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By browsing through the Hodge data of known Calabi-Yau threefolds, I stumbled upon an observation that frequently enough a pair of Hodge numbers $(h^{11},h^{21})$ comes together with the pair $ (h^{11}-240,h^{21}+240)$. This shift, together with mirror symmetry accounts for a certain symmetry in the top portion of the Kreuzer-Skarke plot.

Has anyone observed this shift before?

Is there an empirical explanation for this shift? Is there any relation to the roots of $E_8$?

Are there other similar but smaller shifts?

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    $\begingroup$ A very interesting observation. $\endgroup$ Commented Mar 22, 2014 at 1:00
  • $\begingroup$ In the sactter plot, do the two parabola-like upper curves match up if one is translated to the other? $\endgroup$
    – Mark Gross
    Commented Mar 22, 2014 at 7:55
  • $\begingroup$ I believe so. I zoomed in on the picture near the vertex of the parabola and it was completely symmetric. Also notice that the top points are $(11,491),(251,251),(491,11)$ which exhibits the same shift twice. $\endgroup$ Commented Mar 22, 2014 at 14:36
  • $\begingroup$ Certainly very intriguing. I didn't realise that the curve was symmetric. $\endgroup$
    – Mark Gross
    Commented Mar 22, 2014 at 17:45
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    $\begingroup$ Look at this image: benjaminjurke.net/dynamic/cyexp/… There is almost a perfect symmetry in the top ranges, as if there was some sort of mechanism to get the new Hodge pair if the original numbers are large enough. $\endgroup$ Commented Mar 22, 2014 at 22:00

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In the mathoverflow answer Today's world record on the Betti numbers of Calabi-Yau three-folds., the paper http://arxiv.org/abs/arXiv:1207.4792 is cited. In this paper, the shift of Hodge numbers by (-240,240) is mentionned and the symmetry is explained, at least for the Calabi-Yau threefolds obtained from reflexive polytopes by the Batyrev's construction. E_8 plays indeed an important role (via Calabi-Yau which are K3 fibrations).

EDIT: a video conference of P.Candelas on this paper : https://www.youtube.com/watch?v=4996MUz25vg

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  • $\begingroup$ Thank you! This makes sense to me at least at the combinatorial level. $\endgroup$ Commented Mar 27, 2014 at 13:14

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