In 4 dimensions, for a conformal change of metric $g=e^{2u}g_0$ the Ricci curvature tensor $\operatorname{Ric}$ satisfies the transformation law

\begin{equation}\tag{1} \operatorname{Ric}_g = \operatorname{Ric}_0 - 2\nabla_0^2 u + 2du\otimes du - (\Delta_0 u)g_0 -2|\nabla_0 u|_0^2g_0 \end{equation}

where a subscript $0$ means w.r.t. the metric $g_0$, and a subscript $g$ means w.r.t. $g=e^{2u}g_0$. I am trying to compute an analogue to (1) but for the quantity \begin{equation}\tag{2} |\operatorname{Ric}_g|_g^2 = (\operatorname{Ric}_g)^{ij}(\operatorname{Ric}_g)_{ij} = g^{ia}g^{jb}(\operatorname{Ric}_g)_{ab}(\operatorname{Ric}_g)_{ij} = e^{-4u}g_0^{ia}g_0^{jb}(\operatorname{Ric}_g)_{ab}(\operatorname{Ric}_g)_{ij} = \cdots \end{equation}

i.e. a transformation law for $|\operatorname{Ric}_g|_g^2$ in terms of the metric $g_0$. Aside from continuing the calculation in (2) using coordinates (which gives a whole load of terms, but doesn't seem to simplify to anything nice) I am not really sure how to go about this. I feel like it should be easy given the transformation law in (1), so apologies if I'm missing something obvious. Thanks in advance!

  • $\begingroup$ Why don't you just plug the result from (1) in? It doesn't look that horrible. $\endgroup$ – Willie Wong Oct 9 '18 at 5:00
  • $\begingroup$ @WillieWong So if I write out (1) in coordinates and continue, I end up with terms such as $e^{-4u}g_0^{ia}g_0^{jb}(\nabla_0)_a\partial_b u\cdot\partial_i u\cdot\partial_j u$ - this comes from the $\nabla_0^2 u$ and $du\otimes du$ terms in coordinates being multiplied together (i.e. the $(\nabla_0)_a\partial_b$ and $\partial_i u\cdot\partial_j u$ being multiplied together). Does this, for instance, have an expression without coordinates? It looks sort of like a contraction $:$ between two matrices. $\endgroup$ – jl2 Oct 11 '18 at 13:33
  • $\begingroup$ You are looking at the double contraction of $\nabla^2 u$ against $\nabla u \otimes \nabla u$. This can be rewritten as a multiple of $g( \nabla u , \nabla |\nabla u|^2)$ (for clarity I omit all the ${}_0$ subscripts). $\endgroup$ – Willie Wong Oct 11 '18 at 13:55
  • $\begingroup$ In any case, you have a thing with five terms, you square it, you should expect to pick up something with 15 terms. There will be some simplifications due to some of the trace structure leaving you with I think 10 terms. Why do you think there should be something "nicer"? $\endgroup$ – Willie Wong Oct 11 '18 at 14:07
  • 2
    $\begingroup$ that's in general the best. $\endgroup$ – Willie Wong Oct 11 '18 at 18:02

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