6
$\begingroup$

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?

$\endgroup$
6
  • 2
    $\begingroup$ $\nabla_bV_c$ is a tensor, call it $A_{bc}$. Then how do you find $\nabla_aA_{bc}$? $\endgroup$
    – MBN
    May 17, 2016 at 15:06
  • $\begingroup$ By using the Leibniz rule? Which to prove we needed to use the fact that $A_{bc}$ splits into (0,1) tensors. $\endgroup$
    – Wooster
    May 17, 2016 at 15:08
  • $\begingroup$ Why is that a problem? $\endgroup$
    – MBN
    May 18, 2016 at 5:20
  • 1
    $\begingroup$ Well $A_{bc}$ doesn't split into two (0,1) tensors, because $\nabla_b$ is not a tensor? I may well be being very stupid and missing something here! $\endgroup$
    – Wooster
    May 18, 2016 at 8:15
  • 1
    $\begingroup$ $\nabla V$ splits as $\sum_{ij}V^i{}_{;j}\partial_i\otimes\mathrm{d}x^j$. $\endgroup$
    – Ryan Unger
    May 21, 2016 at 1:42

2 Answers 2

3
$\begingroup$

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 $$ \nabla_m \nabla_b V_c \equiv \partial_m \Big( \nabla_b V_c \Big)- \Gamma_m{}^d{}_b\nabla_d V_c-\Gamma_m{}^e{}_c \nabla_b V_e $$

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.