# Relation between different connections on the same manifold

As I was thinking about this other problem, I stumbled upon another issue (maybe easier to solve).

Suppose we have got a smooth manifold $M$, endowed with two different metrics ($g_1$,$g_2$). Then we have got two riemannian manifolds $M_1 := (M,g_1)$ and $M_2 := (M,g_2)$.

What is the Levi-Civita connection of $M_2$ when viewed in $M_1$? What is its meaning?

Example: $M_1 = (H^2,dx^2+dy^2)$ and $M_2 = (H^2,\frac{dx^2+dy^2}{y^2})$. What's the meaning of the connection of $M_2$ in the euclidean space?

Let $M_1=(M,g)$ be a smooth Riemannian manifold. Let $\lambda:M\to\mathbb{R}$ be a smooth function, define $\tilde{g}=e^{2\lambda}g$, and consider the pointwise conformally equivalent Riemannian manifold $M_2=(M,\tilde{g})$. Then we have $$\nabla^{\tilde{g}}_XY=\nabla^g_XY+(X\lambda)Y+(Y\lambda)X-g(X,Y)\mathrm{grad}(\lambda),$$ where $\nabla^g$ and $\nabla^{\tilde{g}}$ are the Levi-Civita connections of $M_1$ and $M_2$, respectively. This is Exercise 8.5 (has lots of hints) in do Carmo's Riemannian Geometry In terms of usefulness, we can use this formula to compute the Levi-Civita connection for a large class of smooth manifolds easily. I'm not sure what to say about it's "meaning", since that seems very problem specific / it depends on what the manifold and metrics represent.
• Thank you for your answer! Yes, maybe you're right, I didn't quite explain thoroughly what I was looking for. Let's take my example with the euclidean and hyperbolic metrics, for instance, and say we take a straight line $c(t) = vt + x_0$. Then one can compute $\nabla^{H^2}_{c'(t)}c'(t)$, and what one finds out is that this vector field resembles very much a force that prevents the path to bend into a geodesic of the hyperbolic plane: if for example $v=(1,0)$, my calculation yields $\nabla^{H^2}_{v}v = (0,1/y)$, a field pointing upward (thus not making the path bend into a circumference). – Alessio Di Lorenzo Aug 5 '17 at 20:10
• If, on the other hand, $\nabla^{H^2}_{c'(t)}c'(t) = 0$, there is no "force" trying to resist the "bending", and then the path is indeed a geodesic. Anyway, in this case your answer is very useful and I will try and put it to good use! – Alessio Di Lorenzo Aug 5 '17 at 20:12