Parallel Transport

Parallel transport is the operation of moving a vector, phase, or section along a path so that it stays covariantly constant with respect to a chosen connection.

In one line:

$$\nabla_{\dot\gamma}s=0.$$

This equation is the precise version of “keep the object parallel while the base point moves.” It is not defined by the path alone. It requires a connection, because the fibers over different points are separate spaces until a comparison rule is supplied.

Helpful background. Connections and Curvature introduces the covariant derivative used in the transport equation.

Keeping vectors parallel along a path

Think of moving a tangent vector on a sphere while avoiding any turn within the local tangent plane. Its Cartesian components may change even though it is parallel in the geometric sense. The same distinction appears when a quantum eigenvector changes because its phase convention or local basis changes.

Required Data

To speak about parallel transport, one needs:

• a base manifold $M$;
• a path $\gamma:[0,T]\to M$;
• a vector space or phase line attached to each point of $M$;
• a connection $\nabla$ on those fibers;
• an initial value $s(0)$ in the fiber over $\gamma(0)$.

Given this data, parallel transport produces an object $s(t)$ in the fiber over $\gamma(t)$. Equivalently, it defines a linear transport map

$$U_\gamma: E_{\gamma(0)}\to E_{\gamma(T)}.$$

Here $E_{\gamma(t)}$ denotes the fiber over the point $\gamma(t)$. The notation is bundle language; see Fiber Bundles, First Look for the base, fiber, and section definitions.

Transport Equation in Components

Choose local coordinates $x^i$ on the base and a local frame $e_a$ for the fibers. Write

$$s(t)=s^a(t)e_a(\gamma(t)).$$

If the connection coefficients are

$$\nabla_i e_a=\Gamma^b{}_{ia}e_b,$$

then the parallel-transport equation becomes

$$\frac{ds^b}{dt} + \dot x^i(t)\Gamma^b{}_{ia}(x(t))s^a(t) = 0.$$

It is often useful to define the connection matrix along the path:

$$\Gamma_t(t) = \dot x^i(t)\Gamma_i(x(t)).$$

Then the transport equation is

$$\frac{ds}{dt} = -\Gamma_t(t)s.$$

This is a first-order linear ordinary differential equation. Once $s(0)$ is given, the transported section is locally unique.

Transport Operator

The same equation can be written for the transport operator $U(t,0)$:

$$\frac{dU(t,0)}{dt} = -\Gamma_t(t)U(t,0), \qquad U(0,0)=I.$$

Then

$$s(t)=U(t,0)s(0).$$

If all connection matrices along the path commute, the solution is the ordinary exponential

$$U(t,0) = \exp \left( - \int_0^t\Gamma_\tau(\tau)\,d\tau \right).$$

In general, matrices at different parameter values need not commute. The formal solution is the path-ordered exponential

$$U(t,0) = \operatorname{P}\exp \left( - \int_0^t\Gamma_\tau(\tau)\,d\tau \right).$$

Path ordering places factors from later points on the path in the correct order. This is the same structural reason time-evolution operators require time ordering when Hamiltonians at different times do not commute.

Reparametrization

Parallel transport is a property of the oriented path, not of the speed used to traverse it. If $t=t(u)$ is an increasing reparametrization, then

$$\frac{ds}{du} + \frac{dt}{du}\Gamma_t(t(u))s = \frac{dt}{du} \left( \frac{ds}{dt}+\Gamma_t s \right).$$

Thus the equation equals zero in the $u$ parameter exactly when it equals zero in the $t$ parameter. The direction of traversal matters; the arbitrary clock used along the path does not.

Composition and Reversal

If a path is followed by another path, the transport maps compose. If $\gamma_1$ goes from $p$ to $q$ and $\gamma_2$ goes from $q$ to $r$, then

$$U_{\gamma_2\circ\gamma_1} = U_{\gamma_2}U_{\gamma_1}.$$

Transport along the reversed path gives the inverse map:

$$U_{\bar\gamma} = U_\gamma^{-1}.$$

These properties follow from uniqueness of the first-order transport equation.

Abelian Line Transport

For a one-dimensional fiber, the connection along the path is an ordinary scalar function $\Gamma_t(t)$. The transport equation

$$\frac{ds}{dt}+\Gamma_t(t)s=0$$

has solution

$$s(t) = s(0) \exp \left( - \int_0^t\Gamma_\tau(\tau)\,d\tau \right).$$

For a unitary U(1) phase convention, it is common to write the connection as

$$\Gamma=-iA,$$

where $A$ is a real one-form. Then

$$s(t) = s(0) \exp \left( i\int_{\gamma|_{[0,t]}} A \right).$$

This sign convention is the one that makes a Berry connection $A_n=i\langle n\rvert dn\rangle$ produce a phase factor $e^{i\int A_n}$.

Flat Transport and Curved Transport

For a smooth flat connection on a simply connected patch, transport is path independent inside that patch and a parallel frame can be chosen. In such a frame, parallel transport looks like keeping components constant.

If curvature is nonzero, two paths between the same endpoints can transport the same initial vector to different final vectors. In the abelian case, a small loop with enclosed surface $\Sigma$ gives a phase controlled by the curvature flux:

$$\exp \left( i\oint_{\partial\Sigma}A \right) = \exp \left( i\int_\Sigma F \right), \qquad F=dA,$$

when a smooth gauge exists on $\Sigma$. The integration and Stokes-theorem assumptions are treated in Integration on Manifolds. The general loop classification belongs to Holonomy.

Sphere Example

On the unit sphere, the Levi-Civita connection parallel transports tangent vectors. Along the equator, a vector pointing toward the north pole remains parallel as it moves eastward: as an ambient vector in $\mathbb R^3$, it is simply the constant vector pointing in the $z$ direction, and it stays tangent to the sphere at every point of the equator. It is perpendicular to the equatorial path, not tangent to that path.

Along more general curves, the transported vector must be continually projected back into the local tangent plane in the connection-prescribed way. Around a closed loop on the sphere, the vector may return rotated. The important lesson is not the sphere formula itself, but the mechanism: local “do not turn relative to the surface” rules can produce global mismatch after a loop.

This is the geometric analogy behind Berry phase. The analogy is structural, not literal: Berry transport moves phase data over parameter space, while Levi-Civita transport moves tangent vectors on a curved surface.

Berry Parallel-Transport Gauge

Let $R(t)$ be a path in parameter space and let $\lvert n(R(t))\rangle$ be a local choice of normalized eigenvector. The Berry connection along the path is

$$A_t(t) = i\langle n(R(t))\rvert\frac{d}{dt}n(R(t))\rangle.$$

Along an open path, one can choose a phase convention so that

$$A_t(t)=0.$$

This is called a parallel-transport gauge. It says that the local eigenvector is being chosen so that its change has no pure phase component along the path.

If

$$\lvert n'(R(t))\rangle = e^{i\chi(t)}\lvert n(R(t))\rangle,$$

then

$$A_t'(t) = A_t(t)-\frac{d\chi}{dt}.$$

Choosing

$$\chi(t) = \int_0^t A_\tau(\tau)\,d\tau$$

makes $A_t'=0$ along the path.

For a closed loop, this gauge choice may fail to return the vector representative to the original phase. That mismatch is the geometric phase. The physical adiabatic phase and sign conventions are treated in Berry Phase.

Open Paths versus Closed Loops

Parallel transport along an open path gives a comparison between the initial fiber and the final fiber. The result depends on the path and the connection.

A closed loop is special because the initial and final base points are the same. Then the transport map acts on one fiber:

$$U_\gamma:E_p\to E_p.$$

The resulting loop transport is called holonomy. It is the next step toward understanding loop phases and why flat connections can retain global information.

Common Mistakes

• Saying that a vector was transported “without change” before specifying a connection.
• Treating parallel transport as ordinary component constancy in every frame.
• Forgetting that transport depends on the path, not only on endpoints, when curvature or topology matters.
• Confusing the transported vector with the tangent vector of the path itself.
• Dropping path ordering for matrix-valued connections whose values do not commute.
• Assuming that an open-path phase is automatically gauge invariant.
• Treating the Berry parallel-transport gauge as a global gauge that must work on every closed loop.

Exercises

1. Show that the equation $\frac{ds}{dt}+\Gamma_t s=0$ is invariant under increasing reparametrization of the path.

Solution

Let $t=t(u)$ with $dt/du>0$. Then

$$\frac{ds}{du} = \frac{dt}{du}\frac{ds}{dt}.$$

The connection component along the $u$-parametrized path is

$$\Gamma_u = \frac{dt}{du}\Gamma_t.$$

Therefore

$$\frac{ds}{du}+\Gamma_u s = \frac{dt}{du} \left( \frac{ds}{dt}+\Gamma_t s \right).$$

Since $dt/du\ne0$, one equation vanishes exactly when the other does.

2. Let $\Gamma_t=\Gamma_0$ be a constant matrix. Solve the transport equation.

Solution

The equation is

$$\frac{ds}{dt} = -\Gamma_0s.$$

Since $\Gamma_0$ is constant, the solution is

$$s(t)=e^{-\Gamma_0t}s(0).$$

Equivalently,

$$U(t,0)=e^{-\Gamma_0t}.$$

3. In a $U(1)$ convention with $\Gamma=-iA$, derive the phase factor for line transport.

Solution

Along the path, the transport equation is

$$\frac{ds}{dt}-iA_t s=0.$$

Thus

$$\frac{1}{s}\frac{ds}{dt} = iA_t.$$

Integrating gives

$$s(t) = s(0) \exp \left( i\int_0^t A_\tau(\tau)\,d\tau \right).$$

Since $A_\tau d\tau$ is the pullback of $A$ to the path, this is

$$s(t) = s(0) \exp \left( i\int_{\gamma|_{[0,t]}}A \right).$$

4. Let $A=\frac{B}{2}(-y\,dx+x\,dy)$ on the plane. What phase factor is obtained by $U(1)$ transport around a counterclockwise circle of radius $R$ with $\Gamma=-iA$?

Solution

The curvature is

$$F=dA=B\,dx\wedge dy.$$

By Stokes theorem,

$$\oint_C A = \int_{\text{disk}}F = B\pi R^2.$$

With $\Gamma=-iA$, the phase factor is

$$\exp \left( iB\pi R^2 \right).$$

5. Along an open path, show how to choose a Berry gauge with $A_t'=0$.

Solution

Under

$$\lvert n'\rangle=e^{i\chi(t)}\lvert n\rangle,$$

the Berry connection component transforms as

$$A_t'=A_t-\frac{d\chi}{dt}.$$

Choose

$$\chi(t)=\int_0^t A_\tau(\tau)\,d\tau.$$

Then $d\chi/dt=A_t$, so

$$A_t'=0.$$

6. Why is an open-path transported phase not automatically a gauge-invariant observable?

Solution

An open path has different initial and final base points. A gauge transformation can change the phase convention independently at those endpoints, so the transported vector’s final phase relative to a chosen local representative changes with the gauge. Gauge-invariant quantities require endpoint reference data, an interferometric comparison, or a closed-loop phase factor where endpoint conventions cancel modulo the relevant phase ambiguity.

References

• T. Frankel, The Geometry of Physics, 3rd ed., Cambridge University Press, 2011.
• J. M. Lee, Introduction to Smooth Manifolds, 2nd ed., Springer, 2013.
• M. Nakahara, Geometry, Topology and Physics, 2nd ed., CRC Press, 2003.