Homework 5. Solutions 1 Calculate the Christoffel Symbols of The

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) Remark One can calculate Christoffel symbols using Levi-Civita Theorem (Homework 5). There is a third way to calculate Christoffel symbols: It is using approach of Lagrangian. This is may be the easiest and most elegant way. (see the Homework 6)

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

x = r cos ϕ r = x2 + y2 y = r sin ϕ and ϕ = arctan y (  p 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

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

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

We already know the fast way to calculate Christoffel symbol using Lagrangian of free particle and this method work for a flat connection since flat connection is a Levi-Civita connection for Euclidean metric So perform now brute force calculations only for some components. (Then later (in homework 6) we will calculate using very quickly Lagrangian of free particle. )

1 2 r ∂ xi 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

∂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 i i a) Let Γkm =Γmk. We have to prove that Γk′m′ =Γm′k′

2 We have ′ ′ ′ i k m r i i ∂x ∂x ∂x i ∂x ∂x Γ ′ ′ = ′ ′ Γ + ′ ′ . (1) k m ∂xi ∂xk ∂xm km ∂xk ∂xm ∂xr

Hence ′ ′ ′ i m k r i i ∂x ∂x ∂x i ∂x ∂x Γ ′ ′ = ′ ′ Γ + ′ ′ m k ∂xi ∂xm ∂xk mk ∂xm ∂xk ∂xr i i ∂xr ∂xr ′ ′ ′ ′ But Γkm =Γmk and ∂xm ∂xk = ∂xk ∂xm . Hence

′ ′ ′ ′ ′ i m k r i i m k r i ′ i ∂x ∂x ∂x i ∂x ∂x ∂x ∂x ∂x i ∂x ∂x i Γ ′ ′ = ′ ′ Γ + ′ ′ = ′ ′ Γ + ′ ′ =Γ ′ ′ . m k ∂xi ∂xm ∂xk mk ∂xm ∂xk ∂xr ∂xi ∂xm ∂xk km ∂xk ∂xm ∂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 X ∂ Y ∂ (see the next part). Consider = ∂xi , = ∂xj then since [∂i,∂j ] = 0 we have that

k k k k XY YX [X, Y]= i∂j j ∂i =Γ ∂k Γ ∂k = (Γ Γ )∂k = 0 ∇ − ∇ − ∇ − ∇ ij − ji ij − ji

k k We see that commutator for basic ﬁelds XY YX [X, Y]=0 if and only if Γ Γ = 0. ∇ − ∇ − ij − ji 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) ∇ − ∇ − ∇ − ∇ −

5 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 E2 ∂ ϕ induced by the canonical flat connection on . It suffices to define ∂ =Γϕϕ∂ϕ. ∇ ∂ϕ ∂ϕ Recall the general rule. Let r(uα): xi = xi(uα) is embedded surface in Euclidean space En. The basic ∂ ∂r(u) vectors = . To take the induced covariant derivative XY for two tangent vectors X, Y we take ∂uα ∂uα ∇ 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) ∂ =Γαβ = ∂ = ∇ ∂uα ∂uβ ∂uγ ∇ α ∂uβ ∂uα∂uβ tangent tangent 3 ∂ ( ∂ β ) is just usual derivative in Euclidean space since for canonical connection all Christoﬀel ∇canonical α ∂u symbols vanish.) In the case of 1-dimensional manifold, curve it is just tangential acceleration!:

2r ∂ u ∂ (canonical) ∂ d (u) a ∂ =Γuu = ∂ = = tangent ∇ ∂u ∂u ∂u ∇ u ∂u du2 tangent tangent

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

∂ ∂x ∂ ∂y ∂ ∂ ∂ rϕ = = + = R sin ϕ + R cos 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 − ∂x − ∂y tangent tangent ∂ ∂ since the vector R cos ϕ R sin ϕ is orthogonal to the tangent vector rϕ. In other words it means that − ∂x − ∂y 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 2 ∂ ∂x ∂ ∂y ∂ 2R 2 2 ∂ ∂ rt = = + = (R t ) + 2tR , ∂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 ∂ ∂ rt + 2t + 2R −t2 + R2 (R2 + t2)2 − ∂x ∂y tangent

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

(rtt) = rtt rtt, nt n tangent −h i We have r 1 2 2 nt = = 2tR∂x +(t R )∂y , r R2 + t2 − | | 3 4R where rt, n = 0. Hence rtt, nt = 2− 2 2 and h i h i (t +R )

(rtt) = rtt rtt, nt n = tangent −h i

2 3 4t 2R ∂ ∂ 4R 1 2 2 2t rt + 2t + 2R + 2tR∂x +(t R )∂y = − rt −t2 + R2 (R2 + t2)2 − ∂x ∂y (t2 + R2)2 · R2 + t2 − t2 + R2 We come to the answer: 2t t 2t ∂ ∂t = − ∂t, i.e.Γ = − ∇ t t2 + R2 tt t2 +R2

4 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

It is easy to see that t = R tan π + ϕ , i.e. ϕ = 2arctan t π and 4 2 R − 2 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) we need to extract its orthogonal component: (rtt) = rtt rtt, nt n, where n tangent tangent −h i is an orthogonal unit vector: n, rt = 0, n, n = 1: h i h i 1 nt = ( 2t∂x + ∂y) . √1 + 4t2 −

We have

1 1 (rtt) = rtt rtt, nt n = 2∂y 2∂y, ( 2t∂x + ∂y) ( 2t∂x + ∂y)= tangent −h i − √ t2 − √ t2 − 1 + 4 1 + 4 4t 8t2 4t 4t ∂ + ∂ = (∂ + 2t∂ )= ∂ 1 + 4t2 x 1 + 4t2 y 1 + 4t2 x y 1 + 4t2 t We come to the answer: 4t t 4t ∂ ∂t = ∂t, i.e.Γ = ∇ t 1 + 4t2 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     2r h ϕ ∂ ∂ ∂h =Γ ∂h +Γ ∂ϕ = = 0since rhh = 0. ∇ h hh hh ∂h2 tangent h ϕ Hence Γhh =Γhh = 0

2r h ϕ ∂ ∂ ∂ϕ = ∂ ∂h =Γ ∂h +Γ ∂ϕ = = 0since rhϕ = 0 ∇ h ∇ ϕ hϕ hϕ ∂h∂ϕ tangent 5 h h ϕ ϕ Hence Γhϕ =Γϕh =Γhϕ =Γϕh = 0.

2r a cos ϕ h ϕ ∂ − ∂ϕ ∂ϕ =Γϕϕ∂h +Γϕϕ∂ϕ = = a sin ϕ = 0 ∇ ∂ϕ∂ϕ tangent  −  0 tangent   a cos ϕ r − h h ϕ since the vector ϕϕ = 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 =0. i.e. Γ =Γ = 0. ∇ h hh hh k sin ϕ r r We have that r = r = −k cos ϕ = ϕ , i.e. ∂ = ∂ = ϕ : hϕ ϕh   h ∂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  r n kh cone: 0 = ϕϕ, = 2 . Hence we have consider its decomposition: 6 h i − √1+k

rϕϕ = rϕϕ rϕϕ, n n + rϕϕ, n n −h i h i tangent component orthogonal component | {z } | {z } Hence we have kh ϕ∂ϕ =(rϕϕ) = rϕϕ rϕϕ, n n = rϕϕ + n = ∇ tangent −h i √1+ k2 kh ϕ ϕ k ϕ cos kh cos hk2 cos 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

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

and ﬁnally 2r θ ϕ ∂ ∂ ∂ϕ =Γ ∂θ +Γ ∂ϕ = . ∇ ϕ ϕϕ ϕϕ ∂ϕ2 tangent We have R θ ϕ ∂2r sin cos = 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 = rϕϕ, n = R sin θ. We decompose the 6 h i − vector rϕϕ on the sum of tangent vector and orthogonal vector:

rϕϕ = rϕϕ n rϕϕ, n +n rϕϕ, n , − h i h i tangent vector | {z } 7 We see that

2r R sin θ cos ϕ sin θ cos ϕ ∂ 2 − 2 = rϕϕ n rϕϕ, n = rϕϕ + R sin θn = R sin θ sin ϕ + R sin θ sin θ sin ϕ = ∂ϕ2 − h i  −    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 2r θ ϕ ∂ ∂ ∂ϕ =Γ ∂θ +Γ ∂ϕ = = sin θ cos θrθ, i.e. ∇ ϕ ϕϕ ϕϕ ∂ϕ∂ϕ − tangent Γθ = sin θ cos θ, Γϕ = 0. ϕϕ − ϕϕ 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:   2r u v ∂ ∂ ∂u =Γ ∂u +Γ ∂v = =(ruu) = 0since ruu = 0. ∇ u uu uu ∂u2 tangent tangent u v Hence Γuu =Γuu = 0. u v r Analogously Γvv =Γvv = 0 since vv = 0. u v u v Now calculate Γuv, Γuv, Γvu, Γvu:

0 u v ∂ ∂v = ∂ ∂u =Γ ∂u +Γ ∂v =(ruv) = 0 ∇ u ∇ v uv uv tangent   1 tangent   u v Using normal unit vector n we have: (ruv) = ruv ruv, n n =Γ ∂u +Γ ∂v = tangent −h i uv uv 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. 6 Let , be two different connections. Let (1)Γi and (2)Γi be the Christoffel symbols of connec- ∇1 ∇2 km km tions and respectively. ∇1 ∇2 a) Find the transformation law for the object : T i = (1)Γi (2)Γi under a change of coordinates. km km − km 1 Show that it is tensor. 2 8 ? b)∗ Consider an operation on vector fields and find its properties. ∇1 − ∇2 Christoffel symbols of both connections transform according the law (1). The second term is the same. Hence it vanishes for their difference:

′ ′ ′ ′ ′ i k m i k m i (1) i (2) i ∂x ∂x ∂x (1) i (2) i ∂x ∂x ∂x i T ′ ′ = Γ ′ ′ Γ ′ ′ = ′ ′ Γ Γ = ′ ′ T k m k m − k m ∂xi ∂xk ∂xm km − km ∂xi ∂xk ∂xm km ′ i 1 We see that T′ ′ transforms as a tensor of the type . km 2 b) One can do it in invariant way. Using axioms of connection study T = is a vector ﬁeld. ∇1 − ∇2 Consider T (X, Y)= XY XY ∇1 − ∇2 Show that T (fX, Y)= fT (X, Y) for an arbitrary (smooth) function, i.e. it does not possesses derivatives:

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

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

′ ∂ xm ∂2xr ∂xk t ′ = ′ t + ′ ′ . m ∂xm m ∂xm ∂xk ∂xr

b) † Show that this law can be written as

∂ xm ∂ ∂x t ′ = ′ t + ′ log det . m ∂xm m ∂xm ∂x ′

Solution. Using transformation law (1) we have

′ ′ ′ i k m r i i ∂x ∂x ∂x i ∂x ∂x t ′ =Γ ′ ′′ = ′ ′ Γ + ′ ′ m i m ∂xi ∂xi ∂xm km ∂xi ∂xm ∂xr

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

′ ′ ′ ′ ′ i k m r i m r i m r i i ∂x ∂x ∂x i ∂x ∂x k ∂x i ∂x ∂x ∂x ∂x ∂x t ′ =Γ ′ ′′ = ′ ′ Γ + ′ ′ = δ ′ Γ + ′ ′ = ′ t + ′ ′ . m i m ∂xi ∂xi ∂xm km ∂xi ∂xm ∂xr i ∂xm km ∂xi ∂xm ∂xr ∂xm m ∂xi ∂xm ∂xr

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

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

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

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