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, 2016 1 / 17 Dual Space Let V ; W be a f.d.v.s and let Hom(V ; W ) = ff j f : V ! W is linearg. V V Then the dual vector space V to V is Hom(V ; R). The elements of V are called 1-covectors or covectors on V . P k If we let e1; :::; en be a basis for V then if v 2 V we have v = k v ek . i i i i 1 n Suppose α (ej ) = δj then α (v) = v where v = (v ; :::; v ). Therefore we i i V have α : V ! R is a linear map and so α 2 V . It follows that the fαi : i = 1; :::; ng 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, 2016 2 / 17 Example 2 1 2 2 Let V = R and x ; x be the standard coordinates on R i.e if i T T p = (p1; p2) then x (p) = pi . Let e1 = (1; 0) ; e2 = (0; 1) denote the i i standard basis then x (ej ) = δj . 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, 2016 3 / 17 Multilinear Functions Let V k = V × · · · × V be the k-product of a real vector space V . We say k f : V ! 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 Lk (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, 2016 4 / 17 Example 2 2 2 Let v; w 2 TpR and let f = h·; ·i : TpR × TpR ! R defined by; X hv; wi = v k w k k 2 2 then f is a 2-linear or bilinear function on R . Here TpR denoted the 2 tangent space on R i.e v; w are understood to be vectors. Faraad M Armwood (North Dakota State) Exterior Algebra ! Differential Forms September 4, 2016 5 / 17 Symmetric and Alternating Tensors Let f 2 Lk (V ) then we say f is symmetric if; f (vσ(1); :::; vσ(k)) = f (v1; :::; vk ); 8σ 2 Sk We say f is alternating if; f (vσ(1); :::; vσ(k)) = sgn(σ) · f (v1; :::; vk ); 8σ 2 Sk Faraad M Armwood (North Dakota State) Exterior Algebra ! Differential Forms September 4, 2016 6 / 17 Example 2 (1) The inner product define in the last example is symmetric on TpR n and even in the extension to TpR due to the commutability of R. (2) A simple function f (x; y) = x + y is also symmetric 3 (3) The cross product ~v × w~ on R is alternating Faraad M Armwood (North Dakota State) Exterior Algebra ! Differential Forms September 4, 2016 7 / 17 Symmetrizing and Alternatrizing Operators Let f (v1; :::; vk ) 2 Lk (V ) and σ 2 Sk . Then we can define an action of a perm. on f by σf := f (vσ(1); :::; vσ(k)). We now demonstrate how to get an alternating and symmetric frunction from f ; X Sf = σf (symmetric) σ2Sk X Af = (sgnσ) σf (alternating) σ2Sk Faraad M Armwood (North Dakota State) Exterior Algebra ! Differential Forms September 4, 2016 8 / 17 Example Let f 2 L3(V ) and suppose v1; v2; v3 2 V then; X Af (v1; v2; v3) = (sgnσ) σf σ2S3 Recall that S3 = f(1); (12); (13); (123); (132)g and to determine their sign we observe that (123) = (13)(12); (132) = (12)(13) and so; Af = f (v1; v2; v3) − f (v2; v1; v3) − f (v3; v2; v1) + f (v3; v2; v1) + f (v2; v3; v1) (12)Af = f (v2; v1; v3)−f (v1; v2; v3)−f (v3; v1; v2)+f (v3; v1; v2)+f (v1; v3; v2) Faraad M Armwood (North Dakota State) Exterior Algebra ! Differential Forms September 4, 2016 9 / 17 Tensor Product Let f 2 Lk (V ) and g 2 Ll (V ) then we defined their tensor product; (f ⊗ g)(v1; :::; vk+l ) = f (v1; :::; vk )g(vk+1; :::; vk+l ) 2 Lk+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, 2016 10 / 17 Example Let e1; :::; en be a basis for V and f = h·; ·i : V × V ! R. Define P i P j hei ; ej i = gij and v = i v ei ; w = j w ej then by the previous remarks, αi (v) = v i ; αj (w) = w j and so; X j i X j i hv; wi = gij (α ⊗ α )(w; v) ) h; i = gij α ⊗ α i;j i;j Faraad M Armwood (North Dakota State) Exterior Algebra ! Differential Forms September 4, 2016 11 / 17 The Wedge Product I Let f 2 Ak (V ) and g 2 Al (V ) then we define their exterior product or wedge product to be; 1 f ^ g = A(f ⊗ g) k!l! or explicitely; 1 X (f ^ g)(v) = (sgnσ) f (v ; :::; v )g v ; :::; v k!l! σ(1) σ(k) σ(k+1) σ(k+l) σ2Sk+l where v = (v1; :::; vk+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, 2016 12 / 17 Discussion (1) If σ(k + j) = k + j; 8j = 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; 8j = 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 2 A0(V ) and g 2 Al (V ) then f is a constant function say c 2 R and; 1 X c ^ g(v ; :::; v ) = (sgnσ)2 cg(v ; :::; v ) = cg 1 l l! 1 l σ2Sl Faraad M Armwood (North Dakota State) Exterior Algebra ! Differential Forms September 4, 2016 13 / 17 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 2 Sk+l is such a permutation, we say that σ is a (k; l) shuffle. Therefore we have; X (f ^ g)(v) = (sgnσ) f (vσ(1); :::; vσ(k))g(vσk+1; :::; vσk+l ) (k;l)−shuffles σ The two definitions are not the same! Faraad M Armwood (North Dakota State) Exterior Algebra ! Differential Forms September 4, 2016 14 / 17 Example To demonstrate that the two definitions differ, we take f 2 A1(V ) and 3 g 2 A2(V ) then the permutation group in discussion is S , 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, 2016 15 / 17 Properties of Wedge Product kl (1) If f 2 Ak (V ) and g 2 Al (V ) then f ^ g = (−1) g ^ f . (2) If f 2 A2k+1(V ) then f ^ f = 0 (3) (f ^ g) ^ h = f ^ (g ^ h); 8g; h; f 2 A∗(V ) i 1 k i (4) If α 2 L1(V ) and vi 2 V then; (α ^ · · · ^ α )(v1; :::; vk ) = det[α (vj )] Faraad M Armwood (North Dakota State) Exterior Algebra ! Differential Forms September 4, 2016 16 / 17 Basis for k-Covectors 1 n Lemma: Let e1; :::; en be a basis for a v.s V and α ; :::; α be its dual basis V in V . If I = (1 ≤ i1 < ··· < ik ≤ n) and J = (1 ≤ j1 < ··· < jk ≤ n) are strictly ascending multi-indices of length k then; I I α (eJ ) = δJ I Proposition: The alternating k-linear function α ; I = (i1 < ··· < ik ), form a basis for the space Ak (V ) of alternating k-linear functions on V i.e if f 2 Ak (V ) then; X I f = aI α I . Corollary: If k > dim(V ) then Ak (V ) = 0. Faraad M Armwood (North Dakota State) Exterior Algebra ! Differential Forms September 4, 2016 17 / 17.