# Covariant derivative of metric tensor

#### Damidami

Hi, I'm trying to verify that the covariant derivative of the metric tensor is D(g) = 0.
But I have a few questions:
1) This is a scalar 0 or a tensorial 0? Because it is suposed that the covariant derivative of a (m,n) tensor is a (m,n+1) tensor, and g is a (0,2) tensor so I think this 0 should be a (0,3) tensor. Am I right?

2) Following the MTW book in the example of page 341 we have:
$$\Gamma^{\theta}_{\phi\phi} = -sin(\theta)cos(\theta)$$
$$\Gamma^{\phi}_{\phi\theta} = cos(\theta)/sin(\theta)$$
and all the other $$\Gamma = 0$$

But when I try to verify the covariant derivative of the metric tensor for the component $$g_{\phi\phi;\theta}$$ it doesn't give me 0, but instead:

$$g_{\phi\phi;\theta} = g_{\phi\phi,\theta} - \Gamma^{k}_{\theta\phi} g_{k\phi} - \Gamma^{k}_{\theta\phi} g_{\phi k} = 2a^2sin(\theta)cos(\theta) - (0+0) - (0+0) = 2a^2sin(\theta)cos(\theta) \neq 0$$

I checked it a lot of times and am not sure if this is a conceptual error or a procedure error.
Can someone clarify this to me?

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#### cristo

Staff Emeritus
Hi, I'm trying to verify that the covariant derivative of the metric tensor is D(g) = 0.
But I have a few questions:
1) This is a scalar 0 or a tensorial 0? Because it is suposed that the covariant derivative of a (m,n) tensor is a (m,n+1) tensor, and g is a (0,2) tensor so I think this 0 should be a (0,3) tensor. Am I right?
Yup, your reasoning is correct: it would be the type (0,3) tensor with every component zero.

2) Following the MTW book in the example of page 341 we have:
$$\Gamma^{\theta}_{\phi\phi} = -sin(\theta)cos(\theta)$$
$$\Gamma^{\phi}_{\phi\theta} = cos(\theta)/sin(\theta)$$
and all the other $$\Gamma = 0$$

But when I try to verify the covariant derivative of the metric tensor for the component $$g_{\phi\phi;\theta}$$ it doesn't give me 0, but instead:

$$g_{\phi\phi;\theta} = g_{\phi\phi,\theta} - \Gamma^{k}_{\theta\phi} g_{k\phi} - \Gamma^{k}_{\theta\phi} g_{\phi k} = 2a^2sin(\theta)cos(\theta) - (0+0) - (0+0) = 2a^2sin(\theta)cos(\theta) \neq 0$$

I checked it a lot of times and am not sure if this is a conceptual error or a procedure error.
Can someone clarify this to me?
I don't know what the actual question is (since I don't have the book to hand) but I can't quite follow your last line. I get $$g_{\phi\phi;\theta}=g_{\phi\phi,\theta}-g_{k\phi}\Gamma^{k}_{\phi\theta} - g_{\phi k}\Gamma^{k}_{\phi\theta}= g_{\phi\phi,\theta}-2g_{\phi\phi}\Gamma^\phi_{\phi\theta}$$

Is this what you get before plugging in the values?

#### kdv

Hi, I'm trying to verify that the covariant derivative of the metric tensor is D(g) = 0.
But I have a few questions:
1) This is a scalar 0 or a tensorial 0? Because it is suposed that the covariant derivative of a (m,n) tensor is a (m,n+1) tensor, and g is a (0,2) tensor so I think this 0 should be a (0,3) tensor. Am I right?

2) Following the MTW book in the example of page 341 we have:
$$\Gamma^{\theta}_{\phi\phi} = -sin(\theta)cos(\theta)$$
$$\Gamma^{\phi}_{\phi\theta} = cos(\theta)/sin(\theta)$$
and all the other $$\Gamma = 0$$

But when I try to verify the covariant derivative of the metric tensor for the component $$g_{\phi\phi;\theta}$$ it doesn't give me 0, but instead:

$$g_{\phi\phi;\theta} = g_{\phi\phi,\theta} - \Gamma^{k}_{\theta\phi} g_{k\phi} - \Gamma^{k}_{\theta\phi} g_{\phi k} = 2a^2sin(\theta)cos(\theta) - (0+0) - (0+0) = 2a^2sin(\theta)cos(\theta) \neq 0$$

I checked it a lot of times and am not sure if this is a conceptual error or a procedure error.
Can someone clarify this to me?

I do get zero. Note that $$\Gamma^\phi_{\theta \phi}$$ is not zero, it is equal to
$$\Gamma^\phi_{\phi \theta}$$ !

#### Damidami

Indeed that was the mistake, I forgot the symetry of the chistoffel symbols, it gives 0 now!
:)

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