Relation between two Riemannain connections

The formula for the conformal rescaling of the Levi-Civita connection is an essential tool in Riemannian geometry, and its derivation is given in many sources.

As Isaac Solomon and Ted Shifrin have mentioned in the comments, a slick way to derive it is to consider $f = e^{2 \omega}$ and use the Koszul formula. The result will be in the form: $$ \nabla' _X Y = \nabla _X Y + (X \omega )Y + (Y \omega )X - g(X,Y) \operatorname{grad}\omega \tag{1} $$

Proof. The Koszul formula (see e.g. here) gives the following expression for the Levi-Civita connection $\nabla$ of the metric $g$: $$ \begin{align} 2 g(\nabla_X Y, Z) & = X \, g(Y,Z) + Y \, g(Z,X) - Z \, g(X,Y) \\ \tag{2} &- g(X,[Y,Z]) + g(Y,[Z,X]) + g(Z,[X,Y]) \end{align} $$

Let $\nabla'$ be the Levi-Civita connection for the metric $g' = e^{2\omega}g$. Substituting these objects into (2) $$ \begin{align} 2 e^{2 \omega} g(\nabla'_X Y, Z) & = X \left( e^{2 \omega} g(Y,Z) \right) + Y \left( e^{2 \omega} g(Z,X) \right) - Z \left( e^{2 \omega} g(X,Y) \right) \\ &- e^{2 \omega} g(X,[Y,Z]) + e^{2 \omega} g(Y,[Z,X]) + e^{2 \omega} g(Z,[X,Y]) \end{align} $$ and computing the derivatives using the product rule, we obtain $$ \begin{align} 2 e^{2 \omega} g(\nabla'_X Y, Z) & = e^{2 \omega} X g(Y,Z) + e^{2 \omega} Y g(Z,X) - e^{2 \omega} Z g(X,Y) \\ & + 2 e^{2 \omega} g(Y,Z) \, X \omega + 2 e^{2 \omega} g(Z,X) \, Y \omega - 2 e^{2 \omega} g(X,Y) \, Z \omega \\ &- e^{2 \omega} g(X,[Y,Z]) + e^{2 \omega} g(Y,[Z,X]) + e^{2 \omega} g(Z,[X,Y]) \end{align} $$

In the last display we can divide both sides of the equation by $e^{2 \omega}$, which is a strictly positive function, to get $$ \begin{align} 2 g(\nabla'_X Y, Z) & = X g(Y,Z) + Y g(Z,X) -Z g(X,Y) \\ & + 2 g(Y,Z) \, X \omega + 2 g(Z,X) \, Y \omega - 2 g(X,Y) \, Z \omega \\ &- g(X,[Y,Z]) + g(Y,[Z,X]) + g(Z,[X,Y]) \end{align} $$

Using the Koszul formula (2) again we rewrite the above expression as $$ \begin{align} 2 g(\nabla'_X Y, Z) & = 2 g(\nabla_X Y, Z) + 2 g(Y,Z) \, X \omega + 2 g(Z,X) \, Y \omega - 2 g(X,Y) \, Z \omega \end{align} $$ which is equivalent to (1) because vector field $Z$ is arbitrary, $g$ is non-degenerate, and $Z \omega = \mathrm{d} \omega (Z)$. Recall also that $\operatorname{grad} \omega = (\mathrm{d} \omega)^{\sharp}$.

The version of this formula in terms of coordinates and the Christoffel symbols is obtained by a similar calculation, the result will be $$ '\Gamma^{k}_{ij}=\Gamma^{k}_{ij} + \delta_{i}^{k} \partial_j \omega + \delta_{j}^{k} \partial_i \omega - g_{i j} g^{k l} \partial_{l} \omega $$

This can be also obtained as a consequence of (1).