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

1. May 1, 2013

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.

2. May 1, 2013

### George Jones

Staff Emeritus
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.

3. May 1, 2013

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 $V_{μ}$?

4. May 1, 2013

### George Jones

Staff Emeritus
Use the product rule for partial derivatives.

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

5. May 1, 2013

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

6. May 1, 2013

### George Jones

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

7. May 1, 2013

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:

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