I'm currently reading about complex geometry in order to get a feeling of how to determine the Hodge numbers, e.g. of certain orbifold constructions. Since I'm a physicist with no deeper mathematical knowledge in algebraic geometry and complex geometry, I apologize for the following possibly trivial questions.

In the following I'm referring to two examples of the string theory book by Becker², Schwarz.

Example 1 (p. 368):

$T^4/\mathbb{Z}_2$, i.e. $z_1 \sim z_1+1$ and $z_2 \sim z_2 + i$, where the $\mathbb{Z}_2$ isometry is generated by $\mathcal{I}:(z_1,z_2) \to (-z_1,-z_2)$. First of all, it is easy to see that there is no invariant $(1,2)-$form, hence $h^{1,2}=0$ and that $dz_1 \wedge dz_2$ is the only $(2,0)-$form, i.e. $h^{2,0}=1$. Moreover, there are four invariant $(1,1)-$forms (they're obvious, so I'm not gonna write them down) which contribute to $h^{1,1}$. But there are in total also $16$ fixed points of the orbifold which can be blown up à la Eguchi-Hanson. It is claimed that $h^{1,1}=4+1 \times 16 =20$ because there is one $(1,1)-$cycle for every blown up singularity.

Sorry for this questions: How do I see that there is just one $(1,1)-$form for every singularity? And why aren't there cycles from the EH-Space that contribute to $h^{2,2}$ and the other Hodge-numbers?

Example 2 (pp. 372):

Now we consider $T^2 \times T^2/ \mathbb{Z}_3$ with the action $(z_1,z_2) \to (\omega z_1, \omega^{-1}z_2)$ with $\omega =\exp(2\pi i/3)$. One has then $9$ singularities.
In this case there are $2$ invariant $(1,1)-$forms. It is claimed that there are now $2$ $(1,1)-$cycles for every blow-up, i.e. $h^{1,1}=2+ 2 \times 9 =20$.

Why are there now two $(1,1)-$cycles for each singularity? And why aren't there other cycles contributing to the remaining Hodge-numbers?

I would be happy if someone could explain me the intuition and the maths behind this counting.



1 Answer 1


I'll add a small disclaimer as well, I am a mathy with little to no physics background, so if any of the below needs expanding, feel free to ask! (Though I'll mention that I felt a lot more comfortable with this stuff when I first computed the induced actions I'll mention below and really got my hands dirty verifying everything.)

To see that there is only one $(1,1)$-form coming from each fixed point in the first example, you need to get a little into how the blowing up actually adds anything to the cohomology. For this, there are two steps.

(1) What is the blowup?

(2) How are we adding classes in cohomology?

For (1), note we are taking the quotient of $T^4$ by $\mathbb{Z}_2$ and the problem is this generates 16 singularities. You can indeed blowup these singularities, but there is no guarantee this actually resolves them, so after one simple blowup you may not have a smooth object yet (i.e., a resolution!) and you may need to continue blowing up to actually resolve the singularities. (This is why example 2 has a more exotic resolution.)

To see that the resolution is complete after the first blowup, you have to somehow note the action of $\mathbb{Z}_2$, induced on the $\mathbb{P}^1$s lying above each fixed point is trivial (so the blownup quotient is smooth). A simple way for this is to note the involution acts as $-1$ on the differentials, (cotangent space) so acts as $-1$ on the tangent space at each singular point, and the blowup replaces each point with the projectivization of the normal bundle at that point, but the projectivation of the action by $-1$ is trivial and the $\mathbb{P}^1$s are indeed fixed by the action of $\mathbb{Z}_2$.

For problem 2, we are looking at the same situation except the action is not $-1$, and does not projectivize to something trivial. Hence, after the first blowup, the $\mathbb{P}^1$s are not fixed and you'll need to blowup the fixed points (there are two) on each $\mathbb{P}^1$ lying above a fixed point. If you carry the action up on the $\mathbb{P}^1$ and once more on the blowup of the fixed points on the $\mathbb{P}^1$ you see that the action is trivial on the second blowup, so you've got 2 fixed $\mathbb{P}^1$s lying above each original fixed point.

Now, the fun comes in with part (2). Why does this add $(1,1)$-forms and nothing else? For this, perhaps the original source is Grothendieck's SGA V, in which (if I recall correctly) section VII is largely devoted to proving the following result (which I've simplified greatly for the case with dimensional examples):

Let $X$ be a two dimensional smooth projective irreducible variety with $Y$ a finite set of points on $X$. Let $\widetilde{X}$ be the blowup of $X$ along $Y$. Then $$ H^k\left(\widetilde{X}\right) \simeq H^{k-2}(Y) \oplus H^k(X). $$

I'll be present the full version below, but it's far easier for the two-folds, and just follows from topological Poincare duality, so lets finish up the examples first! With duality, we need only work out

$$ H^0\left(\widetilde{X}\right) \simeq H^{-2}(Y) \oplus H^0(X), $$ $$ H^1\left(\widetilde{X}\right) \simeq H^{-1}(Y) \oplus H^1(X), $$ and $$ H^2\left(\widetilde{X}\right) \simeq H^{0}(Y) \oplus H^2(X). $$

Now $H^{-2}(Y) = H^{-1}(Y) = 0$, so the only change the blowup has in cohomology is on $H^2$. (As per your question, note that $H^{2,2}\left(\widetilde{X}\right) \simeq H^{0,0}\left(\widetilde{X}\right)$ so this shows there are no $(2,2)$-classes coming from the blowup.)

For $H^0(Y)$, note that each point has a $(0,0)$-class, so each point contributes 1 class in $H^{1,1}\left(\widetilde{X}\right)$ as desired.

In example two, recall we had two fixed points to blowup, lying above the original fixed points, so there are 18 $(0,0)$-classes coming in to the final numbers.

The full SGA result is as follows:

Let $X$ be a smooth projective irreducible variety of dimension $n$, and let $Y_1,\ldots,Y_r \subset X$ be mutually disjoint closed irreducible subvarieties of $X$ of codimension $d\geq 2$. Let $Y$ be the union of the $Y_i$ and let $\widetilde{X}\to X$ be the blowup of $X$ along $Y$. Then $$ H^k\left(\widetilde{X}\right) \simeq H^{k-2}(Y) \oplus \cdots \oplus H^{k-2(d-2)}(Y)\xi^{d-2} \oplus H^k(X), $$ where $\xi$ represents the line bundle $\mathcal{O}(1)$. (This is how you can define Chern classes, using the $\xi^i$ as a basis for the cohomology of the blowup - or any vector bundle in general - if you've seen these before.)

So above, with the nice two-fold examples, the $\xi$ terms were not necessary as the exponents aren't big enough.

Note though, that as any $n$-fold product of elliptic curves (or curves in general) only has fixed points in the singular locus, so you'll only get $(1,1)$-classes in the resolution. You have to look at more exotic construction, like $E\times S/\mathbb{Z}_2$ where $E$ is an elliptic curve and $S$ is a surface, where the fixed locus on $S$ may now have curves, so you'll find $(2,1)$-classes in the resolution, and more and more complicated fixed loci (can) give higher classes in the cohomology. (Because of your tags, I'll mention the $E\times S$ case is the classic Borcea-Voisin construction when $S$ is a K3 surface and the involutions are non-symplectic.)

Added: Let's look at the second blowup explicitly. We know the action of $\mathbb{Z}_3$ on the two tori induces an action on the respective $1$-forms of multiplication by $\omega$ and $\omega^{-1}$. To find how this affects the blowup, we need to see what the action on the tangent space of a fixed point is. Note these 1-forms give us a basis for the cotangent space of each torus, which is isomorphic to the tangent space (which in general takes the transpose action, but in our one dimensional case that means it's the same action) so the $\mathbb{Z}_3$ action induces multiplication by $\omega$ and $\omega^{-1}$ on the respective tangent spaces.

To make things a little simpler, lets give our (complex) two-fold a name, let $$ X = T^2\times T^2/\mathbb{Z}_3. $$

Thus, the blowup of an affine chart about a fixed point (which we can set as the origin) can be written as $$ \widetilde{X} = \left\{\Big((x,y),[u:v]\Big) \mid xu = vy\right\} \subset \mathbb{A}^2\times \mathbb{P}^1, $$ the $\mathbb{P}^1$ being what we've added in the blowup. To double check this is correct, note that any point $(x,y)\neq (0,0)$ has a single pair $[u:v]$ satisfying the condition $xu=vy$, whereas the origin $(x,y)=(0,0)$ has the entire $\mathbb{P}^1$ satisfying $xu=vy$, so we really do have a manifold that corresponds one-to-one with $X$, away from the singularity, and the $\mathbb{P}^1$ lies above the fixed point.

Aside: This $\mathbb{P}^1$ arises because it is the (projectivization of the) normal bundle of the singular point on $X$. In general, if you blowup a $d$-dimensional submanifold on an $n$-dimensional manifold, you get a $\mathbb{P}^{d-1}$ lying above the locus you are blowing up. (E.g., on a surface, blowing up a curve gives a $\mathbb{P}^1$ lying above each point on the curve, whereas blowing up a point gives a $\mathbb{P}^2$ lying above the point.) The idea of blowing up in general is (loosely) you want to add all the information `coliding' at the singularity. If you've never seen much about the resolution of singularities, I recommend glossing the Wiki page as it gives a nice description of blowing up projective space, affine space, complex manifolds, and (perhaps best to ignore) the heavy schemes way.

Back to work: But this means we know the action on the $\mathbb{P}^1$ that lies over the fixed point, since we know the action of $\mathbb{Z}_3$ on the tangent space of the fixed point above. We have $$ [u:v] \mapsto [\omega u:\omega^{-1}v]. $$ This action is not trivial, so the $\mathbb{P}^1$ is not fixed by the action, so we've simply added a $\mathbb{P}^1$ but not changed the fixed locus at the point, so the quotient still has a singularity here!

Things get worse though, as this $\mathbb{P}^1$ actually has two fixed points! The origin $[0:v]$ and the point at infinity $[u:0]$. (Our manifold has 18 fixed points now, but they are all isolated fixed points, so they are all the same type of singularity as before.)

We must now try to resolve both. We can take the chart $v\neq 0$ and note that we can write $$ \widetilde{X} = \left\{\left((x,\frac{xu}{v}),\left[\frac{u}{v}:1\right]\right) \right\}, $$ i.e., there are only two degrees of freedom. This means we can write our affine chart (still on a two-fold) in the coordinates $$ Y := \{(x,z)\} $$ where $z = u/v$. Note that action of $\mathbb{Z}_3$ on $z$ is $$ z = \frac{u}{v} \mapsto \frac{\omega u}{\omega^{-1} v} = \omega^{-1}\frac{u}{v} = \omega^{-1}z. $$

Blowing up this we have the same story as above, and $$ \widetilde{Y} = \left\{\Big((x,z),[s:t]\Big) \mid xs = zt\right\}. $$ This time we don't have the action on the tangent space about the fixed point $(0,0)$ in $\widetilde{X}$, but we can actually decipher it from above. We know the action on $[u:v]$ in the first blowup, so the condition $xu = vy$ tells us $$ xu \mapsto \omega xu $$ and $$ vy \mapsto \omega^{-1}vy, $$ so we have $(x,y)\mapsto (\omega^{-1}x,\omega y)$. Thus, our action on the $\mathbb{A}^2$ in the second blowup is $(x,z)\mapsto (\omega^{-1} x,\omega^{-1} z)$. Hence, the condition $xs = ut$ in the second blowup means the (projective!) action of $\mathbb{Z}_3$ on this $\mathbb{P}^1$ is trivial, and our fixed locus now has a whole $\mathbb{P}^1$ instead of just a point.

Thus, at least at this point, $\widetilde{Y}$ is smooth. A similar calculation shows the $\mathbb{P}^1$ lying above the point at infinity, say in the chart $u\neq 0$, is also fixed. This means every fixed point on $X$ is only resolved after the two blowups, where we've actually resolved 18 fixed points all in all.

  • $\begingroup$ Dear Alex! Thank you very much for your kind help! Clearly, in the book the authors pretended that there was no deeper reasoning behind these calculations.... However, I believe I didn't quite get the point why two blowups are sufficient for resolution in the second case. Hence, I would be happy if you could make the steps more explicit. Moreover, do the $\mathbb{C}P^1$s arise because the singularities shall be blownup to something of topology $S^2$ (which is basically $\mathbb{C}P^1$)? Sorry for the level of my questions... $\endgroup$
    – psm
    Feb 24, 2014 at 15:38
  • 1
    $\begingroup$ @psm, No need to apologize! I've added some detail on the second example. As you can see, it's a little bit messy to go through everything (and this is for just the one example) so that may be why most authors tend to just assume/hope you've seen it before and not bother filling in the details themselves! $\endgroup$
    – Alex
    Feb 25, 2014 at 3:22
  • $\begingroup$ Thanks again! Your additional comments are very illuminating! $\endgroup$
    – psm
    Feb 28, 2014 at 19:43
  • $\begingroup$ @Alex excellent analysis. Just a stupid question. In the action of $\mathbb{Z}_3$ on z it is $\omega^{-1}$ z, but this should be $\omega^2$ z, right? $\endgroup$
    – user172341
    Apr 5, 2020 at 15:06
  • 1
    $\begingroup$ @Darth_Bane, yes, but they are the same since $\omega^3 = 1$ so you can write it with either. I had been writing stuff like this a lot back when I wrote this answer, I probably got used to many small shortcuts to keep the notation looking more consistent. $\endgroup$
    – Alex
    Apr 7, 2020 at 22:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.