Levi-Civita connection under a rescaled metric let $(M,g)$ be a Riemannian manifold and $g'=c.g$ for some $ c > 0$. Then show that the Levi-Civita connections for $g$ and $g'$ are same.
I was trying to solve this by using Christoffel symbols but I got stuck.Need some help.
 A: The Levi Civita is the unique connection without torsion for which the Riemann tensor has  $\nabla g=0$. Of course, if $\nabla g=0, \nabla c.g=0$.
A: Consider a generalized situation. Let $\bar{D}$ be the Levi-Civita connection of $\phi g$, where $\phi$ is a positive smooth function. Let $D$ be the Levi-Civita connection of $g$. It is clear that
$$
D_{\frac{\partial}{\partial x^i}}\frac{\partial}{\partial x^j}= \Gamma_{ij}^k \frac{\partial}{\partial x^k} = \frac{1}{2} g^{kl}(\frac{\partial g_{lj}}{\partial x^i}+ \frac{\partial g_{li}}{\partial x^j}-\frac{\partial g_{ij}}{\partial x^l}) \cdot \frac{\partial}{\partial x^k}.
$$
Now, consider
$$
\begin{aligned}
\bar{D}_{\frac{\partial}{\partial x^i}}\frac{\partial}{\partial x^j} = &\frac{1}{2} (\phi g)^{kl}(\frac{\partial \phi g_{lj}}{\partial x^i}+ \frac{\partial \phi g_{li}}{\partial x^j}-\frac{\partial \phi g_{ij}}{\partial x^l}) \cdot \frac{\partial}{\partial x^k}\\
= &\frac{1}{2} (\phi g)^{kl} (\phi \frac{\partial g_{lj}}{\partial x^i}+ \phi \frac{\partial g_{li}}{\partial x^j}- \phi \frac{\partial g_{ij}}{\partial x^l}) \cdot \frac{\partial}{\partial x^k} \\
&+ \frac{1}{2} (\phi g)^{kl}(g_{lj}\frac{\partial \phi}{\partial x^i}+ g_{li} \frac{\partial \phi}{\partial x^j}-g_{ij} \frac{\partial \phi}{\partial x^l}) \cdot \frac{\partial}{\partial x^k}\\
= &D_{\frac{\partial}{\partial x^i}}\frac{\partial}{\partial x^j}+ \frac{1}{2} (\phi g)^{kl}(g_{lj}\frac{\partial \phi}{\partial x^i}+ g_{li} \frac{\partial \phi}{\partial x^j}-g_{ij} \frac{\partial \phi}{\partial x^l}) \cdot \frac{\partial}{\partial x^k}.
\end{aligned}
$$
Let $\phi=c$ be a constant. The second summand is obviously zero. So we have
$$
\bar{D}_{\frac{\partial}{\partial x^i}}\frac{\partial}{\partial x^j}=D_{\frac{\partial}{\partial x^i}}\frac{\partial}{\partial x^j}.
$$
