Covariant derivative of a covariant derivative

Question

I'm trying to find the covariant derivative of a covariant derivative, i.e. $\nabla_a (\nabla_b V_c)$.

This is something I've taken for granted a lot in calculations, namely I though that by the Leibniz rule we just have:

$$\nabla_a (\nabla_b V_c) = \partial_a(\nabla_b V_c) - \Gamma_{ab}^{d}\nabla_c V_d - \Gamma_{ac}^{d} \nabla_d V_c$$

However when we prove that the covariant derivative of a $(0,2)$ tensor is the above, we use the fact that the covariant derivative satisfies a Leibniz rule on $(0,1)$ tensors: $\nabla_a(w_b v_c) = v_c\nabla_a(w_b) + w_b\nabla_a(v_c)$. However $\nabla_a$ on it's own is not a tensor so how do we have the above formula for it's covariant derivative?

Answer 1

The term $\nabla_b V_c$ is a (0,2) tensor writing in the abstract index notation, when writing in full basis form it reads \begin{equation} \nabla_b V_c \;dx^b\otimes dx^c\;, \end{equation} Now the status of $\nabla_b V_c $ is a components it is a scalar function while $dx^b\otimes dx^c$ is a basis of (0,2)-tensor.

Then the double covariant derivative reads \begin{equation} \nabla \Big( \nabla_b V_c \;dx^b\otimes dx^c \Big)\;, \end{equation} where $$\nabla_b V_c \equiv \partial_b V_c -\Gamma_b{}^q{}_c V_q\;.$$ The Leibniz rule is needed in this step \begin{eqnarray} \nabla \Big( \nabla_b V_c \;dx^b\otimes dx^c \Big)&=& \nabla \Big( \nabla_b V_c \Big)\;dx^b\otimes dx^c + \nabla_b V_c \;\nabla \Big( dx^b \Big)\otimes dx^c + \nabla_b V_c \;dx^b\otimes \nabla \Big( dx^c \Big)\;,\\ &=& \nabla_m \Big( \overbrace{ \nabla_b V_c}^{a\; scalar} \Big)\;dx^m\otimes dx^b\otimes dx^c + \nabla_b V_c \;\times \Big(-\Gamma^b{}_n dx^n \Big)\otimes dx^c \\&&+ \nabla_b V_c \;dx^b\otimes \times \Big( -\Gamma^c{}_p dx^p \Big)\;,\\ &=& \nabla_m \Big( \overbrace{ \nabla_b V_c}^{a\; scalar} \Big)\;dx^m\otimes dx^b\otimes dx^c -\Gamma^b{}_n \nabla_b V_c \; dx^n \otimes dx^c \\&& -\Gamma^c{}_p \nabla_b V_c \;dx^b\otimes dx^p \;,\\ &=&\partial_m \Big( \overbrace{ \nabla_b V_c}^{a\; scalar} \Big)\;dx^m\otimes dx^b\otimes dx^c -\Gamma_r{}^b{}_n dx^{r} \otimes \nabla_b V_c \; dx^{n} \otimes dx^c \\&&-\Gamma_s{}^c{}_p dx^{s} \otimes \nabla_b V_c \;dx^b \otimes dx^{p} \;,\\ &=&\partial_m \Big( \nabla_b V_c \Big)\;dx^m\otimes dx^b\otimes dx^c - \Gamma_r{}^b{}_n\nabla_b V_c \; dx^r \otimes dx^n \otimes dx^c \\&&-\Gamma_s{}^c{}_p \nabla_b V_c \;dx^s\otimes dx^b\otimes dx^p \;,\\ &=&\Big[\partial_m \Big( \nabla_b V_c \Big)- \Gamma_m{}^d{}_b\nabla_d V_c-\Gamma_m{}^e{}_c \nabla_b V_e\Big ]\;dx^m\otimes dx^b\otimes dx^c \;. \end{eqnarray} Then we define $$ \nabla \Big( \nabla_b V_c \;dx^b\otimes dx^c \Big)=:\nabla_m \nabla_b V_c \;dx^m \otimes dx^b\otimes dx^c $$ Finally, in abstract index notation we have $$ \nabla_m \nabla_b V_c \equiv \partial_m \Big( \nabla_b V_c \Big)- \Gamma_m{}^d{}_b\nabla_d V_c-\Gamma_m{}^e{}_c \nabla_b V_e $$

Answer 2

Easy way Let me first state the straight-forward way to do this computation. $$ \langle \nabla_a \nabla_b V, \partial_c\rangle = \partial_a \langle \nabla_b V, \partial_c \rangle - \langle \nabla_aV, \nabla_a \partial_c\rangle = \partial_a (\nabla_bV)_c - (\nabla_bV)_d \Gamma_{ac}^d $$

First equality follows from compatibility, second equality uses definition of Levi-Civita symbols.

Hard way You are suggesting a roundabout way to do this, which formalizes to the following: $$ \nabla_a\nabla_bV = \nabla_a\left[~(\nabla_cV\otimes dx^c)[\partial_b]~\right] = \nabla_a\left[~C(\nabla_cV\otimes dx^c \otimes \partial_b)~\right] = C [\nabla_a (\nabla_cV\otimes dx^c \otimes \partial_b)] $$

where $$ C: T_pM \otimes T_pM \otimes T^*_pM \to T_pM, ~~ w\otimes z\otimes V \to z[V]w $$

is the contraction map of the last two arguments. Covariant derivative on mixed-type tensors commute with contractions (used in the last equality). Observe the expression within $C[ \cdots ]$ is a covariant derivative of a mixed tensor, which you can compute with the Leibneiz rule, and use your favorite component-wise formulas.