Vanishing covariant derivative of a vector field

Question

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...

Answer 1

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 $[\nabla_b,\nabla_c]v^d=\nabla_b\nabla_cv^d-\nabla_c\nabla_bv^d=0$