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Dhaka Univ. J. Sci. 60(2): 259-263, 2012 (July)

On Some Characterizations of Vector Fields on Khondokar M. Ahmed1, Md. Raknuzzaman2 and Md. Showkat Ali1 1Department of Mathematics, Dhaka University, Dhaka-1000, Bangladesh. 2Department of Arts and Sciences, Ahsanullah University of Sciences and Technology, 141-142 Love Road, Tejgaon Industrial Area, Dhaka-1208, Bangladesh. [email protected] Received on 10. 12. 11. Accepted for Publication on 22. 01.12 Abstract The basic geometry of vector fields and definition of the notions of tangent bundles are developed in an essential different way than in usual . -related vector fields are studied and some related properties are developed in our paper. Finally, a theorem 5.04 on our natural injection of submanifolds which is -related to vector is treated.

I. Introduction agree on the intersection ∩ of their domains and so together they define a vector fields on [10]. We have The word bundle occurs in with various words defined a vector at a point to be derivation on (). It is before it there are vector bundles, tangent bundles and possible to characterize a vector field in similar way by its fibre bundles [2] to mention the more commonly action on real valued functions. occurring usages. The most general vector bundles and tangent bundles [2] are just certain sort of fibre bundles. Definition 1.03 Let denote the set of all differentiable However, we introduce the theory of tangent bundles, functions [1] : → ℝ whose domains meet a given open vector fields and -related vector fields. We study subset of a . A differentiable (of first orientable manifolds, independent vector fields, order) on is a global -linear independent global vector fields, transformation group and . In this paper some necessary propositions : → related to geometry of vector fields are treated and the such that , ∈ then theorem 5.04 has been derived. () the domain of in the intersection with domain of Definition 1.01 A section of the function ∶ → associates with each point of its domain a tangent () () = () + () vector at . When the domain meets the domain of a where we supposed that the domains of the function involved : → ℝ, we define a function are not empty. If is a vector field on , the function : → ℝ on the intersection by ⟼ is clearly a differential operator on . ⟼ () Proposition 1.04 Any differential operator on an open The section is called a vector field [7] in if all such subset of arises from a unique vector field on . functions are differentiable. Its domain is necessarily Proof. Choose any point ∈ . Then if ℎ ∈ () it follows an open subsets of . from () that the ( ) is defined. Condition () Example 1.02 Suppose that , are the charts of the implies that ℎ ⟼ ( ) is a derivation on () and we stereographic atlas on . On the intersection ∩ of denote it by . The function is defined on by ⟼ their domains and is therefore a section of . If ∈ , is equal to

and so it is differentiable. Consequently is a vector field on = = . □ [()()] so that Proposition 1.05 If , are vector fields in and is an open subset of , then [|, |] = [, ]|. = = − ; Proof. Suppose that , have domains , respectively. [()()] Since | − is the zero vector field on ∩ , [| − , ] = = is the zero vector field 0 on ∩ ∩ and consequently [()()] [|, ] − [, ] = 0. It can be verified that the vector fields It follows that [|, ] = [, ]|. A similar argument shows ( − ) + ( + ) that [, |] = [, ]| and together these give the result required. □ (− − ) + ( − ) * Correspondence author : E-mail: [email protected] 260 Khondokar M. Ahmed, Md. Raknuzzaman and Md. Showkat Ali

II. The ∑ + ⟼ ∑ , ∑ (,) If is a chart of with domain , any vector ∈ can be expressed uniquely as ∑ (/) where where = 1, … … . . , . Using the above standard chart and the = (, … … … , ) ∈ ℝ. We therefore have an product of the standard charts associated with and ′, injection has coordinate representative (, ): → ℝ (, , , ) ⟼ (, ,, ) defined by ⟼ (, ), whose domain is and where ∈ , ′ ∈ ′′ and , ∈ ℝ . This function is whose range is the open set × ℝ. This will be the differentiable. □ standard chart of associated with the chart of . Proposition 2.03 A section : → of is a vector field The coordinate functions are defined by iff it is differentiable [1]. = (), = ( = 1, … … … . , ). Proof. Suppose that is a chart of whose domain meets Let ′ be a chart of with domain ′ and suppose that the domain of . Then | = ∑ , where the the domain of the corresponding standard chart ( , ′) of functions : → ℝ have domain ∩ . | is a vector meets the domain of (, ). Then the domains of , ′ field iff = (, … … . , ): → ℝ is differentiable. is a intersect and the change of coordinates = ′ ∘ is a vector field iff | is a vector field for each such chart . diffeomorphisn. It follows that the corresponding change of coordinates on is defined on ( ∩ ′) × ℝ by On the other hand, using the chart and the corresponding standard chart of , the coordinate representative for is (, ) ⟼ (, []) defined on ( ∩ ) by ⟼ (, ()) and this is and so it is differentiable. The standard charts therefore differentiable iff is differentiable. is differentiable iff is form an atlas which defines a structure of dimension differentiable for each such chart . □ 2 on the tangent bundle . III. Independent Vector Fields Using any chart of and the corresponding standard A set , … … . , of a vector fields on a given open subset chart of , is represented by a function (, ) ⟼ of is said to be independent if, at each point ∈ , whose domain is × ℝ. It follows that is a , … … . , are linearly independent vectors of . In submersion of onto [5]. particular, a single vector field is independent iff it is nowhere Proposition 2.01 If : → ′ is a differentiable zero [9]. function, its differential ∗: → ′ is also For example, the vector fields , … … . . , on the domain differentiable [7]. of a chart are an independent set; consequently independent Proof. Choose a point in the domain of ∗ and charts vector fields certainly exist on any coordinate domain. , ′ of , ′ whose domains include , () respectively. Let Φ = ′ ∘ ∘ . Using the Proposition 3.01 If v: ℝ × ℝ → ℝ is a -linear corresponding standard charts of , ′ it follows from function such that the representative for ∗ is defined on the neighbourhood () v(, ) = 0 implies that = 0 or = 0 of (, ) by () there exists ∈ ℝ such that v(, ) = for all (, ) ⟼ (Φ, []) ∈ ℝ then admits independent vector fields. and is therefore differentiable at the point. □ Proof. Given an element ∈ ℝ, a vector field ′ is defined Proposition 2.02 If , ′ are the projections of the product on ℝ by manifold × ′ onto , ′ respectively, the function ⟼ (v(, )) = ( , ) is a of ( × ′) onto ∗ ∗ () × (). where is the standard isomorphism of ℝ onto ℝ . If : ℝ → is the differentiable function defined for all Proof. is clearly differentiable and it is also bijection. ≠ 0 by ⟼ /|| and if : → ℝ is the natural We have only to prove that is differentiable. Suppose injection, the composition that , ′ are charts of , ′ with domain , ′. On the domain of the standard chart associated with the chart = ∗ ∘ ′ ∘ = × ′ of × ′, is given by is easily seen to be a vector field on . Choose elements , … … … … . , of ℝ so that, together with , they form a On Some Characterizations of Vector fields on Manifolds 261 basis for ℝ. We shall show that the corresponding partitions the -frames into two equivalence classes is called vector field , … … , on are independent. an orientation of . An orientation in an a is a function : ⟼ , where is an Suppose that the real numbers ( = 1, … … … … , ) orientation of , which satisfies the following exist so that for some ∈ . differentiability condition. Each point in the domain of ∑ () = 0 admits a neighbourhood on which the values of are determined by the values of a parallelization on [8]. If a As we saw that the kernel of ∗ is the subspace of manifold admits a global orientation it is said to be orientable. ℝ generated by . Consequently Proposition 4.01 Let be an orientation on a manifold and (∑ ( )) = let ′ be an orientation in with domain . If is connected for some ∈ ℝ. Replacing by v(, ) and using the then ′ is a restriction of either or – . definition of the vector fields , we obtain the equation Proof. Let be the set of points ∈ for which = . v(∑ − , ) = 0. Choose a point ∈ . The functions , ′ are determined on neighbourhoods , ′ of by parallelizations , ( = Since ≠ 0, this implies that 1, … . . , ) respectively. On ∩ ′ ∑ − = 0. = ∑ But , , … … . , are linearly independent and consequently = 0 ( = 1, … … , ). □ where det : → ℝ is a differentiable function such that (det ) > 0. Thus det > 0 on some neighbourhood of and so Proposition 3.02 If manifolds , ′ admit , ′ = ′ on this neighbourhood. Consequently is an open independent vector fields respectively, then × ′ admit subset of and hence of the subspace . The set − is the + ′ independent global vector fields [8]. set of points ∈ for which = − and the same argument shows that this set is open in . Because is Proof. Suppose that ( = 1, … … , ) and ( = 1, … … , ′) are independent vector fields on , ′ connected, is either the empty set or itself. Consequently respectively and that 0, 0′ are the zero vector fields on ′ is a restriction either of – or of . □ these manifolds. We shall show that the vector fields Proposition 4.02 If : → ′ is locally a diffeomorphism, ∘ ( × 0′) and ∘ (0 × ) on × ′ are an orientation ′ in ′ with domain ′ determines an independent. orientation in with domain = ′. If (, ′) is any given point of × ′, the function Proof. Suppose that and ′ have dimension . Given a point ∈ , choose in -frame , … … . , at = ⊕ ′ → ,( × ′) which agrees with ′. Since ∗ is an isomorphism, the induced by is an isomorphism. Consequently if the vectors ∗ form an -frame at and so they determine an real numbers , exit such that orientation at . is clearly independent of the choice of the -frame at ′. ∑ (, 0 ) + ∑ (0, ′) We shall show that the function : ⟼ is an orientation is the zero vector of ,( × ′) then in with domain . Given ∈ , choose a neighbourhood of such that = | is a diffeomorphism and a (∑ (), ∑ ( ′)) parallelization , … … . . , in ′ which agrees with ′ an a neighbourhood of . The vector fields ∗ ∘ ∘ form a must be the zero vector of ⊕ ′ and so = 0 ( = 1, … … , ) and ( = 1, … … , ′). □ parallelization which determines on a neighbourhood of . □ IV. Orientable Manifolds V. -related Vector Fields An -frame in a real vector is an ordered set If : → ′ is a global differential function then the vector , … … , of linearly independent vectors of . When fields , ′ on , ′ respectively are said to be -related if has dimension , any two such frames , ′ are related by ∗ ∘ = ′ ∘ . = ∑ (, = 1,2, … … . . , ) We may express this condition in terms of the action of these where = [] is an non-singular matrix. The vector fields on differentiable functions [6]. equivalence relation Proposition 5.01 Vector fields , ′ on , ′ are -related iff ~ ′ iff det > 0 ( ∘ ) = (′) ∘ 262 Khondokar M. Ahmed, Md. Raknuzzaman and Md. Showkat Ali where : → ℝ is any differentiable function which We complete the proof by showing that this vector field is composes with . projectable and that its projection is ′. Choose any point ∈ and denote by the finite subset of such that Proof. Suppose that , ′ are -related. The functions ≠ 0 iff ∈ . Then ( ∘ ), (′) ∘ both have domain (domain ). If is any point of this domain = ∑ = ∑() ( ∈ ) ( ∘ ) = (∗ ) = . and so, because is finite

The condition is therefore necessary. ∗() = ∑ ∗ = ∑ (). Suppose that it is satisfied. Choose a point ∈ and Since ∑ = ∑ = 1 it follows that ∗() = ∈ where = . Since (). ( ∘ ) = ( ∘ ) = (′)′, Proposition 5.03 is a transformation group acting on a It follows that (∗) = for all ∈ and so manifold and \ is a quotient manifold. A sufficient ∗ ∘ = ′ ∘ at . , ′ are therefore -related. condition for a vector field on to be projectable is that it □ is under all the transformations where ∈ . If Proposition 5.02 Every vector field on a quotient and \ have the same dimension this condition is manifold ′ of a paracompact manifold is the necessary. projection of a projectable vector field on . Proof. Since ∘ = it follows that ∗ ∘ ∗ = ∗ and so Proof. If ′ is a given vector field on ′ and is any is invariant under , chosen point of it is easy to construct a vector field ∘ ∘ = ∘ ∘ = ∘ . on a sufficiently small neighbourhood of with the ∗ ∗ ∗ ∗ required property that on , ∗ ∘ = ′ ∘ . To do this, If this is true for all ∈ it implies that ∗ ∘ is an invariant choose charts , of , ′ whose domains , include of the equivalence relation on and so is projectable. , respectively so that the function ∘ ∘ is defined on by (, ) ⟼ . That such a choice is Conversely, suppose that is projectable. Then ∗ ∘ ∘ = possible is consequence on ∗ ∘ and therefore ∗ ∘ ∘ = ∗ ∘ ∗ ∘ for all ∈ . Thus the two vectors () and () at have the ∗ ∘ = ∘ ( = 1,2, … … . , ) ∗ same images under ∗. But if and \ have the same dimension, ∗ induces an isomorphism on each tangent vector and so, if = ∑ on , our requirements are space and so satisfied by the vector field () = ∗() = ∑( ∘ ) . for all ∈ . This mean that ∘ = ∗ ∘ for all ∈ and so is invariant under all the transformation . The collection of such neighbourhoods corresponding □ to all points of is a covering of . Since is a Theorem 5.04 ′ is a submanifold of and : ′ → is paracompact, we can choose a partition of unity the natural injection. A vector field in is tangent to ′ { } ( ∈ ) subordinate to this covering. The iff there exists a vector field ′ in ′ which is -related to of is contained in a neighbourhood and we . denote the corresponding vector field on by . The vector field on and the zero vector field on Proof. We denote the domain of by . Suppose that each a − agree on the intersection − of their vector field exists, so that ∗ ∘ = ∘ . Then if ∈ domains and so together they define a vector field on ∩ . Since the collection { } is locally finite, any given point has a neighbourhood on which all except a finite = ∗(′) ∈ ∗( ′) number of these vector fields are zero. Consequently, and so is tangent to ′. the sum Conversely, suppose that is tangent to ′. Since ∗ is an = ∑ ( ∈ ) isomorphism, we can define a section ′ of : ′ → ′ by is defined and is differentiable on and so it is a vector ⟼ (∗) () field on . On Some Characterizations of Vector fields on Manifolds 263

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