Curvature and parallel transport

In the transport experiment, the plane and rolled sheet returned the arrow unchanged, while the sphere could return it rotated. Both directions are compared at the same starting point, so changing coordinate labels cannot remove their mismatch. This return transformation is called holonomy. For the sphere’s tangent plane, it is a rotation.

Shrink the sphere’s triangle. The return angle becomes smaller, approaching zero with the enclosed area. What remains after dividing by that small area is a local measure of curvature. On a sphere of radius $R$, its magnitude is $1/R^2$. Increasing $R$ while keeping the same angular triangle does not change the rotation: the physical area grows as $R^2$ while the curvature falls as $1/R^2$.

There is another way to make the comparison: start with two identical arrows at A, carry them along different routes, and let them meet at C. Now both arrows live in C’s tangent plane. Their angle can be measured there with no further transport.

GEOMETRY LAB / 05

Two routes. One destination.

Start two arrows at A, pointing the same way. Carry one straight to C and the other via B, without twisting either arrow within the surface. Compare their directions where they meet.

Direct A → CVia BTangent plane

Blue / pink vector length1.00 / 1.00

Angle when they meet at C-90.0°

Area between the routes2.655 m²

Both arrows have arrived at C, but their directions differ by 90.0°. Neither was deliberately twisted. The difference comes from the routes they took across the curved surface. Choose “Look straight at C” to see the angle without foreshortening.

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 to compare the directions

During the journey, the arrows occupy different tangent planes. The readout reports the angle they will make at C, where a comparison is possible without choosing another transport route. “Look straight at C” views that tangent plane face on. Blue is the direct route; pink goes via B. A positive signed angle turns from blue toward pink by the right-hand rule about the outward normal.

Each spherical leg is a great-circle arc. The arrow and the point rotate about that arc’s axis together, which solves the no-twist rule exactly. The largest triangle covers one eighth of the sphere and gives a 90° mismatch. Shrinking the triangle shrinks the angle. At a fixed angular size, increasing the sphere’s radius increases the physical area but leaves the angle unchanged: curvature falls as $1/R^2$.

To connect this experiment to the previous chapter, follow the pink route from A to C and then follow the blue route backward from C to A. This closes a loop. Backward transport preserves angles, so the mismatch at C equals the return rotation at A.

Both arrows have unit length. The radius is in metres. Progress is the fraction of each route’s length; the longer route is played faster so both arrows meet together. This is a comparison of directions, not a model of two particles moving at equal speeds. The camera uses a fixed scale for each surface. Small visual offsets keep arrows and paths legible.

If 3D is unavailable, the static diagram and controls still show both directions; dashed paths and arrows lie on the far side. Exact transport is checked against an independent numerical integration of the no-twist equation. See David Tong’s explanation of path-dependent parallel transport.

The arrows meet at the same point in every version of this experiment. On the plane and rolled sheet, their directions agree. On the sphere, they generally disagree. This differs from the two-flow experiment: there, changing instructions carried the points to different destinations even on a flat plane. Here, the destinations are fixed; it is the transported direction that remembers the route.

Following the route via B from A to C and the direct route backward from C to A makes a closed loop. Transport along that last leg preserves the angle between the arrows, so the mismatch at C is exactly the return rotation at A. The open-route comparison and the closed-loop experiment measure the same effect.

To calculate this effect for a small loop, we need to compare derivatives taken in two different orders.

8.1 Comparing two orders of differentiation#

For an ordinary smooth scalar in ordinary coordinates, mixed partial derivatives commute:

But a vector transported and compared through changing tangent spaces contains extra structure. Differentiating first in one direction and then another need not give the same answer as reversing the order.

Using the torsion-free Levi-Civita connection, define curvature by

The brackets denote a commutator: the first operation order minus the reverse. The placement of indices encodes the roles. The final pair $\mu,\nu$ identifies the two differentiation directions; $\sigma$ receives the original vector; $\rho$ labels the output vector.

Let us derive the components. Because $\nabla_\nu V^\rho$ has both an upper index and a lower index,

The last line is the commonly forgotten correction for the derivative index $\nu$. Swap $\mu$ and $\nu$, then subtract.

The ordinary second derivatives cancel because mixed partials commute. The final line cancels because $\Gamma^\lambda{}_{\mu\nu}$ is symmetric. The terms containing first derivatives of $V$ also cancel in pairs. What survives is proportional to $V$ itself, with no derivatives of $V$ remaining:

That cancellation is conceptually important. The mismatch depends on the vector’s value at the event, not on how you happened to extend it into a vector field around the event. Curvature is a local multilinear geometric object: a tensor.

The derivative terms measure variation of the connection; the quadratic terms account for successive changes of basis acting on each other. Schematically,

where the schematic final expression suppresses index contractions. On the flat polar plane, the derivative and product terms will cancel. Keeping both is necessary to recover zero curvature.

8.2 Curvature takes three vector inputs#

We can express the same operation using vector fields $X,Y,Z$ rather than coordinate directions:

Why subtract the Lie-bracket term? If $X$ and $Y$ themselves do not commute, their flows reach slightly different points when taken in opposite orders. We must remove that displacement effect before interpreting the remaining mismatch as curvature. In a coordinate basis the bracket vanishes, giving the component formula above.

The inputs have different jobs. $X$ and $Y$ specify the two directions of comparison, while $Z$ is the vector being compared. The output $R(X,Y)Z$ is a vector, and it depends linearly on each input at the point. The sign relating it to an actual transport loop depends on the order in which we walk the loop, as we now specify.

For a covector the curvature acts with the opposite sign:

For a tensor, one curvature term acts on each index, with the same upper-plus/lower-minus pattern as covariant differentiation. The scalar pairing is again the consistency check.

8.3 Parallel transport around a loop#

Transport a vector around a small coordinate parallelogram. Let $a^\mu$ and $b^\mu$ be the small side displacements, and traverse the sides in the order $+a,+b,-a,-b$. With our curvature convention and the transport equation $dV=-\Gamma_\mu V\,dx^\mu$, the returned vector satisfies

where all side lengths scale with a small parameter $\ell$. Reverse the loop and the leading sign reverses. Defining the difference by subtracting the two open-path results in the opposite order also reverses it. These orientation choices explain many apparent sign disagreements in pictures of holonomy.

The net transformation obtained around a closed loop is called holonomy. Because the initial and final vectors live in the same tangent space, their mismatch is a genuine comparison. There is no need to argue about how to compare vectors at different endpoints.

For a concrete loop, start on the equator, follow a meridian to the north pole, descend along a meridian a quarter-turn farther east, and return along the equator. These three great-circle arcs form a triangle with three right angles. A great circle is the intersection of the sphere with a plane through its center; its arcs are geodesics of the sphere.

Carry an arrow initially pointing north along the first arc. Keep it continuous at each corner, without rotating it to follow the next side. The worked example below calculates all three legs. On returning, the arrow points east: a right-angle change at the very same point. The loop encloses one eighth of the sphere, with area $\pi a^2/2$ for sphere radius $a$.

In the calculation we draw the sphere in three dimensions. The vector must remain tangent to it. Its ordinary three-dimensional derivative can point normally to the surface; parallel transport requires that derivative to have no tangential part. This is a convenient way to implement the intrinsic comparison rule on this particular surface. The rule itself was already defined without an embedding.

Further calculation: why spherical area determines the transport angle

Use the spherical angles from §4.7: $\theta$ measures angle down from the north pole and $\phi$ measures angle around the axis. In fixed Cartesian components the two unit tangent vectors are

$$\begin{aligned} \hat e_\theta&=(\cos\theta\cos\phi,\cos\theta\sin\phi,-\sin\theta),\\ \hat e_\phi&=(-\sin\phi,\cos\phi,0). \end{aligned}$$

Differentiate these expressions and retain only the tangent part, calling that change $D$. In the $\theta$ direction the change of $\hat e_\theta$ is entirely normal, while $\hat e_\phi$ is constant. In the $\phi$ direction, projecting onto the two unit tangents gives

$$D\hat e_\theta=\cos\theta\,d\phi\,\hat e_\phi, \qquad D\hat e_\phi=-\cos\theta\,d\phi\,\hat e_\theta.$$

For a unit arrow $V=\cos\psi\,\hat e_\theta+\sin\psi\,\hat e_\phi$, the product rule then gives $DV=0$ precisely when $d\psi=-\cos\theta\,d\phi$.

Walk a small coordinate rectangle in the positive order: increasing $\theta$, increasing $\phi$, decreasing $\theta$, decreasing $\phi$. The two meridian sides have $d\phi=0$. The other two give

$$\Delta\psi=[\cos\theta-\cos(\theta+\Delta\theta)]\Delta\phi \simeq\sin\theta\,\Delta\theta\,\Delta\phi.$$

Tile a region with these rectangles and add their contributions. Shared interior edges cancel, leaving its boundary integral. The sphere’s area element is $dA=a^2\sin\theta\,d\theta\,d\phi$, so

$$\Delta\psi=-\oint\cos\theta\,d\phi =\frac{\text{oriented area}}{a^2}.$$

This expression applies directly to a loop bounding a region inside the chosen frame patch. Changing patches lets us describe other loops, with the final rotation defined modulo $2\pi$. The octant example in §8.3 independently verifies a rotation magnitude $\pi/2$ for area $\pi a^2/2$.

For a positively oriented geodesic triangle, the path’s tangent turns by $\pi-\alpha_i$ at a corner with interior angle $\alpha_i$, while the transported arrow stays continuous. Comparing after the three corners gives a rotation equivalent to $-\sum_i(\pi-\alpha_i)$, or $\alpha_1+\alpha_2+\alpha_3-\pi$ modulo a full turn. This angular excess is therefore related to the triangle’s area. For an ordinary convex spherical triangle, choosing the area between zero and $2\pi a^2$ gives the familiar equality $\text{area}=a^2(\alpha_1+\alpha_2+\alpha_3-\pi)$.

Local and global comparisons. A loop is contractible if it can be shrunk continuously to a point while staying in the region under discussion. A flat connection has no net transport change around such a loop. A loop surrounding an excluded point may behave differently, even when the geometry is flat everywhere along it.

Further example: a flat cone with its tip removed

Cut a wedge of angle $0<\delta<2\pi$ from a flat sheet and join its edges into a cone. Away from the tip, the sheet has the same local lengths as before. If we exclude the tip, a loop around it cannot shrink to a point without leaving the surface.

Use distance $r$ from the tip along the sheet and angle $\phi$ with period $2\pi$. Put $\alpha=1-\delta/(2\pi)$. A full circle has circumference $2\pi\alpha r$, giving the metric

$$ds^2=dr^2+\alpha^2r^2d\phi^2.$$

The calculation from Chapter 7 now gives $\Gamma^r{}_{\phi\phi}=-\alpha^2r$ and $\Gamma^\phi{}_{r\phi}=1/r$. Along a circle, write $W=\alpha rV^\phi$ for the angular component in a unit basis. Parallel transport becomes

$$\frac{dV^r}{d\phi}=\alpha W,\qquad \frac{dW}{d\phi}=-\alpha V^r.$$

These are the sine-and-cosine equations for component rotation by $-\alpha\phi$. After a full circuit, the rotation is $-2\pi\alpha=\delta-2\pi$, equivalent to $\delta$ for the final arrow. The missing wedge can therefore be detected by a loop, although every small patch away from the tip is flat.

WORKED EXAMPLE

Carry a direction home—and find it changed

How can an arrow turn after a journey if you never turn it within the surface?

Builds onParallel transport as a differential equation Metric compatibility Tangent spaces

See the idea

At one point of a sphere, a tangent arrow lies in a flat plane touching that point. At the next point, the tangent plane is different. Keeping the same three-dimensional arrow will generally make it stick out of the surface. The relevant rule is local: allow the change needed to remain tangent, but remove any turning within the surface. Then carry out that rule around a whole loop.

Work it out

1. 01
   Split an ambient change into normal and tangent parts

   Let $n$ be the outward unit normal. Any three-dimensional change $W$ has normal part $(W\cdot n)n$. Subtracting that part leaves its tangent projection $P W$.

   $$P=I-nn^{\mathsf T},\qquad PW=W-(W\cdot n)n.$$

   Why this step works The projection subtracts the component perpendicular to the tangent plane. Because the normal has unit length, the subtraction removes that component completely.

2. 02
   State the local transport rule

   For a tangent vector $V(s)$ along a path, require its derivative to have no tangent component. Here $s$ measures distance along the path. This is the Levi-Civita parallel-transport rule for the sphere with its usual induced metric.

   $$\nabla_{\dot\gamma}V=P\frac{dV}{ds}=0.$$

   Why this step works The remaining ambient change points normally to the surface. It adjusts tangency without adding an intrinsic turn.

3. 03
   Verify one great-circle arc

   For this unit-sphere calculation, measure lengths in units of the sphere radius, making $s$ dimensionless arc length. Take $\gamma(s)=(\cos s,0,\sin s)$ and carry its unit tangent $V=(-\sin s,0,\cos s)$. Its derivative is $-\gamma(s)$, entirely normal, so its tangent projection vanishes. The arrow changes in three-dimensional space while remaining parallel intrinsically.

   $$\frac{dV}{ds}=(-\cos s,0,-\sin s)=-n,\qquad P\frac{dV}{ds}=0.$$

   Why this step works A great circle bends toward the sphere center, with no left-or-right turn inside the tangent plane.

4. 04
   Join three verified arcs without twisting at the corners

   Follow $A=(1,0,0)$ to $N=(0,0,1)$, then to $B=(0,1,0)$, then back to $A$. Start with $V=(0,0,1)$. It reaches $N$ as $(-1,0,0)$, stays that same ambient vector on the second arc, and reaches $A$ as $(0,1,0)$ after the final arc. At a corner, keep the arrow continuous even though the path changes direction.

   $$(0,0,1)\ \longrightarrow\ (-1,0,0)\ \longrightarrow\ (-1,0,0)\ \longrightarrow\ (0,1,0).$$

   Why this step works Each arc obeys the tangent-projection rule. Initial and final vectors now share one tangent plane and differ by a right angle.

5. 05
   Read the loop as a curvature experiment

   The triangle occupies one octant of the sphere. Its area is $\pi a^2/2$ and its rotation magnitude is $\pi/2$. More generally, the sphere gives signed rotation equal to oriented area divided by $a^2$, modulo a full turn, as derived above. For a small loop the magnitude is directly area divided by $a^2$. Reversing the journey reverses its sign.

   $$|\Delta\psi|=\frac{\text{area}}{a^2}=\frac{\pi}{2}.$$

   Why this step works The explicit transport calculation establishes the octant result. It compares vectors at the same endpoint, so the mismatch cannot be blamed on different endpoint coordinates.

Go deeper

The projection applies to the derivative. A single finite projection between distant tangent planes is not exact parallel transport and usually shortens an arrow. Repeated tiny projections approach the transport rule as the steps shrink. Norm preservation follows directly: $d(V\cdot V)/ds=2V\cdot dV/ds=0$, because a tangent vector is perpendicular to a normal derivative. The calculation proves the octant example. The further calculation in §8.3 derives the sphere’s signed area-to-rotation relation by adding the contributions from small coordinate rectangles. For a positively oriented convex geodesic triangle, the angular excess is its angle sum minus $\pi$; it equals its area divided by $a^2$. The final arrow records rotation modulo a full turn, so the area formula should not be interpreted as the smallest unsigned angle for an arbitrarily large region.

Further readingUniversity of Mannheim · Metrics and connections, induced connection David Tong · Metrics, connections, and transport, §§3.1–3.3 Tristan Needham · Visual Differential Geometry and Forms

A STATE YOU CAN CHECK

After the three great-circle arcs, the vector returns to its initial point. Its final direction is perpendicular to its initial direction (0, 0, 1).

Final point

$(X,Y,Z)=(1,0,0)$

Final vector

$(V^X,V^Y,V^Z)=(0,1,0)$

Turn relative to the initial vector

$\Delta\alpha=\pi/2$

The worked steps explain these measurements. Interactive controls appear when available.

Open the reference diagram

Test the idea

FIRST, PREDICT

Could we reproduce this experiment by keeping the original ambient arrow fixed, moving it straight to the final point, and projecting once there?

Yes. Parallel transport depends only on the two endpoints.Yes, provided the arrow is made longer before projection.No. That ignores the local tangent-plane comparison along the chosen route.

Compare the reasoning

Yes. Parallel transport depends only on the two endpoints.

Endpoint-only comparison is exactly what this loop disproves. The route matters in curved geometry.

Yes, provided the arrow is made longer before projection.

Rescaling can repair a length, but it cannot supply the missing route-dependent direction.

No. That ignores the local tangent-plane comparison along the chosen route.

Right. At the end of this closed loop the tangent plane is the starting plane, so one final projection would return the original arrow and miss the holonomy.

A hint

The initial and final tangent planes are identical. What would a single projection at that final point do?

NOW CHANGE THE EXAMPLE

A small geodesic triangle on a sphere of radius $a=2\,\mathrm m$ encloses area $\pi/3\,\mathrm m^2$. What is the parallel-transport rotation magnitude in degrees?

A hint

Divide area by $a^2$ to get radians, then multiply by $180/\pi$.

Work through the solution

The rotation is $(\pi/3)/2^2=\pi/12$ radians, or $15^\circ$. The length units cancel. Doubling the sphere radius at fixed physical loop area reduces the rotation by a factor of four.

Curvature can reveal itself as a mismatch after a direction is transported locally without turning and brought back to its starting tangent plane.

8.4 The curvature of the polar plane#

Using the polar connection from Chapter 7, calculate

In two dimensions the Riemann symmetries leave only one independent curvature component, so this establishes that the plane is flat wherever the polar chart is valid. The apparent singularity at $r=0$ is a failure of the chart; Cartesian coordinates cover the origin smoothly.

The derivative term was $-1$; the connection-product term was $+1$. Their cancellation agrees with the Cartesian calculation, where every connection coefficient and curvature component is zero.

8.5 Counting independent curvature components#

Lower the first index with the metric:

For the Levi-Civita connection,

The second symmetry follows directly from reversing the derivative order. To check the others without a long expansion, we can choose coordinates with $\Gamma=0$ at the point under study. Here is why those coordinates exist locally. Put the old coordinate origin at that point and set

The Jacobian is the identity at the origin. The inverse map has second derivatives $-\Gamma^\rho{}_{\mu\nu}(0)$ there, which cancel the old coefficients in the transformation law from §7.4. The invertible Jacobian makes this a valid chart sufficiently near the origin. Metric compatibility then gives $\partial g=0$ at the point as well.

At that point the curvature reduces to second derivatives of the metric:

Exchanging $\rho,\sigma$ reverses the sign. Exchanging the two index pairs preserves it, using $g_{ab}=g_{ba}$ and commuting partial derivatives. Adding the three cyclic terms makes each second derivative cancel against its opposite. This last relation is called the algebraic Bianchi identity. All are tensor equations, so a check in this convenient chart establishes them in every chart at the same point.

In four dimensions, an antisymmetric pair has six choices: $01,02,03,12,13,23$. Reversing a pair changes only a sign; repeating an index gives zero. Treat each pair as one label. Pair exchange makes curvature a symmetric $6\times6$ array, with six diagonal entries and $6\times5/2=15$ entries above the diagonal: 21 in total.

For four distinct indices, the algebraic Bianchi identity supplies one independent relation,

Its other versions follow by relabeling and the pair symmetries; versions with repeated indices already follow from those symmetries. Thus 20 independent components remain. In two dimensions there is only one pair, $12$, and one independent curvature entry. This justifies the single-component flatness check in §8.4.

Twenty counts independent curvature values at one event. It does not tell us how many independent wave patterns can propagate. Answering that requires the field equations; Chapter 18 derives the two independent polarizations of a gravitational wave, meaning its two independent transverse stretching patterns.

ONE EVENT · FOUR DIMENSIONS

Find the twenty independent pieces.

Each tile names a component, not a blank to fill. Select a tile to see which other entries it determines.

$1$Pair the indices$2$Match the mirror entries$3$Use the cyclic identity

$$6\times6=36$$

Equal indices inside either pair give zero. Reversing a pair changes only a sign. Six increasing-order choices remain for each pair, giving thirty-six slots.

$R_{\rho\sigma\mu\nu}$row pair, then column pair

	$01$	$02$	$03$	$12$	$13$	$23$
$01$	$m_{11}$	$m_{12}$	$m_{13}$	$m_{14}$	$m_{15}$	$m_{16}$
$02$	$m_{21}$	$m_{22}$	$m_{23}$	$m_{24}$	$m_{25}$	$m_{26}$
$03$	$m_{31}$	$m_{32}$	$m_{33}$	$m_{34}$	$m_{35}$	$m_{36}$
$12$	$m_{41}$	$m_{42}$	$m_{43}$	$m_{44}$	$m_{45}$	$m_{46}$
$13$	$m_{51}$	$m_{52}$	$m_{53}$	$m_{54}$	$m_{55}$	$m_{56}$
$23$	$m_{61}$	$m_{62}$	$m_{63}$	$m_{64}$	$m_{65}$	$m_{66}$

Six choices for each pair; one symbol per slot.

This tile is$R_{0123}$

$$R_{0123}=R_{2301}$$

The reflected tile has exactly the same value. Swapping complete pairs adds no minus sign.

$$R_{0123}=-R_{1023}$$$$R_{0123}=-R_{0132}$$

Reversing either pair changes the sign. Reversing both changes it twice.

Conventions and the scope of this count

All indices are lowered before applying these symmetries. The labels $0,1,2,3$ refer to one spacetime basis. The Levi-Civita connection is metric-compatible and torsion-free. Exchanging pairs means transposing this component array; it is not raising indices with a Euclidean metric on two-forms.

The symbols name potentially independent numbers at one event. Field equations and differential identities add other restrictions. Twenty is not the number of propagating modes.

Source: Tong, General Relativity, §3.4

Reference diagram

8.6 Calculating curvature on a sphere#

Now take a sphere of fixed radius $a$. As in §4.7, $\theta$ is the angle down from the north pole and $\phi$ is the angle around the axis. A meridian step has length $a\,d\theta$ and a latitude step has length $a\sin\theta\,d\phi$, giving

Insert $g_{\theta\theta}=a^2$, $g_{\phi\phi}=a^2\sin^2\theta$, and $\partial_\theta g_{\phi\phi}=2a^2\sin\theta\cos\theta$ into the Christoffel formula. The nonzero coefficients are

Here $\cot\theta=\cos\theta/\sin\theta$. We work away from the poles, where this angular chart is valid. The circumference scale is $a\sin\theta$, instead of the plane’s linear factor $r$; the two terms in the curvature calculation will no longer cancel.

Compute one component:

Thus

This component has units of length squared because all its slots refer to angular coordinate vectors. In a unit orthonormal basis the corresponding component is $1/a^2$. In two dimensions, we can express the one independent curvature entry per unit area. This scalar is the Gaussian curvature:

We can summarize some of this curvature by contracting indices, using the operation from Chapter 2. Define the Ricci tensor by $R_{ab}=R^c{}_{acb}$ and the Ricci scalar by $R=g^{ab}R_{ab}$. Here each index runs over $\theta,\phi$. For example, $R_{\theta\theta}=g^{\phi\phi}R_{\phi\theta\phi\theta}=(a^2\sin^2\theta)^{-1}(a^2\sin^2\theta)=1$. The complete result is

This also fixes our sign convention operationally: a round sphere has positive scalar curvature. Some books reverse the sign of the Riemann tensor. Equations borrowed from them must be translated consistently, especially geodesic deviation and the Einstein equation.

An inhabitant of the sphere need not see its embedding. Draw a circle of geodesic radius $s$ around a point. Its circumference is

It is shorter than the Euclidean prediction. Curvature can be discovered with intrinsic distance measurements alone.

A cylinder illustrates the opposite lesson. A sheet can be rolled into a cylinder without locally stretching it. Its embedding looks bent, but its intrinsic metric is locally flat. Extrinsic bending describes the relation to an ambient space; intrinsic curvature describes the metric geometry experienced by inhabitants. Spacetime curvature does not demand an invisible higher-dimensional room into which spacetime bends.

WORKED EXAMPLE

Measure curvature without leaving the surface

Could inhabitants of a sphere discover its curvature using only lengths measured on their world?

Builds onLocal approximation Intrinsic versus extrinsic geometry The polar metric

See the idea

Fix one point and walk the same surface distance $\rho$ from it in every direction. The resulting geodesic circle consists of points at that distance from the center. Measure its circumference $C$. In a flat plane, $C=2\pi\rho$. On a sphere, the circumference is smaller. The comparison uses distances on the surface; no view from outside is needed to perform it.

Work it out

1. 01
   Distinguish the two radii

   Let $a$ be the sphere radius and put the chosen center at its north pole. A meridian distance $\rho$ subtends angle $\theta=\rho/a$. The circular latitude at that distance has ambient radius $a\sin\theta$. This ambient radius helps calculate the answer; the inhabitants measure $\rho$ along the surface.

   $$C(\rho)=2\pi a\sin(\rho/a),\qquad 0<\rho<\pi a.$$

   Why this step works Great-circle arc length is the sphere radius times the angular distance. The latitude circle instead uses the radius projected onto the equatorial plane.

2. 02
   Compare with the flat prediction

   For a small circle, use the Taylor expansion $\sin z=z-z^3/6+O(z^5)$. Here $O(z^5)$ collects terms no larger than a constant times $|z|^5$ as $z$ tends to zero. Divide by the flat circumference.

   $$\frac{C(\rho)}{2\pi\rho}=1-\frac{\rho^2}{6a^2}+O\!\left(\frac{\rho^4}{a^4}\right).$$

   Why this step works The leading term is the flat-plane result. The first correction measures a deficit that becomes easier to interpret as the circle shrinks.

3. 03
   Extract a quantity that does not vanish with the circle

   The relative deficit shrinks like $\rho^2$. Divide by that known scale before taking the limit. On the round sphere this gives its Gaussian curvature.

   $$K=\lim_{\rho\to0}\frac{6}{\rho^2}\left(1-\frac{C(\rho)}{2\pi\rho}\right)=\frac{1}{a^2}.$$

   Why this step works Dividing out the area scale isolates a local property of the surface instead of a property of the size of our measuring circle.

4. 04
   Separate bending from intrinsic curvature

   A sheet rolled into a cylinder without stretching has the same local length measurements as the sheet before rolling. A sufficiently small geodesic circle, contained in an unwrapped patch, still satisfies $C=2\pi\rho$. Its Gaussian curvature is zero, even though the cylinder bends in three-dimensional space.

   Why this step works Unrolling preserves all distances inside the patch. The sphere’s circumference deficit cannot be removed by an equally distortion-free flattening.

Go deeper

This is an exact derivation of the round sphere’s circumference and its small-circle limit. The theorem that the same limit equals Gaussian curvature on any smooth Riemannian surface is a broader local result; it is not proved here. For the sphere, the value $K=1/a^2$ agrees with the curvature tensor calculation in this section. Curvature has units of inverse length squared. Finite circles give an estimate whose corrections are controlled by $\rho/a$; a large circle should not be put into the limiting formula as if it were exact. This circumference deficit prevents a spherical patch from being flattened while preserving its local lengths.

Further readingAndreas Kriegl · Riemannian Geometry, Theorem 11.1 David Tong · Metrics, connections, and transport, §§3.1–3.3 Tristan Needham · Visual Differential Geometry and Forms

Test the idea

FIRST, PREDICT

For the same small geodesic radius $\rho$, double the sphere radius $a$. What happens to the leading relative circumference deficit?

It becomes four times as large.It becomes one quarter as large.It becomes half as large.

Compare the reasoning

It becomes four times as large.

The sphere bends more gently when its radius increases. At fixed $\rho$, the discrepancy from a flat circle decreases.

It becomes one quarter as large.

Yes. The leading deficit is $\rho^2/(6a^2)$. A larger sphere looks flatter over the same measured neighborhood.

It becomes half as large.

The effect depends on inverse radius squared. Curvature is measured per area scale, not per length scale.

A hint

Keep $\rho$ fixed in $1-C/(2\pi\rho)\approx\rho^2/(6a^2)$.

NOW CHANGE THE EXAMPLE

On an unknown round sphere, a geodesic circle has $\rho=0.12\,\mathrm m$ and measured ratio $C/(2\pi\rho)=0.9994001$. Use the leading small-circle formula to estimate the sphere radius $a$.

A hint

Call the measured deficit $\delta$. Solve $\delta\approx\rho^2/(6a^2)$ for $a$ before inserting the numbers.

Work through the solution

Here $\delta=0.0005999$, so $a\approx\rho/\sqrt{6\delta}\approx2.00017\,\mathrm m$, or about $2\,\mathrm m$. The inferred ratio $\rho/a\approx0.06$ is small, making the neglected next term consistent with this estimate.

Intrinsic curvature can be detected by comparing a small circle’s measured circumference with the flat prediction for its measured radius.