3.7 The Wedge Product 25 3.6 The Tensor Product Let f be a k-linear function and g an -linear function a vector space V.Their tensor product is the (k+)-linear function fg defined by (f⑧g)(1,+)=f(v1,)8(+,+) Example 3.17 (Euclidean inner product).Let e.,en be the standard basis for Rm and leta,.a be its dual basis.The Euclidean inner product on R"is the bilinear function(,:R"×R"-R defined by (v.w)=>vwi for v=ve;and w=wei.We can express(.)in terms of the tensor product: (,w)='w=∑ad'(w)a'(w) =(a⑧a)u,w). Exercise 3.18(Associativity of the tensor product).Check that the tensor product of multi- linear functions is associative:if f.g.and h are multilinear functions on V.then (f®g)®h=f⑧(g⑧h). 3.7 The Wedge Product If two multilinear functions f and g on a vector space V are alternating,then we would like to have a product that is alternating as well.This motivates the definition of the wedge product:for f Ak(V)and g A(V). fN8=CAC®8 (3.1) or explicitly, (fA8)(v1,k+) n)f(((). (3.2) GESk+1 By Proposition 3.13.f A g is alternating When k =0.the elem nt fA(V)is simply a constant c.In this case.the wedge product cg is sca rmultiplication,since the right-hand side of (3.2)is
3.7 The Wedge Product 25 3.6 The Tensor Product Let f be a k-linear function and g an -linear function on a vector space V . Their tensor product is the (k + )-linear function f ⊗ g defined by (f ⊗ g)(v1,.,vk+ ) = f (v1,.,vk)g(vk+1,.,vk+ ). Example 3.17 (Euclidean inner product). Let e1,.,en be the standard basis for Rn and let α1,.,αn be its dual basis. The Euclidean inner product on Rn is the bilinear function , : Rn × Rn −→ R defined by v, w = vi wi for v = vi ei and w = wi ei. We can express , in terms of the tensor product: v, w = i vi wi = i αi (v)αi (w) = i (αi ⊗ αi )(v, w). Hence, , = i αi ⊗ αi . This notation is often used in differential geometry to describe an inner product on a vector space. Exercise 3.18 (Associativity of the tensor product). Check that the tensor product of multilinear functions is associative: if f, g, and h are multilinear functions on V , then (f ⊗ g) ⊗ h = f ⊗ (g ⊗ h). 3.7 The Wedge Product If two multilinear functions f and g on a vector space V are alternating, then we would like to have a product that is alternating as well. This motivates the definition of the wedge product: for f ∈ Ak(V ) and g ∈ A (V ), f ∧ g = 1 k! ! A(f ⊗ g); (3.1) or explicitly, (f ∧ g)(v1,.,vk+ ) = 1 k! ! σ∈Sk+ (sgn σ )f (vσ (1),.,vσ (k))g(vσ (k+1),.,vσ (k+ )). (3.2) By Proposition 3.13, f ∧ g is alternating. When k = 0, the element f ∈ A0(V ) is simply a constant c. In this case, the wedge product c ∧ g is scalar multiplication, since the right-hand side of (3.2) is
26 3 Alternating k-Linear Functions 1 >(sgna)cg(va().va()=cg(v1.vt). !aese Thus cAg=cg forc ER and g E Ae(V). The coefficient 1/(k!e!)in the definition of the wedge product compensates for repetitions in the sum:for every permutation o Sk+,there are k!permutations in S that permute the first k arguments v().v)and leave the arguments of g alone;for all r in S,the resulting permutations ot in S+contribute the same term to the sum since (sgnat)f(var(D).vat())=(sgnat)(sgn t)f(va(D).va(k)) =(sgna)f(va(D).v(k)). where the first equality follows from the fact that((1).(k))is a permutation of (1.k).So we divide by k!to get rid of the k!repeating terms in the sum coming from the permutations of the k arguments of f:similarly.we divide by e!on account of the e arguments of g. Example3.19.Forf∈A2(V)andg∈A1(V, A(f⑧g)(U1,2,v3)=f(U1,2)g(3)-f(v1,3)g(v2)+f(2,v3)g(v1) -f(2,)g()+f(3,)g()-f(3,2)g(u1). Among these 6 terms,there are three pairs of equal terms: f(v1,2)g(v3)=-f(2,v1)g(3),ands0on Therefore,after dividing by 2. (fAg)(v1,2,3)=f(w1,2)g(3)-f(v1,3)g(v2)+f(2,3)g(w1). One way to avoid redundancies in the definition of fAg is to stipulate that in the sum (3.2),(1),.(k)be in ascending order and a(+1),. o(k+)also be in ascending order.We call a permutationS+a(k.()-shuffle if o(1)<.··<o(k)ando(k+1)<.··<o(k+) Then one may rewrite(3.2)as (fAg(,+) (sgna)f(va(1).Va(k))8(va(+I).va(+)). (33) (k.()-shuffles Written this way,the definition of (fAg)(vI,.v+)is a sum of (terms, instead of (k+)!terms. Example 3.20(Wedge product of two covectors).If f and g are covectors on a vector space V and vI,v2 E V,then by (3.3) (jAg)(v1,2)=f(u)g(U2)-f(2)g(U). Exercise 3.21(Wedge product oftwo2-covectors).For.gA(V).writeouthe definition of fg using (2.2)-shuffles
26 3 Alternating k-Linear Functions 1 ! σ∈S (sgn σ )cg(vσ (1),.,vσ ( )) = cg(v1,.,v ). Thus c ∧ g = cg for c ∈ R and g ∈ A (V ). The coefficient 1/(k! !) in the definition of the wedge product compensates for repetitions in the sum: for every permutation σ ∈ Sk+ , there are k! permutations τ in Sk that permute the first k arguments vσ (1),.,vσ (k) and leave the arguments of g alone; for all τ in Sk, the resulting permutations σ τ in Sk+ contribute the same term to the sum since (sgn σ τ )f (vστ(1),.,vσ τ (k)) = (sgn σ τ )(sgn τ )f (vσ (1),.,vσ (k)) = (sgn σ )f (vσ (1),.,vσ (k)), where the first equality follows from the fact that (τ (1), . . . , τ (k)) is a permutation of (1,. , k). So we divide by k! to get rid of the k! repeating terms in the sum coming from the permutations of the k arguments of f ; similarly, we divide by ! on account of the arguments of g. Example 3.19. For f ∈ A2(V ) and g ∈ A1(V ), A(f ⊗ g)(v1, v2, v3) = f (v1, v2)g(v3) − f (v1, v3)g(v2) + f (v2, v3)g(v1) − f (v2, v1)g(v3) + f (v3, v1)g(v2) − f (v3, v2)g(v1). Among these 6 terms, there are three pairs of equal terms: f (v1, v2)g(v3) = −f (v2, v1)g(v3), and so on. Therefore, after dividing by 2, (f ∧ g)(v1, v2, v3) = f (v1, v2)g(v3) − f (v1, v3)g(v2) + f (v2, v3)g(v1). One way to avoid redundancies in the definition of f ∧ g is to stipulate that in the sum (3.2), σ (1), . . . , σ (k) be in ascending order and σ (k + 1), . . . , σ (k + ) also be in ascending order. We call a permutation σ ∈ Sk+ a (k, )-shuffle if σ (1) < ··· < σ (k) and σ (k + 1) < ··· < σ (k + ). Then one may rewrite (3.2) as (f ∧ g)(v1,.,vk+ ) = (k, )-shuffles σ (sgn σ )f (vσ (1),.,vσ (k))g(vσ (k+1),.,vσ (k+ )). (3.3) Written this way, the definition of (f ∧ g)(v1,.,vk+ ) is a sum of k+ k terms, instead of (k + )! terms. Example 3.20 (Wedge product of two covectors). If f and g are covectors on a vector space V and v1, v2 ∈ V , then by (3.3) (f ∧ g)(v1, v2) = f (v1)g(v2) − f (v2)g(v1). Exercise 3.21 (Wedge product of two 2-covectors). For f, g ∈ A2(V ), write out the definition of f ∧ g using (2, 2)-shuffles.
3.8 Anticommutativity of the Wedge Product 27 3.8 Anticommutativity of the Wedge Product It follows directly from the definition of the wedge product (3.2)that fAg is bilinear in f and in g. Proposition 3.22.The wedge product is anticommutative:if f E Ak(V)and g A(V).then fΛg=(-1)“gΛf Proof.Define Sk+to be the permutation 「1.ee+1.+k] r=k+1.k+01.k This means that t(I)=k+1,t()=k+(,t(+1)=1,t(+k)=k. Then 0I)=0t(t十1I),.,o(k)=t(t十k) o(k+1)=ot(1),(k+)=ot(). For any v,g+t∈V, A(f®g(U1,k+) =∑(sgno)f(va,vak)8(vak+),ak+O） GES =∑(sgno)f(vgr(+1.vgr+)g(Uar,at(O） GES4 =(sgn)>(sgnat)g(var(1).var())f(var(+1).v(+)) =(sgn r)A(g f)(v1.,v+). The last equality follows from the fact that as runs through all permutations inS+ so does or. We have proved A(f⑧g)=(sgnt)A(g⑧f). Dividing by ke!gives fAg=(sgn t)g A f. Exercise 3.23 (The sign of a permutation).Show that sgn=(-1) ◇ Corollary 3.24.If f is a k-covector on V and k is odd,then ff=0. Proof.By anticommutativity. ff=(-I)ff =-fΛf since k is odd.Hence,2f f =0.Dividing by 2 gives ff=0
3.8 Anticommutativity of the Wedge Product 27 3.8 Anticommutativity of the Wedge Product It follows directly from the definition of the wedge product (3.2) that f ∧g is bilinear in f and in g. Proposition 3.22. The wedge product is anticommutative: if f ∈ Ak(V ) and g ∈ A (V ), then f ∧ g = (−1) k g ∧ f. Proof. Define τ ∈ Sk+ to be the permutation τ = 1 ··· + 1 ··· + k k + 1 ··· k + 1 ··· k . This means that τ (1) = k + 1, . . . , τ ( ) = k + , τ ( + 1) = 1, . . . , τ ( + k) = k. Then σ (1) = σ τ ( + 1), . . . , σ (k) = σ τ ( + k), σ (k + 1) = στ(1), . . . , σ (k + ) = σ τ ( ). For any v1,.,vk+ ∈ V , A(f ⊗ g)(v1,.,vk+ ) = σ∈Sk+ (sgn σ )f (vσ (1),.,vσ (k))g(vσ (k+1),.,vσ (k+ )) = σ∈Sk+ (sgn σ )f (vσ τ ( +1),.,vσ τ ( +k))g(vστ(1),.,vσ τ ( )) = (sgn τ ) σ∈Sk+ (sgn σ τ )g(vστ(1),.,vσ τ ( ))f (vσ τ ( +1),.,vσ τ ( +k)) = (sgn τ )A(g ⊗ f )(v1,.,vk+ ). The last equality follows from the fact that as σ runs through all permutations in Sk+ , so does σ τ . We have proved A(f ⊗ g) = (sgn τ )A(g ⊗ f ). Dividing by k! ! gives f ∧ g = (sgn τ )g ∧ f. Exercise 3.23 (The sign of a permutation). Show that sgn τ = (−1)k . Corollary 3.24. If f is a k-covector on V and k is odd, then f ∧ f = 0. Proof. By anticommutativity, f ∧ f = (−1) k2 f ∧ f = −f ∧ f, since k is odd. Hence, 2f ∧ f = 0. Dividing by 2 gives f ∧ f = 0.
28 3 Alterating k-Linear Functions 3.9 Associativity of the Wedge Product have defined their wedge producto be the (k+)-covector fA8=kOAG⑧8. To prove the associativity of the wedge product,we will follow Godbillon [7]by first proving the following lemma on the alternating operator A. Lemma 3.25.Suppose f is ak-linear function and g an t-linear function on a vector space V.Then ①A(A(f)⑧g)=k:A(f8g,and iA(f⑧A(g)=:A(f⑧g): Proof.(i)By definition. A(AD©g)=∑(sgno)a∑(sgnx(r©8 OESt+L ATESE We can as a permutation in S+such that (i)=i fori=k+1.k+. For sucha. (Tf)⑧g=tfg) Hence. A(A(f⑧g)=∑∑(sgna)sgnt)(ar)f⑧g) GESK+E TESE Let u=o E Sk+.For each ue Sk+,there are k!ways to write u=ar with g∈Sk+t and r∈Sk,because eachr∈S&determines a unique o by the formula =ur-.So the double sum above can be rewritten as A(A(f)⑧g)=:(sgu)μ(f⑧g) L∈St+e =k1A(f8g): The equality in(ii)is proved in the same way Proposition 3.26 (Associativity ofthe wedge product).Let V be arealvector space mghanagmhmarmtosonvyigs长.&mepemah (f g)h=f(gAh)
28 3 Alternating k-Linear Functions 3.9 Associativity of the Wedge Product If f is a k-covector and g is an -covector, we have defined their wedge product to be the (k + )-covector f ∧ g = 1 k! ! A(f ⊗ g). To prove the associativity of the wedge product, we will follow Godbillon [7] by first proving the following lemma on the alternating operator A. Lemma 3.25. Suppose f is a k-linear function and g an -linear function on a vector space V . Then (i) A(A(f ) ⊗ g) = k!A(f ⊗ g), and (ii) A(f ⊗ A(g)) = !A(f ⊗ g). Proof. (i) By definition, A(A(f ) ⊗ g) = σ∈Sk+ (sgn σ )σ ⎛ ⎝ τ∈Sk (sgn τ )(τf ) ⊗ g ⎞ ⎠ . We can view τ ∈ Sk as a permutation in Sk+ such that τ (i) = i for i = k + 1,.,k + . For such a τ , (τf ) ⊗ g = τ (f ⊗ g). Hence, A(A(f ) ⊗ g) = σ∈Sk+ τ∈Sk (sgn σ )(sgn τ )(σ τ )(f ⊗ g). Let µ = σ τ ∈ Sk+ . For each µ ∈ Sk+ , there are k! ways to write µ = σ τ with σ ∈ Sk+ and τ ∈ Sk, because each τ ∈ Sk determines a unique σ by the formula σ = µτ −1. So the double sum above can be rewritten as A(A(f ) ⊗ g) = k! µ∈Sk+ (sgn µ)µ(f ⊗ g) = k!A(f ⊗ g). The equality in (ii) is proved in the same way. Proposition 3.26 (Associativity of the wedge product). LetV be a real vector space and f, g, h alternating multilinear functions on V of degrees k, , m, respectively. Then (f ∧ g) ∧ h = f ∧ (g ∧ h).
3.9 Associativity of the Wedge Product 29 Proof.By the definition of the wedge product. 1 +(A(8⑧ (k+) =k+mA⑧8g)8)（by Lemma3.256) =kmAG⑧8)® Similarly, fe创=C+A(@赢Ag©创 1 =kmAU®g多h》. Since the tensor product is associative,we conclude tha (f Ag)h=f(gAh). ▣ By associativity,we can omit the parentheses in a multiple wedge product such as (fg)Ah and write simply fgh. Corollary 3.27.Under the hypotheses of the proposition. f入8Ah=mAG⑧8⑧h, This corollary easily generalizes to an arbitrary number of factors:if fi Ad(V),then iA.A斤=d.dA8.@f 3.4) In particular,we have the following proposition.We use the notation [to denote the matrix whose(i.j)-entry is b. Proposition 3.28(Wedge product of 1-covectors).Ifa.are linearfunctions on a vector space V and v.v e V,then (a.)(vI.v)=det[a'(vj)]. Proof.By (3.4). (aA.Aa)(u1,w)=A(a⑧.⑧a)U1,w) =∑(sgno)a'(oa)）.a*(a) dES =detfa'(vj)]. 0
3.9 Associativity of the Wedge Product 29 Proof. By the definition of the wedge product, (f ∧ g) ∧ h = 1 (k + )!m! A((f ∧ g) ⊗ h) = 1 (k + )!m! 1 k! ! A(A(f ⊗ g) ⊗ h) = (k + )! (k + )!m!k! ! A((f ⊗ g) ⊗ h) (by Lemma 3.25(i)) = 1 k! !m! A((f ⊗ g) ⊗ h). Similarly, f ∧ (g ∧ h) = 1 k!( + m)! A f ⊗ 1 !m! A(g ⊗ h) = 1 k! !m! A(f ⊗ (g ⊗ h)). Since the tensor product is associative, we conclude that (f ∧ g) ∧ h = f ∧ (g ∧ h). By associativity, we can omit the parentheses in a multiple wedge product such as (f ∧ g) ∧ h and write simply f ∧ g ∧ h. Corollary 3.27. Under the hypotheses of the proposition, f ∧ g ∧ h = 1 k! !m! A(f ⊗ g ⊗ h). This corollary easily generalizes to an arbitrary number of factors: if fi ∈ Adi(V ), then f1 ∧···∧ fr = 1 (d1)! . . . (dr)! A(f1 ⊗···⊗ fr). (3.4) In particular, we have the following proposition. We use the notation [bi j ] to denote the matrix whose (i, j )-entry is bi j . Proposition 3.28 (Wedge product of 1-covectors).If α1,.,αk are linear functions on a vector space V and v1,.,vk ∈ V , then (α1 ∧···∧ αk)(v1,.,vk) = det[αi (vj )]. Proof. By (3.4), (α1 ∧···∧ αk)(v1,.,vk) = A(α1 ⊗···⊗ αk)(v1,.,vk) = σ∈Sk (sgn σ )α1(vσ (1))··· αk(vσ (k)) = det[αi (vj )].