W if p = 0; ; W ) if p 1. p times

Alternating and symmetric multilinear functions. Suppose and W are normed vector spaces. For each integer p we set {0} if p < 0; W if p = 0; ( ; W = L( }... {{... } ; W if p 1. p times We say µ p ( ; W is symmetric if either p 1 or p > 1 and the value of µ does not change if any pair of its arguments are transposed, and we say µ is antisymmetric or alternating if either p 1 or the value of µ is multiplied by 1 if any pair its arguments are transposed. We let ( ; W and ( ; W be the sets of symmetric and alternating members of p ; W, respectively. p ( ; W are linear subspaces of p ( ; W. If U is a vector spaces and l L(U, we define the linear map Evidently, p ( ; W and (l; W : ( ; W (U; W by setting (l; W (ϕ(u1,..., u p = ϕ(l(u 1,..., l(u p, ϕ ( ; W, u1,..., u p U. This extends the notion of adjoint encountered previously. We note that p (l; W preserves symmetry and antisymmetry and we set (l; W = (l; W ( ; W and (l; W = (l; W ( ; W. One easily verifies that if Z is a vector space and m L( ; Z then (m l; W = (l; W (m; W and that similar formulae hold with p ( ; W replaced by p ( ; W and p ( ; W. Interior multiplication. For each integer p we define the bilinear map ( ; W 1( ; W as follows: Given ϕ p ( ; W and v we set ϕ = 0 in case p 0, we set ϕ v = ϕ(v in case p = 1 and, in case p > 1, we set ϕ v(v 2,..., v p = ϕ(v, v 2,..., v p, v 2,..., v p. We call ϕ v interior multiplication or contraction of ϕ by v. by requiring that Note that interior multiplication by v preserves the subspaces of symmetric and alternating functions. Suppose B is a basis for and ϕ, ψ p ( ; W. Then ϕ = ψ if and only if ϕ(v(1,..., v(p = ψ(v(1,..., v(p 1

for any map v : {1,..., p} B. Proof. We hope this intuitively obvious to the reader. A short formal proof my proceed by induction on p using the fact that if p 1 and ϕ v = ψ v for all v then ϕ = ψ. Suppose B is a basis for and ϕ, ψ p ( ; W. Then ϕ = ψ if and only if for each p element subset B of B there is a univalent mapping v from {1,..., p} onto B such that ϕ(v(1,..., v(p = ψ(v(1,..., v(p. Proof. We hope this intuitively obvious to the reader. A short formal proof could proceed by induction on p using the fact that if p 1 and ϕ v = ψ v for all v then ϕ = ψ. For each integer p we let = ( ; R, = ( ; R, = ( ; R and we let l = (l; R, l = (l; R, l = (l; R. Exterior algebra. Let be a vector space, let n = dim and suppose n <. Suppose p 2, v i, i = 1,..., p and dim span{v i : i {1,..., p}} < p. Then ϕ(v 1,..., v p = 0 for any ϕ. Proof. For some j {1,..., p} and for some scalars c i, i I, with I = {1,..., p} {j} we have v j = i I c i v i. For each i I we define w (i 1,..., w(i p by setting For each i I we have w (i j = w (i i w (i k = so ϕ(w (i = 0. Thus ϕ(v 1,..., v p = i I { vi if k = j, v k if k I,, i I. c i ϕ(w (i 1,..., w(i p = 0. Corollary. = {0} if p > n. The Contravariant Exterior Product. There is one and only one map such that, if ϕ p and ψ q then q 2 +q

(1 ϕ ψ = ϕ ψ if p = 0 and q = 0; (2 (ϕ ψ v = (ϕ v ψ + ( 1 p ϕ (ψ v for all v in. This mapping is bilinear. Remark. Because (2 holds we say ϕ ϕ v is a skewderivation. Proof. The statement holds trivially if p < 0 or q < 0 so suppose p 0 and q 0 and induct on r = p + q. It is evident by induction on r that there is unique map q +q such that (1 and (2 are satisfied and that this map is bilinear. We need to show that if ϕ p and ψ q then ϕ ψ is alternating. This is trivially the case if r = 0 so assume r > 0 and that the Theorem holds for smaller r. Since (ϕ ψ v = (ϕ v ψ + ( 1 p ϕ (ψ v for any v the inductive hypothesis implies that ϕ ψ is alternating in its last r 1 slots. To complete the proof it will suffice to show that it alternating in its first two slots. That is, given v, w we need to show that ((ϕ ψ v w is alternating in v, w. But ((ϕ ψ v w = ( (ϕ v ψ + ( 1 p ϕ (ψ v w = ( (ϕ v w ψ + ( 1 p 1 (ϕ v (ψ w + ( 1 p (ϕ w (ψ v + ( 1 p ( 1 p ϕ ( (ψ v w. The sum of the second and third terms in this sum is clear alternating and v and w and the first and fourth terms in the summand are alternating by the inductive hypothesis. Suppose ϕ p, ψ q and ζ r. Then (That is, exterior multiplication is associative. (ϕ ψ ζ = ϕ (ψ ζ. Proof. The Theorem holds trivially if any of p, q, r are negative. So we assume that p, q, r are nonnegative and induct on s = p + q + r. The Theorem holds trivially if s = 0 so suppose s > 0 and that Theorem holds for smaller s. Given v we calculate ( (ϕ ψ ζ v = ( (ϕ ψ v ζ + ( 1 p+q (ϕ ψ (ζ v = ( (ϕ v ψ ζ + ( 1 p( ϕ (ψ v ζ + ( 1 p+q (ϕ ψ (ζ v; ( ϕ (ψ ζ v = (ϕ v (ψ ζ + ( 1 p ϕ ( (ψ ζ v Now apply the inductive hypothesis. = (ϕ v (ψ ζ + ( 1 p ϕ ( (ψ v ζ + ( 1 p ( 1 q ϕ ( ψ (ζ v. Suppose ϕ p and ψ q. Then ϕ ψ = ( 1 pq ψ ϕ. 3

(That is, exterior multiplication is commutative in the graded sense. Proof. The Theorem holds trivially if either p or 1 is negative. Induct on r = p + q. If r = 0 this amounts to the commutative law for multiplication of real numbers so suppose r > 0 and that the Theorem holds for smaller r. For any v in we have (ϕ ψ v = (ϕ v ψ + ( 1 p ϕ (ψ v; Now apply the inductive hypothesis. ( 1 pq (ψ ϕ v = ( 1 pq (ψ v ϕ + ( 1 pq ( 1 q ψ (ϕ v. Corollary. Suppose p is odd and ϕ p. Then ϕ ϕ = 0. Remark. Let = R 4. We have As we shall soon see, e 1 e 2 e 3 e 4 0. (e 1 e 2 + e 3 e 4 (e 1 e 2 + e 3 e 4 = 2 e 1 e 2 e 3 e 4. For each p {0, 1,..., n} we let Λ(p, n equal the empty set if p = 0 and we let it equal the set of increasing maps from {1,..., p} into {1,..., n} if p > 0. Let n Λ(n = Λ(p, n. Evidently, card Λ(p, n = ( n p p=0 whenever p {0,..., p}. For each integer p 1, i {1,..., p} and λ Λ(p, n we let λ i be that member of Λ(p 1, n such that rng λ i = rng λ {λ(i}. Bases and Dimension. Let be a vector space, let n = dim and let v 1,..., v n and v 1,..., v n be dual basics sequences for and, respectively. For each p in {1,..., n} and for each λ in Λ(p, n let v [λ] ( be such that v [λ] (ϕ = ϕ(v λ(1,..., v λ(p, 4 whenever ϕ p

and let { v λ(1 v λ = if p = 1, v λ(1 v λ(p else. Suppose m is a positive integer; p 1,..., p m are integers; φ i i, i = 1,..., m; v ; and φ i v = 0, i = 1,..., m. Then (φ 1 φ m v = 0. Proof. Induct on m. The assertion holds trivially if m = 1. Suppose that m > 1 and that the Proposition holds for smaller m. We use the skewderivation property of and the inductive hypothesisto obtain (φ 1 φ m v = (φ 1 v (φ 2 φ m + ( 1 p 1 φ 1 ( (φ 2 φ m v = 0. Suppose p {1,..., n}, j {1,..., n} and λ Λ(p, n. Then v λ v j = { ( 1 i 1 v λ i if λ(i = j for some i {1,..., p}, 0 else. Proof. If j rng λ then v λ v j = 0 by the preceding Proposition so let us suppose that j = λ(i for some i {1,..., p}. We induct on p. The Lemma holds trivially if p = 1. Assume that p {2,..., n} and that the Lemma holds for smaller p. Then ( 1 i 1 (v λ v j = (v j v λi v j = (v j v j v λi v j (v λi v j = v λi where we have used the inductive hypothesis to infer that v λi v j = 0. Suppose p in {1,..., n}. Then v [µ] (v λ = δ µ λ, λ, µ Λ(p, n. Proof. Suppose there is i {1,..., p} such that µ(i rng λ. Then v [µ] (v λ = v [µi ](v λ v µ(i = 0 so the Proposition holds in this case. We show the Proposition holds in case rng λ = rng µ by induction on p. In case p = 1 the assertion to be proved holds trivially so suppose p > 1 and that the assertion to be proved holds for smaller p. Now (v λ(1 v λ 1 v λ(1 = (v λ(1 v λ(1 v λ 1 v λ(1 (v λ 1 v λ(1 = v λ 1 and λ = µ so v [µ] (v λ = v [λ] (v λ = v [λ1 ]( (v λ(1 v λ 1 v λ(1 = v[λ1 ](v λ 1 = 1. Corollary. Moreover, For each p {1,..., n} the set {v λ : λ Λ(p, n} is a basis for p and has cardinality ( n p. (1 ϕ = λ Λ(p,n v [λ] (ϕ v λ whenever ϕ p. 5

Proof. Suppose c λ, λ Λ(p, n, are such that λ Λ(p,n c λ v λ = 0. Let µ Λ(p, n. Applying v [µ] to both sides of the preceding equation to and making use of the previous Theorem we infer that c µ = 0. It follows that {v λ : λ Λ(p, n} is independent and has cardinality ( n p. That (1 holds follows from the fact that for any µ Λ(p, n both sides have the same value on (v µ(1,..., v µ(p. That {v λ : λ Λ(p, n} spans p follows immediately from (1. Remark. A special case of the preceding Theorem is the all important dim n = 1. The Exterior Power of a Linear Map. Recall that if and W are vector spaces and l L( ; W we have the linear map l : W defined by setting l (ϕ(v1,..., v p = ϕ(l(v 1,..., l(v p whenever ϕ p W and v 1,..., v p. Suppose ϕ p and ψ q. Then (+q ( ( ( q ( l (ϕ ψ = l ϕ l ψ. Proof. Induct on p + q. We leave the details as an exercise for the reader. Suppose W and Z are vector spaces, l : W, m : W Z and l and m are linear. Then m l = l m. Proof. This is straight forward calculation. The Covariant Exterior Product. Suppose is a vector space and n = dim <. For each v we let v be such that v (ω = ω(v whenever ω and note that v v is a linear isomorphism from onto. frequently identify v with v. In what follows we will For each integer p we let p =. Whenever p is a integer not less than 2 we and v 1,..., v p we let p (v 1,..., v p = v 1 v p = v 1 v p ( 6

thereby defining Note that for any integers p, q p ( ; p (. p q +q is defined. Note that the associativity and anticommutativity of covariant exterior multiplication follow from the corresponding properties of contravariant exterior multiplication. Bases. Suppose v 1,..., v n and v 1,..., v n are dual basic sequence for and, respectively. Suppose p {1,..., n}. For each λ in Λ(p, n let v λ = { vλ(1 if p = 1, v λ(1 v λ(p else. Note that and that its cardinality is {v λ : λ Λ(p, n} is a basis for ( n. p For each λ in Λ(p, n let v [λ] ( be such that v [λ] (v µ = δ λ µ, µ Λ(p, n. by letting Suppose W is a finite dimensional vector space and l L( ; W. We define p l L( ; W p l = l. Suppose l L( ; W, p is an integer at least 2 and v 1,..., v p. Then (p l (v 1 v p = l(v 1 l(v p. Proof. We have p l(v 1 v p = l (v 1 v p = v1 l vp l = l(v 1 l(v p = l(v 1 l(v p. Suppose l L( ; W, ξ and η. Then ( (p+q l (ξ η = (p l (ξ (η. q 7

Proof. This is a consequence of facts already established for p. Suppose l L( ; W and m L(W ; Z. Then p m l = m l. Proof. This is a consequence of facts already established for p. The universal property of. Suppose p is an integer not less than 2. Then L( ( ; W φ φ ( ; W carries L( ; W isomorphically onto p ( ; W. In fact, if v 1,..., v n is a basis for and W = R this map carries v [λ] = v λ ( p to v λ p for λ Λ(p, n. Proof. Straighforward exercise. Remark. Taking W = R above we have that ( carries p isomorphically onto p. Moreover, (p φ φ p p = ; in fact, if ω 1,..., ω p then ω 1 ω p corresponds to ω 1 ω p p under this isomorphism. Suppose l L( ;. Keeping in mind that n is 1-dimensional we let det l, the determinant of l, be such that n l is multiplication by det l. Suppose l, m L( ;. Proof. Let ξ n {0}. We calculate det(l m = det(l det(m. det(l m ξ = (n l m (ξ = (n l m (ξ n = (n (n l m (ξ = (n l (det(m ξ = det(m (n l (ξ = det(m det(l ξ. 8

Corollary. Suppose l L( ;. Then l is invertible if and only if det l 0 in which case det l 1 = (det l 1. Corollary. Suppose W is a vector space, i carries isomorphically onto W, l L( ;, k L(W ; W and the following diagram is commutative: l i W i k W. Then Proof. Suppose ξ n. Then det k = det l. det l (n i (ξ = i((det lξ n = ( i l(ξ n n = n i l(ξ = n k i(ξ = n k ( n i(ξ = det k (n i (ξ. Let S n be the group of permutations of {1,..., n}. Let s : S n GL(R n be defined by the requirement that Note that s is a homomorphism. We the map and call it the signature or index. Moreover, s(σ(e j = e σ(j, σ S n, j = 1,..., n. sgn = det s : S n { 1, 1} sgn(σ τ = sgn(σsgn(τ, σ, τ S n. sgn(σ = card {(i, j {1,..., n} {1,..., n} : σ(i > σ(j} for any σ S n. Proof. The first assertion follows directly from the product rule for determinants. The second assertion is proved by induction on n. Observe that if σ S n and k {1,..., n} is such that σ(k = n then card {(i, n {1,..., n} : σ(i > σ(n} 9

and v σ(1 v σ(n = ( 1 n i v τ(1 v τ(n 1 v n where τ S n 1 is such that { σ(k if k < i, τ(k = k {1,..., n 1}. σ(k 1 if k i, Inner Products. Suppose has an inner product. Let be the corresponding polarity: β : β(v(w = v w whenever v, w. Suppose p is an integer not less than 2. Then there is one and only one inner product on such that (1 v 1 v p w 1 w p = β(v 1 β(v p (w 1,..., w p whenever v i, w i, i = 1,..., p. Moreover, if v 1,..., v n is an orthonormal basic sequence for then {v λ : λ Λ(p, n} is an orthonormal basis for. Proof. By virtue of earlier work we have an isomorphism β p = ( p p p which has the property that ( if v 1,..., v n is an orthonormal basic sequence for then v λ, λ Λ(p, n is carried to the the member of p which is dual to vλ. The Theorem now follows easily. with equality only if Proof. Write Suppose p is an integer not less than 2 and v 1,..., v p. Then v 1 v 2 v p v 1 v 2 v p v 1 span {v 2,..., v p }. v 1 = u 1 + w 1 where u 1 span {v 2,..., v p } and w 1 span {v 2,..., v p }. Then, as w 1 p i=2 ker β(v i, so β(v 1 β(v 2 β(v p w 1 = β(v 1 (w 1 β(v 2 β(v p = w 1 2 β(v 2 β(v p v 1 v 2 v p 2 = β(v 1 β(v 2 β(v p (w 1, v 2,..., v p = (β(v 1 β(v 2 β(v p w 1 (v 2,..., v p = w 2 2 β(v 2 β(v p (v 2,..., v p = w 2 2 v 2 v p 2. Suppose W is a finite dimensional inner product space and l L(, W. Then ( l = l. Proof. This will come later. 10