Curvature

The Riemann curvature tensor $R$ is defined by:

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

In local coordinates: $R^i_{\,jkl}\, \frac{\partial}{\partial x^i} = R\left(\frac{\partial}{\partial x^k}, \frac{\partial}{\partial x^l}\right)\frac{\partial}{\partial x^j}$.

Symmetries of the Riemann tensor:

1. $R_{ijkl} = -R_{jikl}$ (anti-symmetric in first two indices).
2. $R_{ijkl} = -R_{ijlk}$ (anti-symmetric in last two indices).
3. $R_{ijkl} = R_{klij}$ (pair symmetry).
4. $R_{ijkl} + R_{iklj} + R_{iljk} = 0$ (first Bianchi identity).

For linearly independent $u, v \in T_p M$, the sectional curvature of the 2-plane $\mathrm{span}\{u, v\}$ is:

$K(u, v) = \frac{\langle R(u, v)v, u\rangle}{\|u\|^2 \|v\|^2 - \langle u, v\rangle^2}$

Example. $\mathbb{R}^n$ has $K \equiv 0$ (flat). $S^n$ has $K \equiv 1$ (constant positive curvature). $H^n$ (hyperbolic space) has $K \equiv -1$ (constant negative curvature).

The Ricci tensor is the trace of the Riemann tensor:

$\mathrm{Ric}(X, Y) = \mathrm{tr}(Z \mapsto R(Z, X)Y) = \sum_i \langle R(e_i, X)Y, e_i\rangle$

In coordinates: $\mathrm{Ric}_{jk} = R^i_{\,jik}$.

The scalar curvature is the trace of the Ricci tensor:

$S = \mathrm{tr}_g(\mathrm{Ric}) = g^{jk}\, \mathrm{Ric}_{jk}$

In general relativity, spacetime is a 4-dimensional Lorentzian manifold $(M, g)$. The Einstein field equations relate the curvature of spacetime to the stress-energy tensor:

$R_{\mu\nu} - \frac{1}{2} R g_{\mu\nu} + \Lambda g_{\mu\nu} = \frac{8\pi G}{c^4} T_{\mu\nu}$

where $R_{\mu\nu}$ is the Ricci tensor, $R$ is the scalar curvature, $\Lambda$ is the cosmological constant, and $T_{\mu\nu}$ is the stress-energy tensor.

Theorem 7.1 (Second Bianchi Identity). The covariant derivative of the Riemann tensor satisfies:

$\nabla_{[i} R_{jk]lm} = 0$

Taking a contraction gives the contracted Bianchi identity:

$\nabla^i \mathrm{Ric}_{ij} = \frac{1}{2} \nabla_j S$

This implies that the Einstein tensor $G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2}R g_{\mu\nu}$ is divergence-free: $\nabla^\mu G_{\mu\nu} = 0$, which is consistent with the conservation of energy-momentum.

The Riemann tensor can be decomposed into the Ricci part and the Weyl tensor $W$:

$R_{ijkl} = C_{ijkl} + \frac{1}{n-2}(g_{ik}R_{jl} - g_{il}R_{jk} - g_{jk}R_{il} + g_{jl}R_{ik}) - \frac{R}{(n-1)(n-2)}(g_{ik}g_{jl} - g_{il}g_{jk})$

where $C_{ijkl}$ is the Weyl tensor. The Weyl tensor is trace-free (all contractions vanish) and has the same symmetries as the Riemann tensor. In dimension $n \geq 4$, $C = 0$ if and only if the manifold is conformally flat.

A Riemannian manifold is an Einstein manifold if the Ricci tensor is proportional to the metric:

$\mathrm{Ric} = \lambda g$

for some constant $\lambda$. In this case, the scalar curvature $S = n\lambda$ is constant. Examples include space forms (constant sectional curvature) and Calabi-Yau manifolds (Ricci-flat, $\lambda = 0$).

In local coordinates, the components of the Riemann tensor are expressed in terms of Christoffel symbols:

$R^i_{\,jkl} = \partial_k \Gamma^i_{jl} - \partial_l \Gamma^i_{jk} + \Gamma^i_{km} \Gamma^m_{jl} - \Gamma^i_{lm} \Gamma^m_{jk}$

Worked example. For the 2-sphere $S^2$ with metric $g = d\theta^2 + \sin^2\theta\, d\phi^2$, the only non-zero Christoffel symbols are $\Gamma^\phi_{\theta\phi} = \cot\theta$ and $\Gamma^\theta_{\phi\phi} = -\sin\theta\cos\theta$. The Riemann tensor has a single independent component:

$R^\theta_{\,\phi\theta\phi} = \sin^2\theta$

from which $R_{\theta\phi\theta\phi} = \sin^2\theta$ and $K = 1$.

Problem 1. Compute the Riemann tensor for the Poincaré half-plane $\mathbb{H}^2 = \{(x, y) : y > 0\}$ with metric $g = (dx^2 + dy^2)/y^2$.

Problem 2. Show that the scalar curvature of a product manifold $M \times N$ is the sum of the scalar curvatures: $S_{M\times N} = S_M + S_N$.

Problem 3. Prove that if $\mathrm{Ric} = \lambda g$ on a connected manifold, then $\lambda$ is constant.

Problem 4. For the Schwarzschild metric $ds^2 = -(1 - 2M/r) dt^2 + (1 - 2M/r)^{-1} dr^2 + r^2 d\Omega^2$, verify that the Ricci tensor vanishes (vacuum solution).

The Gauss equation relates the curvature of a submanifold $N \subseteq M$ to the curvature of $M$ and the second fundamental form $II$:

$\langle R_N(X, Y)Z, W\rangle = \langle R_M(X, Y)Z, W\rangle + \langle II(X, Z), II(Y, W)\rangle - \langle II(X, W), II(Y, Z)\rangle$

For a surface in $\mathbb{R}^3$, the Gauss equation gives the Gaussian curvature $K$ as the product of principal curvatures: $K = \kappa_1 \kappa_2$.

The holonomy group of a connection measures how parallel transport around closed loops changes vectors. The Riemann curvature tensor is the infinitesimal holonomy: for an infinitesimal parallelogram spanned by $X, Y$, parallel transport around the loop rotates a vector $Z$ by $R(X, Y)Z$.

Example. On $S^2$, parallel transport around a spherical triangle rotates a vector by an angle equal to the area of the triangle (Gauss-Bonnet).

Theorem 7.2 (Gauss-Bonnet). For a compact, oriented Riemannian 2-manifold $M$:

$\int_M K\, dA = 2\pi \chi(M)$

where $\chi(M)$ is the Euler characteristic. This deep result links local curvature to global topology.

Problem 5. Show that for a surface of revolution obtained by rotating $y = f(x)$ around the $x$-axis, the Gaussian curvature is $K = -f''(x)/(f(x)(1 + f'(x)^2)^2)$.

Problem 6. Compute the Ricci tensor and scalar curvature of $S^2 \times S^2$ with the product metric.

Problem 7. Prove that a Riemannian manifold with constant sectional curvature $\kappa$ is Einstein with $\mathrm{Ric} = (n-1)\kappa\, g$.