lecture22

lecture22 - Lecture 22 In R3 we had the operators grad,...

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Unformatted text preview: Lecture 22 In R3 we had the operators grad, div, and curl. What are the analogues in Rn ? Answering this question is the goal of today’s lecture. 4.9 Tangent Spaces and k ­forms Let p ∈ Rn . Definition 4.36. The tangent space of p in Rn is Tp Rn = {(p, v ) : v ∈ Rn }. We identify the tangent space with Rn via the identification Tp Rn ∼ Rn = (p, v ) → v . (4.145) (4.146) (4.144) Via this identification, the vector space structure on Rn gives a vector space structure on Tp Rn . Let U be an open set in Rn , and let f : U → Rm be a C 1 map. Also, let p ∈ U and define q = f (p). We define a linear map dfp : Tp Rn → Tq Rm according to the following diagram: Tp Rn − − Tq Rm −→ ⏐ � ∼⏐ ∼⏐ =� =⏐ Rn So, dfp (p, v ) = (q, Df (p)v ). Definition 4.37. The cotangent space of Rn at p is the space ∗ Tp Rn ≡ (Tp Rn )∗ , dfp (4.147) (4.148) −− − → Rm . (4.149) Df (p) (4.150) which is the dual of the tangent space of Rn at p. Definition 4.38. Let U be an open subset of Rn . A k ­form on U is a function ω ∗ which assigns to every point p ∈ U an element ωp of Λk (Tp Rn ) (the k th exterior power ∗n of Tp R ). 1 Let us look at a simple example. Let f ∈ C ∞ (U ), p ∈ U , and c = f (p). The map dfp : Tp Rn → Tc R = R (4.151) ∗ is a linear map of Tp Rn into R. That is, dfp ∈ Tp Rn . So, df is the one­form on U which assigns to every p ∈ U the linear map ∗ ∗ dfp ∈ Tp Rn = Λ1 (Tp R). (4.152) As a second example, let f, g ∈ C ∞ (U ). Then gdf is the one­form that maps ∗ p ∈ U → g (p)dfp ∈ Λ1 (Tp Rn ). (4.153) As a third example, let f, g ∈ C ∞ (U ). Then ω = df ∧ dg is the two­form that maps p ∈ U → dfp ∧ dgp . (4.154) ∗ ∗ Note that dfp , dgp ∈ Tp R, so dfp ∧ dgp ∈ Λ2 (Tp Rn ). As a fourth and final example, let f1 , . . . , fk ∈ C ∞ (U ). Then df1 ∧ · · · ∧ dfk is the k ­form that maps (4.155) p ∈ U → (df1 )p ∧ · · · ∧ (dfk )p . ∗ ∗ Note that each (dfi )p ∈ Tp Rn , so (df1 )p ∧ · · · ∧ (dfk )p ∈ Λk (Tp Rn ). Let us now look at what k ­forms look like in coordinates. Let e1 , . . . , dn be the standard basis of Rn . Let p ∈ U and let vi = (p, ei ) for each i. Then, the vectors v1 , . . . , vn are a basis of Tp Rn . Suppose we have a map f ∈ C ∞ (U ). What is dfp (vi )? dfp (vi ) = Dei f (p) = ∂f (p). ∂xi (4.156) In particular, letting xi be the ith coordinate function, � 1 if i = j , ∂xi (dxi )p (vj ) = = ∂xj 0 if i = j . � ∗ So, (dx1 )p , . . . , (dxn )p is the basis of Tp Rn dual to v1 , . . . , vn . ∞ For any f ∈ C (U ), (4.157) dfp (vj ) = = ∂f (p) ∂xj � � ∂f i � (p)(dxi )p (vj ) (4.158) ∂xi � ∂f =⇒ dfp = (p)(dxi )p ∂xi � ∂f =⇒ df = dxi . ∂ xi 2 ∗ Since (dx1 )p , . . . , (dxn )p is a basis of Tp Rn , the wedge products (dxI )p = (dxi1 )p ∧ · · · ∧ (dxik )p , 1 ≤ i1 < · · · < ik ≤ n, ∗ (I strictly increasing) are a basis of Λk (Tp Rn ). ∗ Therefore, any element wp of Λk (Tp Rn ) can be written uniquely as a sum (4.159) ωp = � aI (p)(dxI )p , aI (p) ∈ R, (4.160) where the I ’s are strictly increasing. Hence, any k ­form can be written uniquely as a sum � ω= aI dxI , I strictly increasing, (4.161) where each aI is a real­valued function on U . That is, aI : U → R. Definition 4.39. The k ­form ω is C r (U ) if each aI ∈ C r (U ). Just to simplify our discussion, from now on we will always take k ­forms that are C ∞. Definition 4.40. We define Ωk (U ) = the set of all C ∞ k ­forms. � So, ω ∈ Ωk (U ) implies that ω = aI dxI , where aI ∈ C ∞ (U ). Let us now study some basic operations on k ­forms. (4.162) 1. Let ω ∈ Ωk (U ) and let f ∈ C ∞ (U ). Then f ω ∈ Ωk (U ) is the k ­form that maps ∗ p ∈ U → f (p)ωp ∈ Λk (Tp Rn ). (4.163) 2. Let ωi ∈ Ωk (U ), i = 1, 2. Then ω1 + ω2 is the k ­form that maps ∗ p ∈ U → (ω1 )p + (ω2 )p ∈ Λk (Tp Rn ). (4.164) 3. Let ωi ∈ Ωki (U ), i = 1, 2, and k = k1 + k2 . Then w1 ∧ ω2 ∈ Ωk (U ) is the k ­form that maps ∗ p ∈ U → (ω1 )p ∧ (ω2 )p ∈ Λk (Tp Rn ), (4.165) ∗ since (ωi )p ∈ Λki (Tp Rn ). ∗ Definition 4.41. We find it convenient to define Λ0 (Tp Rn ) = R. 3 A zero­form f on U is just a real­valued function, so Ω0 (U ) = C ∞ (R). Take f ∈ C ∞ (U ) and df ∈ Ω1 (U ). This gives an operation d : Ω0 (U ) → Ω1 (U ), f → df. (4.166) (4.167) Let f, g ∈ C ∞ (U ) (that is, take f, g to be zero­forms). Then d(f g ) = gdf + f dg . We can think of this operation as a slightly different notation for the gradient operation. The maps d : Ωk (U ) → Ωk+1 (U ), k = 0, . . . , (n − 1) give n vector calculus operations. If ω ∈ Ωk (U ), then ω can be written uniquely as the sum � ω= aI dxI , I strictly increasing, (4.168) where each aI ∈ C ∞ (U ). We define dω = � daI ∧ dxI . (4.169) This operator is the unique operator with the following properties: � ∂f dxi . 1. For k = 0, this is the operation we already defined, df = ∂xi 2. If ω ∈ Ωk , then d(dω ) = 0. 3. If ωi ∈ Ωki (U ), i = 1, 2, then d(ω1 ∧ ω2 ) = dω1 ∧ ω2 + (−1)k1 ω1 ∧ dω2 . Let a ∈ C ∞ (U ), and adxI ∈ Ωk (U ), I strictly increasing. Then d(adxI ) = da ∧ dxI . Suppose that I is not strictly increasing. Then dxI = dxi1 ∧ · · · ∧ dxik = 0 if ir = is . (4.171) (4.170) If there are no repetitions, then there exists σ ∈ Sk such that J = I σ is strictly increasing. Then (4.172) dxJ = (−1)σ dxI , so d(adxI ) = (−1)σ d(adxJ ) = (−1)σ da ∧ dxJ = da ∧ dxI . Putting this altogether, for every multi­index I of length k , d(adxI ) = da ∧ dxI . (4.174) (4.173) 4 ...
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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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