3 0. Their translates for p, denoted by (e i ) p, form a Basis for R 3 p. Consider now the dual (R 3 p) of R 3 p. This space is called the Co-Tangent Space. We obtain a basis for this space by taking dx i, i = 1, 2, 3, where x i : R 3 R is the map assigning a point p its i-th coordinate. Proposition 1.2. The set {(dx i ) p ; i = 1, 2, 3} is the dual basis of {(e i ) p } Proof. (dx i ) p (e j ) = x i x j = { 0 if i j 1 if i = j Let us now introduce two essential maps working on those spaces: Definition 1.3. A map v : R 3 R 3 p, given by v(p) = 3 a i (p)e i where a i : R 3 R is called Vector Field. t is called differentiable, if the a i are. Definition 1.4. A map ω : R 3 (R 3 p), ω(p) = i=1 3 a i (p)(dx i ) p is called exterior form of degree 1 and Differential 1-Form, if the a i are differentiable. i=1 As the name suggests, there are forms of higher degree: Definition 1.5. Denote by Λ 2 (R 3 p) the set of all alternating, bilinear maps R 3 p R 3 p R. f ω 1, ω 2 (R 3 p), define ω 1 ω 2 Λ 2 (R 3 p) as. ω 1 ω 2 (v 1, v 2 ) = det(ω i (v j )) We need a basis for this space, conveniently the set {(dx i dx j ) p ; i = 1, 2, 3, i < j} is one. This result will be proven later in a more general manner. Also notice the following as an imediate result of this definition: Proposition 1.6. and Definition 1.7. A map ω : R 3 Λ 2 (R 3 p), (dx i dx j ) p = (dx j dx i ) p ; i j ω(p) = i<j dx i dx i = 0 a ij dx i dx j ; i, j = 1, 2, 3 where a ij are real functions, is called exterior form of degree 2 or differential 2-form, if the a ij are differentiable. Sven Grützmacher 1

4 DFFERENTAL FORMS N R N Stephan Schmitt Differential Forms in R n We will now generalize these concepts to R n. The definitions of tangent and co-tangent space are the exact same with n in place of 3. We will start with the n-dimensional variant of Λ 2 (R n p ) : Definition 2.1. Let Λ k (R n p ) denote the space of alternating, k-linear maps from R n p R n p R. Analogous to 3 dimensions if ω i (R n p ) ; i = 1,..., k we get an element ω 1 ω k Λ k (R n p ) by setting ω 1 ω k (v 1... v k ) = det(ω i (v j ))); i, j = 1,..., k Again we need a Basis for this space: Proposition 2.2. The Set forms a Basis for Λ k (R n p ). {dx i 1 dx i k ; i 1 < i 2 < < i k ; i j {1,..., n}} Proof. The elements of this Set are linearly independet. To see this, take: i 1 < <i k a i1...i k dx i1 dx ik = 0 and apply it to this gives (e j1... e jk ), j 1 < < j k, j l {1,..., n} i 1 < <i k a i1...i k dx i1 dx ik (e j1,..., e jk ) = a j1...j k = 0 Because: Start by looking at the determinant ρ = dx i l(e jn ). Assuming i 1 > j 1 it follows that i k > > i 1 > j 1. This in turn implies dx i l(e j1 ) = 0; l = 1,..., k. n turn assuming i 1 < j 1 gives j k > > j 1 > i 1, implying: dx i 1 (e jn ) = 0; n = 1,..., k. Both giving us ρ = 0 if i 1 j 1. Now assume i 1 = j 1 and j 2 i 2. Using the exact same Argument as above it follows that ρ = 0 if i 2 j 2. terating gives us the desired (dx i 1 dx i k )(e j1,..., e jk ) = 0 if j l i l. t remains to see that the above expression gives 1 if i 1 = j 1,..., i k = j k. But this is an immediate result of the definition of forms. We will now show f Λ k (R n p ) f = i 1 < <i k a i1...i k dx i 1 dx i k Define g = Clearly: i 1 < <i k f(e i1,..., e ik )dx i1 dx ik g(e i1... e ik ) = f(e i1... e ik ); i 1,... i k t follows that f = g. Setting f(e i1,..., e ik ) = a i1...i k Definition 2.3. A map ω : R n Λ k (R n p ), given by f = the result follows. i 1 < <i k a i1...i k (p)(dx i1 dx ik ) p ; i j {1,..., n} 2 Sven Grützmacher

5 Stephan Schmitt DFFERENTAL FORMS N R N where the a i1...i k are real functions in R n, if they are differentiable, this map is called differential k-form. n the following we will denote (i 1,..., i k ), i 1 < < i k ; i k {1,..., n} with. Therefore ω can be written ω = a dx Further we will set the convention to refer to differentiable functions as differential 0-forms. Let us now look at an example for forms in R 4 Example 2.4. Let a i, a ij,... denote real functions. 0-forms: functions in R 4 1-forms: a 1 dx 1 + a 2 dx 2 + a 3 dx 3 + a 4 dx 4 2-forms: a 12 dx 1 dx 2 + a 13 dx 1 dx 3 + a 14 dx 1 dx 4 + a 23 dx 2 dx 3 + a 24 dx 2 dx 4 + a 34 dx 3 dx 4 3-forms: a 123 dx 1 dx 2 dx 3 + a 124 dx 1 dx 2 dx 4 + a 134 dx 1 dx 3 dx 4 + a 234 dx 2 dx 3 dx 4 4-forms: a 1234 dx 1 dx 2 dx 3 dx 4 From now on we will restrict ourself to differential k-forms, or simply k-forms. n the following we will define operations of k-forms to see their structure. Exterior Product Definition 2.5. Let α = a dx, β = b dx denote two k-forms, their sum is definded to be α + β = (a + b )dx. Now let α = a dx be a k-form and β = J b Jdx J an s-form. Their exterior product is defined to be: α β = J a b J dx dx J, which is an (s+k)-form. Notice that this agrees, by definition of the determinant, with our previous definition of. Example 2.6. Let us look at the explicit exterior product of two forms, therefore let α = x 1 dx 1 + x 2 dx 2 + x 3 dx 3 denote a 1-form in R 3 and β = x 1 dx 1 dx 2 + dx 1 dx 3 a 2-form. Using dx i dx i = 0 and dx i dx j = (dx j dx i ), i j, we obtain: α β = x 2 dx 2 dx 1 dx 3 + x 3 x 1 dx 3 dx 1 dx 2 = (x 1 x 3 x 2 )dx 1 dx 2 dx 3 Proposition 2.7. Let α be a k-form, β a s-form and γ an r-form, then: a) α (β γ) = (α β) γ b) α β = ( 1) ks (β α) c) α (β + γ) = α β + α γ, if r = s Proof. a) and c) trivially follow from definition, let us look at b): Denote α = a dx, β = b J dx J J. Then: α β = J a b J dx i 1 dx i k dx j 1 dx js Switching one indice, we obtain α β = J b J a ( 1)dx i 1 dx i k 1 dx j 1 dx i k dx js Sven Grützmacher 3

6 DFFERENTAL FORMS N R N Stephan Schmitt repeating this k-times, we get: b J a ( 1) k dx j 1 dx i 1 dx i k dx j 2 dx js J Since J has s elements, we repeat this process for all j l J and obtain α β = ( 1) ks (β α) Pullback So far we introduced the tangent space and co-tangent space with maps giving you elements of those spaces. Another important question is, given R n and R m can we somehow get differential forms in R n from those in R m. The next definition will give us a possibility to do so in a very natural way. All we need is a differentiable function. Definition 2.8. Let f : R n R m be a differentiable function. This induces a map, denoted by f, called the pullback of f, that takes k-forms in R m to k-forms in R n. For a k-form ω in R m (f w) p (v 1,..., v k ) = ω(f(p))(df p (v 1 ),..., df p (v k )). Where df p : R n p R m f(p) is the differential of f at point p R n and v 1,..., v k R n p. n the special case of g being a 0-form(a differentiable function), we set: f g = g f As always, we need some properties for this new map, after that we will try to make more sense of this map. Proposition 2.9. Let f : R n R m be a differentiable function, α, β k-forms on R m and g : R m R a 0-form in R m, then: a) f (α + β) = f α + f β b) f (gα) = f (g)f (α) c) f α 1,..., α k are 1-forms in R m, f (α 1 α k ) = f (α 1 ) f (α k ) Proof. These proofs will be rather short: a) f (α + β)(p)(v 1,..., v k ) = (α + β)(f(p))(df p (v 1 ),..., df p (v k )) = (f α)(p)(v 1,..., v k ) + (f β)(p)(v 1,..., v k ) = (f α + f β)(p)(v 1,..., v k ) b) f (gα)(p)(v 1... v k ) = (gα)(f(p))(df p (v 1 ),..., df p (v k )) = (g f)(p)f α(p)(v 1,..., v k ) = f g(p)f α(p)(v 1,..., v k ) c) f (α 1 α k ) = (α 1 α k )(df(v 1 ),..., df(v k )) = det(α i (df(v j )) = det(f α i (v j )) = (f α 1 f α k )(v 1,..., v k ) 4 Sven Grützmacher

7 Stephan Schmitt DFFERENTAL FORMS N R N Note that the last part of this proposition is also true for k-forms, and will be proven so later. For now we try to make sense of the pullback by interpreting it as a substitution of variables, to justify this interpretation: Example Let (x 1,..., x n ) be coordinates in R n and (y 1,..., y m ) in R m, define f : R n R m by y 1 = f 1 (x 1,..., x n ),..., y m = f m (x 1,..., x n ). Now let ω = a dy be a k-form in R m. Using the last Proposition we obtain f ω = f (a )(f dy i 1 ) (f dy i k ) since we finally get f ω = f (dy i )(v) = dy i (df(v)) = d(y i f(v)) = df i (v) a (f 1 (x 1,..., x n ),..., f m (x 1,..., x n ))df i 1 df i k where f i and df i are functions of x j. Thus, applying f to ω is equivalent to substitue the variables y i and their differentials in ω by x k and dx k obtained abvore. Another thing we have yet to mention: Subsets of R n. Up until now we only referred to R n directly, sometimes however it is more convenient to use the concepts we introduced only on Subsets of R n. As long as these are open, everything done up to now extends trivially. To showcase these last two facts see: Example Let ω be the 1-form in R 2 {0, 0} given by Next define and let f : U R 2 be the map ω = y x 2 + y 2 dx + x x 2 + y 2 dy U = {r > 0; 0 < θ < 2π} f(r, θ) = We can now compute f ω. For that, notice which gives us f ω = rsinθ r 2 { x y = rcosθ = rsinθ dx = cosθdr rsinθdθ dy = sinθdr + rcosθdθ (cosθdr rsinθdθ) + rcosθ r 2 (sinθdr + rcosθdθ) = dθ We can now proof another important fact, the Pullback actually commutes with the exterior product: Proposition Let f : R n R m be a differentiable map. Then a) f (α β) = (f α) (f β), where α is a k-form and β a s-form in R m b) (f g) α = g (f α), g : R p R n differentiable Proof. Setting (y 1,..., y m ) = (f 1 (x 1,..., x n ),..., f m (x 1,..., x n )) R m,(x 1,..., x n ) R n, α = a dy, β = J b Jdy J We get: Sven Grützmacher 5

8 DFFERENTAL FORMS N R N Stephan Schmitt a) f(α β) = f ( J a b J dy dy J ) = J a (f 1,..., f m )b J (f 1,..., f m )df df J We can now seperate this and get: a (f 1,..., f m )df J b J (f 1,..., f m )df J = f α f β b) (f g) α = = a ((f g) 1,..., (f g) m )d(f g) a (f 1 (g 1,..., g n ),..., f m (g 1,..., g n ))df (dg 1,..., dg n ) = g (f (α)) exterior derivative Unitl now we have seen some properties of k-forms in R n. Yet we haven t covered why those forms are also called : differential forms. This name implies some form of differentiability we haven t seen so far. To motivate the following definition we take a look at 0-forms which we understand quite well. Example Let g : R n R be a 0-form (and thus a smooth function). Then its differential is given by dg = n i=1 g x i dx i which is a 1-form We generalize this idea to k-forms now Definition Let w = a dx be a k-form in R n. The exterior differential is dw = da dx We will see later that this generalizes some concepts of differential calculus. To show this we will take a look at a little example und some properties of exterior differentiation: Example Let w = xydx + e z dy + xdz. Then dw = d(xy) dx + d(e z ) dy + d(x) dz = (ydx + xdy) dx + e z dz dy + dx dz = xdx dy + dx dz e z dy dz Proposition Let w be a k-form, ϕ a s-form in R n and f : R m R n smooth. Then a) d is R-linear b) d(w ϕ) = dw ϕ + ( 1) k w dϕ 6 Sven Grützmacher

9 Stephan Schmitt DFFERENTAL FORMS N R N c) d(dw) = d 2 w = 0 d) d(f w) = f (dw) Proof. a) follows directly from the definition b) Sei w = a dx und ϕ = b J dx J. Dann gilt ( ) d(w ϕ) = d (a b J )dx dx J J = d(a b J ) dx dx J J = J b J da dx dx J + J a db J dx dx J = dw ϕ + ( 1) k J a dx db J dx J = dw ϕ + ( 1) k w dϕ c) we will give this proof in two steps. First let w be a 0-form and thus a smooth function R n R; x f(x 1,..., x n ). Then ( n ) f n ( ) f n d(df) = d dx i = d dx i 2 f n = dx j dx i = ji fdx j dx i x i x i x j x i i=1 i=1 i,j=1 Since f is smooth the order of differtiation does not matter, e.g. ij f = ji f. Furthermore dx j dx i = dx i dx j. Thus we get d(df) = i<j ( ij f ji f)dx i dx j = 0 i,j=1 Now let w = a dx. Through linearity it suffices to consider w = a dx where a 0. Using b) we get while d(dx ) = d(1) dx = 0 we can compute dw = da dx + a d(dx ) d(dw) = d(da dx ) = d(da ) dx + da d(dx ) = 0 since we have already shown d 2 = 0 in the cases above. d) we use the same method as in c). So let g : R n R; y g(y 1,..., y n ) be a smooth function. Then ( ) f (dg) = f g dy i = g f i dx j = (g f) dx j = d(g f) = d(f g) y i i y ij i x j x j j Now again let w = a dx be a k-form. Since f commutes with the exterior product we obtain ( ) d(f w) = d f (a )f (dx ) = d(f (a )) f (dx ) = f (da ) f (dx ) = f ( da dx ) = f (dw) Sven Grützmacher 7

10 DFFERENTAL FORMS N R N Stephan Schmitt The first properties certainly justify the name differential form since we got a product rule and linearity. Property c) is probably the most important one which can be easily seen if you have some basic knowledge in Ko-/Homology-theory. Since in R n we can identify smooth vector fields with smooth 1-forms via the scalarproduct induced isomorphism we can take a look at some concepts of differential calculus to see in which way they re connected to 1-forms. divergence and gradient The divergence and the gradient are commonly known objects in the first calculus lectures. There they re often defined by concrete formulas. Yet they can naturally be defined using the exteroir derivative Definition Let v be a vector field in R n. We can identify v with a map R n R n. Then the divergence of v is a function R n R defined as follows: div(v)(p) = trace(dv) p Since we know what dv p looks like concerning the standard basis we deduce the known formula for v = a i e i div(v) = a i x i Definition Given a smooth function f : R n R we define a vector field grad(f) : R n R n through f we choose the standard basis for R n p we get < grad(f)(p), u >= df p (u), u R n p graf(f) = f x i e i which can be easily derived from the initial definition by < grad(f)(p), u >= df p (u) = f x i dx i (u) = f x i u i laplacian A combination of the two concepts above is the laplacian which involves second derivatives. Definition Let f : R n R be a smooth function. Then the lapacian is f = div(grad(f)); R n R As before we can describe it in local coordinates: ( ) f f = div e i = x i i ij 2 f x i x j 8 Sven Grützmacher