A question about the formula for the Christoffel symbol

Question

This refers to the following formula on this wiki page:

$$\Gamma^i_{ki}=\frac{1}{2}g^{im}\frac{\partial g_{im}}{\partial x^k}$$

Shouldn't it be $$\Gamma^{i}_{ki}=\frac{1}{2}g^{im}(\frac{\partial g_{mi}}{\partial x_k}+\frac{\partial g_{mk}}{\partial x_i}-\frac{\partial g_{ki}}{\partial x_m})$$ Why do the other terms disappear?

Answer 1

In your expression : $\Gamma^{i}_{ki}=\frac{1}{2}g^{im}(\frac{\partial g_{mi}}{\partial x^k}+\frac{\partial g_{mk}}{\partial x^i}-\frac{\partial g_{ki}}{\partial x^m})$ the terms $\frac{\partial g_{mk}}{\partial x^i}-\frac{\partial g_{ki}}{\partial x^m}$ cancel each other out because of the symmetry of the metric tensor.

So (using Einstein summation convention):

$g^{im}\frac{\partial g_{mk}}{\partial x^i}-g^{im}\frac{\partial g_{ki}}{\partial x^m}=g^{im}\frac{\partial g_{mk}}{\partial x^i}-g^{mi}\frac{\partial g_{ki}}{\partial x^m}=$ (relabeling dummy indices) $\implies =g^{im}\frac{\partial g_{mk}}{\partial x^i}-g^{im}\frac{\partial g_{km}}{\partial x^i}=g^{im}\frac{\partial g_{mk}}{\partial x^i}-g^{im}\frac{\partial g_{mk}}{\partial x^i}=0$

giving the formula.

Answer 2

Yes that's true. But note that by the symmetry of $g$, $$g^{im}\partial_ig_{mk}=g^{mi}\partial_mg_{ik}=g^{im}\partial_mg_{ki}$$ so the last two terms cancel.

Edit: To clarify, the $i,m$ indices are just labels for summation. They can be relabeled at will to any other distinct letters. For example, $$g^{im}\partial_ig_{mk}=g^{ab}\partial_ag_{bk}$$ $$g^{mi}\partial_mg_{ik}=g^{ab}\partial_ag_{bk}$$ where in the first identity $i\to a$, $m\to b$, while in the second, $i\to b$, $m\to a$.