Transformation rule for Levi-Civita connection on differential forms under conformal metric change

It is well-known, and shown e.g. here or here, how the Levi-Civita connection behaves under a conformal metric change acting on vector fields. What I want to deduce is the representation for differential forms.

Let $$(M,g)$$ be a Riemannian manifold of dimension $$n$$. Considering a conformal metric change, say $$\tilde g = \sigma^{-2}g$$, we get the following transformation rules for the Levi-Civita connection.

\begin{align} \tilde\nabla_XY &= \nabla_XY - \sigma^{-1} \left( X(\sigma)Y + Y(\sigma)X - \langle X,Y \rangle_g \operatorname{grad}_g \sigma\right) \tag{1}\\ \tilde\nabla_X\omega &= \nabla_X\omega - \sigma^{-1} \left( \omega(\operatorname{grad}_g\sigma)X^\flat - X(\sigma)\omega - \omega(X) \mathrm d\sigma \right).\tag{2} \end{align}

To deduce (2) from (1), my idea was to use dualisation since the Levi-Civita connection commutes with it, i.e. \begin{align*} \tilde\nabla_X\omega = \left( \tilde\nabla_X\omega^{\tilde\sharp}\right)^{\tilde\flat}. \end{align*}

Clearly, we get $$X^{\tilde\flat} = \sigma^{-2}X^\flat$$ and $$\eta^{\tilde\sharp} = \sigma^2\eta^\sharp$$.

And we should use the definitions given by the musical isomorphisms, i.e.

• $$\omega^\sharp$$ is a vector field dual of a one-form $$\omega \in \Omega^1(M)$$, defined by $$g(\omega^\sharp,Y) = \omega(Y)$$,
• $$X^\flat$$ is a one-form dual to a vector field $$X$$, defined by $$X^\flat(Y) = g(X,Y)$$.
• the gradient is defined as $$\operatorname{grad}_g\sigma = \sharp(\mathrm d\sigma)$$

My questions

1. Using the characterisation $$g(\omega^\sharp,Y)$$ and $$\sharp(\mathrm d\sigma)$$ for the gradient, the last summand in the brackets is clear. But I do not see how to get the first two.
2. I want to generalise the formula to $$p$$-forms and deduce the well-known representation for the codifferential $$\mathrm \delta = \iota_{e_j}\nabla_{e_j}$$ as the trace of the covariant derivative.

In order to deduce $$(2)$$ from $$(1)$$ we usually simply observe, that for any $$1$$-form $$\omega$$ and any vector field $$Y$$, the expression $$\omega(Y)$$ is just a (smooth) function, so that both derivatives agree on it: $$\tilde{\nabla}_X \big( \omega(Y) \big) = \nabla_X \big( \omega(Y) \big)$$
Using the Leibniz rule, we can unwrap both sides of the above equation: $$\big( \tilde{\nabla}_X \omega \big)(Y) + \omega(\tilde{\nabla}_X Y) = \big( \nabla_X \omega \big) (Y) + \omega(\nabla_X Y)$$ and substitute the equation $$(1)$$ to express $$\tilde{\nabla}_X Y$$ in the LHS: $$\big( \tilde{\nabla}_X \omega \big)(Y) + \omega \bigg( \nabla_XY - \sigma^{-1} \big( X(\sigma)Y + Y(\sigma)X - \langle X,Y \rangle_g \operatorname{grad}_g \sigma \big) \bigg) \\ = \big( \nabla_X \omega \big) (Y) + \omega(\nabla_X Y)$$
Since $$\omega$$ is a $$1$$-form, that is linear, we can expand the second term in the LHS further as $$\big( \tilde{\nabla}_X \omega \big)(Y) + \omega \big( \nabla_XY \big) - \sigma^{-1} X(\sigma) \omega (Y) - \sigma^{-1} Y(\sigma) \omega(X) + \sigma^{-1} \langle X,Y \rangle_g \omega (\operatorname{grad}_g \sigma ) \\ = \big( \nabla_X \omega \big) (Y) + \omega(\nabla_X Y)$$
Simplifying the above equation and factoring $$Y$$ away, we obtain $$(2)$$.