Christoffel symbols equality

Question

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 $\Xi^i_{jk}[\phi_0](y_0) = 0$ and $(\partial_a g_{bc})[\phi_0](y_0)=0$. Show $\Xi^i_{jk}[\phi] =\Gamma^i_{jk}[\phi]$ for all $\phi \in \mathrm{Diff}(\Omega)$.

First thank you for all your answers.

I know now, that $T^{i}_{\ jk}$ is a Tensor field and that $T^{i}_{\ jk}[\phi_0](y_0) = 0$. But I'm struggling with the last step. In the second answer,there is a $x_0$ defind by $x_0 = \phi^{-1}(y_0)$. I don't think that one can reason that $\forall x_0 \in \Omega \ \exists \phi_0: T^{i}_{\ jk}[\phi_0](\phi_0(x_0))$, because not every point $x$ in $\Omega$ has to fulfill $x = \phi_0^{-1}(y_0)$. Every $y_0$ can have a different diffeomorphism $\phi_0$, with which it fulfills $\Xi^i_{jk}[\phi_0](y_0) = 0$. So $x_0 = \phi_0(y_0)$ is not bijective.

I've already done different attempts, but i don't get rid of the dependency of $y_0$ and $\phi_0$.

Answer 1

I tried to solve the exercise by following the hint, could someone please tell me if it is correct? Thank you

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:

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

AFTER THIS POINT THE PROOF IS WRONG, SEE ANSWER ON BOTTOM FOR THE CONTINUATION

I still leave it here so that you can see what my original idea was (before they changed the exercise guidelines), not more:

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

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

So we get for our given $y_0$ and $\phi_0$ :

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

Since $T$ is a tensor this Implies:

$(\Xi^r_{pq}-\Gamma^r_{\,pq})[\mathrm{id}](\phi_0^{-1}(y_0))\overset{!}{=}0$

Now we notice that since $\phi \in \mathrm{Diff}(\Omega)$ we have that $y_0\in\Omega$ this implies that for each $y_0$ we can find an $x_0\in\Omega$ s.t $y_0=\phi_0(x_0)$ so the constraint is equal to:

$\forall x_0\in\Omega$ $\exists$ $\phi_0$ s.t $T^{i}_{\,jk}[\phi_0](\phi_0(x_0))=0$

We can now take any $\phi \in \mathrm{Diff}(\Omega)$ and have $\phi=\phi\circ\phi_0^{-1}\circ\phi_0:=\tilde{\phi}\circ\phi_0$

So eventually we get: $\forall x_0\in\Omega$

$T^{i}_{\,jk}[\phi](\phi(x_0))=T^{i}_{\,jk}[\phi](\phi\circ\phi_0^{-1}(y_0))=T^{i}_{\,jk}[\tilde{\phi}\circ\phi_0](\tilde{\phi}(y_0))= \frac{\partial x^p}{\partial y^j}(\tilde{\phi}(y_0)) \frac{\partial x^q}{\partial y^k}(\tilde{\phi}(y_0) ) (\Xi^r_{pq}-\Gamma^r_{\,pq})[\mathrm{id}]((\tilde{\phi}\circ\phi_0)^{-1}(\tilde{\phi}(y_0)))\frac{\partial y^i}{\partial x^r}((\tilde{\phi}\circ\phi_0)^{-1}(\tilde{\phi}(y_0)))=\frac{\partial x^p}{\partial y^j}(\tilde{\phi}(y_0)) \frac{\partial x^q}{\partial y^k}(\tilde{\phi}(y_0) ) (\Xi^r_{pq}-\Gamma^r_{\,pq})[\mathrm{id}]((\phi_0^{-1})(\tilde{\phi}^{-1}\circ\tilde{\phi}(y_0)))\frac{\partial y^i}{\partial x^r}((\phi_0^{-1})(\tilde{\phi}^{-1}\circ\tilde{\phi}(y_0)))= \frac{\partial x^p}{\partial y^j}(\tilde{\phi}(y_0)) \frac{\partial x^q}{\partial y^k}(\tilde{\phi}(y_0) ) (\Xi^r_{pq}-\Gamma^r_{\,pq})[\mathrm{id}](\phi_0^{-1}(y_0))\frac{\partial y^i}{\partial x^r}(\phi_0^{-1}(y_0))=0$

$\forall \phi\in Diff(\Omega)$

This implies:

$\forall \phi\in Diff(\Omega)$ => $T^{i}_{\,jk}[\phi] = \Xi^i_{\,jk}[\phi] - \Gamma^i_{\,jk}[\phi]=0$

$\Xi^i_{\,jk}[\phi] = \Gamma^i_{\,jk}[\phi]$

$\blacksquare$

Answer 2

For the beginning you could define the following:

$$ T^{i}_{\,jk}[\phi](y) := \Xi^i_{\,jk}[\phi](y) - \Gamma^i_{\,jk}[\phi](y) $$ From the premises you know that there exists $\phi_0 \in$ Diff$(\Omega)$ and $y_0 \in \Omega$ so that $\Xi^i_{\,jk}[\phi_0](y_0) = 0$ and you can also show, that the Christoffel-Symbol is zero in those coordinates.

Then show that $T$ transforms the same way as a tensor, i.e. is a Tensor.

Now you only have to prove the invariance under diffeomorphisms and you are done!

Answer 3

The final/right solution should be the following:

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 $\phi\in Diff(\Omega)$ :

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

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 $\phi\in Diff(\Omega)$ is the one which maps to zero, we conclude for all of $\Omega$:

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

$\blacksquare$