In the Chow's book there is a question that I can't solve, the question is
Let $\nabla^g$ denote the Levi-Civita connection of the metric $g$. Show that for any constant $c>0$ and metric $g$, $\nabla^{cg}=\nabla^g$.
I don't know what exactly $cg$ means. Someone has a hint?
If $g$ is a metric on a manifold $M$, then $cg$ is the metric given by : $(cg)_m(X,Y)=c\times g_m(X,Y)$ for any $X,Y\in T_mM$.
To really answer the question, you can use the existence/uniqueness property of the Levi-Cevita connection: for any $X,Y\in T_mM$, $\nabla_XY$ is uniquely determined by the equality $$\forall Z\in T_mM, \qquad 2g(\nabla_XY,Z)=X (g(Y,Z)) + Y (g(Z,X)) - Z (g(X,Y)) + g([X,Y],Z) - g([Y,Z], X) - g([X,Z], Y).$$