is the desired collar neighbourhood. Corollary Suppose M1 n, M 2 and f : N1 n 1 N2 n 1 is a diffeomorphism between some connected components

Transcription

1 1. Collar neighbourhood theorem Definition Let M n be a manifold with boundary. Let p M. A vector v T p M is called inward if for some local chart x: U V where U subsetm is open, V H n is open and x(p) H n when we write v as v = n i=1 v i x i p it holds that v n > 0. Lemma If v T p M is inward with respect to a chart x then it s inward with respect to any other chart y. Proof. This was proved in homework 5, problem (5). Lemma Let M n be a manifold with boundary. Then there exists a smooth vector field V on M such that V (p) is inward for any p M. Proof. Observe that if v 1,..., v k T p M are inward, λ 1,..., λ k 0 and i λ i > 0 then v = i λ iv i is also inward. Now the result follows by an easy application of partition of unity. Remark If M is compact it s easy to construct a vector field V as above such that supp V is a compact neighbourhood on M. In particular integral flow ϕ t of V is defined for all t, i.e. V is complete. Theorem (Collar neighbourhood theorem). Let M n be a smooth manifold with boundary. Then there exists a neighbourhood U M of M diffeomorphic to M [0, 1) Proof. We will only give a proof in case M is compact. Let V be a complete vector field on M which is inward along M provided by the lemma above Let ϕ: R M M be the flow of V restricted to the boundary and let p M. Since ϕ 0 (x) = x for any x we have that dϕ (0,p) (0, v) = v for any v T p M. Also, by definition of the flow we have that dϕ (0,p) ( t, 0) = V (p). Since V (p) is inward and dim M = dim R M = n this implies that dϕ (0,p) is an isomorphism. Therefore, by the Inverse Function Theorem, there is an open neighbourhood U p M containing p and ε p > 0 such that ϕ [0,εp) Up is a diffeomorphism onto its image which is an open neighbourhood of p in M. Since M is compact we can choose a finite subcover {U i N i=1 from the open cover M = p M U p. Then for ε = min i ε i we have that ϕ [0,ε) M is a local diffeomorphism from [0, ε) M onto its image. Using compactness of M and arguing by contradiction it s easy to see that there exists 0 < ε 1 < ε such that ϕ [0,ε1 ) M is 1-1. This means that W = ϕ([0, ε 1 ) M) is the desired collar neighbourhood. Corollary Suppose M1 n, M 2 n are smaooth manifolds with boundary and f : N1 n 1 N2 n 1 is a diffeomorphism between some connected components of M 1 and M 2 respectively. Then the space obtained by gluing M 1 to M 2 along f is an n-dimensional manifold (possibly with boundary). 2. Tensors on vector spaces Let V be a finite dimensional vector space over R. 1

2 2 Definition A tensor of type (k, l) on V is a map which is linear in every variable. Example T : V V R k times l times 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 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 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 pullback of T by L. It s easy to see that L : T k (W ) T k (V ) is linear. k times which we ll call the Definition Let T T k (V ), S T l (V ). product T S T k+l (V ) by the formula We define their tensor 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)

3 Let us construct a basis of T k (V ) and compute its dimension. Let e 1,..., e n be a basis of V and let e 1,... e n be the dual basis of V, i.e. 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 { (2.0.1) 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 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 The set {ϕ I I=(i1,...,i k ) is a basis of T k (V ). In particular, dim T k (V ) = n k 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 (2.0.1) we obtain 3 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 and hence T belongs to the span of {ϕ I I. 3. Alternating tensors Definition 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 T (v 1,..., v i,..., v j,..., v k ) = T (v 1,..., v j,..., v i,..., v k ) Example Any 1-tensor on V is alternating. The determinant tensor defined in Example 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.

4 4 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 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 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 (3.0.1) e I (e J ) = δ IJ Lemma 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 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 but using Lemma instead of Lemma and (3.0.1) instead of (2.0.1).

5 Example 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 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 3.0.6, 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 (e 12...n ) = λe 12...n where λ = L (e 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 ). 4. Wedge product Lemma Let V be a finite-dimensional vector space. There exists a unique operation : A k (V ) A l (V ) A k+l (V ) satisfying the following conditions i) (λ 1 ω 1 + λ 2 ω 2 ) η = λ 1 ω 1 η + λ 2 ω 2 η ii) (ω η) ζ = ω (η ζ) iii) ω η = ( 1) ω η η ω iv) 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 )) Sketch of proof. Let e 1,..., e n be a basis of V and let e 1,..., e n be the dual basis of V Let ω = ω I e I, η = ω J e J. Define ω η by the formula (4.0.1) ω η := I,J ω I η J e IJ It s easy to see that so defined wedge product satisfies i)-iii). To see that it satisfies iv) check it for ω 1 = e i 1,... ω k = e i k, v 1 = e j1,... v k = e jk. The general case follows by linearity. This proves that a wedge product operation satisfying i)-iv) exists. It is also easy to see that because of iv) any such operation must satisfy the following: For any I = (i 1,... i k ), J = (j 1,..., j l ) it holds that e I = e i 1... e i k and e I e J = e IJ. Therefore, there is only one possible wedge product operation satisfying i)-iv). In particular, wedge product is well-defined and formula (4.0.1) produces the same answer irrespective of which basis of V we use. 5