k times l times n times

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1 1. Tensors on vector spaces Let V be a finite dimensional vector space over R. Definition 1.1. A tensor of type (k, l) on V is a map T : V V R k times l times which is linear in every variable. Example 1.2. Let v V be a vector. Then v defines a tensor T v of type (0, 1) with the map T v : V R given by T v (f) = f(v). The map v T v gives a linear isomorphism from V onto (V ) = space of all tensors of type (0, 1). Let, be an inner product on V. Then it is a tensor of type (2, 0). Let V = R n and let T : V R be given by T (v 1,..., v n ) = n times det A where A is the n n matrix with columns v 1,..., v n. Then T is a tensor of type (n, 0). From now on we will only consider tensors of type (k, 0) which we ll refer to as simply k-tensors. Let T k (V ) be the set of all k tensors on V. It s obvious that T k (V ) is a vector space and T 1 (V ) = V. Also T 0 (V ) = R. Definition 1.3. Let V, W be vector spaces and let L: V W be a linear map. Let T be a k-tensor on W. Let L (T ): V R be defined by the formula k times L (T )(v 1,..., v k ) := T (L(v 1 ),..., L(v k )) Then it s immediate that L (T ) is a k-tensor on V which we ll call the pullback of T by L. It s easy to see that L : T k (W ) T k (V ) is linear. Definition 1.4. Let T T k (V ), S T l (V ). We define their tensor product T S T k+l (V ) by the formula T S(v 1,..., v k+l ) = T (v 1,..., v k ) S(v k+1,..., v k+l ) It s obvious that T S is a tensor. The following properties of tensor product are obvious from the definition Tensor product is associative: (T S) R = T (S R) tensor product is linear in both variables: (λ 1 T 1 + λ 2 T 2 ) R = λ 1 T 1 R + λ 2 T 2 R and the same holds for R. tensor product commutes with pullback, i.e. if L: V W is a linear map between vector spaces and T, S are tensors on W then L (T S) = L (T ) L (S) 1

2 2 Let us construct a basis of T k (V ) and compute its dimension. e 1,..., e n be a basis of V and let e 1,... e n be the dual basis of V, i.e. Let e i (e j ) = δ ij For any multi-index I = (i 1,..., i k ) with 1 i j n define φ I as φ I = e i 1... e i k. Also, we will denote the k-tuple (e i1,... e ik ) by e I. It s immediate from the definition that { (1) φ I 1 if I = J (e J ) = δ IJ = 0 if I J For example e 1 e 2 (e 2, e 1 ) = e 1 (e 2 ) e 2 (e 1 ) = 0. Lemma 1.5. Let T, S T k (V ) then T = S iff T (e I ) = S(e I ) for any multi-index I = (i 1,... i k ). Proof. This follows immediately from multi-linearity of T and S. Lemma 1.6. The set {φ I I=(i1,...,i k ) is a basis of T k (V ). dim T k (V ) = n k In particular, Proof. Let us first check linear independence of φ I s. Suppose I λ Iφ I = 0. Let J = (j 1,..., j k ) be a multi-index of length k. Using (1) we obtain 0 = ( I λ I φ I )(e J ) = I λ I φ I (e J ) = I λ I δ IJ = λ J Since J was arbitrary this proves linear independence of {φ I I=(i1,...,i k ). Let us show that they span T k (V ). Let T T k (V ). Let S = I T (e I)φ I. Then S belongs to the span of {φ I I. For any J we have that S(e J ) = I T (e I)φ I (e J ) = I T (e I)δ IJ = T (e J ). Therefore, T = S by Lemma 1.5 and hence T belongs to the span of {φ I I. 2. Alternating tensors Definition 2.1. Let V be a finite dimensional vector space. A k-tensor T on V is called alternating if for any v 1,..., v k and any 1 i < j k we have Example 2.2. T (v 1,..., v i,..., v j,..., v k ) = T (v 1,..., v j,..., v i,..., v k ) Any 1-tensor on V is alternating. The determinant tensor defined in Example 1.2 is alternating. More generally, let e 1,... e n be a basis of V. Let I = (i 1,..., i k ) be a k-multi-index. Define a k tensor e I as follows.

3 Let v 1,..., v k V. Let A be the n k matrix whose i-th column is given by the coordinates of v i with respect to the basis e 1,... e n. Let A I be the k k matrix made of rows i 1,..., i k of A. Define e I by the formula e I (v 1,..., v k ) = det A I It s immediate that e I is alternating because the determinant of a matrix changes sign if two of its columns are switched. It s also obvious that if I has some repeating indices then e I = 0. The following properties of alternating tensors are immediate from the definition Let A k (V ) be the set of all alternating k tensors on V. Then A k (V ) is a vector subspace of T k (V ), i.e. a linear combination of alternating tensors is alternating. If L: V W is linear and ω A k (W ) then L (ω) A k (W ) Remark 2.3. In the notations of the book A k (V ) = Λ k (V ). Let σ S k be a permutation and let T T k (A) be a k-tensor. Define σ T by σ T (v 1,..., v k ) = T (v σ(1),..., v σ(k) ) The following properties are immediate from the definition σ (λ 1 T 1 + λ 2 T 2 ) = λ 1 σ T 1 + λ 2 σ T 2 στ T = σ ( τ T ) Lemma 2.4. Let T T k (V ). Then TFAE (i) T (v 1,..., v k ) = 0 if v i = v j for some i j (ii) T is alternating; (iii) T (v 1,..., v k ) = 0 if v 1,..., v k are linearly dependent. (iv) σ T = sign σ T for any σ S k. Let e 1,... e n be a basis of V. Let I = (i 1,... i k ), J = (j 1,..., j k ) be two multi-indices with i s i t, j s j t for all s t. It s easy to see from the definition of e I that e I (e J ) = 0 if {i 1,..., i k = {j 1,..., j k and e I (e J ) = sign σ if {i 1,..., i k = {j 1,..., j k and σ S k is the unique permutation satisfying I = σ(j) = (j σ(1),..., j σ(k) ) In particular, if I = (i 1 < i 2 <... < i k ), J = (j 1 < j 2 <... < j k ) then (2) e I (e J ) = δ IJ Lemma 2.5. Let α, β A k (V ). Then α = β iff α(e I ) = β(e I ) for any I = (i 1 < i 2 <... < i k ). Lemma 2.6. The set {e I I=(i1 <i 2 <...<i k ) is a basis of A k (V ). In particular dim A k (V ) = ( n k) for k n and dim A k (V ) = 0 if k > n. Proof. The proof is the same as the proof of Lemma 1.6 but using Lemma 2.5 instead of Lemma 1.5 and (2) instead of (1). 3

4 4 Example 2.7. Any element of A k (V ) can be written as a linear combination of the standard basis {φ I of T k (V ). For example, if n = dim V = 4 then e 13 = e 1 e 3 e 3 e 1 = φ 13 φ 31. Lemma 2.8. Let L: V V be a linear map and let A = [L] be the matrix of L with respect to the basis e 1,..., e n of V. Then for any w A n (V ) we have that L (w) = (det A)w. Proof. By Lemma 2.6, A n (V ) is 1-dimensional with basis given by e 12...n. Therefore, it s enough to prove the lemma for ω = e 12...n. We have L ( 12...n ) = λ 12...n where λ = L ( 12...n )(e 1,..., e n ) = e 12...n (L(e 1 ), L(e 2 ),..., L(e n )) = det A by definition of e 12...n and because columns of A are given by L(e 1 ), L(e 2 ),..., L(e n ) written in the basis (e 1,..., e n ). 3. Wedge product Let T T k (V ). Define Alt(T ) as Alt(T ) = 1 sign σ σt k! σ S k Lemma 3.1. a) Alt(T ) is alternating for any T. b) If ω A k (V ) is alternating then Alt(ω) = ω c) Alt: T k (V ) A k (V ) is linear. d) Alt( σ T ) = sign σ σ Alt(T ) e) Alt commutes with pullbacks: Let L: V W be linear and let T T k (W ). Then Alt(L (T )) = L (Alt(T )) Definition 3.2. Let ω A k (V ), η A l (V ). Then we define their wedge product ω η by the formula Lemma 3.3. ω η = (k + l)! k! l! Alt(ω η) i) e I e J = e IJ for any I = (i 1,..., i k ), J = (j 1,... j l ) ii) (λ 1 ω 1 + λ 2 ω 2 ) η = λ 1 ω 1 η + λ 2 ω 2 η iii) (ω η) ζ = ω (η ζ) iv) ω η = ( 1) ω η η ω v) Let I = (i 1,..., i k ) then e I = e i 1... e i k vi) For any ω 1,..., ω k V, v 1,..., v k V we have ω 1... ω k (v 1,..., v k ) = det(ω i (v j )) vii) let ω 1,..., ω k V. Then ω 1... ω k = 0 iff ω 1,..., ω k are linearly dependent.

5 Proof. Parts i)-vi) are proved in the book (Propositions and 14.11). Let s prove vii). Note that for any ω V by part iv) we have that ω ω = ( 1)ω ω and hence ω ω = 0. Suppose ω 1,..., ω k are linearly dependent. Then we can write one of them as a linear combination of the others. WLOG we can assume ω k = k 1 i=1 λ iω i. Therefore, k 1 ω 1... ω k = ω 1... ω k 1 ( λ i ω i ) = i=1 5 k 1 λ i ω 1... ω k 1 ω i = 0 because each summand contains a repeated factor. Now suppose ω 1,..., ω k are linearly independent. Then we can find ω k+1,..., ω n V such that ω 1,..., ω n is a basis of V. Let v 1,..., v n be the dual basis of V, i.e. ω i (v j ) = δ ij. Then by vi) we have ω 1... ω k (v 1,..., v k ) = det(ω i (v j )) = 1 and hence ω 1... ω k 0. i=1