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I need to prove the following identity. $\Omega \subset \mathbb{R}^d$ is open and $g$ is a metric field on $\Omega$. Further $\Gamma_{\,kl}^j$ denotes the Christoffel symbol of second kind. $g^{ij}$ is the inverse matrix to $g_{ij}$ and $g$ denotes the determinant.

$$ \Gamma_{\,jl}^j = \frac{\partial}{\partial x^l} \log \sqrt{\vert g \vert} $$

Here is what I have so far: $$ \Gamma^j_{\;jl} = \frac{1}{2} g^{jk} \left( \frac{\partial }{ \partial x^l} g_{kj} + \frac{\partial}{ \partial x^j} g_{kl} - \frac{\partial}{ \partial x^k} g_{jl} \right) = \frac{1}{2} g^{jk} \frac{\partial }{ \partial x^l} g_{kj} $$

The second and the third term cancel each other out. Now as stated here (it's about the sixth equation, after the line starting with "The contracting relations..."

$$ \frac{1}{2} g^{jk} \frac{\partial }{ \partial x^l} g_{kj} = \frac{1}{2g} \frac{\partial}{ \partial x^l} g = \frac{\partial}{ \partial x^l} \log \sqrt{\vert g \vert} $$

I don't get the second last step. I suspect it has got something to do with the Laplace formula and the Cramer Rule. However i seem not to be able to figure it out, the derivation is bugging me.

Any hints are appreciated!

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First the slick method: write $\bar g_{ik}$ for the value of $g_{ik}$ at the specific $x^l = \bar x^l$ (so $\bar g_{ik}$ is constant). Let $x^l$ be a function of $\epsilon$ that is a perturbation of $\bar x^l$, so that $x^l = \bar x^l$ at $\epsilon=0$.

Let $$h_{ij} = \bar g^{ik} g_{kj} . \tag1$$ Then $h_{ij}$ is a perturbation of the identity: $$h_{ij} = \delta_{ij} + \epsilon \frac{\partial h_{ij}}{\partial \epsilon} + o(\epsilon).$$ Then it is easily computed that $$ \frac{\partial h}{\partial \epsilon} = \frac{\partial h_{ii}}{\partial \epsilon} + o(\epsilon)$$ Differentiate $(1)$ with respect to $\epsilon$ $$ \frac{\partial h_{ii}}{\partial\epsilon} = \bar g^{ik} \frac{\partial g_{ki}}{\partial \epsilon} $$ Also $h = \bar g^{-1} g$. Hence $$ \bar g^{-1} \frac{\partial g}{\partial \epsilon} = \bar g^{ik} \frac{\partial g_{ki}}{\partial \epsilon} + o(\epsilon). $$ In particular, at $\epsilon = 0$ we can remove the overlines, and we get: $$ g^{-1} \frac{\partial g}{\partial \epsilon} = g^{ik} \frac{\partial g_{ki}}{\partial \epsilon}. $$

Next the direct way: so $$g = \sum_\pi\sigma(\pi) \prod_i g_{i\pi(i)}.$$ Here the sum is over permutations $\pi$ of $\{1,\dots,n\}$. Then $$\frac{\partial g}{\partial x^l} = \sum_\pi \sigma(\pi) \sum_i \frac{\partial g_{i\pi(i)}}{\partial x^l} \prod_{j \ne i} g_{j\pi(j)} \\ =\sum_{ik} \frac{\partial g_{ik}}{\partial x^l}\sum_{\pi:\pi(i)=k} \prod_{j\ne i}\sigma(\pi) g_{j\pi(j)} \\ =\sum_{ik} \frac{\partial g_{ik}}{\partial x^l} g \ g^{ki}. $$

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  • $\begingroup$ Hello Stephen, thanks for your answer! I'm still not sure whether I completely understood your direct solution. In order for it to work I would need to move the $g g^{ki}$ in the last equation to the left. But it seems they depend on $x^l$ so i would not be allowed to do that. In other words: what is the derivation acting on? $\endgroup$ – stebu92 Dec 28 '13 at 13:28
  • $\begingroup$ When I write $\frac{\partial A}{\partial b} c$, I mean $\left(\frac{\partial A}{\partial b}\right) c$. $\endgroup$ – Stephen Montgomery-Smith Dec 28 '13 at 15:21
  • $\begingroup$ $\sum_{\pi:\pi(i)=k} \prod_{j\ne i}\sigma(\pi) g_{j\pi(j)}$ is the formula for the transpose of the adjoint matrix - en.wikipedia.org/wiki/Adjugate_matrix - and hence it is $g g^{ki}$. $\endgroup$ – Stephen Montgomery-Smith Dec 28 '13 at 15:25
  • $\begingroup$ I rewrote the slick proof using a more easy to understand notation here: math.stackexchange.com/questions/620771/… $\endgroup$ – Stephen Montgomery-Smith Dec 28 '13 at 21:35
  • $\begingroup$ I have one more question: Is it trivial to state that this equation is invariant under diffeomorphisms i.e. $$ \Gamma_{\,jl}^j[\phi] = \frac{\partial}{\partial x^l} \log \sqrt{\vert g[\phi] \vert} $$ where $\phi \in$ Diff$(\Omega)$? If not, how would I go about proving that? My suspicion is that since when I write down the def. of the CS in terms of $g_ij$ two terms cancel out due to the symmetry and I'm left with one, so i can just state it holds for diffeomorphisms. But I'm not sure. $\endgroup$ – stebu92 Jan 19 '14 at 7:38

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