Christoffel Symbols Equality Solution?

Question

They changed the exercise, so I tried to solve it again:

I have to prove the following:

Let $\Omega \subseteq \mathbb{R}^d$ be open and $g$ a metric field on $\Omega$. For every $\phi \in \mathrm{Diff}(\Omega)$ let $\Xi^i_{jk}[\phi]$ be functions on $\Omega$ that transform in the same way as the Christoffel symbols $\Gamma^i_{jk}[\phi]$ $$ \Xi^i_{jk}[\phi](y) = \frac{\partial x^p}{\partial y^j}(y) \frac{\partial x^q}{\partial y^k}(y) \Xi^r_{pq}[\mathrm{id}](\phi^{-1}(y))\frac{\partial y^i}{\partial x^r}(\phi^{-1}(y)) + \frac{\partial y^i}{\partial x^m} (\phi^{-1}(y)) \frac{\partial^2 x^m}{\partial y^j \partial y^k}(y). $$ Assume that for each $y_0 \in \Omega$ there is a $\phi_0 \in \mathrm{Diff}(\Omega)$ such that $\Xi^i_{jk}[\phi_0](\phi_0(y_0)) = 0$ and $(\partial_a g_{bc})[\phi_0](\phi_0(y_0))=0$. Show $\Xi^i_{jk}[\phi] =\Gamma^i_{jk}[\phi]$ for all $\phi \in \mathrm{Diff}(\Omega)$.

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Here is my try:

We define:

$T^{i}_{\,jk}[\phi](y) := \Xi^i_{\,jk}[\phi](y) - \Gamma^i_{\,jk}[\phi](y)$

Thus we see that this transforms as a Tensor:

$T^{i}_{\,jk}[\phi](y)= \Xi^i_{jk}[\phi](y) - \Gamma^i_{\,jk}[\phi](y) = \bigg(\frac{\partial x^p}{\partial y^j}(y) \frac{\partial x^q}{\partial y^k}(y) \Xi^r_{pq}[\mathrm{id}](\phi^{-1}(y))\frac{\partial y^i}{\partial x^r}(\phi^{-1}(y)) + \frac{\partial y^i}{\partial x^m} (\phi^{-1}(y)) \frac{\partial^2 x^m}{\partial y^j \partial y^k}(y) \bigg)- \bigg(\frac{\partial x^p}{\partial y^j}(y) \frac{\partial x^q}{\partial y^k}(y) \Gamma^r_{pq}[\mathrm{id}](\phi^{-1}(y))\frac{\partial y^i}{\partial x^r}(\phi^{-1}(y)) + \frac{\partial y^i}{\partial x^m} (\phi^{-1}(y)) \frac{\partial^2 x^m}{\partial y^j \partial y^k}(y)\bigg) = \frac{\partial x^p}{\partial y^j}(y) \frac{\partial x^q}{\partial y^k}(y) (\Xi^r_{pq}-\Gamma^r_{\,pq})[\mathrm{id}](\phi^{-1}(y))\frac{\partial y^i}{\partial x^r}(\phi^{-1}(y))$

Now we see that for a given pair $y_0$, $\phi_0$:

$\Gamma^i_{\,jk}[\phi_0](\phi_0(y_0))=\frac{1}{2} g^{il}(\partial_k g_{lk} + \partial_j g_{lk} - \partial_l g_{jk})[\phi_0](\phi_0(y_0))= \frac{1}{2} g^{il}\left(\partial_k g_{lk}[\phi_0](\phi_0(y_0)) + \partial_j g_{lk}[\phi_0](\phi_0(y_0)) - \partial_l g_{jk}[\phi_0](\phi_0(y_0))\right)=0$

So for $y_0$ and $\phi_0$ :

$T^{i}_{\,jk}[\phi_0](\phi_0(y_0)) := \Xi^i_{\,jk}[\phi_0](\phi_0(y_0)) - \Gamma^i_{\,jk}[\phi_0](\phi_0(y_0))=0$

Thus

$\frac{\partial x^p}{\partial y^j}(\phi_0(y_0)) \frac{\partial x^q}{\partial y^k}(\phi_0(y_0)) (\Xi^r_{pq}-\Gamma^r_{\,pq})[\mathrm{id}](\phi_0^{-1}(\phi_0(y_0)))\frac{\partial y^i}{\partial x^r}(\phi_0^{-1}(\phi_0(y_0)))= \frac{\partial x^p}{\partial y^j}(\phi_0(y_0))\frac{\partial x^q}{\partial y^k}(\phi_0(y_0)) \left({(\Xi^r_{pq}-\Gamma^r_{\,pq})[\mathrm{id}](y_0)}\right)\frac{\partial y^i}{\partial x^r}(y_0)\overset{!}{=}0$

Now my BIG question is: can I say that: Because $T^{i}_{\,jk}$ is a Tensor:

$(\Xi^r_{pq}-\Gamma^r_{\,pq})[\mathrm{id}](y_0)\overset{???}{=}0$

Because IF so then the exercise becomes trivial since: $T^{i}_{\,jk}[\phi](\phi(y_0))=\frac{\partial x^p}{\partial y^j}(\phi(y_0))\frac{\partial x^q}{\partial y^k}(\phi(y_0)) \overset{\underbrace{=0}}{\left({(\Xi^r_{pq}-\Gamma^r_{\,pq})[\mathrm{id}](y_0)}\right)}\frac{\partial y^i}{\partial x^r}(y_0)=0$

for all $\phi\in Diff(\Omega)$ and we are done

If not my old proof/idea does not hold anyway and I have got no other clue on how to overcome the problem...

Answer 1

I've finally managed to get a solution which does not imply any assumption which was not given in the statement of task, does it make sense to you guys?

We define:

$T^{i}_{\,jk}[\phi](y) := \Xi^i_{\,jk}[\phi](y) - \Gamma^i_{\,jk}[\phi](y)$

Thus we see that this transforms as a Tensor, hence is a Tensor:

$T^{i}_{\,jk}[\phi](y)= \Xi^i_{jk}[\phi](y) - \Gamma^i_{\,jk}[\phi](y) = \bigg(\frac{\partial x^p}{\partial y^j}(y) \frac{\partial x^q}{\partial y^k}(y) \Xi^r_{pq}[\mathrm{id}](\phi^{-1}(y))\frac{\partial y^i}{\partial x^r}(\phi^{-1}(y)) + \frac{\partial y^i}{\partial x^m} (\phi^{-1}(y)) \frac{\partial^2 x^m}{\partial y^j \partial y^k}(y) \bigg)- \bigg(\frac{\partial x^p}{\partial y^j}(y) \frac{\partial x^q}{\partial y^k}(y) \Gamma^r_{pq}[\mathrm{id}](\phi^{-1}(y))\frac{\partial y^i}{\partial x^r}(\phi^{-1}(y)) + \frac{\partial y^i}{\partial x^m} (\phi^{-1}(y)) \frac{\partial^2 x^m}{\partial y^j \partial y^k}(y)\bigg) = \frac{\partial x^p}{\partial y^j}(y) \frac{\partial x^q}{\partial y^k}(y) (\Xi^r_{pq}-\Gamma^r_{\,pq})[\mathrm{id}](\phi^{-1}(y))\frac{\partial y^i}{\partial x^r}(\phi^{-1}(y))$

Now we have that for our given $y_0$ and $\phi_0$:

$\Gamma^i_{\,jk}[\phi_0](\phi_0(y_0))=\frac{1}{2} g^{il}(\partial_k g_{lk} + \partial_j g_{lk} - \partial_l g_{jk})[\phi_0](\phi_0(y_0))= \frac{1}{2} g^{il}\left(\partial_k g_{lk}[\phi_0](\phi_0(y_0)) + \partial_j g_{lk}[\phi_0](\phi_0(y_0)) - \partial_l g_{jk}[\phi_0](\phi_0(y_0))\right)=0$

Now take any $y_0$ and any diffeomorphism $\phi\in \operatorname{Diff}(\Omega)$ :

$T^{i}_{\,jk}[\phi_0](\phi(y_0))=\frac{\partial x^p}{\partial y^j}(\phi(y_0))\frac{\partial x^q}{\partial y^k}(\phi(y_0)) {\left({(\Xi^r_{pq}-\Gamma^r_{\,pq})[\mathrm{id}](y_0)}\right)}\frac{\partial y^i}{\partial x^r}(y_0)$

By the statement of task we can always find a $\phi_0$ such that:

$T^{i}_{\,jk}[\phi_0](\phi_0(y_0))=0=\frac{\partial x^p}{\partial y^j}(\phi_0(y_0))\frac{\partial x^q}{\partial y^k}(\phi_0(y_0)) {\left({(\Xi^r_{pq}-\Gamma^r_{\,pq})[\mathrm{id}](y_0)}\right)}\frac{\partial y^i}{\partial x^r}(y_0)$

for each $y_0\in\Omega$.

Since for any diffeomorphism we have that

$\frac{\partial x^p}{\partial y^j}(\phi(y))\frac{\partial x^q}{\partial y^k}(\phi(y))\neq 0$

We see that the transformed part which is invariant under any $\phi\in \operatorname{Diff}(\Omega)$ is the one which maps to zero, we conclude for all of $\Omega$:

$T^i_{jk}[\phi]=\Xi^i_{jk}[\phi]-\Gamma^i_{jk}[\phi]\overset{\forall \phi}{=}0$ $$\Rightarrow \Xi^i_{jk}[\phi]=\Gamma^i_{jk}[\phi]$$

$\blacksquare$