Timeline for Does $S^n\times H^k$ have non-isometric conformal transformations?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jan 6, 2016 at 11:56 | answer | added | Holonomia | timeline score: 4 | |
| Jan 6, 2016 at 9:55 | comment | added | Misha | @RenatoG.Bettiol: This is just Liouville's theorem (conformal = restriction of Moebius), since your dimension $k+n$ is at least 3. | |
| Jan 6, 2016 at 5:21 | comment | added | Renato G. Bettiol | I was also thinking the same, and counting dimensions everything seems to match. Namely, the dimension of the group of conformal transformations of $S^{n+k}$ that leaves $S^{k-1}$ invariant is $k(k-1)/2+n(n+1)/2+k$, where the last $k$ come from dilations. At the same time, the dimension of the isometry group of $S^n\times H^k$ is $n(n+1)/2+k(k+1)/2$, so these numbers agree. But there are still some gaps in my understanding of the situation: (1) why do these conformal transformations extend across $S^{k-1}$? (2) Why aren't there other connected components? | |
| Jan 6, 2016 at 4:24 | comment | added | Igor Belegradek | My reading of what Misha said is that $S^{k-1}$ is removable, i.e. the conformal maps extend to conformal automorphisms of $S^{k+n}$ that stabilize $S^{k-1}$, and aren' those exactlly the isometries of $S^n\times H^k$? | |
| Jan 6, 2016 at 1:12 | comment | added | Renato G. Bettiol | @Misha: I understand that $(M,g)$ is conformally flat and also conformally equivalent to the open manifold $S^{k+n}\setminus S^{k-1}$ with the round (incomplete) metric. But how does this imply that all of its conformal transformations are isometric? Can you please elaborate? | |
| Jan 6, 2016 at 0:03 | comment | added | Misha | They are all isometric, you can see this by observing that the product metric is conformally flat with conformal structure realized as the complement of a round k-1 sphere in the round k+n sphere. | |
| Jan 5, 2016 at 22:27 | history | asked | Renato G. Bettiol | CC BY-SA 3.0 |