$\nabla_a \nabla_b v^c$ abstract index notation

Question

I am having some trouble with the equivalence between abstract index notation (AIN) and standard tensor component notation (TCN, for short).

Let us consider a covariant derivative $\nabla$. In TCN, we can define it, for a given basis of tangent vectors $e_i$, by the relation $$ \nabla_i e_j = \Gamma_{ij}^k e_k\,, $$ where $\nabla_i=\nabla_{e_i}$. Then, for any vector field $v=v^i e_i$, we have $$ \nabla_i v = (\partial_iv^j + \Gamma^j_{ik}v^k)e_j \implies (\nabla_i v)^j=\partial_iv^j + \Gamma^j_{ik}v^k $$ since $\nabla_i(v^j e_j)=(\nabla_i v^j) e_j + v^j(\nabla_i e_j)$ and $\nabla_i v^j=\partial_i v^j$. In AIN, this translates to $$ \nabla_a v^b = \partial_a v^b + \Gamma_{ac}^b v^c\,, $$ where however $a, b,c,\ldots$ are just abstract indices and not components with respect to a specific basis.

Let us now consider the second derivative. In TCN, we have $$ \nabla_i \nabla_j v = \nabla_i((\partial_j v^k+ \Gamma^k_{jl}v^l)e_k)=(\partial_i(\partial_j v^k+ \Gamma^k_{jl}v^l)+(\partial_j v^m+ \Gamma^m_{jl}v^l)\Gamma_{im}^k)e_k\,. $$ Therefore, $$ (\nabla_i \nabla_j v)^k=\partial_i \partial_j v^k + v^l \partial_i \Gamma^k_{jl}+\Gamma^k_{jl}\partial_i v^l + \Gamma_{il}^k \partial_j v^l+ \Gamma^k_{im}\Gamma^{m}_{jl}v^l\,. $$ On the other hand, in AIN, $$ \nabla_a \nabla_b v^c = \partial_a \nabla_b v^c + \Gamma^c_{ad}\nabla_b v^d-\Gamma_{ab}^d \nabla_d v^c\,, $$ because we need to treat the lower $b$ index in $\nabla_b$ according to its covariant nature, and therefore $$ \nabla_a \nabla_b v^c = \partial_a \partial_b v^c + v^d\partial_a\Gamma^c_{bd} + \Gamma^c_{bd}\partial_a v^d + \Gamma^c_{ad}\partial_b v^d + \Gamma^c_{ad}\Gamma^{d}_{be}v^e-\Gamma_{ab}^d(\partial_d v^c + \Gamma_{de}^c v^e)\,. $$ But the last term is not there in TCN!

Do I have to conclude that somehow these two notations are not really equivalent? (I always assumed they were...) The unwanted piece cancels in the calculation of the commutator $[\nabla_a,\nabla_b]v^c$, provided the connection is symmetric, so this is not really an issue in GR, but still I would like to understand what goes wrong.

Answer 1

If we define the dual basis $\tilde e^i(e_j)=\delta^i_j$ and we ask that this relation is preserved by the covariant derivative, then $$ (\nabla_k \tilde e^i)(e_j) = - \Gamma^l_{kj}\tilde e^i(e_l) = - \Gamma^i_{kj} \implies (\nabla_k \tilde e^i)_j = - \Gamma^i_{kj}\,. $$ Then $$ \nabla_{i}\left(\nabla_{j} v \otimes \tilde e^j\right)=\left(\partial_{i}\left(\partial_{j} v^{k}+\Gamma_{j l}^{k} v^{l}\right)+\left(\partial_{j} v^{m}+\Gamma_{j l}^{m} v^{l}\right) \Gamma_{i m}^{k}\right) e_{k}\otimes \tilde e^j - (\partial_m v^k+\Gamma^k_{ml}v^l)e_k\otimes \Gamma^m_{ij}\tilde e^{j} $$ so we may write $$ \left(\nabla_{i}\left(\nabla v\right)\right)^k_j = \partial_{i} \partial_{j} v^{k}+v^{l} \partial_{i} \Gamma_{j l}^{k}+\Gamma_{j l}^{k} \partial_{i} v^{l}+\Gamma_{i l}^{k} \partial_{j} v^{l}+\Gamma_{i m}^{k} \Gamma_{j l}^{m} v^{l} - (\partial_m v^k+\Gamma^k_{ml}v^l) \Gamma^m_{ij}\,. $$ So it seems that in order to obtain the same result in both notations it is important to keep track of the transformation law of dual basis elements as well.