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edit approved Dec 4, 2023 at 14:07

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No in dimensions $\geq 3$. To be conformally flat in 3 dimensions, the Cotton tensor Cotton tensor must to vanish, and in dimensions $\geq 4$, the Weyl tensor Weyl tensor must vanish.

Maybe the point of your question though is to ask why is the Cotton or Weyl tensor non-vanishing? I don't have a good explanation for this.

Here's a special example in 3 dimensions. Consider the metric $$dr^2+e^{2k_1r}dx_1^2+e^{2k_2r}dx_2^2.$$ This metric is homogeneous, and when $k_1>0, k_2>0, k_1\neq k_2$, the metric is not conformally flat. The 3 principal sectional curvatures are $-k_1^2, -k_2^2, -k_1k_2$, and the isometry group is a solvable group. If the metric were conformally flat, then this solvable group would embed into $O(3,1)$ by Liouville's theorem Liouville's theorem. However, one can check that this solvable group does not embed by analyzing the Lie algebra and comparing it to the Lie algebras of the solvable subgroups of $O(3,1)$.

No in dimensions $\geq 3$. To be conformally flat in 3 dimensions, the Cotton tensor must to vanish, and in dimensions $\geq 4$, the Weyl tensor must vanish.

Maybe the point of your question though is to ask why is the Cotton or Weyl tensor non-vanishing? I don't have a good explanation for this.

Here's a special example in 3 dimensions. Consider the metric $$dr^2+e^{2k_1r}dx_1^2+e^{2k_2r}dx_2^2.$$ This metric is homogeneous, and when $k_1>0, k_2>0, k_1\neq k_2$, the metric is not conformally flat. The 3 principal sectional curvatures are $-k_1^2, -k_2^2, -k_1k_2$, and the isometry group is a solvable group. If the metric were conformally flat, then this solvable group would embed into $O(3,1)$ by Liouville's theorem. However, one can check that this solvable group does not embed by analyzing the Lie algebra and comparing it to the Lie algebras of the solvable subgroups of $O(3,1)$.

No in dimensions $\geq 3$. To be conformally flat in 3 dimensions, the Cotton tensor must to vanish, and in dimensions $\geq 4$, the Weyl tensor must vanish.

Maybe the point of your question though is to ask why is the Cotton or Weyl tensor non-vanishing? I don't have a good explanation for this.

Here's a special example in 3 dimensions. Consider the metric $$dr^2+e^{2k_1r}dx_1^2+e^{2k_2r}dx_2^2.$$ This metric is homogeneous, and when $k_1>0, k_2>0, k_1\neq k_2$, the metric is not conformally flat. The 3 principal sectional curvatures are $-k_1^2, -k_2^2, -k_1k_2$, and the isometry group is a solvable group. If the metric were conformally flat, then this solvable group would embed into $O(3,1)$ by Liouville's theorem. However, one can check that this solvable group does not embed by analyzing the Lie algebra and comparing it to the Lie algebras of the solvable subgroups of $O(3,1)$.

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created Sep 20, 2012 at 22:32

Ian Agol

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No in dimensions $\geq 3$. To be conformally flat in 3 dimensions, the Cotton tensor must to vanish, and in dimensions $\geq 4$, the Weyl tensor must vanish.

Maybe the point of your question though is to ask why is the Cotton or Weyl tensor non-vanishing? I don't have a good explanation for this.

Here's a special example in 3 dimensions. Consider the metric $$dr^2+e^{2k_1r}dx_1^2+e^{2k_2r}dx_2^2.$$ This metric is homogeneous, and when $k_1>0, k_2>0, k_1\neq k_2$, the metric is not conformally flat. The 3 principal sectional curvatures are $-k_1^2, -k_2^2, -k_1k_2$, and the isometry group is a solvable group. If the metric were conformally flat, then this solvable group would embed into $O(3,1)$ by Liouville's theorem. However, one can check that this solvable group does not embed by analyzing the Lie algebra and comparing it to the Lie algebras of the solvable subgroups of $O(3,1)$.