Vector space and Dual vector space Let V be a finite dimensional vector space over R. Suppose v1; :::; vn is a basis for V then the dual vector space to V is denoted V ∗ = ff : V ! Rg where f is a R-linear map. The next question might be, what is ∗ i ∗ i j i 1 n ∗ ∗ a basis for V ? Suppose α 2 V with α (v ) = δj. We claim that α ; :::; α is a basis for V . As V is P i also a vector space over R, suppose i ciα = 0, it then follows that: ! X i j ciα (v ) = cj = 0 : 8j i Now let f 2 V ∗ then we have by direct computation: ! X j X j X j j f(v) = f cjv = cj · f(v ) = α (v) · f(cjv ) j j j i.e α1; :::; αn are a basis for V ∗ and so dim(V ) = dim(V ∗) = n. 1 0 Example: Let V = 2 with scalar field with; e1 = e2 = and αi : 2 ! defined by R R 0 1 R R i j i 2 1 2 1 2 α (e ) = δj. Suppose v 2 R then v = αe + βe ) α (v) = α and α (v) = β. And so, the basis elements are precisely the projections onto the x; y axis respectively. Multilinear Functions Let V be a vector space and consider V × · · · × V (k-copies). We say f is a multilinear function if f is linear in each component i.e f(::::; av + bw; :::) = af(:::; v; :::) + bf(:::; w; :::). We say f is k-linear, f is a k-tensor or f is a k-form and denote Lk(V ) to be the set of all k-linear functions on V . n n n P j j Example: Let v; w 2 R and h·; ·i : R × R ! R defined by hv; wi = j v · w . Let f 2 Lk(V ). We say f is symmetric if given any τ 2 Sk : f(vτ(1); :::; vτ(k)) = f(v1; :::; vk). We say f is alternating if f(vτ(1); :::; vτ(k)) = (sgn τ)f(v1; :::; vk). Example: For any example of a symmetric function, consider the dot product given above. Example: Let g : R3 ! R be defined by g(v; w) = v × w which is the cross-product from multivariable calculus. Example: Let f = det : Mat(n; ) ! defined by f(A) = P (sgn τ) a a ··· a . If we let R R τ2Sn 1,τ(1) 1,τ(2) 1,τ(n) n A = [v1 v2 ··· vn] then f 2 An(R ). 1 Actions of Permutations on Functions Let f 2 Lk(V ) and σ 2 Sk then we define (σf)(v1; :::; vk) = f(vσ(1); :::; vσ(n)). With this action we have f is symmetric () σf = f and f is alternating () σf = (sgn σ)f. We now introduce a very useful lemma; P P Lemma: If f 2 Lk(V ) and σ 2 Sk then Sf = σ σf is symmetric and Af = σ(sgn σ) σf is alternating. Proof: Let τ 2 Sk then; ! X X X X τ(Sf) = τ σf = τ(σf) = (τσ)f = σf σ2Sk σ2Sk σ2Sk σ2Sk −1 The last inequality comes from the fact that for any σ 2 Sk we have σf = ττ σf. For the next result, notice that sgn(τσ) = (sgn τ)(sgn σ) ! X X X τAf = τ (sgn σ) σf = (sgn σ) τσf = (sgn τσ)(sgn τ) τσf σ2Sk σ2Sk σ2Sk X = (sgn τ) (sgn τσ) τσf σ2Sk X = (sgn τ) (sgn σ) σf σ2Sk = (sgn τ) Af Example: Let f 2 L3(V ) and suppose v1; v2; v3 2 V . Using the short hand notation for permutations, we have S3 = f(1); (12); (13); (23); (123); (132)g. All of the transpositions i.e cycles of length two, have by definition sign of −1. To determine the others, note that (123) = (13)(12) and (132) = (12)(13) i.e that have sign +1. 2 Tensor Product Let f 2 Lk(V ) and g 2 Lj(V ) then we define the tensor product; f ⊗ g :(v1; :::; vk; vk+1; :::; vk+1) 7! f(v1; :::; vk)g(vk+1; :::; vk+j) The juxtaposition above denotes the standard multiplication which makes sense due to the fact that f; g are functionals. In this case, we can think of the tensor ⊗ as a bilinear operator; ⊗ : Lk(V ) × Lj(V ) ! Lk+j(V ) ⊗(f; g) = f ⊗ g The operation above can be shown to be associative. The next example will show that we can express the inner product on a vector space, as a linear combination of tensor products. Example: We will take a special case where h·; ·i : Rn × Rn ! R. If we take e1; :::; en to be the standard i j basis and he ; e i = gij then we have; * + X i i X j j X i j i j X i j X i j X 1 j hv; wi = v e ; b e = v b he ; e i = v b gij = α (v)α (w)gij = gij(α ⊗ α )(v; w) i j i;j i;j i;j i;j In the above we have αi(v) = vi and αj(w) = wj by our comment on the maps in the dual vector space being the coordinates with respect to the standard basis. The Wedge Product Let f 2 Ak(V ) and g 2 Al(V ) then we define their exterior product or wedge product; 1 f ^ g = A(f ⊗ g) k!l! In coordinates, if we let v1; :::; vk+l 2 V then; 1 X (f ^ g)(v ; :::; v ) = (sgn σ) f(v ; :::; v ) · g(v ; :::; v ) 1 k+1 k!l! σ(1) σ(k) σ(k+1) σ(k+l) σ2Sk+l 3 The division by k!l! comes from the fact that in Sk+l there are permutations σ such that σ(j) = j for all j > k. For such σ we have σ(f ⊗ g)(v1; :::; vk+l) = (sgn σ)f(v1; :::; vk) · σg(vk+1; :::; vk+l). Since jSkj = k!, this value is repeated k!-times. Similarly we have permutations τ 2 Sk+l with τ(j) = j for all j < k + 1 and so τ(f ⊗ g)(v) is repeated l!-times. Our last note is that by definition f ^ g is alternating. Properties of Wedge Product We will prove a few of these properties however some are a bit messy and will be reserved for k-forms on tangent spaces. kl • If f 2 Ak(V ); g 2 Al(V ) then f ^ g = (−1) g ^ f • If f 2 A2k+1 then f ^ f = −f ^ f (by above) and so f ^ f = 0. 1 • If f; g; h 2 A (V ) then (f ^ g) ^ h = f ^ (g ^ h) = A(f ⊗ g ⊗ h) ∗ k!l!h! i • If α 2 L1(V ) and vi 2 V then; 1 k X 1 2 k i (α ^ · · · ^ α )(v1; :::; vk) = (sgn σ) α (v1) · α (v2) ··· α (vk) = det[α (vj)] σ2Sk Multi-Index Notation 1 n P i j i j Let e ; :::; e be a basis for V . Consider f 2 L2(V ) then f(v; w) = i;j v w f(e ; e ) then f is determined j by its values on e . Since we will be only considering forms in Ak(V ) we make another observation; n n X X X viwjf(ei; ej) = viwjf(ei; ej) i;j j=1 i=1 In any case when i > j we have f(ei; ej) = −f(ej; ei). Therefore, we can get the same sum if we sum over 1 ≤ i < j ≤ n since the only thing we are doing is using the property of f to establish an identity, still keeping all summands. X f(v; w) := f(ei; ej)(αi ⊗ αj)(v; w) 1≤i<j≤n I i1 ik We introduce the multi-index notation I = (i1; :::; ik) and write e = (e ; :::; e ). By the above, since f 2 Ak(V ) it is enough to consider strictly increasing indices. 4 n Remark: If we defined A∗(V ) = ⊕k=0Ak(V ) then (A∗(V ); ^) is an anti-commutative graded algebra, called the exterior algebra or the Grassmann algebra of multicovectors. Basis for k-covectors Let e1; :::; en be a basis for V and α1; :::; αn be its dual basis in V ∗. Consider the strictly increasing multi-indices I = (1 ≤ i1 < ··· < ik ≤ n) and J = (1 ≤ j1 < ··· < jk ≤ n) then; I i1 ik ir α (eJ ) = (α ^ · · · ^ α )(ej1 ; :::; ejk ) = det[α (ejs )] ir If I = J then [α (ejs )] = I and so the determinant is 1. If I 6= J we look at the indices (is; js) until ir 6= jr. Without loss of generality, suppose ir < jr then ir 6= jr; jr+1; :::; jn. We also have ir 6= j1; :::; jr−1 ir since we have equality of pairs until ir i.e det[α (ejs )] = 0 since the r-th row has all zeros. I I We will now show that the alternating k-linear functions α , for a basis for Ak(V ). Since α is also a vector space over R suppose we have; ! X I X I J aI α = 0 () aI α (e ) = aJ = 0 I I I The above shows that α are linearly independent. To show that they span Ak(V ), let f 2 Ak(V ). We P I ∗ claim f = f(eI )α . This consideration comes from when we showed the one forms spanned V . Let P I I J P I I J v 2 V and define g = f(e )α then g(e ) = f(e )α (e ) = f(eJ ) ) f = g.