Riemann Curvature Tensor

Definition of Riemann Curvature Tensor

Let $M$ be a manifold equipped with a covariant derivative $\nabla$.

A parametrized surface in $M$ is a smooth function that maps $\mathbb{R}$ to $M$ $(\sigma,\tau)\mapsto p=p(\sigma,\tau)$ and is defined on some rectangle $\{(\sigma,\tau)||\sigma - \sigma_0|\leq a, |\tau - \tau_0|\leq b\}$

Let $X = X(\sigma,\tau)$ be a smooth vector field along this parametrized surface $p(\sigma,\tau)$. $U=\partial p(\sigma,\tau)/\partial \sigma$ and $V=\partial p(\sigma,\tau)/\partial\tau$ are parameter vector fields, which are tangential to the surface. The covariant derivatives along the $\sigma$-lines and $\tau$-lines are denoted as $\nabla/\partial \sigma$ and $\nabla/\partial\tau$ respectively. The curvature tensor of the connection $\nabla$ is defined as $$R(U,V)X := \frac{\nabla}{\partial \tau}\frac{\nabla X}{\partial \sigma} - \frac{\nabla}{\partial\sigma}\frac{\nabla X}{\partial \tau}$$

Geometrical Meaning

Let $U_0$, $V_0$ and $W$ be three vectors at the point p_0, such that $\partial p/\partial\sigma|_{(\sigma_0,\tau_0)} = U_0$, $\partial p/\partial\tau|_{(\sigma_0,\tau_0)}=V_0$ and $W = X(p_0)$. Let $T=T(\sigma_0,\tau_0;\Delta\sigma,\Delta\tau)$ be a linear transformation operates on a vector such that $TW = \text{parallel transport of }W$ along the boundary $\{p(\sigma,\tau)|\sigma_0\leq\sigma\leq\sigma_0+\Delta\sigma,\tau_0\leq\tau\leq\Delta\tau\}$

The vector after parallel transport along the boundary after one loop can be broken down by the four edges of the rectangle $$TW = T(\sigma_0,\tau\to\tau_0)T(\sigma\to\sigma_0,\tau)T(\sigma,\tau_0\to\tau)T(\sigma_0\to\sigma,\tau_0)W$$

Since $T(\sigma\to\sigma_0,\tau)T(\sigma_0\to\sigma,\tau)=I$, an identity, $$\lim_{\Delta\sigma,\Delta\tau\to 0}\frac{1}{\Delta\sigma\Delta\tau}(W-TW) =\lim_{\Delta\sigma,\Delta\tau\to 0}\frac{1}{\Delta\sigma\Delta\tau} \left[T(\sigma_0,\tau\to\tau_0)T(\sigma\to\sigma_0,\tau)\right] \left[T(\sigma_0\to\sigma,\tau_0)T(\sigma_0,\tau_0\to\tau)W-T(\sigma,\tau_0\to\tau)T(\sigma_0\to\sigma,\tau_0)W\right]\tag{1}$$ Since $X(\sigma_0,\tau_0) = W$, after adding some terms, $$\lim_{\Delta\sigma,\Delta\tau\to 0}\frac{1}{\Delta\sigma\Delta\tau}(W-TW) = \lim_{\Delta\sigma,\Delta\tau\to 0}\frac{1}{\Delta\sigma\Delta\tau} T(\sigma_0\to\sigma,\tau)\left[T(\sigma_0,\tau_0\to\tau)X(\sigma_0,\tau_0)-X(\sigma_0,\tau)\right] + \left[T(\sigma_0\to\sigma,\tau)X(\sigma_0,\tau)-X(\sigma,\tau)\right] - T(\sigma,\tau_0\to\tau)\left[T(\sigma_0\to\sigma,\tau_0)X(\sigma_0,\tau_0) - X(\sigma,\tau_0)\right] - \left[T(\sigma,\tau_0\to\tau)X(\sigma,\tau_0) - X(\sigma,\tau)\right]$$

Since $X(t)= \xi^i(t)E_i(t)$, where $E_1(t),...,E_n(t)$ are of a Euclidean frame, $$\begin{align} \frac{\nabla X}{dt} &= \frac{d\xi^i}{dt}E_i(t)\\ &=\lim_{\epsilon\to 0}\frac{1}{\epsilon}\left(\xi^i(t+\epsilon)-\xi^i(t)\right)E_i(t)\\ &= \lim_{\epsilon\to 0}\frac{1}{\epsilon}\left(\xi^i(t+\epsilon)-\xi^i(t)\right)E_i(t+\epsilon)\\ &= \lim_{\epsilon\to 0}\frac{1}{\epsilon}\left(\xi^i(t+\epsilon)E_i(t+\epsilon)-\xi^i(t)E_i(t+\epsilon)\right)\\ &= \lim_{\epsilon\to 0}\frac{1}{\epsilon}\left(X(t+\epsilon)-T(t\to t+\epsilon)X(t)\right) \end{align}$$ we have $$\begin{align} \lim_{\Delta\sigma,\Delta\tau\to 0}\frac{1}{\Delta\sigma\Delta\tau}(W-TW) &= \lim_{\Delta\sigma,\Delta\tau\to 0}\frac{1}{\Delta\sigma\Delta\tau} \left[-\frac{1}{\Delta\sigma}T(\sigma_0\to\sigma,\tau)\left(\frac{\nabla X}{\partial\tau}\right)_{\sigma_0,\tau_0} -\frac{1}{\Delta\tau}\left(\frac{\nabla X}{\partial\sigma}\right)_{\sigma_0,\tau} +\frac{1}{\Delta\tau}T(\sigma,\tau_0\to\tau)\left(\frac{\nabla X}{\partial \sigma}\right)_{(\sigma_0,\tau_0)} +\frac{1}{\Delta\sigma}\left(\frac{\nabla X}{\partial\tau}\right)_{(\sigma,\tau_0)}\right]\\ &= \lim_{\Delta\sigma,\Delta\tau\to 0} \left[\frac{1}{\Delta\sigma} \left[ \left(\frac{\nabla X}{\partial\tau}\right)_{(\sigma,\tau_0)} - T(\sigma_0\to\sigma,\tau)\left(\frac{\nabla X}{\partial \tau}\right)_{(\sigma_0,\tau_0)} \right] + \frac{1}{\Delta\tau} \left[ T(\sigma,\tau_0\to\tau)\left(\frac{\nabla X}{\partial\sigma}\right)_{(\sigma_0,\tau_0)} -\left(\frac{\nabla X}{\partial \sigma}\right)_{(\sigma_0,\tau)} \right] \right]\\ \end{align}$$

Since the 1st term is the covariant derivative of $\frac{\nabla X}{\partial\tau}$ along the $\sigma$-lines and the 2nd term is the covariant derivative of $\frac{\nabla X}{\partial\sigma}$ along the $\tau$-lines, we have

$$\begin{align} \lim_{\Delta\sigma,\Delta\tau\to 0}\frac{1}{\Delta\sigma\Delta\tau}(W-TW) &= \left(\frac{\nabla}{\partial\sigma}\frac{\nabla X}{\partial\tau} - \frac{\nabla}{\partial\tau}\frac{\nabla X}{\partial\sigma}\right)_{(\sigma_0,\tau_0)}\\ &= R(U_0,V_0)W. \end{align}$$ We conclude that the Riemann curvature tensor of a vector $W$ at a point $p_0$ is the difference between $W$ and its parallel transport along an infinitely small rectangle loop. $$\bbox[5px,border:2px solid #666] { \lim_{\Delta\sigma,\Delta\tau\to 0}\frac{1}{\Delta\sigma\Delta\tau}(W-TW) = R(U_0,V_0)W. }$$ If the manifold is flat, parallel transport of a vector does not change the vector at all, hence has zero curvature. The more curved the manifold, the larger the difference after transporting the loop. This is a way to measure how curved a manifold is, without knowing how its normal vector changes w.r.t the Euclidean space. We can still tell how curved is the space even with the set of changing coordinate system.

Component Form

$$\begin{align} \frac{\nabla W}{\partial\sigma} &= \frac{\partial W^i}{\partial\sigma}\partial_i + W^i\frac{\nabla\partial_i}{\partial\sigma}\\ \frac{\nabla}{\partial\tau}\frac{\nabla W}{\partial\sigma} &= \frac{\partial^2W}{\partial\tau\partial\sigma} + \frac{\partial W^i}{\partial\sigma}\frac{\nabla\partial_i}{\partial\tau} + \frac{\partial W^i}{\partial\tau}\frac{\nabla\partial_i}{\partial\sigma} + W^i\frac{\nabla^2\partial_i}{\partial\tau\partial\sigma}. \end{align}$$ Similarly, $$\frac{\nabla}{\partial \sigma}\frac{\nabla W}{\partial\tau} = \frac{\partial^2 W}{\partial\tau\partial\sigma}\partial_i + \frac{\partial W}{\partial \tau}\frac{\nabla\partial_i}{\partial\sigma} + \frac{\partial W}{\partial \sigma}\frac{\nabla \partial_i}{\partial\tau} + W^i\frac{\nabla^2\partial_i}{\partial\sigma\partial\tau}.$$ Since partial derivatives commute, subtracting will give $$\frac{\nabla}{\partial \tau}\frac{\nabla W}{\partial \sigma} - \frac{\nabla}{\partial\sigma}\frac{\nabla W}{\partial \tau} = W^i\frac{\nabla^2\partial_i}{\partial\tau\partial\sigma} + W^i\frac{\nabla^2\partial_i}{\partial\sigma\partial\tau}.$$ Since $\frac{\nabla}{\partial\tau}\partial_i = \frac{\partial x^j}{\partial \tau}\frac{\nabla}{\partial x^j}\partial_i$ and $\frac{\nabla}{\partial x^j}\partial_i = \Gamma^k_{ij}\partial_k$, $$\begin{align} \frac{\nabla\partial_i}{\partial\sigma} &= \frac{\partial x^j}{\partial\sigma}\frac{\nabla}{\partial x^j} = \frac{\partial x^j}{\partial\sigma}\Gamma^k_{ij}\partial_k\\ \frac{\nabla^2\partial_i}{\partial\tau\partial\sigma} &= \frac{\nabla}{\partial\tau}\left(\frac{\partial x^j}{\partial\sigma}\Gamma^k_{ij}\partial_k\right)\\ &= \frac{\partial^2x^j}{\partial\tau\partial\sigma}\Gamma^k_{ij}\partial_k + \frac{\partial x^j}{\partial\sigma}\left(\frac{\partial\Gamma^k_{ij}}{\partial\tau}\partial_k + \Gamma^k_{ij}\frac{\nabla\partial_k}{\partial\tau}\right)\\ &= \frac{\partial^2x^j}{\partial\tau\partial\sigma}\Gamma^k_{ij}\partial_k + \frac{\partial x^j}{\partial\sigma}\left(\frac{\partial\Gamma^k_{ij}}{\partial\tau}\partial_k + \Gamma^k_{ij}\frac{\partial x^l}{\partial\tau}\frac{\nabla}{\partial x^l}\partial_m\right)\\ &= \frac{\partial^2x^j}{\partial\tau\partial\sigma}\Gamma^k_{ij}\partial_k + \frac{\partial x^j}{\partial\sigma}\left(\frac{\partial\Gamma^k_{ij}}{\partial\tau}\partial_k + \Gamma^k_{ij}\frac{\partial x^l}{\partial\tau}\Gamma^n_{kl}\partial_n\right) \\ &= \frac{\partial^2x^j}{\partial\tau\partial\sigma}\Gamma^k_{ij}\partial_k + \left(\frac{\partial x^j}{\partial\sigma}\right)\left(\frac{\partial x^l}{\partial\tau}\right)\left(\frac{\partial\Gamma^k_{ij}}{\partial x^l} + \Gamma^n_{ij}\Gamma^k_{nl}\right)\partial_k.\\ \end{align}$$ Similarly for the 2nd term, $$\frac{\nabla\partial_i}{\partial\sigma\partial\tau} = \frac{\partial^2x^j}{\partial\sigma\partial\tau}\Gamma^k_{ij}\partial_k + \left(\frac{\partial x^j}{\partial\tau}\right)\left(\frac{\partial x^l}{\partial\sigma}\right)\left(\frac{\partial\Gamma^k_{ij}}{\partial x^l} + \Gamma^n_{ij}\Gamma^k_{nl}\right)\partial_k.$$ Replacing dummy variable $l$ with $j$ and $j$ with $l$, $$\frac{\nabla\partial_i}{\partial\sigma\partial\tau} = \frac{\partial^2x^j}{\partial\sigma\partial\tau}\Gamma^k_{ij}\partial_k + \left(\frac{\partial x^l}{\partial \tau}\right)\left(\frac{\partial x^j}{\partial\sigma}\right)\left(\frac{\partial\Gamma^k_{il}}{\partial x^j} + \Gamma^n_{il}\Gamma^k_{nj}\right)\partial_k.$$ Subtracting the two terms, $$\frac{\nabla}{\partial \tau}\frac{\nabla W}{\partial \sigma} - \frac{\nabla}{\partial\sigma}\frac{\nabla W}{\partial \tau} = W^i\left(\frac{\partial x^j}{\partial\sigma}\right)\left(\frac{\partial x^l}{\partial\tau}\right)\left(\partial_l\Gamma^k_{ij} + \Gamma^n_{ij}\Gamma^k_{nl} - \partial_j\Gamma^k_{il} - \Gamma^n_{il}\Gamma^k_{nj}\right)\partial_k.$$ Since $U^j = \frac{\partial x^j}{\partial\sigma}$ and $V^l = \frac{\partial x^l}{\partial\tau}$, $$R(U,V)W = R^k_{ijl}U^jV^lW^i\partial_k,$$ where $$\bbox[5px,border:2px solid #666] { R^k_{ijl} = \partial_l\Gamma^k_{ij} + \Gamma^n_{ij}\Gamma^k_{nl} - \partial_j\Gamma^k_{il} - \Gamma^n_{il}\Gamma^k_{nj}. }$$

Bianchi Identities

By Riemann tensor defintion, $$R_{\alpha\beta\mu\nu,\lambda} = \frac{1}{2}(g_{\alpha\nu,\beta\mu\lambda}-g_{\alpha\mu,\beta\nu\lambda}+g_{\beta\mu,\alpha\nu\lambda}-g_{\beta\nu,\alpha\mu\lambda}).$$ Since the metric tensor is symmetric and the partial derivatives commute, $$R_{\alpha\beta\mu\nu,\lambda}+R_{\alpha\beta\lambda\mu,\nu}+R_{\alpha\beta\nu\lambda,\mu}=0.$$ Since the Christoffel symbols vanish locally, $$R_{\alpha\beta\mu\nu;\lambda}+R_{\alpha\beta\lambda\mu;\nu}+R_{\alpha\beta\nu\lambda;\mu} = 0.$$ This is known as the Bianchi identities.

Ricci Tensor

The Ricci tensor is defined as the trace of the linear transformation $u \to R(u,v)w$, i.e. $$\text{Ric}(v,w)=R^k_{qkl}v^l w^q,$$ which is the contraction of $R^\mu_{\alpha\nu\beta}$ on the first and third indices. So, one may write the Ricci tensor as $$R_{\alpha\beta} := R^\mu_{\alpha\mu\beta} = R_{\beta\alpha}.$$ Since the Riemann tensor is antisymmetric on the first and second indices and the third and fourth indices, the contraction on these pairs are zero. Other contraction will reduce to $\pm R_{\alpha\beta}$ because of the symmetry of the Riemann tensor in other indices. Therefore, the Ricci tensor actually tells all the contraction of a Riemann tensor. Define the Ricci scalar as $$R := g^{\mu\nu}R_{\mu\nu} = g^{\mu\nu}g^{\alpha\beta}R_{\alpha\mu\beta\nu}$$

Einstein Tensor

By the Bianchi identities, $$g^{\alpha\mu}(R_{\alpha\beta\mu\nu;\lambda}+R_{\alpha\beta\lambda\mu;\nu}+R_{\alpha\beta\nu\lambda;\mu})=0.$$ As $g^{\alpha\beta}_{;\mu}=0$, the metric tensor can be moved into the covariant derivative, $$R_{\beta\nu;\lambda}+(-R_{\beta\lambda;\nu})+R^\mu_{\beta\nu\lambda;\mu}=0.$$ Contracting again $$\begin{align} g^{\beta\nu}(R_{\beta\nu;\lambda}-R_{\beta\lambda;\nu}+R^\mu_{\beta\nu\lambda;\mu})&=0\\ R_{;\lambda}-R^\mu_{\lambda;\mu}+(-R^\mu_{\lambda;\mu}) &=0. \end{align}$$ Since $R$ is a scalar, $R_{;\lambda}=R_{,\lambda}$, $$(2R^\mu_\lambda - \delta^\mu_\lambda R)_{;\lambda}=0.$$ Define the Einstein tensor as $$G^{\alpha\beta}\equiv R^{\alpha\beta}-\frac{1}{2}g^{\alpha\beta}R = G^{\beta\alpha}.$$ Then, we have $$G^{\alpha\beta}_{;\beta}=0$$

Reference: Lectures on Differential Geometry by Wulf Rossmann.