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1 Multilinear Forms Linear Algebra Fall 2013 Multilinear Excursions 1 Multilinear forms In these notes V is a finite dimensional vector space of dimension n ≥ 1 over a field F . We write V 0 to denote the dual vector space. I will use a semi-standard notation, namely T k(V 0) to denote the space of all k-forms (to be defined in a second!) of V . • A 0-form is just an element of F ; that is T 0(V 0) = F . • A 1-form is a linear functional on V ; that is T 1(V 0) = V 0. • ≥ k 2 k 0 × · × ! If k 2 then a k-form is a multilinear map from V to F . That is, ! T (V ) iff ! : |V {z V} F and ktimes satisfies that for each choice of x1; : : : xk−1 2 V the maps x 7! !(x; x1; : : : ; xk−1); x 7! !(x1; x; x2; : : : ; xk−1); ::: ::: x 7! !(x1; : : : ; xk−1; x); are linear. One makes each T k(V 0) into a vector space over F in the obvious way. 0 K Let !1;:::;!k 2 V . We define !1 ⊗ · · · ⊗ !k 2 T (V ) by Yk !1 ⊗ · · · ⊗ !k(x1; : : : ; xk) = !1(x1) ··· !k(xk) = [xk;!k] j=1 So, for example, !1 ⊗ !2(x1; x2) = !1(x1)!2(x2); which can also be written as !1 ⊗ !2(x1; x2) = [x1;!1][x2;!2]: 0 Some notation could now be of some use. Let Jn = f1; : : : ; ng. We will fix a basis fβ1; : : : ; βng of V ; we may f g 2 k ≥ suppose that it is dual to a basis e1; : : : ; en of V . If i = (i1; : : : ; ik) Jn with k 1 we define i ⊗ · · · ⊗ β = βi1 βik : 2 1 − i This makes sense also if k = 1, then i Jn = Jn simply means that i is a number in the range 1 n and β = βi. 0 For the sake of completeness we also define J0 = f0g and β = 1 2 F . 2 N N [ f g f i 2 kg k Theorem 1 Let k 0 = 0 . Then β : i Jn is a basis of T (V ) ≥ 2 k Proof. The cases k = 0; 1 are obvious; assume k 2. Notice first that if i = vei1k, j = (j1; : : : ; jk) Jn , then { Yk 1; if i = j for ν = 1; : : : ; k, βi(e ; : : : ; e ) = δ = ν ν j1 jk iν jν 0; otherwise: ν=1 2 ALTERNATING LINEAR FORMS 2 P k i k Linear Independence. Let ci 2 F for i 2 J and 2 k ciβ = 0. That means that as a map from V ! F this n i Jn form is 0, in particular X i 0 = ciβ (ej1 ; : : : ; ejk ) = cj 2 k i Jn 2 k for all j = (j1; : : : ; jk) Jn . k 0 Spanning property. Assume ! 2 T (V ). If x1; : : : xk 2 V we can write Xn xj = ξjiei i=1 for elements ξji 2 F for 1 ≤ j ≤ k, 1 ≤ i ≤ n. Then multilinearity implies Xn ··· !(x1; : : : ; xk) = ξ1;i1 ξk;ik !(ei1 ; : : : ; eik ): (1) i1;:::;ik=1 One sees that X i ! = !(ei1 ; : : : ; eik )β : k i=(i1;:::;ik)2J In particular the dimension of T k(V 0) = nk. 2 Alternating linear forms We keep assuming V is an n-dimensional vector space over the field F al to the basis fe1; : : : ; eng of V . k 0 k Let ! 2 T (V ), k ≥ 2, and assume σ 2 Sk. We define σ · ! : V ! F by σ · !(x1; : : : ; xk) = !(xσ(1); : : : ; xσ(k)): It is easy to see, one just has to look, that σ· is a linear map from T k(V 0) onto itself. Exercise 1 Prove: If σ; τ 2 SK , then σ · (τ · !) = (τσ) · ! (2) for all ! 2 T k(V 0). Definition 1 Let k ≥ 2. We say that a k-form ! is symmetric iff σ·! = ! for all σ 2 Sk. It is skew symmetric iff σ · ! = ϵ(σ)! for all σ 2 Sk, where ϵ(σ) is the sign of σ; ϵ(σ) = 1 is σ is even, −1 if σ is odd. It is an immediate{ consequence of Exercise} 1 (and the{ fact that all permutations} are products of transpositions) symmetric σ · ! = ! that a k-form ! is if and only if for all transpositions σ 2 S . skew symmetric σ · ! = −! k Definition 2 Let k ≥ 2.A k-form ! is alternating iff !(x1; : : : ; xk) = 0 whenever x1; : : : ; xk 2 V and there exist 1 ≤ i =6 j ≤ k such that xi = xk. Here is the silly situation one encounters when working with fields of characteristic 2. As someone who works in analysis, I have had no encounter with these fields outside of algebra courses. But finite fields in general play an important role in algebra, fields of characteristic 2 play important roles in coding theory and cryptography, so we must either include them, or point out why we exclude them. Suppose F has characteristic 2 and V is a vector space over F . If x 2 V , then x + x = 1 · x + 1 · x = (1 + 1)x = 0 · x = 0; that is, in such vector space x + x = 0 for all x in the space. Another way of phrasing this is by −x = x for all x in the space. In conclusion all k-forms are symmetric and all are skew-symmetric; there is no difference between the two. But not all k-forms are alternating. The relation between the two concepts is given by the following simple lemma. 2 ALTERNATING LINEAR FORMS 3 Lemma 2 Let k ≥ 2 and let ! 2 T k(V 0). If ! is alternating, then ! is skew symmetric. The converse is also true if the characteristic of F is different from 2. Proof. Assume ! is alternating and let σ = (ij) 2 Sk (the transposition exchanging i; j; leaving all other numbers fixed. Assume as we may that i < j. I am going to assume a bit more to avoid too messy notation; it should be clear that what I do works in general. That is, I will assume that i = 1; j = 2. Let x1; : : : ; xk 2 V . Then, by multilinearity, 0 = !(x1 + x2; x1 + x2; x3 : : : ; xk) = !(x1; x1; x3; : : : ; xk) + !(x1; x2; x3; : : : ; xk) + !(x2; x1; x3; : : : ; xk) + !(x2; x2; x3; : : : ; xk) = 0 + !(x1; x2; x3; : : : ; xk) + !(x2; x1; x3; : : : ; xk) + 0; that is, !(x1; x2; x3; : : : ; xk) = −!(x2; x1; x3; : : : ; xk) For general i < j one uses the same idea. On applies ! to the vectors y1; : : : ; yk where y` = x` if ` =6 i; j and yi = yj = xi + xj. By multilinearity one gets σ · !(x1; : : : ; xk) = −!(x1; : : : ; xk). Conversely, assume that ! is skew symmetric. If x1; : : : ; xk 2 V and xi = xj for some j =6 i, then it is clear that for every k-form one has !(x1; : : : ; xk) = σ · !(x1; : : : ; xk). Since ! is skew symmetric, σ · !(x1; : : : ; xk) = −!(x1; : : : ; xk), hence !(x1; : : : ; xk) = −!(x1; : : : ; xk). If the characteristic of the field is 2, this implies nothing. But if it is different from 2 it implies !(x1; : : : ; xk) = 0. In fact, if F is a field of characteristic different from 2, then 2 = 1 + 1 2 F , 2 =6 0, so that 2−1 exists in F . If W is a vector space over such a field, and if x 2 W and x + x = 0, then 0 = 1 · x + 1 · x = (1 + 1)x = 2 · x, and we can multiply by 2−1 to get x = 0. Definition 3 (or notation) If k ≥ 2, the subset of all alternating k-forms is denoted by Λk(V 0). We supplement this defining Λ1(V 0) = V 0 and Λ0(V 0) = F . It is easy to see that Λk(V 0) is a subspace of T k(V 0). In a previous version of these notes I had an alternation map that allowed one to proceed in a very elegant fashion. Unfortunately, that map doesn't work for fields of characteristic p =6 0. That is, it doesn't work well. So let us bite the bullet and proceed in a generally valid way. 0 Up to a point, I do want to use an analog of this alternation map. Let !1;:::;!k 2 V with k ≥ 2. I will define !1 ^ · · · ^ !k by X !1 ^ · · · ^ !k = ϵ(σ)σ · (!1 ⊗ · · · ⊗ !k): σ2Sk Notice that Yk Yk σ · (!1 ⊗ · · · ⊗ !k)(x1; : : : ; xk) = !1 ⊗ · · · ⊗ !k(xσ(1); : : : ; xσ(k)) = !i(xσ(i)) = !σ−1(i)(xi) i=1 i=1 = !σ−1(1) ⊗ · · · ⊗ !σ−1(k)(x1; : : : ; xk); that is, σ · (!1 ⊗ · · · ⊗ !k) = !σ−1(1) ⊗ · · · ⊗ !σ−1(k): −1 −1 Notice also that as σ ranges through Sk, so does σ , and ϵ(σ) = ϵ(σ ). So we can also define X !1 ^ · · · ^ !k = ϵ(σ)!σ(1) ⊗ · · · ⊗ !σ(k): (3) σ2Sk Let us see a few examples. If k = 2, then !1 ^ !2 = !1 ⊗ !2 − !1 ⊗ !2 so !1 ^ !2(x1; x2) = !1(x1) ⊗ !2(x2) − !1(x2) ⊗ !2(x1): If k = 3, !1 ^ !2 ^ !3 = !1 ⊗ !2 ⊗ !3 + !2 ⊗ !3 ⊗ !1 + !3 ⊗ !1 ⊗ !3 − !1 ⊗ !3 ⊗ !2 − !3 ⊗ !2 ⊗ !1 − !2 ⊗ !1 ⊗ !3 We will need the following theorem. 2 ALTERNATING LINEAR FORMS 4 0 Theorem 3 Let k ≥ 2 and let !1;:::;!k 2 V 1. Let τ 2 Sk. Then !τ(1) ^ · · · ^ !τ(k) = ϵ(τ)!1 ^ · · · ^ !k. 2. If there exists i; j, 1 ≤ i =6 j ≤ k such that !i = !j, then !1 ^ · · · ^ !k = 0. Proof. 1. Let vi = !τ(i) for i = 1; : : : ; k. Then X !τ(1) ^ · · · ^ !τ(k) = ϵ(σ)vσ(1) ⊗ · · · ⊗ vσ(k): σ2Sk Now vi = !τ(i) implies vσ(i) = !τ(σ(i)) = !τσ(i).
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