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Let $(r,\theta,\phi)$ be the spherical coordinates on $\mathbb{R}^3$ where $\theta \in (0,\pi)$ and $\phi\in (0,2\pi)$.

Does there exist an asymptotically flat metric $g$ on $\mathbb{R}^3\setminus B_1$ such that ${\rm Ric}_{r\phi} = \frac{\sin\theta}{r^2}$? Or even $\frac{f}{r^2}$ where $f \in C^{\infty}(S^2)$ satisfying $\int_{S^2} f \neq 0$? Is there an easy way to find an example?

$g$ is asymptotically flat if there exists coordinates such that $g_{ij} - \delta_{ij} = O_{2}(r^{-1})$.

It is easy to see that $g$ cannot be conformal to $g_{euc}$ since ${\rm Ric}_{r\phi}$ will be equal to $r^{-2}\partial_{\phi} A$ for some function $A$ on the sphere, and so integrates to $0$.

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  • $\begingroup$ what exactly do you mean by the equality ${\rm Ric}_{r\phi} = \frac{\sin\theta}{r^2}$? that Ricci curvature is pointwise constant? then the answer is No by Schur's Lemma. $\endgroup$ Commented Apr 16, 2022 at 2:20
  • $\begingroup$ I mean the function ${\rm Ric}(\frac{\partial}{\partial r}, \frac{\partial}{\partial \phi})$ on $\mathbb{R}^3\setminus B_1$ is the function $\frac{\sin \theta}{r^2}$. $\endgroup$
    – Laithy
    Commented Apr 16, 2022 at 2:23
  • $\begingroup$ I am not making any other assumptions about the rest of the Ricci components. So the Ricci curvature need not be (and cannot be) pointwise constant. $\endgroup$
    – Laithy
    Commented Apr 16, 2022 at 2:27
  • $\begingroup$ ah, ok, sorry, I thought subindices were coordinates of a point. $\endgroup$ Commented Apr 16, 2022 at 2:28

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