# Is $\Gamma_{jk}^{t}=\frac{1}{2}g^{it}g_{ij,k}$?

Here $\Gamma_{jk}^{t}$ is the Christoffel symbol of the second kind, and $g$ is the Riemann metric on a Riemann manifold.When learning Riemann Geometry, we are usually introduced to the following equation:

$g_{ij,k}-g_{lj}\Gamma_{ik}^{l}-g_{il}\Gamma_{jk}^{l}=0$.

Then to get an expression of $\Gamma$ in terms of $g_{ij}$, we will replace $(ijk)$ with $(jki),(kij)$, and so on...

But what about the following way:

Suppose $(A^{i})$ is any tensor, behaving like $(dx^i)$.By contraction, we will get

$A^{i}A^{j}(g_{ij,k}-g_{lj}\Gamma_{ik}^{l}-g_{il}\Gamma_{jk}^{l})=0$,

$A^{i}A^{j}g_{ij,k}=A^{i}A^{j}g_{lj}\Gamma_{ik}^{l}+A^{i}A^{j}g_{il}\Gamma_{jk}^{l}= A^{j}A^{i}g_{li}\Gamma_{jk}^{l}+A^{i}A^{j}g_{il}\Gamma_{jk}^{l}=2A^{i}A^{j}g_{li}\Gamma_{jk}^{l}$.

Since $(A^{i})$ is arbitary , we then get that

$g_{ij,k}=2g_{li}\Gamma_{jk}^{l}$.i.e.$\Gamma_{jk}^{t}=\frac{1}{2}g^{it}g_{ij,k}$.

It is a bit weird, isn't it?Is the above calculation right?

Will someone be kind enough to point out the flaws in the above reasoning or give me some hints on this problem?Thank you very much!

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## 1 Answer

The problem is that there is a symmetry in $A^i A^j$, i.e., $A^i A^j = A^j A^i$. So, you can not get $g_{ij,k}=2g_{li} \Gamma^{l}_{jk}$ from the above equation. You can just get $\Sigma_{i>j} A^i A^j (g_{ij,k}-2g_{li}\Gamma^l_{jk}+g_{ji,k}-2g_{lj}\Gamma^l{ik})=0$ and $g_{ij,k}-2g_{li}\Gamma^l_{jk}+g_{ji,k}-2g_{lj}\Gamma^l{ik}=0$, which is just the equation showed in the first paragraph.

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Thank you very much for your concise answer! – user14242 Dec 2 '11 at 6:11