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Some Jet Bundle Structure Andrew D Some jet bundle structure Andrew D. Lewis∗ 26/10/2005 Abstract Two results concerning the structure of jet bundles are presented and explained. Introduction. We present, discuss, and hopefully make clear, two results stated in [Pom- maret 1978] as Lemma 9.12 and Proposition 9.13 in Chapter 1. The setup is a fibred manifold π : Y → X. By Vπ = ker(T π) we denote the vertical subbundle of TY. Let us denote ν = π ◦ (πTY|Vπ) the natural projection from Vπ to X. (By πTM : TM → M we denote the tangent bundle projection.) We denote by Jkπ the bundle of k k-jets of sections of Y, and we denote by πk : Jkπ → X and πl : Jkπ → Jlπ (k ≥ l) the natural projections. If (ξ, U) is a local section, then jkξ : U → Jkπ denotes the corresponding local section of Jkπ. First result. We consider the fibred manifold ν : Vπ → X and denote by Jkν the bundle of k-jets of sections. We first make an observation about this manifold being, in fact, the total space of a vector bundle over Jkπ. Let us define the projection, which we denote by αk : Jkν → Jkπ. For a local section (η, U) of ν : Vπ → X, denote by (ξη, U) the local section of π : Y → X defined by ξη(x) = πTY(η(x)). Let us define αk(jkη(x)) = jkξη(x). We then have the following result. 1 Lemma: (The vector bundle structure of Jkν) The map αk : Jkν → Jkπ is a surjective submersion, and resulting fibred manifold has a vector bundle structure. Proof: That αk is a surjective submersion is easily checked in local coordinates. We exhibit the vector bundle structure of αk : Jkν → Jkπ by indicating how one defines vector addition and scalar multiplication in the fibres of αk. Let (η1, U) and (η2, U) be local sections of ν : Y → X such that jkξη1 (x) = jkξη2 (x) for all x ∈ U. Thus jkη1(x) and jkη2(x) are in the same fiber of αk. Moreover, η1(x) and η2(x) are in the same fibre of πTY|Vπ for each x ∈ U. Thus we may define the local section (η1 +η2, U) by asking that (η1 +η2)(x) = η1(x)+η2(x), where addition on the right takes place in V π. We then define ξη1 (x) (jkη1 + jkη2)(x) = jk(η1 + η2)(x), so defining addition in the fibres of αk. Scalar multiplication is defined, for a ∈ R, by a(jkη(x)) = jk(aη)(x), where (aη, U) is the local section defined by aη(x) = a(η(x)) with scalar multiplication on the right being that in Vξη(x)π. Now one should really verify that there is a vector bundle structure by defining a vector bundle atlas. This is done easily in natural coordinates, however. We leave the working out of the details of this to the reader. ∗Associate Professor, Department of Mathematics and Statistics, Queen’s University, Kingston, ON K7L 3N6, Canada Email: [email protected], URL: http://penelope.mast.queensu.ca/~andrew/ 1 2 A. D. Lewis To get an understanding of the vector bundle structure, let us consider in coordinates n m the case when k = 1. We denote adapted coordinates for Y by (x, y) ∈ U × V ⊂ R × R . m Coordinates for Vπ are denoted by (x, y, v) where v ∈ R . Coordinates for J1ν are then n m denoted by (x, y, v, p1, p2) where pa ∈ L(R ; R ) for a ∈ {1, 2}. Thus p1 is to be thought of as the partial derivatives of y with respect to x, and p2 is to be thought of as the partial derivatives of v with respect to x. The local representative of α1 is then easily seen to be (x, y, v, p1, p2) 7→ (x, y, p1). Thus, imagining how this looks for arbitrary k, the fibres of αk are formed by v and its partial derivatives with respect to x. The vector bundle structure is obtained from the fact that v is to be thought of as a vertical vector, and so is an element of a vector space. Addition and scalar multiplication are then just addition and scalar multiplication of these vectors and their derivatives. Now we consider another vector bundle over Jkπ, namely the vertical bundle Vπk of the fibred manifold πk : Jkπ → X. This is obviously a vector bundle over Jkπ. The final construction preliminary to the statement of the result is the following. Let η : X → Vπ be a section of ν : Vπ → X with ξη the corresponding section of π : Y → X. Let x0 ∈ X, let U be a neighbourhood of x0, and define ση,x0 : U × I → Y such that 1. I ⊂ R is an interval for which 0 ∈ int(I), 2. ση,x0 (x, 0) = ξη(x) for all x ∈ U, ◦ 3. π ση,x0 (x, t) = x for all (x, t) ∈ U × I, and 4. d σ (x , t) = η(x ) ∈ V π. dt t=0 η,x0 0 0 ξη(x0) Thus ση,x0 is a variation of ξη at x0 in the direction of η(x0). Define jkση,x0 : U × I → Jkπ to be such that x 7→ jkση,x0 (x, t) is the k-jet of the local section x 7→ ση,x0 (x, t). Thus −1 t 7→ jkση,x0 (x, t) is a curve in πk (x) for each x ∈ U, and so has a tangent vector field that is vertical with respect to the projection πk. We now have the following result. d 2 Proposition: (Vπk ' Jkν) The map jkη(x) 7→ dt t=0jkση,x(x) is an isomorphism of vector bundles over Jkπ. Proof: This is easily, but tediously, checked in local coordinates. Let us illustrate the result in the case where k = 1. We use, as above, local coordinates for J1ν denoted by m n m n m (x, y, v, p1, p2) ∈ U × V × R × L(R ; R ) × L(R ; R ). Local coordinates for Vπ1 are denoted by n m m n m (x, y, p1, v, A) ∈ U × V × L(R ; R ) × R × L(R ; R ). The vector bundle isomorphism of the proposition looks like (x, y, v, p1, p2) 7→ (x, y, p1, v, p2). The idea, roughly speaking, is that jkη(x) is mapped to the vertical vector consisting of the derivatives of the vertical part of η. Some jet bundle structure 3 Second result. The second result we state has essentially to do with fixed point sets for the symmetric group. So let us first consider some seemingly pointless constructions along these lines. For m ∈ N denote by Sm the group of bijections of {1, . , m}, i.e., the symmetric L group of order m. For k, l ∈ N, let us define group monomorphisms ιk : Sk → Sk+l and R ιl : Sl → Sk+l by (σ(j), j ∈ {1, . , k}, ιL(σ)(j) = k j, j ∈ {k + 1, . , k + l}, and (j, j ∈ {1, . , k}, ιR(σ)(j) = l σ(j − k) + k, j ∈ {k + 1, . , k + l}. L R Thus image(ιk ) consists of the permutations of only the first k elements and image(ιl ) L L consists of the permutations of only the last l elements. Let us denote Sk = image(ιk ) and R R L R Sl = image(ιl ). Note that Sk ×Sl then consists of those permutations that independently permute the first k terms and the last l terms. If Φ: G × A → A is an action of a group G on a set A, let Fix(G, A) = {x ∈ A | Φ(g, x) = x, g ∈ G} be the fixed point set. We now have the following result. 3 Lemma: (Fixed point sets for actions of the symmetric group) Let k, l ∈ N and let Φ: Sk+l × A → A be an action of the symmetric group on the set A. Then L R L R Fix(Sk+l,A) = Fix(Sk × Sl ,A) ∩ Fix(Sk+1 × Sl−1,A). L R Proof: If x ∈ Fix(Sk+l,A) then Φ(σ1, x) = x and Φ(σ2, x) = x for all σ1 ∈ Sk × Sl and L R σ2 ∈ Sk+1 × Sl−1. Thus L R L R x ∈ Fix(Sk × Sl ,A) ∩ Fix(Sk+1 × Sl−1,A) and so L R L R Fix(Sk+l,A) ⊂ Fix(Sk × Sl ,A) ∩ Fix(Sk+1 × Sl−1,A). To prove the opposite inclusion we note that, since every element of Sk+l is the product of a finite number of transpositions, it suffices to prove that if L R L R x ∈ Fix(Sk × Sl ,A) ∩ Fix(Sk+1 × Sl−1,A) then Φ(σ, x) = x for every transposition σ ∈ Sk+l. Indeed, if σ ∈ Sk+l is arbitrary, then write σ = σ1 ◦ ··· ◦ σr for transpositions σ1, . , σr. Then, if Φ(σa, x) = x for each a ∈ {1, . , r}, we have Φ(σ, x) = Φ(σ1 ◦ ··· ◦ σr, x) = Φ(σ1, Φ(σ2,..., Φ(σr, x))) = x. Thus we let L R L R x ∈ Fix(Sk × Sl ,A) ∩ Fix(Sk+1 × Sl−1,A) 4 A. D. Lewis L R L R and we let σ ∈ Sk+l be a transposition. If σ ∈ Sk × Sl or if σ ∈ Sk+1 × Sl−1 then we immediately have Φ(σ, x) = x. Thus we may as well suppose that L R L R σ ∈ Sk+l \ ((Sk × Sl ) ∪ (Sk+1 × Sl−1)). If σ is a transposition of the elements j1, j2 ∈ {1, . , k +l}, this means that we may assume that j1 ∈ {1, .
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