Let $(M, g)$ be a pseudo-Riemannian manifold and $f \in \mathcal{C}^{\infty}(M)$ a positive function. Find a relation between the two covariant derivatives $\nabla^g$ and $\nabla^h$ of $g$ and $h= f\cdot g$. Hint: The formula is nicer, if you write $f = e^{2u}$ for some $u \in \mathcal{C}^{\infty}(M)$.
For this problem, I have complete no clue. It would be nice if you could show me how to prove this
I write $g=g$ and $h=\tilde{g}$ where $\tilde{g}=e^{2f}g$ for $f$ a diferentiable map. Then you have: $\tilde{\nabla}_{X}Y=\nabla_{X}Y+(Xf)Y+(Yf)X-g(X,Y)\nabla f$
To prove it, just note:
Using the Koszul formula for $g$ \begin{equation*} \label{eq:Koszul} \begin{array}{rcl} 2g(\nabla_X Y,Z)&=& X g(Y,Z) + Y g(X,Z) - Z g(X,Y) \\ && + g([Z,X],Y) + g([Z,Y],X) + g([X,Y],Z). \end{array} \end{equation*} And using again the Koszul formula for $\tilde{g}$, we get \begin{equation*} \begin{array}{rcl} 2e^{2f}g(\tilde{\nabla}_X Y,Z)&=& X( e^{2f}g(Y,Z)) + Y (e^{2f}g(X,Z)) - Z (e^{2f}g(X,Y)) \\ && + e^{2f}g([Z,X],Y) + e^{2f}g([Z,Y],X) + e^{2f}g([X,Y],Z), \end{array} \end{equation*} if we write the three first terms in the form $X( e^{2f}g(Y,Z))=2(Xf)e^{2f}g(Y,Z)+e^{2f}Xg(Y,Z)$ we arrive to \begin{align*} 2e^{2f}g(\tilde{\nabla}_X Y-\nabla_X Y,Z)&=2e^{2f}\{(Xf)g(Y,Z)+(Yf)g(X,Z)-(Zf)g(X,Y)\} \end{align*} and this finishes our prove due to $Zf=g(\nabla f,Z)$.