Timeline for Find a surface or 3-manifold whose fundamental group is $(\mathbb{Z}/n\mathbb{Z}) \rtimes (\mathbb{Z}/2\mathbb{Z})$
Current License: CC BY-SA 4.0
7 events
when toggle format | what | by | license | comment | |
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Aug 24, 2018 at 16:08 | vote | accept | john mangual | ||
Aug 2, 2018 at 8:06 | history | edited | HJRW | CC BY-SA 4.0 |
Corrected a mistake imported from Scott's paper.
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Jul 31, 2018 at 10:23 | comment | added | HJRW | @AllenHatcher: thanks for the useful clarification. I’ll correct the answer shortly. | |
Jul 30, 2018 at 21:29 | comment | added | Allen Hatcher | This seems to be a mistake in Scott's paper. He says that certain subgroups of the unit quaternion group $S^3$ are either cyclic, dihedral, or generalized quaternion (also known as binary dihedral). From this he deduces in particular that dihedral groups act freely on $S^3$. However, $S^3$ has a unique element of order $2$, the quaternion $-1$, so $S^3$ cannot contain dihedral subgroups. On Scott's webpage there is a list of errata for this paper but it does not include this one. | |
Jul 30, 2018 at 8:58 | comment | added | HJRW | @AllenHatcher: Then I must have misunderstood the nomenclature. What should one call the quotients of the 3-sphere by the free dihedral actions that Scott describes on page 451 of his article? | |
Jul 29, 2018 at 20:37 | comment | added | Allen Hatcher | The fundamental groups of prism manifolds are not the dihedral groups themselves but the {\it binary\/} dihedral groups $D^*_{2n}$ which are the preimages of the subgroups $D_{2n}\subset SO(3)$ under the projection $S^3\to SO(3)$. Thus $D^*_{2n}$ maps onto $D_{2n}$ with a ${\mathbb Z}/2$ kernel. | |
Jul 29, 2018 at 10:46 | history | answered | HJRW | CC BY-SA 4.0 |