Manifolds (CM437Z/CMMS18)

Transcription

1 Manifolds (CM437Z/CMMS18) Neil Lambert Deartment of Mathematics King s College London Strand London WC2R 2LS, U. K. lambert@mth.kcl.ac.uk Contents 1 Manifolds Elementary Toology and Definitions Manifolds The Tangent Sace Mas from M to IR Tangent Vectors Curves and their Tangents Mas Between Manifolds Diffeomorhisms Push Forward Vector Fields Vector Fields Integral and Local Flows Lie Derivatives

2 5 Tensors Co-Tangent Vectors Pull-back and Lie Derivative of a co-vector Tensors Differential Forms Forms Exterior Derivative Integration on Manifolds de Rahm Cohomology and Homology Connections, Curvature and Metrics Connections, Curvature and Torsion Riemannian Manifolds Acknowledgement These notes are indebted to the notes of A. Rogers. The introduction is largely taken from the book by Messiah. 2

3 1 Manifolds 1.1 Elementary Toology and Definitions This section should be a review of concets (hence it is all definitions and no theorems). Definition: A toological sace X is set whose elements are called oints together with a collection U = {U i } of subsets of X which are called oen sets and satisfy i) U is closed under finite intersections and arbitrary unions. ii), X U Definition: A set is closed if its comlement in X is oen. Definition: An oen cover of X is a collection of oen sets whose union is X. A toology allows us to define a notion of local meaning that something is true or exists in some oen set about a oint, but not necessarily the whole sace or even all oen sets about that oint. Some other imortant concets are: Definition: X is connected if it is not the union of disjoint oen sets. Definition: A ma f : X Y between two toological saces is continuous iff f 1 (V ) = {x X f(x) V } is oen for any oen set V Y. Definition: If X Y where Y is a toological sace then X can be made into a toological sace too by considering the induced toology: oen sets in X are generated by U X X where U is an oen set of Y. Problem: Show that the induced toology indeed satisfies the definition of a toology. Definition: A Hausdorff sace is a toological sace with the additional roerty that oints can be seerated : for any two distinct oints x, y X there exists oen sets U x and U y such that x U x, y U y and U x U y =. Hausdorff saces are also known as T 2 saces (as there are infact weaker notions of searability). Non-Hausdorff saces have various athologies that we do not want to consider. Therefore in what follows we will take all toological saces to be Hausdorff unless otherwise mentioned. Definition: A function f : X Y is one-to-one iff f(x) = f(y) imlies x = y. f 1 L This guarantees the existance of a left inverse fl 1 : f(x) Y X such that f(x) = x for all x X, since every element in the image f(x) comes from a unique oint in X. Definition: A function f : X Y is onto iff f(x) = Y, i.e. for all y Y there exists an x X such that f(x) = y. 3

4 This guarantees the existance of a right inverse f 1 R : Y X such that f fr 1 (y) = y for all y Y, since every element in Y has some x X (not necessarily unique) which is maed to it by f. Definition: A bijection is a ma which is both one-to-one and onto. Definition: A homeomorhism is a bijection f : X Y which is continuous and whose inverse is continuous. We often have much more structure. For examle if there is a notion of distance then the usual toology is that defined by the oen balls. Definition: A metric sace is a oint set X together with a ma d : X X IR such that i) d(x, y) = d(y, x) ii) d(x, y) 0 with equality iff x = y iii) d(x, y) d(x, z) + d(z, y) The metric toology is then defined by the oen balls U ɛ (x) = {y X d(x, y) < ɛ} (1.1) where ɛ > 0, along finite intersections and aribtrary unions of oen balls. Note that it follows that X is oen since it is a union of oen balls X = x X U ɛ (x) (1.2) for any ɛ > 0 of your choosing. These saces are always Hausdorff (roerty (ii) ensures that any two distinct oints have a finite distance between them and hence oen balls with ɛ taken to be half this distance will seerate them) and this also imlies that is oen. In articular we will heavily use IR n viewed as a metric toological sace (with the usual Pythagorian definition of distance). 1.2 Manifolds Rougly seaking a manifold is a toological sace for which one can locally make charts which iece together in a consistent way. Definition: An n-dimensional chart on M is a air (U, φ) where U is an oen subset of M and φ : U IR n is a homeomorhism onto its image φ(u) IR n Definition: A n-dimensional differentiable structure on M is a collection of n-dimensional charts (U i, φ i ), i I such that i) M = i I U i 4

5 ii) For any air of charts U i, U j with U i U j the ma φ j φ 1 i : φ i (U i U j ) φ j (U i U j ) is C, i.e. all artial derivatives exist u to any order. iii) We always take a differentiable structure to be a maximal set of charts, i.e. the union over all charts which satisfy (i) and (ii). N.B.: The functions φ j φ 1 i are called transition functions. Theorem: If M is connected then n is well defined, i.e. all charts have the same value of n. Proof: Suose that two charts had different values of n then it is clear that they can t intersect because the ma φ j φ 1 i takes a subset of IR n i to a subset of IR n j is a diffeomorhism and hence reserves the dimension. Since this is true for all charts we see that M must slit into disjoint charts, at least one for each different value of n. Since a chart is an oen set we can therefore write M as a union over disjoint oen sets, one for each value of n. Thus if M is connected it must have a unique value of n. Henceforth we will only consider connected toological saces. Definition: A differentiable manifold M of dimension n is a toologcal sace together with an n-dimensional differentiable structure. N.B.: One can also study differentiable structures where the transition functions are only C k for some k > 0. Alternatively one could relace IR n by C n and demaned that the transition functions are holomorhic. These are therefore a secial case of 2n-dimensional real manifolds. This leads to the beautiful and rich subject of comlex differential geometry which we will not have time to consider here. Examle: Trivially IR n is a n-dimensional manifold. A single chart that covers the whole of IR n is (IR n, id) where id is the identity ma id(x) = x Examle: Any oen subset U IR n is a n-dimensional manifold. Again the single chart (U, id) is sufficient. In fact any oen subset U of a manifold M with charts (U i, φ i ) is also a manifold since one can use the charts (U U i, φ i ). Problem: Why aren t closed subsets of IR n, e.g. a disk with boundary or a line in IR 2, along with the identity ma charts (note that in its own induced toology any subset of IR n is an oen set)? Examle: The Circle is a one-dimensional manifold. Let M = {(x, y) IR 2 x 2 + y 2 = 1}. We need at least two charts say U 1 = {(x, y) IR 2 x 2 + y 2 = 1, y > 1/ 2} and U 2 = {(x, y) IR 2 x 2 + y 2 = 1, y < 1/ 2}. We let φ 1 : U 1 IR φ 1 (x, y) = θ ( π 4, 5π 4 ) where (x, y) = (cos θ, sin θ) φ 2 : U 2 IR φ 2 (x, y) = θ ( 5π 4, π 4 ) where (x, y) = (cos θ, sin θ ) (1.3) 5

6 Now U 1 U 2 = {(x, y) IR < y < 1 2 } = V L V R (1.4) where and V L = {(x, y) IR < y < 1 2, x < 0} (1.5) V R = {(x, y) IR < y < 1 2, x > 0} (1.6) Now on φ 1 (V L ) = ( 3π, 5π), φ φ 1 1 (θ) = θ 2π whereas on φ 1 (V R ) = ( π, π), φ φ 1 1 (θ) = θ. Similarly φ 1 φ 1 2 (θ ) = θ +2π on φ 2 (V L ) = ( 5π, 3π) and φ φ 1 2 (θ ) = θ on φ 2 (V L ) = ( π, π). These mas are 4 4 C and hence we have a manifold. Here on the circle we see that locally we can defined a single coordinate, θ which we think of as an angle. But θ is not defined over the whole circle θ = 0 and θ = 2π are the same. This illustrates a key oint: The mas φ i rovide coordinates, just like a ma in an atlas rovides coordinates in the form of longitude and lattitude. However the coordinates will not in general work over the whole of the manifold. For examle the surface of the earth can be maed in an atlas but the notion of latitude and longitude will break down somewhere; at the oles longitude is not defined. Mas that one sees hanging on a wall always break down somewhere (usually at both oles) but an atlas can smoothly cover the whole earth. Often the U i are called coordinate neighbourhoods or atches and the φ i are coordinate mas. If M is some oint contained in a given atch U i then the local coordinates of are φ i () = (x 1 (), x 2 (), x 3 (),..., x n ()) IR n (1.7) Clearly there is a huge amount of choice of local coordinates. Tyically in any given atch U i we could choose from an infinite number of different functions φ i. Furthermore for a given manifold there will be infinitely many choices of oen sets U i which we use to cover it with. Differential geometry is the study of manifolds and uses tensoriol objects which take into account this huge redundancy in the actual way that we may choose to describe a given manifold. This is the so-called coordinate free aroach. Often, esecially in older texts, one fixes a covering and coordinate atches and writes any tensor in terms of it values in some given local coordinate system. This may be convenient for some calculational uroses but it obscures the true coordinate indeendent meaning of the imortant concets. In addition it should always be ket in mind when using exlicit coordinates that they are almost certainly not valid everywhere. One might often need to change coordinates, either because we refer to use a different choice of coordinates 6

7 valid in the same atch, or because we need to transform to a new atch which covers a different ortion of the manifold. In this course we will use the coordinate free aroach as much as ossible. Examle: Let us consider IRP n = (IR n+1 {0})/ where is the equivalence relation (x 1, x 2, x 3,..., x n+1 ) λ(x 1, x 2, x 3,..., x n+1 ) (1.8) for any λ IR {0}. We denote an element of the equivalence class by [x 1, x 2, x 3,..., x n+1 ]. Let us choose for the charts U 1 = {[x 1, x 2, x 3,..., x n+1 ] IRP n x 1 0} U 2 = {[x 1, x 2, x 3,..., x n+1 ] IRP n x 2 0} U 3 = {[x 1, x 2, x 3,..., x n+1 ] IRP n x 3 0}... U n+1 = {[x 1, x 2, x 3,..., x n+1 ] IRP n x n+1 0} (1.9) with the functions φ 1 ([x 1, x 2, x 3,..., x n+1 ]) = φ 2 ([x 1, x 2, x 3,..., x n+1 ]) = φ 3 ([x 1, x 2, x 3,..., x n+1 ]) =... φ n+1 ([x 1, x 2, x 3,..., x n+1 ]) = ( ) x 2 x, x3 1 x, x4 xn+1,..., IR n 1 x1 x 1 ( ) x 1 x, x3 2 x, x4 xn+1,..., IR n 2 x2 x 2 ( ) x 1 x, x2 3 x, x4 xn+1,..., IR n 3 x3 x 3 ( x 1 x n+1, x 2 x n+1, x 3 x n+1,..., x n x n+1 ) IR n (1.10) Clearly n+1 i=1 U i = IRP n and each φ i is a homeomorhism (without loss of generality we can take x i = 1 in U i ). Consider the intersection of U 1 and U 2. Here we can take x 1 = 1 and x 2 = v 0 so that φ 1 1 (u 1, u 2, u 3,..., u n ) = [1, u 1, u 2, u 3,..., u n ] (1.11) and hence ( 1 φ 2 φ 1 1 (u 1, u 2, u 3,..., u n ) = v, u 2 v, u 3 v,..., u ) n IR n (1.12) v 7

8 Since v 0 this ma is in C on φ 1 (U 1 U 2 ). All the other intersections follow the same way. Thus IRP n is an n-dimensional manifold. Problem: What is IRP 1? Theorem: If M and N are m and n-dimensional manifolds resectively then M N is an (m + n)-dimensional manifold. Proof: Let (U i, φ i ), i I be an m-dimensional differential structure for M and (V a, ψ a ) a A be an n-dimensional differential structure for N. We can construct a differential structure for M N by taking the following charts: W ia = U i V a χ ia : W ia IR n+m, χ ia (x, y) = (φ i (x), ψ a (y)) i I a J (1.13) where x M and y N. Clearly ia W ia = M N and χ ia are homeomorhims. It is also clear that the transition functions are C. χ jb χ 1 ia Problem: Show that the following: = (φ j φ 1 i, ψ b ψb 1 ) (1.14) U 1 = {(x, y) S 1 y > 0}, φ 1 (x, y) = x U 2 = {(x, y) S 1 y < 0}, φ 2 (x, y) = x U 3 = {(x, y) S 1 x > 0}, φ 3 (x, y) = y U 4 = {(x, y) S 1 x < 0}, φ 4 (x, y) = y (1.15) are a set of charts which cover S 1. Problem: Show that the 2-shere S 2 = {(x, y, z) IR 3 x 2 + y 2 + z 2 = 1} is a 2- dimensional manifold. 2 The Tangent Sace An imortant notion in geometry is that of a tangent vector. This is intuitively familiar for a curve in IR n. But the elementary defintion of a tangent vector, or indeed any vector, relies on secial roerties of IR n such a fixed coordinate system and its vector sace structure. Once given a vector v = (v 1, v 2, v 3 ) IR 3 for examle, we can consider the derivative in the direction of v: v 1 x + 1 v2 x + 2 v3 (2.16) x 3 8

9 Here we view this exression as an oerator acting on functions f : IR 3 IR. Changing coordinates, for examle by erforming a rotation, we also must change the coefficients v 1, v 2, v 3 in an aroriate way however the action on a function remains the same. We need to generalise the notion of a tangent vector to manifolds in a coordinate free way. There are several equiavalent ways to do this but here we will use the identification of a vector field with an oerator acting on functions. Our first ste is to introduce differentiable functions on manifolds. We will then roceed to understand vectors as oerators on differentiable functions. 2.1 Mas from M to IR Definition: A function f : M IR is C iff for each chart (U i, φ i ) in the differentiable structure of M f φ 1 i : φ i (U i ) IR n IR (2.17) is C. The set of such functions on a manifold M is denoted C (M). Note that if f φ 1 i is C and (U j, φ j ) is another chart with U i U j then f φ 1 j is C over a set of charts will be C on U i U j. Thus we need only check that f φ 1 i that covers M. Problem: Consider the circle S 1 as above. Show that f : S 1 IR with f(x, y) = x 2 +y is C. Definition: An algebra V is a real vector sace along with an oeration : V V V such that i) v 0 = 0 v = 0 ii) (λv) u = v (λu) = λ(v u) iii) v (u + w) = v u + v w iv) (u + w) v = u v + w v for all u, v, w in V and λ IR. Theorem: C (M) is an algebra with addition and multilication defined ointwise (f + g)() = f() + g() (λf)() = λf() (f g)() = f()g() (2.18) Proof: Let us show that f g is in C (M). Now (f g) φ 1 i = f φ 1 i 9 g φ 1 i (2.19)

10 where f φ 1 i and g φ 1 i are C. Therefore their ointwise roduct is too. Hence (f g) φ 1 i is C which is what we needed to show. The other roerties can be shown in a similar manor. Definition: For a oint M we let C () be the set of functions such that i) f : U IR where U M and U is an oen set. ii) f C (U) (recall that an oen subset of a manifold is a manifold) 2.2 Tangent Vectors We can now state our main definition: Definition: A tangent vector at a oint M is a ma X satisfies i) X (f + g) = X (f) + X (g) ii) X (constant ma) = 0 iii) X (fg) = f()x (g) + X (f)g() Leibniz rule : C () IR which The set of tangent vectors at a oint M is called the tangent sace to M at and is denoted by T M. The union of all tangent saces to M is called the tangent bundle T M = M T M (2.20) This is an examle of a fibre bundle and is itself a 2n-dimensional manifold. N.B.: In general objects that satisfy these roerties are called derivations. N.B.: With this definition a tangent vector is a linear ma from C () to IR. Since C () is a vector sace (it is an algebra) the tangent vectors are therefore elements of the dual vector sace to C (). However C () and hence its dual are infinite dimensional. The conditions (i),(ii) and (iii) restrict the ossible linear mas that we identify as tangent vectors and in fact we will see that they become a finite dimensional vector sace. Examle: Consider IR n as a manifold with its obvious chart U = IR n, φ : U IR n taken to be the identity. Then n X = λ µ (2.21) x µ is a tangent vector. In fact we will learn that all tangent vectors have this form. In what follows we will denote f x µ = µf, µ=1 2 f x µ x ν = µ ν f etc.. (2.22) 10

11 where f is defined on some oen set in IR n. We can extend this to manifolds by the following: Definition: Let (x 1,..., x n ) be local coordinates about a oint M. That is there exists a chart (U i, φ i ) with U i and φ i (q) = (x 1 (q), x 2 (q),..., x n (q)) for all q U i. We define x µ : C () IR (2.23) by Theorem: x µ f = µ (f φ 1 i )(x 1 (),..., x n ()) = µ (f φ 1 i ) φ i () (2.24) is a tangent vector to M at. x µ Proof: Let f, g C () be defined on a common oen set U in M that contains then x µ (f + g) = µ ((f + g) φ 1 i )(x 1 (),..., x n ()) = µ (f φ 1 i + g φ 1 i )(x 1 (),..., x n ()) = µ (f φ 1 i )(x 1 (),..., x n ()) + µ (g φ 1 i )(x 1 (),..., x n ()) = x µ (f) + x µ (g) (2.25) It is clear that (constant ma) = 0. x µ x µ (fg) = µ ((fg) φ 1 i )(x 1 (),..., x n ()) = µ (f φ 1 i g φ 1 i )(x 1 (),..., x n ()) = µ (f φ 1 i +(f φ 1 i )(x 1 (),..., x n ())(g φ 1 i (x 1 (),..., x n ()) µ (g φ 1 i = x µ (f)g() + f() x µ (g) )(x 1 (),..., x n ()) )(x 1 (),..., x n ()) (2.26) Of course we will show that all tangent vectors arise in this way. Examle: Consider IRP 1 = (IR 2 {0})/ where (x, y) λ(x, y), λ 0. We have two charts U t = {(x, y) x 0}, t = φ t ([x 1, x 2 ]) = x2 x 1 U s = {(x, y) y 0}, s = φ s ([x 1, x 2 ]) = x1 x 2 (2.27) 11

12 thus on the intersection U s U t, s = 1/t. Let = [1, 3] and consider the tangent vector X : C () IR, X(f) = f = d (f φ 1 t t )(t = 3) (2.28) dt How does X act in the other coordinate system (where they overla)? Recall that Now s(t) = φ s φ 1 t d ds(t) (f(t)) = dt dt and t(s) = φ t φ 1 s d (f(t(s))) (2.29) ds so we have that X(f) = d (f φ 1 t )(t = 3) dt d = ds(t) dt = 1 d t 2 ds ds = 1 d 9 ds (f φ 1 t (φ t φ 1 s ))(s = 1 3 ) (f φ 1 s ) s= 1 3 (f φ 1 s ) s= 1 3 (2.30) Thus the tangent vector can look rather different deending on the coordinate system one choses. However its definition as a linear ma from C () to IR is indeendent of the choice of coordinates, i.e. (2.28) and (2.30) will agree on any function in C (). Lemma: Let (x 1,..., x n ) be a coordinate system about M. Then for every function f C () there exist n functions f 1,..., f n in C () with and, where defined, f µ () = f(q) = f() + µ x µ f (2.31) (x µ (q) x µ ())f µ (q) (2.32) Proof: Let F = f φ 1 i which is defined on V = φ i (U U i ) where U is the domain of f. Let B be an oen ball in V IR n centred on v = φ i (). Take y B F (y 1,..., y n ) = F (y 1,..., y n ) F (y 1,..., y n 1, v n ) +F (y 1,..., y n 1, v n ) F (y 1,..., v n 1, v n ) F (y 1, v 2..., v n ) F (v 1,..., v n ) + F (v 1,..., v n ) = F (v 1,..., v n ) + ( F (y 1,..., y µ, v µ+1,..., v n ) F (y 1,..., v µ, v µ+1,..., v n ) ) µ = F (v 1,..., v n ) + µ F (y 1,..., v µ + t(y µ v µ ), v µ+1,..., v n ) t=1 t=0 12

13 = F (v 1,..., v n ) + µ = F (v 1,..., v n ) + µ d dt F (y1,..., v µ + t(y µ v µ ), v µ+1,..., v n )dt µ F (y 1,..., y µ 1, v µ + t(y µ v µ ), v µ+1,..., v n )(y µ v µ )dt (2.33) If we let F µ (y 1,..., y n ) = then we have that 1 0 µ F (y 1,..., v µ + t(y µ v µ ), v µ+1,..., v n )dt (2.34) F (y 1,..., y n ) = F (v 1,..., v n ) + µ (y µ v µ )F µ (y 1,..., y n ) (2.35) Finally we recall that F = f φ 1 i then we have f φ 1 i φ i (q) = f φ 1 i φ i () + µ so that if we let (y 1,..., y n ) = φ i (q) = (x 1 (q),..., x n (q)) (x µ (q) x µ ())F µ φ i (q) (2.36) thus f(q) = f() + µ (x µ (q) x µ ())f µ (q) (2.37) where we identify f µ = F µ φ i. It also follows that (f φ 1 i ) x µ f = φ x µ i () = F y φ i() µ = ( F (v 1,..., v n ) + ) (y ν v ν )F y µ ν (y 1,..., y n ) ν ( ) = F µ (y 1,..., y n ) + ν = F µ (φ i ()) = f µ () (y ν v ν ) µ F ν (y 1,..., y n ) y=φ i () y=φ i () (2.38) Theorem: T (M) is an n-dimensional real vector sace and a set of basis vectors is { } x µ, µ = 1,..., n 13 (2.39)

14 i.e. a general element of T (M) can be written as X = µ λ µ x µ (2.40) Proof: First we note that if X and Y are two tangent vectors at a oint M then we can add them or multily them by a number λ IR: i) (X + Y ) : C IR, f X f + Y f ii) (λx ) : C IR, f λ(x f) (Convince yourself of this.) Thus T M is a real vector sace. From the above lemma we have that X (f) = X ( f() + µ (x µ x µ ())f µ ) (2.41) Now X (f()) = X (x µ ()) = 0 since f() and x µ () are constants. Thus we have that X (f) = µ = µ = µ ((x µ (q) x µ ())X (f µ ) + X (x µ )f µ (q)) q= X (x µ )f µ () X (x µ ) x µ f (2.42) where we used (2.38). This shows that the elements (2.39) san the tangent sace. We must now show that they are linearly indeendent. To this end suose that The coordinate functions are in C () so we may consider λ µ µ x µ = 0 (2.43) 0 = µ λ µ x µ x ν = µ λ µ x µ (xν ) = µ λ µ δ ν µ = λ ν (2.44) Thus λ µ = 0 for all µ. 14

15 So why are they called tangent vectors? First we consider IR n consider a curve C : (0, 1) IR n. We recall from elementary geometry that the tangent vector to a oint = C(t 1 ) is a line through in the direction (i.e. with the sloe) ( ) dc 1,..., dcn dt t=t1 dt t=t1 where C(t) = (C 1 (t),..., C n (t)) IR n is C. Now if f C () then, by our definition, X : C () IR defined by (2.45) X (f) = d dt f(c(t)) t=t1 (2.46) is a tangent vector to IR n at = C(t 1 ), i.e. it acts linearly on the function f, vanishes on constant functions and satisfies the Leibniz rule. On the other hand we also have X (f) = dcµ dt Thus dc µ /dt are the comonents of X in the basis So we have suceeded in generalising the definition (2.16) 2.3 Curves and their Tangents We can now discuss curves on manifolds and their tangents. µ (f(c(t)) t=t1 (2.47) t=t1 x µ (2.48) Definition: Consider an oen interval (a, b) IR a ma C : (a, b) M to a manifold M is called a smooth curve on M if φ i C is C whereever it is definined for any chart (U i, φ i ) of M (i.e. with U i C((a, b)) ). Definition: For a oint t 1 (a, b) with C(t 1 ) = we can define the tangent vector T (C) T (M) to the curve C at by T (C)(f) = d dt f(c(t)) t=t1 (2.49) It should be clear that T (C) T M. Let M be a oint on a curve C at t = t 1 which is covered by a chart (U i, φ i ). Then there is some ɛ > 0 such that C((t 1 ɛ, t 1 + ɛ)) U i (2.50) 15

16 We can exress the tangent to C at as T (C)(f) = df(c(t)) dt t=t1 = d ( f φ 1 i φ i C(t) ) t=t1 dt n d(φ µ C(t)) = µ (f φ 1 dt t=t1 i )(φ i (C(t 1 )) µ=1 (2.51) Here we have slit u f C : (t 1 ɛ, t 1 + ɛ) IR as the comosition of a function φ i C : (t 1 ɛ, t 1 + ɛ) IR n and a function f φ 1 i : IR n IR and used the chain rule. Thus we have n d(φ µ C(t)) T (C)(f) = dt t=t1 x µ f (2.52) so that T (C) = µ=1 n µ=1 d(φ µ C(t)) dt t=t1 x µ (2.53) Conversely suose that T is a tangent vector to M at. We will now show that we can construct a curve through such that T is its tangent at. Let (x 1,..., x n ) = φ i (q) be local coordinates about defined on an oen set U i. Hence we may write for some numbers T µ. We define C : ( ɛ, ɛ) M by n T = T µ µ=1 x µ (2.54) φ µ i C(t) = x µ () + tt µ (2.55) where we ick ɛ sufficiently small so that φ i (C( ɛ, ɛ)) U i. It follows that d dt f(c(t)) t=0 = d dt ( (f φ 1 i ) (φ i C)(t) ) t=0 = d(φµ i C) µ (f φ 1 dt t=0 i )(φ i (C(0)) = T µ x µ f (2.56) We have therefore shown that all curves through M define a tangent vector to M at and that conversely all tangent vectors to M at can be realised as the tangent vector to some curve C. However this corresondence is not unique. Clearly many distinct curves may have the same tangent vector at and conversely the construction of the curve through with tangent T was not unique. But this does lead to the following theorem: 16

17 Theorem: T M is isomorhic to the set of all curves through M modulo the equivalence relation that C(t) C (t) iff d(f C) dt for all f C () where C(t 1 ) = C (t 1) =. t=t1 = d(f C ) t=t dt 1 (2.57) Proof: It is easy to check that the construction above rovides a bijection between these two saces. The more difficult art, which we won t go into here, is to show that the vector sace structure is reseved. Indeed to do this we d need to give a vector sace structure to the equivalance class of curves through. Theorem: Let (x 1,..., x n ) = φ µ 1 and (y 1,.., y n ) = φ µ 2 be two coordinate systems at a oint M with U 1 U 2 and suose that X T M. If then n X = A µ n µ=1 x µ and X = B µ µ=1 y µ (2.58) n B µ = A ν ν=1 x ν y µ (2.59) where y µ (x 1,..., x n ) = φ 2 φ 1 1 is a smooth function from φ 1 (U 1 ) IR n φ 2 (U 2 ) IR n Proof: We have in the second coordinate system that but in the first coordinate system we see that since these must agree we rove the theorem. N.B.: This formula is often simly written as or even n X (y µ ) = B ν ν=1 y ν y µ = B µ (2.60) n X (y µ ) = A ν ν=1 x ν y µ (2.61) B µ = n ν=1 A ν yµ x ν (2.62) A µ = A ν x µ x ν (2.63) with a sum over ν understood and a rime denoting quantities in the new coordinate system. 17

18 3 Mas Between Manifolds 3.1 Diffeomorhisms Definition: Suose that f : M N where M, (U i, φ i ), i I and N, (V a, ψ a ), a A are two manifolds. We say that f is C iff is C for all i I and a A. ψ a f φ 1 i : φ i (f 1 (V a )) IR n (3.64) Problem: Show that f : S 1 S 1 defined by f(e 2πiθ ) = e 2πinθ is C for any n. Theorem: Suose that M, N and P are manifolds with f : M N and g : N P C. Then g f : M P is also C. Proof: this follows from the chain rule. Definition: If f : M N is a bijection with both f and f 1 C then f is called a diffeomorhism. Two manifolds with a diffeomorhic, i.e. for which there exists a difformorhism between them, are equivalent geometrically. Problem: Show that the charts of two diffeomorhic manifolds are in a one to one corresondence. Problem: Show that the set of diffeomorhisms from a manifold to itself forms a grou under comosition. 3.2 Push Forward Theorem: Let f : M N be C. If X T M then is a tangent vector to N at f(). f X : C (f()) IR defined by g X (g f) (3.65) Proof: Let g 1, g 2 C (f()) and λ IR. Then f X f() (g 1 + g 2 ) = X ((g 1 + g 2 ) f) = X (g 1 f + g 2 f) = X (g 1 f) + X (g 2 f) = f X f() (g 1 ) + f X f() (g 2 ) (3.66) 18

19 and Finally we have f X f() (λ) = X (λ f) = X (λ) = 0 (3.67) f X f() (g 1 g 2 ) = X ((g 1 g 2 ) f) = X ((g 1 f) (g 2 f)) = X (g 1 f)(g 2 f) + (g 1 f)x (g 2 f) = f X f() (g 1 )g 2 (f()) + g 1 (f())f X f() (g 2 ) (3.68) Definition: f X is called the ush forward of X. Theorem: Suose that f : M N is C, (x 1,..., x m ) are local coordinates for a oint M and y 1,..., y n are local coordinates for the image f() N. If m X = A µ µ=1 x µ (3.69) is in T M then f X = m n A µ µ=1 ν=1 x µ (y ν f) y ν (3.70) f() Proof: Let g C (f()). Then f X(g) = X (g f) = m A µ x µ (g f) = = = = µ=1 m µ=1 m µ (g f φ 1 i ) A φ x µ i () µ (g ψ 1 a A ψ a f φ 1 i ) φ x µ i () µ=1 m n A µ (ψν a f φ 1 i ) (g ψ 1 a ) (φ µ=1 ν=1 x µ i ()) (ψ y ν a (f()) m n A µ µ=1 ν=1 x µ (y ν f) y ν (g) f() (3.71) 19

20 4 Vector Fields 4.1 Vector Fields Next we consider vector fields Definition: A vector field is a ma X : M T M such that X() = X T M and for all f C (M) the maing X(f)() (4.72) is C. For vector fields we will dro the exlicit suscrit. Thus a vector field assigns, in a smooth way, a vector in T M of to each oint M. N.B.: As we defined it a vector field is valid over all of M however it can also be defined over an oen subset U M. Is the roduct of two vector fields a vector feild? To check this we consider two vector fields X, Y and f, g C (M). With multilication of vectors taken to mean X Y (f) = X(Y (f)) then we see that indeed X(Y (f + g)) = X(Y (f) + X(g)) = X(Y (f)) + X(Y (g)) X(Y (constant ma)) = X(0) = 0 (4.73) where 0 is the constant ma that takes all oints to zero. However we find X(Y (f g)) = X(Y (f)g + fy (g)) = X(Y (f))g + Y (f)x(g) + X(f)Y (g) + fx(y (g)) (4.74) and this is not a vector field (due to the middle two terms). But we can construct a vector field by taking [X, Y ](f) = X(Y (f)) Y (X(f)) (4.75) since we see that [X, Y ](f g) = X(Y (f))g + Y (f)x(g) + X(f)Y (g) + fx(y (g)) so that [X, Y ] is a vector field. Y (X(f))g X(f)Y (g) Y (f)x(g) fy (X(g)) = [X, Y ](f)g + f[x, Y ](g) (4.76) Definition: [X, Y ] is called the commutator of two vector fields. Problem: What goes wrong if try to define (X Y )(f) = X(f) Y (f)? 20

21 Problem: Show that, if in a articular coordinate system, X = µ X µ (x) x µ, Y = µ Y µ (x) x µ (4.77) then [X, Y ] = µ (X µ µ Y ν Y µ µ X ν ) ν x ν (4.78) Theorem: With the roduct of two vector fields defined as the commutator the sace of vector fields is an algebra. Proof: We have already seen that T M is a vector sace for a articular M. This clearly ensures that the sace of vector fields is a vector sace (with addition and scalar mutlilication defined ointwise). We need to check the conditions (i) - (iv) in the definition of an algebra. Conditions (i) and (ii) are obviously satisfied. In addition since [X, Y ] = [Y, X] we need only check that [X, Y + Z](f) = X(Y + Z)(f) (Y + Z)(X)(f) = X(Y )(f) + X(Z(f) Y (X)(f) Z(X)(f) = [X, Y ](f) + [X, Z](f) (4.79) Problem: Show that for three vector fields X, Y, Z on M the jacobi identity holds: [X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y ]] = 0 (4.80) Problem: Consider a manifold with a local coordinate system U i, φ i = (x 1,..., x n ). Therefore, in U i we can simly write x µ = x µ (4.81) i) Show that [ ], x µ x = 0 ν ii) Evaluate [ ], ϕ(x 1, x 2 ) x 1 x where ϕ(x 1, x 2 ) is a C function of x 1, x Integral and Local Flows Given a vector field X we can construct curves that ass through M for which the tangent vector at is X. Definition: Let X be a vector field on M and consider a oint M. An integral curve of X assing through is a curve C(t) in M such that ) C(0) = and C ( d dt = X C(t) (4.82) 21

22 for all t in some oen interval ( ɛ, ɛ) IR. Here we are viewing d/dt as a vector field on IR so that ) C ( d dt (f) = d dt f(c(t)) = T (C) (4.83) is just the tangent vector to C(t) at = C(t 1 ). If we introduce a local coordinate system so that, in a oen set about M, X = µ X µ (x) x µ φ 1 (x) (4.84) then we find the integral curve is ( ) d C (f) dt = d dt f(c(t)) = d (f φ 1 i φ i C(t)) dt = dc µ µ dt (t) x µ (f) C(t) (4.85) where, as before, we have C µ = φ µ i C. On the other hand we have X(C(t))(f) = µ X µ (C(t)) x µ (f) (4.86) C(t) Thus we see that the condition for an integral curve is a first order differential equation for the coordinates of the curve C µ (t) d dt Cµ (t) = X µ (C(t)) (4.87) with the initial condition x µ (C(0)) = x µ (). This is a first order differential equation and, as such it has a unique solution with the given initial condition. However it is not at all clear whether or not the solution can be extended to all values of t. In artcular even if there is a solution to the differential equation (4.87) for al t one must worry about atching solutions together over the different coordinate atches. This leads to Definition: A vector field X on M is comlete if for every oint M the integral curve of X can be extended to a curve on M for all values of t. Theorem: If M is comact, i.e. all oen covers have a finite subcover, then all vector fields on M are comlete. Proof: We will not take the time to rove this theorem in this course. 22

23 Let σ(t, ) be an integral curve of a vector field X that asses through at t = 0. This therefore satisfies d dt σµ (t, ) = X µ (σ(t, )) (4.88) along with the initial condition σ(0, ) =. Thus we have found the following, at least locally (and globally for comlete vector fields): Definition: The flow generated by a vector field X is a differentiable ma σ : IR M M (4.89) such that i) at each oint M the tangent to the curve C (t) = σ(t, ) at is X ii) σ(0, ) = iii) σ(t + s, ) = σ(t, σ(s, )) We must show that the third roerty holds. To do this we simly note that since (4.88) is a first order ordinary differential equation it has a unique solution for a fixed initial condition. So consider d dt σµ (t + s, ) = dσµ (t + s, ) d(t + s) = X µ (σ(t + s, )) which satisfies the intial condition σ(0 + s, ) = σ(s, ). On the other hand we have (4.90) d dt σµ (t, σ(s, )) = X µ (σ(t, σ(s, ))) (4.91) with the intial condition σ(0, σ(s, )) = σ(s, ). Thus both σ(t + s, ) and σ(t, σ(s, )) satisfy (4.88) with the same boundary condition and hence they must be equal. We see that each oint M a flow defines a curve C (t) = σ(t, ) in M whose tangent is X and such that C (0) =. Thinking of things the other way we have that, for each t, the flow defines a ma ψ t : M M such that ψ t+s = ψ t ψ s Now for t = ɛ small we have, from (4.88), ψ µ ɛ () = σ µ (ɛ, ) = x µ () + ɛx µ () + O(ɛ 2 ) (4.92) Thus at least for a small enough value of t ψ t () is 1-1 and C. Hence it is a diffeomorhism onto its image (at least for an oen set in M). This romts another defintion Definition: A one arameter grou of diffeomorhism of M is collection of diffeomorhisms σ t : M M with t IR such that i) σ t (σ s ) = σ t+s 23

24 ii) σ 0 is the identity ma iii) σ t = σt 1 So in effect we can think of vector fields as generating infinitessimal diffeomorhism, through their flows. In this sense vector fields can be identified with the Lie algebra of the diffeomorhism grou. 4.3 Lie Derivatives A vector field allows us to introduce a notion of a derivative on a manifold. The roblem with the usual derivative f = lim f( + ɛ) f() (4.93) ɛ 0 ɛ is that we don t know how to add two oints on a manifold, i.e. what is + ɛ? However we saw that, at least locally, a vector field generates a unique integral flow about any given oint. Therefore we can use the flow to take us to a nearby oint and hence form a derivative. This is the notion of a Lie deriviative Definition: Let X be a vector field on M and f C (M) we define the Lie derivative of f along X to be f(σ(ɛ, )) f() L X f() = lim (4.94) ɛ 0 ɛ where σ(ɛ, ) is the flow generated by X at. Theorem: L X f = X(f) Proof: We have from the definition that L X f() = lim ɛ 0 f(σ(ɛ, )) f(σ(0, )) ɛ = d dt f((σ(t, )) t=0 ) = σ(t, ) ( d dt = X (f) (f) t=0 (4.95) where we used the defining roerty of the flow, namely that its tangent at is X. We can also define the Lie derivative of a vector field Y along X by Definition: σ( ɛ) Y σ(ɛ) Y L X Y = lim (4.96) ɛ 0 ɛ where again σ(ɛ) is the flow generated by X and we have suressed the deendence on the oint. 24

25 Theorem: L X Y = [X, Y ] (4.97) Proof: Let us introduce coordinate about M such that X = µ We start with the fact that X µ (x) x µ, Y = µ Y µ (x) x µ (4.98) σ ν (ɛ, ) = x ν () + ɛx ν () + O(ɛ 2 ) (4.99) and note that, for any f, Y σ(ɛ) f = µ = µ = µ = µ Y µ (σ(ɛ)) x µ f σ(ɛ) (Y µ + ɛx λ λ Y µ ) x µ f + O(ɛ 2 ) σ(ɛ) (Y µ + ɛx λ λ Y µ )( µ f + ɛx λ µ λ f) + O(ɛ 2 ) (Y µ µ + ɛx λ λ Y µ µ + ɛy µ X λ µ λ )f + O(ɛ 2 ) (4.100) Therefore σ( ɛ) Y σ(ɛ) (f) = Y σ(ɛ) (f σ( ɛ)) = Y µ (f σ(ɛ)) µ x µ (f σ( ɛ)) = ( ) Y µ µ + ɛx λ λ Y µ µ + ɛy µ X λ µ λ (f ɛν fx ν ) + O(ɛ 2 ) µ = µ Y µ µ f + ɛ( λ Y µ X λ Y λ λ X µ ) µ f + O(ɛ 2 ) From which it follows that and we are done. σ(ɛ) Y σ(ɛ) (f) Y (f) ɛ = ( λ Y µ X λ Y λ λ X µ ) µ f + O(ɛ) = [X, Y ] µ µ f + O(ɛ) = [X, Y ](f) + O(ɛ) 25 (4.101) (4.102)

26 5 Tensors 5.1 Co-Tangent Vectors Recall that the tangent sace T M at a oint M is a vector sace. For any vector sace there is a natural notion of a dual vector sace which is defined as the sace of a linear mas from the vector sace to IR. Problem: Show that the dual sace is a vector sace and has the same dimension as the original vector sace (if it is finite dimensional). Thus we can define Definition: The co-tangent sace to M at M is the dual vector sace to T M and is denoted by T M. In other words ω T M iff ω : T M IR is a linear ma. We denote the action of ω on a vector X T M by Since ω is a linear ma we have ω (X ) = ω, X (5.103) ω, X + λy = ω (X + λy ) = ω (X ) + λω (Y ) = ω, X + λ ω, Y (5.104) and we may also take, in effect by definition, ω + λη, X = (ω + λη )(X ) = ω (X ) + λ η(x ) = ω, X + λ η, X (5.105) Thus, is linear in each of its entries. Now the dual of the dual of a vector sace is just the orginal sace itself. Why? Well for a fixed vector X we can construct the ma: ω ω (X ) IR (5.106) The roerties of dual vectors ensure that this is a linear ma. Thus we can view vectors X as linear mas acting on co-vectors ω via X (ω ) = ω, X (5.107) Just as for the tangent bundle we defined the co-tangent bundle to be which is a 2n-dimensional manifold. T M = M T M (5.108) Definition: A smooth co-vector field is a ma ω : M T M such that 26

27 i) ω() T M ii) ω(x) : M IR is in C (M) for all smooth vector fields X We saw that a chart (U i, φ i ) defines a natural basis of T M for U i : x µ (5.109) where φ i () = (x 1 (),..., x n ()). This allows us to define a natural basis for T M by dx µ, x ν = δ µ ν (5.110) i.e. dx µ is a linear ma from T M to IR that mas the vector to v µ. Thus if in a local coordinate chart we have a vector n v ν ν=1 x ν (5.111) n V () = V µ () µ=1 x ν (5.112) and a co-vector Then n ω() = ω ν ()dx ν (5.113) ν=1 ω(), V () = ω ν ()dx ν, V µ () ν=1 µ=1 x µ = ω ν ()V µ () dx ν, µ=1 ν=1 x µ = ω ν ()V µ ()δµ ν µ=1 ν=1 = µ=1 ω µ ()V µ () (5.114) Theorem: Let (x 1,..., x n ) = φ µ 1 and (y 1,.., y n ) = φ µ 2 be two coordinate systems at a oint M with U 1 U 2 and suose that ω T M. If n ω = A µ dx µ n and ω = B µ dy µ (5.115) µ=1 µ=1 27

28 then n A µ = B ν ν=1 x µ y ν (5.116) where y µ (x 1,..., x n ) = φ 2 φ 1 1 is a smooth function from φ 1 (U 1 ) IR n φ 2 (U 2 ) IR n. Proof: We have in the first coordinate system that ( ) n ω x µ = A ν dx ν, x µ = A µ (5.117) ν=1 but in the second coordinate system we see that ( ) n ω x µ = B ν dy ν, x µ (5.118) now we saw in the case of vectors that x µ = λ Thus we find ( ) ω x µ = = = ν=1 ( ) x µ y λ y λ (5.119) n B ν dy ν, ( ) ν=1 λ x µ y λ y λ ( ) n B ν ν=1 x µ y λ dy ν, λ y λ n B ν x µ y ν (5.120) ν=1 since these must agree we rove the theorem. N.B.: We can also think in terms of x µ (y) = φ 1 φ 1 2. We can see that x µ y ν and y µ x ν (5.121) are inverses of each other (when view as matrices): δµ ν = xν x µ = ( ) φ ν x µ 1 φ 1 1 (x) = ( ) φ ν x µ 1 φ 1 2 φ 2 φ 1 1 (x) = ( ) φ ν λ y λ 1 φ 1 ( ) 2 (y) φ λ x µ 2 φ 1 1 (x) = ( ) ( ) λ y λ x ν x µ y λ (5.122) 28

29 In this case the formula becomes n B µ = A ν ν=1 y µ x ν (5.123) N.B.: This formula is often simly written as B µ = n ν=1 A ν x ν y µ (5.124) or even A x ν µ = A ν (5.125) x µ with a sum over ν understood and a rime denoting quantities in the new coordinate system. Note the different ositions of the rime and unrimed coordinates as comared to the analogous formula for a vector! 5.2 Pull-back and Lie Derivative of a co-vector Suose we have smooth ma f : M N. We saw that we could ush-forward a vector X T M to a vector f X f() T f() N by f X f() (g) = X (g f) (5.126) We therefore can consider the dual ma f : T f() N T M defined by f ω(x ) = f ω, X = ω, f X f() (5.127) Note that for each co-vector ω T f() N this defines a linear ma f ω : T M IR and hence an element of T M. Theorem: Let f : M N, (y 1,..., y n ) be local coordinates on V N and (x 1,..., x m ) local coordinates on U f 1 (V ) M. If then f ω = n ω = ω ν dy ν f() (5.128) ν=1 m n ω ν µ=1 ν=1 x µ (yν f)dx µ (5.129) Proof: We have that ω, f X f() = = n ω ν (f X f() ) ν ν=1 n m ω ν X µ ν=1 µ=1 x µ (y ν f) 29 (5.130)

30 where we used our earlier result for the comonents of the ush forward of a vector. On the other hand we have m f ω, X = (f ω) µ X µ (5.131) µ=1 and since, by definition, f ω, X = ω, f X f() we rove the theorem. We can also extend the definition of the Lie Derivitive to co-vector fields (and in fact all tensor fields). Definition: If X is a smooth vector field and ω a smooth co-vector field on M then L X ω = d dt ( σ(t, ) ω t=0) (5.132) where σ(t, ) is the flow generated by X. Theorem: If, in a local coordinate system (x 1,..., x n ), X = µ X µ (x) x µ, ω = µ ω µ dx µ (5.133) then L X ω = µ ( ν ω µ X ν + ω ν µ X ν ) dx µ (5.134) ν Problem: Prove this theorem. 5.3 Tensors We can now definition the notition of a (r, s)-tensor. First we need to recall the definition of the tensor roduct. If V and W are two vector saces, with basis {v i i = 1,..., n} and {w a a = 1,..., m} resectively, then the vector sace sum is an n + m dimensional vector sace which is sanned by {v i, w a i = 1,..., n, a = 1,..., n} i.e. so a general element is V W = San i,a {v i, w a } (5.135) n m a i v i + b a w a (5.136) i a where the sum is interreted as a formal sum. Hence V W is an (n + m)-dimensional vector sace. On the other hand we can also contruct the tensor roduct which sanned by {v i w a i = 1,..., n, a = 1,..., m}, i.e. V W = San i,a {v i w a } (5.137) 30

31 where v i w a is a formal roduct. A general element is m i m c ia v i w a (5.138) a Therefore is V W an nm-dimensional vector sace. Definition: An (r, s)-tensor T at a oint M is an element of T (r,s) M = ( r T M) ( s T M) (5.139) where r denotes the rth tensor roduct. It follows that, given a local coordinate system, a local basis for (r, s)-tensors is given by x µ... dx ν 1... dx νs (5.140) 1 x µr and from the definitions earlier we have that if T = T µ 1µ 2...µ rν1 ν 2...ν s x µ... 1 then the comonents can be comuted as x µr ( T µ 1µ 2...µ rν1 ν 2...ν s = T dx µ 1,..., dx µr, x ν,... 1 dx ν 1... dx νs (5.141) x νs ) (5.142) Problem: Show that if, in a local coordinate system T = A µ 1µ 2...µ r ν1 ν 2...ν s x µ... 1 is an (r, s)-tensor, are s vectors and are r co-vectors, then X i = X µ i x µr dx ν 1... dx νs (5.143) x µ, i = 1,..., s (5.144) ω I = ω I νdx ν, I = 1,..., r (5.145) T (ω 1,..., ω r, X 1,..., X s ) = A µ 1...µ r ν 1...ν s ω 1 µ 1...ω r µ r X ν X νs 1 (5.146) Problem: Show that if T is an (r, s)-tensor, (U 1, φ 1 = (x 1,..., x n )) and (U 2, φ 2 = (y 1,..., y n )) are two local coordinates charts with U 1 U 2 such that T = A µ 1µ 2...µ r ν1 ν 2...ν s x µ x µr dx ν 1... dx νs (5.147)

32 and then T = B µ 1µ 2...µ sν1 ν 2...ν s y µ... 1 B µ 1µ 2...µ rν1 ν 2...ν s = y µr dy ν 1... dy νs (5.148) n n n n A ρ 1ρ 2...ρ s λ1 λ 2...λ s λ 1 =1 λ r=1 ρ 1 =1 ρ s=1 ( ) ( ) x ρ y µ 1... y µr 1 x ρr ( ) ( ) y ν x λ 1... x λs 1 y νs (5.149) where y µ (x) is understood to be the comonent of the transition function φ 2 φ 1 1 and x µ (y) is understood to be the comonents of the transition function φ 1 φ 1 Definition: An (r, s)-tensor field is a ma T : M ( r T M) ( s T M) such that T () ( r T M) ( s T M) (5.150) which smooth in that, for any choice of r smooth co-vector fields ω 1,..., ω r and s smooth vector fields V 1,..., V s, the ma T (ω 1,..., ω r, V 1,..., V s )() : M IR is C. Some tensors and tensor fields have secial names: A (0, 0)-tensor is a scalar. As a field it assigns a number to each oint in M. A (1, 0)-tensor is a vector. As a field it assigns a tangent vector to each oint in M. 2. A (0, 1)-tensor is a 1-form. As a field it assigns a co-vector to each oint in M. Sometimes, esecially in older books, (r, 0)-tensors are called covariant and (0, s)- tensors contravariant. An (r, 0)-tensor is called symmetric if Similarly and (0, s) tensor is called symmetric if T ( ω P (1),..., ω P (r)) = T ( ω 1,..., ω r) (5.151) T ( V P (1),..., V P (s)) = T ( V 1,..., V s) (5.152) On the other hand they are called anti-symmetric if T ( ω P (1),..., ω P (r)) = sgn(p )T ( ω 1,..., ω r) (5.153) 32

33 or T ( V P (1),..., V P (s)) = sgn(p )T ( V 1,..., V s) (5.154) Here P is a ermutation and sgn(p ) is its sign. Recall that a ermutation P is a bijection P : {1,.., n} {1,..., n} and can always be written in terms of either an even (sgn(p ) = 1) or odd (sgn(p ) = 1) number of interchanges (where two neighbouring integers are ermuted). Similarly one can talk about the symmetry roerties of the (r, 0) and (0, s) comonents of a mixed (r, s)-tensor seerately. 6 Differential Forms N.B.: Conventions about form maniulations can vary from book to book (by various factors of! and minus signs). So be careful when comaring two sources. 6.1 Forms Definition: A -form on a manifold M is a smooth anti-symmetric (0, )-tensor field on M. In articular if ω is a -form then ω ( X P (1),..., X P () ) = sgn(p )ω (X1,..., X ) (6.155) Definition: A 0-form on M is a function in C (M). Theorem: If M is n-dimensional then all -forms with > n vanish. Proof: First note that a -form acting on a set of vectors with the same vector aearing twice vanishes: ω (X 1,..., Y, X 2,..., Y, X 3,...) = ( 1) n ω (Y, Y, X 1,...) = ( 1) n ω (Y, Y, X 1,...) = 0 (6.156) where in the first line we used a ermutation to lace the same vectors next to each other and in the enultimate line we used an interchange (which of course has no effect). It is now easy to see that if ω is a -form with > n then in any collection of basis vectors at least two must be the same. Hence ω vanishes. Since it vanishes on any set of basis vectors it vanishes identically. The sace of -forms on M is denoted Ω (M, IR) and we let Ω(M) = Ω 0 (M, IR) Ω 1 (M, IR)... Ω n (M, IR) (6.157) 33

34 Here we have included an exlicit reference to the field IR over which the manifold is defined. Note that if ω and η are a -form and q-form resectively then ω η will be a (0, +q) tensor field but not a ( + q)-form since it will not be anti-symmetric. To correct for this we consider the so-called wedge roduct Definition: If ω ω (M) and η ω q (M) we define (ω η)(x 1,..., X +q ) = P sgn(p )(ω η)(x P (1),..., X P (+q) ) (6.158) If should be clear that this defines a ( + q)-form by construction. Let us work out a few examles Examle: dx µ dx ν = dx µ dx ν dx ν dx µ (6.159) Examle: dx µ (dx ν dx λ ) = dx µ dx ν dx λ dx µ dx λ dx ν +dx ν dx λ dx µ dx ν dx µ dx λ +dx λ dx µ dx ν dx λ dx ν dx µ (6.160) Problem: If ω = A 12 dx 1 dx 2 + A 34 dx 3 dx 4 η = B 123 dx 1 dx 2 dx 3 + B 125 dx 1 dx 2 dx 5 (6.161) then what is ω η? Theorem: If ω Ω (M) and η Ω q (M) then ω η = ( 1) q η ω Problem: Prove this! Theorem: A basis for Ω (M) at a oint M is given by dx µ 1... dx µ (6.162) 34

35 Proof: Any ω Ω (M) is defined by its action on a set of basis vectors ( ) ω x µ,..., = ω 1 x µ µ1...µ (6.163) Since ω is antisymmetric we have that ( ) ω µp (1)...µ P () = ω x µ,..., P (1) x µ P () ( ) = sgn(p )ω x µ,..., 1 x µ so 1! ωµ1...ν dx µ 1... dx µ = 1! = sgn(p )ω µ1...µ (6.164) sgn(p )ω µ1...µ dx µ P (1)... dx µ P () P = 1 ω µp! (1)...µ P () dx µ P (1)... dx µ P () P = ω µ1...ν dx µ 1... dx µ = ω (6.165) This shows that dx µ 1... dx µ sans the sace of -forms. But this formula shows that they are linearly indedent too (since we know that dx µ 1... dx µ are linearly indeendent). 6.2 Exterior Derivative We can define a notion of a derivative on -forms by Definition: If ω Ω (M) then we define dω(x 1,..., X +1 ) = i ( 1) i+1 X i (ω(x 1,.., X i 1, X i+1,..., X +1 )) + i<j( 1) i+j ω([x i, X j ], X 1,.., X i 1, X i+1,...,, X j 1, X j+1,..., X +1 ) (6.166) Why on earth is this called a derivative? Theorem: If in a local coordinate system we have that ω = 1 ωµ1,...,µ! dx µ 1... dx µ (6.167) 35

36 then dω = 1 ν ω µ1,...,µ! dx ν dx µ 1... dx µ (6.168) Proof: We need only evaluate dω at the usual choice of basis of vectors ( ) dω x µ,..., 1 x µ = ( 1) ν+1 ( +1 ν x ν (ω x 1,..., x ν 1 )) x ν+1,... x +1 (6.169) where we have used the fact that [ ] x µ, x ν = 0 (6.170) so that the second term in the definition of dω vanishes. Thus if write we find dω = 1 ( + 1)! (dω)νµ1...µ dx ν dx µ 1... dx µ (6.171) (dω) νµ1...µ +1 = ν ( 1) ν+1 ν ω µ1...ν 1,ν+1...µ = ( + 1) [ν ω µ2...µ +1 ] (6.172) where we have used the antisymmetry of ω µ2...µ +1 and the square brakets are defined by X [µ1...µ ] = 1 sgn(p )X µp! (1)...µ P () (6.173) However we also need to show that (we imlicitly used it above): Theorem: If ω Ω (M) then dω Ω +1 (M) P Proof: It is clear that this defines something that is anti-symmetric under an interchange of any two vectors X µ X ν. Why does it have this funny form? Let us exand it in terms of some basis. The first term gives X 1 (ω(x 2, X 3,..., X )) X 2 (ω(x 1, X 3,..., X )) +... = µν... X µ 1 µ (ω ν... X ν 2...) X µ 2 µ (ω ν... X ν 1...) +... = µν...(x µ 1 X ν 2... X µ 2 X ν 1...) µ ω ν... + νλ... ω νλ... (X µ 1 µ X ν 2 X µ 2 µ X ν 1 )X λ (6.174)

37 These derivative terms on X ν need to be cancelled in order for dω to be a form. This is the role of the second term since ( 1) i+j ω([x i, X j ], X 1,.., X i 1, X i+1,...,, X j 1, X j+1,..., X +1 ) i<j = µν... ω νλ... [X 1, X 2 ] ν X λ = µν... ω νλ... (X µ 1 µ X ν 2 X µ 2 µ X ν 1 )X λ (6.175) Examle: Consider IR 3. A 0-form is just a function and df = µ fdx µ (6.176) is just the gradient of f. A 1-form ω = ω µ dx µ has dω = ν ω ν dx ν dx µ = ( 1 ω 2 2 ω 1 )dx 1 dx 2 + ( 2 ω 3 3 ω 2 )dx 2 dx 3 + ( 3 ω 1 1 ω 3 )dx 3 dx 1 (6.177) whose comonents are just those of curl(ω). Lastly for a 2-form ω = 1 ωµν dx µ dx ν 2 we have dω = 1 λ ω µν dx λ dx µ dx ν 2 = 1 2 ( 1ω ω ω 12 )dx 1 dx 2 dx 3 (6.178) and these are the comonents of div( ω) where ω = 1 2 ω 23dx ω 31dx ω 12dx 3 (6.179) Next we rove the most imortant roerty of the exterior derivative: Theorem: d 2 = 0 Proof: Let us choose a coordinate system as above so that dω = 1 ν ω µ1,...,µ! dx ν dx µ 1... dx µ (6.180) Then it follows that d 2 ω = 1 λ ν ω µ1,...,µ! dx λ dx ν dx µ 1... dx µ (6.181) 37

38 but this must vanish as λ ν ω µ1,...,µ = ν λ ω µ1,...,µ (6.182) Theorem: If ω Ω (M) and η Ω q (M) then d(ω η) = (dω) η + ( 1) ω dη Problem: Prove this. Let us return to the notion of a ull-back. This can be easily extended to any (0, )-tensor (not necessarily a -form). Consider a C ma F : M N between two manifolds and let ω be a (0, )-tensor on N. Then we define f ω(x 1,..., X ) = ω (f X 1,..., f X 1 ) (6.183) Just as with a co-vector this defines a (0, )-tensor on M. Clearly f ω will be antisymmetric if ω. Thus if ω Ω (N, IR) then f ω Ω (M, IR). We can also define the ull back of a 0-form or function g : N IR by the rule so that f g : M IR. f g = g f (6.184) Theorem: Let f : M N, (y 1,..., y n ) be local coordinates on V N and (x 1,..., x m ) local coordinates on U f 1 (V ) M. If then n ω = ω ν1,...,ν (f())dy ν 1... dy ν f() (6.185) f() ν=1 m n f ω = ω ν1...ν (f()) µ=1 ν=1 x µ (yν 1 f)... 1 x µ (yν f)dx µ 1... dx µ (6.186) Problem: Prove this too. Theorem: The exterior derivative and the ull back commute: d(f ω) = f dω. Proof: In a local coordinate system we saw that that f ω = 1! ων1...ν (f(x)) f ν 1 x µ 1...f ν x µ dxµ... dx µ (6.187) where we have used the short hand f ν = y ν f Therefore we have that df ω = 1! λ (ω ν1...ν (f(x)) f ν 1 x µ 1 ) ν...f dx λ dx µ 1... dx µ x µ 38