c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011Chapter 1 Topology We start by defining a topological space. • A topological space is a set S together with a collection O of subsets called open sets such that the following are true: i) the empty set ∅ and S are open, ∅, S ∈ O ii) the intersection of a finite number of open sets is open; if U1 , U2 ∈ O, then U1 ∩ U2 ∈ O iii) S the union of any number of open sets is open, if Ui ∈ O, then Ui ∈ O irrespective of the range of i. 2 i It is the pair {S, O} which is, precisely speaking, a topological space, or a space with topology. But it is common to refer to S as a topological space which has been given a topology by specifying O. Example: S = R, the real line, with the open sets being open intervals ]a, b[ , i.e. the sets {x ∈ R | a < x < b} and their unions, plus ∅ and R itself. Then (i) above is true by definition. For two such open sets U1 = ]a1 , b1 [ and U2 = ]a2 , b2 [ , we can suppose a1 < a2 . Then if b1 6 a2 , the intersection U1 ∩ U2 = ∅ ∈ O . Otherwise U1 ∩ U2 = ]a2 , b1 [ which is an open interval and thus U1 ∩ U2 ∈ O. So (ii) is true. And (iii) is also true by definition. 2 Similarly Rn can be given a topology via open rectangles, i.e. via the sets {(x1 , · · · , xn ) ∈ Rn | ai < xi < bi }. This is called the standard or usual topology of Rn . • The trivial topology on S consists of O = {∅, S}. 2 1 c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 2 Chapter 1. Topology • The discrete topology on a set S is defined by O = {A | A ⊂ S}, i.e., O consists of all subsets of S. 2 • A set A is closed if its complement in S, also written S\A or as A{ , is open. 2 Closed rectangles in Rn are closed sets as are closed balls and single point sets. A set can be neither open nor closed, or both open and closed. In a discrete topology, every set A ⊂ S is both open and closed, whereas in a trivial topology, any set A 6= ∅ or S is neither open nor closed. The collection C of closed sets in a topological space S satisfy the following: i) the empty set ∅ and S are open, ∅, S ∈ C ii) the union of a finite number of open sets is open; if A1 , A2 ∈ C , then A1 ∪ A2 ∈ C iii) the intersection of any number of open sets is open, if Ai ∈ C , T then Ai ∈ C irrespective of the range of i. i Closed sets can also be used to define a topology. Given a set S with a collection C of subsets satisfying the above three properties of closed sets, we can always define a topology, since the complements of closed sets are open. (Exercise!) • An open neighbourhood of a point P in a topological space S is an open set containing P . A neighbourhood of P is a set containing an open neighbourhood of P . Neighbourhoods can be defined for sets as well in a similar fashion. 2 Examples: For a point x ∈ R, and for any > 0, ]x − , x + [ is an open neighbourhood of x, [x − , x + [ is a neighbourhood of x, {x − 6 y < ∞} is a neighbourhood of x, [x, x + [ is not a neighbourhood of x. 2 • A topological space is Hausdorff if two distinct points have disjoint neighbourhoods. 2 Topology is useful to us in defining continuity of maps. • A map f : S1 → S2 is continuous if given any open set U ⊂ S2 its inverse image (or pre-image, what it is an image of) f −1 (U ) ⊂ S1 is open. 2 m n When this definition is applied to functions from R to R , it is c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 3 the same as the usual − δ definition of continuity, which says that • f : Rm → Rn is continuous at x0 if given > 0, we can always find a δ > 0 such that |f (x) − f (x0 )| < whenever |x − x0 | < δ. 2 For the case of functions from a topological space to Rn , this definition says that • f : S → Rn is continuous at s0 ∈ S if given > 0, we can always find an open neighbourhoood U of s0 such that |f (s)−f (s0 )| < whenever s ∈ U. 2 • If a map f : S1 → S2 is one-to-one and onto, i.e. a bijection, and both f and f −1 are continuous, f is called a homeomorphism and we say that S1 and cs2 are homeomorphic. 2 Proposition: The composition of two continuous maps is a continuous map. Proof: If f : S1 → S2 and g : S2 → S3 are continuous maps, and U is some open set in S3 , then its pre-image g −1 (U ) is open in S2 . So f −1 (g −1 (U )), which is the pre-image of that, is open in S1 . Thus (g ◦ f )−1 (U ) = f −1 (g −1 (U )) is open in S1 . Thus g ◦ f is continuous. 2 c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 Chapter 2 Manifolds Now that we have the notions of open sets and continuity, we are ready to define the fundamental object that will hold our attention during this course. • A manifold is a topological space which is locally like Rn . 2 That is, every point of a manifold has an open neighbourhood with a one-to-one map onto some open set of Rn . • More precisely, a topological space M is a smooth ndimensional manifold if the following are true: i) We can cover the space with open sets Uα , i.e. every point of M lies within some Uα . ii) ∃ a map ϕα : Uα → Rn , where ϕα is one-to-one and onto some open set of Rn . ϕα is continuous, ϕα −1 is continuous, i.e. ϕα → Vα ∈ Rn is a homeomorphism for Vα . (Uα , ϕα ) is called a chart (Uα is called the domain of the chart). The collection of charts is called an atlas. iii) In any intersection Uα ∩ Uβ , the maps ϕα ◦ ϕβ −1 , which are called transition functions and take open sets of Rn to open sets of Rn , i.e. ϕα ◦ ϕβ −1 : ϕβ (Uα ∩ Uβ ) → ϕα (Uα ∩ Uβ ), are smooth maps. 2 • n is called the dimension of M. 2 We have defined smooth manifolds. A more general definition is that of a C k manifold, in which the transition functions are C k , i.e. k times differentiable. Smooth means k is large enough for the purpose at hand. In practice, k is taken to be as large as necessary, up to C ∞ . We get real analytic manifolds when the transition functions 4 c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 5 are real analytic, i.e. have a Taylor expansion at each point, which converges. Smoothness of a manifold is useful because then we can say unambiguously if a function on the manifold is smooth as we will see below. • A complex analytic manifold is defined similarly by replacing Rn with Cn and assuming the transitions functions ϕα ◦ ϕβ −1 to be holomorphic (complex analytic). 2 • Given a chart (Uα , ϕα ) for a neighbourhood of some point P, the image (x1 , · · · , xn ) ∈ Rn of P is called the coordinates of P in the chart (Uα , ϕα ). A chart is also called a local coordinate system.2 In this language, a manifold is a space on which a local coordinate system can be defined, and coordinate transformations between different local coordinate systems are smooth. Often we will suppress U and write only ϕ for a chart around some point in a manifold. We will always mean a smooth manifold when we mention a manifold. Examples: Rn (with the usual topology) is a manifold. 2 The typical example of a manifold is the sphere. Consider the sphere S n as a subset of Rn+1 : (x1 )2 + · · · + (xn+1 )2 = 1 (2.1) It is not possible to cover the sphere by a single chart, but it is possible to do so by two charts.1 For the two charts, we will construct what is called the stereographic projection. It is most convenient to draw this for a circle in the plane, i.e. S 1 in R2 , for which the equatorial ‘plane’ is simply an infinite straight line. Of course the construction works for any S n . consider the ‘equatorial plane’ defined as the x1 = 0, i.e. the set {(0, x2 , · · · , xn+1 )}, which is simply Rn when we ignore the first zero. We will find homeomorphisms from open sets on S n to open sets on this Rn . Let us start with the north pole N , defined as the point (1, 0, · · · , 0). We draw a straight line from N to any point on the sphere. If that point is in the upper hemisphere (x1 > 0) the line is extended till it hits the equatorial plane. The point where it hits the plane is the 1 The reason that it is not possible to cover the sphere with a single chart is that the sphere is a compact space, and the image of a compact space under a continuous map is compact. Since Rn is non-compact, there cannot be a homeomorphism between S n and Rn . c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 6 Chapter 2. Manifolds image of the point on the sphere which the line has passed through. For points on the lower hemisphere, the line first passes through the equatorial plane (image point) before reaching the sphere (source point). Then using similarity of triangles we find (Exercise!) that the coordinates on the equatorial plane Rn of the image of a point on S n \{N } is given by x2 xn+1 1 2 n+1 ϕN : x , x , · · · , x 7→ . (2.2) ,··· , 1 − x1 1 − x1 Similarly, the stereographic projection from the south pole is ϕS : S n \{S} → Rn , 1 2 n+1 x ,x ,··· ,x 7→ x2 xn+1 , · · · , 1 + x1 1 + x1 . (2.3) If we write z= xn+1 x2 , · · · , 1 − x1 1 − x1 , (2.4) we find that 2 n+1 2 x2 x 1 − (x1 )2 1 + x1 2 |z| ≡ + · · · + = = (2.5) 1 − x1 1 − x1 (1 − x1 )2 1 − x1 The overlap between the two sets is the sphere without the poles. Then the transition function between the two projections is ϕS ◦ ϕN : Rn \{0} → Rn \{0}, z 7→ z . |z|2 (2.6) These are differentiable functions of z in Rn \{0}. This shows that the sphere is an n-dimensional differentiable manifold. 2 • A Lie group is a group G which is also a smooth (real analytic for the cases we will consider) manifold such that group composition written as a map (x, y) 7→ xy −1 is smooth. 2 Another way of defining a Lie group is to start with an nparameter continuous group G which is a group that can be parametrized by n (and only n) real continuous variables. n is called the dimension of the group, n = dim G. (This is a different definition of the dimension. The parameters are global, but do not in general form a global coordinate system.) g. the group of rotations plus reflections in three dimensions. the only subsets of a connected manifold which are both open and closed are ∅ and the manifold itself. linear combinations of solutions of Schr¨ odinger equation which vanish outside some region form a manifold. if we are talking about finite dimensional Lie groups. b)) where φ = (φ1 . O(3). The full Lorentz group is a 6-dimensional manifold. 2 The space of functions with some specified properties is often a manifold. the functions φ are smooth (real analytic) functions of a and b. 2 The M¨ obius strip is a 2-dimensional manifold. These definitions of a Lie group are equivalent. 2 L↑+ . because all such groups are also real analytic manifolds. an n-dimensional Lie group is an n-dimensional manifold. · · · . i.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 7 Then any element of the group can be written as g(a) where a = (a1 . is a 6-dimensional connected manifold. For example. Any of these definitions makes a Lie group a smooth manifold. is a 3dimensional connected manifold. Apparently there is another definition of a Lie group as a topological group (like n-parameter continuous group. . · · · . but without an a priori restriction on n. 3N coordinates and 3N momenta. 2 The phase space of N particles is a 6N -dimensional manifold. in which the composition map (x. φn ) are n functions of a and b.e. 2 Infinite dimensional vector spaces with finite norm (e. Further. Then for a Lie group. an ) . 2 • A connected manifold cannot be written as the disjoint union of open sets. Since the composition of two elements of G must be another element of G. Hilbert spaces) are manifolds. y) 7→ xy −1 is continuous) in which it is always possible to find an open neighbourhood of the identity which does not contain a subgroup. define the same objects. 2 SO(3). but is not connected since it can be written as the disjoint union SO(3)∪PSO(3) where P is reflection. it is sufficient to define them as smooth manifolds if we are interested only in finite dimensions. 2 Finite dimensional vector spaces are manifolds. is also a 3-dimensional manifold. the group of rotations in three dimensions. Alternatively. we can write g(a)g(b) = g(φ(a. the group of proper (no space reflection) orthochronous (no time reflection) Lorentz transformations. Reflection is represented by the matrix P = −I. 2 . The space of 3 × 3 real orthogonal matrices is a connected manifold. 2 The space of all n× real non-singular matrices is called GL(n. Manifolds not connected.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 8 Chapter 2. R). This is an n2 -dimensional Lie group and connected manifold. 2 Rotations in three dimensions can be represented by 3 × 3 real orthogonal matrices R satisfying RT R = I. And motion means comparing things at nearby points along the trajectory. ψ) around Q.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 Chapter 3 Tangent vectors Vectors on a manifold are to be thought in terms tangents to the manifold. which is its trajectory. Then ψ ◦ f ◦ ϕ−1 is a map from Rm → Rn and represents f in these local charts. let us first start with the definitions of these things. ϕ) around P and (W. which is a generalization of tangents to curves and surfaces. one-to-one and onto) and f and f are 9 . f is differentiable at P if the coordinates y i = f i (xµ ) of Q are differentiable functions of the coordinates xµ of P .e. the function f ◦ ϕ−1 : Rn → R is differentiable at ϕ(P ). ϕβ ) because f ◦ ϕβ −1 = (f ◦ ϕα −1 ) ◦ (ϕα ◦ ϕβ −1 ) (3. and will be defined shortly. Consider two manifolds M and N of dimension m and n. This should be thought of as a special case of functions from one manifold to another. 2 −1 • If f is a bijection (i. And comparing functions at nearby points leads to differentiation. Consider local charts (U. P 7→ Q. So in order to get to vectors. which of course comes from motion along the curve. 2 −1 This definition does not depend on the chart. In other words. But a tangent to a curve is like the velocity of a particle at that point. ϕα ) at P . • f is differentiable at P if ψ ◦ f ◦ ϕ−1 is differentiable at ϕ(P ).1) and the transition functions (ϕα ◦ ϕβ −1 ) are differentiable. If f ◦ ϕα is differentiable at ϕα (P ) in a chart (Uα . the f ◦ ϕβ −1 is differentiable at ϕβ (P ) for any chart (Uβ . and a mapping f : M → N . • A function f : M → R is differentiable at a point P ∈ M if in a chart ϕ at P. for some γ. Then a curve is a map γ : I → M. Note that the definition of a curve implies that it is parametrized. coordinate transformation. There are different ways of defining tangent vectors. (Uα .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 10 Chapter 3. 2 k In all of these definitions. We are now ready to define tangent vectors and the tangent space to a manifold. So the same collection of points in M can stand for two different curves if they have different parametrizations. i. differentiable can be replaced by C or smooth. Given a curve γ : I → M. iii) Curves approach: A vector tangent to a manifold is tangent to a curve on the manifold. Tangent vectors both differentiable. 1] ⊂ R. This is thinking of a vector as defining a “directional derivative”. a curve γ is called smooth iff its image in a chart is smooth in Rn . ϕα ) → (Uβ . 2 • A curve in a manifold M is a map γ of a closed interval R to M. If γ(0) = P and γ(1) = P 0 . (This definition can be given also when M is a topological space. i. we say that f is a diffeomorphism and that M and N are diffeomorphic. • A manifold M is connected (actually arcwise connected)1 if any two points in it can be joined by a continuous curve in M.) 2 We will take this interval to be I = [0. We will follow the third approach. ϕβ ).. with a well-defined 1 It can be shown that an arcwise connected space is connected. we say that γ joins P and P 0 . • Two Lie groups are isomorphic if there is a diffeomorphism between them which is also a group homomorphism. ii) Derivation approach: A vector is defined as a derivation of functions on the manifold. the map f ◦ γ : I → R is well-defined. or perhaps a mix of the second and the third approaches. iff ϕ ◦ γ : I → Rn is smooth in Rn . Later we will briefly look at the derivation approach more carefully and compare it with the way we have defined tangent vectors. The approaches are equivalent in the sense that they end up defining the same objects and the same space.e. 2 As for any map. . Consider a smooth function f : M → R.e. i) Coordinate approach: Vectors are defined to be objects satisfying certain transformation rules under a change of chart. such that . The rate of change of f along γ is written as . dt Suppose another curve another curve µ(s) meets γ(t) at some point P .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 11 df derivative. where s = s0 and t = t0 . . d d . . (f ◦ γ). = (f ◦ µ). so that the maps are f ◦ ϕ−1 : Rn → R. we can say that two curves γ. In any chart ϕ containing the point P. Let us introduce a convenient notation. ∀f ∈ C ∞ (M) (3. for the curve µ we find (3. n ϕ◦γ : I →R . · · · . let us write ϕ(P ) = (x1 . (3.2) dt ds P P That is. Using the chain rule for differentiation.6) ds ds ∂xi ds Since f is arbitrary.4) The last are the coordinates of the curve in Rn . (3. xn ).3) (3. we find d ∂f dxi (γ(t)) d (f ◦ γ) = f (x(γ(t))) = . µ have the same tangent vector at the point P ∈ M (where t = t0 and s = s0 ) iff .5) d ∂f dxi (µ(s)) d (f ◦ µ) = f (x(µ(s))) = . we are considering a situation where two curves are tangent to each other in geometric and parametric sense. dt dt ∂xi dt Similarly. x 7→ f (x) or f (xi ) i t 7→ {x (γ(t))}. Let us write f ◦ γ = (f ◦ ϕ−1 ) ◦ (ϕ ◦ γ). . dxi (γ(t)) . . dxi (µ(s)) . . 7) . (3. = . . (3. dt ds t=t0 s=s0 We can say that these numbers completely determine the rate of change of any function along the curve γ or µ at P. • The tangent vector to a curve γ at a point P on it is defined as the map d (f ◦ γ)|P . So we can define the tangent to the curve.8) dt As we have already seen. in a chart with coordinates {xi } we can write using chain rule . dxi (γ(t)) ∂f . . 9) dt ∂xi . γ˙ P (f ) = (3. ϕ(P ) γ˙ P : C ∞ (M) → R. f 7→ γ˙ P (f ) ≡ . Tangent vectors .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 12 Chapter 3. dxi (γ(t)) . . The numbers are thus the components of γ˙ P . We will . ∼ is an equivalence relation. It is easy to see.e. if vP = γ˙ P . and symmetric. using Eq. 2 The earlier definition is related to this by saying that if a vector vP is tangent to some curve γ at P . (3. We note here that there is another description of tangent vectors based on curves. i. • A tangent vector at P ∈ M is an equivalence class of curves under the above equivalence relation. In other words. . we can write vP = [γ]. reflexive. that this relation ∼ is transitive. for which the equivalence class [γ] contains all curves tangent to γ (as well as to one another) at P .7) for example. dt ϕ(P ) often write a tangent vector at P as vP without referring to the curve it is tangent to. Let us write γ ∼ µ if γ and µ are tangent to each other at the point P . e. we need a curve λ passing through P such that λ˙ P (f ) = XP (f ) + YP (f )∀f ∈ C ∞ (M). (4. 2 Note: we cannot define λ = γ + µ − P because addition does not make sense on the right hand side. Then take n curves λk such that ϕ ◦ λk (t) = x1 (P ). Define λ : I → Rn in some chart ϕ around P by λ = ϕ ◦ γ + ϕ ◦ µ − ϕ(P ). XP + YP ∈ TP M . i. (Exercise!) To see that TP M has n basis vectors. we consider a chart ϕ with coordinates xi . YP = µ˙ P . The proof of the other part works similarly. (4. µ passing through P such that XP = γ˙ P .1) aXP ∈ TP M . · · · . · · · .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 Chapter 4 Tangent Space • The set of all tangent vectors (to all curves) at some point P ∈ M is the tangent space TP M at P. That is. So λk is like the axis of the k-th coordinate (but only in some open neighbourhood of P ). given curves γ. 13 . YP ∈ TP M. a ∈ R.3) is a curve with the desired property. Proof: We need to show that TP M is a vector space. xk (P ) + t.e. (4. and λ=ϕ◦λ:I →M (4.2) ∀XP . xn (P ) . Then λ is a curve in Rn . 2 Proposition: TP M is a vector space with the same dimensionality n as the manifold M.4) i. only the k-th coordinate varies along t.. Tangent Space ∂ Now denote the tangent vector to λk at P by .e. i.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 14 Chapter 4.. ∂xk P . . . ∂ d . ˙ . 5) . = (f ◦ λ ) f = λ (f ) (4. . k k . dt ∂xk P P P This notation makes sense when we remember Eq.9). Using it we can write . (3. ˙λk (f ). . 6) ∂xk P P ∂ is notation. = ∂f ∀f ∈ C ∞ (M). (4. We should understand this as Note that ∂xk P . . . ∂f . . ∂ ∂ −1 . ≡ f= f ◦ϕ . 7) ∂xk P ∂xk ∂xk . (4. not on Rn . which is the tangent vector to some curve γ at P . Take any vector vP ∈ TP M .ϕ(P ) ϕ(P ) ∂ in a chart around P . The are defined only when this chart ∂xk P is given. but these are vectors on the manifold at P .) Then . Let us now show that the tangent space at P has λ˙ k |P as a basis. (We may sometimes refer to P as γ(0) or as t = 0. . d vP (f ) = (f ◦ γ). . 8) dt t=0 . (4. . . . d . −1 . = ((f ◦ ϕ ). ◦ (ϕ ◦ γ)). (4. .9) . t 7→ (x1 (γ(t)). dt ϕ(P ) t=0 Note that ϕ ◦ γ : I → Rn . xn (γ(t))) are the coordinates of the curve γ. · · · . so we can use the chain rule of differentiation to write . . . . ∂ d i −1 . vP (f ) = (f ◦ ϕ ). (x ◦ γ). . 10) i ∂x t=0 ϕ(P ) dt . (4. . ∂ −1 . (4. vP (xi ) .11) = (f ◦ ϕ ). i ∂x ϕ(P ) The first factor is exactly as shown in Eq.7). (4.12) ∂xk P P . so we can write ∂ vP (f ) = f v (xi ) ∀f ∈ C ∞ (M) (4. e. so ∂x∂ k P form a basis of TP M and vPi are the components of vP in that basis. Thus the vectors ∂x∂ k P span TP M . g ∈ C (M) and α ∈ R (4.e.14) (4. ∂ o∂xk P are called coordinate basis vectors and the set n The ∂ is called the coordinate basis.15) (4.. i. we can write i vP = vP ∂ ∂xk ∀vP ∈ TP M (4. vP (f + αg) = vP (f ) + αvP (g) vP (f g) = vP (f )g(P ) + f (P )vP (g) 1 ∀f. satisfies linearity and Leibniz rule.. ∂xk P It can be shown quite easily that for any smooth (actually C 1 ) function f a vector vP defines a derivation f 7→ vP (f ) .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 15 i. These can be shown to be linearly independent as well. These are to be thought of as tangents to the coordinate curves inϕ.16) .13) P where vPi = vP (xi ). 1) ∀uP . (5. 2 A dual space can be defined for any vector space V as the space of linear mappings V → R (or V → C if V is a complex vector space). 2 We will write the action of ω on vP ∈ TP M as ω(vP ) or sometimes as hω | vP i . cotangent vectors etc. The dual space is a vector space under the operations of vector addition and scalar multiplication defined by a1 ω1 + a2 ω2 : vP 7→ a1 ω1 (vP ) + a2 ω2 (vP ) . etc. every vector at P produces a number. Example: • Vector column vectors kets |ψi functions Dual vector row vector bras hφ| linear functionals.2) The elements of TP∗ M are called dual vectors. Linearity of the mapping ω means ω(uP + avP ) = ω(uP ) + aω(vP ) . covectors. Thus f defines a 16 . vP ∈ TP M and a ∈ R . (5.2 • Given a function on a manifold f : M → R .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 Chapter 5 Dual space • The dual space TP∗ M of TP M is the space of linear mappings ω : TP M → R. vP (f ) ∈ R ∀vP ∈ TP M . c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 17 covector df , given by df (vP ) = vP (f ) called the differential or 2 Since vP is linear, so is df , gradient of f . df (vP + awP ) = (vP + awP )(f ) = vP (f ) + awP (f ) (5.3) ∀vP , wP ∈ TP M, a ∈ R . Thus df ∈ TP∗ M . Proposition: TP∗ M is also n-dimensional. Proof: Consider a chart ϕ with coordinate functions xi . Each xi is a smooth function xi : M → R . then the differentials dxi satisfy ∂ ∂ ∂ i i −1 i = δji .4) = (x ) = x ◦ϕ . (5. e. since the form a basis. Thus ∂xj P the dxi span TP∗ M . (5. it must vanish on every one of the basis vectors. Next consider a linear combination of these covectors. Then letting this act on a coordinate basis vector.6) j ∂x P ∂xi P j ∂ So λ vanishes on all vectors. consider the covector λ = ω − ∂ i ω ∂xi P dx . we get ∂ λ ∂xj P ∂ ∂ ∂ i =ω −ω dx ∂xj P ∂xi P ∂xj P ∂ ∂ =ω −ω δ i = 0∀j (5. given any covector ω . ω = ωi dxi . dx ∂xj ∂xj ∂xj P P ϕ(P ) The differentials dxi are covectors. ∂ ω=0 ⇒ ω =0 ∂xj P ∂ i ⇒ ωi dx =0 ∂xj P ⇒ ωi δji = 0 i. . as we already know. If this vanishes. so TP∗ M is n-dimensional. Finally. In other words. ωj = 0 . So we have constructed n covectors in TP∗ M .5) So the dxi are linearly independent. Dual space Also. Given a vector v.8) This justifies the name gradient. i. from which it follows that λ0a = (A−1 )ba λb . (5.e.12) . It is straightforward to calculate the effect of switching to another overlapping chart. (5.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 18 Chapter 5.10) ∂y i P ∂y i P ∂xj P These formulae can be generalized to arbitrary bases. but given a basis {ea }.11) with A a non-singular matrix. (5. we can define its dual basis {ω a } by ω a (eb ) = δba . so that ω 0a (e0b ) = δba . (5. ω a 7→ ω 0a = Aab ω b . ea 7→ e0a = (A−1 )ba eb . any covector ω ∈ TP∗ M can be written as ω = ωi dx i where ωi = ω ∂ ∂xi . as we have just seen. ∂ Since is the dual basis in TP M to {dxi }.7) P so in particular for ω = df .8) to write the gradient of y i as i ∂y i dy = dxj (5. λ = λa ω a = λ0a ω 0a = λ0a Aab ω a . In a new chart ϕ0 where the coordinates are y i (and the transition functions are thus y i (x)) we can use Eq. we get ∂ ∂f ωi ≡ (df )i = df = ∂xi P ∂xi ϕ(P ) (5. in order i P ∂x ∂ to be the dual basis to {dy i } we must have for ∂y i P j ∂ ∂x ∂ = (5. We can make a change of bases by a linear transformation. a coordinate transformation.9) ∂xj P This is the result of coordinate transformations on a basis of covectors. Given a 1-form λ we can write it in both bases. it is not meaningful to talk about its dual. we can write v = v a ea = v 0a e0a = v 0a (A−1 )ba eb . • Quantities which transform like λa are called covariant. if v is a vector.13) and it follows that v a = Aab v b . while those transforming like v a are called contravariant.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 19 Similarly. 2 . (5. c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 Chapter 6 Vector fields • Consider the (disjoint) union of tangent spaces at all points. the theory of differential 20 .2) ∂xi P • The vector field v is smooth if the functions v i = v(xi ) are smooth for any chart (and thus for all charts). v has components v i in the chart. If v i are smooth. we can write γ(t) ˙ =v⇒ d i x (γ(t)) = v i (x(t)) . (6. ∂ i vP = v . (6. (One curve need not pass through all P ∈ M. 2 • Given a smooth vector field v (actually C 1 is sufficient) we can define an integral curve of v. Then in a chart containing P . This is a set of ordinary first order differential equations. v : P 7→ v(P ) ≡ vP ∈ TP M. 2 • A vector field v chooses an element of TP M for every P .3) with initial condition xi (0) = xi |P . 2 We will often write v(f )|P = vP (f ). [ TM = TP M .1) P ∈M This is called the tangent bundle of M. with γ(0) = P. (v i )P = vP (xi ) . which is a curve γ in M such that γ(t)| ˙ P = vP at every P ∈ γ.e. dt (6. i. 2 ∗ • A rule that selects a covector from TP M for each P is called a one−form (often written as a 1−form). Given a chart.) 2 Suppose γ is an integral curve of a given vector field v. Given a smooth vector field v such that v|P 6= 0. the pre-image π −1 (P ) is TP M.5) We could also choose γ(t0 ) = P. locally). d i x (γ(t)) = v i (γ(t)) ≡ v(xi (γ(t))) . 2 Given a smooth vector field v. the integral curve through Q. φt takes a point by a parameter distance t along the curve γQ (t). One use of integral curves is that they can be thought of as coordinate lines. for given t. This φt is called the local flow of v. the existence of exactly one solution. Q) = γQ (t) where γQ (t) satisfies d i x (γQ (t)) = v(xi γQ (t)) .. v) where P ∈ M and v ∈ TP M. it is possible to define a coordinate system {xi } in a neighbourhood ∂ around P such that v = . Then a vector field can be thought of as a section of the tangent bundle. It is called the fiber over P . i. The uniqueness of these solutions implies that the integral curves of a vector field do not cross. dt γ(0) = P . Then in any neighbourhood U of P we also have γQ . • The map π : T M → M. Q) = γQ (t) . 2 The local flow has the following properties: i) φ0 is the identity map of U .e. v) 7→ P (where v ∈ TP M) is called the canonical projection (or simply projection). we can define an integral curve γ through any point P by γ(t) ˙ = v . (P.7) • This φ defines a map φt : U → M at each t by φt (Q) = φ(t.6) (6.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 21 equations guarantees. ∂xi • A vector field v is said to be complete if at every point P ∈ M the integral curve the integral curve γ(t) of v passing through P can be extended to all t ∈ R .e.e. So we can define a map φ : I × U → M given by φ(t. i.4) (6. The topological structure and differential structure are given appropriately. 2 The tangent bundle T M is a product manifold. (6. 2 • For each P ∈ M. (6. at least for small t (i. a point in T M is an ordered pair (P. dt γQ (0) = Q . i. .e.. so that moving a parameter distance t from γQ (s) finds the same point on M as by moving a parameter distance s + t from γQ (0) ≡ Q . v are vector fields on M and α is now a smooth function on M. while the second property follows from the uniqueness of integral curves. • A ring R is a set or space with addition and multiplication defined on it. (6. 1x = x1 = x . iii) each flow is a diffeomorphism with φt −1 = φ−t . . x(y + z) = xy + xz . and two special elements 0 and 1. If u. The map v : f 7→ v(f ) has the following properties: v(f + αg) = v(f ) + αv(g) (6. s + t ∈ U . The first property is obvious. with scalar multiplication by elements of a ring defined on it. satisfying (xy)z = x(yz) . A vector field can also be thought of as a map from the space of differentiable functions to itself v : C ∞ (M) → C ∞ (M). v ∈ V (M). (6. Often v(f ) is called the Lie derivative of f along v and denoted £v f . 0 + x = x + 0 = x . g ∈ C (M). A module X is an Abelian group under addition. define u + αv by (u + αv)P (f ) = uP (f ) + α(P )vP (f ) ∀f ∈ C ∞ (M). i. u. α ∈ R.9) ∞ ∀f.10) It is possible to replace α by some function on C ∞ (M). P 7→ vP (f ) . with v(f ) : M → R. P ∈ M. Then the integral curve passing through the point γQ (s) is the same as the integral curve passing through Q. of solutions to first order differential equations. f 7→ v(f ) . (x + y)z = xz + yz .e.8) v(f g) = f v(g) + v(f )g (6. the additive and multiplicative identity elements. t. Vector fields ii) φs ◦ φt = φs+t for all s.11) This looks like a vector space but actually it is what is called a module.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 22 Chapter 6. α∈R The set of all (real) vector fields V (M) on a manifold M has the structure of a (real) vector space under vector addition defined by (u + αv)(f ) = u(f ) + αv(f ). and an inverse exists for every element except 0. Given a vector field v. in an open neighbourhood of some P ∈ M and in a chart. we have . and for any f ∈ C ∞ (M) . in general α−1 ∈ / C ∞ (M).e. i. when the ring multiplication is commutative. so the space of vector fields on M is in general a module. xy = yx.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 23 • A module becomes a vector space when this ring is a commu tative division ring. not a vector space. Given a smooth function α. ∂f . i v(f ). 13) as an obvious generalization of vector space expansion to the module V (M). it may not be possible in general to define the coordinate vector fields globally. Note that this is correct only in some open neighbourhood on which a chart can be defined.e. i. ∂ The v i are now the components of the vector field v. which we will call the coordinate vector fields. and ∂x i are now vector fields. and thus the components v i may not be defined globally either. (6. (6. . = vP (f ) = vP . In particular. where vPi = vP (xi ) .12) i ∂x P P Thus we can write v = vi ∂ ∂xi with v i = v(xi ) . everywhere on M. f ∗g = g ◦ f .3) 2 (7. Thus we can write f∗ v(g) = v(f ∗ g) . • Given two manifolds M1 and M2 with a smooth map f : M1 → M2 . M2 .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 Chapter 7 Pull back and push forward Two important concepts are those of pull back (or pull-back or pullback) and push forward (or push-forward or pushforward) of maps between manifolds.1) 2 So in particular. if M1 and M2 are two manifolds with a map f : M1 → M2 and g : M2 → R is a function on M2 . 24 (7. (7. M3 and maps f : M1 → M2 .4) . (7.2) While this looks utterly trivial at this point. this concept will become increasingly useful later on. the pullback of g under f is a function on M1 . the pullback of g under f is the map f ∗ g : M1 → M3 defined by f ∗g = g ◦ f . g : M2 → M3 . • Given manifolds M1 . P 7→ Q the pushforward of a vector v ∈ TP M1 is a vector f∗ v ∈ TQ M2 defined by f∗ v(g) = v(g ◦ f ) for all smooth functions g : M2 → R . xm ) .8) Note that for this pushforward to be defined. In particular. M3 are manifolds with maps f : M1 → M2 . ∀v ∈ TP M1 .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 25 The pushforward is linear. Let us find the components of the pushforward f∗ v in terms of the components of v for any vector v. M2 . 4. so we can expand it in the basis f∗ ∂ ∂xi P µ ∂ ∂ = f∗ ∂xi P ∂y µ Q ∂ ∂y i Q . the two manifolds may have different dimensions. which is a vector in TQ M2 . The pushforward of an equivalence class of curves is f∗ v = f∗ [γ] = [f ◦ γ] (7. given charts ϕ : P 7→ (x1 .9) In any coordinate basis. · · · .5) f∗ (λv) = λf∗ v . g : M2 → M3 . ∂ ∂ For the basis vector ∂x i P .10) (7. Let us in fact consider.7) Remember that we can think of a vector v as an equivalence class of curves [γ].6) And if M1 . · · · . P 7→ Q . (7. ψ : Q 7→ (y 1 .11) . (7. (g ◦ f )∗ v = g∗ f∗ v i. the components of a vector are given by the action of the vector on the coordinates as in Chap. Suppose M1 and M2 are two manifolds with dimension m and n respectively. So for a map f : M1 → M2 . (7. f∗ (v1 + v2 ) = f∗ v1 + f∗ v2 (7. vPµ = vP (y µ ) Thus we can write µ ∂ ∂ f∗ = f∗ (y µ ) ∂xi P ∂xi P (7. we do not need the original maps to be 1-1 or onto. it follows that (g ◦ f )∗ = g∗ f∗ .e. the pushforward f∗ will not have an inverse if m 6= n . we want the pushforward f∗ ∂xi P . y n ) the pushforward of the basis vectors. So in the respective tangent spaces TP M1 and TQ M2 are also of dimension m and n respectively. 12) so f∗ ∂ ∂xi µ (y ) = P ∂ ∂xi (y µ ◦ f ) .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 26 Chapter 7. i. Pull back and push forward But f∗ v(g) = v(g ◦ f ) . coordinates around the point f (P ) = Q . So we can write y µ ◦ f as y µ (x) .e..13) P But y µ ◦f are the coordinate functions of the map f . which is what we understand by this. (7. Thus . (7. µ ∂ ∂y µ (x) . . (7.14) f∗ ∂xi P ∂xi P ∂xi . ∂ µ = (y ◦ f ) = . so the chart maps are hidden in this equation. these are actually done in charts around P and Q = f (P ) . this matrix need not be invertible and a determinant may not be defined for it. 2 For the basis vectors. we can then write . Note that since m and n may be unequal. • The right hand side is called the Jacobian matrix (of y µ (x) = y µ ◦ f with respect to xi ).P Because we are talking about derivatives of coordinates. ∂ ∂y µ (x) . . 15) ∂xi P ∂xi . ∂ f∗ = (7. P ∂y µ f (P ) Since f∗ is linear. we can use this to find the components of (f∗ v)Q for any vector vP . ∂ i f∗ vP = f∗ vP ∂xi P ∂ i = vP f∗ ∂xi P . µ . ∂ i ∂y (x) . 16) ∂xi . = vP (7. P ∂y µ f (P ) . µ . µ i ∂y (x) . (7.17) i ∂x . ⇒ (f∗ vP ) = vP . so we can . we know that the components of f∗ v should be linear combinations of the components of v .P Note that since f∗ is linear. of derivatives on I . different points P and P 0 may have the same image. 2 Even though in order to define the pushforward of a vector v under a map f . Remember that tangent vectors are derivatives along curves. Another example of the pushforward map is the following. The matrix is made of first derivatives because vectors are first derivatives.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 27 already guess that (f∗ vP )µ = Aµi vPi for some matrix Aµi . Suppose we have a vector field v on M . Suppose vP ∈ TP M is the derivative along γ . we do not need f to be invertible. Thus for γ : I → M . so f∗ v will not be a vector field on N . the pushforward of a vector field can be defined only if f is both one-to-one and onto.18) so γ∗ d dt = vP (7.21) which may not be true. t 7→ γ(t) = P . If f is one-to-one and onto. (7. (7. and for some g : M → R . we can consider pushforwards under γ . f (P ) = Q = f (P 0 ) . And if f : M → N is not onto. d d γ∗ g = (g ◦ γ)|t=0 dt t=0 dt = γ˙ P (g)|t=0 = vP (g) .20) dt t for all t in some interval containing P . Then for the same vector field v we must have f∗ v|Q = f∗ (vP ) = f∗ (vP 0 ) . in which case vector fields can be pushed forward. Since γ : I → M is a map. (7. by the rule (f∗ v)f (P ) = f∗ (vP ) . Then the integral curve of v passing through P ∈ M is a curve γ : t 7→ γ(t) such that γ(0) = P and d γ∗ = v|γ(t) (7.19) t=0 • We can use this to give another definition of integral curves. f∗ v will be meaningless outside some region f (M) . it is a diffeomorphism.22) . If f is not one-to-one. (8.3) Thus (uv − vu)(f g) = f (uv − vu)(g) + (uv − vu)(f )g . Then u(v(f )) is also a smooth function. Suppose we consider two vector fields u . given a smooth function f . (8. linear in f . But is uv ≡ u ◦ v a vector field? To find out. v . 28 . (8. In other words.1) We reorder the terms to write this as uv(f g) = f uv(g) + uv(f )g + u(f )v(g) + v(f )u(g) . But if we also consider the combination vu . (8. v : f 7→ v(f ) .5) is a vector field on M . we get vu(f g) = f (vu(g) + vu(f )g + v(f )u(g) + u(f )v(g) . we consider u(v(f g)) = u(f v(g) + v(f )g) = u(f )v(g) + f u(v(g)) + u(v(f ))g + v(f )u(g) .4) which means that the combination [u . with the product uv signifying successive operation on any smooth function on M .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 Chapter 8 Lie brackets A vector field v is a linear map C ∞ (M) → C ∞ (M) since it is basically a derivation at each point.2) so Leibniz rule is not satisfied by uv . v(f ) is a smooth function on M . v] := uv − vu (8. [u . u] .. w] . ∂x ∂x ∂x x j (8.9) • The commutator is antisymmetric. ∂xi (8.e. Then the coordinate lines are the integral . Any set of n linearly independent vector fields may be chosen as a basis. (8. v] . ∂ ∂ . In a coordinate system. v] = 0 . but they need not form a coordinate system. w] + [[v .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 29 • This combination is called the commutator or Lie bracket of the vector fields u and v .10) The commutator is useful for the following reason: Once we have a ∂ chart. we can write a vector field in local coordinates v(f ) = v i ∂f .8) from which we can read off the components of the commutator. So n vector fields will form a coordinate system only if they commute.7) Subtracting. u] . 2 In any chart around the point P ∈ M . u] + [[w . ∂x ∂x ∂x x i ∂u ∂f ∂2f v(u(f )) = v j j i + uj v i j i . = 0.6) so that ∂ i ∂f u(v(f )) = u v ∂xj ∂xi ∂v i ∂f ∂2f = uj j i + uj v i j i . and satisfies the Jacobi identity [[u . we get u(v(f )) − v(u(f )) = uj i ∂v i ∂f j ∂u ∂f − v . i.11) ∂xi ∂xj because partial derivatives commute. we can use as a basis for vector fields in a neighbour∂xi hood. ∂xj ∂xi ∂xj ∂xi (8. have vanishing commutators with one another. [u . v]i = uj i ∂v i j ∂u − v ∂xj ∂xj (8. v] = −[v . (8. r = x2 + y 2 (8. For analytic manifolds. A simple example is the polar coordinate system in R2 .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 30 Chapter 8. so {er . . eθ } do not form a coordinate basis.12) ∂ ∂ and ey = being the Cartesian coordinate basis ∂x ∂y vectors. (8. Lie brackets curves of the vector fields. it is easy to show that [er . sin θ = . The unit vectors are er = ex cos θ + ey sin θ eθ = −ex sin θ + ey cos θ . this condition is sufficient as well. and p x y cos θ = .13) r r with ex = Using these expressions. eθ ] 6= 0 . 1) 2 The Jacobi identity is not really an identity — it does not hold for an arbitrary algebra — but it must be satisfied by an algebra for it to be called a Lie algebra. ii) The same space Mn of all n × n matrices as above. If λ. z ∈ A . µ ∈ R . µ are complex numbers and A is a complex vector space. but now 31 . y. we get a complex algebra. (AB)C = A(BC) . i. and ii) satisfies the Jacobi identity .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 Chapter 9 Lie algebra • An (real) algebra is a (real) vector space equipped with a bilinear operation (product) under which the algebra is closed. x • y = −y • x .e. for an algebra A i) x • y ∈ A ∀x. 2 • A Lie algebra is an algebra in which the operation is i) antisymmetric. y ∈ A ii) (λx + µy) • z = λx • z + µy • z x • (λy + µz) = λx • y + µx • z ∀x. A • B = AB . λ. (9.. (x • y) • z + (y • z) • x + (z • x) • y = 0 . This is an associative algebra since matrix multiplication is associative. Example: i) The space Mn = {all n × n matrices} under matrix multiplication. If Li are the angular momentum operators with [Li .3) L= a= ζi Li |ζi ∈ C i If a = P P ai Li and b = bi Li .2) This product is antisymmetric and satisfies Jacobi identity. This is a Lie algebra. their product is X X a • b ≡ [a .5) As a vector space it is infinite- v) Vector fields on a manifold form a real Lie algebra under the commutator bracket.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 32 Chapter 9. b] = ai bj [Li . f • g = [f . since the Jacobi identity is a genuine identity. Lie algebra with matrix commutator as the product. so Mn with this product is a Lie algebra. . iv) The Poisson bracket algebra of a classical dynamical system consists of functions on the phase space.B. a] = 0 and the Jacobi identity is satisfied. i. Dif f (M)) . dimensional. as we have seen in the previous chapter. (9. with the product defined by the Poisson bracket. Lj ] = i ijk ai bj Lk . (It can be thought of as the Lie algebra of the group of diffeomorphisms. g]P.e. B] = AB − BA .4) This is a Lie algebra because it [a . (9. . (9. iii) The angular momentum algebra in quantum mechanics. This algebra is infinite-dimensional. A • B = [A . Lj ] = iijk Lk . automatically satisfied. we can write the elements of this algebra as ( ) X (9. Since φt : U → M. diffeomorphism for sufficiently small values of t . Q 7→ γQ (t) is local diffeomorphism . • The collection φt for t < (for some > 0.1) and ϕ∗ f ∈ C ∞ (M1 ) if ϕ is C ∞ . 33 (10. ϕ∗ v(f )|Q = v (f ◦ ϕ)|ϕ−1 Q = v (ϕ∗ f )|ϕ−1 Q . (10. We push forward a vector field at t = to t = 0 and compare with the vector field at t = 0 .4) .e. i. 2 Consider the vector field in a neighbourhood U of a point Q ∈ M .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 Chapter 10 Local flows We met local flows and integral curves in Chapter 6. we can use φt to push forward vector fields. The pushforward of a vector vP is defined by ϕ∗ vP (f ) = vP (f ◦ ϕ) = vP (ϕ∗ f ) vP ∈ TP M1 . At some point P we have the curve φt (P ) . We recall that for a map ϕ : M1 → M2 the pullback of a function f ∈ C ∞ (M2 ) is defined as ϕ∗ f = f ◦ ϕ : M1 → R . or alternatively for t < 1) is a one−parameter group of local diffeomorphisms .e. (10. Given a vector field v .2) (10. ϕ∗ vP ∈ Tϕ(P ) M2 . write its local flow as φt . we can define the pushforward of a vector field v by ϕ∗ v(f )|ϕ(P ) = v (f ◦ ϕ)|P i.3) If ϕ is a diffeomorphism. ϕ∗ v is not a vector field on M2 . But there are some ϕ and some v such that ϕ∗ v is a differentiable vector field.9) So if v is an invariant vector field. (10. ϕ∗ v] . if ϕ∗ (vP ) = vϕ(P ) for all P ∈ M . Then v and ϕ∗ v are said to be ϕ−related. 2 We can write for any f ∈ C ∞ (M) (ϕ∗ v) (f ) = ϕ−1 ⇒ ⇒ ∗ (v (ϕ∗ f )) ϕ∗ ((ϕ∗ v) (f )) = v (ϕ∗ f ) .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 34 Chapter 10. i. even if ϕ is not invertible or ϕ−1 is not differentiable. ϕ∗ v is not differentiable.6) Proof: ϕ∗ [u . Local flows We can rewrite this definition in several different ways. ϕ∗ [u . we can write ϕ∗ ◦ v = v ◦ ϕ∗ .10) .e. ϕ∗ ◦ ϕ∗ v = v ◦ ϕ∗ . If ϕ−1 exists but is not differentiable. (10.5) • If ϕ : M1 → M2 is not invertible. v] = [ϕ∗ u . (10. ϕ∗ v](f ) = ϕ∗ u (ϕ∗ v (f )) − u ↔ v while = u (ϕ∗ v (f ) ◦ ϕ) ◦ ϕ−1 − u ↔ v = u v (f ◦ ϕ) ◦ ϕ−1 ◦ ϕ ◦ ϕ−1 − u ↔ v = u (v (f ◦ ϕ)) ◦ ϕ−1 − u ↔ v .8) 2 • A vector field v is said to be invariant under a diffeomorphism ϕ : M → M if ϕ∗ v = v .e.7) [ϕ∗ u . (10. (10. (ϕ∗ v)(f ) = v (f ◦ ϕ) ◦ ϕ−1 = (ϕ−1 )∗ (v (f ◦ ϕ)) = (ϕ−1 )∗ (v (ϕ∗ f )) . v] (f ◦ ϕ) ◦ ϕ−1 = u (v (f ◦ ϕ)) ◦ ϕ−1 − u ↔ v . (10. 2 Proposition: Given a diffeomorphism ϕ : M1 → M2 (say both C ∞ manifolds) the pushforward ϕ∗ is an isomorphism on the Lie algebra of vector fields. i. v](f ) = [u . γ˙ Q (t) = u(γQ (t)) . φt (Q) = γQ (t) . and is satisfied by all differential operators invariant under ϕ .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 35 This expresses invariance under ϕ . d f ◦ γQ (t) dt d (f ◦ φt (Q)) = dt .11) But for any f ∈ C ∞ (M) . (10. and the local flow (or one-parameter diffeomorphism group) φt corresponding to u . Consider a vector field u . d ∗ . (φt (f )) = uγQ (t) (f ) ≡ u(f ). = dt γQ (t) γ˙ Q (f ) = (10.12) At t = 0 we get the equation . . d ∗ . . (φt (f )) . = u(f ). (10. dt t=0 Q (10. t) = (10. We can treat v i (x) as components of a vector field v . Then a solution to this equation is f (x. Proof: ∂ d ∗ ∂f f (x. This is an equation on Rn+1 .15) v i (x) i f (x.17) . t) ∂t ∂x i=1 with initial condition f (x.13) We can also write d ∗ (φ f ) (Q) = u(f ) (φt (Q)) = φ∗t u(f )(Q) . so it can be on a chart for a manifold as well. t) = φ∗t g(x) ≡ g (φt (x)) ≡ g ◦ φt (x) . dt t (10. t) = (φt g) = v(f ) ≡ v i i . 0) = g(x) and everything smooth. ∂t dt ∂x (10.16) where φt is the flow of v .14) This formula can be used to solve linear partial differential equations of the form n X ∂ ∂ f (x. (10. 0) = x2 + y 2 .18) f (x.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 36 Chapter 10. y) in our example is given by γ(t) = ((x − y)t + x. t) = (x − y) − ∂t ∂x ∂y with initial condition f (x. y) . y) = [(x − y)t + x]2 + [(−x + y)t + y]2 = (x − y)2 t2 + x2 + 2(x − y)xt + (x − y)2 t2 + y 2 − 2(x − y)yt = 2(x − y)2 t2 + (x2 + y 2 )(1 + 2t) − 4xyt . Example: Consider the equation in 2+1 dimensions ∂ ∂f ∂f (10. −x + y) . y0 ) is given by the coordinates γ(t) = (vx (P )t + x0 . 0) = f (x. The integral curve passing through the point P = (x0 . Local flows using Eq. 0) ◦ Φt (x. (−x + y)t + y) (10.19) so the integral curve passing through (x. t) = Φ∗t f (x. (10. The corresponding vector field is v(x) = (x − y.13) . It can be shown. the flow of v. that this solution is also unique. using well-known theorems about the uniqueness of solutions to first order partial differential equations. 2 Thus the partial differential equation can be solved by finding the integral curves of v (the flow of v) and then by pushing (also called dragging) g along those curves. vy (P )t + y0 ) . (10. So the solution is f (x.20) = Φt (x.21) . defined by . ∗ (ϕ∗ v)f = ϕ−1 v (ϕ∗ (f )) .1) We will apply this to the flow φt of a vector field u .5) for pushforwards and pullbacks. we have Eq. (11. (10.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 Chapter 11 Lie derivative Given some diffeomorphism ϕ. d ∗ . . . (φt f ) . = u(f ). we get ∗ φ−t∗ v(f ) = φ−1 v φ∗−t (f ) −t = φ∗t v φ∗−t (f ) . (11.2) dt t=0 Q Applying this at −t . (11. .3) where we have used the relation φ−1 t = φ−t . . Let us differentiate this equation with t . . . d d . (φ−t∗ v)(f ). = φ∗t v φ∗−t (f ) . Then the right hand side is of the form . (11. so we can imagine At = φ∗t v as a kind of linear operator acting on the function ft = φ∗−t f .4) dt dt t=0 t=0 On the right hand side. φ∗t acts linearly on vectors and v acts linearly on functions. . . . . . d d d At ft . . = At ft . . + At ft . . dt dt dt t=0 t=0 . t=0 . . . d ∗ d ∗ . + At = φt v ft . φ−t (f ) . . dt dt t=0 t=0 . . . . = u (v(f )) . − v (u(f )) . t=0 . t=0 . = [u . v](f ). 5) t=0 37 . (11. . Lie derivative The things in the numerator are numbers. appropriately defined. £u f = 0 means φ∗t f = f . Also. w] .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 38 Chapter 11. and it can be shown to have the properties of a derivation on the module of vector fields.7) The derivation on functions by a vector field u : C ∞ (M) → C ∞ (M) .10) . So the Lie bracket is also called the Lie derivative. (11. We can also write this as lim t→0 φt∗ vφt (P ) − vP t = [u . For functions. and written as £u f .9) So £u is a derivation on the space C ∞ (M) . we see that £u (v • w) = (£u v) • w + v • (£u w) . (11. so £u is a derivation on the Lie algebra of vector fields. (11. So £u is a derivation on the module of vector fields. f 7→ u(f ) . v] . or leaves f invariant.8) • So this can also be called the Lie derivative of f with respect to u . Lie derivatives are useful in physics because they describe invariances. unlike vectors which may be compared only on the same space. (11. So the flow of u preserves f .11) where v • w = [v .6) • This has the look of a derivative. and £u (f v) = (£u f ) v + f £u v ∀f ∈ C ∞ (M) (11. so the function does not change along the flow of u . can be defined similarly as φ∗t f − f . . t→0 t u(f ) = lim (11. and written as £u v = [u . using Jacobi identity. so they can be compared at different points. v] . and £u (f + ag) = £u f + a£u g . Also. 2 Then it is easy to see that £u (f g) = (£u f ) g + f (£u g) . £u (v + aw) = £u v + a£u w . (11. (11. a vector field is invariant under a diffeomorphism ϕ if ϕ∗ v = v .v] w .14) Thus again the vector fields leaving w invariant form a Lie algebra.15) P Alternatively. a 1−form is a section of the cotangent bundle T ∗M = [ TP∗ M .8). as mentioned earlier.12) So the vector fields which preserve f form a Lie algebra. for convenience). (11. df (u + av) = u(f ) + av(f ) .v] f = 0 . As we mentioned in Chap. (11. But we know from the Eq. ∀a ∈ R (11. (11. we find that a vector field v is invariant under the flow of u if φ−t v = v ⇒ £u v = v . £v ]f = £[u . the pullback ϕ∗ ω is defined by (ϕ∗ ω) (v) = ω (ϕ∗ ω) . ω : v 7→ ω(v) ∈ C ∞ (M). we find that 0 = £u £v w − £v £u w = £[u . Using the flow of u .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 39 If there are two vector fields u and v which leave f invariant. ω(u + av) = ω(U ) + aω(v) . 2 • Given a smooth map ϕM1 → M2 (say a diffeomorphism. (11. 6.16) A 1-form is a rule that (smoothly) selects a cotangent vector at each point.17) • We have already seen the gradient 1-form for a function f : M → R . i.e. if £u w = 0 = £v w . Similarly. which is a linear map from the space of vector fields to functions.18) . which defines the Lie derivative of a function that £u+av f = £u f + a£v = 0 and [£u . a 1-form is a smooth linear map from the space of vector fields on M to the space of smooth functions on M . • Let us also define the corresponding operations for 1-forms. (11. £u f = 0 = £v f .13) So if a vector field w is invariant under the flows of u and v . we can multiply both sides of the last equation by A−1 and write j (11. it follows that 0 ωi0 Aij v j = ωj v j ⇒ 0 ωi0 Aij = ωj (11.21) Note that the notation is somewhat ambiguous here – i0 also runs from 1 to n .25) j ∂x ∂x0i0 We can define the Lie derivative of a 1-form very conveniently by going to a chart. 0 For coordinate transformations from a chart {xi } to a chart {x0i } . 2 For an arbitrary 1-form ω . so is ωI v i .20) All the components ωi .22) Since coordinate transformations are invertible. we can find the components ωi0 of ω in a new chart by noting that 0 ω(v) = ωi v i = ω 0i v i . v = vi ∂ . Since the function ω(v) is chartindependent. and treating the components of 1-forms and vector fields as functions. ω = ωi dxi . rather than the index i . Lie derivative and which can be written as df = ∂f i dx ∂xi (11. If the components of v in the new chart are related to those in 0 0 the old one by v i = Aij v j . we can write in a chart and for any vector field v . ∂xi ω(v) = ωi v i . ωi0 = ωj . 0 Aij = ∂ £u ω(v) = £u ωi v i = uj j ωi v i ∂x ∂ωi ∂v i = uj j v i + uj ωi j .23) ωi0 = A−1 i0 ωj . 0 j ∂xj ∂x0i .24) j ∂x ∂x0i0 0 ∂x0i j ∂xj 0 so vi = v . A−1 i0 = (11. ∂x ∂x (11. and the prime actually distinguished the chart.26) .19) in some chart. v i are smooth functions. or the coordinate system. (11. (11. The space of 1-forms is a module. (11.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 40 Chapter 11. For the sake of convenience.31) Similarly.33) .26). (11. let us write down the Lie derivatives of the coordinate basis vector fields and basis 1-forms. and the right hand side can be calculated in a chart as (£u ω) (v) + ω (£u v) = (£u ω)i v i + ωi (£u v)i = (£u ω)i v i + ωi [u .29) These are the components of £u ω in a given chart {xi } . ∂xj ∂xj (11. we get v= £u ∂ ∂ = [u . The coordinate basis vector corresponding to the i-th coordinate is ∂ ⇒ v j = δij . i ∂x ∂xj (11.27) We already know the left hand side of this equation from Eq. (11. (11. (11. ∂xi Putting this into the formula for Lie derivatives.29) we get £u dxi = δki ∂ui j ∂uk j dx = dx . we can write (£u ω)i = uj ∂uj ∂ωi + ω .26).32) Using this in the formula Eq.e. dxi j = δji . j ∂xj ∂xi (11. i.28) ∂xj ∂xj Equating the right hand side of this with the right hand side of Eq. v]i i i i j ∂v j ∂u = (£u ω)i v + ωi u −v . (11.30) (11.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 41 But we want to define things such that £u ω(v) = (£u ω) (v) + ω (£u v) . the 1-form corresponding to the i-th basis coordinate is dxi = δji dxj . v]j j ∂xi ∂x j j ∂ k ∂v k ∂u = u −v k k ∂xj ∂x ∂x ∂uj ∂ = 0 − δik k ∂xj ∂x j ∂u ∂ =− . (11. i 1h ∗ £u ω|P = lim φt ω|φ (P ) − ωP t t→0 t .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 42 Chapter 11. Lie derivative There is also a geometric description of the Lie derivative of 1forms. d ∗ . . 34) dt t . = φ ω . (11. P We will not discuss this in detail.29). (11. and the same description in terms of components as in Eq. . (11.27). but only mention that it leads to the same Leibniz rule as in Eq. λc ) . AP : (uP . ∗ ∗ Ap. TP∗ M . A tensor of type (p.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 Chapter 12 Tensors So far.q : TP M × · · · × TP M × TP M × · · · × TP M P | {z } | {z } q times p times 43 → R (12. AP (uP . ωc λc = ua v b ωc Acab . ωP ) 7→ AP (uP . and also vector fields and 1-forms.5) . (12. 2) tensor AP at P ∈ M is a map AP : TP M × TP M × TP∗ M → R (12. • A (1. Write Acab = AP (ea . vP = v a ea .3) Then for arbitrary vectors uP = ua ea . and the convention varies between books. ωP ) ∈ R .2) Suppose {ea }. We will now define tensors. ωP ) = AP ua ea . we have defined tangent vectors. eb . 2) tensor or a (2. 1) tensor. and covector ωP = ωa λa we get using linearity of the tensor map.1) 2 which is linear in every argument. So it is best to specify the tensor by writing indices as there is no confusion about Acab . (12. vP .4) It is a matter of convention whether A as written above should be called a (1. v b eb . vP . vP and a covector ωP . We will do this by starting with the example of a specific type of tensor. vP . {λa } are bases for TP M. (12. cotangent vectors. So given two vectors uP . q) can be defined in the same way. ω) = Akij ui v j ωk . 0) tensor has components with p upper indices. It is called a covariant q−tensor. 0) tensor is a linear map AP : TP∗ M → R . A (0. ∂xk (12. (12. ∂ i j dx ⊗ dx ⊗ k (u. A tensor field is a rule giving a tensor at each point. A(u. 2) tensor. • Alternatively. q) tensor has components with q lower indices. 2) tensor field A . Tensors in such a way that the map is linear in every argument.6) We can define the components of this tensor in the same way that we did for the (1. dxk ) . Then a (p. 2 We can now define the Lie derivative of a tensor field by using Leibniz rule in a chart. q) tensors have special names. (12. 2 It is possible to add tensors of the same type. a ···a a ···a a ···a (12. the components in a chart are ∂ ∂ Akij = A( i . j . A (1.11) . A (0. in agreement with the earlier definition. Thus we can write this tensor field as • A = Akij dxi ⊗ dxj ⊗ ∂ .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 44 Chapter 12.7) Ab11···bqp + Bb11···bqp = (A + B)b11···bqp . AP is an element of the tensor product space ∗ ∗ AP ∈ TP M ⊗ · · · ⊗ TP M ⊗ TP M ⊗ · · · ⊗ TP M | {z } | {z } p times q times (12. so it is a tangent vector. • Some special types of (p. It is called a contravariant p−tensor. in the sense that its action on two vectors and a 1-form is a product of the respective components. A (p.8) ∂x ∂x The components are functions of x in a chart.9) where the × indicates a ‘product’. Let us first consider the components of a tensor field in a chart. v. v.10) ∂x Thus we find. but not of different types. For a (1. q) tensor has components which can be written as a ···a Ab11···bqp . (12. ω) = ui v j ωk . 1) tensor is a cotangent vector. We can calculate the Lie derivative of a tensor field (with respect to a vector field u.19) .e. we can write 0 Akij 0 = 0 ∂x0i ∂x0j ∂xk ∂xi ∂xj ∂x0k0 0 ∂xi ∂xj ∂x0k .13) ∂xk ∂ ∂ ∂x0i i ∂x0j 0 dx ⊗ dxj ⊗ 0k0 k . = ∂xi ∂x0i0 ∂x0i0 ∂xi 0 (i and i are not equal in general). we will use the notation ∂i for and ∂i f for ∂xi ∂xi unless there is a possibility of confusion. we have m···n m···n £u Ta···b = ui ∂i Ta···b .12) 0 ∂x0i i ∂ ∂xi ∂ dx . we get 0 dx0i = 0 0 Akij dxi ⊗ dxj ⊗ (12. ∂x0i0 ∂x0j 0 ∂xk 0 Aki0 j 0 Aki0 j 0 = Akij (12. (12. (12.16) ∂ ∂f From now on.14) Equating components. say) by using the fact that £u is a derivative on the modules of vector fields and 1-forms.18) where the dots stand for the terms involving all the remaining upper and lower indices.15) (12. Since the components of a tensor field are functions on the manifold. coordinate system xi → x0i . = Aki0 j 0 i j k ∂x ∂x ∂x ∂x ∂x (12. This will save some space and make the formulae more readable. Consider a tensor field m···n T = Ta···b ∂m ⊗ · · · ⊗ ∂n ⊗ dxa ⊗ · · · ⊗ dxb . + · · · (12. i.17) Then m···n £u T = (£u Ta···b ) ∂m ⊗ · · · ⊗ ∂n ⊗ dxa ⊗ · · · ⊗ dxb m···n +Ta···b (£u ∂m ) ⊗ · · · ⊗ ∂n ⊗ dxa ⊗ · · · ⊗ dxb + · · · m···n +Ta···b ∂m ⊗ · · · ⊗ ∂n ⊗ (£u dxa ) ⊗ · · · ⊗ dxb . the components of the tensor field change according to A = Akij dxi ⊗ dxj ⊗ Since ∂ ∂ 0 0 0 = Aki0 j 0 dx0i ⊗ dx0j ⊗ 0k0 ∂xk ∂x (12.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 45 0 Under a change of charts. and by assuming Leibniz rule for tensor products. + Ti···b (12. i m···n (£u T )m···n a···b = u ∂i Ta···b i···n m···i − Ta···b ∂i um − · · · − Ta···b ∂i un m···n m···n ∂a ui + · · · + Ta···i ∂b ui . we find the components of the Lie derivative. ∂xm £u dxa = ∂ua i dx .20) Putting these into the expression for the Lie derivative for T and relabeling the dummy indices. ∂xi (12.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 46 Chapter 12. Tensors and we also know that £u ∂m = − ∂ui ∂i .21) . · · · . · · · . ω (v1 . On the other hand. v2 .2) Similarly. ∂j ) = −Aji . an element of C (M) .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 Chapter 13 Differential forms There is a special class of tensor fields. So p 6 n . vj . given vector fields v1 . Consider a 2-form A . for a p-form ω . • A p−form is a (0. vi . the components are ωi1 ···ip . p) tensor which is completely antisymmetric. · · · . v2 ) = −A(v2 . i.e. vp ) = −ω (v1 . · · · . vp . · · · . (13. it is possible to construct a 2-form ω 47 . 2 ∞ A 0-form is defined to be a function.1) for any pair i. j .. v1 ) . i. Any 1-form produces a function when acting on a vector field.e. vj . There are called differential p−forms or p−forms for short. · · · . · · · . and components are multiplied by (−1) whenever any two indices are interchanged. Then the components of A in a chart are Aij = A (∂i . The antisymmetry of any p-form implies that it will give a nonzero result only when the p vectors are linearly independent. n It follows that a p-form has independent components in np dimensions. no more than n vectors can be linearly independent in an n-dimensional manifold. and a 1-form is as defined earlier. So given a pair of 1-forms A. B. which is so useful as to have a separate treatment. we have A(v1 . Given any two vector fields v1 . vi . vp ) (13. so that A(v) = Ai v i . ω(u. and 1-forms. so that ω(u. {∂i } .7) Then a 2-form can be expanded in this basis as ω= 1 ωij dxi ∧ dxj . A 1-form A can be written as A = Ai dxi . Then for the ω defined above and for any pair of vector fields u. {dxi } . (13. v j ∂j ) = ωij ui v j . 2! (13. 2! ω(u. Clearly.4) and called the wedge product . v. v) 2! 1 = ωij ui v j − uj v i = ωij ui v j . The coordinate bases for the vector fields. 2 Let us work in a coordinate basis. We can now construct a basis for 2-forms. ∀u. satisfy dxi (∂j ) = δji . Differential forms by defining ω(u. ω is a 2-form.9) . (13. dxi ∧ dxj = dxi ⊗ dxj − dxj ⊗ dxi . and a vector field v can be written as v = v i ∂i . v) = A(u)B(v) − B(u)A(v) = Ai ui Bj v j − Bi ui Aj v j = (Ai Bj − Bi Aj ) ui v j . but the results we find can be generalized to any basis.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 48 Chapter 13. v) = A(u)B(v) − B(u)A(v). which we write as dxi ∧ dxj .8) because then 1 ωij dxi ⊗ dxj − dxj ⊗ dxi (u. v) = (13. (13. (13. (13. where ⊗ is called the outer product. ∂j ) . v .5) The components of ω are ωij = ω(∂i .6) Then ωij = Ai Bj − Bi Aj for the 2-form defined above. v) = ω(ui ∂i .3) • This is usually written as ω = A ⊗ B − B ⊗ A . 2 • Then the above construction defines a product written as ω = A ∧ B = −B ∧ A . so that the factorial goes away if we write each basis 3-form up to permutations. a basis for p−forms is dxi1 ∧ · · · ∧ dxip = dx[i1 ⊗ · · · ⊗ dxip ] .14) 2 A 2-form in three dimensions can be written as 1 ωij dxi ∧ dxj 2! = ω12 dx1 ∧ dx2 + ω23 dx2 ∧ dx3 + ω31 dx3 ∧ dx1 (13. a basis is dxi ∧ dxj ∧ dxk = dxi ⊗ dxj ⊗ dxk − dxj ⊗ dxi ⊗ dxk +dxj ⊗ dxk ⊗ dxi − dxk ⊗ dxj ⊗ dxi +dxk ⊗ dxi ⊗ dxj − dxi ⊗ dxk ⊗ dxj .15) ω = . (13.11) Then an arbitrary 3-form Ω can be written as Ω= 1 Ωijk dxi ∧ dxj ∧ dxk . (13. 3! (13.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 49 Similarly.10) where the square brackets stand for total antisymmetrization: all even permutations of the indices are added and all the odd permutations are subtracted. (Caution: some books define the ‘square brackets’ as antisymmetrization with a factor 1/p! . for a 3-form. i. p! 1 p (13. Thus a p−form α can be written in terms of its components as α= 1 αi ···i dxi1 ∧ · · · ∧ dxip .13) Examples: A 2-form in two dimensions can be written as 1 ωij dxi ∧ dxj 2! 1 = ω12 dx1 ∧ dx2 + ω21 dx2 ∧ dx1 2! 1 (ω12 − ω21 ) dx1 ∧ dx2 = 2! = ω12 dx1 ∧ dx2 .12) Note that there is a sum over indices. ω = (13. treating different permutations as equivalent.) For example.e. It can be shown by explicit calculation that wedge products are associative.17) Here P stands for a permutation of the vector fields. α acts on the first p vector fields in a given permutation P . Differential forms 2 In three dimensions. · · · . and deg P is 0 or 1 for even and odd permutations. So we can think of the wedge product as a generalization of the cross product. and β acts on the remaining q vector fields. · · · . vp+q . α ∧ (β ∧ γ) = (α ∧ β) ∧ γ . • We can also define the wedge product of a p−form α and a q−form β as a (p + q)−form satisfying. and that a term in which some i is equal to some j must vanish because of the antisymmetry of the wedge product. vp+q )) . respectively.18) Note that α ∧ β = 0 if p + q > n . In the outer product on the right hand side. vp+q ) = (−1)deg P α ⊗ β (P (v1 . 1 X α ∧ β (v1 .16) The components are like the cross product of vectors in three dimensions. (13. for any p + q vector fields v1 . 1 α= αi ···i dxi1 ∧ · · · ∧ dxip p! 1 p 1 βj ···j dxj1 ∧ · · · ∧ dxjq β = q! 1 q 1 α∧β = αi1 ···ip βj1 ···jq dxi1 ∧ · · · ∧ dxip ∧ dxj1 ∧ · · · ∧ dxjq .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 50 Chapter 13. p!q! (13. p!q! P (13. (13. consider two 1-forms α = αi dxi . Then 1 α∧β = (αi βj − αj βi ) dxi ∧ dxj 2! = αi βj dxi ∧ dxj = (α1 β2 − α2 β1 ) dx1 ∧ dx2 + (α2 β3 − α3 β2 ) dx2 ∧ dx3 + (α3 β1 − α1 β3 ) dx3 ∧ dx1 .19) . β = βi dxi . · · · . 2 The wedge product above can also be defined in terms of the components of α and β in a chart as follows. (13.22) Continuing this process.25) with p − 1 empty slots. we find α ∧ β = (−1)pq β ∧ α (13.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 51 Cross-products are not associative. Then contraction by v gives ιv ω = ω(v. (13. This is a (p − 1)-form. ω = λ ∧ µ = λ ⊗ µ − µ ⊗ λ . In fact. It is called a graded commutative algebra . (13. so there is a distinction between cross-products and wedge products. · · · ) (13. Note that the position of v only affects the sign of the contracted form. (13. a × (b × c) = (a × b) × c . 2 • Given a vector field v . 2 Example: Consider a 2-form made of the wedge product of two 1-forms. v) . 2 • The wedge product defines an algebra on the space of differential forms. for 1-forms in three dimensions. we can define its contraction with a p-form by ιv ω = ω(v. Then we exchange the basis 1-forms. • ) = λ(v)µ − µ(v)λ = −ω( • .23) Putting back the components. we find α ∧ β = (−1)pq β ∧ α . we get dxi1 ∧ · · · ∧ dxip ∧ dxj1 ∧ · · · ∧ dxjq = (−1)p dxj1 ∧ dxi1 ∧ · · · ∧ dxip ∧ dxj2 ∧ · · · ∧ dxjq = ··· = (−1)pq dxj1 ∧ · · · ∧ dxjq ∧ dxi1 ∧ · · · ∧ dxip .20) For a p-form α and q-form β .26) . One exchange gives a factor of −1 . the above equation is analogous to the identity for the triple product of vectors.21) Proof: Consider the wedge product written in terms of the components.24) as wanted. dxip ∧ dxj1 = −dxj1 ∧ dxip . We can ignore the parentheses separating the basis forms since the wedge product is associative. (13. e. but we get the same factor by rearranging the indices of ω . v) . 2 ∗ i i Then we can consider the pullback ϕ dx of a basis 1-form dx . (13. we have ϕ∗ λ = ϕ∗ (λi dxi ) . so we consider the action of each basis 1-form dxik on ∂i by carrying dxik to the first position and then writing a δiik . v) = (λ ∧ µ)(ϕ∗ u . • given a diffeomorphism ϕ : M1 → M2 . defined by ϕ∗ λ(v) = λ(ϕ∗ v) (13.27) ιv ω = (p − 1)! 2 p 1 . ϕ∗ v) = λ(ϕ∗ u)µ(ϕ∗ v) − µ(ϕ∗ u)λ(ϕ∗ v) = ϕ∗ λ(u)ϕ∗ µ(v) − ϕ∗ µ(u)ϕ∗ λ(v) = (ϕ∗ λ ∧ ϕ∗ µ)(u . thus getting a +1 for each index.33) .32) So we can write ϕ∗ λ = (ϕ∗ λi ) ϕ∗ dxi . (13. ϕ∗ v) − µ ⊗ λ(ϕ∗ u .28) p! Since the contraction is done in the first slot. (13. For the wedge product of two 1-forms. (13. ϕ∗ (λ ∧ µ)(u. Differential forms If we have a p-form ω = p!1 ωi1 ···ip dxi1 ∧ · · · ∧ dxip . i. To see how the factor becomes (p−1)! write the contraction as 1 ιv ω = ωi1 ···ip dxi1 ∧ · · · ∧ dxip v i ∂i . This leads to an overall factor of p . But ϕ∗ λ(v) = λ(ϕ∗ v) = λi dxi (ϕ∗ v) .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 52 Chapter 13. the pullback of a 1form λ (on M2 ) is ϕ∗ λ .31) where ϕ∗ λi are now functions on M1 . we Note the sum over indices. This gives a factor of (−1) for each exchange. so we can write this as ϕ∗ λ(v) = (ϕ∗ λi )ϕ∗ dxi (v) . its contraction with a vector field v = v i ∂i is 1 ωii ···i v i dxi2 ∧ · · · ∧ dxip . dxi (ϕ∗ v) = ϕ∗ dxi (v) and the thing on the right hand side is a function on M1 .30) Now.29) for any vector field v on M1 . ϕ∗ v) = λ ⊗ µ(ϕ∗ u . (ϕ∗ λi )|P = λi |ϕ(P ) (13. For a general 1-form λ = λi dxi . (13. we can write (by assuming an obvious generalization of the above formula) ϕ∗ dxi ∧ dxj ∧ dxk = ϕ∗ dxi ∧ dxj ∧ dxk = ϕ∗ dxi ∧ dxj ∧ ϕ∗ dxk = ϕ∗ dxi ∧ ϕ∗ dxj ∧ ϕ∗ dxk . .36) and in terms of components. ϕ∗ vp ) . (13. · · · . Then for any p-form α and q-form β we can calculate from this that ϕ∗ (α ∧ β) = ϕ∗ α ∧ ϕ∗ β .34) Since the wedge product is associative. but it is not necessary to make that assumption – it can be shown explicitly by taking three vector fields and following the arguments used earlier for the wedge product of two 1-forms. So for any p-form ω . p! (13. let us define the pullback ϕ∗ ω by ϕ∗ ω(v1 . · · · .38) Thus pullbacks commute with (are distributive over) wedge products.37) We assumed above that the pullback of the wedge product of a 2-form and a 1-form is the wedge product of the pullbacks of the respective forms. v are arbitrary vector fields it follows that ϕ∗ (λ ∧ µ) = ϕ∗ λ ∧ ϕ∗ µ ϕ∗ (dxi ∧ dxj ) = ϕ∗ dxi ∧ ϕdxj . (13. by ϕ∗ ω = 1 ϕ∗ ωi1 ···ip ϕ∗ dxi1 ∧ · · · ∧ dxip .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 53 Since u. (13. vp ) = ω (ϕ∗ v1 . (13.35) and we can continue this for any number of basis 1-forms. . (14. Let us define the exterior derivative of a p-form ω in a chart as dω = 1 ∂i ωi1 ···ip dxi1 ∧ · · · ∧ dxip p! 54 (14.4) We will try to define the exterior derivative in a way such that it has these properties.1) This should also satisfy Leibniz rule. d(α + ω) = dα + dω (14.. It is a map from p-forms to (p + 1)-forms.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 Chapter 14 Exterior derivative The exterior derivative is a generalization of the gradient of a function. The two formulas are identical by virtue of the fact that dβ is a (q + 1)-form. functions. Note that it gives the correct result when one or both of α. i. so it should be linear. (14. ω . but the algebra of p-forms is not a commutative algebra but a graded commutator algebra.5) . So we need d(α ∧ β) = dα ∧ β + (−1)pq dβ ∧ α . so that α ∧ dβ = (−1)p(q+1) dβ ∧ α .e.e.3) or alternatively. involves a factor of (−1)pq for exchanges. i. β are 0-forms.2) d(α ∧ β) = dα ∧ β + (−1)p α ∧ dβ . (14. ∀p-forms α . This will be the Leibniz rule for wedge products. This should be a derivation. Consider a 1-form A = Aµ dxµ where Aµ are smooth functions on M . So d2 = 0 on all forms. dxj ∧ dxi = −dxi ∧ dxj . p! (14. and the indices are summed over. we write 1 d ∂i ωi1 ···ip dxi ∧ dxi1 ∧ · · · dxip p! 1 = ∂j ∂i ωi1 ···ip dxj ∧ dxi ∧ dxi1 ∧ · · · dxip .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 55 This clearly has the first property of linearity.6) d(α ∧ β) = A third property of the exterior derivative immediately follows from here. (14. (14. d2 = 0 .10) . Note that we can also write dω = 1 dωi1 ···ip ∧ dxi1 ∧ · · · dxip . so the above object must be antisymmetric in ∂j . (14. But that vanishes. ∂i . Then using this definition we can write dA = (dAν ) ∧ dxν ⇒ (dA)µν = ∂µ Aν dxµ ∧ dxν 1 = (∂µ Aν − ∂ν Aµ ) dxµ ∧ dxν 2 = ∂ µ Aν − ∂ ν Aµ . To check the (graded) Leibniz rule. p! d(dω) = (14.8) But the wedge product is antisymmetric. Then 1 ∂i αi1 ···ip βj1 ···jq dxi ∧ dxi1 ∧ · · · ∧ dxjq p!q! 1 = ∂i αi1 ···ip βj1 ···jq + αi1 ···ip ∂i βj1 ···jq dxi ∧ dxi1 ∧ · · · ∧ dxjq p!q! 1 = ∂i αi1 ···ip βj1 ···jq dxi ∧ dxi1 ∧ · · · ∧ dxip ∧ dxj1 ∧ · · · dxjq p!q! 1 + (−1)p αi1 ···ip ∂i βj1 ···jq dxi1 ∧ · · · ∧ dxip ∧ dxi ∧ dxj1 ∧ · · · dxjq p!q! = dα ∧ β + (−1)p α ∧ dβ .9) where the object in parentheses is a gradient 1-form corresponding to the gradient of the component.7) To see this. let us write α ∧ β in components. can we say that there must be some (p − 1)-form β such that α = dβ ? • This is a good place to introduce some terminology. This statement is known as the Poincare´ lemma. So if α = dβ . call it α. we cannot uniquely specify the (p − 1)-form β since for any (p − 2)-form γ . Consider the 1-form α= xdy − ydx . but only on forms. where β 0 = β + dγ . Note that if a p-form α = dβ . 2 So every exact form is closed. in a sufficiently small neighbourhood. in agreement with our previous calculation. We say that every closed form is locally exact. for a 1-form A we get from this formula (dA)µν = ∂µ Aν − ∂ν Aµ . For p = 2 we have a 2-form. 2 Example: In R2 remove the origin. Then using this formula we get (dα)µνλ = ∂[µ ανλ] = ∂µ ανλ − ∂ν αµλ + ∂ν αλµ − ∂λ ανµ + ∂λ αµν − ∂µ αλν .11) p! 1 p 1 dα = ∂µ αµ1 ···µp dxµ1 ∧ · · · ∧ dxµp p! 1 = ∂ α dxµ ∧ dxµ1 ∧ · · · ∧ dxµp (p + 1)! [µ µ1 ···µp ] (dα)µµ1 ···µp = ∂[µ αµ1 ···µp ] (14. it follows that dα = 0 . Thus a more precise statement is that given any p-form α such that dα = 0 in a neighbourhood of some point P . 2 By definition.14) . Exterior derivative We can generalize this result to write for a p-form.e.12) α= ⇒ Example: For p = 1 i. there is some neighbourhood of this point and some (p−1)-form β such that α = dβ in that neighbourhood. whereas any form α such that α = dβ is called exact. But given a p-form α for which dα = 0 .13) Note that d is not defined on arbitrary tensors. Any form ω such that dω = 0 is called closed. x2 + y 2 (14. But this may not be true globally. d2 = 0 on any p-form. 1 αµ ···µ dxµ1 ∧ · · · ∧ dxµp (14. Is every closed form exact? The answer is yes. we can always write α = dβ 0 . (14.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 56 Chapter 14. 15) dx ∧ dy − 2 x + y2 (x2 + y 2 )2 Introduce polar coordinates r. (14. (14. . but θ is multivalued so there is no function f such that α = df everywhere.16) r2 α= Thus α is exact. y = r sin θ . In other words. Then dx = dr cos θ − r sin θdθ dy = dr sin θ + r cos θdθ r cos θ (sin θdr + r cos θdθ) r sin θ (cos θdr − r sin θdθ) − r2 r2 2 2 2 r cos θ + sin θ dθ = = dθ .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 57 Then 2x2 1 2y 2 1 − dx ∧ dy − − dy ∧ dx dα = x2 + y 2 (x2 + y 2 )2 x2 + y 2 (x2 + y 2 )2 2 x2 + y 2 = 2 dx ∧ dy = 0 . θ with x = r cos θ . α = dθ is exact only in a neighbourhood small enough that θ remains single-valued. Once we have identified these two classes.e.. they are independent of ω . • An orientable manifold is called oriented once an orientation has been chosen. Choose an n-form field. 2 Euclidean space is orientable. Suppose ω 6= 0 at some point P . 2 • A manifold is called orientable if it is possible to define a continuous n-form field ω which is non-zero everywhere on the manifold. if ω 0 is another n-form which is non-zero at P . That is. Call it ω. once we have decided to choose basis vectors with the same handedness everywhere on the manifold. Then it is possible to choose a basis with the same handedness everywhere on the manifold continuously. These are called righthanded and lefthanded. On an n-dimensional manifold. one for which ω(e1 . So the p space of n-forms in n dimensions is 1-dimensiona. the M¨obius band is not. 2 • It is necessary to choose an oriented manifold when we discuss the integration of forms. Thus all vector bases at P fall into two classes.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 Chapter 15 Volume form n The space of p-forms in n dimensions is dimensional. we have ω(e1 . · · · en ) 6= 0 since ω 6= 0 . · · · en ) > 0 and the other for which it is < 0 . • So every basis (set of n linearly independent vectors) is a member of one of the two classes. there must be some function f 6= 0 such that ω 0 = f ω . i. i. there is only one independent component. Then given any basis {eµ } of TP M . Two bases which gave positive numbers under ω will give the same sign — both positive or both negative — under ω 0 and therefore will be in the same class. a set of n 58 .e. and all n-forms are scalar multiples of one another. but also a redefinition of the region ϕ(U ) . multiplied by infinitesimal volumes. Similarly. we define the integral of a function f on an oriented manifold as the sum of the values of f . define an n-form in a chart by ω = f dx1 ∧· · ·∧ dxn . The right hand side can be seen to be independent of the choice of coordinate system. we get a Jacobian. ∆x2 2 . Then the integral of f over one such cell is approximately f ∆x1 ∆x2 · · · ∆xn = f dx1 ∧ · · · ∧ dxn ∆x1 ∂1 . ∆xn ∂n = ω(cell) . Given a function f . divide it up into infinitesimal ‘cells’.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 59 linearly independent vectors define an n-dimensional parallelepiped. This ω is called a volume form. any set of n linearly independent vectors will define a positive or negative volume. If we choose a different coordinate system. . We will do this in two dimensions with ω = f dx1 ∧ dx2 . where the ∆xi are small numbers. If we define an n-form ω 6= 0 we can think of the value of these vectors as the volume of this parallelepiped. The way to do that is the following. To integrate over an open set U . 2 Once a volume form has been chosen.1) Adding up the contributions from all cells and taking the limit of cell size going to zero. (15. The integral of a function f on Rn is the sum of the values of f . multiplied by infinitesimal volumes of coordinate elements. and the left hand side is our notation which we are defining. ∆xn n 1 ∂x ∂x ∂x 1 . · · · . Let us check that the left hand side is also invariant of the choice of the coordinates.2) ϕ(U ) The right hand side is the usual integration in calculus of n variables. · · · . (15. spanned by vectors ∂ ∂ ∂ ∆x . we find Z Z ω= U f dn x . if we choose a basis with the opposite orientation. y 2 )Jdy 1 ∧ dy 2 U Z = f (y 1 .3) dx1 = ⇒ and J is the Jacobian. Given the same f . (15. . Volume form In another coordinate system (y 1 . x2 )dx1 ∧ dx2 U U Z = f (y 1 .4) ϕ0 (U ) so we get the same result both ways. (15. Manifolds become even more interesting if we define a metric. y 2 ) corresponding to ϕ0 (U ) ∂x1 1 ∂x1 2 dy + 2 dy ∂y 1 ∂y 2 ∂x ∂x2 2 1 dx2 = dy + dy ∂y 1 ∂y 2 1 2 ∂x1 ∂x2 ∂x ∂x 1 2 − 2 2 dy 1 ∧ dy 2 dx ∧ dx = ∂y 1 ∂y 1 ∂y ∂y 1 2 = Jdy ∧ dy .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 60 Chapter 15. y 2 )Jd2 y . So what we have here is Z Z ω = f (x1 . This is why the choice of orientation has to be made before integration. the integral of ω will have the opposite sign. g(v. and has the properties (16. we say that v. w) = 0 . (16. we find that g(v. eν ) = 0 if µ 6= ν and ±1 if µ = ν . (16. we can always find an orthonormal basis {eµ } such that g(eµ . w) = 0 ⇒ v = 0. 2 61 . • A metric g on a manifold M is a (0. v). ii) symmetric: g(v. w) g(v.. 2 • Given a metric g on V . w) = g(w.2) tensor. w) = ag(v1 . (16. w) is a smooth function on M .3) mentioned earlier.e. w1 ) + ag(v.1). 2) tensor field such that if (v.1) i. We can generalize this definition to a manifold M by the following.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 Chapter 16 Metric tensor • A metric on a vector space V is a function g : V × V → R which is i) bilinear: g(av1 + v2 . We have defined a metric for a vector space. w1 + aw2 ) = g(v. g is a (0. 2 • If for some v. w) are smooth vector fields.3) iii) non-degenerate: g(v.2) ∀w (16. q) . w2 ) . 2 • If the number of (+1)’s is p and the number of (−1)’s is q . w) + g(v2 . w 6= 0 .2) and (16. we say that the metric has signature (p. w are orthogonal. g(v. Metric tensor It is possible to show that smoothness implies that the signature is constant on any connected component of M . e. the corresponding vector field is Aµ eµ . In a basis.6) ←→ Tµνρ ··· ←→ · · · (16. w 7→ g(v.g. if we know one set of components. Non-degeneracy implies that this matrix is invertible. where Aµ = g µν Aν . (16. and we will assume that it is constant on all of M . Let g µν denote the inverse matrix. and {λµ } is the dual basis to {eµ } . This is an n × n matrix in an n-dimensional manifold. Thus g(v. w) ∈ R .4) Then the linear isomorphism takes the following form. i) If v = v µ eµ is a vector field in a chart. say T µνρ ··· . so the map V → V ∗ defined by g(v.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 62 Chapter 16. and the metric. ·) ∈ V ∗ where V ∗ is the dual space of V . Then. by definition of an inverse matrix. ·) is linear. What it means is that. this map is an isomorphism. ·) is itself linear in v . in terms of components. T µν ←→ T µν ←→ Tµ ν ←→ Tµν T µνρ ··· ←→ T µν ρ ··· ←→ T µν ρ ··· (16. This is the isomorphism between vector fields and 1-forms. ii) If A = Aµ λµ is a 1-form written in a basis {λµ } . and vector v defines a linear map g(v. ·) = vµ λµ .5) where vµ = gµν v ν . we also know every other set of components. We can thus write g(v. ·) : V → R . the components of the metric are gµν = g(eµ . Given a vector space V with metric g . (We could of course define a similar isomorphism between vectors and covectors without referring to a manifold. It then follows that on a manifold we can use the metric to define a linear isomorphism between vectors and 1-forms. But g(v.7) These correspondences are not equalities — the components are not equal. eν ) . we have gµν g νλ = δµλ = g λν gµν . w) = gµν v µ wν in terms of the components. A vector space becomes related to its dual space by the metric.) A similar isomorphism holds for tensors. (16. . Since g is non-degenerate. 11) We wish to show that dV 0 = dV . i. We need to show that this definition is independent of the chart. by hA | Bi = g µν Aµ Bν (16. Take an overlapping chart. independent of the coordinate system. hv | wi = g(v. just like the inner product of vector fields. there is a canonical volume form dV (sometimes written as vol) . (16. On the other hand.e.13) . dV 0 = | det gµν (16. we can define an inner product of 1−forms.12) Then dx01 ∧ · · · ∧ dx0n = (det A)dx1 ∧ · · · ∧ dxn . dx0µ = ∂x0µ ν dx = Aµν dxν (say) ∂xν (16. which in a coordinate chart reads q dV = | det gµν |dx1 ∧ · · · ∧ dxn . as gµν is non-degenerate. ∂ν0 ) α ∂x ∂xβ = ∂ .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 63 • Using the fact that a non-degenerate metric defines a 1-1 linear map between vectors and 1-forms. ∂ α β ∂x0µ ∂x0ν α β = g A−1 µ ∂α . (16. Then in the new chart. (16.9) 2 Given a manifold with metric. if we look at the components of the metric tensor in the new chart. 0 gµν = g(∂µ0 .8) for 1-forms A. B . w) = gµν v µ wν . This result is independent of the choice of basis.10) Note that despite the notation. this is not a 1-form. the corresponding volume form is q 0 |dx01 ∧ · · · ∧ dx0n . This is clearly a volume form because it is an n-form which is non-zero everywhere. nor the gradient of some function V . A−1 ν ∂β α β = A−1 µ A−1 ν gαβ . In the overlap. and sometimes we mean |g|dn x . (16. 2 When we write dV . sometimes we mean the n-form as defined p above. Thus |g| µ1 ···µn are the components of the volume form. • This is called the metric volume form and written as p dV = |g|dx1 ∧ · · · ∧ dxn (16.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 64 Chapter 16. Another way of writing the volume form in a chart is in terms of its components. (16. we find Thus 0 det gµν = (det A)−2 (det gµν ) . with 12···n = +1 .17) n! where is the totally p antisymmetric Levi-Civita symbol.16) in a chart. Metric tensor Taking determinants. the measure for the usual integral.15) and so dV 0 = dV . .14) q q −1 0 | det gµν | = |det A| | det gµν | . p |g| dV = µ1 ···µn dxµ1 ∧ · · · ∧ dxµn (16. 1) . 1. So (?dx)ij = ijk g kl (dx)l = ijk g kl δl1 = ijk g k1 1 1 ⇒ ?dx = (?dx)ij dxi ∧ dxj = ijk g k1 dxi ∧ dxj 2! 2! g k1 = 1 for k = 1 . dy. i. gµν = diag(1. ?dz = dx ∧ dy . g µν diag(1. We will defineit in a chart rather than abstractly. Example: Consider R3 with metric +++. (dy)i = δi2 . (17.1) p! where ω is a p-form. denoted ? in an n-dimensional manifold is a map from p-forms to (n − p)-forms given by (?ω)µ1 ···µn−p p |g| ≡ µ1 ···µn g µn−p+1 ν1 · · · g µn νp ων1 ···νp . 1.2) the δ’s on the right hand sides are Kroenecker deltas. 2 The ? operator acts on forms.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 Chapter 17 Hodge duality We will next define the Hodge star operator. dz . 65 2 . 1) . 2! (17. (17.3) Similarly. • The Hodge star operator. Write the coordinate basis 1-forms as dx. (dz)i = δi3 . 0 otherwise 1 ⇒ ?dx = dx2 ∧ dx3 − dx3 ∧ dx2 = dx2 ∧ dx3 = dy ∧ dz . not on components.e. ?dy = dz ∧ dx . Then |g| ≡ g = 1 . Their components are clearly (dx)i = δi1 . in four dimensional Minkowski space.(17.. i.5) For the volume form. it can be shown in the same way that (?)2 = (−1)p(n−p)+s . p |g| dV = µ1 ···µn dxµ1 ∧ · · · ∧ dxµn n! p (dV )ν1 ···νn = |g|ν1 ···νn |g| µ ···µ g µ1 ν1 · · · g µn νn ν1 ···νn (?dV ) = n! 1 n |g| |g| n! = n!(det g)−1 = = sign(g) = (−1)s . n − s) .9) In particular. p (?ω)µ1 ···µn = |g|µ1 ···µn ω p ⇒ (?1)µ1 ···µn = |g|µ1 ···µn p |g| ⇒ (?1) = µ1 ···µn dxµ1 ∧ · · · ∧ dxµn n! = dV (17. on a p-form in an n-dimensional manifold with signature (s. s = 1.6) n! n! g where s is the number of (−1) in gµν . n = 4 . So we find that 2 ?(?1) = ?dV = (−1)s . Then p |g| µ1 ···µn g µ1 ν1 · · · g µn νn ων1 ···νn . (?ω) = n! (17. In general.10) .e. (17. (?)2 = (−1)s on 0-forms and n-forms.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 66 Chapter 17. (17.7) ?(?dV ) = (−1)s (?1) = (−1)s dV . (17.4) 2 Example: p = n . a 0-form ω in n dimensions. so (?)2 = (−1)p(4−p)+1 . (17. Hodge duality Example: Consider p = 0 (scalar).e.8) and i. we get X ?ω = ωI1 ···Ip ? (dxI1 ∧ · · · ∧ dxIp ) I = X I p |g| ν ···ν µ ···µ g µ1 I1 · · · g µp Ip ωI1 ···Ip dxν1 ∧ · · · ∧ dxνn−p .13) The sum over I is a sum over different index sets as before. (13. but there is no sum over the indices {I1 .11) We will use this to calculate ?ω ∧ ω . (n − p)! 1 n−p 1 p (17. Thus we calculate p |g| X ?ω ∧ ω = ν1 ···νn−p µ1 ···µp g µ1 I1 · · · g µp Ip ωI1 ···Ip × (n − p)! I. we have 1 ωµ ···µ dxµ1 ∧ · · · ∧ dxµp p! 1 p X = ωI1 ···Ip dxI1 ∧ · · · ∧ dxIp ω = (17. Then its components are I p!δµI11 · · · δµpp . and the Greek indices are summed over as usual.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 67 It is useful to work out the Hodge dual of basis p-forms. Ip } themselves.12) I where the sum over I means a sum over all possible index sets I = I1 < · · · < Ip . (17.J dxν1 ∧ · · · ∧ dxνn−p ∧ ωJ1 ···Jp dxJ1 ∧ · · · ∧ dxJp p |g| X = ν1 ···νn−p µ1 ···µp g µ1 I1 · · · g µp Ip ωI1 ···Ip ωJ1 ···Jp × (n − p)! I.13). · · · . Suppose we have a basis p-form dxI1 ∧ · · · ∧ dxIp . where the indices are arranged in increasing order Ip > · · · > I1 . in a given index set the Ik are fixed.14) . For a p-form ω . and Eq. Using the dual of basis p−forms. So ? dxI1 ∧ · · · ∧ dxIp ν1 ···νn−p p |g| 0 0 I = ν1 ···νn−p µ1 ···µp g µ1 µ1 · · · g µp µp p! δµI10 · · · δµp0 p 1 p! p µ1 I1 µp Ip = |g| ν1 ···νn−p µ1 ···µp g ···g .J dxν1 ∧ · · · ∧ dxνn−p ∧ dxJ1 ∧ · · · ∧ dxJp (17. · · · . · · · . g µν is diagonal. Jp } . νn−p } cannot have any overlap with {µ1 . it follows that we can replace J by I. p |g| X ν1 ···νn−p I1 ···Ip g I1 I1 · · · g Ip Ip ωI1 ···Ip ωI1 ···Ip × ?ω ∧ ω = (n − p)! I dxν1 ∧ · · · ∧ dxνn−p ∧ dxI1 ∧ · · · ∧ dxIp p |g| X = ν1 ···νn−p I1 ···Ip ω I1 ···Ip ωI1 ···Ip × (n − p)! I dxν1 ∧ · · · ∧ dxνn−p ∧ dxI1 ∧ · · · ∧ dxIp . µp } because is totally antisymmetric in its indices. · · · . the indices {ν1 · · · νn−p } are completely determined. So the set {µ1 . the indices {I1 · · · Ip } must be the same as {J1 · · · Jp } because both sets are totally antisymmetrized with the indices {ν1 · · · νn−p }. Then g µk Ik = g Ik Ik etc. But this sign is the same as that for the permutation to bring the basis to the order dx1 ∧ · · · ∧ dxn . The indices on this are a permutation of {1. · · · . νn−p } cannot have any overlap with the set J = {J1 . · · · .16) In each term of this sum.J dxν1 ∧ · · · ∧ dxνn−p ∧ dxJ1 ∧ · · · ∧ dxJp . · · · . Now consider the case where the basis is orthogonal. and we can write p |g| X ?ω ∧ ω = ν1 ···νn−p I1 ···Ip g I1 I1 · · · g Ip Ip ωI1 ···Ip ωJ1 ···Jp × (n − p)! I. Hodge duality We see that the set {ν1 . (17. µp } must have the same elements as the set J = {J1 . so that p X ?ω ∧ ω = |g| K1 ···Kn−p I1 ···Ip ω I1 ···Ip ωI1 ···Ip × I p dxK1 ∧ · · · ∧ dxKn−p ∧ dxI1 ∧ · · · ∧ dxI(17. On the other hand. Jp }.17) . but they may not be in the same order.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 68 Chapter 17.e. so we can replace them by the corresponding ordered set K = K1 < · · · < Kn−p . · · · . because of the wedge product. Since both sets are ordered. i. (17. so is ±1.15) We see that in each term of the sum. which is completely determined by the set I . {ν1 . so the overall sign to get both to . n} . So we can diagonalize it locally by going to an appropriate basis. the components of ω may be ωµ0 1 ···µp . it is still symmetric. or set of coordinates. .18) If we are in a basis where the metric is not diagonal.18) even when the metric is not diagonal. so we can write 1 µ01 ···µ0p ωµ01 ···µ0p (vol0 ) (17.19) ?ω ∧ ω = ω p! But both factors are invariant under a change of basis. (17. In this basis. Note that the metric may not be diagonalizable globally or even in an extended region. at each point.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 69 the standard order is positive. So we can now change back to our earlier basis. and find Eq. Thus we get p X I ···I ?ω ∧ ω = |g| ω 1 p ωI1 ···Ip 1···n dx1 ∧ · · · ∧ dxn I p 1 = |g| ω µ1 ···µp ωµ1 ···µp dx1 ∧ · · · ∧ dxn p! 1 µ1 ···µp = ω ωµ1 ···µp (vol) p! (17. (18. 1) .5) For the electric field define a 1-form E = Ex dx + Ey dy + Ez dz . 1.4) The electric and magnetic fields are all vectors in three dimensions. ∇·E ∂E ∇×B− ∂t ∇·B ∂B ∇×E+ ∂t =ρ (18. 1.2) =0 (18. but these equations are Lorentz-invariant. with c = 1 .1) =j (18.3) = 0. Consider R4 with Minkowski metric gµν = diag(−1. We will write 1. (18. (18. Let us calculate dF = d(B + E ∧ dt) = dB + dE ∧ dt . As usual. For the magnetic field define a 2-form B = Bx dy ∧ dz + By dz ∧ dx + Bz dx ∧ dy .6) Combine these two into a 2-form F = B + E ∧ dt .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 Chapter 18 Maxwell equations We will now consider a particular example in physics where differential forms are useful. 3 70 . The Maxwell equations of electrodynamics are. We will write these equations in terms of differential forms. 2. 11) for dual basis forms. (18. (18. For the other two equations we need ?F . (18.9) Thus two of Maxwell’s equations are equivalent to dF = 0 . we can combine these two to get dF = (∂t B1 + ∂2 E3 − ∂3 E2 ) dt ∧ dy ∧ dz + (∂t B2 + ∂1 E3 − ∂3 E1 ) dt ∧ dz ∧ dx + (∂t B3 + +∂1 E2 − ∂2 E1 ) dt ∧ dx ∧ dy + (∂1 B1 + ∂2 B2 + ∂3 B3 ) dx ∧ dy ∧ dz = (∂t B1 + (∇ × E)1 ) dt ∧ dy ∧ dz + (∂t B2 + (∇ × E)2 ) dt ∧ dz ∧ dx + (∂t B3 + (∇ × E)3 ) dt ∧ dx ∧ dy + (∇ · B) dx ∧ dy ∧ dz . ?(dz ∧ dt) = dx ∧ dy . Using the formula (17. it is easy to calculate that ?(dx ∧ dy) = dt ∧ dz .8) Thus.11) . dB = d(B1 dy ∧ dz + B2 dz ∧ dx + B3 dx ∧ dy) = ∂t B1 dt ∧ dy ∧ dz + ∂1 B1 dx ∧ dy ∧ dz +∂t B2 dt ∧ dz ∧ dx + ∂2 B2 dy ∧ dz ∧ dx +∂t B3 dt ∧ dx ∧ dy + ∂3 B3 dz ∧ dx ∧ dy . (18. y. ?(dy ∧ dz) = dt ∧ dx . ?(dy ∧ dt) = dz ∧ dx .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 71 for the component labels x. (18. z . ?(dx ∧ dt) = dy ∧ dz . ?(dz ∧ dx) = dt ∧ dy . remembering that the wedge product changes sign under each exchange.10) We use these to calculate ?F = ?(B + E ∧ dt) = B1 dt ∧ dx + B2 dt ∧ dy + B3 dt ∧ dz +E1 dy ∧ dz + E2 dz ∧ dx + E3 dx ∧ dy .7) And d(E ∧ dt) = d(E1 dx ∧ dt + E2 dy ∧ dt + E3 dz ∧ dt) = ∂2 E1 dy ∧ dx ∧ dt + ∂3 E1 dz ∧ dx ∧ dt +∂1 E2 dx ∧ dy ∧ dt + ∂3 E2 dz ∧ dy ∧ dt +∂1 E3 dx ∧ dz ∧ dt + ∂2 E3 dy ∧ dz ∧ dt . 17) 2 This expression holds in both flat and curved spacetimes. (18. So in terms of components. jµ dxµ = −ρdt + j1 dx1 + j2 dx2 + j3 dx3 . x.14) Then using Eq. (18. with local coordinates (t. (18. using Eq. (18. (17.12) We need to relate this to the charge-current.18).16) Finally. (18. y. we see that the action of electromagnetism can be written as Z 1 − F ∧ ?F (18. z) we find √ F ∧ ?F = (B 2 − E 2 ) −g dt ∧ dx ∧ dy ∧ dz .18) .13) Then there is a corresponding one-form jµ dxµ with jµ = gµν j ν .12) we find that the other two Maxwell equations can be written as d?F = −?j . Define the current four-vector as j µ ∂µ = ρ∂t + j 1 ∂1 + j 2 ∂2 + j 3 ∂3 .15) Comparing this equation with Eq. (18. Maxwell equations Then in the same way as for the previous calculation. For the latter. (17.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 72 Chapter 18.11) it is easy to calculate that ?j = −ρ dx ∧ dy ∧ dz + j1 dt ∧ dy ∧ dz +j2 dt ∧ dz ∧ dx + j3 dt ∧ dx ∧ dy . (18. we find d?F = (∇ · E) dx ∧ dy ∧ dz + (∂t E1 − (∇ × B)1 ) dt ∧ dy ∧ dz + (∂t E2 − (∇ × B)2 ) dt ∧ dz ∧ dx + (∂t E3 + (∇ × B)3 ) dt ∧ dx ∧ dy . If ω vanishes outside some compact region we can again set U = M . • Consider a smooth (n − 1) form in M .1) U ∂U If M is a compact manifold with boundary ∂M . 73 . like a 2-surface in a 3-manifold. Stokes’ formula says that Z Z dω = ω . • A submanifold S is a subset of points in M such that any point in S has an open neighbourhoood in M for which there is some chart where (n − m) coordinates vanish. but we will not prove it. Then ∂U is automatically an oriented manifold. 1] . 2 Example: Let U = [0. Also. This is actually a theorem. this formula can be applied to all of M .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 Chapter 19 Stokes’ theorem We will next discuss a very beautiful result called Stokes’ formula. only state the result and discuss its applications. U can be a submanifold in another manifold. S is then m-dimensional. Now suppose U has an oriented smooth boundary ∂U . by considering the restrictions of the charts on U to ∂U . Then a function f : M → R is a 0-form. So for us it is only a formula. 2 • Suppose U is a region of an oriented manifold M . but still deep and beautiful. (19. and df = f 0 (x)dx is a 1-form. Take the orientation of M to be from 0 to 1. The bound ary ∂U of U is a submanifold of dimension n − 1 which divides M in such a way that any curve joining a point in U with a point in U c must contain a point in ∂U . Then ∂M consists of the points x = 0 and x = 1 . n! (19. We can write this in a chart as ω = f dx1 ∧ · · · ∧ dxn 1 = f µ1 ···µn dxµ1 ∧ · · · ∧ dxµn . Before getting to Gauss’ theorem. we need to make a new definition.e. (19. f ∂M f 0 (x)dx = f (1) − f (0) . Consider an n-form ω 6= 0 on an n-dimensional manifold.2) 0 Example: Consider a 2-d disk D in R2 . So we can write A = Ai dt .3) D ∂D Let us seee this equation in a chart.5) ϕ(D) Similarly for higher forms on higher dimensional manifolds. We can write A = Ai dxi dA = ∂i Aj dxi ∧ dxj (19. Then Stokes’ formula says Z Z A = dA . • Gauss’ divergence theorem is a special case of Stokes’ theorem.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 74 Chapter 19. (19. (19.6) . Take a 1-form A . Stokes’ theorem and Stokes’ formula says that Z Z df = M Z1 i. and dt dt Z Z i Ai dx = ∂i Aj dxi ∧ dxj D ∂D Z = (∂1 A2 − ∂2 A1 ) dx1 ∧ dx2 D Z = (∂1 A2 − ∂2 A1 ) d2 x .4) d d A evaluated on ∂D can be written as A where is tangent dt dt d dxi to ∂D . with boundary ∂D . · · · ) = ∂1 (f v 1 ) dx1 ∧ dx2 ∧ · · · ∧ dxn +∂2 (f v 2 ) dx1 ∧ dx2 ∧ · · · ∧ dxn + · · · + ∂n (f v n ) dx1 ∧ dx2 ∧ · · · ∧ dxn = ∂µ (f v µ ) dx1 ∧ dx2 ∧ · · · ∧ dxn 1 = ∂µ (f v µ ) ω .12) . if ω is the volume form. n! p 1 d(ιv (vol)) = p ∂µ (v µ |g|)(vol) . vµ = gµµ0 v µ 0 (19.11) p |g| ⇒ ?v = µ ···µ µ v µ dxµ1 ∧ · · · ∧ dxµn−1 (n − 1)! 1 n−1 ! p |g| d?v = ∂µn µ ···µ µ v µ dxµn ∧ dxµ1 ∧ · · · ∧ dxµn−1 (n − 1)! 1 n−1 p (−1)n−1 = µ1 ···µn−1 µ ∂µn |g|v µ dxµ1 ∧ · · · ∧ dxµn (n − 1)! (19. (19. · · · ) = (19. |g| (19.8) (19.10) Consider ?v . Remember that given a vector field v . we can write p |g| ω = µ1 ···µn dxµ1 ∧ · · · ∧ dxµn . with components defined with the help of the metric. f In particular. also called v .9) • This is called the divergence of the vector field v . its contraction with ω is 1 ωµ µ ···µ v µ1 dxµ2 ∧ · · · ∧ dxµn (n − 1)! 1 2 n = f v 1 dx2 ∧ · · · ∧ dxn − f v 2 dx1 ∧ dx3 ∧ · · · ∧ dxn + · · · ιv ω = ω(v. we can define a one-form.7) Then we can calculate d(ιv ω) = dω(v. There is another expression for the divergence.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 75 Given a vector field v . which has components p 0 (?v)µ1 ···µn−1 = |g|µ1 ···µn−1 µ g µµ vµ0 p = |g|µ1 ···µn−1 µ v µ . |g| (19. And b 6= 0 on ∂U because ∂U is defined as the submanifold where one coordinate is d constant.e. so that one component of vanishes dt at any point on ∂U .14) where as before s is the signature of the manifold. For the same reason. α always exists. · · · . so a sum over (µ1 . and therefore the corresponding component of b can be chosen to be non-zero. i. if α acts only on tangent . · · · . and α is an (n − 1)-form such that b ∧ α = dt (vol) . it is unique up to a factor. if b 6= 0 on ∂U . so µ = µn . µn−1 ) automatically selects µn (= µ) . Further. µn ) overcounts n times. i. Then we find Z Z p 1 p ∂µ ( |g|v µ )(vol) = d(ιv (vol)) |g| U U Z = ιv (vol) . But we can always scale α → α0 = f α so that b0 ∧ α0 = b ∧ α . Stokes’ theorem Both µ and µn must be different from (µ1 . µn−1 ) .e. the choice of (µ1 . Let us now go back to Stokes’ formula.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 76 Chapter 19. the number of negative entries in the metric in a locally diagonal form. usually set to zero. i.15) ∂U d Now suppose b is a 1-form normal to ∂U . Since all n-forms are proportional. So b is unique up to a rescaling b → b0 = f b for some nonvanishing −1 function f . b = 0 for any dt d vector tangent to ∂U . |g| Since this is an n-form in n dimensions. we can calculate from here that ?d?v = p (−1)n+s−1 p ∂µ ( |g|v µ ) .13) = (−1)n−1 p ∂µ ( |g|v µ )(vol) . So we can write p (−1)n−1 d?v = µ1 ···µn ∂µ |g|v µ dxµ1 ∧ · · · ∧ dxµn (n − 1)! p 1 (19.e. · · · . if we restrict α to ∂U . Take a region U of M which is covered by a single chart and has an orientable boundary ∂U as before. (19. Thus in each term of the sum. for any vector v . so it is unique up to scaling. α is unique once b is given. Therefore. .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 77 vectors to ∂U . Finally. we find that α is an (n − 1)-form on an (n − 1)dimensional manifold. . . . (19.16) ιv (vol). = ιv (b ∧ α). Then . ∂U ∂U is an (n − 1)-form on ∂U which acts only on vectors tangent to ∂U . . . . 17) ιv (b ∧ α). (19. = b(v)α. Then α is the volume form on ∂U . (19. Then we have Z Z p 1 µ p ∂µ ( |g|v )(vol) = b(v)α |g| U ∂U Z = (nµ v µ ) α . ∂U ∂U because all terms of the form b ∧ ιv α gives zero for any choice of (n − 1) vectors on ∂U .18) ∂U Usually b is taken to have norm 1. and we can write Z Z q p 1 p ∂µ ( |g|v µ )(vol) = (nµ v µ ) |g(∂U ) |dn−1 x . (19.19) |g| U ∂U . . with djp being defined by djp : Tp P → Tj(p) M such that djp (v)(g) = v(g · jp ) .e. it is always possible to find a set of coordinate charts covering G such that the overlap functions are real analytic. So is Cn .→ M is smooth and at each point p ∈ P its differential djp is one to one. Example: Rn is a Lie group under addition. The definition above comes from a theorem that given a continuous group G in which the group product and group inverse are C ∞ functions of the group parameters. a submanifold of G . P is an immersed submanifold of M if the inclusion map j : P . Let us mention some more examples.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 Chapter 20 Lie groups We start a brief discussion on Lie groups. • A Lie group is a group which is also an analytic manifold. or alternatively the group product and group inverse are both C ∞ . but said that a Lie group was a manifold in which the group product is analytic in the group parameters. are C ∞ and their Taylor series at any point converge to their respective values. 2 We did not define a Lie group in this way in Chap. and is a topological group. 2 • Sometimes this expressed in terms of another definition. a topological space in which the group product and group inverse are continuous maps. i. 2 We have mentioned some specific examples of Lie groups earlier. 2 . 78 . i. mainly with an eye to their structure as manifolds and also application to the theory of fiber bundles.e. • A Lie subgroup of G is a subset H of G which is a subgroup of G . where J= 0 1n×n −1n×n 0 2 Example: O(p. We can similarly define GL(n.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 79 Example: Rn \{0} is a Lie group under multiplication. the Special unitary group SU (n) is the subgroup of U (n) with determinant +1.q) and SU (p. This is also the space of all invertible linear maps of Rn to itself.q U = ηp. R)| det A = 1} . R) . Example: The Special Linear group SL(n. R) | RT ηp. Similarly for SO(p. q) . called the General Linear group GL(n. R) . These are the only two spheres (other than the point S 0 ) which admit a Lie group structure. 2 Example: The Symplectic group Sp(n) . R) = {A ∈ GL(n. . h2 ) = (g1 g2 . h1 h2 ) . q) = {R ∈ GL(p+q. R) is the subset of GL(n.q R = ηp. The group U (1) is the group of phases U (1) = {eiφ |φ ∈ R} .q = 1p×p 0 0 −1q×q 2 Example: U (p. 2 The next few examples are Lie subgroups of GL(n. R) | R T R = I} . The group SU (2) is isomorphic as a manifold to a three-sphere S 3 . C) . Example The Special Orthogonal group SO(n) is the subgroup of O(n) for which determinant is +1. R) for which all the matrices have determinant +1 . C) | U † ηp.. Similarly. SL(n. this is isomorphic to a circle S 1 . One can define SL(n. So is Cn \{0} . C) | U U = I} . with multiplication (g1 . i.q } .q } . Example: The direct product of two Lie groups is itself a Lie group. h1 )(g2 . defined as the subgroup of U (2n) given by AT JA = J . q) = {U ∈ GL(p + q. where ηp. As a manifold. 2 Example: The set of all n × n real invertible matrices forms a group under matrix multiplication. 2 † Example: The Unitary group U (n) = {U ∈ GL(n. C) in a similar manner.e. 2 Example: The Orthogonal group O(n) = {R ∈ GL(n. The left and right translations lead to diffeomorphisms which relate the tangent space at any point to this Lie algebra. as we shall see now. the isomorphism between the tangent spaces is induced by group operations. called • Left translation lg : G → G g 0 7→ gg 0 . lg−1 (g) = e and rg−1 (g) = e . 2 • Right translation rg : G → G g 0 7→ g 0 g . Of course. For Lie groups. . 2 These can be defined for any group. We see that lg−1 lg (g 0 ) = lg−1 (gg 0 ) = g −1 gg 0 = g 0 ⇒ (lg )−1 = lg−1 rg−1 rg (g 0 ) = rg−1 (g 0 g) = g 0 gg −1 = g 0 ⇒ (rg )−1 = rg−1 . The tangent space at the identity forms a Lie algebra.1) It is easy to check that lg1 lg2 = lg1 g2 rg1 rg2 = rg2 g1 (20. we can define diffeomorphisms of G labelled by elements g ∈ G . For any Lie group G .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 80 Chapter 20. so is in some sense natural. the tangent space at any pooint of a manifold is isomorphic to the tangent space at any other point. but are diffeomorphisms for Lie groups. Lie groups An important property of a Lie group is that the tangent space of any point is isomorphic to the tangent space at the identity by an appropriate group operation. as we shall see. so any element of G can be moved to the identity by a diffeomorphism.2) Further. (20. X = lg∗ (X) (21. (21. lg∗ (Xg0 ) = Xgg0 • (21. ∀g. lg∗ (Xe ) = Xg (21.e. we can consider the pushforwards due to them. since ∀g ∈ G . a right-invariant vector field X is defined by X = rg∗ (X) i. Since left and right translations are diffeomorphisms. • A left-invariant vector field X is invariant under left ttranslations. Write the set of all left-invariant vector fields on G as L(G) .e. Since the push-forward is linear..3) A left or right invarian vector field has the important property that it is completely determined by its value at the identity element e of the Lie group. rg∗ (Xg0 ) = Xg0 g ∀g ∈ G . So a vector field on the Lie group selects a vector at each g ∈ G . g 0 ∈ G . the vector (field) at g 0 is pushed forward by lg to the same vector (field) at lg (g 0 ): ∀g. 81 (21.2) Similarly.4) and similarly for right-invariant vector fields. ∀g ∈ G .1) In other words.5) .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 Chapter 21 Tangent space at the identity A point on the Lie group is a group element. we get lg∗ (aX + Y ) = alg∗ X + lg∗ Y . g 0 ∈ G . i. c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 82 Chapter 21. Y ∈ L(G) .. Proof: We will show that left translation leads to an isomorphism. Thus the set of all left-invariant vector fields on G forms a Lie algebra. (21. For X ∈ Te G . Tangent space at the identity so that if both X and Y are left-invariant. Y ] ∈ L(G) . Y ] . [lg∗ X . (21.8) so [X . Y ] = [lg∗ X . (21. We also know that push-forwards leave the Lie algebra invariant.7) Thus if X. • This L(G) is called the Lie algebra of G . lg∗ Y ] = [X .e. lg∗ [X .6) so the set of left-invariant vector fields form a real vector space. for lg∗ . lg∗ Y ] = lg∗ [X . 2 The dimension of this Lie algebra is the same as that of G because of the Theorem: L(G) as a real vector space is isomorphic to the tangent space Te G to G at the identity of G . lg∗ (aX + Y ) = aX + Y . i. Y ] . define the vector field LX on G by . ∀g ∈ G (21.9) LX . (21. g 0 ∈ G .10) Note that for two diffeomorphisms ϕ1 .g ≡ LX g := lg∗ X Then for all g.11) Since left translation is a diffeomorphism.12) . ϕ2 . we can write (ϕ1∗ (ϕ2∗ v))(f ) = (ϕ2∗ v)(f ◦ ϕ1 ) = v(f ◦ ϕ1 ◦ ϕ2 ) = ((ϕ1 ◦ ϕ2 )∗ v)(f ) ⇒ ϕ1∗ (ϕ2∗ v) = (ϕ1 ◦ ϕ2 )∗ v (21. X lg0 ∗(LX g ) = lg 0 ∗ (lg∗ X) = lg 0 g∗ X = Lg 0 g . lg0 ∗ (lg∗ X) = (lg0 ◦ lg )∗ X = (lg0 g∗ )X (21. 17) Then So the map X 7→ LX is onto. (21. we have Y LX g = Lg ∀g ∈ G . define Xe ∈ Te G by Xe = lg−1 ∗ LX g for any g ∈ G . If LX = LY . We need to prove that this map is 1-1 and onto.14) So the map X → LX is 1-1. (21. Then we can define a Lie bracket on Te G by . so is the map X → LX . (21. Since the pushforward is a linear map. (21. (21. and we have a map Te G → L(G) .15) We can also write Xe = LX e .16) X lg∗ Xe = lg∗ lg−1 ∗ LX g = Lg . Now given LX .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 83 So it follows that LX is a left-invariant vector field.13) so Y lg−1 ∗ LX g = lg −1 ∗ Lg ⇒ X=Y (∈ Te G) . v] = [Lu . Lv ]. [u . tn } is a set of basis vectors on Te G ' L(G) . It follows that dim L(G) = dim Te G = dim G . {t1 . · · · . (21.19) Note that since commutators are defined for vector fields and not vectors. the Lie bracket on Te G has to be defined using the commutator of left-invariant vector fields on G and the isomorphism Te G ↔ L(G) . the Lie bracket of any pair of these vectors must be a linear combination of them. so X k [ti . tj ] = Cij tk (21.18) The Lie algebra of vectors in Te G based on this bracket is thus the Lie algebra of the group G .20) k . • If for an n-dimensional Lie group G . (21.e . 23) . (21. tj ] = [tj . Tangent space at the identity k . 2 Since L(G) is a Lie algebra. the Lie bracket is antisymmetric. These numbers are known as the for some set of real numbers Cij structure constants of the Lie group or algebra. [tj . ⇒ [ti .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 84 Chapter 21. ti ]] + [tk . [ti .22) A similar construction can be done using a set of right-invariant vector fields defined by RgX := rg∗ X and its ‘inverse’ Xe = rg−1 ∗ RgX . with the Lie bracket as the product. Cjl Cilm + Cki Cij Ckl + Cjk (21. tj ]] = 0 ⇒ l m m l l = 0. for X ∈ Te G (21. [tk . tk ]] + [tj . ti ] X X k k Cij tk = Cji tk k ⇒ k k k Cij = Cji .21) and the structure constants satisfy the Jacobi identity [ti . c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 Chapter 22 One parameter subgroups There is another characterization of Te G for a Lie group G as the set of its one parameter subgroups. γ(t) = − sin t cos t 0 . which we will now define. since this is a homomorphism. Then γ : R → G is a curve such that γ(s + t) = γ(s)γ(t) . R) . γ : (R. γ(0) = e . +) → G . This is also called the “infinitesimal” description of a Lie group. and γ(−t) = γ(t)−1 . and what Lie called an infinitesimal group. γ(t) = − sin t cos t cos t sin t 0 Example: G = GL(3. 2 Also. Example: G = SU (2) . cos t sin t . 0 0 et The relation between 1-p subgroups and . Example: For G = (R\{0}. the one parameter subgroup is Abelian. Example: G = U (1) . ×) the multiplicative group of nonzero real numbers. γ(t) = eit. • A one parameter subgroup of a Lie group G is a smooth homomorphism from the additive group of real numbers to G . γ(t) = et is a 1-p subgroup. and Te G is given by the Theorem: The map γ 7→ γ(0) ˙ = γ˙ . We need to find a smooth homomorophism from R to G using LX . This homomorphism is provided by the flow or integral curve of LX .e defines a 1-1 correspondence between 1-p subgroups of G. 85 . and Te G . but let us work this out in more detail. Proof: For any X ∈ Te G define LX = lg∗ X as the corresponding left-invariant vector field. c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 86 Chapter 22. γ X (0) = LX e =X and γ˙ X (t) = LX γ(t) = lγ(t)∗ X (22. One parameter subgroups Denote the integral curve of LX by γ X (t) .e..1) X Since LX is left-invariant. i. Consider the equation . lg0 ∗ LX g = Lg 0 g . d . . d X γ(t) = Lγ(t) = lγ(t)∗ X ≡ γ∗ . (22.2) dt dt . and at t = 0 both curves are at the point γ(τ ) . there are elements in G which do not lie on a 1-p subgroup. every 1-p subgroup is of the form . 2 In a compact connected Lie group G . both satisfy the equation of the integral curve of LX . However.6) where.3) Remember that γ(t) is an element of the group for each t . i. Thus for each X ∈ Te G we find a 1-p subgroup γ(t) given by the integral curve of LX . i. γ(t) ˙ = LX γ(t) = lγ(t)∗ X . every element lies on some 1-p subgroup. (22. (γ(0) ˙ = X) . as mentioned earlier. replace γ(t) by γ(τ + t) to get γ(τ + t) = lγ(τ +t)∗ X . γ(τ + t) = γ(τ )γ(t) . (22.2) by γ(τ )γ(t) to get d γ(τ )γ(t) ∗ (22. For matrix groups.e. This is not true in a non-compact G .t Given some τ .4) = LX γ(τ )γ(t) dt We see that γ(t + τ ) and γ(τ )γ(t) are both integral curves of LX . Now replace γ(t) in Eq. an Abelian non-compact group will always have a 1-p subgroup.e. (22. (22. so this remark applies only to non-Abelian non-compact groups. Thus by uniqueness these two are the same curve.5) and t 7→ γ(t) is the homomorphism R → G that we are looking for. n o . γ(t) = etM . M fixed. (22. t ∈ R .7) . (We will not a give a proof of this here. (22.e.8) Differentiate with respect to s and set s = 0 .9) Write γ(0) ˙ = M . The allowed matrices {M } for a given matrix group G thus form a Lie algebra with the Lie bracket being given by the matrix commutator. so these are in fact the tangent vectors at the identity. the Lie algebra is spanned by N × N real antisymmetric matrices. Since G is a matrix group. Then γ(0)γ(t) ˙ = γ(t) ˙ . Then γ(t) is a matrix for each t .) We can find the Lie algebra of a matrix group by considering elements of the form γ(t) = etM for small t . Then the unique solution for γ is γ(t) = etM . det R = 1 .12) • This is the infinitesimal group. for compact connected groups this is identical to the Lie group itself. i. Let us construct a basis for this algebra. γ(t) = I + tM (22. . the entire group can be generated by exponentiating the Lie algebra. and γ(s)γ(t) = γ(s + t) . So in such cases. a Lie algebra with basis {ti } . once we are given. The allowed matrices {M } for a given group G are the {γ(0)} ˙ for all the 1-p subgroups γ(t) .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 87 Let us see why. then AT = −A . not every matrix M will generate any group. or have found. M is a matrix.10) The properties of M are determined by the properties of the group and vice versa. Conversely. the group of N ×N real orthogonal matrices R with RT R = I . i. 2 Example: Consider SO(N ) . Noncompact groups cannot be written as the exponential of the Lie algebra in general.11) for small t . (22.. we can exponentiate the Lie algebra to find the set of 1-p subgroups γ(a) = exp ai ti (22. Suppose {γ(t)} is a 1-p subgroup of the matrix group. (22. This Lie algebra is isomorphic to the Lie algebra of the group G . Write R = I + A .e. (22. ν are not matrix indices ρ . Mαβ ] = δνα Mµβ − δµα Mνβ + δµβ Mνα − δνβ Mµα .14) Then the commutators are calculated to be [Mµν . Often the basis is multiplied by i to write γ(a) = exp(iaj tj ) .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 88 Chapter 22. (22. · · · . σ are matrix indices . and det(I + tM ) = 1 ⇒ Tr M = 0 . (22. Mµν = −Mνµ (Mµν )ρσ = (Mµν )σρ . where tj are now Hermitian matrices. labelled by µ. the group of N × N unitary matrices U with U † U = I . So we define N (N − 1)/2 independent antisymmetric matrices.15) This defines the Lie algebra.16) . Example: For SU (N ) . N . the 1-p subgroups are given by γ(t) = etM with M † + M = 0 in the same way as above.13) A convenient choice for the basis is given by (Mµν )ρσ = δµρ δνσ − δµσ δνρ . with [ti . tj ] = ifabc tc . (22. ν = 1. det U = 1 . So the SU (N ) Lie algebra consists of traceless antihermitian matrices. One parameter subgroups An N × N antisymmetric matrix has N (N − 1)/2 independent elements. µ . 2 E is called the total space. Then in the neighbourhood of any point P . i. even though the bundle is actually the triple (E. i. 2 • If E can be written globally as a product space. TP M 7→ P . it is called a trivial bundle. the pre-image π −1 (Ui ) is homeomorphic to Ui ×F . vP ) .e. vP ) . Thus there is a projection map π : T M → M . as the set of ordered pairs (P.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 Chapter 23 Fiber bundles Consider a manifold M with the tangent bundle T M = S P ∈M TP M . Locally a fiber bundle is a product manifold E = B × F with the following properties. Given an open set Ui of B . We express this in the following way. and if P ∈ B .e. π. we can think of T M as a product manifold. • E is locally a product space. • There is a projection map π : E → B . i. the preimage π −1 (P ) is homeomorphic. This is generalized to the definition of a fiber bundle. • B is a manifold called the base manifold. The set {Ui .e. ϕi } is called a local trivializa tion of the bundle. bicontinuously isomorphic. which associates the point P ∈ M with TP M . and F is another manifold called the typical fiber or the standard fiber. to the standard fiber. or in other words there is a bicontinuous isomorphism ϕi : π −1 (Ui ) → Ui × F . T M can be thought of as the original manifold M with a tangent space stuck at each point P ∈ M . B) . 2 89 . Then we can say that T M consists of pooints P ∈ M and vectors v ∈ TP M as an ordered pair (P. Let us look at this more closely. but usually it is also called the bundle. E = B × F . Locally for some open set U ( S 1 we can still write a segment of the M¨obius strip as U × I . f : . π −1 (P ) . the structure group of T M is GL(n. globally a product space. F : E1 → E2 . Consider a covering of S 1 by open sets Ui . So B = S 1 . π2 . v) 7→ P . • A fiber bundle where the standard fiber is a vector space is called a vector bundle .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 90 Chapter 23. On the other hand. dλi So we can define a homeomorphism hi (P ) : TP S 1 → R. ai and aj are also independent. So the structure group is R\{0} with aj multiplication. Fiber bundles • This description includes a homeomorphism π −1 (P ) → F for each P ∈ Ui . those which map fibers into fibers. For an n-dimensional manifold M . where P ∈ S 1 . They are called bundle morphisms. 2 Let us consider an example. which in this case relates the component of the vector in two coordinate systems. a M¨obius strip is obtained by twisting the strip and then glueing. and since λi and λj are independent. i. π1 . called the structure group of F . the relevant or useful maps between these are those which preserve the bundle structure locally. B1 ) and (E2 . has homeomorphisms hi (P ) and hj (P ) onto F . but the total space is no longer a product space. Then the tangent bundle E = T S 1 has F = R and π(P. 2 A cylinder can be made by glueing two opposite edges of a flat strip of paper. R) . Then hi (P ) · hj (P )−1 : F → F (or R → R) maps aj to ai . and let the coordinates of Ui ⊂ S 1 be denoted by λi . 2 A bundle morphism is a pair of maps (F. v 7→ ai (fixed i). If P ∈ Ui ∩ Uj there are two such homeomorphisms T S 1 → R . B2 ) . F = I and this is a trivial bundle. • Given two bundles (E1 . the M¨ obius strip is non-trivial. Then in some overlap Ui ∩ Uj the fiber on P . This is then a Cartesian product of acircle S 1 with a line segment I . It follows that hj (P ) · hi (P )−1 is a homeomorphism F → F . Then any vector at d TP S 1 can be written as v = ai (no sum) for P ∈ Ui . Suppose B = S 1 . These are called transition functions. Let us denote this map by hi (P ) . is simply multiplication by the ai number rij = ∈ R\{0} .e. i. f ) . The homeomorphism. As a bundle.e. The transition functions F → F form a group. v ∈ T S 1 . F = {AP } . (23. 2 • Given a manifold M the tangent space TP M . and G acts on F by left translation is called a principal fiber bundle. Example: A function on M is a section of the bundle which locally looks like M × R (or M × C if we are talking about complex functions). So we can also say π ◦ s = identity. · · · . R) . • A fiber bundle in which the typical fiber F is identical (or homeomorphic) to the structure group G. any basis AP may be expressed as ei = aji ej . structure group SU (2) . R) .) Not all systems of coordinates are appropriate for a bundle. a set of n linearly independent vectors at P . v : P 7→ vP . Thus. (This is of course better understood in terms of a commutative diagram. Clearly the structure group of the typical fiber of the frame bundle is also GL(n . starting from any one basis. which must be invertible so that we can recover the original basis from the new one. 2 Example: 1. Typical fiber = S 3 . The typical fiber in the frame bundle is the set of all bases. 3. π. en ) . 2 Given a particular basis AP = (e1 . any other basis can be reached by an n×n invertible matrix. for which the typical fiber is GL(n . where p ∈ B . the coordinates of the bundle are given by bundle morphisms onto open sets of Rn × Rp . Typical fiber = S 1 . 2. R) . · · · . s(p) ∈ π −1 (p) . such that π2 ◦ F = f ◦ π1 . AP is a basis in TP M . So there is a bijection between the set of all frames in TP M and GL(n . and any n×n invertible matrix produces a new basis.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 91 B1 → B2 . 2 Example: A vector field is a section of the tangent bundle. Given a differentiable fiber bundle with n-dimensional base manifold B and p-dimensional fiber F . en ) . . p 7→ s(p) . • A section of a fiber bundle (E. consider AP = (e1 . But it is possible to define a set of fiber coordinates in the following way. Bundle of frames. structure group U (1) .1) The numbers aji can be thought of as the components of a matrix. as is the structure group. B) is a mapping s : B → E . • These relations are what are called gauge transformations in physics. 2 • A vector bundle (E. there are . G) of a principal bundle (P. and G is called the gauge group. we must have vi = D (gij (x)) vj . their observations are relatted by a group transformation (p. which is essentially the same as writing Φi. usually a Lie group. let us consider an associated vector bundle (E. Because the fiber carries a representation {D(g)} of G . This is written as T ∗ M . x ∈ Ui . π. π e. π e. Fiber bundles • Starting from the tangent bundle we can define the cotangent bundle. B. which characterizes the theory. G) by the representation {D(g)} of G on F if its transition functions are the images under D of the transition functions of P. 2 Fields appearing in various physical theories are sections of vector bundles. Then the transition functions are in some representation of the group G . F.x : π −1 (x) → G . V carries a representation of some group G . 2 That is. in which the typical fiber is the dual space of the tangent space. Usually G is a Lie group for reasons of continuity. B. G) with typical fiber F and structure group G is said to be associated to the principal bundle (P. which in some trivialization look like Uα × V where Uα is some open neighborhood of the point we are interested in. (23. and local trivialization of P with respect to this covering is Φi : π −1 (Ui ) → Ui × G .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 92 Chapter 23. G). 2 π • Remember that a vector bundle F → E → B is a bundle in which the typical fiber F is a vector space. v) ' (p. To discuss this a little more concretely. That is. a section of T ∗ M is a 1-form field on M . suppose we have a covering {Ui } of B . B. and V is a vector space. As we have seen before. Then the transition functions of P are of the form gij = Φi ◦ Φj −1 : Ui ∩ Uj → G .3) A more physical way of saying this is that if two observers look at the same vector at the same point. B.2) The transition functions of E corresponding to the same covering of B are given by φi : π e−1 (Ui ) → Ui × F with φi ◦ φj −1 = D(gij ) . F. if vi and vj are images of the same vector vx ∈ Fx under overlapping trivializations φi and φj . D(gij v) . π. (23. 9) where now g 0 = gβα ggβα −1 . gv)α . v)α 7→ (x.5) . gv)α . a gauge transformation is a representation-valued linear transformation of the sections at each point of the base space. In other words.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 93 always linear transformations Tx : Ex → Ex which are members of the representation {D(g)} . To see this. if Tx : (x. (23. (23.8) can be written also as Tx : (x. (23.7) But we also have using Eq. gβα gv)β (23. and vβ = D(gβα )vα . gβα v)β . A gauge transformation T acts as Tx : (x. So it is also true that Tx : (x.5) In the other notation we have been using. (23. gβα v)β 7→ (x. v)α 7→ (x.e. i. Tx ∈ {D(g)} for some g . i. consider a point x ∈ Uα ∩ Uβ .e.6) (x. v)α ∈ Uα × F . vα and vβ are images of the same vector vx ∈ Vx . • We say that a linear map T : Γ(E) → Γ(E) is a gauge trans formation if at each point x of the base space. g 0 v 0 )β . Such a map assigns an element of V to each point of the base space.4) for some g ∈ G and for (x. Let us write the space of all sections of this bundle as Γ(E) . Then Eq. The right hand side is often written as (x. So T is a gauge transformation in Uβ as well. v 0 )β 7→ (x. define v 0 = gβα v . (23. gv)α = (x. v)α = (x. The definition of a gauge transformation is independent of the . (23. (23. 2 This definition is independent of the choice of Uα . An element of Γ(E) is a map from the base spacce to the bundle. we can think of gv as a change of variables. D(g)v)α . gβα gv)β . Then (x.8) Since F carries a representation of G . (g −1 )(x) = (g(x)) −1 . but T itself is not.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 94 Chapter 23. (23. Fiber bundles choice of Uα .10) • The groups G and G arre both called the gauge group by different people. 2 . with (gh)(x) = g(x)h(x) . The set of all gauge transformations G is a group. (24. s. v. w are vector fields on B .e. So there is no unique way of adding or subtracting points on different fibers. Thus there are many ways of differentiating sections of fiber bundles. whenever we refer to a connection D .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 Chapter 24 Connections There is no canonical way to differentiate sections of a fiber bundle. Let us consider a bundle π : E → B . In other words. t ∈ Γ(E) . f ∈ C ∞ (B) and α is a real number (or complex. • Each way of taking a derivative. Each fiber is isomorphic to the standard fiber but the isomorphism is not canonical or unique.2) where s. i. Let us see why. not some unique connection.1) But a section of a bundle assigns an element of the fiber at any point P ∈ M to the base point P . →0 f 0 (x) = lim (24. t are sections of the bundle. we need to choose a connection before we can talk about it. we 95 . no unique derivative arises from the definition of a bundle. The usual derivative of a function on R is of the form f (x + ) − f (x) . In what follows. is called a connection. where Γ(E) is the space of all sections. That is to say. of comparing. depending on what the manifold is). Then a connection D on B assigns to every vector field v on B a map Dv : Γ(E) → Γ(E) satisfying Dv (s + αt) = Dv s + α Dv t Dv (f s) = v(f )s + f Dv s Dv+f w (s) = Dv s + f Dw s . 2 Note that this is a connection. Connections mean that we have chose a connection D and that is what we are discussing. • We call Dv s the covariant derivative of s . Consider coordinates xµ in an open set U ⊂ M . which may be a principal fiber bundle such as a frame bundle. like a set of basis vector fields. but along the fiber at M . (24. (x. we can write Dv s = Dvµ ∂µ s = v µ Dµ s . Remember that a gauge transformation is a linear map T : E → E .e. then {ei (x)} is a basis for the fiber at P ∈ M . 2 To be specific. with ∂µ the coordinate basis vector fields. (24. (24. with {x} being the set of coordinates at P . Let us apply this idea to the section s . i. So we should be able to talk about gauge transformations.3) • These Aiµj are called components of the vector potential or the connection one-form. we can say (Dµ s)i = ∂µ si + Aµj i sj . Dµ s = Dµ (si ei ) = ∂µ si ei + si Dµ ei = ∂µ si ei + si Aµij ej = ∂µ si + Aµj i sj ei . Call this basis {ei } . Dµ ej = Aµj i ei . but the vectors are not along M .4) Also. let us consider the bundle to be a vector bundle on a manifold M . and try to understand the meaning of D by going to a chart. Also choose a basis for sections.7) . there is a connection D0 on E such that D0 (gφ) = gDv φ .6) We have considered connections on an associated vector bundle. But then Dµ ej can be expressed uniquely as a linear combination of the ei . We claim that given a connection D .5) so writing Dµ s = (Dµ s)i ei . (24. 2 Given a section s = si ei with si ∈ C ∞ (M) . Any element of V ' Fx can be written uniquely as a linear combination of ei (x) . which is like a basis for the fiber (vector space) at each point of M . (24.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 96 Chapter 24. Write Dµ = D∂µ . gv) for some g ∈ G and for all v ∈ V ' Fx . v) 7→ (x. and thus Dv g −1 φ ∈ Γ(E) and also gDv g −1 φ ∈ Γ(E) . So Dv0 (φ) = Dv0 gg −1 φ = gDv g −1 φ . 0 Dv+αw φ = gDv+αw g −1 φ = g Dv g −1 φ + αDw g −1 φ = gDv (g −1 φ) + αgDw g −1 φ 0 = Dv0 φ + αDw φ. φ = s). We can now write (24. so is g −1 φ.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 97 where v is a vector field on M and φ ∈ Γ(E) (i. (24. This completes the definition of the gauge transformation of the connection. Dv0 : Γ(E) → Γ(E) . i. We have of course assumed that g(x) is differentiable as a function of x. Since φ is a section. (24.11) So D0 is a connection. (24. we know that g −1 (x) exists for all x. (24. there is a connection D0 satisfying Dv0 (gφ) = g (Dv φ) .10) Similarly.e.12) Dµ0 φ = ∂µ φi + A0µij φj ei . Dv0 maps sections to sections.9) And it satisfies Leibniz rule because Dv0 (f φ) = gDv g −1 f φ = gDv f g −1 φ = gv(f )g −1 φ + gf Dv g −1 φ = v(f )φ + f gDv g −1 φ = v(f )φ + f Dv0 (φ) . φ ∈ Γ(E).e.e. Let us now check that D0 is a connection according to our definitions.8) and thus D0 is defined on all φ for which Dv φ is defined. Therefore. Let us first check if the definition makes sense. i. Since g(x) ∈ G for all x ∈ M .e. i. D0 exists because D does. . D0 is linear since Dv0 (φ1 + αφ2 ) = gDv g −1 (φ1 + αφ2 ) = gDv g −1 φ1 + αgDv g −1 φ2 = Dv0 φ1 + αDv0 φ2 . . so that the fiber over any point p ∈ M is C . We can make E into a U (1) bundle by thinking of the fiber C as the fundamental representation space of U (1) .16) A connection which transforms like this is also called a G2 Example: Consider G = U (1) .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 98 Chapter 24.13) where θi is the dual basis to {ei } . Connections Using the dual space. as always. Suppose E is a trivial complex line bundle over M . (24.15) = ∂µ φi + g∂µ g −1 + gAµ g −1 j φj ei .14) where. 2 connection. From this we can read off A0µ = gAµ g −1 + g∂µ g −1 . The gauge transformation is then given by Dv0 φ = gDv g −1 φ ⇒ Dµ0 φ = gDµ g −1 φ h i j i ⇒ ∂µ φi + A0µij ei = g ∂µ g −1 φ + Aµij g −1 φ ei . • (24. E = M×C . A connection D on E may be written as Dµ = ∂µ + Aµ . Then sections are complex functions. i. (24. A0µ = A0µij ei ⊗ θj .e. Then we can write the right hand side as h i i i ∂µ φi + g∂µ g −1 j φj + gAµ g −1 j φj ei h i i (24. and a gauge transformation is multiplication by a phase. let us write Aµ = Aµij ei ⊗ θj . the g’s are in some appropriate representation of G. and a section s .1) Remember that D v s = D v µ ∂µ s = v µ D µ s = v µ ∂µ si + Aµij sj ei = v(si )ei + v µ Aµij sj ei .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 Chapter 25 Curvature We start with a connection D . (25.3) Since the connection components Aν j k are functions. Dw are both maps Γ(E) → Γ(E) . (25. (25. Then Dv . (25. we can write v wν Aν j k sk = v(wν )Aν j k sk +wν v(Aν j k )sk +wν Aν j k v(sk ) . w)s = Dv Dw s − Dw Dv s − D[v. w) : Γ(E) → Γ(E) by F (v.4) 99 .2) Dµ s is again a section. all on some associated vector bundle of some principal G-bundle E . w . two vector fields v.w] s . produces a linear map F (v. w on B . We will define the curvature of this connection D as a rule F which. given two vector fields v . So we can act with D on it and write h i Dv Dw s = Dv w(sj )ej + wν Aν j k sk ej = v w sj ej + v wν Aν j k sk ej + w sj + wν Aν j k sk v µ Aµij sj ei . w)s = v µ wν ∂µ Aν ij − ∂ν Aµij + Aµik Aν kj − Aν ik Aµkj sj ei . F (u.5) Also. ∂ν ] = 0 .11) where f ∈ C ∞ (B) .13) Since F (∂µ .7) Thus we can define Fµν by F (∂µ .12) For coordinate basis vector fields [∂µ .10) • It is not very difficult to work out that the curvature acts linearly on the module of sections. we find Dv Dw s − Dw Dv s = [v . F (u. • (25. ∂ν ] = 0 .c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 100 Chapter 25. so is Dλ (Fµν s) = Dλ [Dµ . w]µ Aµij sj ei + wν v Aν ij − v µ w Aµij sj ei + v µ wν Aµij Aν j k − Aν ij Aµj k sk ei . w)s . Dν ]s . v)(s1 + f s2 ) = F (u. (25. (25. ∂ν )s is a section. Dν ]s . (25. v + f w) s = F (u. D[v .8) (Fµν )ij = ∂µ Aν ij − ∂ν Aµij + Aµik Aν kj − Aν ik Aµkj . w]µ Aµij sj ei . (25. v)s1 + f F (u.6) so that F (v. (25. w](si )ei + [v . ∂ν )s = Dµ Dν s − Dν Dµ s = [Dµ . Dν ]s . so Fµν s = F (∂µ .w] s = [v . Curvature Inserting this into the previous equation and writing Dw Dv s similarly. (25. (25. (25. w](si )ei + [v . (25. [∂µ . v)s + f F (u.14) . ∂ν )s = Dµ Dν s − Dν Dµ s = [Dµ . ∂ν )s = Fµν s = (Fµν s)i ei = (Fµν )ij sj ei .9) so that Note that since coordinate basis vector fields commute. Also. v)s2 . Fµν s = F (∂µ . and noting that maps are associative under map composition. [Dµ .19) where we have defined (Dλ Fµν ) by (Dλ Fµν )ij in this. Then this is a Leibniz rule. Dν ]Dλ s .18) On the other hand. (25. Dλ (Fµν s) = (Dλ Fµν ) s + Fµν (Dλ s) . Dν ]]s .17) Fµν s = (Fµν s)i ei = Fµν ij sj ei .16) Thus Considering C ∞ sections. we have D given by Dv0 φ = gDv g −1 φ . So we can write Dλ (Fµν s) = ∂λ Fµν ij sj + Fµν ij sj Dλ ei = ∂λ Fµν ij sj ei + Fµν ij ∂λ sj ei + Fµν ij sj Aλki ek = ∂λ Fµν ij + Fµν kj Aλik sj ei + Fµν ij ∂λ sj ei − Fµν ik Aλkj sj ei + Fµν ij Aλj k sk ei = (Dλ Fµν )ij sj ei + Fµν ij (Dλ s)j ei . (25.21) This is known as the Bianchi identity . where Fµν ij and si are in C ∞ (B) .20) Then we can write Dλ (Fµν s) − Fµν (Dλ s) + cyclic = 0 ⇒ (Dλ Fµν ) s + cyclic = 0 ⇒ Dλ Fµν + cyclic = 0 . so is Fµν Dλ s = [Dµ . 2 0 Given D and g such that g(x) ∈ G . (25.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 101 Similarly. (25.22) . (25. ∀s (25. (25.15) Dλ (Fµν s) − Fµν Dλ s = [Dλ . (25. [Dµ . Dν ]]s + cyclic = 0 . since Dλ s is a section. we find that [Dλ . g is in some representation of G .v] φ = gDu Dv g −1 φ − Dv Du g −1 φ − gD[u . This is the meaning of the statement that the curvature is gauge covariant . (25. Curvature Then Du0 Dv0 φ = Du0 gDv g −1 φ = gDu Dv g −1 φ .24) As before.v] g −1 φ = gF (u.c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 102 Chapter 25. 2 . (25. v) φ ≡ Du0 Dv0 − Dv0 Du0 − D[u .23) and thus 0 F 0 (u. and D (and thus F ) acts on the same representation. v)g −1 φ ⇒ 0 Fµν = g ◦ Fµν ◦ g −1 .
Report "Amitabha Lahiri_Lecture Notes on Differential Geometry for Physicists 2011.pdf"