# Codifferential / divergence of differential form under conformal metric change

I have a question related to this and a second post. I want to calculate the codifferential under a conformal metric change, $$g_\psi = e^{2\psi} g$$. By Besse's book on Einstein manifolds, or an obvious calculation, we have that (acting on $$p$$-forms)

$$*_{g_\psi} = e^{(n-2p)\psi} *_g.$$

The codifferential is then given by

$$\delta^{g_\psi} = e^{-2\psi} \left( \delta - (n-2p) \iota_{\nabla \psi}\right).$$

But I don't end up with the correct factor in front of the brackets.

By definition, $$\delta = *^{-1} \mathbf d \ *$$. So, for some $$\alpha \in \Omega^p(M)$$ locally (consider a fixed multi-index for the moment), \begin{align*} \delta^{g_\psi} \alpha &= *_{g_\psi}^{-1} \mathbf d *_{g\psi} \left(\alpha_i \mathrm dx^i\right)\\ &= *_{g_\psi}^{-1} \mathbf d \left( e^{(n-2p)\psi} \alpha_{i} \mathrm dx^{1} \wedge ... \wedge \widehat{\mathrm dx^{i}} \wedge ... \wedge \mathrm dx^{p} \right)\\ &= *_{g_\psi}^{-1} \Big( (n-2p)e^{(n-2p)\psi} \alpha_{i} \mathrm d\psi \wedge \mathrm dx^{1} \wedge ... \wedge \widehat{\mathrm dx^{i}} \wedge ... \wedge \mathrm dx^{p}\\ &\qquad+ e^{(n-2p)\psi} (\partial_j\alpha_i) \mathrm dx^j \wedge \mathrm dx^{1} \wedge ... \wedge \widehat{\mathrm dx^{i}} \wedge ... \wedge \mathrm dx^{p} \Big) \\ &\overset?= \underbrace{*_{g_\psi}^{-1}}_{= e^{-(n-2p)\psi} *_g} \Big[ e^{(n-2p)\psi} (n-2p) \mathrm d\psi \wedge \left(\alpha_{i} \mathrm dx^{1} \wedge ... \wedge \widehat{\mathrm dx^{i}} \wedge ... \wedge \mathrm dx^{p}\right)\\ &\qquad + (\partial_i\alpha_{i}) \mathrm dx^{1} \wedge ... \wedge \mathrm dx^{p} \Big] \\ &= -(n-2p) \ \iota_{\nabla\psi}\alpha + \delta, \\ \end{align*} where $$\widehat{\mathrm dx^{i}}$$ denotes the element that was left out and using that $$*(X^\flat \wedge \alpha) = (-1)^p \iota_X(*\alpha)$$.

(1) Where is the error?

(2) Is there an easy argument to deduce the formula from (cf. here) $$\text{div}_{g_\psi} X = \text{div}_{g} X + n e^{-2\psi} X(f)?$$

## 1 Answer

(1) The error indeed appears to occur at the equality marked with a $$?$$: If we denote $$\hat\ast := \ast_{g_\psi},$$ then in the calculation $$\hat\ast^{-1}$$ is applied to $$d[e^{(n - 2 p) \psi} \ast \alpha]$$, which is an $$(n - p + 1)$$-form. Thus, the inverse Hodge star operators $$\ast^{-1}, \hat\ast^{-1}$$ are related by $$\hat\ast^{-1} = e^{[n - (2 (n - p + 1))] \psi} \ast^{-1} = e^{(-n + 2p - 2) \psi} \ast^{-1}.$$ Thus, the total conformal factor that appears is $$e^{(-n + 2p - 2) \psi} \cdot e^{(n - 2p) \psi} = e^{-2 \psi}$$ as claimed. There's no need here, by the way, to resort to a coordinate computation here. Using the transformation rule for $$\ast$$, we can write the first few steps of the computation, again using hats $$\hat\cdot$$ to denote objects constructed from $$\hat g := e^{2 \psi} g$$, as $$\hat\delta \alpha = \hat\ast^{-1} d \hat\ast \alpha = \hat\ast^{-1} d [ e^{(n - 2 p) \psi} \ast \alpha] .$$

(2) The formula for the change of the divergence of a vector field under a conformal transformation is essentially equivalent to the specialization to $$p = 1$$ of the formula in this question: The vector field formula gives $$\widehat{\operatorname{div}} \,{\alpha^{\hat\sharp}} = \operatorname{div} \alpha^{\hat\sharp} + n \alpha^{\hat\sharp} \cdot f .$$ Unwinding definitions gives that the index-raising operators $$\sharp, \hat\sharp$$ are related by $$\alpha^{\hat\sharp} = e^{-2 \psi} \alpha^\sharp.$$ Then, substituting, applying the Leibniz rule for $$\operatorname{div}$$, rearranging, and replacing $$\operatorname{div} \alpha^{\sharp}$$ with $$\delta \alpha$$ and $$\widehat{\operatorname{div}} \,{\alpha^{\hat\sharp}}$$ with $$\hat\delta \alpha$$ recovers the $$p = 1$$ formula.