What is the relation between crepant resolutions and minimal resolutions? Are they the same thing?
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$\begingroup$ Certainly not every minimal resolution is crepant (not every singularity admits a crepant resolution, consider $x^4 + y^4 + z^4$). $\endgroup$– Karl SchwedeCommented Nov 12, 2010 at 13:56
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2 Answers
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As my comment up there says, not every minimal resolution is crepant. However, something like the reverse direction is true.
Suppose you have a projective crepant resolution $Y \to X$. If you have a factorization $Y \to Z \to X$ where $Z$ is also a resolution of singularities, then $Z \to X$ is also necessarily crepant (the same computation on relative canonical divisors holds). Thus $Y \to Z$ is projective birational and $Z$ is smooth, so $Y \to Z$ can't be small.
Thus there are some new divisors on $Y$, and we know $K_{Y/Z} > 0$ again since $Z$ is smooth, but $K_Z$ is the same as the pullback of $K_X$ (since $Z \to X$ is crepant) so $K_{Y/X}$ is also $> 0$, contradicting the crepant assumption.
EDIT: I just added the projective assumption on $Y \to X$, I don't think it's necessary, but it makes $Y \to Z$ much clearer that it's a blow-up of something.
answered Nov 12, 2010 at 14:02
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$\begingroup$ Should it read K_{Y/Z} < 0, and the same for K_{Y/X}? $\endgroup$– user5117Commented Nov 12, 2010 at 14:10
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3$\begingroup$ I don't think so, am I being dumb for some reason? If I blow up a point on ${\mathbb A}^2=X$, obtaining $f:{\widetilde{X}} \to X$, then $K_{\widetilde X}−f^*K_X=K_{\widetilde{X}/X}= $+1 copy of the exceptional divisor. $\endgroup$ Commented Nov 12, 2010 at 14:17
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$\begingroup$ Of course you're right; for some reason when I read "K_{Y/Z} > 0" I interpreted it as "K_{Y/Z} is numerically positive". In retrospect, I have no idea why... $\endgroup$– user5117Commented Nov 12, 2010 at 14:22
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They are definitely not the same thing, far from it:
Minimal resolution usually refers to surfaces, and every surface has one: you can get it from an arbitrary resolution by consecutively blowing down $(-1)$-curves. In general, one usually speaks about minimal models. Actually, one could say minimal resolution in arbitrary dimension as well, but have to keep in mind that that is usually a) not actually a resolution (minimal models are not necesssarily smooth) b) not unique.
Crepant resolution exists in every dimension, but not for everything. In fact, admitting a crepant resolution is a very special property of the variety. In particular, it means that it has (strictly) canonical singularities, that is, as soon as it has either worse or better (say $\mathbb Q$-factorial terminal, but not smooth) singularities, then there is no crepant resolution.
As Karl shows, not every minimal resolution of surfaces is crepant, but it is easy to see that every crepant resolution of a surface is necessarily minimal.
EDIT: Added "$\mathbb Q$-factorial" to rule out small resolutions following Karl's comment.
EDIT2: Added comment about minimal resolutions in arbitrary dimension.
answered Nov 12, 2010 at 14:24
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2$\begingroup$ Sandor, aren't there terminal singularities admitting small resolutions, and thus they are crepant? (ie $xy - uv$). So terminal doesn't imply non-existence of crepant resolutions (Long recently pointed this out to me when I said the same thing to him). $\endgroup$ Commented Nov 12, 2010 at 14:45
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2$\begingroup$ Karl, you are right, I was thinking surfaces when I wrote that. $\endgroup$ Commented Nov 12, 2010 at 14:56