Homework 5. Solutions 1 Calculate the Christoffel Symbols of The

Home , Christoffel symbols

Homework 5. Solutions 1 Calculate the Christoﬀel symbols of the canonical ﬂat connection in E3 in a) cylindrical coordinates (x = r cos ϕ, y = r sin ϕ, z = h), b) spherical coordinates. r r r r r (For the case of sphere try to make calculations at least for components Γrr, Γrθ, Γrϕ, Γθθ, ..., Γϕϕ)

In cylindrical coordinates (r, ϕ, h) we have

(  p 2 2 x = r cos ϕ  r = x + y y = r sin ϕ and y ϕ = arctan x z = h  h = z

We know that in Cartesian coordinates all Christoﬀel symbols vanish. Hence in cylindrical coordinates (see in detail lecture notes): ∂2x ∂r ∂2y ∂r ∂2z ∂r Γr = + + = 0 , rr ∂2r ∂x ∂2r ∂y ∂2r ∂z ∂2x ∂r ∂2y ∂r ∂2z ∂r Γr = Γr = + + = − sin ϕ cos ϕ + sin ϕ cos ϕ = 0 . rϕ ϕr ∂r∂ϕ ∂x ∂r∂ϕ ∂y ∂r∂ϕ ∂z ∂2x ∂r ∂2y ∂r ∂2z ∂r x y Γr = + + = −x − y = −r . ϕϕ ∂2ϕ ∂x ∂2ϕ ∂y ∂2ϕ ∂z r r ∂2x ∂ϕ ∂2y ∂ϕ ∂2z ∂ϕ Γϕ = + + = 0 . rr ∂2r ∂x ∂2r ∂y ∂2r ∂z ∂2x ∂ϕ ∂2y ∂ϕ ∂2z ∂ϕ −y x 1 Γϕ = Γϕ = + + = − sin ϕ + cos ϕ = ϕr rϕ ∂r∂ϕ ∂x ∂r∂ϕ ∂y ∂r∂ϕ ∂z r2 r2 r ∂2x ∂ϕ ∂2y ∂ϕ ∂2z ∂ϕ −x y Γϕ = + + = −x − y = 0 . ϕϕ ∂2ϕ ∂x ∂2ϕ ∂y ∂2ϕ ∂z r2 r2 · · All symbols Γ·h, Γh· vanish

r r r r r ϕ Γrh = Γhr = Γhh = Γϕh = Γhϕ = Γhr = dots = 00

∂2x ∂2y ∂2z since ∂h∂... = ∂h∂... = ∂h∂... = 0 h h ∂2z ∂h ∂h ∂h ∂2z For all symbols Γ·· Γ·· = ∂·∂· since ∂x = ∂y = 0 and ∂y = 1. On the other hand all ∂·∂· vanish. Hence h all symbols Γ·· vanish. b) spherical coordinates

 p ( r = x2 + y2 + z2 x = r sin cos ϕ  θ = arccos √ z y = r sin sin ϕ x2+y2+z2 z = r cos θ  y ϕ = arctan x

Perform now brute force calculations only for some components ∗ r ∂2xi Γrr = 0 since ∂2r = 0.

∂2x ∂r ∂2y ∂r ∂2z ∂r x y z Γr = Γr = + + = cos θ cos ϕ + cos θ sin ϕ − sin θ = 0 , rθ θr ∂r∂θ ∂x ∂r∂θ ∂y ∂r∂θ ∂z r r r

∗ they can be quickly calculated using Lagrangian of free particle.

1 ∂2x ∂r ∂2y ∂r ∂2z ∂r x y z Γr = + + = −r sin θ cos ϕ − r sin θ sin ϕ − r cos θ = −r θθ ∂2θ ∂x ∂2θ ∂y ∂2θ ∂z r r r ∂2x ∂r ∂2y ∂r ∂2z ∂r x y Γr = Γr = + + = − sin θ sin ϕ + sin θ cos ϕ = 0 rϕ ϕr ∂r∂ϕ ∂x ∂r∂ϕ ∂y ∂r∂ϕ ∂z r r and so on.... 2 Let ∇ be an affine connection on a 2-dimensional manifold M such that in local coordinates (u, v) it u v is given that Γuv = v, Γuv = 0. ∂ Calculate the vector field ∇ ∂ u . ∂u ∂v Using the properties of connection and definition of Christoffel symbols have

∂ ∂ ∂ ∇ ∂ u = ∂ ∂ (u) + u∇ ∂ = ∂u ∂v ∂u ∂v ∂u ∂v

∂ ∂ ∂ ∂ ∂ ∂ ∂ + u Γu + Γv = + u v + 0 = + uv . ∂v uv ∂u uv ∂v ∂v ∂u ∂v ∂u

3 Let ∇ be an aﬃne connection on the 2-dimensional manifold M such that in local coordinates (u, v) ∂ 2 ∂ ∂ ∇ ∂ u = (1 + u ) + u . ∂u ∂v ∂v ∂u

u v Calculate the Christoﬀel symbols Γuv and Γuv of this connection. ∂ ∂ Using the properties of connection we have ∇ ∂ u = u∇ ∂ + ∂u ∂v ∂u ∂v ∂ u ∂ v ∂ ∂ v ∂ u ∂ 2 ∂ ∂ ∂ ∂ (u) = u Γuv + Γuv + 1 · = (1 + uΓuv) + uΓuv = 1 + u + u . ∂u ∂v ∂u ∂v ∂v ∂v ∂u ∂v ∂u

2 v v v u Hence 1 + u = 1 + uΓuv and uΓuv = u, i.e. Γuv = u and Γuv = 1. 4 a) Consider a connection such that its Christoﬀel symbols are symmetric in a given coordinate system: i i Γkm = Γmk. Show that they are symmetric in an arbitrary coordinate system. b∗) Show that the Christoﬀel symbols of connection ∇ are symmetric (in any coordinate system) if and only if

∇XY − ∇YX − [X, Y] = 0 , for arbitrary vector ﬁelds X, Y. c)∗ Consider for an arbitrary connection the following operation on the vector ﬁelds:

S(X, Y) = ∇XY − ∇YX − [X, Y] and ﬁnd its properties. Solution i i i0 i0 a) Let Γkm = Γmk. We have to prove that Γk0m0 = Γm0k0 We have i0 k m r i0 i0 ∂x ∂x ∂x i ∂x ∂x Γ 0 0 = Γ + . (1) k m ∂xi ∂xk0 ∂xm0 km ∂xk0 ∂xm0 ∂xr Hence i0 m k r i0 i0 ∂x ∂x ∂x i ∂x ∂x Γ 0 0 = Γ + m k ∂xi ∂xm0 ∂xk0 mk ∂xm0 ∂xk0 ∂xr 2 i i ∂xr ∂xr But Γ = Γ and 0 0 = 0 0 . Hence km mk ∂xm ∂xk ∂xk ∂xm

i0 m k r i0 i0 m k r i0 i0 ∂x ∂x ∂x i ∂x ∂x ∂x ∂x ∂x i ∂x ∂x i0 Γ 0 0 = Γ + = Γ + = Γ 0 0 . m k ∂xi ∂xm0 ∂xk0 mk ∂xm0 ∂xk0 ∂xr ∂xi ∂xm0 ∂xk0 km ∂xk0 ∂xm0 ∂xr k m

b) The relation

∇XY − ∇YX − [X, Y] = 0 holds for all ﬁelds if and only if it holds for all basic ﬁelds. One can easy check it using axioms of connection ∂ ∂ (see the next part). Consider X = ∂xi , Y = ∂xj then since [∂i, ∂j] = 0 we have that

k k k k ∇XY − ∇YX − [X, Y] = ∇i∂j − ∇j∂i = Γij∂k − Γji∂k = (Γij − Γji)∂k = 0

k k We see that commutator for basic ﬁelds ∇XY − ∇YX − [X, Y] = 0 if and only if Γij − Γji = 0. c) One can easy check it by straightforward calculations or using axioms for connection that S(X, Y) is a vector-valued bilinear form on vectors. In particularly S(fX,Y ) = fS(X, Y) for an arbitrary (smooth) function. Show this just using axioms deﬁning connection:

S(fX,Y ) = ∇fXY − ∇Y(fX) − [fX, Y] = f∇X Y − f∇YX − ∂YfX + [Y, fX] =

f∇X Y − f∇YX − (∂Yf)X + ∂YfX + f [Y, X] = f(∇X Y − ∇YX − [X, Y]) = fS(X, Y)

(1) i (2) i 5 Let ∇1, ∇2 be two different connections. Let Γkm and Γkm be the Christoffel symbols of connections ∇1 and ∇2 respectively. i (1) i (2) i a) Find the transformation law for the object : Tkm = Γkm − Γkm under a change of coordinates. 1 Show that it is tensor. 2 ∗? b) Consider an operation ∇1 − ∇2 on vector fields and find its properties. Christoffel symbols of both connections transform according the law (1). The second term is the same. Hence it vanishes for their difference:

i0 k m i0 k m i0 (1) i0 (2) i0 ∂x ∂x ∂x (1) i (2) i ∂x ∂x ∂x i T 0 0 = Γ 0 0 − Γ 0 0 = Γ − Γ = T k m k m k m ∂xi ∂xk0 ∂xm0 km km ∂xi ∂xk0 ∂xm0 km i0 1 We see that T0 0 transforms as a tensor of the type . km 2

b) One can do it in invariant way. Using axioms of connection study T = ∇1 − ∇2 is a vector ﬁeld. Consider

T (X, Y) = ∇1XY − ∇2XY

Show that T (fX, Y) = fT (X, Y) for an arbitrary (smooth) function, i.e. it does not possesses derivatives:

T (fX, Y) = ∇1fX Y − ∇2fX Y = (∂Xf)Y + f∇1XY − (∂Xf)Y − f∇2XY = fT (X, Y).

∗ i 6 a) Consider tm = Γim. Show that the transformation law for tm is

0 ∂ xm ∂2xr ∂xk t 0 = t + . m ∂xm0 m ∂xm0 ∂xk0 ∂xr 3 b) † Show that this law can be written as

∂ xm ∂ ∂x t 0 = t + log det . m ∂xm0 m ∂xm0 ∂x0

Solution. Using transformation law (1) we have

i0 k m r i0 i0 ∂x ∂x ∂x i ∂x ∂x t 0 = Γ 0 00 = Γ + m i m ∂xi ∂xi0 ∂xm0 km ∂xi0 ∂xm0 ∂xr

0 ∂xi ∂xk k We have that 0 = δ . Hence ∂xi ∂xi i

i0 k m r i0 m r i0 m r i0 i0 ∂x ∂x ∂x i ∂x ∂x k ∂x i ∂x ∂x ∂x ∂x ∂x t 0 = Γ 0 00 = Γ + = δ Γ + = t + . m i m ∂xi ∂xi0 ∂xm0 km ∂xi0 ∂xm0 ∂xr i ∂xm0 km ∂xi0 ∂xm0 ∂xr ∂xm0 m ∂xi0 ∂xm0 ∂xr

† ∂ ∂x b) When calculating 0 log det use very important formula: ∂xm ∂x0

δ det A = det ATr (A−1δA) → δ log det A = Tr (A−1δA) .

Hence 0 ∂ ∂x ∂xi ∂2xr log det = ∂xm0 ∂x0 ∂xr ∂xi0 ∂xm0 and we come to transformation law for (1). To deduce the formula for δ det A notice that

det(A + δA) = det A det(1 + A−1δA) and use the relation: det(1 + δA) = 1 + Tr δA + O(δ2A)

7 Consider the surface M in the Euclidean space En. Calculate the induced connection in the following cases a) M = S1 in E2, b) M— parabola y = x2 in E2, c) cylinder in E3. d) cone in E3. e) sphere in E3. f) saddle z = xy in E3 Solution. a) Consider polar coordinate on S1, x = R cos ϕ, y = R sin ϕ. We have to define the connection on S1 2 ∂ ϕ induced by the canonical flat connection on E . It suffices to define ∇ ∂ = Γ ∂ϕ. ∂ϕ ∂ϕ ϕϕ Recall the general rule. Let r(uα): xi = xi(uα) is embedded surface in Euclidean space En. The basic ∂ ∂r(u) vectors ∂uα = ∂uα . To take the induced covariant derivative ∇XY for two tangent vectors X, Y we take a usual derivative of vector Y along vector X (the derivative with respect to canonical flat connection: in Cartesian coordiantes is just usual derivatives of components) then we take the tangent component of the answer, since in general derivative of vector Y along vector X is not tangent to surface:

2 ∂ γ ∂ (canonical) ∂ ∂ r(u) ∇ ∂ = Γ = ∇ = ∂uα β αβ γ ∂α β α β ∂u ∂u ∂u tangent ∂u ∂u tangent

4 ∂ (∇canonical ∂α ∂uβ ) is just usual derivative in Euclidean space since for canonical connection all Christoﬀel symbols vanish.) In the case of 1-dimensional manifold, curve it is just tangential acceleration!:

2 ∂ u ∂ (canonical) ∂ d r(u) ∇ ∂ = Γ = ∇ = = atangent ∂u uu ∂u 2 ∂u ∂u ∂u tangent du tangent

For the circle S1,(x = R cos ϕ, y = R sin ϕ), in E2. We have

∂ ∂x ∂ ∂y ∂ ∂ ∂ r = = + = −R sin ϕ + R cos ϕ , ϕ ∂ϕ ∂ϕ ∂x ∂ϕ ∂y ∂x ∂y

∂ ϕ (canonic.) ∂ ∇ ∂ = Γ ∂ϕ = ∇ ∂ϕ = rϕ = ∂ϕ ϕϕ ∂ϕ ∂ϕ tangent ∂ϕ tangent ∂ ∂ ∂ ∂ ∂ ∂ (−R sin ϕ) + (R cos ϕ) = −R cos ϕ − R sin ϕ = 0, ∂ϕ ∂x ∂ϕ ∂y tangent ∂x ∂y tangent

∂ ∂ since the vector −R cos ϕ ∂x −R sin ϕ ∂y is orthogonal to the tangent vector rϕ. In other words it means that acceleration is centripetal: tangential acceleration equals to zero. ϕ We see that in coordinate ϕ,Γϕϕ = 0. Additional work: Perform calculation of Christoﬀel symbol in stereographic coordinate t:

2tR2 R(t2 − R2) x = , y = . R2 + t2 t2 + R2

In this case ∂ ∂x ∂ ∂y ∂ 2R2 ∂ ∂ r = = + = (R2 − t2) + 2tR , t ∂t ∂t ∂x ∂t ∂y (R2 + t2)2 ∂x ∂x ∂ t (canonic.) ∂ ∇ ∂ = Γ ∂t = ∇ ∂t = rt = (rtt) = ∂t tt ∂t tangent ∂t tangent ∂t tangent 4t 2R2 ∂ ∂ − 2 2 rt + 2 2 2 −2t + 2R t + R (R + t ) ∂x ∂y tangent

In this case rtt is not orthogonal to velocity: to calculate (rtt)tangent we need to extract its orthogonal component:

(rtt)tangent = rtt − hrtt, ntin We have r 1 n = = 2tR∂ + (t2 − R2)∂ , t |r| R2 + t2 x y

−4R3 where hrt, ni = 0. Hence hrtt, nti = (t2+R2)2 and

(rtt)tangent = rtt − hrtt, ntin =

4t 2R2 ∂ ∂ 4R3 1 −2t − r + −2t + 2R + · 2tR∂ + (t2 − R2)∂ = r t2 + R2 t (R2 + t2)2 ∂x ∂y (t2 + R2)2 R2 + t2 x y t2 + R2 t We come to the answer: −2t −2t ∇ ∂ = ∂ , i.e.Γt = ∂t t t2 + R2 t tt t2 + R2

5 Of course we could calculate the Christoﬀel symbol in stereographic coordinates just using the fact that we ϕ already know the Christoﬀel symbol in polar coordinates: Γϕϕ = 0, hence

dt dϕ dϕ d2ϕ dt d2ϕ dt Γt = Γϕ + = tt dϕ dx dx ϕϕ dt2 dϕ dt2 dϕ

π ϕ t π It is easy to see that t = R tan 4 + 2 , i.e. ϕ = 2 arctan R − 2 and

2 2 d ϕ d ϕ dt 2 2t Γt = = dt = − . tt dt2 dϕ dϕ t2 + R2 dt

b) For parabola x = t, y = t2

∂ ∂x ∂ ∂y ∂ ∂ ∂ r = = + = + 2t , t ∂t ∂t ∂x ∂t ∂y ∂x ∂y

∂ t (canonic.) ∂ ∂ ∇ ∂ = Γ ∂t = ∇ ∂t = rt = (rtt) = 2 ∂t tt ∂t tangent ∂t tangent ∂t tangent ∂y tangent

To calculate (rtt)tangent we need to extract its orthogonal component: (rtt)tangent = rtt − hrtt, ntin, where n is an orthogonal unit vector: hn, rti = 0, hn, ni = 1:

1 nt = √ (−2t∂x + ∂y) . 1 + 4t2

We have

1 1 (rtt) = rtt − hrtt, ntin = 2∂y − 2∂y, √ (−2t∂x + ∂y) √ (−2t∂x + ∂y) = tangent 1 + 4t2 1 + 4t2

4t 8t2 4t 4t ∂ + ∂ = (∂ + 2t∂ ) = ∂ 1 + 4t2 x 1 + 4t2 y 1 + 4t2 x y 1 + 4t2 t We come to the answer: 4t 4t ∇ ∂ = ∂ , i.e.Γt = ∂t t 1 + 4t2 t tt 1 + 4t2 Remark Do not be surprised by resemblance of the answer to the answer for circle in stereographic coordinates.

c) Cylinder ( x = a cos ϕ r(h, ϕ): y = a sin ϕ . z = h  0   −a sin ϕ  ∂h = rh =  0 , ∂ϕ = rϕ =  a cos ϕ  1 0 Calculate 2 h ϕ ∂ r ∇∂h ∂h = Γhh∂h + Γhh∂ϕ = 2 = 0 since rhh = 0. ∂h tangent

h ϕ Hence Γhh = Γhh = 0

2 h ϕ ∂ r ∇∂h ∂ϕ = ∇∂ϕ ∂h = Γhϕ∂h + Γhϕ∂ϕ = = 0 since rhϕ = 0 ∂h∂ϕ tangent

6 h h ϕ ϕ Hence Γhϕ = Γϕh = Γhϕ = Γϕh = 0.   2 −a cos ϕ h ϕ ∂ r ∇∂ϕ ∂ϕ = Γϕϕ∂h + Γϕϕ∂ϕ = =  −a sin ϕ  = 0 ∂ϕ∂ϕ tangent 0 tangent

 −a cos ϕ  h h ϕ since the vector rϕϕ =  −a sin ϕ  is orthogonal to the surface of cylinder. Hence Γhϕ = Γϕh = Γhϕ = 0 ϕ Γϕh = 0 We see that for cylinder all Christoﬀel symbols in cylindrical coordinates vanish. This is not big surprise: in cylindrical coordinates metric equals dh2 = a2dϕ2. This due to Levi-Civita theorem one can see that Levi- Civita connection which is equal to induced connection vanishes since all coeﬃcients are constants.

d) Cone ( x = kh cos ϕ For cone: x2 + y2 = k2z2 we have r(h, ϕ) = y = kh sin ϕ z = h

 k cos ϕ   −kh sin ϕ   cos ϕ  ∂ ∂ 1 = rh =  k sin ϕ  , = rϕ =  kh cos ϕ  , n = √  sin ϕ  ∂h ∂ϕ 2 1 0 1 + k −k

h ϕ We have rhh = 0, hence ∇∂h ∂h = 0. i.e. Γhh = Γhh = 0.  −k sin ϕ  rϕ rϕ We have that rhϕ = rϕh =  k cos ϕ  = h , i.e. ∇∂h ∂ϕ = ∇∂ϕ ∂h = h : 0

1 Γϕ = Γϕ = , Γh = Γh . hϕ ϕ,h h hϕ ϕh

 −kh cos ϕ  Now calculate rϕϕ: rϕϕ =  −kh sin ϕ . This vector is neither tangent to the cone nor orthogonal to the 0 cone: 0 6= hr , ni = − √ kh . Hence we have consider its decomposition: ϕϕ 1+k2

rϕϕ = rϕϕ − hrϕϕ, nin + hrϕϕ, nin | {z } | {z } tangent component orthogonal component

Hence we have kh ∇ϕ∂ϕ = (rϕϕ) = rϕϕ − hrϕϕ, nin = rϕϕ + √ n = tangent 1 + k2  −kh cos ϕ   cos ϕ   k cos ϕ  kh hk2 hk2 −kh sin ϕ + sin ϕ = − k sin ϕ = − r ,   1 + k2   1 + k2   1 + k2 h 0 −k 1 i.e. hk2 Γh = − , Γϕ = 0 . ϕϕ 1 + k2 ϕϕ

e) Sphere

7 ( x = R sin θ cos ϕ For the sphere r(θ, ϕ): y = R sin θ sin ϕ , we have z = R cos θ  R cos θ cos ϕ   −R sin θ sin ϕ   sin θ cos ϕ  ∂ ∂ = r = R cos θ sin ϕ , = r = R sin θ cos ϕ , n = sin θ sin ϕ ∂θ θ   ∂ϕ ϕ     −R sin θ 0 cos θ

Calculate 2 θ ϕ ∂ r ∇∂θ ∂θ = Γθθ∂θ + Γθθ∂ϕ = 2 = 0 ∂θ tangent

∂2r θ ϕ since ∂θ2 = −Rn is orthogonal to the sphere. Hence Γθθ = Γθθ = 0. Now calculate 2 θ ϕ ∂ r ∇∂θ ∂ϕ = Γθϕ∂θ + Γθϕ∂ϕ = . ∂θ∂ϕ tangent We have ∂2r = cotan θr , ∂θ∂ϕ ϕ hence 2 θ ϕ ∂ r ∇∂θ ∂ϕ = Γθϕ∂θ + Γθϕ∂ϕ = = cotan θrϕ, i.e. ∂θ∂ϕ tangent θ ϕ Γθϕ = 0, Γθϕ = cotan θ Now calculate 2 θ ϕ ∂ r ∇∂ϕ ∂θ = Γϕθ∂θ + Γϕθ∂ϕ = . ∂ϕ∂θ tangent We have ∂2r = cotan θr , ∂ϕ∂θ ϕ hence 2 θ ϕ ∂ r ∇∂θ ∂ϕ = Γθϕ∂θ + Γθϕ∂ϕ = = cotan θrϕ, i.e. ∂θ∂ϕ tangent θ ϕ Γϕθ = 0, Γϕθ = cotan θ. Of course we did not need to perform these calculations: since ∇ is symmetric connection and ∇∂ϕ ∂θ = ∇∂θ ∂ϕ, i.e.

θ θ ϕ ϕ Γϕθ = Γθϕ = 0 Γϕθ = Γθϕ = cotan θ . and ﬁnally 2 θ ϕ ∂ r ∇∂ϕ ∂ϕ = Γϕϕ∂θ + Γϕϕ∂ϕ = 2 . ∂ϕ tangent We have  −R sin θ cos ϕ  ∂2r = r = −R sin θ sin ϕ . ∂ϕ2 ϕϕ   0

The vector rϕϕ is not proportional to normal vector n, i.e. it is not orthogonal to the sphere; the vector rϕϕ 2 is not tangent to sphere, i.e. it is not orthogonal to vector n: 0 6= hrϕϕ, ni = −R sin θ. We decompose the vector rϕϕ on the sum of tangent vector and orthogonal vector:

rϕϕ = rϕϕ − nhrϕϕ, ni +nhrϕϕ, ni, | {z } tangent vector

8 We see that

 −R sin θ cos ϕ   sin θ cos ϕ  ∂2r = r − nhr , ni = r + R sin2 θn = −R sin θ sin ϕ + R sin2 θ sin θ sin ϕ = ∂ϕ2 ϕϕ ϕϕ ϕϕ     tangent 0 cos θ

 −R cos2 θ sin θ cos ϕ   cos θ cos ϕ  2  −R cos θ sin θ sin ϕ  = − sin θ cos θ  cos θ sin ϕ  = − sin θ cos θrθ . R sin2 θ cos θ − sin θ We have 2 θ ϕ ∂ r ∇∂ϕ ∂ϕ = Γϕϕ∂θ + Γϕϕ∂ϕ = = − sin θ cos θrθ, i.e. ∂ϕ∂ϕ tangent

θ ϕ Γϕϕ = − sin θ cos θ, Γϕϕ = 0.

f) Saddle ( x = u  1   0  For saddle z = xy: We have r(u, v): y = v , ∂u = ru =  0 , ∂v = rv =  1  It will be useful also z = uv v u  −v  to use the normal unit vector n = √ 1 −u . 1+u2+v2   1 Calculate:

2 u v ∂ r ∇∂u ∂u = Γuu∂u + Γuu∂v = 2 = (ruu)tangent = 0 since ruu = 0. ∂u tangent

u v Hence Γuu = Γuu = 0. u v Analogously Γvv = Γvv = 0 since rvv = 0. u v u v Now calculate Γuv, Γuv, Γvu, Γvu:

 0  u v ∇∂u ∂v = ∇∂v ∂u = Γuv∂u + Γuv∂v = (ruv)tangent =  0  1 tangent

u v Using normal unit vector n we have: (ruv)tangent = ruv − hruv, nin = Γuv∂u + Γuv∂v =

 0   0  * 0   −v +  −v  1 1  0  =  0  −  0  , √  −u  √  −u  = 1 + u2 + v2 1 + u2 + v2 1 tangent 1 1 1 1

 v   1   0  1 v u vr + ur u = 0 + u = u v . 1 + u2 + v2   1 + u2 + v2   1 + u2 + v2   1 + u2 + v2 u2 + v2 v u

u u v v v u Hence Γuv = Γvu = 1+u2+v2 and Γuv = Γvu = 1+u2+v2 . Sure one may calculate this connection as Levi-Civita connction of the induced Riemannian metric using explicit Levi-Civita formula or using method of Lagrangian of free particle.