Ricci’s Theorem

Relevant parts to questions...

• $g_{ik,\ell }=0$: the covariant derivative of the Metric Tensor vanishes - the metric is covariantly constant.
• Useful rearrangement::$\frac{\partial g_{ik}}{\partial x^{\ell }}={\Gamma }_{ik\ell }+{\Gamma }_{ki\ell }$ - expresses metric derivatives via Christoffel Symbols.
• In orthogonal coordinates, this forces ${\Gamma }_{ik\ell }=-{\Gamma }_{ki\ell }$ for $i\ne k$ - first-kind antisymmetry.
• Applies only to first-kind symbols; second-kind do not enjoy the same sign antisymmetry.

Ricci's Theorem

The covariant derivative of the Metric Tensor vanishes:

$g_{ik,\ell }=0$

Equivalently, the metric is covariantly constant. Covariant Differentiation is compatible with the inner-product structure defined by $g_{ik}$.

Proof

Apply the rank-2 covariant-derivative formula to $T_{ik}=g_{ik}$:

$$g_{ik,\ell }=\frac{\partial g_{ik}}{\partial x^{\ell }}-{\Gamma }_{i\ell }^m{g}_{mk}-{\Gamma }_{k\ell }^m{g}_{im}$$

Using the metric to lower an index on $\Gamma$: ${\Gamma }_{i\ell }^m{g}_{mk}={\Gamma }_{ki\ell }$ and ${\Gamma }_{k\ell }^m{g}_{im}={\Gamma }_{ik\ell }$. So:

$$g_{ik,\ell }=\frac{\partial g_{ik}}{\partial x^{\ell }}-{\Gamma }_{ki\ell }-{\Gamma }_{ik\ell }$$

Now apply the first-kind metric formula to each Christoffel symbol and add them. The cross-terms cancel in pairs (using $g_{ki}=g_{ik}$), leaving:

$${\Gamma }_{ik\ell }+{\Gamma }_{ki\ell }=\frac{\partial g_{ik}}{\partial x^{\ell }}$$

Substituting back gives $g_{ik,\ell }=\frac{\partial g_{ik}}{\partial x^{\ell }}-\frac{\partial g_{ik}}{\partial x^{\ell }}=\boxed{0}$. ✓

Consequences

Useful Formula for Metric Derivatives

Rearranging the proof:

$$\boxed{\frac{\partial g_{ik}}{\partial x^{\ell }}={\Gamma }_{ik\ell }+{\Gamma }_{ki\ell }}$$

This lets you read metric derivatives off Christoffel symbols (or, conversely, constrain $\Gamma$ from known metric derivatives).

Antisymmetry in Orthogonal Coordinates

In orthogonal coordinates, $g_{ik}=0$ for $i\ne k$, so $\frac{\partial g_{ik}}{\partial x^{\ell }}=0$ whenever $i\ne k$. The useful formula then forces:

$$\boxed{{\Gamma }_{ik\ell }=-{\Gamma }_{ki\ell }(\text{orthogonal coords},i\ne k)}$$

First-kind Christoffel symbols are antisymmetric in their first two indices when the coordinates are orthogonal. This cuts compute time in half for worked examples.

Second-kind symbols don't obey this

${\Gamma }_{jk}^i$ does not satisfy ${\Gamma }_{jk}^i=-{\Gamma }_{ik}^j$ in general, because raising the first index with $g^{i\ell }$ destroys the simple sign relation (the two raised symbols get multiplied by different metric entries).

Properties

• Coordinate-invariant statement::$g_{ik,\ell }=0$ holds in every coordinate system - not just orthogonal ones.
• Raises and lowers indices commute with $\nabla$::because $g$ is covariantly constant, $(g^{i\ell }{A}_{\ell }{)}_{,k}=g^{i\ell }{A}_{\ell ,k}$. You can raise/lower indices through a covariant derivative freely.
• Ricci’s theorem for $g^{ik}$::$g_{,\ell }^{ik}=0$ too, by the same calculation applied to the contravariant metric.

Applications

1. Computing Christoffel symbols efficiently, by combining Ricci’s antisymmetry (orthogonal case) with the lower-index symmetry ${\Gamma }_{ijk}={\Gamma }_{ikj}$.
2. Pulling the metric through covariant derivatives, to simplify tensor-calculus expressions.
3. Setting up the Riemann-Christoffel Tensor::the derivation in lecture 18 uses Ricci’s theorem to cancel certain metric-derivative terms.

Orthogonal $(\rho ,\tau ,\theta )$ with $g_{22}={\rho }^2-{\tau }^2$

Direct calculation gives ${\Gamma }_{122}=-\frac{1}{2}\frac{\partial g_{22}}{\partial \rho }=-\rho$.

Ricci (orthogonal, $i=2,k=1,\ell =2$): ${\Gamma }_{212}=-{\Gamma }_{122}=\rho$. Then lower-index symmetry gives ${\Gamma }_{221}={\Gamma }_{212}=\rho$.

$\boxed{{\Gamma }_{122}=-\rho ,{\Gamma }_{212}=\rho ,{\Gamma }_{221}=\rho }$ - three symbols from a single derivative.

Metric is covariantly constant. Orthogonal coords ⇒ first-kind antisymmetry in the first two indices.