Timeline for Conformally flat manifold with zero scalar
Current License: CC BY-SA 3.0
19 events
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| Jan 13, 2016 at 11:30 | comment | added | Holonomia | @katz: I learn that $S^2 \times H^2$ is conformally flat time ago in the nice paper by Simon Salamon: Complex structures and conformal geometry. In: BOLLETTINO DELLA UNIONE MATEMATICA ITALIANA, vol. 9, pp. 199-224. The issue is related to the twistor of $S^4$ and there are there is also a nice picture. | |
| Jan 13, 2016 at 11:08 | comment | added | Holonomia | So the original Katz answer was correct. | |
| Jan 13, 2016 at 9:59 | comment | added | Robert Bryant | @ThomasRichard: Indeed, you are correct. The product of the unit $2$-sphere (with $K\equiv1$) with the hyperbolic metric (with $K\equiv-1$) on any compact Riemann surface of genus $g\ge2$ gives a conformally flat metric with zero scalar curvature that is not flat. I forgot about that example when I was proposing a general method to construct families. | |
| Jan 13, 2016 at 9:39 | comment | added | Thomas Richard | In general a product of l.c.f manifolds is not l.c.f. But if I recall correctly there is a small miracle that happens when you take the product of the sphere and the hyperbolic space (with curvature +1 and -1, and this works only with these values). I'll try to post the computation later if anybody is interested. Anyway, if I correctly understand the question, this give that $\mathbb{S}^n\times \mathbb{H}^n$ answers the question in even dimensions. | |
| Jan 13, 2016 at 8:48 | comment | added | Mikhail Katz | You wrote above that "it is well-known that $S^ 2 \times H ^2$ is conformally flat where $H ^2 $ is the unit disk with the hyperbolic metric" but I don't think this is correct. Each factor is obviously conformally flat but the product is not. | |
| Jan 13, 2016 at 8:45 | comment | added | Mikhail Katz | @Holonomia, that metric is indeed of zero scalar curvature, but it is not conformally flat. Do you have reason to think it is? | |
| Jan 13, 2016 at 8:34 | comment | added | Holonomia | @katz I am sure that, with the standard definition (I assume it is the one given by Robert Bryant), your initial answer solves the OP question. Namely, a product $S^2 \times X$ where $S^2$ is endowed with its canonical metric of constant curvature 1 and $X$ is a compact quotient of the unit disk endowed with its canonical metric of constant curvature -1 i.e. a compact hyperbolic surface. | |
| Jan 13, 2016 at 8:29 | comment | added | Mikhail Katz | The metric on $S^2\times X$ is a direct sum of the metrics on the factors. The above metric is a product in a certain sense. Did I get it wrong again? I am having trouble waking up... | |
| Jan 13, 2016 at 6:37 | comment | added | Holonomia | @katz But now, with this local definition of conformally flat, what is wrong with your initial answer $S^2 \times X$ where $X$ is a hyperbolic compact surface? | |
| Jan 13, 2016 at 3:09 | comment | added | Mikhail Katz | Locally one can always write down $e^{x^2+y^2-z^2-w^2}$ and it will be conformally flat but not flat. Certainly globally one needs to use compactness as pointed out by @Holonomia. | |
| Jan 13, 2016 at 0:15 | comment | added | Robert Bryant | @Holonomia: A metric $g$ on $M$ is conformally flat if each point $p\in M$ has an open neighborhood $U$ on which there exists a function $f$ such that $e^f g$ is flat on $U$. This condition really should be called 'locally conformally flat', but the common usage is to simply say 'conformally flat', because, in that way, conformally flat becomes something that one can test locally. Another criterion that is used when $n>3$ is to say that a metric $g$ is conformally flat if its Weyl tensor vanishes. (In dimension $3$, the condition is that its Cotton tensor vanishes.) | |
| Jan 13, 2016 at 0:13 | comment | added | Deane Yang | "Conformally flat" = "locally conformally flat". | |
| Jan 12, 2016 at 23:21 | comment | added | Holonomia | @Robert Bryant: just to avoid further mistakes: What is the definition of conformally flat ? | |
| Jan 12, 2016 at 23:06 | comment | added | Robert Bryant | @Holonomia: I'm afraid that you are still making a mistake. Having $g$ be conformally flat does not imply that there is an $f$ such that $e^fg$ has constant sectional curvature. Again the natural metric $g$ on $M = S^1\times S^{n-1}$ is conformally flat, but, when $n>2$, there is no function $f$ on $M$ such that $e^f g$ has constant sectional curvature. | |
| Jan 12, 2016 at 22:32 | history | edited | Holonomia | CC BY-SA 3.0 |
I add more details to my partial answer.
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| Jan 12, 2016 at 21:18 | comment | added | Holonomia | @ Robert Bryant: You are right I forgot that the conformal metric can also have constant but non zero sectional curvature. For example, the sphere $S^2$ is conformally flat but has no flat metric... | |
| Jan 12, 2016 at 20:42 | comment | added | Robert Bryant | @Holonomia: Your argument is not right either, for, along the way, you argue that a compact conformally flat manifold carries a flat metric (i.e., you do not use the hypothesis that the scalar curvature is zero when you derive this). However, this statement is false for dimensions bigger than $2$: $S^1\times S^{n-1}$ has a conformally flat metric but has no flat metric when $n>2$. | |
| Jan 12, 2016 at 20:33 | comment | added | Holonomia | @ katz I agree that this is too complicated. But your proof does not seems to use the compactness of the manifold,isn'it? | |
| Jan 12, 2016 at 19:23 | history | answered | Holonomia | CC BY-SA 3.0 |