Formula for divergence of conformal metric

Question

If we have riemannian metrics $g, \tilde{g}$ on a 3-manifold $M$ and a differentiable function $\psi$ such that $g_{ij} = \psi^4 \tilde{g}_{ij}$, then for any symmetric and traceless tensor field $A^{ij}$ it holds that $$D_jA^{ij} = \psi^{-10}\tilde{D}_j(\psi^{10}A^{ij}).$$ $D$ denotes the covariant derivatives. This identity should easily follow from known identities for conformal metrics, but my computations just don't add up. Does anyone know how to do the calculations?

Answer 1

Let us change the notation slightly, so that $\widehat{g} = \Omega^2 g$ is the rescaled Riemannian metric, where $\Omega = e^{\omega}$ and $\omega \in C^{\infty}(M)$. In particular, $\Omega > 0$ everywhere on $M$. For the meantime, let us work in a greater generality, assuming that $M$ is a closed smooth manifold of dimension $n = \dim M$. As in the question, $g$ is a Riemannian metric on $M$.

Extending this notation, we write $\nabla$ and $\widehat{\nabla}$ for the Levi-Civita connections of the metrics $g$ and $\widehat{g}$, respectively.

As we know, $$ \widehat{\nabla}_X Y = \nabla_X Y + (X \omega )Y + (Y \omega )X - g(X,Y) \operatorname{grad}\omega \tag{1} $$

Using the abstract index notation, we can rewrite the above equation in the following form: $$ X^a \widehat{\nabla}_a Y^b = X^a \nabla_a Y^b + (X^a \nabla_a \omega) Y^b + (Y^a \nabla_a \omega) - X^a Y^c g_{a c} g^{b d} \nabla_d \omega \tag{2} $$

Recall that $\omega = \log \Omega$ and introduce the notation: $$ \Upsilon_a := \nabla_a \log \Omega \tag{3} $$

It is also customary in Riemannian geometry to lower and raise the indices using the Riemannian metric $g_{a b}$ and its inverse $g^{a b}$ without mentioning this explicitly, so, for instance, $\Upsilon^a = g^{a b} \Upsilon_b$.

Keeping in mind that $X^a$ is arbitrary, we can factor it away from equation $(2)$, and renaming along the way $Y$ to $X$, we obtain: $$ \widehat{\nabla}_a X^b = \nabla_a X^b + \Upsilon_a X^b + \delta_a{}^b \Upsilon_c X^c - X_a \Upsilon^b \tag{4} $$

Here $\delta^a{}_b$ is the identity operator, also known as the Kronecker symbol.

It is convenient to introduce the following difference tensor $$ S_a{}^b{}_c := \Upsilon_a \delta^b{}_c + \delta_a{}^b \Upsilon_c - \Upsilon^b g_{a c} \tag{5} $$

Using $(5)$ we can rewrite $(4)$ simply as $$ \widehat{\nabla}_a X^b = \nabla_a X^b + S_a{}^b{}_c X^c \tag{6} $$

For any tensor $A^{a b}$, we can calculate the conformal rescaling rule of its covariant derivative using the Leibniz rule in disguise: $$ \widehat{\nabla}_a A^{b c} = \nabla_a A^{b c} + S_a{}^b{}_d A^{d c} + S_a{}^c{}_d A^{b d} \tag{7} $$

The divergence $\nabla_a A^{a b}$ is then rescaled according to the identity $$ \widehat{\nabla}_a A^{a b} = \nabla_a A^{a b} + S_a{}^a{}_c A^{c b} + S_a{}^b{}_c A^{a c} \tag{8} $$

We can simplify the terms of the above equation, using $(5)$, as follows: $$ S_a{}^a{}_c A^{c b} = (\Upsilon_c + n \Upsilon_c - \Upsilon_c) A^{c b} = n \Upsilon_c A^{c b} \tag{9} $$ and $$ S_a{}^b{}_c A^{a c} = (\Upsilon_a \delta^b{}_c + \delta_a{}^b \Upsilon_c - \Upsilon^b g_{a c}) A^{a c} = \Upsilon_a A^{a b} + \Upsilon_c A^{b c} - \Upsilon^b \mathrm{tr} A \tag{10} $$

And now, if $A^{a b}$ is symmetric ($A^{a b} = A^{b a}$) and trace-free ($\mathrm{tr}A = 0$), the last equation simplifies even further $$ S_a{}^b{}_c A^{a c} = 2 \Upsilon_c A^{c b} \tag{11} $$

Thus, for a symmetric trace-free $A^{a b}$, we have the following formula for the conformal rescaling of its divergence: $$ \widehat{\nabla}_a A^{a b} = \nabla_a A^{a b} + (n + 2) \Upsilon_a A^{a b} \tag{12} $$

This means that for any (trace-free symmetric) tensor $A^{a b}$ its divergence is not conformally invariant, unless $A^{a b}$ is identically zero (everywhere vanishing).

We can improve our situation by considering weighted versions of $A^{a b}$. Namely, let $w \in \mathbb{R}$ be any real number, so that we can calculate $$ \widehat{\nabla}_a (\Omega^w A^{a b}) = (\widehat{\nabla}_a \Omega^w) A^{a b} + \Omega^w \widehat{\nabla}_a A^{a b} \\ = (w \Omega^{w - 1} \nabla_a \Omega) A^{a b} + \Omega^w (\nabla_a A^{a b} + (n + 2) \Upsilon_a A^{a b}) \\ = (w \Omega^{w} \nabla_a \log \Omega) A^{a b} + \Omega^w (\nabla_a A^{a b} + (n + 2) \Upsilon_a A^{a b}) \\ = \Omega^w \big( \nabla_a A^{a b} + (n + w + 2) \Upsilon_a A^{a b} \big) \tag{13} $$

The weight is something which we can choose, so that if $\dim M = n$, choosing $w = - n - 2$, we can achieve the following conformal covariance property: $$ \widehat{\nabla}_a (\Omega^{ - n - 2} A^{a b}) = \Omega^{ - n - 2} \nabla_a A^{a b} \tag{14} $$

In particular, if $n = 3$, we can use the weight $-5$, and $\psi^{- 2} = \Omega$, to recover the result in OP.