Consider the following complex (complete intersection) variety, $$ f_1: x_0^2 + x_1^2 + x_2^2 + x_3^2 = 4x_4x_5,$$ $$ f_2: x_4^4 + x_5^4 = 2x_0x_1x_2x_3,$$ in $\mathbb{P}^5$. This is the first example in Chapter 5 of Christian Meyer's Modular Calabi-Yau Threefolds, and in computing their Euler characteristic he has the following two lines, $$ \chi(X) = -176 + 32 + 12\cdot 9 = -36,$$ and letting $\tilde{X}$ be a small resolution of $X$, $$ \chi(\tilde{X}) = -36 + 32 + 12(4-1) = 32.$$

This is my first pass at resolution of singularities, though I think I've been able to explain everything except the $(4-1)$ at the very end, and I'm not entirely sure how that comes in to play.

To start, one has that $X$ without its singularities has Euler characteristic $-176$ by a couple adjunctions. Then one can find the singularities fairly easily, and we have 32 nodes (orbit of $(1:1:1:1:1:1)$ under the symmetries of the curve) and 12 type $(2,2,4,4)$ singularities (orbit of $(1:i:0:0:0:0)$ under the same symmetries). Given a node is a type $(2,2,2,2)$ singularity, it's easy to see the Milnor numbers for the nodes are 1 and for the higher singularities, the Milnor numbers are 9. This explains the first equation completely. Now taking a small resolution of $X$ at each of the nodes simply adds one to the Euler characteristic, for each node, as a $\mathbb{P}^1$ adds two, and removing the node itself subtracts one.

Now we take a small resolution of each $(2,2,4,4)$ singularity, so as before, the Milnor number of each singularity is 9, and we add an exceptional $\mathbb{P}^1$ in place of each node, so... this gives $12(2-9) = -84...$ of a change to the Euler characteristic?

I don't see how to get $12(4-1)$.. Because of the 4 I figured maybe we could think about each small resolution as a blow up followed by a blow down, since every blow up adds a $\mathbb{P}^1\times\mathbb{P}^1$ which is 4-dimensional, but I haven't found any way to simply get $(4-1)$.

I'm guessing I'm just fudging up something silly, so hopefully someone can easily remedy the confusion. I would also happy if anyone could recommend any good references to the related material. Meyer's book is very nice! However the $\sim$ 1.5 page treatment of resolution of singularities gives me the impression I was supposed to have seen it before. Thanks!


1 Answer 1


Thanks for praising my book... :) The example applies Section 1.6.3: "[...] the resolving curve has Euler number n+1" (= 4 in this case), and it replaces one point (-> -1). Proofs should be in the Brieskorn paper [16], which should be available online from the Göttinger Digitalisierungszentrum.

  • $\begingroup$ I did notice in 1.6.3, the bit on singularities of type (2,2,n+1,h(n+1)), which works for us, though having the paper where this is shown will be great! Thanks! $\endgroup$
    – Alex
    May 26, 2012 at 21:38

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