# Infinitely many minimal models

There are examples of elliptic fiber spaces over a two-dimensional base which have infinitely many relative minimal models (where two abstractly isomorphic models connected by flops are counted separately). The one I know is given by Reid and Kawamata and works by repeatedly flopping two rational curves in a singular fiber. In Matsuki's "Introduction to the Mori Program" he indicates (pg. 366) that this construction can be extended to a non-relative setting to yield a variety $X$ with infinitely many minimal models over Spec k.

I haven't managed to make this extension or find it written down, so a couple questions: where can I find an explicit example of a variety with infinitely many minimal models (over Spec k)? What is the Kodaira dimension in this case? Is it possible to find a Calabi-Yau threefold with infinitely many minimal models?

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Calabi-Yau varieties have, by definition, (numerically) trivial canonical bundles so are their own unique minimal models. Perhaps you meant to ask something different? – ulrich Sep 15 '11 at 5:57
Dear ulrich, I'm not sure I understand your comment. (Terminal) Calabi--Yaus are certainly their own minimal models, but there's no reason they should be unique: if $f:X \dashrightarrow X'$ is a birational map of Calabi--Yaus, then each one is a minimal model of the other, but they need not be isomorphic. – user5117 Sep 15 '11 at 13:55
Also, note that the OP is asking about marked minimal models, which introduces even more non-uniqueness into the picture! – user5117 Sep 15 '11 at 13:56
@Artie: By minimal model of $X$ I assumed one means a variety $Y$ that is the end result of running the MMP on $X$. If $X$ is Calabi-Yau then the canonical bundle is nef so the MMP ends at $X$ itself. But having read the question again I agree that this is probably not the definition the OP has in mind. – ulrich Sep 15 '11 at 14:24
Dear ulrich: you're right, people aren't always very careful to say exactly what they mean by "minimal model". (Here by "people" I mean "authors", not the OP.) – user5117 Sep 15 '11 at 14:33