MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?)

share|cite|improve this question
What does "isomorphic" mean in this context? – Igor Rivin Sep 9 '12 at 4:22
Presumably it means "biholomorphic". – YangMills Sep 9 '12 at 4:46
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. – user5117 Sep 9 '12 at 18:40
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.) – user5117 Sep 9 '12 at 18:41
Ah, a quick search shows it is proved for any dimension by Kawamata in 2007. – temp Sep 9 '12 at 23:08

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).

share|cite|improve this answer

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.

share|cite|improve this answer
Why are they not isomorphic to each other? – temp Sep 9 '12 at 23:02
Usually some of them are projective and the other are not. It depends on existence of Weil divisors passing through all singular points. – Sasha Sep 10 '12 at 2:52
@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: $$ c_2(X')\cdot D = c_2(X)\cdot D + 2 $$ – Rhys Davies Sep 10 '12 at 7:26
@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. – Sasha Sep 10 '12 at 10:21
What if I want to find projective examples that are birational but not isomorphic? – temp Sep 20 '12 at 7:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.