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.

1 Answer 1


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 $(2)$.


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