Timeline for Higher homotopy groups of Calabi-Yaus
Current License: CC BY-SA 4.0
4 events
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| Mar 14, 2019 at 17:18 | comment | added | ssx | (I guess for you as a physicist CY means holonomy $=\mathrm{SU}(3)$ and so $\pi_1=0$. But many mathematicians consider something like $\mathrm{K3}\times\mathrm{E}$, $\mathrm{E}$ an elliptic curve, to be a perfectly fine CY 3-fold, and for these you could of course use the result you mention together with $\pi_k(\mathrm{E})=0$ for $k>1$. There are also CY 3-folds with nonabelian $\pi_1$: math.unice.fr/~beauvill/pubs/C-Y.pdf . But in 6 dimensions one cannot expect something as simple as in the 4 dimensional case, topology in 4 dimensions is very special.) | |
| Mar 14, 2019 at 16:50 | comment | added | Aleksandar Milivojević | If you know the cohomology ring of your Calabi-Yau, you can use the result of T. Miller that simply-connected 6-manifolds are formal (or the DGMS result that Kahler manifolds are formal) to calculate the rational homotopy groups from the cohomology ring. I don't know if there are any restrictions known on the cohomology rings of Calabi-Yaus, though. | |
| Mar 14, 2019 at 16:43 | comment | added | Aleksandar Milivojević | I listed the result of a paper by Babenko that gives a nice formula for the ranks of the rationalized homotopy groups of hypersurfaces (and more generally complete intersections) here mathoverflow.net/questions/300317/…. For simply connected six-manifolds, the second Betti number won't be enough to determine the rational homotopy groups. For example, $S^2\times S^2 \times S^2$ has the same Betti numbers as $CP^3 \# CP^3 \# CP^3$, but the first has only finitely many nonzero Q-hmtpy groups, while the second has inf. many. | |
| Mar 14, 2019 at 16:27 | history | asked | Federico Carta | CC BY-SA 4.0 |