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I'm asked to prove the following statement in my physics book:

A vector field with covariant components $v^b$, in order to have a vanishing covariant derivative everywhere in a manifold, must satisfy: $$(\partial_b\Gamma^{d}{}_{ac}-\partial_c\Gamma^{d}{}_{ab}+\Gamma^{e}{}_{ac}\Gamma^{d}{}_{eb}-\Gamma^{e}{}_{ab}\Gamma^{d}{}_{ec})v^a=0.$$

Edit: This is what I tried after @PraharMitra's suggestion:

Since $\nabla_a v^b=0$, clearly $[\nabla_b,\nabla_c]v^d=\nabla_b\nabla_cv^d-\nabla_c\nabla_dv^d=0$.

As the covariant derivative of a contravariant component is defined as $\nabla_bv^d=\partial_bv^d+\Gamma^d{}_{eb}v^e$ and the covariant derivative of a covariant component as $\nabla_bv_d=\partial_bv_d+\Gamma^e{}_{db}v_e$, I got to:

$$ \nabla_b(\nabla_cv^d)=\partial_b(\nabla_cv^d)-\Gamma^e{}_{cb}(\nabla_ev^d)+\Gamma^d{}_{eb}(\nabla_cv^e) $$

Now let's plug in the covariant derivatives with respect to c:

$$ \nabla_b(\nabla_cv^d)=\partial_b(\partial_cv^d+\Gamma^d{}_{ac}v^a)-\Gamma^e{}_{cb}(\partial_ev^d+\Gamma^d{}_{ae}v^a)+\Gamma^d{}_{eb}(\partial_cv^e+\Gamma^e{}_{ac}v^a) $$ $$=\partial_b\partial_cv^d+\partial_b\Gamma^d{}_{ac}v^a-\Gamma^e{}_{cb}\partial_ev^d+\Gamma^e{}_{cb}\Gamma^d{}_{ae}v^a+\Gamma^d{}_{eb}\partial_cv^e+\Gamma^d{}_{eb}\Gamma^e{}_{ac}v^a$$

From this expression it's straightforward to obtain the $\nabla_c(\nabla_bv^d)$ term by exchanging the b and c indexes. I got:

$$ \nabla_c(\nabla_bv^d)=\partial_c\partial_bv^d+\partial_c\Gamma^d{}_{ab}v^a-\Gamma^e{}_{bc}\partial_ev^d+\Gamma^e{}_{bc}\Gamma^d{}_{ae}v^a+\Gamma^d{}_{ec}\partial_bv^e+\Gamma^d{}_{ec}\Gamma^e{}_{ab}v^a $$

Now, on substracting:

$$ \nabla_b(\nabla_cv^d)-\nabla_c(\nabla_bv^d)=\partial_b\partial_cv^d+\partial_b\Gamma^d{}_{ac}v^a-\Gamma^e{}_{cb}\partial_ev^d+\Gamma^e{}_{cb}\Gamma^d{}_{ae}v^a+\Gamma^d{}_{eb}\partial_cv^e+\Gamma^d{}_{eb}\Gamma^e{}_{ac}v^a-[\partial_c\partial_bv^d+\partial_c\Gamma^d{}_{ab}v^a-\Gamma^e{}_{bc}\partial_ev^d+\Gamma^e{}_{bc}\Gamma^d{}_{ae}v^a+\Gamma^d{}_{ec}\partial_bv^e+\Gamma^d{}_{ec}\Gamma^e{}_{ab}v^a] $$

The terms with both the partial derivatives vanish, since those can be exchanged:

$$=\partial_b\Gamma^d{}_{ac}v^a-\Gamma^e{}_{cb}\partial_ev^d+\Gamma^e{}_{cb}\Gamma^d{}_{ae}v^a+\Gamma^d{}_{eb}\partial_cv^e+\Gamma^d{}_{eb}\Gamma^e{}_{ac}v^a-[\partial_c\Gamma^d{}_{ab}v^a-\Gamma^e{}_{bc}\partial_ev^d+\Gamma^e{}_{bc}\Gamma^d{}_{ae}v^a+\Gamma^d{}_{ec}\partial_bv^e+\Gamma^d{}_{ec}\Gamma^e{}_{ab}v^a] $$

In this last expression, the first, sixth, eighth and tenth terms can be put together to be $(\partial_b\Gamma^{d}{}_{ac}-\partial_c\Gamma^{d}{}_{ab}+\Gamma^{e}{}_{ac}\Gamma^{d}{}_{eb}-\Gamma^{e}{}_{ab}\Gamma^{d}{}_{ec})v^a$, which is recognised to be the term we wanted to prove to be 0 in order to have vanishing covariant derivative (note that the terms with two connection coefficients commute). We can tell further that $(\partial_b\Gamma^{d}{}_{ac}-\partial_c\Gamma^{d}{}_{ab}+\Gamma^{e}{}_{ac}\Gamma^{d}{}_{eb}-\Gamma^{e}{}_{ab}\Gamma^{d}{}_{ec})v^a=R^d{}_{abc}v^a$, the riemannian curvature. On rewriting, it remains:

$$\nabla_b(\nabla_cv^d)-\nabla_c(\nabla_bv^d)=R^d{}_{abc}v^a+[-\Gamma^e{}_{cb}\partial_ev^d+\Gamma^e{}_{cb}\Gamma^d{}_{ae}v^a+\Gamma^d{}_{eb}\partial_cv^e]-[-\Gamma^e{}_{bc}\partial_ev^d+\Gamma^e{}_{bc}\Gamma^d{}_{ae}v^a+\Gamma^d{}_{ec}\partial_bv^e] $$

Assuming the manifold is torsionless, we would have $\Gamma^a{}_{bc}=\Gamma^a{}_{cb}$, and so both the first and second therm in each bracket would cancel, leaving us with:

$$\nabla_b(\nabla_cv^d)-\nabla_c(\nabla_bv^d)=0=R^d{}_{abc}v^a+\Gamma^d{}_{eb}\partial_cv^e-\Gamma^d{}_{ec}\partial_bv^e$$

In order to prove our initial statement, it seems like those two terms with the connection coefficients must vanish, but I don't see how do they or where did I make a mistake...

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    $\begingroup$ Since $D_a v^b = 0$, we also have $[D_b , D_c ]v^d = D_b D_c v^d - D_c D_b v^d = 0$. However, there is a well known formula (prove this!) which states $[D_b , D_c ] v^d = R^d{}_{abc} v^a$. Therefore, $D_a v^b = 0 \implies R^d{}_{abc} v^a = 0$. The term in the bracket in your equation is precisely $R^d{}_{abc}$. $\endgroup$
    – Prahar
    Mar 11, 2021 at 19:28
  • $\begingroup$ That seems really nice, I'll give it a try, thanks! $\endgroup$ Mar 11, 2021 at 19:32
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    $\begingroup$ Nice reply, @PraharMitra. Should you consider writing an answer? I am new here, so pardon me if this would not be useful. I also wondered about the converse: if $ R^d_{\ abc} v^a = 0$ implied $\nabla_a v^b = 0$, but it is false just by considering a flat space with a non-constant vector v. $\endgroup$
    – megaleo
    Mar 11, 2021 at 19:56
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    $\begingroup$ @LeonardoLessa - I put up an answer. $\endgroup$
    – Prahar
    Mar 11, 2021 at 20:02
  • $\begingroup$ @JorgeCasajus - I've found your mistake and edited my answer! $\endgroup$
    – Prahar
    Mar 15, 2021 at 10:49

1 Answer 1

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Here, are the steps you will need to follow to prove the result:

  1. Since $D_a v^b = 0$, we also have $[D_b , D_c ] v^d = D_b D_c v^d - D_c D_b v^d = 0$.

  2. Next, prove that for any vector field, we have $[D_b , D_c ] v^d = R^d{}_{abc} v^a$ where $R^d{}_{abc} = \partial_b \Gamma^d_{ca} - \partial_c\Gamma^d_{ba} + \Gamma^d_{be} \Gamma^e_{ca} - \Gamma^d_{ce} \Gamma^e_{ba}$.

  3. Conclude that your equation holds from 1) and 2).

The inverse statement does not hold.

I'll give a minor hint for the calculation: \begin{align} D_b D_c v^d &= \partial_b ( D_c v^d ) - \Gamma^e_{bc} D_e v^d + \Gamma^d_{be} D_c v^e \\ &= \partial_b ( \partial_c v^d + \Gamma^d_{ca} v^a ) \\ &\qquad - \Gamma^e_{bc} ( \partial_e v^d + \Gamma^d_{ea} v^a ) \\ &\qquad + \Gamma^d_{be} ( \partial_c v^e + \Gamma^e_{ca} v^a ) \end{align} Now, expand and simplify. Then, interchange $b \leftrightarrow c$ and subtract.


EDIT:

  1. The mistake in OPs calculation is in the following equation $$ \nabla_b(\nabla_cv^d)=\partial_b(\partial_cv^d+\Gamma^d{}_{ac}v^a)-\Gamma^e{}_{cb}(\partial_ev^d+\Gamma^d{}_{ae}v^a)+\Gamma^d{}_{eb}(\partial_cv^e+\Gamma^e{}_{ac}v^a) $$ $$=\partial_b\partial_cv^d+\partial_b\Gamma^d{}_{ac}v^a-\Gamma^e{}_{cb}\partial_ev^d+\Gamma^e{}_{cb}\Gamma^d{}_{ae}v^a+\Gamma^d{}_{eb}\partial_cv^e+\Gamma^d{}_{eb}\Gamma^e{}_{ac}v^a$$ In doing so, they are assuming $$\partial_b(\partial_cv^d+\Gamma^d{}_{ac}v^a)=\partial_b\partial_cv^d+\partial_b\Gamma^d{}_{ac}v^a$$ whereas the correct equation is $$\partial_b(\partial_cv^d+\Gamma^d{}_{ac}v^a)=\partial_b\partial_cv^d+\partial_b\Gamma^d{}_{ac}v^a+\Gamma^d{}_{ac}\partial_bv^a$$

  2. In addition to this there is also a sign mistake. They write $$-\Gamma^e{}_{cb}(\partial_ev^d+\Gamma^d{}_{ae}v^a)=-\Gamma^e{}_{cb}\partial_ev^d+\Gamma^e{}_{cb}\Gamma^d{}_{ae}v^a$$ which should be $$-\Gamma^e{}_{cb}(\partial_ev^d+\Gamma^d{}_{ae}v^a)=-\Gamma^e{}_{cb}\partial_ev^d-\Gamma^e{}_{cb}\Gamma^d{}_{ae}v^a$$

  3. There's also what is most likely a typo in the original question. OP writes $\nabla_bv_d=\partial_bv_d+\Gamma^e{}_{db}v_e$ which should be $\nabla_bv_d=\partial_bv_d-\Gamma^e{}_{db}v_e$ This is most likely just a typo since it does not propagate to the rest of the answer.

  4. Another typo is in the start of the calculation: $[\nabla_b,\nabla_c]v^d=\nabla_b\nabla_cv^d-\nabla_c\nabla_dv^d=0$ should be $[\nabla_b,\nabla_c]v^d=\nabla_b\nabla_cv^d-\nabla_c\nabla_bv^d=0$

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    $\begingroup$ Those terms are not zero. You have made a calculation error. $\endgroup$
    – Prahar
    Mar 12, 2021 at 11:53
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    $\begingroup$ This does not vanish. $\endgroup$
    – Prahar
    Mar 12, 2021 at 12:23
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    $\begingroup$ @JorgeCasajus - I added a hint to my answer. Hope it helps you. $\endgroup$
    – Prahar
    Mar 14, 2021 at 2:26
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    $\begingroup$ @JorgeCasajus - There is really not much else I can say. This is not a homework help site. There is clearly some mistakes in your calculation. Do it again and do it slowly. If you still do not have it, post another answer to this question with ALL the details completely laid out and I will help you. $\endgroup$
    – Prahar
    Mar 14, 2021 at 11:23
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    $\begingroup$ @JorgeCasajus - I've found your mistake and edited my answer! $\endgroup$
    – Prahar
    Mar 15, 2021 at 10:49

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