Take the 2-minute tour ×

I stumbled upon this isolated singularity of a Calabi-Yau fourfold: \begin{equation} x_1x_2+x_3x_4+x_5^2=0 \end{equation} as a hypersurface in $\mathbb{C}^5$.

Clearly, I can resolve this by a simple blow-up. This does not seem to be a crepant resolution however.

So my question is: Is there some simple criterion which allows me to tell if there exists a crepant resolution ? In case the answer is yes, how would one go about constructing one ?

In the corresponding threefold case there even exists a small resolution. A second question is hence: Are there results on the existence of small resolutions of isolated fourfold singularities ?

share|improve this question
This singularity has a noncommutative crepant resolution. –  Sasha Aug 24 '12 at 8:20
@Sasha: Could you please explain your comment or give a reference? –  Jason Starr Aug 24 '12 at 11:58
Dear Sasha, I don't know anything about NCCR's, but this paper of Hailong Dao seems to say that it doesn't, see theorem 1.2(2) unless I am mistaken: math.ku.edu/~hdao/Crepant101109 –  Karl Schwede Aug 24 '12 at 12:11
@Jason: For details see arxiv.org/abs/math/0609240. Shortly, a noncommutative (or categorical) resolution of $X$ is a smooth triangulated category $T$ with an adjoint pair of triangulated functors $\pi^*:D(X) \to T$ and $\pi_*:T \to D(X)$ such that $\pi_*\circ\pi^* = 1$ plus some technical conditions. Such a resolution is weakly crepant if the right adjoint $\pi^!:D(X) \to T$ of $\pi_*$ coincides with $\pi^*$ on perfect complexes. –  Sasha Aug 24 '12 at 14:31
@Hailong: Yes, but I think that the advantage of noncommutative resolutions is in the fact that one can find a noncommutative resolution with nice properties (e.g. crepant) when there is no such commutative resolutions. –  Sasha Aug 24 '12 at 16:41

2 Answers 2

up vote 9 down vote accepted

That particular singularity only can have a crepant resolution if it has a small resolution.

Terminal singularities

If a variety has terminal singularities (like yours does), the then every further blow-up also has discrepancies $> 0$. (In other words, whether or not the variety has terminal singularities can be read off from a single log resolution). In particular, every discrete valuation has discrepancy $ > 0$. It follows that:

Fact: If a variety has terminal singularities, then it only has a (commutative) crepant resolution if it has a small resolution

In your example above, the variety clearly has terminal singularities, which can be read off from the blowup you mentioned.

Small resolutions

Ok, so now you want to check whether your variety has a small resolution. The first thing you can do is look at THIS ANSWER by Sándor Kovács. The upshot is there is only hope for a small resolution if your singularity is not factorial.

After staring at your singularity for 4 minutes, I don't see why its not a UFD (hopefully someone, ie Hailong Dao, can tell us easily -- i have a recollection that higher dimensional isolated hypersurface singularities are factorial).

EDIT: Olivier Benoist points out in a comment below that this singularity is factorial due to Samuel's conjecture (proven by Grothendieck).

However, if you see any given singularity that is not factorial, here's what I would try:

Find a Weil divisor that is not Cartier (the existence of such is equivalent to it being not factorial). Blowup the ideal of that Weil divisor. (Maybe do this more than once). The reason I suggest this is because:

Fact: The only way to get a small resolution is to blowup ideals whose vanishing locus has codimension one.

Here's a quick proof, Blowup an ideal $J$ on a (Noetherian!) variety who vanishing locus $W = V(J)$ has codimension $\geq 2$. The blowup turns that ideal $J$ into a Cartier divisor (it principalizes the ideal). In particular, there is now a divisor lying over $W$. Thus this isn't a small resolution.

Of course, blowing up Weil divisors that are not Cartier is how you get the small resolutions in the $xy-uv = 0$ example.

share|improve this answer
Locally complete intersection + locally factorial in codimension 3 implies locally factorial : this is Samuel's conjecture proven by Grothendieck (see SGA 2). This shows that this singularity is indeed locally factorial. –  Olivier Benoist Aug 24 '12 at 14:13
Thanks! I've edited to reflect this. –  Karl Schwede Aug 24 '12 at 14:25
Karl, in the edit you wrote: "not factorial" (-: –  Hailong Dao Aug 24 '12 at 14:38
Whoops, thanks. –  Karl Schwede Aug 24 '12 at 14:40
Karl, thanks a lot for this very helpful and precise answer ! –  Salix Aug 29 '12 at 2:34

Actually, there is a rather straightforward way to see that this singularity is factorial:

Let $X=\mathrm{Spec}\ k[x_1,x_2,x_3,x_4,x_5]/(x_1x_2+x_3x_4+x_5^2)$ and let $D=(x_2=0)\subseteq X$ be a principal divisor. Notice that then $D= \mathrm{Spec}\ k[x_1,x_3,x_4,x_5]/(x_3x_4+x_5^2)$, so it is a cone over a quadric cone. In particular, it is irreducible and reduced and hence one has an exact sequence: $$ \mathbb Z\cdot D \to \mathrm{Cl}(X) \to \mathrm{Cl}(X\setminus D)\to 0. $$

Since $D$ is principal, it is actually zero in $\mathrm{Cl}(X)$, so this says that $$ \mathrm{Cl}(X) \simeq \mathrm{Cl}(X\setminus D) $$

On the other hand, it is easy to see that $$ X\setminus D=\mathrm{Spec}\ k[x_1,x_2,x_2^{-1},x_3,x_4,x_5]/(x_1x_2+x_3x_4+x_5^2)\simeq \mathrm{Spec}\ k[x_2,x_2^{-1},x_3,x_4,x_5]. $$ Clearly, $k[x_2,x_2^{-1},x_3,x_4,x_5]$ is a UFD, so $\mathrm{Cl}(X) \simeq \mathrm{Cl}(X\setminus D)=0$. Therefore the singularity in question is indeed factorial. Then, as Karl has already pointed out, it follows for instance from the argument here that it does not admit a small resolution of singularities and since it is terminal then also all of its resolutions are non-crepant (I guess perhaps the correct word would be "discrepant").

In fact, the above argument shows the same statement for any singularity defined by a non-degenerate quadric equation in at least 5 variables.

It may be worth noting that the argument fails for fewer number of variables because $D$ is either not irreducible (in 4 variables) or reduced (in 3 variables). In 2 variables X itself is not irreducible.

Finally, let me add that I find it odd to call this a Calabi-Yau 4fold. I know that this terminology is used by others (especially physicists) as well, but I think it is rather misleading. This is at best a log Calabi-Yau.

share|improve this answer

Your Answer


