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.

  • 6
    $\begingroup$ That $\lambda_{a;b}$ really is an object with two lower indices is the whole point of the covariant derivative! $\endgroup$
    – ACuriousMind
    Commented Nov 21, 2014 at 16:57
  • 3
    $\begingroup$ $\lambda_{a;b}$ is a rank-2 tensor. If you take its covariant derivative you'll get two connection terms by definition of the covariant derivative. $\endgroup$
    – Jold
    Commented Nov 21, 2014 at 16:58
  • $\begingroup$ Oh. I didn't realise you treat $\lambda_{a;b}$ as having two lower indices. That does tend to put things in a new light. $\endgroup$
    – Peter4075
    Commented Nov 21, 2014 at 17:11
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    $\begingroup$ Indeed. As ACuriousMind mentioned, this is the whole point of defining the covariant derivative: so that things remain covariant. I.e. tensors go to tensors under the operation. $\endgroup$
    – Jold
    Commented Nov 21, 2014 at 17:19
  • $\begingroup$ The semi-colon threw me. $\endgroup$
    – Peter4075
    Commented Nov 21, 2014 at 17:53

3 Answers 3


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,

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

  • $\begingroup$ Thanks. Does $T^{d...a_{n}}$ mean $T^{da_{2}...a_{n}}$ etc? Do you have a reference for the derivation of this equation? $\endgroup$
    – Peter4075
    Commented Nov 21, 2014 at 21:10
  • $\begingroup$ @Peter4075: Yes, precisely, you'll have $a_2,a_3,\dots, a_n$. For a derivation, look up any GR book. $\endgroup$
    – JamalS
    Commented Nov 21, 2014 at 21:13
  • $\begingroup$ OK. It was that particular detailed equation for a general tensor I was interested in, but I'll take a look around. $\endgroup$
    – Peter4075
    Commented Nov 21, 2014 at 21:22
  • $\begingroup$ @Peter4075: Misner, Thorne and Wheeler's book will definitely have it... somewhere... $\endgroup$
    – JamalS
    Commented Nov 21, 2014 at 21:51
  • $\begingroup$ Carroll gives it on p58 here: arxiv.org/pdf/gr-qc/9712019.pdf. I notice that the upper index on the negative Gammas and the lower indices on the positive Gammas are constant, whilst the lower indices on the negative Gammas and the upper index on the positive Gammas are not. $\endgroup$
    – Peter4075
    Commented Nov 22, 2014 at 7:35

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.


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:

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


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