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Homework 9. Solutions. 1 Let V Be a Connection on N-Dimensional

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Homework 9. Solutions. 1 Let V Be a Connection on N-Dimensional Homework 9. Solutions. r f i g 1 Let be a connection on n-dimensional manifold M and R rmn be the components of the curvature tensor of a connection r in local coordinates (x1; x2; : : : ; xn). a) For arbitrary vector fields A; B and D calculate the vector field (rArB − rBrA) D − rCD ; where the vector field C is a commutator of vector fields A and B: ( ) @ @Bi(x) @Ai(x) @ C = Ci = [A; B] = Am − Bm : (1:0) @xi @xm @xm @xi b) Calculate the vector field (rArB − rBrA) D in the case if for vector fields A and B components Ai and Bm are constants (in the local coordinates (x1; : : : ; xn)) (You have to express the answers in terms of components of the vector fields and components of the i curvature tensor R rmn.) m m a) According to the definition of the curvature tensor for every vector fields X = X @m; Y = Y @m m and Z = Z @m we have that R(X; Y)Z = ( ) R m n r r r − r r − r r i m n (X @m;Y @n)(Z @r) = X Y Y X [X;Y] Z = Z R rmnX Y @i : Hence for vector fields A; B; C = [A; B] and we have that ( ) r r − r r − r r r − r r − r R r i m n A B B A [A;B] D = ( A B B A) D CD = (A; B)D = D R rmnA B @i : (1:1) b) in the case if in the local coordinates (x1; : : : ; xn) for vector fields A and B components Ai and Bm are constants then the commutator of these vector fields C = [A; B] vanishes: C = 0 (see the formula (1.0)). Hence according to the formula (1.1) above we have that r r − r r − r r r − r r R r i m n ( A B B A) D CD = ( A B B A) D = (A; B)D = D R rmnA B @i : 2) Calculate Riemann curvature tensor for the cylindircal surface( x2 + y2 = a2 in E3. x = a cos ' Solution. Cylinder surface in standard coordinates ('; h) is y = a sin ' and Riemannian metric G = z = h 2 2 2 i dh + a d' . Christoffel symbols Γkm of Levi-Civita connection in coordinates ('; h) vanish since in these coordinates all components of matrix of metric are constants (see Levi-Civita formula!). Hence Riemann curvature tensor i i i p − i − i p R kmn = @mΓnk + ΓmpΓnk @nΓmk ΓnpΓmk vanishes too. Note that the fact that Christoffel symbols vanish in some coordiantes does not mean that they vanish in any coordinates. Riemann curvature is a tensor: if it vanishes in some coordinates then it vanishes in any coordinates. i i 3 We know that If Rkmn is Riemann curvature tensor for Riemannian manifold (M; G) (Rkmn us curvature tensor for Levi-Civita connection on M) then the following identities hold: Rikmn = −Riknm;Rikmn = −Rkimn ;Rikmn = Rmnik (∗) a) Show that Riemann curvature tensor for 2-dimensional Riemannian manifold (M; G) possesses only one non-trivial component. 1 Solution. It follows from identities (*) that all components of curvature tensor may be expressed through component R1212. Indeed if R1212 = a then due to identities: R1212 = −R1221 = −R2112 = R2121 = a and all other components (for 2-dimensional case) vanish: R1111 = R1112 = R1121 = R1122 = R1211 = R1222 = R2111 = R2122 = R2211 = R2212 = R2221 = R2222 = 0 : 4) If (M; G) is surface in E3 then R R K = = 1212 ; 2 det g i where K is Gaussian curvature of the surface, and Rkmn Riemann curvature tensor with respect to induced metric, R R1212 a) Prove by straightforward calculations that 2 = det g ∗ R R1212 f g b ) Prove that K = 2 = det g . (It is convenient to choose the orthonormal basis e; f and use derivation formula.) Solution: Show it. Using results of previous exercise we see that i 2 22 21 22 22 R11 = R 1i1 = R 121 = g R2121 + g R1121 = g R1212 = g a ; i 1 11 12 11 11 R22 = R 2i2 = R 212 = g R1212 + g R2221 = g R1212 = g a i 1 12 − 12 − 12 R12 = R21 = R 1i2 = R 112 = g R2112 = g R1212 = g a Thus ( ) g22 −g12 R = a : ik −g21 g11 Hence i km km 11 12 21 22 R = R kimg = Rkmg = g R11 + g R12 + g R21 + g R22 = 2a 2R (g11g22 − g12g12) = 2a det g−1 = ; 1212 det g ( ) g11 g12 − 2 jj jj jj jj where det g = det = g11g22 g12 (matrices gik and gik are inverse). We see that R = g21 g22 2a 2R1212 det g = det g . ∗ It remains to prove that Gaussian curvature is equal to R=2 or that it is equal to R1212 det g. Choose three unit vector fields e; f; g such that e; f form orthonormal basis on points of surface M and n = e × f. According to the definition of curvature calculate R(e; f)e = rerf e − rf ree − r[e;f]e : (∗∗) Using derivation formulae one can calculate this expression and come to the following answer 1): R(e; f)e = rerf e − rf ree − r[e;f]e = da(e; f)f : 1) These calculations are little bit difficult. They are following: note that since the induced connection is symmetrical connection then: ref − rf e − [e; f] = 0 hence [e; f] = ref − rf e = −a(e)e − a(f)f. Thus we see that R(e; f)e = rerf e − rf ree − r[e;f]e = re (a(f)f) − rf (a(e)f) + ra(e)e+a(f)f e = @ea(f)f + a(f)ref − @f a(e)f − a(e)rf f + a(e)ree + a(f)rf e = @ea(f)f − a(f)a(e)e − @f a(e)f + a(e)a(f)e + a(e)a(e)f + a(f)a(f)f = [@ea(f)f − @f a(e)f − a [−a(e)e − a(f)f]] f = = [@ea(f)f − @f a(e)f − a ([e; f])] f = da(e; f)f : 2 Recall that we established that for Gaussian curvature K = b ^ c(e; f) = −da(e; f). Hence we come to the relation: R(e; f)e = da(e; f) = −Kf : This means that 2 − R112 = K ( ) 1 0 Note that in the basis e; f Riemannian metric is . Hence R2 = R = −R , det g = 1. Thus 0 1 112 2112 1212 we come to the relation R R K = 1212 = : det g 2 3.
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