lecture23

lecture23 - Lecture 23 Let U be an open set in Rn . For...

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Unformatted text preview: Lecture 23 Let U be an open set in Rn . For each k = 0, . . . , n − 1, we define the differential operator d : Ωk (U ) → Ωk+1 (U ). (4.175) These maps are the n basic vector calculus operations in n­dimensional calculus. We review how d is defined. For k = 0, Ω0 (U ) = C ∞ (U ). Let f ∈ C ∞ (U ), and let c = f (p), where p ∈ U . ∗ The mapping dfp : Tp Rn → Tc R = R maps Tp Rn to R, so dfp ∈ Tp Rn . The map 1 ∗n df ∈ Ω (U ) is a one­form that maps p ∈ U to dfp ∈ Tp R . A formula for this in coordinates is � ∂f (4.176) df = dxi . ∂xi In k dimensions, d is a map d : Ωk (U ) → Ωk+1 (U ). Given ω ∈ Ωk (U ), ω can be written uniquely as � ω= aI dxI I (4.177) = � I (4.178) aI dxi1 ∧ · · · ∧ dxik , where i1 < · · · < ik and each aI ∈ C ∞ (U ). Then, we define � dω = daI ∧ dxI � ∂ aI = dxi ∧ dxI , ∂xi i,I where each I is strictly increasing. The following are some basic properties of the differential operator d: 1. If µ ∈ Ωk (U ) and ν ∈ Ω� (U ), then dµ ∧ ν = dµ ∧ ν + (−1)k µ ∧ dν. 2. For and ω ∈ Ωk (U ), d(dω ) = 0. Remark. Let I be any multi­index, and let aI ∈ C ∞ (U ). Then d(aI dxI ) = daI ∧ dxI . (4.179) (4.180) (4.181) (4.182) 1 We now prove the above two basic properties of the differential operator. Claim. If µ ∈ Ωk (U ) and ν ∈ Ω� (U ), then dµ ∧ ν = dµ ∧ ν + (−1)k µ ∧ dν. (4.183) � � Proof. Take µ = aI dxI and ν = bJ dxJ , where I, J are strictly increasing. Then � µ∧ν = aI bJ dxI ∧ dxJ . (4.184) � �� � no longer increasing Then d(µ ∧ ν ) = = � ∂ aI b J i,I,J ∂xi dxi ∧ dxI ∧ dxJ (4.185) � ∂ aI bJ dxi ∧ dxI ∧ dxJ (I ) ∂xi � ∂ bJ dxi ∧ dxI ∧ dxJ , (II ) + aI ∂xi We calculate sums (I ) and (II ) separately. (I ) = � ∂ aI i,I,J ∂xi dxi ∧ dxI ∧ bJ dxJ � dxi ∧ dxI ∧ � J = � � ∂ aI i,I ∂xi bJ dxJ (4.186) = dµ ∧ ν. (II ) = ∂ bJ dxi ∧ dxI ∧ dxJ ∂xi i,I,J � ∂ bJ = (−1)k aI dxI ∧ dxi ∧ dxJ ∂xi i,I,J � � � ∂ bJ � dxi ∧ dxJ = (−1)k aI dxI ∧ ∂ xi I i,J � aI = (−1)k µ ∧ dν. So, d(µ ∧ ν ) = (I ) + (II ) = dµ ∧ ν + (−1)k µ ∧ dν. (4.188) (4.187) 2 Claim. For and ω ∈ Ωk (U ), d(dω ) = 0. Proof. Let ω = � aI dxI , so dω = Then, d(dω ) = � ∂aI j,I (4.189) ∂xj dxj ∧ dxI . (4.190) � ∂ 2 aI dxi ∧ dxj ∧ dxI . ∂xi ∂xj i,j,I (4.191) Note that if i = j , then there is a repeated term in the wedge product, so d(dω ) = � ∂ 2 aI dxi ∧ dxj ∧ dxI ∂xi ∂xj i<j � ∂ 2 aI + dxi ∧ dxj ∧ dxI . ∂xi ∂xj i>j (4.192) (4.193) Note that dxi ∧ dxj = −dxj ∧ dxi . We relabel the second summand to obtain ⎛ ⎞ d(dω ) = � ⎜ ∂ 2 aI ∂ 2 aI ⎟ ⎜ ⎟ dxi ∧ dxj ∧ dxI − ⎝ ∂xi ∂xj ∂xj ∂xi ⎠ i<j � �� � 0 (4.194) = 0. Definition 4.42. A k ­form ω ∈ Ωk (U ) is decomposable if ω = µ1 ∧ · · · ∧ µk , where each µi ∈ Ω1 (U ). Theorem 4.43. If ω is decomposable, then dω = k � i=1 (−1)i−1 µ1 ∧ · · · ∧ µi−1 ∧ dµi ∧ µi+1 ∧ · · · ∧ µk . (4.195) Proof. The proof is by induction. 3 The case k = 1 is obvious. We show that if the theorem is true for k − 1, then the theorem is true for k . d((µ1 ∧ · · · ∧ µk−1 ) ∧ µk ) = (d(µ1 ∧ · · · ∧ µk−1 )) ∧ µk + (−1)k−1 (µ1 ∧ · · · ∧ µk−1 ) ∧ dµk = k −1 � i=1 (−1)i−1 µ1 ∧ · · · ∧ dµi ∧ · · · ∧ µk−1 ∧ µk k −1 (4.196) + (−1) = k � i=1 (µ1 ∧ · · · ∧ µk−1 ∧ µk ) (−1)i−1 µ1 ∧ · · · ∧ dµi ∧ · · · ∧ µk . 4.10 Pullback Operation on Exterior Forms Another important operation in the theory of exterior forms is the pullback operator. This operation is not introduced in 18.01 or 18.02, because vector calculus in not usually taught rigorously. Let U be open in Rn and V be open in Rm , and let f : U → V be a C ∞ map. We can write out in components f = (f1 , . . . , fn ), where each fi ∈ C ∞ (U ). Let p ∈ U and q = f (p). The pullback of the map dfp : Tp Rm → Tq Rn is ∗ (dfp )∗ : Λk (Tq∗ Rn ) → Λk (Tp Rm ). (4.197) Suppose you have a k ­form ω on V . ω ∈ Ωk (V ), ωq ∈ Λk (Tq∗ Rn ). Then ∗ (dfp )∗ wq ∈ Λk (Tp Rm ). (4.198) (4.199) (4.200) Definition 4.44. f ∗ ω is the k ­from whose value at p ∈ U is (dfp )∗ ωq . We consider two examples. Suppose φ ∈ Ω0 (V ) = C ∞ (V ). Then f ∗ φ(p) = φ(q ), so f ∗ φ = φ ◦ f , where f : U → V and φ : V → R. Again, suppose that φ ∈ Ω0 (V ) = C ∞ (V ). What is f ∗ dφ? Let f (p) = q . We have the map dφq : Tp Rn → Tc R = R, where c = φ(q ). So, (dfp )∗ (dφ)q = dφq ◦ dfp = d(φ ◦ f )p . (4.201) 4 Therefore, f ∗ dφ = df ∗ φ. Suppose that µ ∈ Ωk (V ) and ν ∈ Ωe ll(V ). Then (f ∗ (µ ∧ ν ))p = (dfp )∗ (µq ∧ νq ) ∗ = (dfp )∗ µq ∧ (dfp )νq . Hence, f ∗ (µ ∧ ν ) = f ∗ µ ∧ f ∗ ν. (4.204) We now obtain a coordinate formula for f ∗ . � Take ω ∈ Ωk (V ). We can write ω = aI dxi1 ∧ · · · ∧ dxik , where each aI ∈ C ∞ (U ). Then � f ∗ω = f ∗ aI f ∗ dxi1 ∧ · · · ∧ f ∗ dxik (4.205) � = f ∗ aI dfi1 ∧ · · · ∧ dfik , where we used the result that f ∗ dxi = dxi ◦ f = dfi . � ∂fi ∂fi Note that dfi = dxj , where ∂xj ∈ C ∞ (U ). Also, f ∗ aI = aI ◦ f ∈ C ∞ (U ), ∂xj which shows that f ∗ ω ∈ Ωk (U ). (4.206) The following theorem states a very useful property of the pullback operator. Theorem 4.45. Let ω ∈ Ωk (V ). Then, df ∗ ω = f ∗ dω. (4.207) (4.203) (4.202) Proof. We have already checked this for ω = φ ∈ C ∞ (V ), k = 0 already. We now prove the general case. � We can write ω = aI dxI . Then � f ∗ω = f ∗ aI dfi1 ∧ · · · ∧ dfik . (4.208) So, df ∗ ω = � df ∗ aI ∧ dfi1 ∧ · · · ∧ dfik � + f ∗ aI ∧ d(dfi1 ∧ · · · ∧ dfik ) (4.209) Note that d(dfi1 ∧ · · · ∧ dfik ) = k � r =1 (−1)r−1 dfi1 ∧ · · · ∧ d(dfir ) ∧ · · · ∧ dfik . (4.210) 5 We know that d(dfir ) = 0, so df ∗ ω = = � I df ∗ aI ∧ dfi1 ∧ · · · ∧ dfik f ∗ daI ∧ f ∗ (dxi1 ∧ · · · ∧ dxik ) (4.211) � I � = f ∗( daI ∧ dxI ) = f ∗ dω. 6 ...
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This note was uploaded on 10/02/2010 for the course MAT unknown taught by Professor Unknown during the Fall '04 term at MIT.

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