Parallel transport on a manifoldSantiago Casas

31.05.2011

AbstractWe find the equations of parallel transport on a 2-dimensional sphere

and solve them for the specific case of a unit vector transported along aclosed path, composed of four different curves. Finally we show how thenon-conservation of the vector components is a measure of the curvatureof the manifold.

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IntroductionIn school we learn how to move a vector in flat 3-dimensional space, by trans-porting it always parallel to itself. By doing this, the vector does not change itsorientation or its magnitude and we can say that the vector is the “same” afterbeing moved a certain distance.

However on a curved surface, or a general manifold, the notion of a vectoris only local. Each vector lives in his own tangent vector space, that can onlybe defined locally for each point on the manifold, so there is no natural way ofrelating or comparing vectors at different points.

That is why we need a general definition of parallel transport, that allowsus to compare different vectors at different points and gives us a prescription onhow to move a vector around a manifold without really altering the vector. Theparallel transport is defined using so called “connections”, that as the namestates, help us to connect vectors at different points. This definition has tobe very general for any curved space, but should also contain the special caseof euclidean geometry, in which moving a vector parallel to itself is a trivialoperation.

On a curved space, maybe against our normal intuition, despite of trans-porting the vector always “parallel” to itself around a closed loop, we find that,when we go back to the point where we started, the vector returns rotated com-pared to the one we started with. This “effect” is due to our lack of intuition forparallel lines and geodesics (lines of shortest distance between two points) on amanifold, and not because we chose a bad definition of parallel transport. Thefact that the vector returns “rotated” to the starting point, is actually deeplyrelated with the definition of curvature on the manifold.

This text is meant to be self-contained, but therefore the reader must havesome background in vector and tensor analysis, calculus, differential equationsand preferably some basics of differential geometry. Some concepts and defini-tions are going to be explained on the way, but many other are going to be onlystated or used, because this text is very far from being a text for learning aboutdifferential geometry or general relativity. The interested reader can check someof the recommended literature.

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1 Metric of the SphereFor a 2-dimensional sphere of unit radius, we have the line element: ds2 =dθ2 + sin2 θdφ2, this allows us to measure infinitesimal distances ds on thesphere. Here we are using common spherical coordinates, where θ and φ are thepolar and azimuthal angles respectively. This line element yields the followingmetric:

gµν =(

1 00 sin2 θ

)(1.1)

gµν =(

1 00 1/ sin2 θ

)(1.2)

The connections on a manifold, also called Christoffel symbols are given by:

Γijk = 12g

lk(gik,j + gjk,l − gij,k) (1.3)

This connections are called Riemannian connections, when they can be ex-pressed as functions of the metric. Later on we are going to see what thisimplies for parallel transport.

Because of the metric being diagonal, we can calculate the Christoffel sym-bols only for l = k = 1, 2 and we find that the non vanishing terms are:

Γ122 = − sin θ cos θ (1.4)

Γ212 = Γ2

21 = cot θ (1.5)

The Christoffel symbols are symmetric in its two lower indices, because we areworking on a torsion-free manifold. For more details on this, see [WB].

2 Definition of Parallel transportWhen we say parallel transport on a manifold, we mean that we want to carry avector v along a curve γ defined on a manifoldM , in such a way that the vectoris parallel to itself when transported an infinitesimal distance. Because we wantthe vector to be unchanged when we perform the transport along the curve, wedefine a derivative operator, which applied to the vector, along the directionof the curve, gives zero. By doing that we are saying that the vector does notchange with respect to a certain operation. This operation is called paralleltransport and at the same time it helps us to define the derivative operator,which is called the covariant derivative ∇. Using this definition of covariantderivative, the equation of parallel transport can be simply stated as:

∇γv = 0 (2.1)

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Where γ is the tangent vector to the curve along which we are transportingthe vector. In terms of local coordinates, this is equivalent to:

vk + Γkij xivj = 0 (2.2)

In our notation each curve γ is parametrized by t, so that for a specific curve,each of the coordinates xi is a function of the parameter t. Therefore xi is thederivative of the coordinate xi with respect to t, and therefore xi represents thetangent vector to the curve γ at the point parametrized by x(t). The vk are thederivatives of the k vector component, with respect to the curve parameter t.

Therefore, this equation is telling us, that we have a differential equationfor each vector component, that depends on the Christoffel symbols and onthe specific path, along which the vector is parallel transported. The theory ofdifferential equations tells us, that there is a unique solution to this equation ifwe specify initial conditions. So by specifying initial components of the vectorv and inserting the final value of the parameter t we can find uniquely how thevector changed along a specific curve, after traveling a distance parametrizedby t.

On a closed loop, formed by different curves, we calculate how the compo-nents change for the first curve and then we use these components as the newvector vk and do the same calculation for the next curve.

The definition of Riemannian connection in equation (1.3) leaves the metricinvariant after applying to it a covariant derivative operator. Because the metricin a Riemannian manifold defines the norm and the scalar product of vectors,this guarantees that the norm of the vector will be unchanged after a paralleltransport on an arbitrary curve. This means that we have to get in the end avector with the same “length” as the one we started with.

The fact that the components of the vector change along the path is also dueto the change of the direction of the basis vectors on the manifold and this isbecause we are dealing with a curved space, in which the vector space, spannedby the basis vectors, is only locally defined at each point.

The concept of parallel transport also allows us to define a geodesic, as thecurves whose tangent vectors remain constant after being parallel-transportedalong the same curve. That means that the covariant derivative of a tangentvector to a geodesic vanishes identically:

∇γ γ = 0 (2.3)

This is called the geodesic equation, and with it we can find the curves of“shortest distance” between two points on the manifold. See [WD] for a detailedderivation of this equation.

In the following section, we will find the solutions to this equations for aspecific closed loop on a 2-dimensional sphere.

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3 Parallel transport of a unit vector along aclosed loop on a sphere

3.1 Parametrized curves and the equation of parallel trans-port

Now we want to calculate how a vector changes when it is parallel-transportedalong a closed loop on a sphere. We are going to consider a unit vector that istangent to the equator at the point θ = π

2 , φ = 0. We are considering sphericalcoordinates, where θ is the polar angle, going from 0 in the “south pole” toπ in the “north pole” and φ the azimuthal angle, going from 0 to 2π. Linesof constant θ are the lines of constant latitude or “latitude circles” and linesof constant φ are the lines of constant longitude or “meridians” in terms ofgeography.

The initial tangent vector has in these coordinates the following componentform:

v = 0∂θ + 1∂φ := (0, 1) (3.1)

Where ∂φ and ∂θ are the unit basis vectors along the θ and φ directionsrespectively. In differential geometry, the basis vectors are the partial derivativeswith respect to the coordinates. This basis vectors, form what its called acoordinate basis.

Starting at φ = 0 and θ = π2 , we are going to consider a closed loop defined in

the following way: First move along the equator in “east” direction a magnitudeof ∆φ = π/4, then move north along the meridian for ∆θ = π/4, now move“west” along the line of constant latitude back to φ = 0 and finally move backto the initial point following the meridian with φ = 0.

Using t as the curve parameter, scaled to run always from 0 at the beginningof the curve, to 1 at its end, we have the following curves γi(t) = (θ(t), φ(t)):

γa(t) = (π2 ,π

4 t) γb(t) = (π2 + π

4 t,π

4 )

γc(t) = (π4 ,π

4 −π

4 t) γd(t) = (π2 −π

4 t,π

4 )

So γa and γc are latitude circles and γb and γd are meridians.Returning back to the equation of parallel transport in local coordinates

(2.2) and using only the non-vanishing Christoffel symbols (1.4) and (1.5) wefind the following system of equations:

v1 − sinθ cos θx2v2 = 0v2 + cot θ(x1v2 + x2v1) = 0

(3.2)

Where v1, v2are the first and second components of the vector v we want totransport, and x1, x2 are the first and second components of the tangent vectorto the curve at the point t, i.e. x(t) = dγ(t)

dt .

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Figure 3.1: Closed loop on a sphere, formed by 4 different curves, each of adifferent color. Curve “a” is blue, “b” red, “c” purple, and “d” green. The unittangent vector parallel to the equator is shown as a black arrow.

3.2 General solution of the system of differential equationsFirst we are going to solve the equations in general and then we will find howthe components of the vector change along each curve.

For the curves along the meridians (γb, γd) we have x2 = 0, because φ(t) =const. along those curves, so using equation (3.2) we find:

v1 = 0 (3.3)v2 = − cot θx1v2 (3.4)

First of all we see that the first component of the vector does not changealong a meridian curve. On the other hand, for the second component we obtainan apparent complicated differential equation, but we can solve this noting thatthe parametrization is done with “constant velocity”, that means γ(t) is linearin t. Therefore x1 = dθ(t)

dt = ω and because the parameter t runs from 0 to 1,we have: ω dt = θ dθ, where θ is the final value of the coordinate “θ”.

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So we can rewrite (3.4) as:

dv2

v2 = − cot θ ωdt

= − cot θ dθ= −d ln(sin θ) (3.5)

⇒ ln v2

v20

= − ln sin θsin θ0

v2 = v20

sin θ0

sin θ (3.6)

This means that transporting the vector along a meridian, will change itssecond component, by a factor that depends on the initial and final polar angles.

Along the latitude circles (γa, γc) we have x1 = 0, because θ(t) = const.

along these curves and we can again interpret x2 = dφ(t)dt = ω as the constant

angular speed the vector has, while being transported along the curve. Usingagain equation(3.2), we find the following system of equations:

v1 = sin θ cos θωv2 (3.7)v2 = − cot θωv1 (3.8)

This is a set of two coupled differential equations (each component dependson the derivative of the other). To solve this we derive (3.7) with respect to theparameter t and insert (3.8), solving for v1 we obtain:

v1 = sin θ cos θω(−cos θsin θ ωv

1)

⇒ v1 = − cos2 θω2v1

⇒ v1 = −ω2v1

⇒ v1 + ω2v1 = 0 (3.9)

Where we have identified ω = ω cos θ, for a constant value of θ, becausewe are moving only along the φ coordinate. We can see that (3.9) is just theequation of an harmonic oscillator, with general solution:

v1 = A cos ωt+B sin ωt (3.10)

where the frequency is ω and A and B are coefficients that we still haveto determine, which depend on the initial conditions. This means that thefirst component of the vector, when parallel-transported along a latitude circle,changes according to a sinusoidal function with period 1/ω = 1/ω cos θ, so thatits change is dependent on the distance traveled and on the final point.

For the second component, we have from (3.7) the following equation:

v2 = v1

ω sin θ cos θ (3.11)

In the next section we will find the specific solutions for the path describedin figure 3.1.

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3.3 Specific solutions for a closed pathWe will now show the solutions of the equations of the previous section, for aunit vector parallel tangent to the equator, that is transported along the paths(γa → γb → γc → γd) of section (3.1). The initial vector is, as we have seen,v = v0 = (0, 1) in our set of coordinates.

For the path “a” we have: x = (0, π4 ) and θ = π/2, so following equations(3.7) and (3.8) we find that:

v1 = v2 = 0So that the vector components do not change along the path “a”.This result is straightforward to interpret, because in this case we are trans-

porting a tangent vector to a curve along itself, and the fact that the vectorremains unchanged, is telling us that the curve is a geodesic curve. So we havefound that the equator is a geodesic on the 2-dimensional sphere.

For the path “b” we can use directly equations (3.3) and (3.6) to find:

v1 = 0

v2 = v20

sin(π2 )sin(π4 ) =

√2

So the vector at the end of the curve “b” is now: vb = (0,√

2).On the path “c”, we are moving along φ, with a constant value of θ = π

4 , sowe have:

ω = ω cos θ = ω cos(π4 ) = ω√2

Now we want to find the values of A and B in equation (3.10). We know thatat the beginning of the curve, that is when t = 0, the vector is still unchangedsuch that:

v1 = A cos ωt+B sin ωt = A cos(0) +B sin(0) = 0⇒ A = 0⇒ v1 = B sin ωt

To find B, we note that from equation (3.7) and for θ = π4 and v2 =

√2, we

have:

v1 = sin θ cos θωv2

= sin θ cos θωv2

= 12ω√

2

but we had

v1 = B sin ωt⇒ v1 = ωB cos ωt

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so for t = 0we find

ωB = Bω√2

and therefore:

⇒ B = 1

So we can solve for the components of the new vector vc (at the end of thepath “c”, where t = 1):

v1 = sin(ωt) = sin( π

4√

2)

v2 = ω cos(ωt)sin θ cos θω =

√2 cos( π

4√

2)

And for the last path “d”, we use equations (3.3) and (3.6) as we did withpath “b”, to find:

v1 = sin ωt = sin( π

4√

2)

v2 = v20

sin θ0

sin θ =√

2 cos(ωt)sin π

4sin π

2= cos( π

4√

2)

Although the path “d” is a geodesic like the curve “a”, the vector componentschange this time, because it is not the tangent vector to the geodesic anymore.

So the final vector vd we obtain at the end of the path “d”, that is afterreturning to the point were we started, is now:

vd ≈ (0.572, 0.850)

And we can see that its norm is conserved after a parallel transport arounda closed loop, as we had demanded above:

‖vd‖ =√

sin2( π

4√

2) + cos2( π

4√

2) = 1

Using the known formula to find the angle between two vectors:

~A · ~B = ‖A‖‖B‖ cosα

We find that the angle between the original vector and its parallel trans-ported version is:

α ≈ 31.7°

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Figure 3.2: Graph showing the vector in a different color after being transportedalong the respective colored curve. The resulting final vector forms an angle ofabout 32° with the original one. The image is a vector graphic, so that it canbe zoomed or rescaled without any loss of quality.

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4 Appendix A: Relation between parallel trans-port and the Riemann tensor

As mentioned in the introduction, there is a deep connection between paralleltransport and the curvature of a manifold. If after being parallel-transportedaround a closed loop, the vector returns “rotated” with respect to itself, this isa clear indication that the manifold is not euclidean.

Now we are going to derive the full expression for parallel transport arounda closed loop, using the fact that the closed loop is infinitesimally small. Weare going to find that the change in the vector components is related to thesubtended area of the loop and the Riemann tensor on the manifold.

Figure 4.1: Closed loop on a manifold, following coordinate curves of constantx1 or x2.

Consider the loop shown in figure 4.1 formed by four intersecting sections ofcoordinate lines, where the first to lines have coordinates: x1 = a, x2 = b andthe other two x1 = a+ δa, x2 = b+ δb, where δa, δb are infinitesimal.

Because this lines follow curves of coordinate lines (for simplicity), the equa-tion of parallel transport (2.2) and the properties of the covariant derivativesyield the following equations, for transporting the vector along the direction ofx1 and x2 respectively:

∂vα

∂x1 = −Γαµ1vµ (4.1)

∂vα

∂x2 = −Γαµ2vµ (4.2)

Integrating this equations on both sides and putting the right limits we findthe following equations:

Trajectory AB :

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ˆ

x2=b

∂vα

∂x1 dx1 = vα(B)− vα(A) =

ˆ

x2=b

Γαµ1vµ dx1 (4.3)

Rearranging for Trajectory BC:

vα(C) = vα(B) −ˆ

x1=a+δa

Γαµ2vµ dx2 (4.4)

Trajectory CD:

vα(D) = vα(C) −ˆ

x2=b+δb

Γαµ1vµ dx1 (4.5)

Trajectory DA:

vα(D) = vα(A) −ˆ

x1=a

Γαµ2vµ dx2 (4.6)

For the entire loop we know that the change of the components of the vectorv are then:

δvα = vα(Afinal)− vα(Ainitial) (4.7)

Therefore we need to sum the individual changes of equations (4.3) ,(4.4),(4.5),(4.6).After rearranging and using the properties of definite integrals, we find:

δvα = δb∂

∂x2

ˆ a+δa

a

Γαµ1vµ dx1 − δa ∂

∂x1

ˆ b+δb

b

Γαµ2vµ dx2 (4.8)

≈ δaδb

[− ∂

∂x1 (Γαµ2vµ) +− ∂

∂x2 (Γαµ1vµ)]

(4.9)

For obtaining equation (4.8) we used the following expansion of the integrals:ˆ

f

x1=b+δb

=ˆ

f

x2=b

+ δb∂

∂x2

ˆf

x2=b

And for equation (4.9) we used the following approximation to first order inδb (for the infinitesimal loop):

ˆ b+δb

b

f dx = F (b+ δb)− F (b) ≈ dF

dxδb ≈ f δb

We leave the explicit derivation as an exercise for the reader.Now we calculate the partial derivatives on equation (4.9) acting on the

Christoffel symbols:

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∂

∂x1 (Γαµ2vµ) = Γαµ2,1v

µ + Γαµ2vµ,1 (4.10)

∂

∂x2 (Γαµ1vµ) = Γαµ1,2v

µ + Γαµ1vµ,2 (4.11)

Where we have just applied the Leibniz rule. Here we use the commonnotation: A,µ = ∂

∂xµ .The terms vµ,1 and vµ,2 can be written as−Γµσ1v

σ and Γµλ2vλ by using equations

(4.1) and (4.2).By inserting this into equation (4.9), rearranging terms and renaming some

dummy indices (exercise for the reader), we obtain:

δvα ≈ δaδb[Γαµ1,2 − Γαλ1Γλµ2 − Γλµ2,1 + Γλν2Γλµ1

]vµ (4.12)

By consulting the reference text books, [WD, WB, CL], we see that theexpression in brackets is just the Riemann tensor R in 2 dimensions, so moregenerally we can write:

δvα ≈ δaδbRαβµν (4.13)

This means that the change in the components of a vector, when paralleltransported along a closed loop, depend on the area of the loop (therefore itspath) and on the curvature of the manifold. In a flat space, the Riemann tensorwould be zero, therefore the operation of parallel transport does not change thecomponents of a vector. In a spherical geometry like the one we used above,the Riemann tensor has some non-zero components and that is why the vectorends up being rotated when it returns to the starting position.

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5 Appendix B: Mathematica code for generat-ing figures.

Algorithm 1 Mathematica code for generating the graphs of this document.sphere1 = ParametricPlot3D[{Cos[\[Theta]] Sin[\[Phi]], Sin[\[Theta]]Sin[\[Phi]], Cos[\[Phi]]}, {\[Theta], 0, 2 Pi}, {\[Phi], 0, Pi}, Boxed ->False, Axes -> False, MeshStyle -> Opacity[0.3]];\[Theta][t_] = (Pi/4)*t; \[Phi][t_] = Pi/2;gamma1[t_] = {Cos[\[Theta][t]] Sin[\[Phi][t]], Sin[\[Theta][t]] Sin[\[Phi][t]],Cos[\[Phi][t]]}; \[Theta]2[t_] = Pi/4; \[Phi]2[t_] = Pi/2 - Pi/4*t;gamma2[t_] = {Cos[\[Theta]2[t]] Sin[\[Phi]2[t]], Sin[\[Theta]2[t]]Sin[\[Phi]2[t]], Cos[\[Phi]2[t]]}; \[Theta]3[t_] = Pi/4 - (Pi/4)*t; \[Phi]3[t_] =Pi/4;gamma3[t_] = {Cos[\[Theta]3[t]] Sin[\[Phi]3[t]], Sin[\[Theta]3[t]]Sin[\[Phi]3[t]], Cos[\[Phi]3[t]]}; \[Theta]4[t_] = 0; \[Phi]4[t_] = Pi/4 +(Pi/4)*t;gamma4[t_] = {Cos[\[Theta]4[t]] Sin[\[Phi]4[t]], Sin[\[Theta]4[t]]Sin[\[Phi]4[t]], Cos[\[Phi]4[t]]};gamma1dot[t_] = D[gamma1[t], t];initialpoint1 = gamma1[0];tangentVect1 = Graphics3D[{Black, Thick, Arrow[{initialpoint1, initialpoint1+ 1/1.33 gamma1dot[0]}]}];gamma2dot[t_] = D[gamma2[t], t];initialpoint2 = gamma2[0];gamma3dot[t_] = D[gamma3[t], t];initialpoint3 = gamma3[0];gamma4dot[t_] = D[gamma4[t], t];initialpoint4 = gamma4[0];finalpoint = gamma4[1];vect2 = Graphics3D[{Blue, Arrow[{initialpoint2, initialpoint2 + 0.6gamma1dot[1]}]}];vect3 = Graphics3D[{Red, Arrow[{initialpoint3, initialpoint3 - Sqrt[2]/1.33gamma3dot[0]}]}];vect4 = Graphics3D[{Purple, Arrow[{initialpoint4, initialpoint4 - Sin[\[Pi]/(4Sqrt[2])]/1.33 gamma4dot[0] - ( Sqrt[2] Cos[\[Pi]/(4 Sqrt[2])])/1.33gamma3dot[1]}]}];finalVect = Graphics3D[{Darker[Green], Arrow[{finalpoint, finalpoint -Sin[\[Pi]/(4 Sqrt[2])]/1.33 gamma4dot[1] + Cos[\[Pi]/(4 Sqrt[2])]/1.33gamma1dot[0]}]}];spherecurve1 = ParametricPlot3D[Evaluate[gamma1[t]], {t, 0, 1}, Boxed ->False, Axes -> False, PlotPoints -> 200, PlotStyle -> {Blue, Thick}];spherecurve2 = ParametricPlot3D[Evaluate[gamma2[t]], {t, 0, 1}, Boxed ->False, Axes -> False, PlotPoints -> 200, PlotStyle -> {Red, Thick}];spherecurve3 = ParametricPlot3D[Evaluate[gamma3[t]], {t, 0, 1}, Boxed ->False, Axes -> False, PlotPoints -> 200, PlotStyle -> {Purple, Thick}];spherecurve4 = ParametricPlot3D[Evaluate[gamma4[t]], {t, 0, 1}, Boxed ->False, Axes -> False, PlotPoints -> 200, PlotStyle -> {Darker[Green], Thick}];solidcombo = Show[sphere1, spherecurve1, spherecurve2, spherecurve3, sphere-curve4,tangentVect1,finalVect, vect2, vect3, vect4];

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References[WD] General Relativity, WALD Robert, 1984, The University of

Chicago Press. 2, 4

[CL] Spacetime and Geometry, CARROLL Sean , 2004, Pearson Edu-cation. 4

[WB] Gravitation and Cosmology, WEINBERG Steven, 1972, John Wi-ley.

1, 4

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