I am trying to prove that the covariant derivative is a tensor (ie it transforms well under a change of coordinates) but I can't succeed to it.
Here is the definition of the covariant derivative :
$$ \nabla_XV= X^{\mu}(\partial_{\mu} V^{\rho}+V^{\nu}\Gamma_{\mu \nu}^{\rho})\partial_{\rho}$$
I write then the $\rho$ component :
$$ \nabla_XV^{\rho}= X^{\mu}(\partial_{\mu} V^{\rho}+V^{\nu}\Gamma_{\mu \nu}^{\rho}) $$
In addition, we have :
$$ \Gamma_{\mu \nu}^{\rho}=\frac{\partial x^{\rho}}{\partial y^{\sigma}}\frac{\partial y^K}{\partial x^{\mu}}\frac{\partial y^{\lambda}}{\partial x^{\nu}}\widetilde{\Gamma}_{K \lambda}^{\sigma}+\frac{\partial^2 y^{\lambda}}{\partial x^{\mu}\partial x^{\nu}}\frac{\partial x^{\rho}}{\partial y^{\lambda}}$$
What we need to prove is that we have :
$$ \nabla_XV^{\nu}=\frac{\partial x^{\nu}}{\partial y^{\rho}} \widetilde{\nabla_XV^{\rho}} $$
But I really don't know how we could end up to such a result.
Indeed, if I basically replace everything in base "tilde" (also called $y$ here), I will have the term : $V^{\nu}\Gamma_{\mu \nu}^{\rho}$ that will give :
$$V^{\nu}\Gamma_{\mu \nu}^{\rho}=\frac{\partial x^{\nu}}{\partial y^{\beta}}\widetilde{V}^{\beta}(\frac{\partial x^{\rho}}{\partial y^{\sigma}}\frac{\partial y^K}{\partial x^{\mu}}\frac{\partial y^{\lambda}}{\partial x^{\nu}}\widetilde{\Gamma}_{K \lambda}^{\sigma}+\frac{\partial^2 y^{\lambda}}{\partial x^{\mu}\partial x^{\nu}}\frac{\partial x^{\rho}}{\partial y^{\lambda}})\\=
\frac{\partial x^{\nu}}{\partial y^{\beta}}\widetilde{V}^{\beta} \frac{\partial x^{\rho}}{\partial y^{\sigma}}\frac{\partial y^K}{\partial x^{\mu}}\frac{\partial y^{\lambda}}{\partial x^{\nu}}\widetilde{\Gamma}_{K \lambda}^{\sigma}+\frac{\partial x^{\nu}}{\partial y^{\beta}}\widetilde{V}^{\beta}\frac{\partial^2 y^{\lambda}}{\partial x^{\mu}\partial x^{\nu}}\frac{\partial x^{\rho}}{\partial y^{\lambda}}
$$
And I don't see how the second part of this last equality could be compensated in any way because there are only "+" (and not "-") in all the equations.
And we need it to go out because the $\widetilde{V}^{\beta}$ should be either multiplied by the $\Gamma$ or it should be present as a derivative of itself, which is not the case in this second part of this last equality.
Could you help me to see how this term would vanish ?
Don't hesitate if anything is not clear (I tried to avoid to write too much calculations but to point out the term that from my perspective should vanish).
Best Answer
Since $\frac{\partial x^\rho}{\partial y^\lambda}\frac{\partial y^\lambda}{\partial x^\nu} = \delta^\rho_\nu$, we obtain \begin{align*} 0 &= \partial_\mu \left(\frac{\partial x^\rho}{\partial y^\lambda}\frac{\partial y^\lambda}{\partial x^\nu} \right)\\ &= \frac{\partial^2 y^\lambda}{\partial x^\mu \partial x^\nu} \frac{\partial x^\rho}{\partial y^\lambda} + \frac{\partial y^\lambda}{\partial x^\nu}\frac{\partial y^\kappa}{\partial x^\mu}\frac{\partial^2 x^\rho}{\partial y^\kappa \partial y^\lambda} \end{align*} It follows that we can rewrite the transformation rule for the Christoffel symbols as \begin{equation} \Gamma^\rho_{\mu\nu} = \frac{\partial x^\rho}{\partial y^\sigma} \frac{\partial y^\kappa}{\partial x^\mu} \frac{\partial y^\lambda}{\partial x^\nu} \widetilde{\Gamma}^\sigma_{\kappa \lambda} - \frac{\partial y^\lambda}{\partial x^\nu}\frac{\partial y^\kappa}{\partial x^\mu}\frac{\partial^2 x^\rho}{\partial y^\kappa \partial y^\lambda} \end{equation} This gets you the "negative part" that you'll need to make everything cancel out appropriately.