Faculty of Engineering, Mathematics and Science School of Mathematics

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1 Faculty of Engineering, Mathematics and Science School of Mathematics JS & SS Mathematics SS Theoretical Physics SS TSM Mathematics Module MA3429: Differential Geometry I Trinity Term 2018???, May??? SPORTS CENTRE??? ??? Prof. Sergey Frolov Instructions to Candidates: Credit will be given for the best 3 questions answered. Each question is worth 20 marks. Materials Permitted for this Examination: Formulae and Tables are available from the invigilators, if required. Non-programmable calculators are permitted for this examination, please indicate the make and model of your calculator on each answer book used. You may not start this examination until you are instructed to do so by the Invigilator. Page 1 of 12

2 1. (a) 5 marks. Let (M, g) be a manifold endowed with a metric. Let be the covariant derivative: j ξ i = j ξ i + ξ k Γ i kj, j ξ i = j ξ i ξ k Γ k ij. Show that a torsion-free connection that is compatible with the metric ( k g ij = 0) defines a unique connection, known as the Levi-Civita (or metric) connection. Obtain an expression for the Levi-Civita connection coefficients Γ k ij. Solution: We have k g ij = k g ij g pj Γ p ik g ipγ p jk = kg ij Γ j,ik Γ i,jk = 0, = j g ki Γ i,kj Γ k,ij = 0, (1) = i g jk Γ k,ji Γ j,ki = 0, where Γ j,ik g pj Γ p ik = g jpγ p ik. Taking the sum of the last two equations and subtracting from the sum the first equation, one gets j g ki + i g jk k g ij 2Γ k,ij = 0, (2) where we took into account that the connection is torsion-free, and therefore Γ k,ij = Γ k,ji. Thus, one finds Γ k,ij = 1 2 ( ig jk + j g ki k g ij ) = Γ k ij = 1 2 gkp ( i g pj + j g ip p g ij ), (3) (b) 5 marks. Let (ξ a ) be a vector field and let (g ab ) be a pseudo-riemann metric on a manifold M. Let u ab = L ξ g ab = ξ c g ab x c + g ξ c cb x + g ξ c a ac x b be the strain tensor, i.e. the Lie derivative of (g ab ) along the vector field ξ. Express the strain tensor in terms of covariant derivatives, and the torsion tensor. In the case of a torsion-free connection compatible with the metric, simplify the expression obtained. Solution: We have u ab = L ξ g ab = ξ c g ab x c + g ξ c cb x + g ξ c a ac x = b ξc c g ab + g cb a ξ c + g ac b ξ c. (4) Page 2 of 12

3 It is clear that we should replace. We get ξ c c g ab + g cb a ξ c + g ac b ξ c = ξ c( c g ab g db Γ d ac g ad Γ d bc) + gcb ( a ξ c + ξ d Γ c da) + gac ( b ξ c + ξ d Γ c db ) (5) = u ab ξ c g db T d ac ξ c g ad T d bc. Thus, u ab = ξ c c g ab + g cb a ξ c + g ac b ξ c + ξ c g db T d ac + ξ c g ad T d bc. (6) If g ab is a metric on the manifold, and the connection is torsion-free, and compatible with the metric, the formula simplifies to u ab = g cb a ξ c + g ac b ξ c = a ξ b + b ξ a, T c ab = 0, c g ab = 0. (7) (c) 5 marks. Let M be a manifold endowed with a torsion-free connection. Prove that given any point p M and a chart (x i ) with connection coefficients Γ i jk, one can find a new chart (ˆx i ) such that The coordinates (ˆx i ) are called normal. connection transforms as ˆΓ i jk(p) = 0. ˆΓ i jk = ˆxi x b x c x a ˆx j ˆx k Γa bc + ˆxi 2 x a x a ˆx j ˆx. k You can use without a proof that the Solution: Let p have coordinates x i = 0, ˆx i = 0, i = 1,..., n. Let us look for (ˆx i ) of the form ˆx i = x i Qi jkx j x k, (8) where Q i jk = Qi kj are some constants to be determined from the condition ˆΓ i jk (p) = 0. For small x = max i x i we can invert (8) x i = ˆx i 1 2 Qi jkˆx j ˆx k + O( x 3 ), (9) Then ˆx i x = j δi j + Q i jkx k, x i ˆx = j δi j Q i jkˆx k + O( x 2 ), 2 x i ˆx j ˆx = k Qi jk + O( x ), Page 3 of 12 ˆx i p = δ i x j j, x i p = δ i ˆx j j, 2 x i p = Q i ˆx j ˆx k jk. (10)

4 The connection transforms as Thus, at p we have ˆΓ i jk = ˆxi x b x c x a ˆx j ˆx k Γa bc + ˆxi 2 x a x a ˆx j ˆx. (11) k ˆΓ i jk(p) = Γ i jk(p) Q i jk. (12) So, if we choose Q i jk = Γi jk (p), then ˆΓ i jk (p) = 0. (d) 5 marks. Let M be an m-dimensional surface in Euclidean n-space R n, let π be the linear operator which projects R n orthogonally onto the tangent space to M at an arbitrary fixed point of M, and let X, Y be vector fields in R n tangent to the surface M. Show that the connection on M compatible with the induced metric on M satisfies X Y = π(x k Y x k ). Solution: Let P be the point on M of dim M = m, and let us choose the origin of our Euclidean coordinate system to be at P and the coordinate axes to be oriented so that the axes x m+1,..., x n be perpendicular to the surface at P. Then in a neighbourhood of P the surface M is described by the system x m+1 = f m+1 (x 1,..., x m ),..., x n = f n (x 1,..., x m ), (13) where the functions f p satisfy f p (0) = 0, i f p (0) = 0, and the linear operator which projects R n orthogonally onto the tangent space to M at P is given by m π(x) = X i n x, X = X k i x. (14) k i=1 The induced metric on M is n m ds 2 = (dx k ) 2 = (dx i ) 2 + k=1 g ij (x) = δ ij + i=1 n p=m+1 n p=m+1 ( k=1 m i f p dx i ) 2 = i=1 m g ij dx i dx j, i,j=1 i f p (x) j f p (x), g ij (0) = δ ij, k g ij (0) = 0. (15) Thus with this choice of coordinates in a neighbourhood of P the connection on M compatible with the induced metric on M satisfies Γ k ij(0) = 0, and therefore m X Y = X i Y Y = π(xk xi x ). (16) k i=1 Page 4 of 12

5 2. (a) 5 marks. The Riemann curvature tensor R i jkl is defined by means of the formula [ k, l ]ξ i = Rjkl i ξ j + T j kl jξ i. Use the formula to find explicit formulae for Rjkl i and T j kl in terms of the connection coefficients Γ i jk. Solution: We have by definition l ξ i = l ξ i + ξ p Γ i pl, k ξ l = k ξ l ξ p Γ p lk, (17) and therefore [ k, l ] ξ i = k ( l ξ i + ξ p Γ i pl) + l ξ p Γ i pk q ξ i Γ q lk (k l) = k ξ p Γ i pl + ξ p k Γ i pl + l ξ p Γ i pk q ξ i Γ q lk (k l) = ( k ξ p ξ q Γ p qk )Γi pl + ξ p k Γ i pl + l ξ p Γ i pk q ξ i Γ q lk (k l) (18) = ξ q Γ p qk Γi pl + ξ p k Γ i pl q ξ i Γ q lk (k l) = ( k Γ i jl l Γ i jk + Γ i pkγ p jl Γi plγ p jk) ξ j + ( Γ j kl Γj lk) j ξ i. Thus and T j kl is the torsion tensor. R i jkl = k Γ i jl l Γ i jk + Γ i pkγ p jl Γi plγ p jk, T j kl = Γj kl Γj lk, (19) (b) 5 marks. Let a manifold be endowed with a metric, and let the connection be torsion-free and compatible with the metric. List the algebraic symmetries the components of the Riemann tensor satisfy. Use normal coordinates to prove the symmetry properties of the Riemann tensor. Solution: We define R ijkl = g iq R q jkl. Then R ijkl satisfies the following algebraic relations R ijkl + R ijlk = 0, R ijkl + R jikl = 0, R ijkl R klij = 0, (20) R ijkl + R iklj + R iljk = 0. Page 5 of 12

6 If the coordinates are normal, then at p we have k g ij (p) = 0, and ( R ijkl (p) = g iq k Γ q jl (p) lγ q jk (p)) = k Γ i,jl (p) l Γ i,jk (p) = 1 ( ) k l g ij + k j g li k i g jl l k g ij l j g ki + l i g jk (21) 2 = 1 ( ) k j g li k i g jl l j g ki + l i g jk. 2 The symmetry properties of the Riemann tensor straightforwardly follow from the formula. (c) 5 marks. Show that a 2-dimensional Riemann manifold with the line-element ds 2 = e 2φ (dx 2 + dy 2 ) has constant scalar curvature R if and only if φ satisfies the Liouville equation 2 φ x φ y 2 + Ke2φ = 0. where K is a constant which has to be related to R. Solution: The connection does not vanish only in the following cases Γ i ii = i φ, Γ k ii = k φ, Γ k ik = i φ = Γ k ki, k i. (22) Thus R = 1 Γ Γ Γ 1 p1γ p 22 Γ 1 p2γ p 21 = 2 1φ 2 2φ ( 1 φ) 2 + ( 2 φ) 2 ( 2 φ) 2 + ( 1 φ) 2 = 2 1φ 2 2φ. (23) Now, taking into account that in 2D R = g 11 R 1212 = 1 2 g11 gr = 1 2 e2φ R one gets the statement. (d) 5 marks. connection Prove Bianchi s identities for the curvature tensor of a symmetric m R n ikl + l R n imk + k R n ilm = 0. Solution: We use normal coordinate, so that Γ k li = 0 at a given point. Then we have at Page 6 of 12

7 this point R n ikl = k Γ n il l Γ n ik + Γ n pkγ p il Γn plγ p ik m R n ikl = m k Γ n il m l Γ n ik l R n imk = l k Γ n im + l m Γ n ik, (24) Thus k R n ilm = k m Γ n il + k l Γ n im. m R n ikl + l R n imk + k R n ilm = m R n ikl + l R n imk + k R n ilm = 0. (25) Since Bianchi s identities are tensor they hold in any coordinate system. 3. (a) 5 marks. Define a general differentiable n-dimensional manifold. Solution: A differentiable n-dimensional manifold is a set M (whose elements we call points ) together with the following structure on it. The set M is the union of a finite or countably infinite collection of subsets U q with the following properties (i) Each subset U q has defined on it coordinates x α q, α = 1,..., n (called local coordinates) by virtue of which U q is identifiable with a region of Euclidean n-space R n with Euclidean coordinates x α q. The U q with their coordinate systems are called charts or local coordinate neighbourhoods. (ii) Each non-empty intersection U p U q of a pair of charts thus has defined on it two coordinate systems, the restrictions of (x α p ) and (x α q ). It is required that under each of these coordinatisations the intersection U p U q is identifiable with a region of R n, and that each of these coordinate systems be expressible in terms of the other in a one-to-one differentiable manner. Thus, if the transition functions from x α q to x α p and back are given by x α p = x α p (x 1 q,..., x n q ), α = 1,..., n, x α q = x α q (x 1 p,..., x n p), α = 1,..., n, (26) then in particular the Jacobian det( x α p / x β q ) is nonzero on U p U q. The general smoothness class of the transition functions for all intersecting pairs U p, U q is called the smoothness class of the manifold M with its accompanying atlas of charts U q. Page 7 of 12

8 (b) Consider a surface M in n-dimensional Euclidean space R n defined by a set of n m equations f i (x 1,..., x n ) = 0, i = 1,..., n m, where f i are smooth functions such that for all x the matrix ( f i x j ) has rank n m. i. 8 marks. Prove that M is a smooth m-dimensional manifold. Solution: Let J j1...j n m be the minor J j1...j n m made up of those columns of the matrix ( f i / x j ) indexed by j 1,..., j n m. Let U j1...j n m be a set of all points of the surface at which the minor J j1...j n m does not vanish. Since the surface is nonsingular, it is covered by the regions U j1...j n m. Let x 0 (x 1 0,..., x n 0) M. If at x 0 the minor J j1...j n m is nonzero, then as local coordinates on a neighbourhood of the surface about the point we take (y 1,..., y m ) = (x 1,..., ˇx j 1,..., ˇx j n m,..., x n ), (27) where the checked symbols are to be omitted. To show that the covering of M by the charts U j1...j n m defines on M the structure of a smooth manifold we need to show that the transition functions are one-to-one and smooth. Since J j1...j n m 0 on U j1...j n m one can solve the equations f i = 0 for x j i x j i = ϕ i (y 1,..., y m ), i = 1,... n m, (28) where ϕ i are smooth functions. Similarly, in the chart U s1...s n m with coordinates (z 1,..., z m ) = (x 1,..., ˇx s 1,..., ˇx s n m,..., x n ), (29) we have x j i = ψ i (y 1,..., y k ), i = 1,... n m, (30) where ψ i are smooth functions. In the intersection U j1...j n m and U s1...s n m we have the following transition functions y z and z y (where for simplicity Page 8 of 12

9 we assume 1 < j 1 < s 1 < j 2 < ; the general case is clear from this): y 1 = z 1 (= x 1 ) y j 1 1 = z j 1 1 (= x j 1 1 ) ϕ 1 (y 1,..., y m ) = z j 1 (= x j 1 ) y j 1 = z j 1+1 (= x j 1+1 ) y s 1 2 = z s 1 1 (= x s 1 1 ) y s 1 1 = ψ 1 (z 1,..., z m ) (= x s 1 ) y s 1 = z s 1 (= x s 1+1 ) y m = z m (= x m ) (31) It si obvious that the two transition functions shown above are mutual inverses, completing the proof. ii. 7 marks. Let ξ and η be two vector fields on R n. Define the commutator (Lie Bracket) of the vector fields ξ and η. Show that if the vector fields ξ and η are both tangent to M, then their commutator is also tangent to M. What does this say about the linear space of vector fields tangent to a surface? Solution: The commutator of two vector fields ξ and η on a manifold is defined by where y i are local coordinates on the manifold. Locally M is given by [ξ, η] i = ξ j ηi ξi ηj yj y, (32) j y m+1 = 0,..., y n = 0, (33) where y 1, y n are local (non-euclidean) coordinates of R n, and y 1, y m serve as local coordinates on M. Any vector field ξ tangent to M has ξ m+1 = 0,..., ξ n = 0 at any point of M. Thus, ξ i y j = 0, i = m + 1,..., n, j = 1,..., m. The commutator of the vector fields ξ and η tangent to M is [ξ, η] i = ξ j ηi ξi ηj yj y = m (ξ j ηi ξi j ηj ) = 0 for i = m + 1,..., n. yj yj j=1 Page 9 of 12 (34)

10 Thus, [ξ, η] is tangent to M. This means that the linear space of vector fields tangent to a smooth surface is a subalgebra of the Lie algebra of all vector fields. 4. (a) 5 marks. Define a group. Define a Lie group. Define a Lie algebra. Solution: Definition 1. A group is a nonempty set G on which there is defined a binary operation (a, b) ab satisfying the following properties. Closure: If a and b belong to G, then ab is also in G. Associativity : a(bc) = (ab)c for all a, b, c G. Identity: There is an element 1 G such that a1 = 1a = a for all a in G. Inverse: If a G, then there is an element a 1 G: aa 1 = a 1 a = 1. Definition 2. A manifold G is called a Lie group if it has given on it a group operation with the property that the maps ϕ : G G, defined by ϕ(g) = g 1 (i.e. the taking of inverses) and ψ : G G G defined by ψ(g, h) = gh (i.e. the group multiplication), are smooth maps. Definition 3. A Lie algebra is a vector space G over a field F with a bilinear operation [, ]: G G G which is called a commutator or a Lie bracket, such that the following axioms are satisfied: It is skew symmetric: [x, x] = 0 which implies [x, y] = [y, x] for all x, y G. It satisfies the Jacobi Identity: [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0. (b) 5 marks. Choose a basis of the Lie algebra sl(2, R) of the Lie group SL(2, R), and calculate the commutation relations the basis matrices satisfy. Solution: The sl(2, R) algebra is the space of real traceless matrices. We choose as its basis the matrices L 3 = , L 1 = , L 2 = (35) Page 10 of 12

11 They satisfy the sl(2, R) Lie algebra commutation relations [L 3, L 1 ] = 2L 2, [L 3, L 2 ] = 2L 1, [L 1, L 2 ] = 2L 3. (36) (c) Consider the set SL(2, R) of all transformations of the real line of the form x x + 2πa + 1 i ln 1 ze ix 1 ze ix, where x R, a R, z C, z < 1 and ln is the main branch of the natural logarithmic function, i.e. the continuous branch determined by ln 1 = 0. i. 5 marks. Show that SL(2, R) is a connected 3-dimensional Lie group. Solution: It is obvious that SL(2, R) is a connected 3-dimensional manifold because it is R D 2 where D 2 is the disk z < 1. To show that it is a Lie group we consider the composition of two transformations with parameters a i, z i, i = 1, 2. We get T g2 T g1 (x) = T g2 (x + 2πa i ln 1 z 1e ix 1 z 1 e ix ) = x + 2πa i ln 1 z 1e ix 1 z 1 e + 2πa ix i ln 1 z 2e i(x+2πa1+ 1 i ln 1 z 1 e ix 1 z 1 e ix ) = x + 2π(a 1 + a 2 ) + 1 i ln 1 z 1e ix 1 z 2 1 z 1 e ix 1 z 1 z 1 e ix 2 1 z 2 e i(x+2πa 1+ 1 i ln 1 z 1 e ix 1 z 1 e ix ) 1 z 1 e ix 1 z 1 e ix e i(x+2πa 1) 1 z 1 e ix e i(x+2πa 1) = x + 2π(a 1 + a 2 ) + 1 i ln 1 z 1e ix z 2 (1 z 1 e ix )e i(x+2πa 1) 1 z 1 e ix z 2 (1 z 1 e ix )e i(x+2πa 1) = x + 2π(a 1 + a 2 ) + 1 i ln 1 + z 2 z 1 e 2πia 1 (z 1 + z 2 e 2πia 1 )e ix 1 + z 2 z 1 e 2πia 1 ( z1 + z 2 e 2πia 1 )e ix = x + 2π(a 1 + a 2 ) + 1 i ln 1 + z 2 z 1 e 2πia 1 (z 1 + z 2 e 2πia 1 )e ix 1 + z 2 z 1 e 2πia 1 ( z1 + z 2 e 2πia 1 )e ix = x + 2π(a 1 + a 2 ) + 1 i ln 1 + z 2 z 1 e 2πia z 2 z 1 e 2πia 1 = x + 2πa i ln 1 z 21e ix 1 z 21 e ix = T g 2 g 1 (x), where the parameters a 21 and z 21 of T g2 g i ln 1 z 1+z 2 e 2πia1 1+z 2 z 1 e 2πia 1 e ix 1 z 1+ z 2 e 2πia 1 1+ z 2 z 1 e 2πia 1 eix are given by (37) a 21 = a 1 + a πi ln 1 + z 2 z 1 e 2πia1, z 1 + z 2 z 1 e 2πia 21 = z 1 + z 2 e 2πia z 2 z 1 e. (38) 2πia 1 Note that z 21 < 1 if z 1 < 1 and z 2 < 1. Page 11 of 12

12 ii. 5 marks. Calculate the Lie algebra of SL(2, R) and show that it is isomorphic to sl(2, R). Solution: Since the identity element of SL(2, R) corresponds to a = z = 0, one gets expanding (38) a 21 = a 1 +a πi (z 2 z 1 z 2 z 1 )+..., z 21 = z 1 +z 2 2πia 1 z (39) Introducing z k = u k + iv k, k = 1, 2, one gets a 21 = a 1 +a π (v 2u 1 u 2 v 1 )+..., u 21 = u 1 +u 2 +2πa 1 v , v 21 = v 1 +v 2 2πa 1 u (40) Now, choosing x 1 = u, x 2 = πa, x 3 = v, as the local coordinates, we get the nonzero structure constants of the Lie algebra c 2 31 = 2, c 1 32 = 2, c 3 12 = 2. (41) Thus, the Lie algebra of SL(2, R) is [J 3, J 2 ] = 2J 1, [J 3, J 1 ] = 2J 2, [J 1, J 2 ] = 2J 3, (42) which the same as (36). Page 12 of 12