First let me give a precise formulation of the question; I'll give some background/motivation at the end.

If X is a projective variety which is Q-factorial (meaning X is normal, and some sufficiently high multiple of every Weil divisor is Cartier), then a small Q-factorial modification or SQM of X means a birational map φ: X --> Y (where Y is another Q-factorial projective variety) which is an isomorphism in codimension 1. (Examples: flips and flops.) The question is then this:

Is there an example of a Q-factorial projective variety X and φ: X --> Y an SQM of X such that the nef cone Nef(X) is rational polyhedral but Nef(Y) is not rational polyhedral? Or (highly unlikely I think) can one prove that this situation is impossible?

And to push my luck:

Give sufficient conditions to ensure that in the above situation, Nef(X) rational polyhedral implies Nef(Y) rational polyhedral.

Really I think only the first question has a hope of being answered positively, but it would be nice to know if I was wrong.

Background: The question was prompted by a theorem of Hu and Keel ("Mori dream spaces and GIT", Michigan Math. J., 2000) which gives a characterisation of so-called Mori dream spaces --- certain varieties which behave very well with respect to the operations of the minimal model program. In particular, if X is a Mori dream space, then Nef(X) is rational polyhedral and so too is Nef(Y) for any SQM Y of X. It then seems natural to ask for an example where the first condition here holds but the second fails.

Hu--Keel's theorem gives one answer to my second question above, since Birkar--Cascini--Hacon--McKernan proved that any Fano variety (or more generally, any variety of Fano type) is a Mori dream space. But it would be great to know of any sufficient condition that applies outside the Fano domain.

Edit (July 23): Balazs' answer settles the first question in the affirmative: flops can change the nef cone from finite to infinite. But it seems to remain open (and interesting to me) whether there are examples $X$ of this phenomenon with Kodaira dimension $kd(X)=-\infty$.


2 Answers 2


In the world of Calabi-Yau (as opposed to Fano) varieties, one does not expect miracles of Mori dream space type. A specific example which gives a positive answer to your first question is described in the paper http://xxx.lanl.gov/abs/math/0102055 of Michael Fryers: a Calabi-Yau threefold (a degenerate quintic) which has some small resolutions having finite polyhedral nef cone, and some having an infinitely generated cone, which is locally rational polyhedral away from a single point on the boundary. The geometry, related to the Horrocks-Mumford bundle and abelian surfaces in projective four-space, is very beautiful.

  • $\begingroup$ Dear Balazs, thanks for your response. I must admit I knew about Fryers' paper, but I hadn't ever read it closely enough to notice that it was an answer to this question. The kind of situation I had in mind when asking the question was kod.dim. X = -\infty. (But I didn't want it to be unnecessarily circumscribed.) It would be interesting to know if that condition is sufficient. $\endgroup$
    – user5117
    Jul 20, 2010 at 6:37

Let me start by a disclaimer: I misread the question and wrote what's below thinking that it was relevant. Then, just before posting it I looked at the question one more time and realized the mistake.

As it happens, the statement and proof below is mathematically correct (I think), except that it does not answer the question, or one could perhaps even say that it does not have much to do with the question and probably everyone before me who have looked at this question knows what I wrote down here.

At the same time, I would hate to just erase what I typed up and perhaps someone finds this useful, so I will just post it. If some MO users think this is inappropriate, I will be happy to erase it, just let me know.

So, here come some mathematical ramblings loosely related to the question above.

(The point is that there does not exist a morphism, but the question was about rational maps).

There does not exist a morphism that is an SQM. This is because if $f:X\to Y$ is a small morphism, then $Y$ is not $\mathbb Q$-factorial. In fact a little more seems to be true:

Proposition. Let $f:X\to Y$ be a proper birational morphism and assume that $X$ is quasi-projective and $Y$ is $\mathbb Q$-factorial. Then the exceptional set of $f$ is of pure codimension $1$.

Proof. Let $H\subset X$ be a very ample Cartier divisor on $X$ and consider $f_*H$ the cycle theoretic image of $H$. This is a priori only a Weil divisor, but since $Y$ is $\mathbb Q$-factorial, it is $\mathbb Q$-factorial and replacing $H$ with an appropriate multiple we may assume that it is in fact Cartier.

Now consider $F:=f^*f_*H-H$ and notice that a) $F$ is effective b) $-F$ is $f$-ample. These show that $F$ must be exceptional. This already proves that $f$ cannot be small, but we can prove the stronger statement stated above.

${\rm Supp}\, F$ is obviously of pure codimension $1$, so if we prove that the exceptional set of $f$ is equal to this, then we are done. We have already seen that ${\rm Supp}\, F$ is contained in the exceptional set, so we only need to demonstrate that $f$ is an isomorphism on the complement of $F$.

Now, observe that by the definition of $F$, we have that $$ \mathcal O_X(-F)\simeq \mathcal O_X(H)\otimes\mathcal O_X(-f^*f_*H) $$ and $$ \begin{multline*} f^*f_* \mathcal O_X(-F)\simeq f^*f_*\mathcal O_X(H-f^*f_*H)\simeq \\ \simeq f^*\left(f_*\mathcal O_X(H)\otimes \mathcal O_Y(-f_*H)\right)\simeq f^*f_*\mathcal O_X(H)\otimes \mathcal O_Y(-f^*f_*H) \end{multline*} $$

Furthermore, since $H$ is very ample, it is generated by global sections, so $$ f^*f_*\mathcal O_X(H) \to \mathcal O_X(H) $$ is surjective. Then it follows from the above displayed lines that $$ f^*f_*\mathcal O_X(-F) \to \mathcal O_X(-F) $$ is also surjective. This implies that $f^{-1}(f({\rm Supp}\, F)) = {\rm Supp}\, F$. It follows then that $H$ and $f^*f_*H$ agree on $X\setminus {\rm Supp}\, F$. In particular, $f^*({\rm some divisor})$ is ample on this $X\setminus {\rm Supp}\, F$. Finally this implies that then $f$ has to be an isomorphism on $X\setminus {\rm Supp}\, F$. Q.E.D.

  • $\begingroup$ Dear S\'andor, You're right that this answer doesn't seem to address my question. But nevertheless, thanks for taking the time to write this out! $\endgroup$
    – user5117
    Oct 21, 2010 at 12:55
  • 1
    $\begingroup$ I don't know what happened with the accent, by the way. (Having seen your homepage, I am now very careful not to be lazy and omit these!) $\endgroup$
    – user5117
    Oct 21, 2010 at 12:57
  • $\begingroup$ Thanks for that. For some reason, MO does not render latex accent commands. Perhaps I should make a meta post on this. On the other hand, many people just use it as you did in situation when accented characters are not available (say a txt only e-mail). Finally, I'm not that picky actually. I would prefer it spelled correctly in published things like a paper or a book, but in day-to-day communication it is all right to drop it. The point I try to make on my webpage is that what is called an "accent" in English is really part of the character in Hungarian and is not there to indicate accent. $\endgroup$ Oct 21, 2010 at 13:55
  • $\begingroup$ So, thanks for the attention to detail. The fact that you cared enough to read what I wrote is good enough/ :) $\endgroup$ Oct 21, 2010 at 13:56

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