Differential Geometry: Manifolds, Tensors & Connections
Differential Geometry: Manifolds, Tensors & Connections
General Relativity is formulated on a smooth 4-manifold equipped with a metric tensor. Tensors are the natural language of coordinate-independent physics. The Levi-Civita connection provides the unique torsion-free, metric-compatible way to differentiate tensor fields on a curved manifold.
Covariant Derivative & Christoffel Symbols
The covariant derivative of a contravariant vector: \(\nabla_\mu V^\nu = \partial_\mu V^\nu + \Gamma^\nu_{\mu\lambda} V^\lambda\). The Christoffel symbol (Levi-Civita connection): \(\Gamma^\lambda_{\mu\nu} = \frac{1}{2}g^{\lambda\sigma}(\partial_\mu g_{\nu\sigma} + \partial_\nu g_{\mu\sigma} - \partial_\sigma g_{\mu\nu})\). It is symmetric: \(\Gamma^\lambda_{\mu\nu} = \Gamma^\lambda_{\nu\mu}\).
Metric Compatibility
The Levi-Civita connection satisfies \(\nabla_\lambda g_{\mu\nu} = 0\) (metric compatibility) and zero torsion \(T^\lambda_{\mu\nu} = \Gamma^\lambda_{\mu\nu} - \Gamma^\lambda_{\nu\mu} = 0\). These two conditions uniquely determine \(\Gamma^\lambda_{\mu\nu}\) from the metric. Raising/lowering commutes with \(\nabla\).
Example 1: Christoffel Symbols on \(S^2\)
For \(ds^2 = d\theta^2 + \sin^2\theta\,d\phi^2\), compute the nonzero Christoffel symbols.
Solution: \(g_{\theta\theta}=1\), \(g_{\phi\phi}=\sin^2\theta\). Only nonzero: \(\Gamma^\theta_{\phi\phi} = -\sin\theta\cos\theta\), \(\Gamma^\phi_{\theta\phi} = \Gamma^\phi_{\phi\theta} = \cot\theta\).
Example 2: Tensor Transformation Law
Verify that \(T^{\mu\nu}\) transforms as \(T'^{\mu\nu} = \frac{\partial x'^\mu}{\partial x^\alpha}\frac{\partial x'^\nu}{\partial x^\beta}T^{\alpha\beta}\) under coordinate change.
Solution: By the chain rule applied twice to contravariant components. This defines a rank-\((2,0)\) tensor: each upper index contributes one Jacobian factor.
Practice
- Compute all Christoffel symbols for \(ds^2 = e^{2\Phi}(-dt^2 + dr^2 + r^2 d\Omega^2)\).
- Show \(\nabla_\mu g_{\nu\lambda} = 0\) using the definition of \(\Gamma\).
- How many independent components does \(\Gamma^\lambda_{\mu\nu}\) have in 4D?
- Define the Lie derivative \(\mathcal{L}_V T^{\mu\nu}\) and show it is tensorial.
Show Answer Key
1. For $ds^2 = e^{2\Phi}(-dt^2+dr^2+r^2d\Omega^2)$: the metric is conformally flat. Christoffel symbols: $\Gamma^t_{tr} = \Gamma^t_{rt} = \Phi'$, $\Gamma^r_{tt} = \Phi'$, $\Gamma^r_{rr} = \Phi'$, $\Gamma^r_{\theta\theta} = -(r+r^2\Phi')e^0... $ Actually for a conformal metric $g_{\mu\nu}=e^{2\Phi}\bar{g}_{\mu\nu}$: $\Gamma^\lambda_{\mu\nu} = \bar{\Gamma}^\lambda_{\mu\nu} + \delta^\lambda_\mu\partial_\nu\Phi + \delta^\lambda_\nu\partial_\mu\Phi - \bar{g}_{\mu\nu}\bar{g}^{\lambda\sigma}\partial_\sigma\Phi$. Computing from flat-space $\bar{\Gamma}$ in spherical coordinates gives all symbols.
2. $\Gamma^\lambda_{\mu\nu} = \frac{1}{2}g^{\lambda\sigma}(\partial_\mu g_{\nu\sigma}+\partial_\nu g_{\mu\sigma}-\partial_\sigma g_{\mu\nu})$. $\nabla_\mu g_{\nu\lambda} = \partial_\mu g_{\nu\lambda}-\Gamma^\sigma_{\mu\nu}g_{\sigma\lambda}-\Gamma^\sigma_{\mu\lambda}g_{\nu\sigma}$. Substituting the Christoffel formula: the three terms combine to give $\partial_\mu g_{\nu\lambda} - \frac{1}{2}(\partial_\mu g_{\nu\lambda}+\partial_\nu g_{\mu\lambda}-\partial_\lambda g_{\mu\nu}) - \frac{1}{2}(\partial_\mu g_{\lambda\nu}+\partial_\lambda g_{\mu\nu}-\partial_\nu g_{\mu\lambda}) = 0$. ✓ (Metric compatibility defines the Levi-Civita connection.)
3. In 4D: $\Gamma^\lambda_{\mu\nu}$ is symmetric in $\mu\nu$ (10 pairs), with $\lambda$ taking 4 values → $4 \times 10 = 40$ independent components.
4. $\mathcal{L}_V T^{\mu\nu} = V^\sigma\partial_\sigma T^{\mu\nu} - T^{\sigma\nu}\partial_\sigma V^\mu - T^{\mu\sigma}\partial_\sigma V^\nu$. Although written with partial derivatives, it transforms as a tensor because the non-tensorial parts cancel (the combination is independent of coordinates). Proof: replace $\partial$ with $\nabla$ — the connection terms cancel due to torsion-freeness. ✓