Math 05 Integration and calculus of several variables week 8 - May 8, 009. Geometry We have developed the calculus of differential forms algebraically, focusing on algebraic manipulations which can be used to calculate the integrals and formulate the basic Stokes s theorem. In doing so, we have never needed to do geometry, i.e. we have never talked about lengths of vectors or angles. (In technical terms, we have worked on a C -manifold, not a Riemannian manifold.) Recall in R n we have the notions of length of a vector and of angle between two vectors. Namely, the length of v = (v,..., v n ) is defined to be v := (v + + vn) / = (v v) /. The angle θ between vectors v and w is determined by cos θ = v w v w ; v w := v w + v w + + v n w n. To a differential -form ω = f dx + + f n dx n on R n we associate the n-tuple of functions (f,..., f n ) which we interpret geometrically as a vector field on R n. A vector field assigns to each point x 0 := (x 0,..., x 0 n) R n a vector, namely (f (x 0 ),..., f n (x 0 )). We can picture a vector field as simply a collection of arrows eminating from points in R n. Here are two examples in R with coordinates x, y. Out[5]= 0 Out[6]= 0 Vector field (y, x) Vector field (y, +x) Exercise. (i) Sketch the vector fields (x, y) and (x +, y ) in R. (ii) Suppose xf(x, y, z) + yg(x, y, z) + zh(x, y, z) = 0. sketch the vector field (f, g, h) in R.
An important example of a vector field is the tangent vector field along a path. Suppose φ : I R n is given by φ(t) = (x (t),..., x n (t)). Then dφ/dt := (dx /dt(t),..., dx n /dt(t)) is the tangent vector field. Notice it is only defined along the path φ. Here is the picture in R when φ(t) = (t, t ). 4 0 Tangent vector field to φ(t) = (t, t ) (Ignore the fact that the arrows seem to curve. This is an artifact of the computer program.) We can now write our typical path integral computation ω = f dx + + f n dx n = φ φ b a (f (φ(t))dx /dt + + f n (φ(t))dx n (t)/dt)dt = b a (f (φ(t)),..., f n (φ(t))) dφ/dtdt In words, we take the dot product of the vector field associated to the -form with the tangent vector field and integrate the resulting function
of t over [a, b]. The picture for ω = ydx + xdy and φ(t) = (t, t ) looks like Here the path φ is in blue, the tangent vector field is in red, and the vector field associated to ω is in green. (I am quite proud of this illustration, by the way.) The expression ds := dφ/dt dt = ((dx /dt) + + (dx n /dt) ) / dt is the traditional notation for the form computing arc length. From our point of view, this is confusing firstly because ds is not d of a function. Secondly, ds depends on the path and is only defined along the path. Our usual situation has the -form independent of the path and defined at least in some neighborhood of the path. On the plus side, φ b ds a is independent of the parametrization as a path integral should be.
4 Namely, writing t = ψ(u), we find ds = ((dx /dt) + + (dx n /dt) ) / )dt = ((dx /dt) + + (dx n /dt) ) / dψ/dudu = ((dx /dt) (dψ/du) + + (dx n /dt) (dψ/du) ) / du = ((dx (ψ(u))/du) + + (dx n (ψ(u))/du) ) / du Note that the bottom line is the expression for ds in the u-coordinate. In Wade, the unit tangent vector field along φ is written T := dφ/dt/ dφ/dt The path integral becomes ω = φ b a (f φ,..., f n φ) T ds. Exercise. Compute the path lengths and the tangent vector fields for the following paths on [0, ]: φ(t) = (t, t ); φ(t) = (t, t, t ); φ(t) = (cos(πt), sin(πt)). You may leave the arc length integrals unevaluated if they seem difficult.. Surface integrals in R There is a classical geometric construction called the cross product of vectors in R. It is defined by x y := (x y x y, x y x y, x y x y ). Lemma. (i). x ( x y) = y ( x y) = 0. (ii) x y equals the (unsigned) area of the parallelogram spanned by x and y. Proof. (i) is straightforward. Geometrically, this means that x y is perpendicular to the plane spanned by x and y. For (ii), note first that even to talk about area in R (as opposed to volume) is something intrinsically new for us. Technically, one would say that volume is invariant under the group of matrices with determinant ±, while area is only invariant under the group O of -matrices preserving lengths and angles. (This is called the orthogonal group.) Once we have lengths and angles, we can define the area of a parallelogram spanned by two vectors to be the volume of the solid parallelogram obtained by thickening the original figure taking the product with a normal vector of unit length. ( Normal means perpendicular to x and y.)
Write x = (x, x, x ) and y = (y, y, y ). The key algebraic fact which you should check is 5 x y = x y ( x y). Once you have this, you can prove (ii) axiomatically. Show x x = 0 and ( x + z) y = x y + z y. Finally, show x y = x y when x is perpendicular to y. Exercise 4. Write out the proof of lemma in full detail. Consider now a surface integral ω in ϕ R. -form and ϕ : I R. Write This means ω is a ω = f dx dx + f dx dx + f dx dx. Substituting as usual dx i = dx i dt + dx i du we find dt du ω = ϕ (f ϕ, f ϕ, f ϕ) Ndt du I Here N = ( x / t x / u x / t x / u, x / t x / u x / t x / u, x / t x / u x / t x / u) = ( x / t, x / t, x / t) ( x / u, x / u, x / u).
6 In the picture, the two red arrows are the tangents ( x / t, x / t, x / t) ~. and ( x / u, x / u, x / u). The chartreuse arrow is the normal N ~ is perpendicular to the tangent vectors by lemma (i). Note that N ~ / N ~ for the unit normal vector, and Following Wade, I write ~n = N ~ dtdu. dσ := N Exercise 5. Argue from lemma (ii) that dσ gives a plausible definition for surface area in R, i.e. that one should define ZZ Area(ϕ(I )) := dσ. I Finally, our surface integral now reads (assuming we have oriented R I so dt du > 0) ZZ ZZ ω= (f ϕ, f ϕ, f ϕ) ~ndσ. ϕ I Exercise 6. Define ϕ(t, u) = (cos(t) cos(u), cos(t) sin(u), sin(t)) for 0 t π, 0 u π. Write down the surface area integral for this surface. Compute ~n(t, u) and draw the picture.
Exercise 7. Let ϕ(t, u) = (t, u, t + u ). Write down the surface area integral for this surface where 0 t, u. Compute n(t, u) and draw the picture. Compute dx dy + dy dz + dz dx. ϕ 7