# A Two Covariant Derivatives (Chain Rule)?

#### berlinspeed

Summary
Failed find information on the internet, really appreciate any help.
Summary: Failed find information on the internet, really appreciate any help.

Can someone tell me what is ∇ϒδ𝒆β? It seems to be equal to 𝒆μΓμβδ,ϒ+(𝒆νΓνμϒμβδ. Is this some sort of chain rule or is it by any means called anything?

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#### Orodruin

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What do you get when you try to compute it yourself?

#### berlinspeed

What do you get when you try to compute it yourself?
At first look I just get the second term, so still trying to figure out where went wrong..

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#### berlinspeed

Okay I think where went wrong is I need to expand ∇ϒ(𝒆μΓμβδ), which now looks like something to apply chain rule to, and if I do that I get the answer except shouldn't the first term be 𝒆μΓμβδ;ϒ? The "," stands for partial derivative?

#### Orodruin

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Yes, the , stands for a regular partial derivative. You are here including the basis vector $e_\mu$ in your notation. When you do that you are using a notation where anything multiplying it is just a scalar function. It holds that $\nabla_a e_b = \Gamma_{ab}^c e_c$ by definition. The covariant derivative is linear and satisfies the product rule (this is not chain rule)
$$\nabla_a (fV) = V \nabla_a f + f \nabla_a V,$$
where $f$ is a scalar field and $V$ is a vector. Note that $\nabla_a f = \partial_a f$ for any scalar field. In your case, therefore
$$\nabla_a(\Gamma^c_{bd}e_c) = \Gamma^c_{bd} \nabla_a e_c + e_c \nabla_a \Gamma^c_{bd} = \Gamma^c_{bd} \Gamma^f_{ac} e_f + e_c \Gamma^c_{bd,a}.$$

Do not confuse this notation with the usual notation without writing out the vector basis, where we typically use the convention
$$\nabla_a V^b \equiv (\nabla_a V)^b,$$
i.e., the LHS is defined as the $b$-component of $\nabla_a V$. Actually writing out the basis in the RHS gives you
$$(\nabla_a V)^b = [\nabla_a (V^c e_c)]^b = [V^c \nabla_a e_c]^b + [e_c \nabla_a V^c]^b = V^c [\Gamma^d_{ac} e_d]^b + [V^c_{,a} e_c]^b = V^c \Gamma^b_{ac} + V^b_{,a},$$
since $[e_c]^b = \delta^b_c$.

#### berlinspeed

Yes, the , stands for a regular partial derivative. You are here including the basis vector $e_\mu$ in your notation. When you do that you are using a notation where anything multiplying it is just a scalar function. It holds that $\nabla_a e_b = \Gamma_{ab}^c e_c$ by definition. The covariant derivative is linear and satisfies the product rule (this is not chain rule)
$$\nabla_a (fV) = V \nabla_a f + f \nabla_a V,$$
where $f$ is a scalar field and $V$ is a vector. Note that $\nabla_a f = \partial_a f$ for any scalar field. In your case, therefore
$$\nabla_a(\Gamma^c_{bd}e_c) = \Gamma^c_{bd} \nabla_a e_c + e_c \nabla_a \Gamma^c_{bd} = \Gamma^c_{bd} \Gamma^f_{ac} e_f + e_c \Gamma^c_{bd,a}.$$

Do not confuse this notation with the usual notation without writing out the vector basis, where we typically use the convention
$$\nabla_a V^b \equiv (\nabla_a V)^b,$$
i.e., the LHS is defined as the $b$-component of $\nabla_a V$. Actually writing out the basis in the RHS gives you
$$(\nabla_a V)^b = [\nabla_a (V^c e_c)]^b = [V^c \nabla_a e_c]^b + [e_c \nabla_a V^c]^b = V^c [\Gamma^d_{ac} e_d]^b + [V^c_{,a} e_c]^b = V^c \Gamma^b_{ac} + V^b_{,a},$$
since $[e_c]^b = \delta^b_c$.
Okay cleared now, thanks much.

Last edited:

#### berlinspeed

Yes, the , stands for a regular partial derivative. You are here including the basis vector $e_\mu$ in your notation. When you do that you are using a notation where anything multiplying it is just a scalar function. It holds that $\nabla_a e_b = \Gamma_{ab}^c e_c$ by definition. The covariant derivative is linear and satisfies the product rule (this is not chain rule)
$$\nabla_a (fV) = V \nabla_a f + f \nabla_a V,$$
where $f$ is a scalar field and $V$ is a vector. Note that $\nabla_a f = \partial_a f$ for any scalar field. In your case, therefore
$$\nabla_a(\Gamma^c_{bd}e_c) = \Gamma^c_{bd} \nabla_a e_c + e_c \nabla_a \Gamma^c_{bd} = \Gamma^c_{bd} \Gamma^f_{ac} e_f + e_c \Gamma^c_{bd,a}.$$

Do not confuse this notation with the usual notation without writing out the vector basis, where we typically use the convention
$$\nabla_a V^b \equiv (\nabla_a V)^b,$$
i.e., the LHS is defined as the $b$-component of $\nabla_a V$. Actually writing out the basis in the RHS gives you
$$(\nabla_a V)^b = [\nabla_a (V^c e_c)]^b = [V^c \nabla_a e_c]^b + [e_c \nabla_a V^c]^b = V^c [\Gamma^d_{ac} e_d]^b + [V^c_{,a} e_c]^b = V^c \Gamma^b_{ac} + V^b_{,a},$$
since $[e_c]^b = \delta^b_c$.
Btw oddly enough it is named "chain rule" in the Charles & Wheeler book..

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