Riemann Tensor from the Covariant Commutator
Statement
For a torsion-free connection compatible with the metric, the failure of two covariant derivatives to commute on a vector field defines the Riemann curvature tensor through \( [\nabla_\mu,\nabla_\nu]V^\rho = R^\rho{}_{\sigma\mu\nu}\,V^\sigma \), where \( R^\rho{}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho{}_{\nu\sigma} - \partial_\nu\Gamma^\rho{}_{\mu\sigma} + \Gamma^\rho{}_{\mu\lambda}\Gamma^\lambda{}_{\nu\sigma} - \Gamma^\rho{}_{\nu\lambda}\Gamma^\lambda{}_{\mu\sigma} \); this object carries the algebraic symmetries \( R_{\rho\sigma\mu\nu}=-R_{\sigma\rho\mu\nu}=-R_{\rho\sigma\nu\mu}=R_{\mu\nu\rho\sigma} \), the first (algebraic) Bianchi identity \( R^\rho{}_{[\sigma\mu\nu]}=0 \), and the second (differential) Bianchi identity \( \nabla_{[\lambda}R^\rho{}_{|\sigma|\mu\nu]}=0 \).
Why it matters
Curvature is what cannot be transformed away. A nonzero Christoffel symbol only signals a bad coordinate choice (polar coordinates on a flat plane already produce them), but the commutator of covariant derivatives is a genuine tensor: if it vanishes in one frame it vanishes in all. The Riemann tensor is therefore the coordinate-independent measure of how spacetime deviates from flatness, and its vanishing is the exact necessary-and-sufficient condition for a metric to be Minkowski in disguise.
Every geometric statement in general relativity is built from this tensor. Geodesic deviation (tidal forces), holonomy around a closed loop, the Einstein field equations through the Ricci contraction, and the conservation of the stress-energy tensor via the contracted second Bianchi identity all descend directly from the object and symmetries derived here.
Assumptions
Derivation
Result
Reading. Parallel-transporting a vector around an infinitesimal coordinate parallelogram spanned by the \(\mu\) and \(\nu\) directions rotates it by an amount \(R^\rho{}_{\sigma\mu\nu}\) per unit area; the tensor packages that rotation. The four algebraic symmetries \(R_{\rho\sigma\mu\nu}=-R_{\sigma\rho\mu\nu}=-R_{\rho\sigma\nu\mu}=R_{\mu\nu\rho\sigma}\) plus the first Bianchi identity cut the naive \(n^4\) components down to \(\tfrac{1}{12}n^2(n^2-1)\) — 20 in four dimensions. The second Bianchi identity is the differential constraint that, once contracted, guarantees \(\nabla_\mu G^{\mu\nu}=0\) and hence local energy-momentum conservation.
Units check. \(\Gamma\) has dimension \([\text{length}]^{-1}\) (it is \(\partial g/g\)), so \(\partial\Gamma\) and \(\Gamma\Gamma\) both carry \([\text{length}]^{-2}\); the Riemann tensor with two lower derivative indices therefore has dimension \([\text{length}]^{-2}\), matching a curvature (inverse radius squared). Both the \(\partial\Gamma\) and \(\Gamma\Gamma\) groups share this dimension, so the sum is homogeneous.
Limiting cases
- Flat spacetime: a global chart exists with \(\Gamma=0\), so \(R^\rho{}_{\sigma\mu\nu}\equiv0\) and covariant derivatives commute — the converse (vanishing Riemann \(\Rightarrow\) flat) also holds locally.
- Weak field \(g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}\), \(|h|\ll1\): the \(\Gamma\Gamma\) terms are second order and drop, giving the linearised \(R^\rho{}_{\sigma\mu\nu}\approx\tfrac12\eta^{\rho\alpha}(\partial_\mu\partial_\sigma h_{\alpha\nu}+\partial_\nu\partial_\alpha h_{\mu\sigma}-\partial_\mu\partial_\alpha h_{\nu\sigma}-\partial_\nu\partial_\sigma h_{\alpha\mu})\), the seed of gravitational waves.
- Two dimensions: only one independent component survives, \(R_{\rho\sigma\mu\nu}=K(g_{\rho\mu}g_{\sigma\nu}-g_{\rho\nu}g_{\sigma\mu})\) with \(K\) the Gaussian curvature; the Ricci scalar is \(R=2K\).
- Maximally symmetric space (sphere, de Sitter): \(R_{\rho\sigma\mu\nu}=\dfrac{R}{n(n-1)}(g_{\rho\mu}g_{\sigma\nu}-g_{\rho\nu}g_{\sigma\mu})\) with constant scalar \(R\).
Breaks when
- Torsion present. For a connection with \(T^\lambda{}_{\mu\nu}=\Gamma^\lambda{}_{\mu\nu}-\Gamma^\lambda{}_{\nu\mu}\neq0\) (Einstein–Cartan theory), the commutator becomes \([\nabla_\mu,\nabla_\nu]V^\rho = R^\rho{}_{\sigma\mu\nu}V^\sigma - T^\lambda{}_{\mu\nu}\nabla_\lambda V^\rho\); Step 8 no longer cancels and the first Bianchi identity gains torsion source terms.
- Non-metric connection. If \(\nabla_\lambda g_{\mu\nu}=Q_{\lambda\mu\nu}\neq0\), Step 13 fails: \(R_{\rho\sigma\mu\nu}\) loses its antisymmetry in the first index pair and the pair-exchange symmetry, so the component-counting formula \(\tfrac{1}{12}n^2(n^2-1)\) no longer applies.
- Curvature singularities. Where a scalar invariant such as the Kretschmann scalar \(R_{\rho\sigma\mu\nu}R^{\rho\sigma\mu\nu}\) diverges (e.g. \(r\to0\) in Schwarzschild), the tensor is not defined and the construction breaks down physically, not merely in coordinates.
- Non-differentiable metric. At a thin shell or cosmic string the connection is discontinuous, \(\partial\Gamma\) is distributional, and the Riemann tensor contains a Dirac-delta piece rather than a smooth field.
Failure modes
- Index-order amnesia. Writing \(R^\rho{}_{\sigma\mu\nu}=\partial_\mu\Gamma^\rho{}_{\nu\sigma}-\partial_\nu\Gamma^\rho{}_{\mu\sigma}+\dots\) with the \(\Gamma\Gamma\) indices in the wrong slots. The upper index of the outer \(\Gamma\) is contracted with the lower of the inner \(\Gamma\); flipping them changes the sign convention or gives a wrong tensor.
- Sign-convention mixing. Different texts (MTW, Weinberg, Wald) differ by overall signs and by whether \(R^\rho{}_{\sigma\mu\nu}\) or \(R^\rho{}_{\sigma\nu\mu}\) is defined. Combining a Christoffel from one book with a Riemann formula from another produces spurious minus signs.
- Treating \(\partial_\mu\partial_\nu V^\rho\) as curvature. Students expect the second partial to be antisymmetric; it is symmetric and cancels (Step 6). Curvature lives entirely in the \(\Gamma\) terms.
- Confusing nonzero \(\Gamma\) with nonzero curvature. Polar coordinates on the flat plane have \(\Gamma^r{}_{\phi\phi}=-r\neq0\) yet \(R^\rho{}_{\sigma\mu\nu}=0\) (see Worked Example 2). Christoffel symbols are not tensors; only their curvature combination is.
- Forgetting the lower-index \(\Gamma\) in Step 2. Omitting the \(-\Gamma^\lambda{}_{\mu\nu}W^\rho{}_\lambda\) term when differentiating the \((1,1)\) tensor. It is precisely this term whose torsion-free symmetry is used in Step 8.
- Applying pair-exchange symmetry without metric compatibility. \(R_{\rho\sigma\mu\nu}=R_{\mu\nu\rho\sigma}\) requires Step 13; invoking it for a general affine connection is invalid.
Discussion
The deepest content of this derivation is that curvature is a local obstruction to integrability. Covariant derivatives commuting would mean that parallel transport is path-independent, that a globally flat frame exists, and that the vector's components can be integrated up unambiguously. The Riemann tensor measures exactly the amount by which that integration fails around an infinitesimal loop. This is the tensorial heart of the equivalence principle: gravity cannot be detected in a single freely-falling frame (where \(\Gamma\) can be set to zero at a point) but shows up in the second derivatives \(\partial\Gamma\) as the relative acceleration of neighbouring geodesics — tidal forces — encoded by \(R^\rho{}_{\sigma\mu\nu}\) in the geodesic deviation equation \(\tfrac{D^2\xi^\rho}{d\tau^2}=-R^\rho{}_{\sigma\mu\nu}u^\sigma\xi^\mu u^\nu\).
The symmetries are not decorative. Antisymmetry in each index pair lets one view \(R_{\rho\sigma\mu\nu}\) as a symmetric matrix on the space of 2-forms (bivectors), which is what makes the sectional-curvature and Petrov-classification pictures possible. The first Bianchi identity is the statement that the totally antisymmetric part vanishes — a purely algebraic constraint tied to the torsion-free condition. The second Bianchi identity, contracted twice, yields \(\nabla_\mu\big(R^{\mu\nu}-\tfrac12 g^{\mu\nu}R\big)=0\); the object in parentheses is the Einstein tensor, and its automatic divergencelessness is why Einstein's equations \(G^{\mu\nu}=8\pi G\,T^{\mu\nu}/c^4\) are consistent with local conservation \(\nabla_\mu T^{\mu\nu}=0\). The mathematics of the identity forces the physics of conservation.
The contractions organise the physical content. The Ricci tensor \(R_{\sigma\nu}=R^\rho{}_{\sigma\rho\nu}\) and scalar \(R=g^{\sigma\nu}R_{\sigma\nu}\) capture how volumes converge (they are sourced directly by matter through Einstein's equations), while the trace-free remainder — the Weyl tensor \(C_{\rho\sigma\mu\nu}\) — captures tidal shape distortion and free gravitational radiation. In vacuum the Ricci part vanishes but Weyl need not, which is how gravitational waves propagate through empty space where \(R_{\mu\nu}=0\).
A more structural view identifies the Riemann tensor as the curvature 2-form \(\mathcal{R}=d\omega+\omega\wedge\omega\) of the Levi-Civita connection 1-form \(\omega\) on the frame bundle, in exact analogy with the Yang–Mills field strength \(F=dA+A\wedge A\). In this language the second Bianchi identity is simply \(D\mathcal{R}=0\), the geometric counterpart of the homogeneous Maxwell equations \(dF=0\); the algebraic first Bianchi identity, by contrast, has no Yang–Mills analogue and is special to the soldered tangent-bundle connection, arising from the torsion-free condition \(D\theta^a=0\) on the canonical solder form. Curvature holonomy — the group element by which a vector is rotated around a loop — is the integral of \(\mathcal{R}\), tying this local tensor to the global topology through Chern–Gauss–Bonnet-type theorems.
Common misconceptions. The Riemann tensor is not the gradient of a force, and its vanishing is a statement about the whole neighbourhood, not one point: at any single point one can always choose Riemann-normal coordinates in which \(g_{\mu\nu}=\eta_{\mu\nu}\) and \(\Gamma=0\), yet \(\partial\Gamma\neq0\) still records curvature. Nonzero Christoffel symbols never by themselves prove curved space; only the specific antisymmetric combination \(R^\rho{}_{\sigma\mu\nu}\) does.
Worked examples
Reading. The Gaussian curvature is the constant \(1/a^2\), independent of \(\theta\) and \(\phi\) as required for a homogeneous sphere; a unit sphere (\(a=1\)) gives \(K=1\). Units check. With \(a\) in metres, \(K\) has units \([\text{m}]^{-2}\), correct for a curvature; a sphere of radius \(a=6.37\times10^6\,\text{m}\) (Earth) gives \(K\approx2.46\times10^{-14}\,\text{m}^{-2}\).
Reading. Every component vanishes: the plane is flat even though the coordinate basis rotates (nonzero \(\Gamma\)). Curvature is the specific combination \(\partial\Gamma-\partial\Gamma+\Gamma\Gamma-\Gamma\Gamma\), not the presence of Christoffels. Units check. The two surviving terms are pure numbers here (dimensionless in these coordinates because \(r\) carries the length); their exact cancellation to \(0\) is dimensionally consistent and coordinate-independent — a tensor that is zero in one chart is zero in all.
Problems
- Show explicitly that \([\nabla_\mu,\nabla_\nu]f = 0\) for any scalar field \(f\) on a torsion-free manifold, and explain why the analogous statement fails for a vector.
Solution
\(\nabla_\nu f=\partial_\nu f\). Then \(\nabla_\mu\nabla_\nu f=\partial_\mu\partial_\nu f-\Gamma^\lambda{}_{\mu\nu}\partial_\lambda f\) (the \(\nabla_\nu f\) is a covector, so a \(-\Gamma\) appears). Antisymmetrising: \(\partial_\mu\partial_\nu f-\partial_\nu\partial_\mu f=0\) and \(\Gamma^\lambda{}_{\mu\nu}=\Gamma^\lambda{}_{\nu\mu}\) makes the Christoffel term cancel. So \([\nabla_\mu,\nabla_\nu]f=0\). For a vector the extra index brings a \(+\Gamma^\rho{}_{\mu\lambda}\) term whose \(\partial\Gamma\) and \(\Gamma\Gamma\) pieces do not cancel, leaving \(R^\rho{}_{\sigma\mu\nu}V^\sigma\). - For a covector \(\omega_\rho\), derive \([\nabla_\mu,\nabla_\nu]\omega_\rho = -R^\sigma{}_{\rho\mu\nu}\,\omega_\sigma\) and confirm the sign relative to the vector case.
Solution
Repeat the vector derivation with the lower-index rule \(\nabla_\nu\omega_\rho=\partial_\nu\omega_\rho-\Gamma^\sigma{}_{\nu\rho}\omega_\sigma\). Differentiating again and antisymmetrising, the symmetric pieces cancel exactly as before, leaving \(-(\partial_\mu\Gamma^\sigma{}_{\nu\rho}-\partial_\nu\Gamma^\sigma{}_{\mu\rho}+\Gamma^\sigma{}_{\mu\lambda}\Gamma^\lambda{}_{\nu\rho}-\Gamma^\sigma{}_{\nu\lambda}\Gamma^\lambda{}_{\mu\rho})\omega_\sigma=-R^\sigma{}_{\rho\mu\nu}\omega_\sigma\). The sign is opposite to the vector case, consistent with contracting: \([\nabla_\mu,\nabla_\nu](V^\rho\omega_\rho)=0\) since \(V^\rho\omega_\rho\) is a scalar, and indeed \(R^\rho{}_{\sigma\mu\nu}V^\sigma\omega_\rho - R^\sigma{}_{\rho\mu\nu}\omega_\sigma V^\rho=0\). - Count the independent components of \(R_{\rho\sigma\mu\nu}\) in \(n=4\) using the symmetries, and separately confirm the general formula \(\tfrac{1}{12}n^2(n^2-1)\) for \(n=2\) and \(n=3\).
Solution
Antisymmetry in each pair reduces each pair to \(\binom{n}{2}=N_p\) values; pair-exchange symmetry makes \(R\) a symmetric \(N_p\times N_p\) matrix with \(\tfrac12 N_p(N_p+1)\) entries. For \(n=4\), \(N_p=6\), giving \(21\). The first Bianchi identity imposes \(\binom{n}{4}=1\) extra constraint, leaving \(20\). General formula: \(\tfrac{1}{12}n^2(n^2-1)\). Check \(n=2\): \(\tfrac{1}{12}\cdot4\cdot3=1\). Check \(n=3\): \(\tfrac{1}{12}\cdot9\cdot8=6\). Check \(n=4\): \(\tfrac{1}{12}\cdot16\cdot15=20\). All agree. - The Ricci tensor is \(R_{\sigma\nu}=R^\rho{}_{\sigma\rho\nu}\). Using the pair-exchange symmetry, prove \(R_{\sigma\nu}=R_{\nu\sigma}\).
Solution
\(R_{\sigma\nu}=R^\rho{}_{\sigma\rho\nu}=g^{\rho\lambda}R_{\lambda\sigma\rho\nu}\). Apply pair-exchange \(R_{\lambda\sigma\rho\nu}=R_{\rho\nu\lambda\sigma}\): \(R_{\sigma\nu}=g^{\rho\lambda}R_{\rho\nu\lambda\sigma}\). Relabel dummies \(\rho\leftrightarrow\lambda\): \(=g^{\lambda\rho}R_{\lambda\nu\rho\sigma}=R^\rho{}_{\nu\rho\sigma}=R_{\nu\sigma}\). Hence the Ricci tensor is symmetric — a consequence that requires metric compatibility (through the pair-exchange symmetry). - Contract the second Bianchi identity twice to obtain \(\nabla_\mu\big(R^{\mu\nu}-\tfrac12 g^{\mu\nu}R\big)=0\).
Solution
Start from \(\nabla_\lambda R_{\rho\sigma\mu\nu}+\nabla_\mu R_{\rho\sigma\nu\lambda}+\nabla_\nu R_{\rho\sigma\lambda\mu}=0\). Contract with \(g^{\rho\mu}\) (using \(\nabla g=0\)): \(\nabla_\lambda R_{\sigma\nu}-\nabla_\mu R^\mu{}_{\sigma\nu\lambda}-\nabla_\nu R_{\sigma\lambda}=0\), where the middle term used antisymmetry. Now contract with \(g^{\sigma\nu}\): \(\nabla_\lambda R-\nabla_\mu R^\mu{}_\lambda-\nabla_\nu R^\nu{}_\lambda=0\), i.e. \(\nabla_\lambda R-2\nabla_\mu R^\mu{}_\lambda=0\). Rearranging, \(\nabla_\mu\big(R^\mu{}_\lambda-\tfrac12\delta^\mu{}_\lambda R\big)=0\); raising the index gives \(\nabla_\mu\big(R^{\mu\nu}-\tfrac12 g^{\mu\nu}R\big)=0\), the contracted Bianchi identity guaranteeing \(\nabla_\mu G^{\mu\nu}=0\).