How do I derive the Covariant Derivative for Covectors? (Lower index)

[LoadedAnvils]

Hello everyone!

I'm trying to learn the derivation the covariant derivative for a covector, but I can't seem to find it.

I am trying to derive this:

\nabla_{α} V_{μ} = \partial_{α} V_{μ} - \Gamma^{β}_{αμ} V_{β}

If this is a definition, I want to know why it works with the definition of the covariant derivative of a vector:

\nabla_{α} V^{μ} = \partial_{α} V^{μ} + \Gamma^{μ}_{αβ} V^{β}

Why is the Christoffel symbol negative for a covector? Can anyone explain this to me?

Thank you all so much.

 

[George Jones]

Try working out \nabla_\alpha \left( V^\mu V_\mu \right) a couple of different ways:

1) Using the product rule for covariant derivatives;

2) using the fact that V^\mu V_\mu is a scalar.

 

[LoadedAnvils]

This is one of the leads I found here, but I don't know how to progress.

Right now, I am here:

V^{μ} \nabla_{α} V_{μ} = \partial_{α} (V^{μ} V_{μ}) - V_{μ} \partial_{α} V^{μ} - V_{μ} \Gamma^{μ}_{αβ} V^{β}

Here are some questions I have:

I understand that V^{μ} V_{μ} is a scalar, so what is the partial derivative of it? I tried to do it with zero if it is independent of α but it doesn't look like it leads to the covariant derivative.

Also, how do I remove the V^{μ} on the LHS? Do I multiply by V_{μ}?

 

[George Jones]

I understand that V^{μ} V_{μ} is a scalar, so what is the partial derivative of it?

Use the product rule for partial derivatives.

Also, how do I remove the V^{μ} on the LHS? Do I multiply by V_{μ}?

You don't have to remove it; you need to relabel indices in the last term on the right side.

 

[LoadedAnvils]

Use the product rule for partial derivatives.

I don't know what two terms I would use in the product rule. Once I separate the vector and covector I can't use the partial derivative anymore (or so I think). Can you explain to me how I would use the product rule in this case?

You don't have to remove it; you need to relabel indices in the last term on the right side.

I see what you mean, however, without knowing the partial derivative I don't know what I can do with this yet.

 

[George Jones]

I don't know what two terms I would use in the product rule. Once I separate the vector and covector I can't use the partial derivative anymore (or so I think). Can you explain to me how I would use the product rule in this case?

I am not sure what you mean. More explicitly: use the product rule for partial derivatives on \partial_\alpha \left( V^\mu V_\mu \right).

 

[LoadedAnvils]

Oh, now I get it!

For those reading this in the future, this is what I got:

\partial_{α} (V^{μ} V_{μ}) = V^{μ} \partial_{α} V_{μ} + V_{μ} \partial_{α} V^{μ}

Thus,

V^{μ} \nabla_{α} V_{μ} = V^{μ} \partial_{α} V_{μ} + V_{μ} \partial_{α} V^{μ} - V_{μ} \partial_{α} V^{μ} - V_{μ} \Gamma^{μ}_{αβ} V^{β}

So V_{μ} \partial_{α} V^{μ} cancels out, and we have V^{μ} \nabla_{α} V_{μ} = V^{μ} \partial_{α} V_{μ} - V_{μ} \Gamma^{μ}_{αβ} V^{β}.

We can rewrite V_{μ} \Gamma^{μ}_{αβ} V^{β} as V^{β} \Gamma^{μ}_{αβ} V_{μ}.

Since μ and β are arbitrary indices, we can swap them around.

Thus, V^{β} \Gamma^{μ}_{αβ} V_{μ} = V^{μ} \Gamma^{β}_{αμ} V_{β}, and

V^{μ} \nabla_{α} V_{μ} = V^{μ} \partial_{α} V_{μ} - V^{μ} \Gamma^{β}_{αμ} V_{β}.

Eliminating V^{μ}, we reach the desired result:

\nabla_{α} V_{μ} = \partial_{α} V_{μ} - \Gamma^{β}_{αμ} V_{β}