Derivation of the Riemann tensor confusion

Question

I'm trying to understand the derivation of the Riemann curvature tensor as given in Foster and Nightingale's A Short Course In General Relativity, p. 102. They start by giving the covariant derivative of a covariant vector field $\lambda_{a}$: $$\lambda_{a;b}=\partial_{b}\lambda_{a}-\Gamma_{ab}^{d}\lambda_{d}.$$ Which is OK. They then do a second covariant differentiation to get$$\lambda_{a;bc}=\partial_{c}\left(\lambda_{a;b}\right)-\Gamma_{ac}^{e}\lambda_{e;b}-\Gamma_{bc}^{e}\lambda_{a;e}.$$ And I'm lost. I can understand the first term on the rhs, but why are there two connection coefficient terms. I would expect two negative connection coefficient terms if they were taking the covariant derivative of $\lambda_{xy}$, but not $\lambda_{a;b}$. Is it correct to treat the covariant derivative $\lambda_{a;b}$ as having two lower indices? Actually, I find the second and third rhs terms completely baffling.

Answer 1

The covariant derivative for a general tensor of the form $T^{a_1\dots a_n}_{b_1 \dots b_n}$ is given by,

$$\nabla_c T^{a_1\dots a_n}_{b_1 \dots b_n} = \partial_c T^{a_1\dots a_n}_{b_1 \dots b_n} + \Gamma^{a_1}_{cd}T^{d\dots a_n}_{b_1 \dots b_n} + \dots - \Gamma^d_{c b_1}T^{a_1\dots a_n}_{d \dots b_n} - \dots$$

Taking the covariant derivative of a covariant field $V_a$, we find,

$$\nabla_b V_a = \partial_b V_a - \Gamma^c_{ba}V_c$$

Now, the object $\nabla_b V_a$ has two lower indices, so taking the covariant derivative again, we find,

$$\nabla_c (\nabla_b V_a) = \partial_c(\nabla_b V_a) - \Gamma^d_{cb} (\nabla_d V_a) - \Gamma^d_{ca}(\nabla_b V_d)$$

Inserting the original covariant derivative, we find explicitly,

$$\nabla_c (\nabla_b V_a) = \partial_c (\partial_b V_a -\Gamma^{e}_{ba}V_e) - \Gamma^d_{cb}(\partial_d V_a - \Gamma^e_{da}V_e) - \Gamma^d_{ca}(\partial_b V_d - \Gamma^e_{bd}V_e)$$

Answer 2

A covariant derivative of a tensor is itself a tensor. Actually, when we say something is covariant (or invariant under coordinate transformation), we mean that thing is a tensor. So, in this case $\nabla_\mu V^\nu\equiv T_\mu{}^\nu$. Now calculate $\nabla_\alpha T_\mu{}^\nu$ easily.

\begin{equation} \nabla_\alpha T_\mu{}^\nu=\partial_\alpha T_\mu{}^\nu+\Gamma^\nu_{\alpha \beta} T_{\mu}{}^\beta -\Gamma^\beta_{\alpha \mu} T_{\beta}{}^\nu \end{equation} Each gamma term in the right hand side is due to one of the indices. Note how the plus or minus sign has appeared in front of each one.

P.S., Do not compare a tensor with a matrix. The meaning of tensor is completely different.

Answer 3

The first term: $$\begin{align} \nabla_i \nabla_j T^k & = \frac{\delta(\nabla_j T^k)}{\delta Z^i} - \Gamma_{ij}^m\nabla_m T^k + \Gamma_{im}^k\nabla_j T^m \\ &= \frac{\delta^2 T^k}{\delta Z^i \delta Z^j} + \frac{\delta(\Gamma^k_{jm}T^m)}{\delta Z^i} - \Gamma_{ij}^m(\frac{\delta T^k}{\delta Z^m} + \Gamma_{ml}^k T^l) + \Gamma_{im}^k(\frac{\delta T^m}{\delta Z^j} + \Gamma_{jl}^m T^l) \\ &= \color{red}{\frac{\delta^2 T^k}{\delta Z^i \delta Z^j}} + \frac{\delta\Gamma^k_{jm}}{\delta Z^i}T^m + \color{green}{\frac{\delta T^m}{\delta Z^i}\Gamma^k_{jm}} - \color{tan}{\Gamma_{ij}^m(\frac{\delta T^k}{\delta Z^m} + \Gamma_{ml}^k T^l)} + \Gamma_{im}^k(\color{plum}{\frac{\delta T^m}{\delta Z^j}} + \Gamma_{jl}^m T^l) \end{align}$$

The second term: $$\begin{align} \nabla_j \nabla_i T^k = \color{red}{\frac{\delta^2 T^k}{\delta Z^j \delta Z^i} } + \frac{\delta\Gamma^k_{im}}{\delta Z^j}T^m + \color{plum}{\frac{\delta T^m}{\delta Z^j}\Gamma^k_{im}} - \color{tan}{\Gamma_{ji}^m(\frac{\delta T^k}{\delta Z^m} + \Gamma_{ml}^k T^l)} + \Gamma_{jm}^k(\color{green}{\frac{\delta T^m}{\delta Z^i}} + \Gamma_{il}^m T^l) \end{align}$$

Subtracting second from first we have something to cancel out. And you left with four terms which aren't colored:

$$\begin{align} \nabla_i \nabla_j T^k - \nabla_j \nabla_i T^k &= \frac{\delta\Gamma^k_{jm}}{\delta Z^i}T^m - \frac{\delta\Gamma^k_{im}}{\delta Z^j}T^m + \Gamma_{im}^k\Gamma_{jl}^m T^l - \Gamma_{jm}^k\Gamma_{il}^m T^l \\ &= \Big(\frac{\delta\Gamma^k_{jl}}{\delta Z^i} - \frac{\delta\Gamma^k_{il}}{\delta Z^j} + \Gamma_{im}^k\Gamma_{jl}^m - \Gamma_{jm}^k\Gamma_{il}^m \Big)T^l \end{align}$$