The connection and parallel transport

7.1 Parallel transport along a path#

Put a small arrow on a flat sheet. Slide its base around a triangle without turning the arrow. At a corner, your route changes direction, but the arrow keeps pointing the same way. Back at the starting point, its direction is unchanged.

Now roll the sheet into a cylinder. The arrow must tip in the surrounding room to stay flat against the sheet. That tipping is unavoidable; an extra twist within the sheet is not. Parallel transport carries a direction while allowing only the change required by the local geometry. On these surfaces, we can see it as keeping the arrow tangent without turning it within its tangent plane.

Try the plane and rolled sheet first. Then try the sphere. Its path follows great circles: intersections of the sphere with planes through its center. The equator is one example. Three arcs make the triangle. The blue arrow records the starting direction at A; the pink arrow travels. The question is what happens when both arrows can be compared at A again.

GEOMETRY LAB / 03

Carry an arrow home.

Keep an arrow flat against the surface without twisting it in its tangent plane. Follow the triangle back to A. Does it return pointing the same way?

Starting directionCarried directionTangent plane

Carried vector length1.000

Rotation on return to A0.0°

Enclosed surface area1.280 m²

Keep the arrow pointing in the same direction while its base follows the path. At a corner, the path turns; the arrow need not turn with it.

What “without twisting” means

The gold patch is the tangent plane: the flat plane touching the surface at the moving point. The arrow stays in this plane. As the plane tips, the arrow must change its direction in the surrounding three-dimensional space. Parallel transport makes only the change needed to stay tangent, with no extra turning within the tangent plane.

Let $n$ be the unit normal, pointing perpendicular to the surface. Tangency says $V\cdot n=0$. Differentiate this dot product along distance $s$ on the path:

$$\frac{dV}{ds}\cdot n=-V\cdot\frac{dn}{ds}$$

With no tangential part of $dV/ds$, its normal part is the whole change. Therefore

$$\frac{dV}{ds}=-\left(V\cdot\frac{dn}{ds}\right)n$$

This is the Levi-Civita transport rule for these surfaces with their ordinary Euclidean measuring rule. Its change is perpendicular to $V$, so $d(V\cdot V)/ds=0$: the arrow keeps its length. The coordinate equation below describes the same rule without needing an outside three-dimensional picture.

How the calculation is checked

On the plane, the direction stays constant. On the cylinder, we keep its components constant in the frame of the unrolled sheet and roll that frame back up. On the sphere, each leg follows a great circle: a circle cut out by a plane through the sphere’s center. Rotating the point and its tangent vector about that arc’s fixed axis gives exact parallel transport. At a corner, the arrow continues from its previous value; it does not turn to follow the next leg.

The largest sphere triangle has three right-angle corners and covers one eighth of the sphere. Its return rotation has magnitude 90°. Reverse the route and the rotation changes sign. Smaller triangles give smaller rotations. Positive angles follow the right-hand rule about the outward normal at A. The Gaussian curvature is the signed return rotation per unit signed area in the limit of a small loop. For these constant-curvature surfaces, multiplying it by the signed enclosed area also gives the exact return angle of the finite loops shown here. On a sphere of radius R, the curvature is $1/R^2$.

The size control shrinks the loop toward A. On the sphere it changes the angular size. Increase the radius at a fixed angular size: the physical area increases, but the return angle stays the same. The cylinder is a rolled piece of the same flat sheet, so its physical loop area stays fixed when its bend changes.

The path, arrow, and model use the same exact positions. The sphere and cylinder radius is in metres; the carried direction has unit length. Arrow glyphs and the tangent patch are drawn slightly above the surface for visibility. Camera rotation does not change the model or its scale. “Look straight at A” views the starting tangent plane face on, so its angles can be read without foreshortening. Playback takes nine seconds per circuit and stops at A.

The static diagram remains available when 3D is unavailable. Its dashed paths and arrows indicate the far side. The mathematical construction is also described in David Tong’s discussion of parallel transport and Keenan Crane’s differential geometry notes.

On the plane and cylinder the arrow returns unchanged. On the sphere, it can return rotated even though we added no local twist along the way. We have separated two questions: how to carry a vector through each small step, and whether carrying it around a whole loop returns it unchanged. The connection answers the first. Curvature, developed in Chapter 8, answers the local version of the second.

To express the carrying rule in coordinates, take a curve $x^\mu(\lambda)$ and a vector $V^\mu(\lambda)$ attached to its points. The covariant derivative along the curve is

We say $V$ is parallel transported when this derivative vanishes:

To carry this out, choose the vector at the starting point and rearrange the equation as $dV^\mu/d\lambda=-\Gamma^\mu{}_{\alpha\beta}\dot x^\alpha V^\beta$. The known path and connection tell us how to update its components at each step. This is a linear first-order differential equation, with a locally unique solution for smooth coefficients. In a flat Cartesian basis, $\Gamma=0$ and the components simply stay constant.

If the transported vector is the tangent to the path itself, the prescription becomes

which is the affinely parameterized geodesic equation. A geodesic transports its own direction. This supplies a precise meaning of “as straight as possible” that does not require drawing the curve inside a larger space.

An ideal gyroscope carried by a freely falling laboratory gives a physical example: its spin orientation is parallel transported when no torque acts on it. This example assumes free fall as well as the absence of torque.

7.2 Two requirements that select ordinary GR’s connection#

PREPARATION FOR THIS SECTION

Do the two moves in the other order

If moving right changes how fast you can move upward, does the order of the moves matter?

Builds onDirectional derivatives

See the idea

Let $X$ say “move right at unit speed” and $Y$ say “move upward at a speed equal to your x coordinate.” Moving right first changes the speed of the upward move. Reversing the order reaches a different point. This is a property of the two vector fields, even on a flat plane.

Work it out

1. 01
   Treat a vector field as differentiation

   For a scalar function $f$, write $Xf=X^a\partial_af$. Define the bracket by the difference between two ordered derivatives. Expanding with the product rule cancels the terms with second derivatives of $f$.

   $$[X,Y]f=X(Yf)-Y(Xf),\quad [X,Y]^a=X^b\partial_bY^a-Y^b\partial_bX^a.$$

   Why this step works Ordinary mixed partial derivatives of a smooth scalar commute in a coordinate chart.

2. 02
   Calculate the simplest nonzero example

   On a plane with dimensionless labels, take $X=\partial_x$ and $Y=x\partial_y$. Only the x derivative of Y’s y component is nonzero.

   $$[\partial_x,x\partial_y]=\partial_y.$$

   Why this step works The coefficient x changes when X moves you.

3. 03
   Check it by actually composing moves

   Starting at $(x,y)$, move along $X$ for parameter $\epsilon$, then along $Y$ for $\epsilon$. The endpoint is $(x+\epsilon,y+\epsilon x+\epsilon^2)$. Reversing these two moves gives $(x+\epsilon,y+\epsilon x)$. Their difference is $\epsilon^2\partial_y$.

   $$\Delta y=\epsilon^2.$$

   Why this step works The endpoint comparison fixes the sign; a differently ordered closed commutator loop may carry the opposite sign.

Go deeper

Coordinate basis fields satisfy $[\partial_a,\partial_b]=0$; an arbitrary frame need not. The torsion of a connection is $T(X,Y)=\nabla_XY-\nabla_YX-[X,Y]$. Torsion-free means this expression vanishes. It does not mean the bracket of every pair of fields vanishes.

Further readingDavid Tong · General relativity

Test the idea

FIRST, PREDICT

On a flat plane, does $[\partial_x,x\partial_y]\ne0$ prove curvature?

No. The field coefficient changes across the plane.Yes. Every noncommuting operation is curvature.It proves torsion even without choosing a connection.

Compare the reasoning

No. The field coefficient changes across the plane.

Noncommuting chosen flows occur in flat geometry; curvature requires the covariant comparison including the bracket correction.

Yes. Every noncommuting operation is curvature.

This example uses only ordinary derivatives on the Euclidean plane.

It proves torsion even without choosing a connection.

Torsion is defined using a connection as well as the bracket.

A hint

Which geometrical structure has actually been specified?

NOW CHANGE THE EXAMPLE

For $X=\partial_x$ and $Y=x^2\partial_y$, find the y component of $[X,Y]$ at $x=3$.

A hint

Differentiate $x^2$ with respect to $x$.

Work through the solution

$[X,Y]=2x\partial_y$, so the component is 6.

A noncommuting frame is not automatically curved or torsionful; calculate the appropriate geometric object.

A manifold can carry many connections. Standard metric GR chooses the Levi-Civita connection, characterized by two conditions.

First, metric compatibility:

This says the connection preserves the metric’s inner products. If $V$ and $W$ are parallel transported along the same curve, the product rule gives

Thus their lengths and mutual inner product remain fixed under transport. In Lorentzian geometry, causal character is preserved too: timelike vectors remain timelike, and null vectors remain null.

The condition does not say $\partial_\lambda g_{\mu\nu}=0$ everywhere. Metric components may vary because the coordinates vary. Compatibility says the connection accounts for that variation consistently.

Second, zero torsion. For vector fields $X,Y$, torsion is

where $[X,Y]$ is their Lie bracket. The subtraction removes the failure of the vector fields themselves to form commuting coordinate directions. In a coordinate basis, $[\partial_\mu,\partial_\nu]=0$, so

Torsion-free therefore means symmetry of the lower two connection indices in a coordinate basis. For a frame whose basis fields have a nonzero bracket, that bracket must still be subtracted. Symmetry of the connection coefficients alone would then be a different condition.

The flat-plane example in §6.6 already has $[\partial_x,x\partial_y]=\partial_y$. Its ordinary Euclidean connection nevertheless has zero torsion: $\nabla_{\partial_x}(x\partial_y)=\partial_y$ and $\nabla_{x\partial_y}\partial_x=0$, so the bracket cancels their difference. A nonzero bracket of chosen fields therefore does not imply torsion. We choose zero torsion in ordinary GR; other gravity theories can make a different choice.

7.3 Deriving the Christoffel symbols#

Expand metric compatibility:

There are two connection terms because the metric has two lower indices. Now write the same statement with the indices permuted:

Add the first two equations and subtract the third. Why this particular maneuver? We want to isolate the connection with lower indices $\mu\nu$. The unwanted terms pair off because torsion-free symmetry lets us exchange the lower two indices of $\Gamma$:

The inverse metric removes $g_{\sigma\lambda}$. Multiply by $\tfrac12g^{\rho\sigma}$:

This proves uniqueness: any torsion-free, metric-compatible connection must have these coefficients. Existence follows by checking the formula: it is symmetric in $\mu\nu$, substitution gives $\nabla g=0$, and the chain rule supplies the connection transformation law below. There is exactly one such connection for every smooth nondegenerate metric.

This is the same formula that appeared in the free-particle equation in §5.5. It now has a second interpretation: it is the comparison rule that preserves inner products and has zero torsion.

Metric compatibility also gives $\nabla_\lambda g^{\mu\nu}=0$. Differentiate $g^{\mu\alpha}g_{\alpha\nu}=\delta^\mu{}_{\nu}$, use the product rule and $\nabla g=0$, and multiply by the inverse metric. Raising and lowering indices now commute with covariant differentiation.

7.4 Why a connection is geometric but its coefficients are not a tensor#

Under a coordinate change,

The first line resembles the transformation of a $(1,2)$ tensor. The second line is the essential extra term.

Here is the product-rule step. Using the Jacobian $J$ and its inverse $K$ from Chapter 6, write $e'_\nu=K^\lambda{}_{\nu}e_\lambda$. Then

Replace $e_\rho$ by $J^\alpha{}_{\rho}e'_\alpha$ and read off its coefficient. The second term gives the first line of the transformation law. In the first term, $K^\sigma{}_{\mu}\partial_\sigma=\partial'_\mu$ differentiates $K^\rho{}_{\nu}=\partial x^\rho/\partial x'^\nu$, producing the second derivative on the second line.

The inhomogeneous term cancels the unwanted second derivatives in $\partial'_\mu V'^\alpha$. This cancellation makes the complete covariant derivative transform as a tensor.

A tensor that vanishes in one coordinate system at a point vanishes in every coordinate system there. Connection coefficients can vanish at a point in one system and be nonzero in another. They therefore cannot themselves be a tensor measuring gravitational curvature.

Nevertheless, the difference of two connections is a tensor. If $A^\rho{}_{\mu\nu}=\Gamma^\rho{}_{\mu\nu}-\widetilde\Gamma^\rho{}_{\mu\nu}$, the second-derivative terms cancel under transformation. This fact underlies comparisons between a background connection and a perturbed connection. Also, $\delta\Gamma$ in a metric variation is tensorial when comparing connections on the same manifold with the same coordinate identification. That observation will become useful in the action derivation.

For additional derivations of connections and their relation to transport, see Sean Carroll’s university lecture notes, “Curvature”. The calculations here use the conventions stated in this book.

7.5 Five calculations on the polar plane#

We can test the formulas by describing a flat plane in polar coordinates. The geometry is already known, so each calculation has an independent Cartesian check. Away from the origin, use

and

Thus $g_{rr}=1$, $g_{\theta\theta}=r^2$, and $g^{\theta\theta}=r^{-2}$. An angular coordinate is dimensionless, so $g_{\theta\theta}$ has units of length squared. Metric components need not all have the same units when their coordinates do not.

Only one metric derivative is nonzero: $\partial_rg_{\theta\theta}=2r$. For example, set the three free indices in the Christoffel formula to $r,\theta,\theta$. Since the inverse metric is diagonal, only its $rr$ entry contributes:

Doing the same substitution for the other indices gives the complete nonzero list:

with every other coefficient zero.

These values have simple origins. Moving around a circle changes the radial direction, while the angular coordinate basis vector $\partial_\theta$ has length $r$. Its scale changes when you move radially. The connection records both effects.

Experiment 1: a vector that is actually constant. Take $V=\partial_x$, a unit vector pointing in the same Cartesian direction everywhere. Its polar components are

The ordinary derivatives are not all zero. But the covariant derivatives are:

Every covariant derivative vanishes, agreeing with the constant Cartesian vector. Here $V^\theta=-\sin\theta/r$, while the unit-frame component in §6.1 was $V^{\hat\theta}=-\sin\theta$. The factor $1/r$ comes from the length of the angular coordinate basis vector.

Experiment 2: compatibility with a visibly changing metric. Although $\partial_rg_{\theta\theta}=2r$,

There is no contradiction between nonconstant metric components and a covariantly constant metric.

Experiment 3: a straight line with coordinate acceleration. The geodesic equations are

where dots mean differentiation with respect to an affine parameter. The second equation says $d(r^2\dot\theta)/d\lambda=0$.

For a concrete check, choose $\lambda$ to be time and take the Cartesian motion $x=v\lambda$, $y=b$, with constant speed $v>0$ and fixed distance $b>0$. Transforming to polar coordinates and differentiating gives

Its radial coordinate accelerates even though the path is perfectly straight. The $r\dot\theta^2$ term is the coordinate accounting needed to express zero geometric acceleration. A nonzero $d^2x^\mu/d\lambda^2$ is not, by itself, a physical acceleration measurement.

Experiment 4: transporting around a full circle. Along $r=r_0$, parallel transport obeys

These are the equations for a rotating pair of components. The vector remains fixed in Cartesian space while the polar basis rotates underneath it. After $2\pi$, the components return to their initial values. The loop causes no net geometric rotation.

Experiment 5: divergence and the Laplacian. Since $\sqrt{g}=r$,

where $a$ here ranges over the two coordinates. The physical angular component in a unit-length basis is $V^{\hat\theta}=rV^\theta$. Replacing $V^\theta$ by $V^{\hat\theta}/r$ gives the angular divergence term $r^{-1}\partial_\theta V^{\hat\theta}$. The factor $1/r$ converts change per angular increment into change per physical length.

In this spatial geometry, the divergence of a gradient is called the Laplacian, written $\Delta$. Insert $(\operatorname{grad}f)^r=\partial_rf$ and $(\operatorname{grad}f)^\theta=r^{-2}\partial_\theta f$ into the divergence formula:

Check it on $f=r^2=x^2+y^2$. The polar formula gives

In Cartesian coordinates, the same operator is $\partial_x^2+\partial_y^2$, giving $2+2=4$. Using just $\partial_r^2f+\partial_\theta^2f$ would instead give 2. The metric and volume factors are needed for the two calculations to agree.

We have nonzero Christoffel symbols, changing basis components, and coordinate acceleration, all in flat space. The final experiment—checking the curvature itself—belongs to the next chapter.