Unit 6: Curvature Tensors: Riemann, Ricci & Bianchi Identities

1. The Non-Commutativity of Covariant Differentiation

In flat Euclidean space with Cartesian coordinates, partial derivatives commute unconditionally: $\partial_j \partial_k A_i = \partial_k \partial_j A_i$. On a curved manifold, however, covariant differentiation is non-commutative:

This commutator measures the failure of infinitesimal loops to close, which is the intrinsic geometric definition of curvature.

2. Derivation of the Riemann Curvature Tensor

Let $A_i$ be an arbitrary smooth covariant vector field. Compute its second covariant derivative:

Differentiating covariantly with respect to $x^j$:

Now interchange indices $j$ and $k$ to compute $\nabla_k (\nabla_j A_i)$:

Subtracting the two equations:

• $\partial_j \partial_k A_i - \partial_k \partial_j A_i = 0$ (commutation of ordinary partials).
• $-\Gamma^m_{ki} \partial_j A_m$ and $-\Gamma^m_{ji} \partial_k A_m$ cancel by subtraction.
• In a torsion-free Levi-Civita connection, $\Gamma^m_{jk} = \Gamma^m_{kj}$, so $-\Gamma^m_{jk} \nabla_m A_i + \Gamma^m_{kj} \nabla_m A_i = 0$.

All derivative terms of $A$ vanish, leaving only terms proportional to $A_l$:

We define the quantity in brackets as the Riemann curvature tensor (Ricci identity):

Definition 6.1 (Riemann Curvature Tensor of Type $(1, 3)$):

where:

(Sign convention note: in modern differential geometry and general relativity, the definition is often written as $[\nabla_j, \nabla_k] A^l = R^l_{\, ijk} A^i$).

1. The Fully Covariant Riemann Tensor $R_{hijk}$

By lowering the contravariant index using the metric tensor $g_{hl}$:

In terms of metric derivatives and Christoffel symbols:

2. Fundamental Algebraic Symmetries

Theorem 6.1 (Algebraic Symmetries of $R_{hijk}$):

1. Antisymmetry in First Pair: $R_{hijk} = -R_{ihjk}$
2. Antisymmetry in Second Pair: $R_{hijk} = -R_{hikj}$
3. Block Symmetry (Pair Exchange): $R_{hijk} = R_{jkhi}$
4. First (Algebraic) Bianchi Identity:

Proof of Antisymmetry in First Pair:

From Ricci's Theorem $\nabla_k g_{hi} = 0$, applying $[\nabla_j, \nabla_k] g_{hi} = 0$:

3. Number of Independent Components in $n$ Dimensions

A rank-4 tensor has $n^4$ total components. How many are algebraically independent?

Theorem 6.2 (Independent Components Count): In an $n$-dimensional Riemannian manifold, the number of algebraically independent components of the Riemann curvature tensor is:

Step-by-Step Combinatorial Proof:

1. $R_{hijk}$ is antisymmetric in $(hi)$ and in $(jk)$.

The number of independent antisymmetric pairs is $M = \frac{n(n-1)}{2}$.

1. Treat $R_{(hi)(jk)}$ as a symmetric $M \times M$ matrix of pairs (due to block symmetry $R_{AB} = R_{BA}$).

The number of independent components in a symmetric $M \times M$ matrix is:

1. The first Bianchi identity $R_{h[ijk]} = 0$ imposes additional non-trivial constraints whenever all 4 indices $h, i, j, k$ are distinct.

The number of completely distinct 4-index combinations is $\binom{n}{4} = \frac{n(n-1)(n-2)(n-3)}{24}$.

1. Subtracting these constraints:

Factoring out $\frac{n(n-1)}{24}$:

Table of Independent Components:

• $n = 1$: $\frac{1(0)}{12} = 0$ (A 1D curve has no intrinsic curvature!).
• $n = 2$: $\frac{4(3)}{12} = 1$ component ($R_{1212} = K g$, where $K$ is the Gaussian curvature).
• $n = 3$: $\frac{9(8)}{12} = 6$ components (identical to the 6 components of the Ricci tensor).
• $n = 4$: $\frac{16(15)}{12} = 20$ components (crucial for Einstein's General Relativity).

1. The Ricci Curvature Tensor $R_{ij}$

By contracting the contravariant index with the third covariant index of the Riemann tensor:

Definition 6.2 (Ricci Tensor):

In terms of Christoffel symbols:

Symmetry of the Ricci Tensor:

Using block symmetry and index lowering:

The Ricci tensor is strictly symmetric: $R_{ij} = R_{ji}$. In $n$ dimensions, $R_{ij}$ has $\frac{n(n+1)}{2}$ independent components (10 components in 4D).

2. The Ricci Scalar Curvature $R$

Contracting the Ricci tensor with the conjugate metric tensor produces the Ricci scalar (scalar curvature):

$R$ is an absolute invariant scalar field.

• In 2D, $R = 2K$, where $K$ is the Gaussian curvature.
• If $R > 0$, the volume of a geodesic ball is smaller than in Euclidean space.
• If $R < 0$, the volume of a geodesic ball is larger than in Euclidean space.

3. The Einstein Tensor $G_{ij}$

Definition 6.3 (Einstein Tensor): The Einstein tensor $G_{ij}$ is defined as the trace-reversed Ricci tensor:

Trace of the Einstein Tensor in $n$ Dimensions:

In $n = 4$ spacetime dimensions:

The Einstein tensor is symmetric ($G_{ij} = G_{ji}$) and plays the central geometric role on the left side of Einstein's field equations.

1. The Second (Differential) Bianchi Identity

Beyond the algebraic symmetries, the Riemann tensor satisfies a fundamental differential relation:

Theorem 6.3 (Second Bianchi Identity): In any Riemannian or Pseudo-Riemannian space:

or in fully covariant form:

Proof using Riemann Normal Coordinates:

At any chosen point $P$, choose Riemann normal coordinates where $\left. \Gamma^k_{ij} \right|_P = 0$. At point $P$, covariant derivatives reduce to ordinary partial derivatives:

Summing over cyclic permutations of $(j, k, m)$:

All 12 terms cancel in pairs because third partial derivatives of $g$ commute! Since this is a tensorial equation valid at an arbitrary point $P$, it holds universally in all coordinate systems! $\blacksquare$

2. The Twice-Contracted Bianchi Identity

Theorem 6.4 (Divergence-Free Property of the Einstein Tensor): The covariant divergence of the Einstein tensor vanishes identically:

Proof:

Start with the second Bianchi identity in mixed form:

Contract indices $l$ and $k$ (setting $k = l$):

Now contract with $g^{ij}$:

By Ricci's Theorem, $\nabla$ passes through $g^{ij}$:

1. $g^{ij} \nabla_m R_{ij} = \nabla_m (g^{ij} R_{ij}) = \nabla_m R$.
2. $g^{ij} R^l_{ijm} = -g^{ij} R^l_{imj} = -R^l_{\ m}$. Contracting with $\nabla_l$ gives $-\nabla_l R^l_m$.
3. $g^{ij} \nabla_j R_{im} = \nabla_j R^j_m = \nabla_l R^l_m$.

Substituting:

Dividing by 2 and moving to one side:

Recognizing $G^l_m = R^l_m - \frac{1}{2} R \delta^l_m$:

This mathematical identity is why Einstein chose $G_{\mu\nu}$ for gravitation: since the stress-energy tensor is conserved ($\nabla_\mu T^{\mu\nu} = 0$), the geometric tensor on the left side of $G_{\mu\nu} = \kappa T_{\mu\nu}$ MUST automatically satisfy $\nabla_\mu G^{\mu\nu} = 0$!

1. Sectional Curvature $K(\Pi)$

Let $\Pi = \text{span}(u, v)$ be a 2-dimensional tangent plane spanned by two linearly independent vectors $u^i, v^i$ at $P \in M$.

Definition 6.4 (Sectional Curvature): The sectional curvature of $M$ associated with the 2-plane $\Pi$ is:

$K(\Pi)$ is the Gaussian curvature of the 2D geodesic surface formed by geodesics emanating from $P$ tangent to $\Pi$.

2. Spaces of Constant Curvature

A manifold is called a space of constant curvature if $K(\Pi) = K_0$ is independent of the choice of 2-plane $\Pi$ and point $P$. In such spaces, the Riemann tensor takes the maximally symmetric form:

Contracting with $g^{hj}$:

Contracting with $g^{ik}$:

3. Schur's Theorem

Theorem 6.5 (Schur's Theorem): If in a connected Riemannian manifold of dimension $n \ge 3$, the sectional curvature $K$ at each point is independent of the 2-plane direction $\Pi$, then $K$ is constant throughout the entire manifold ($\partial_m K = 0$).

Proof:

If $K(\Pi)$ is isotropic at each point, then $R_{hijk} = K(x) (g_{hk} g_{ij} - g_{hj} g_{ik})$. The Ricci tensor is $R_{ij} = (n-1) K(x) g_{ij}$, and scalar curvature is $R = n(n-1) K(x)$. The Einstein tensor is:

Now apply the twice-contracted Bianchi identity $\nabla^i G_{ij} = 0$:

Since $\nabla_i g_{ij} = 0$:

For $n \ge 3$, $(n-1)(2-n) \ne 0$. Therefore, we must have:

This proves that spatial isotropy at every point automatically forces spatial homogeneity across the entire manifold!