Exterior Algebra Differential Forms

Transcription

1 Exterior Algebra Differential Forms Faraad M Armwood North Dakota State September 4, 2016 Faraad M Armwood (North Dakota State) Exterior Algebra Differential Forms September 4, / 17

2 Dual Space Let V, W be a f.d.v.s and let Hom(V, W ) = {f f : V W is linear}. Then the dual vector space V V to V is Hom(V, R). The elements of V V are called 1-covectors or covectors on V. If we let e 1,..., e n be a basis for V then if v V we have v = k v k e k. Suppose α i (e j ) = δ i j then α i (v) = v i where v = (v 1,..., v n ). Therefore we have α i : V R is a linear map and so α i V V. It follows that the {α i : i = 1,..., n} are linearly independent and they span V V i.e dim(v ) = dim(v V ). Faraad M Armwood (North Dakota State) Exterior Algebra Differential Forms September 4, / 17

3 Example Let V = R 2 and x 1, x 2 be the standard coordinates on R 2 i.e if p = (p 1, p 2 ) then x i (p) = p i. Let e 1 = (1, 0) T, e 2 = (0, 1) T denote the standard basis then x i (e j ) = δj i. This example and the above demonstrate that for a f.d.v.s the dual basis is determined by the coordinates of a point in the standard basis. Faraad M Armwood (North Dakota State) Exterior Algebra Differential Forms September 4, / 17

4 Multilinear Functions Let V k = V V be the k-product of a real vector space V. We say f : V k R is k-linear if is linear in each of its k-arguments i.e; f (..., av + bw,...) = af (..., v,...) + bf (..., w,...) A k-linear function on V is also called a k-tensor on V. We will denote by L k (V ) the set of all k-tensors on V. If f is a k-tensor, we also say that f has degree k. Faraad M Armwood (North Dakota State) Exterior Algebra Differential Forms September 4, / 17

5 Example Let v, w T p R 2 and let f =, : T p R 2 T p R 2 R defined by; v, w = k v k w k then f is a 2-linear or bilinear function on R 2. Here T p R 2 denoted the tangent space on R 2 i.e v, w are understood to be vectors. Faraad M Armwood (North Dakota State) Exterior Algebra Differential Forms September 4, / 17

6 Symmetric and Alternating Tensors Let f L k (V ) then we say f is symmetric if; We say f is alternating if; f (v σ(1),..., v σ(k) ) = f (v 1,..., v k ), σ S k f (v σ(1),..., v σ(k) ) = sgn(σ) f (v 1,..., v k ), σ S k Faraad M Armwood (North Dakota State) Exterior Algebra Differential Forms September 4, / 17

7 Example (1) The inner product define in the last example is symmetric on T p R 2 and even in the extension to T p R n due to the commutability of R. (2) A simple function f (x, y) = x + y is also symmetric (3) The cross product v w on R 3 is alternating Faraad M Armwood (North Dakota State) Exterior Algebra Differential Forms September 4, / 17

8 Symmetrizing and Alternatrizing Operators Let f (v 1,..., v k ) L k (V ) and σ S k. Then we can define an action of a perm. on f by σf := f (v σ(1),..., v σ(k) ). We now demonstrate how to get an alternating and symmetric frunction from f ; Sf = σ S k σf (symmetric) Af = σ S k (sgnσ) σf (alternating) Faraad M Armwood (North Dakota State) Exterior Algebra Differential Forms September 4, / 17

9 Example Let f L 3 (V ) and suppose v 1, v 2, v 3 V then; Af (v 1, v 2, v 3 ) = σ S 3 (sgnσ) σf Recall that S 3 = {(1), (12), (13), (123), (132)} and to determine their sign we observe that (123) = (13)(12), (132) = (12)(13) and so; Af = f (v 1, v 2, v 3 ) f (v 2, v 1, v 3 ) f (v 3, v 2, v 1 ) + f (v 3, v 2, v 1 ) + f (v 2, v 3, v 1 ) (12)Af = f (v 2, v 1, v 3 ) f (v 1, v 2, v 3 ) f (v 3, v 1, v 2 )+f (v 3, v 1, v 2 )+f (v 1, v 3, v 2 ) Faraad M Armwood (North Dakota State) Exterior Algebra Differential Forms September 4, / 17

10 Tensor Product Let f L k (V ) and g L l (V ) then we defined their tensor product; (f g)(v 1,..., v k+l ) = f (v 1,..., v k )g(v k+1,..., v k+l ) L k+l (V ) The operation above is associative i.e (f g) h = f (g h). Faraad M Armwood (North Dakota State) Exterior Algebra Differential Forms September 4, / 17

11 Example Let e 1,..., e n be a basis for V and f =, : V V R. Define e i, e j = g ij and v = i v i e i, w = j w j e j then by the previous remarks, α i (v) = v i, α j (w) = w j and so; v, w = i,j g ij (α j α i )(w, v), = i,j g ij α j α i Faraad M Armwood (North Dakota State) Exterior Algebra Differential Forms September 4, / 17

12 The Wedge Product I Let f A k (V ) and g A l (V ) then we define their exterior product or wedge product to be; or explicitely; f g = 1 A(f g) k!l! (f g)(v) = 1 k!l! ( ) (sgnσ) f (v σ(1),..., v σ(k) )g vσ(k+1),..., v σ(k+l) σ S k+l where v = (v 1,..., v k+l ) and dividing out by k!l! compensates for the repetition in the sum. Faraad M Armwood (North Dakota State) Exterior Algebra Differential Forms September 4, / 17

13 Discussion (1) If σ(k + j) = k + j, j = 1,..., l then σf = sgn(σ) f and σg = g and so for each such σ you get fg in the amount of k! times. Similarly if τ(j) = j, j = 1,..., k then you get fg in the amount of l! times. Now convince yourself that there are no other repetitions in the sum. (2) Let f A 0 (V ) and g A l (V ) then f is a constant function say c R and; c g(v 1,..., v l ) = 1 (sgnσ) 2 cg(v 1,..., v l ) = cg l! σ S l Faraad M Armwood (North Dakota State) Exterior Algebra Differential Forms September 4, / 17

14 The Wedge Product II Another way to compensate for the repeated terms in the sum for the tensor product is to arrange that σ(1) < < σ(k) and σ(k + 1) < < σ(k + l). If σ S k+l is such a permutation, we say that σ is a (k, l) shuffle. Therefore we have; (f g)(v) = (k,l) shuffles σ (sgnσ) f (v σ(1),..., v σ(k) )g(v σk+1,..., v σk+l ) The two definitions are not the same! Faraad M Armwood (North Dakota State) Exterior Algebra Differential Forms September 4, / 17

15 Example To demonstrate that the two definitions differ, we take f A 1 (V ) and g A 2 (V ) then the permutation group in discussion is S 3, but the only (1, 2) shuffle is the identity. Now compute f g by the original definition and see that they differ. Faraad M Armwood (North Dakota State) Exterior Algebra Differential Forms September 4, / 17

16 Properties of Wedge Product (1) If f A k (V ) and g A l (V ) then f g = ( 1) kl g f. (2) If f A 2k+1 (V ) then f f = 0 (3) (f g) h = f (g h), g, h, f A (V ) (4) If α i L 1 (V ) and v i V then; (α 1 α k )(v 1,..., v k ) = det[α i (v j )] Faraad M Armwood (North Dakota State) Exterior Algebra Differential Forms September 4, / 17

17 Basis for k-covectors Lemma: Let e 1,..., e n be a basis for a v.s V and α 1,..., α n be its dual basis in V V. If I = (1 i 1 < < i k n) and J = (1 j 1 < < j k n) are strictly ascending multi-indices of length k then; α I (e J ) = δ I J Proposition: The alternating k-linear function α I, I = (i 1 < < i k ), form a basis for the space A k (V ) of alternating k-linear functions on V i.e if f A k (V ) then; f = I a I α I. Corollary: If k > dim(v ) then A k (V ) = 0. Faraad M Armwood (North Dakota State) Exterior Algebra Differential Forms September 4, / 17