Covariant derivative and 'see-saw rule'

[binbagsss]

Homework Statement

Apologies if this is a stupid question but just thinking about the see-saw rule applied to something like:

$w_v \nabla_u V^v = w^v \nabla_u V_v$

It is not obvious that the two are equivalent to me since one comes with a minus sign for the connection and one with a plus sign for the connection.

I guess just to stupidly ask whether you are okay to use the see-saw here, and if there is a quick, obvious way of showing the two are equivalent.

Homework Equations

the see-saw rule

The Attempt at a Solution

$w_v \nabla_u V^v = w_v(\partial_u V^v + \Gamma_{uc}^v V^c)$[1]

$w^v \nabla_u V_v = w^v (\partial_u V_v - \Gamma_{uv}^c V_c)$[1]

i can rename dummy indicies $u$ and $c$ but can't really see this helping?

thanks .

 

[Orodruin]

Yes, it is the same. What you are missing is that $V^c$ and $V_c$ are not the same and neither is $w_v$ and $w^v$ - and that the connection is metric compatible. As long as the connection is metric compatible you can use it to raise and lower indices.
$$
w_a \nabla_b V^a = w^c g_{ca} \nabla_b V^a = w^c \nabla_b g_{ca} V^a = w^c \nabla_b V_c = w^a \nabla_b V_a,
$$
where we have used metric compatibility ($\nabla_b g_{ca} = 0$) when moving the metric inside the derivative.

 

[stevendaryl]

Just to add to Orudruin's comment: Note that the same thing with partial derivatives is not true:

$w_v \partial_u V^v \neq w^v \partial_u V_v$

The difference involves partial derivatives of the metric tensor, which are related to the $\Gamma^v_{uc}$ in exactly the way needed to make

$w_v \nabla_u V^v = w^v \nabla_u V_v$