7
$\begingroup$

Is something known about the higher homotopy groups of Calabi-Yau threefolds? For example, one of the easiest CYs is the quintic, defined as the anticanonical divisor in $\mathbb{CP}_4$. What are its homotopy groups?

In particular, here a similar question was asked for K3. What are the higher homotopy groups of a K3 suface? In the first answer, an interesting theorem is cited. It says that the homotopy group of any simply-connected closed 4-manifold $M$ are determined by the second Betti number of $M$. Call it $k$. Then, if $k\geq 1$ and $j\geq 3$ then $\pi_j(M)=\pi_j(\#^{k-1}S^2\times S^3)$. Does a similar result exist for 6-manifolds?

$\endgroup$
  • 6
    $\begingroup$ 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. $\endgroup$ – Aleksandar Milivojevic Mar 14 at 16:43
  • 5
    $\begingroup$ 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. $\endgroup$ – Aleksandar Milivojevic Mar 14 at 16:50
  • 1
    $\begingroup$ (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.) $\endgroup$ – Einfacher Schreiberling Mar 14 at 17:18

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.