Levi-Civita connection under a rescaled metric

Question

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.

Answer 1

The Levi Civita is the unique connection without torsion for which the Riemann tensor has $\nabla g=0$. Of course, if $\nabla g=0, \nabla c.g=0$.

Answer 2

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 $$ \bar{D}_{\frac{\partial}{\partial x^i}}\frac{\partial}{\partial x^j}=D_{\frac{\partial}{\partial x^i}}\frac{\partial}{\partial x^j}. $$