Math 225B: Differential Geometry, Homework 6

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1 ath 225B: Differential Geometry, Homework 6 Ian Coley February 13, 214 Problem 8.7. Let ω be a 1-form on a manifol. Suppose that ω = for every lose urve in. Show that ω is exat. We laim that this onition is equivalent to the following: if : [, 1] is any urve, then ω epens only on ) an 1), that is, the integral of ω is path inepenent. We will show only the forwar iretion, whih we require. Suppose that, are two paths suh that ) = ) an 1) = 1). Then the omposition is a lose path at ), where t) = 1 t). Hene ω =. But we also have ω = ω ω = = ω = Therefore without loss of generality assume that is onnete, sine the following proof may be repeate for any number of onnete omponents. Then fix p an let, efine a funtion f : R by fx) = ω where γ is a path from p to x. Sine ω is path inepenent, this makes sense for any γ. Further, onsier the funtion gx) onstrute as above with basepoint q an path δ an γ from that q to p. Then we have gx) = ω = ω + ω = C + fx), δ γ γ where C is a onstant. Therefore every hoie of basepoint yiels the same funtion up to positive onstant epening on basepoint. We laim that f = ω. To see this, we will show they take the same value on any vetor fiel X at any p. Let p. Then onsier a urve through p so that ) = p an fx) p = X p f) = t ft)) t=, γ ω. 1

2 whih is always possible by results in previous setions. We parametrise the path between ) an t) the path t s) by t s) = st). Then This ompletes the proof. fx) = t ft)) t= = ω = t t t= t = 1 ) ω t s)) s t s t= = 1 ) ωst)) t st) s t s s = t ) ωu)) u u t t u t= ) = ωt)) t t t t= ) = ω)) = ωp)x p ). t t= 1 t ω Problem 8.8. A manifol is alle simply onnete if is onnete an if every smooth map f : S 1 is smoothly ontratible to a point. a) If is smoothly ontratible to a point, then is simply onnete. b) S 1 is not simply onnete. ) S n is simply onnete for n > 1. ) If is simply onnete an p, then any smooth map f : S 1 is smoothly ontratible to p. e) If = U V, where U an V are simply onnete open subsets with U V simply onnete, then is simply onnete. f) If is simply onnete, then H 1 ) =. a) Let p : be the smooth ontration of to a point, i.e. there is a homotopy H t : so that H = i an H 1 = {p}. Then let f : S 1 be any map. Then the omposition f H t = F t : I S 1 is a smooth homotopy with F = f an F 1 = {p}. Therefore f is smoothly ontratible to a point, so is simply onnete. b) If S 1 were simply onnete, then every map f : S 1 S 1 woul be homotopi to a onstant map. In partiular, the ientity map woul be homotopi to a onstant map, so S 1 woul be ontratible. But S 1 is a ompat manifol of positive imension, so it is not ontratible. Therefore S 1 is not simply onnete. t= 2

3 ) Let f : S 1 S k be a map. We woul like to show that this is homotopi to a onstant. Taking the hint, Sar s theorem allows us to hoose p S k \ fs 1 ) when k > 1. Reall that we showe in that S k \{p} is iffeomorphi to R k for k 1. Further, R k is ontratible, so it is simply onnete. Therefore fs 1 ) is iffeomorphi to a loop in R k, whih is homotopi to a onstant. Therefore this homotopy an be pulle bak along the iffeomorphism so fs 1 ) is simply onnete in S k \ {p}, hene in S k as well. This ompletes the proof. ) Sine is path onnete an embeable into Euliean spae, it is in partiular path onnete via loal harts. Therefore suppose that f : S 1 ontrats to a point q. Then there is a path γ : [, 1] so that γ) = q an γ1) = p. Therefore if H t is the homotopy between f an the point {q}, we have a homotopy F t : given by fx) t = H 2t x) < t < 1/2 F t x) = q t = 1/2. γ2t 1) 1/2 < t < 1 p t = 1 Hene f is ontratible to an arbitrary point p. e) Let f : S 1 be a smooth map. Firstly, if the image of f lans in either U or V alone, then f is ontratible to a point, so we are one. Therefore suppose that the image of f lans in both U an V, an that the image of f is not ontaine in U V. Let us parametrise S 1 by [, 1] for simpliity. Then we may break up [, 1] into isjoint segments X i = [a i, b i ] so that fx i ) U or V. From here, we may onnet the points fa i ) an fb i ) in U V by a path γ i to reate a loop in a simply onnete subset either U or V ). Therefore there is a homotopy of this loop to the loop onsisting of γ i on [, 1/2] an γ 1 i on [1/2, 1]. Deleting the bakwars iretion, we have taken the setion fx i ) of our loop an smoothly ontate it to lie ompletely in U V. Repeat this proess a finite number of times, so that the image of f lies ompletely in U V. Thus we are reue to the first ase, so we are one. f) We verify this via the integral. Assume that ω is a lose 1-form. Note that every lose loop in is the image of S 1 in uner some map f. By assumption, f is homotopi to a onstant map p, so by Theorem 8.13 their pullbaks are also equal. Therefore for a 1-form ω, ω = f ω = p ω = ω = S 1 S 1 {p} This hols for every lose urve. Therefore by Problem 8.7, ω is exat. Therefore H 1 ) = sine every 1-form is exat. 3

4 Problem a) Let be {x, y) R 2 : x, y) < 1} together with a proper portion of its bounary, an let ω = x y. Show that ω ω, even though both sies make sense. b) Similarly, fin an example to Stokes Theorem when =, 1) an ω is a -form whose support is not ompat. ) Examine a partition of unity for, 1) by funtions with ompat support to see just why the proof of Stokes Theorem breaks own in this ase. a) Let us examine D, the unit is in R 2. We may apply Stokes Theorem to this manifol: ω = ω. S 1 D Now let : [, 1] S 1 be an appropriately-oriente parametrisation of S 1. Then 1 ω = ω. D Now uner the assumption that we have only a proper portion of the bounary of the unit isk, we take α < 1 so that we have [, α]) =. Hene α ω = ω. Now we ompute iretly. Let t) = os2πt + ), sin2πt + )), where is an appropriate rotation fator. On one han, ω = x y) = x x x y + x y y = x y. y Sine this is a volume form on, we obtain ω = = π. However, ω = = = α α α x y) = α x ) y ) = os2πt + )2π os2πt + )) t os 2 2πt + )2πt 4 α os2πt + ) sin2πt + ))

5 Using the substitution u = 2πt +, = = 1 2 2πα+ 2πα+ os 2 u)u = 1 2 2πα+ 2 os 2 u) u os2u) + 1 u = 1 2 sinu) osu) + u) 2πα+ = 1 2 sin2πt + ) os2πt + ) + 2πt + ) α = πα + 1 sin2πα + ) os2πα + ) sin) os)) 2 = πα + 1 sin4πα + 2) sin2)). 4 It is lear that, for small values of α, we have ω < π. Therefore we hek the maximum of this integral on [, 1]. It an be verifie by taking the erivative of this integral with respet to α or using Wolfram Alpha that the maximum of this integral ours at α = 1. Hene we always have, for our hoie of α, 1), ω > ω, so in partiular they are not equal. b) Let =, 1), an let ω = x, the oorinate funtion. Sine =, the integral of any form over the bounary is zero. Then 1 x = x = 1 ω. ) The step of the proof where Stokes Theorem fails is that, given a partition of unity Φ, the support of ω nee not be ontaine in finitely many Φ. Therefore we annot onlue ω = ω = Φ Φ sine the marke equality only hols when ω is supporte in finitely many. Problem Suppose is a ompat orientable n-manifol with no bounary), an θ is an n 1)-form on. Show that θ is at some point. First, sine we may put a finite partition of unity on, we reue to the loal ase. We know that θ = θ = θ =. 5

6 Further, we know that θ is a volume form, so it is of the form θ = f x 1 x n for some funtion f. Sine at every point the term x 1 x n ontributes positively to the integral, if f were always positive or always negative, then the entire integral woul be always positive or always negative. Therefore f = ientially, or f is somewhere positive an somewhere negative, hene is somewhere zero by the mean value theorem. Sine the sign of f etermines the sign of θ, we are one. Problem Let 1, 2 R n be ompat n-imensional manifols with bounary with 2 1 \ 1. Show that for any lose n 1)-form ω on 1, ω = ω. 1 2 An n-imensional submanifol of R n takes the inue orientation from R n, so i are orientable. Consier the manifol = 1 \ 2 2, so that = 1 2. Then applying Stokes theorem, = ω = ω = ω 1 2 ω, where the minus sign is introue sine the outwar pointing normal of 2 with respet to is the negation of the outwar pointing normal of 2 with respet to. Therefore ω = ω 1 2 as laime. 6

Chapter 2: One-dimensional Steady State Conduction

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