The Riemann Curvature Tensor

Curvature is the tensorial remainder left when two covariant derivatives are commuted. It measures the local, direction-dependent failure of parallel transport to be path independent; it is not the mere presence of nonzero connection coefficients. This page fixes the operator and component signs, derives the coordinate formula, and checks it in both flat polar coordinates and a curved sphere.

Required background. Connections and Covariant Derivatives provides the connection, its extension to tensors, and the distinction between coefficients and tensor fields.

Helpful background. Calculus and Linear Algebra for Relativity supplies multivariable differentiation and index notation for the component calculation; Exterior Algebra and Differential Forms supplies the alternating two-form and wedge-product algebra used in the curvature-form viewpoint.

Curvature as a commutator

For an affine connection $\nabla$ and vector fields $X,Y,Z$, define

$$R(X,Y)Z = \nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z.$$

The Lie-bracket term is essential: it removes the failure of noncoordinate directions to commute. Expanding the Leibniz rules shows that all derivatives of any multiplying functions cancel, so $R$ is $C^\infty$-linear in $X$, $Y$, and $Z$. Thus curvature at a point depends only on the three tangent vectors there, even though its definition contains second derivatives. It is antisymmetric in its first two operator arguments, $R(X,Y)=-R(Y,X)$.

For the rest of this page, specialize to a Levi–Civita connection. In a coordinate basis, the adopted convention is

$$[\nabla_\mu,\nabla_\nu]V^\rho =R^\rho{}_{\sigma\mu\nu}V^\sigma.$$

Duality fixes the covector sign rather than leaving it to memory. Since $\nabla_\mu(\omega_\rho V^\rho)$ is a scalar derivative, commuting twice gives

$$[\nabla_\mu,\nabla_\nu]\omega_\rho =-R^\sigma{}_{\rho\mu\nu}\omega_\sigma.$$

Every upper tensor index therefore contributes a $+R$ action and every lower index a $-R$ action. The invariant definition uses the same sign as Calegari 2022, Definition 7.1.1 and Proposition 7.1.3, p. 53, PDF; the coordinate convention agrees with Tong 2019, §3.4, eq. (3.39), p. 130, PDF.

The coordinate expression

Set $X=\partial_\mu$, $Y=\partial_\nu$, and $Z=\partial_\sigma$. Because coordinate basis vectors commute and $\nabla_\mu\partial_\sigma=\Gamma^\lambda{}_{\mu\sigma}\partial_\lambda$,

$$\begin{aligned} R^\rho{}_{\sigma\mu\nu} ={}&\partial_\mu\Gamma^\rho{}_{\nu\sigma} -\partial_\nu\Gamma^\rho{}_{\mu\sigma}\\ &+\Gamma^\rho{}_{\mu\lambda}\Gamma^\lambda{}_{\nu\sigma} -\Gamma^\rho{}_{\nu\lambda}\Gamma^\lambda{}_{\mu\sigma}. \end{aligned}$$

This expression contains first derivatives and quadratic products of connection coefficients. Neither group is tensorial separately; their non-tensorial transformation terms cancel in the displayed combination. One lowers the first index with the metric, $R_{\rho\sigma\mu\nu}=g_{\rho\lambda}R^\lambda{}_{\sigma\mu\nu}$.

For a connection with torsion, distinguish the operator composition $[\nabla_{\partial_\mu},\nabla_{\partial_\nu}]V$ from antisymmetrizing the full second covariant derivative, which also differentiates the lower derivative slot. The former still equals $R(\partial_\mu,\partial_\nu)V$; the latter is

$$(\nabla_\mu\nabla_\nu-\nabla_\nu\nabla_\mu)V^\rho =R^\rho{}_{\sigma\mu\nu}V^\sigma -T^\lambda{}_{\mu\nu}\nabla_\lambda V^\rho.$$

Here $T^\lambda{}_{\mu\nu}=\Gamma^\lambda{}_{\mu\nu}-\Gamma^\lambda{}_{\nu\mu}$ in a coordinate basis. The two meanings agree when torsion vanishes. This distinction explains why otherwise compatible references can display an extra derivative term.

Infinitesimal transport and two decisive checks

Choose two independent small coordinate displacements $A^\mu$ and $B^\mu$, and transport a vector around the closed coordinate parallelogram in the order $+A,+B,-A,-B$. With $\Delta V=P_\gamma V-V$, the leading mismatch is

$$\Delta V^\rho =-R^\rho{}_{\sigma\mu\nu}V^\sigma A^\mu B^\nu +O(\epsilon^3), \qquad A,B=O(\epsilon).$$

The minus sign follows from the transport equation $dV=-\Gamma_\mu V\,dx^\mu$. In normal coordinates at the base point, integrating the first derivatives of $\Gamma$ around the loop gives $-(\partial_\mu\Gamma_\nu-\partial_\nu\Gamma_\mu)A^\mu B^\nu$; in a general frame the ordered products add $-[\Gamma_\mu,\Gamma_\nu]A^\mu B^\nu$. Reversing the loop reverses the sign. The two displacements must remain distinct: contracting an antisymmetric curvature pair with $A^\mu A^\nu$ would give zero. Curvature is therefore infinitesimal holonomy, while finite-loop holonomy also retains global and noncommutative information.

The first check prevents a classic coordinate error. Euclidean plane geometry in polar coordinates has $ds^2=dr^2+r^2d\theta^2$ and nonzero coefficients $\Gamma^r{}_{\theta\theta}=-r$ and $\Gamma^\theta{}_{r\theta}=1/r$. Nevertheless,

$$R^r{}_{\theta r\theta} =\partial_r(-r)-\Gamma^r{}_{\theta\theta}\Gamma^\theta{}_{r\theta} =-1+1=0.$$

This polar chart has $r\gt0$. Connection coefficients can be nonzero in flat space. Conversely, on a round two-sphere of radius $a\gt0$, away from the coordinate poles $\theta=0,\pi$,

$$ds^2=a^2(d\theta^2+\sin^2\theta\,d\phi^2), \qquad R_{\theta\phi\theta\phi}=a^2\sin^2\theta,$$

so the invariant Gaussian curvature is $1/a^2$. These examples independently check both cancellation in a curvilinear flat chart and the sign on a positively curved surface. A systematic derivation and the tensoriality proof appear in Calegari 2022, §7.1, pp. 53–54.

Interpretation and boundary of scope

Curvature is local and geometric. It does not by itself assign a gravitational field equation, a stress tensor, or an operational tidal measurement; those require additional physics developed in the General Relativity volume. Nor does $R=0$ on a patch by itself decide the topology or the holonomy of all finite loops. What it does decide is local flatness for a torsion-free metric connection: vanishing Riemann curvature permits local parallel frames and locally constant metric components.

The tensoriality of $R$ also suggests an efficient calculation strategy. At a selected point one may choose normal coordinates for a Levi–Civita connection, making $\Gamma^\rho{}_{\mu\nu}=0$ there. The quadratic $\Gamma\Gamma$ terms then vanish at that point, but the derivatives $\partial\Gamma$ generally remain. Curvature is precisely the obstruction that prevents one from making the connection vanish throughout a neighborhood. This distinction explains why the equivalence principle’s local inertial coordinates do not erase curvature.

For an explicit metric, first record the coordinate domain and compute only the connection coefficients that can contribute to the desired component. Then test the antisymmetry $R^\rho{}_{\sigma\mu\nu}=-R^\rho{}_{\sigma\nu\mu}$ before contracting. Finally perform a coordinate-independent check: a flat limit, a constant-curvature identity, or small-loop transport. This sequence catches swapped derivative indices and missing quadratic terms earlier than computing a large scalar contraction.

Dimensional analysis is another useful guard. If coordinates carry length and the metric components are dimensionless, $\Gamma$ has inverse-length dimension and Riemann has inverse-length squared dimension. On the sphere the only scale is $a$, so the sectional curvature must be proportional to $a^{-2}$; the explicit component contains metric factors and therefore has the coordinate-basis value $a^2\sin^2\theta$. Confusing a component with the invariant sectional curvature would miss exactly those basis factors.

Exercises

1. Dual curvature. Show directly that the commutator on a one-form has the opposite sign from that on a vector.

Solution

For the scalar $f=\omega_\rho V^\rho$, $[\nabla_\mu,\nabla_\nu]f=0$. Apply the Leibniz rule:

$$0=([\nabla_\mu,\nabla_\nu]\omega_\rho)V^\rho +\omega_\rho R^\rho{}_{\sigma\mu\nu}V^\sigma.$$

Relabel $\rho$ and $\sigma$ in the second term. Since $V$ is arbitrary, $[\nabla_\mu,\nabla_\nu]\omega_\rho=-R^\sigma{}_{\rho\mu\nu}\omega_\sigma$.

2. Curvature from a metric. For $ds^2=a^2(d\theta^2+\sin^2\theta\,d\phi^2)$, compute $R^\theta{}_{\phi\theta\phi}$ from the connection coefficients and recover the sectional curvature.

Solution

The needed coefficients are $\Gamma^\theta{}_{\phi\phi}=-\sin\theta\cos\theta$ and $\Gamma^\phi{}_{\theta\phi}=\cot\theta$. Therefore

$$\begin{aligned} R^\theta{}_{\phi\theta\phi} &=\partial_\theta(-\sin\theta\cos\theta) -(-\sin\theta\cos\theta)\cot\theta\\ &=\sin^2\theta. \end{aligned}$$

Lowering the first index gives $R_{\theta\phi\theta\phi}=a^2\sin^2\theta$. Dividing by $g_{\theta\theta}g_{\phi\phi}-g_{\theta\phi}^2=a^4\sin^2\theta$ gives $K=a^{-2}$. The positive sign and inverse-length-squared dimension are separate checks.

References

• Danny Calegari, Introduction to Differential Geometry, preliminary version, May 26, 2022, University of Chicago. PDF.
• David Tong, General Relativity, University of Cambridge Part III Mathematical Tripos lecture notes, 2019. PDF.