Homework 9. Solutions. ∇ { i } 1 Let be a connection on n-dimensional manifold M and R rmn be the components of the curvature tensor of a connection ∇ in local coordinates (x1, x2, . . . , xn). a) For arbitrary vector fields A, B and D calculate the vector field

(∇A∇B − ∇B∇A) D − ∇CD ,

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 (∇A∇B − ∇B∇A) 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 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 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 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 φ Solution. Cylinder surface in standard coordinates (φ, h) is y = a sin φ and Riemannian metric G = z = h 2 2 2 i dh + a dφ . Christoffel symbols Γkm of Levi-Civita connection in coordinates (φ, 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 { } 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 || || || || 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 = ∇e∇f e − ∇f ∇ee − ∇[e,f]e . (∗∗) Using derivation formulae one can calculate this expression and come to the following answer 1):

R(e, f)e = ∇e∇f e − ∇f ∇ee − ∇[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: ∇ef − ∇f e − [e, f] = 0 hence [e, f] = ∇ef − ∇f e = −a(e)e − a(f)f. Thus we see that R(e, f)e =

∇e∇f e − ∇f ∇ee − ∇[e,f]e = ∇e (a(f)f) − ∇f (a(e)f) + ∇a(e)e+a(f)f e =

∂ea(f)f + a(f)∇ef − ∂f a(e)f − a(e)∇f f + a(e)∇ee + a(f)∇f 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