I was told that Calabi-Yau's can be birational to each other but not isomorphic (biholomorphic).
But I've never seen explicit examples. Can anybody here show me one?
(E.g. maybe an explicit example of a flop between Calabi-Yaus?)
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$\begingroup$ What does "isomorphic" mean in this context? $\endgroup$– Igor RivinCommented Sep 9, 2012 at 4:22
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1$\begingroup$ Presumably it means "biholomorphic". $\endgroup$– YangMillsCommented Sep 9, 2012 at 4:46
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$\begingroup$ Try looking at the paper "The movable fan of the Horrocks--Mumford quintic" by Michael Fryers. That gives an explicit example of a CY 3-fold with (IIRC) precisely 8 birational models. $\endgroup$– user5117Commented Sep 9, 2012 at 18:40
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1$\begingroup$ One other small comment is that your "e.g." is really more than an "e.g.": any two birational Calabi--Yaus (as long as they're smooth, or a bit more generally, have only terminal singularieties) are related by a sequence of flops. (Maybe you already knew that.) $\endgroup$– user5117Commented Sep 9, 2012 at 18:41
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1$\begingroup$ Ah, a quick search shows it is proved for any dimension by Kawamata in 2007. $\endgroup$– tempCommented Sep 9, 2012 at 23:08
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2 Answers
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You may be interested in Lee and Oguiso's paper Connecting certain rigid birational non-homeomorphic Calabi--Yau threefolds via Hilbert scheme. This gives a pair of CY3s you want (with additional interesting properties).
answered Sep 9, 2012 at 5:53
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Take a quintic hypersurface in $P^4$ with several (say $n$) ordinary double points. Each of them locally analytically has 2 small resolution. Combining those you can construct $2^n$ global small resolutions. All of them are birational Calabi-Yau threefolds.
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1$\begingroup$ Why are they not isomorphic to each other? $\endgroup$– tempCommented Sep 9, 2012 at 23:02
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1$\begingroup$ Usually some of them are projective and the other are not. It depends on existence of Weil divisors passing through all singular points. $\endgroup$– SashaCommented Sep 10, 2012 at 2:52
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$\begingroup$ @temp: Suppose $X$ and $X'$ are connected by flopping a single rational curve $C$ to $C'$. Since they are birational, the divisor class groups of $X$ and $X'$ are naturally isomorphic, and you can use this to show that they're not biholomorphic. For example, if $D$ is an effective divisor on $X$, satisfying $D\cdot C = 1$, then its proper transform on $X'$ (which I'll also call $D$), satisfies $D\cdot C' = -1$. The second Chern class also changes: $$ c_2(X')\cdot D = c_2(X)\cdot D + 2 $$ $\endgroup$ Commented Sep 10, 2012 at 7:26
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2$\begingroup$ @Rhys Davies: this argument only shows that the birational isomorphism is not biregular, or that $X$ and $X'$ are not isomorphic OVER their common contraction. Sometimes, manifolds $X$ and $X'$ related by a flop are ABSTRACTLY isomorphic, for example two small resolutions of a quadratic cone are. $\endgroup$– SashaCommented Sep 10, 2012 at 10:21
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$\begingroup$ What if I want to find projective examples that are birational but not isomorphic? $\endgroup$– tempCommented Sep 20, 2012 at 7:49