Timeline for when constant scalar curvature implies Einstein?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Nov 29, 2020 at 12:08 | comment | added | Michael Albanese | @R.Rankin: It means that it is flat. Note that $|\operatorname{Riem}|^2 \geq 0$ so $\int_M|\operatorname{Riem}|^2d\mu = 0$ if and only if $|\operatorname{Riem}|^2 \equiv 0$ and hence $\operatorname{Riem} \equiv 0$. | |
| Nov 29, 2020 at 9:40 | history | edited | Benoît Kloeckner | CC BY-SA 4.0 |
The answer was wrong as it was written, salvaged it by adding a precision.
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| Nov 29, 2020 at 7:59 | comment | added | R. Rankin | @MichaelAlbanese Does that really mean that M is flat everywhere, or that the AVERAGE Riemann curvature (squared) vanishes? | |
| Jun 4, 2019 at 15:56 | comment | added | Malkoun | @MichaelAlbanese, yes, thank you. So essentially I was right, up to rearranging curvature terms maybe, and some factor. Thank you. | |
| Jun 4, 2019 at 15:26 | comment | added | Michael Albanese | @Malkoun: Chern-Gauss-Bonnet gives $$\chi(M) = \frac{1}{32\pi^2}\int_M(|\operatorname{Riem}|^2 - 4|\operatorname{Ric}|^2 + R^2)d\mu = \int_M(|\operatorname{Riem}|^2 - 4|\operatorname{Ric}\limits^{\circ}|^2)d\mu$$ where $\operatorname{Ric}\limits^{\circ}$ denotes the trace-free Ricci curvature. So on an Einstein four-manifold, $\chi(M) = \frac{1}{32\pi^2}\int_M |\operatorname{Riem}|^2d\mu$. In particular, $\chi(M) = 0$ if and only if $M$ is flat. | |
| Jun 4, 2019 at 11:24 | comment | added | Malkoun | I think I get it. Is it because the Pfaffian of the curvature must be for a closed oriented Ricci flat $4$-manifold a non-negative function ($|W|^2$) times the volume form? So in this case the curvature must vanish identically, by Gauss-Bonnet? | |
| Jun 4, 2019 at 11:18 | comment | added | Malkoun | @KlausKröncke, can you give a bit more details? Why must the manifold be flat? Why can't it have say a non-zero Weyl tensor? | |
| Dec 12, 2014 at 16:56 | comment | added | Klaus Kröncke | This comment comes very late but it is maybe worth to post it: $S^1\times S^3$ does not admit an Einstein metric because the Euler Characteristic is zero. In this case (this is special in dimension $4$), the manifold must be flat, and since it is simply-connected, it must be $\mathbb{R}^4$. | |
| Apr 18, 2013 at 9:36 | comment | added | Thomas Richard | Just a caveat, for this metric on $\mathbb{S}^2\times\mathbb{S}^2$ not to be Einstein, the two $\mathbb{S}^2$ need to have different radius. And a remark about Agol's comment : what I like in it is that it in fact doesn't admit any Einstein metric, just because in dimension 3 Einstein is equivalent to constant sectional curvature. I wonder wether $\mathbb{S}^1\times\mathbb{S}^3$ enjoys the same property or not... | |
| Apr 18, 2013 at 9:16 | vote | accept | Mathboy | ||
| Apr 17, 2013 at 21:04 | comment | added | Ian Agol | $S^1\times S^2$ works too. | |
| Apr 17, 2013 at 12:46 | comment | added | Mathboy | Yes, I think your example has scalar curvature equal to 4 and Ricci curvature nonnegative. But some sectional curvature is zero. Thank you very much. | |
| Apr 17, 2013 at 12:40 | vote | accept | Mathboy | ||
| Apr 18, 2013 at 9:16 | |||||
| Apr 17, 2013 at 11:56 | history | answered | Benoît Kloeckner | CC BY-SA 3.0 |