Pieter Belmans

New paper: Ample stability implies rigidity for quiver moduli

Mon, 05 Oct 2026 00:00:00 +0000

Ample stability implies rigidity for quiver moduli (with Gianni Petrella) is a new paper that gets rid of an annoying technical condition in our earlier paper Rigidity and Schofield's partial tilting conjecture for quiver moduli (and subsequent papers building on the vanising results in this paper).

Recall that, for a dimension vector $\mathbf{d}$ and a stability parameter $\theta$ with $\theta(\mathbf{d})=0$, we have the following three properties:

Ample stability means that the complement of the stable locus in the space of representations of dimension $\mathbf{d}$ has codimension at least $2$. In this case the moduli space has maximal Picard rank: $\operatorname{rk}\operatorname{Pic}=\#Q_0-1$.
The rigidity inequality asks that, for every nontrivial Harder–Narasimhan type $\mathbf{d}^*=(\mathbf{d}^1,\ldots,\mathbf{d}^\ell)$, where $\mu(\mathbf{e})=\theta(\mathbf{e})/|\mathbf{e}|$ denotes the slope, \[ \mu(\mathbf{d}^1)-\mu(\mathbf{d}^\ell) \lt \sum_{1\leq m\lt n\leq\ell} \bigl(\mu(\mathbf{d}^m)-\mu(\mathbf{d}^n)\bigr) \bigl(-\langle\mathbf{d}^m,\mathbf{d}^n\rangle\bigr). \] This is precisely what is needed to apply Teleman quantization to the bundle $\mathcal{U}_i\otimes\mathcal{U}_j^\vee$. We would like to be able to apply this as widely as possible!
Strong ample stability means that for every proper nonzero subdimension vector $\mathbf{e}$ with $\theta(\mathbf{e})>0$ we have $\langle\mathbf{e},\mathbf{d}-\mathbf{e}\rangle\leq -2$, where $\langle-,-\rangle$ is the Euler form. Strong ample stability implies ample stability, and in fact also implies the rigidity inequalities.

Thus, in the $\theta$-coprime setting, the previously known implications were \begin{equation} \text{ample stability} \Longleftarrow \text{rigidity inequality} \Longleftarrow \text{strong ample stability}. \end{equation} Our new paper proves the converse of the first implication: ample stability implies the rigidity inequality. Consequently, the vanishing results for universal bundles, Schofield's partial tilting conjecture, and rigidity of quiver moduli only require ample stability.

So from now on, strong ample stability is a useful explicit criterion to verify ample stability, but it (or the tedious to verify rigidity inequalities) is no longer needed to apply the results I like to apply. This brings me great relief!

New appendix: An effective expression for the decomposition of Fano schemes of intersections of two quadrics

Fri, 02 Oct 2026 00:00:00 +0000

This post concerns the appendix (in its standalone version, written by me) to Lena Ji and Fumiaki Suzuki's paper on motivic classes of Fano schemes of lines.

A few years ago we wrote a paper on decompositions of Fano schemes of linear subspaces on intersections of two quadrics. For $g\ge 2$ and $0\le k\le g-2$, consider a smooth intersection $Q_1\cap Q_2\subset\mathbb{P}^{2g+1}$ with associated hyperelliptic curve $C$. One of our conjectures predicts the following identity in $\mathrm{K}_0(\mathrm{Var})$: \begin{equation} [\mathrm{F}_k(Q_1\cap Q_2)] = \sum_{i=0}^{k+1}\mathrm{M}_{g,k,i}(\mathbb{L})[\operatorname{Sym}^i C], \end{equation} where

$\mathbb{L}=[\mathbb{A}^1]$,
$\operatorname{Sym}^i C$ is the $i$th symmetric power of $C$, and
for $0\le i\le k+1$, the coefficient is \begin{equation} \begin{aligned} \mathrm{M}_{g,k,i}(\mathbb{L}) ={}&\mathbb{L}^{i(g-k-1)}\Biggl( \binom{2g-k-i}{k+1-i}_{\mathbb{L}}\\ &-\bigl(\mathbb{L}^{g-k-1}+\mathbb{L}^{g+2k-3i}\bigr) \binom{2g-k-i-4}{k-i}_{\mathbb{L}}\\ &-\bigl( \mathbb{L}^{g-k}+\mathbb{L}^{g-i}+\mathbb{L}^{g+k-2i} +\mathbb{L}^{3g-3k-4}+\mathbb{L}^{3g-2k-i-4}+\mathbb{L}^{3g-k-2i-4} \bigr)\binom{2g-k-i-4}{k-i-1}_{\mathbb{L}}\\ &-\bigl( \mathbb{L}^{3(g-k-1)}+\mathbb{L}^{3(g-k-1)+1} +\mathbb{L}^{3g-2k-i-3}+\mathbb{L}^{3g-2k-i-2} \bigr)\binom{2g-k-i-4}{k-i-2}_{\mathbb{L}}\\ &-\mathbb{L}^{4(g-k)-2} \binom{2g-k-i-4}{k-i-3}_{\mathbb{L}} \Biggr). \end{aligned} \end{equation}

Here the subscript $\mathbb{L}$ denotes a Gaussian binomial coefficient, which is zero when its lower index is negative or exceeds the upper index.

Important for today's post is that we also conjectured that these polynomials are effective: their coefficients are nonnegative.

Last year, IMProofBench was looking for interesting benchmark problems (oh, the times have changed) and I submitted this effectivity question as something that seemed to require expertise beyond what I had in store. I did try to solve it myself for some time, but none of my (admittedly limited) bag of tricks seemed to work, a fact related to me not being a combinatorialist.

The first attempts by LLMs as part of IMProofBench were utter nonsense. However, back in March 2026, one LLM managed to produce what looked like a correct proof (reproduced below), using an interesting new ingredient! When I finally had a look at it in June, I wrote up a more human-readable version of the proof, at the very least for my own sake to verify the argument. I shared it with a few people back then, but didn't settle on what to do with it.

However, when Lena (and Fumiaki) reached out about a preprint they had been working on which proves our full conjecture for $k=1$, we agreed that it would be good if the LLM-proof of the effectivity conjecture became an appendix, so that others can use the effectivity part. I rewrote the existing writeup, making sure that its exposition was as good as I could make it, and it is now available as an appendix (the link is to an independent PDF) to Lena Ji and Fumiaki Suzuki's paper on motivic classes of Fano schemes of lines. The appendix proves effectivity for all the polynomials in the conjecture; the motivic identity itself remains conjectural in general.

The original answer

For full disclosure, here is the original LLM-proof of the conjecture. I can imagine that for some mathematicians, this abbreviated and terse proof is enough. However, if the goal is that I (and maybe other non-combinatorialists) understand what's going on, I do think (and hope) that my writeup is more useful.

Let $a=g-k-1\ge 1$ and $m=k-i\in\{-1,0,\dots,k\}$. The polynomial from the conjecture can then be written as \begin{equation} M_{g,k,i}(t)=t^{ia}B_{a,m}(t), \end{equation} where $B_{a,m}(t)$ is the following expression in $a$ and $m$: \begin{equation} \begin{aligned} B_{a,m}(t)={}&\binom{2a+m+2}{m+1}_t -(t^a+t^{a+1+3m})\binom{2a+m-2}{m}_t\\ &-(t^{a+1}+t^{a+1+m}+t^{a+1+2m}+t^{3a-1}+t^{3a-1+m}+t^{3a-1+2m})\binom{2a+m-2}{m-1}_t\\ &-(t^{3a}+t^{3a+1}+t^{3a+m}+t^{3a+m+1})\binom{2a+m-2}{m-2}_t -t^{4a+2}\binom{2a+m-2}{m-3}_t. \end{aligned} \end{equation}
Here $\binom{n}{r}_t$ is the Gaussian binomial coefficient, taken to be zero for $r<0$ or $r>n$. It is enough to prove that $B_{a,m}(t)\in\mathbb{Z}_{\ge 0}[t]$ for all $a\ge 1$ and $m\ge -1$.
1. Generating function of $B_{a,m}$

Write \begin{equation} (x;t)_n:=\prod_{r=0}^{n-1}(1-t^r x). \end{equation}
We use the standard Gaussian binomial generating series \begin{equation} \sum_{r\ge 0}\binom{n+r}{r}_t y^r=\frac1{(y;t)_{n+1}} \qquad (n\ge 0). \end{equation}
Define \begin{equation} F_a(x):=\sum_{m\ge -1} B_{a,m}(t)x^{m+1}. \end{equation}
Applying the above identity term by term gives \begin{equation} \begin{aligned} F_a(x)={}&\frac1{(x;t)_{2a+2}} -\frac{t^a x}{(x;t)_{2a-1}} -\frac{t^{a+1}x}{(t^3x;t)_{2a-1}}\\ &-x^2\left( \frac{t^{a+1}+t^{3a-1}}{(x;t)_{2a}} +\frac{t^{a+2}+t^{3a}}{(tx;t)_{2a}} +\frac{t^{a+3}+t^{3a+1}}{(t^2x;t)_{2a}} \right)\\ &-x^3\left( \frac{t^{3a}+t^{3a+1}}{(x;t)_{2a+1}} +\frac{t^{3a+2}+t^{3a+3}}{(tx;t)_{2a+1}} \right) -\frac{t^{4a+2}x^4}{(x;t)_{2a+2}}. \end{aligned} \end{equation}
Now multiply by $(x;t)_{2a+2}$. Using the quotient identities \begin{equation} \frac{(x;t)_{2a+2}}{(x;t)_{2a-1}}=(1-t^{2a-1}x)(1-t^{2a}x)(1-t^{2a+1}x), \end{equation} \begin{equation} \frac{(x;t)_{2a+2}}{(t^3x;t)_{2a-1}}=(1-x)(1-tx)(1-t^2x), \end{equation} \begin{equation} \frac{(x;t)_{2a+2}}{(x;t)_{2a}}=(1-t^{2a}x)(1-t^{2a+1}x), \quad \frac{(x;t)_{2a+2}}{(tx;t)_{2a}}=(1-x)(1-t^{2a+1}x), \end{equation} \begin{equation} \frac{(x;t)_{2a+2}}{(t^2x;t)_{2a}}=(1-x)(1-tx), \quad \frac{(x;t)_{2a+2}}{(x;t)_{2a+1}}=1-t^{2a+1}x, \quad \frac{(x;t)_{2a+2}}{(tx;t)_{2a+1}}=1-x, \end{equation}
and collecting coefficients of $x^0,x^1,x^2,x^3,x^4$, one gets \begin{equation} (x;t)_{2a+2}F_a(x) =(1-t^a x)(1-t^{a+1}x)(1-t^{2a+1}x^2). \end{equation}
Hence \begin{equation} F_a(x) =\frac{(1-t^a x)(1-t^{a+1}x)(1-t^{2a+1}x^2)}{(x;t)_{2a+2}} =\frac{1-t^{2a+1}x^2}{(x;t)_a\,(t^{a+2}x;t)_a}. \end{equation}
So we have proved the compact identity \begin{equation} \sum_{m\ge -1} B_{a,m}(t)x^{m+1} =\frac{1-t^{2a+1}x^2}{(x;t)_a\,(t^{a+2}x;t)_a}. \end{equation}
2. Manifestly effective form

Now use the elementary decomposition \begin{equation} 1-t^{2a+1}x^2=(1-t^{a-1}x)+t^{a-1}x(1-t^{a+2}x). \end{equation}
Substituting this into the previous formula gives \begin{equation} F_a(x) =\frac{1}{(x;t)_{a-1}(t^{a+2}x;t)_a} +\frac{t^{a-1}x}{(x;t)_a(t^{a+3}x;t)_{a-1}}. \end{equation}
This is manifestly positive, because every factor \begin{equation} \frac1{1-t^j x}=\sum_{r\ge 0} t^{jr}x^r \end{equation}
has coefficients in $\mathbb{Z}_{\ge 0}[t]$, so both summands lie in $\mathbb{Z}_{\ge 0}[t][[x]]$. Therefore every coefficient of $F_a(x)$, and hence every $B_{a,m}(t)$, lies in $\mathbb{Z}_{\ge 0}[t]$. Since $M_{g,k,i}(t)=t^{ia}B_{a,m}(t)$ is a monomial multiple, its coefficients are nonnegative as well.
3. The final formula in the original variables

Returning to $a=g-k-1$ and $m=k-i$, we obtain \begin{equation} \begin{aligned} M_{g,k,i}(t) =t^{i(g-k-1)}[x^{k+1-i}]\Biggl(& \prod_{j=0}^{g-k-3}\frac1{1-t^j x} \prod_{j=g-k+1}^{2g-2k-1}\frac1{1-t^j x}\\ &+t^{g-k-2}x \prod_{j=0}^{g-k-2}\frac1{1-t^j x} \prod_{j=g-k+2}^{2g-2k-1}\frac1{1-t^j x} \Biggr), \end{aligned} \end{equation}
An empty product equals $1$. The formula makes effectivity explicit.
Equivalently, if $h_r$ denotes the complete homogeneous symmetric polynomial, then \begin{equation} \begin{aligned} M_{g,k,i}(t)=t^{i(g-k-1)}\Bigl( &h_{k+1-i}(1,t,\dots,t^{g-k-3},t^{g-k+1},\dots,t^{2g-2k-1})\\ &+t^{g-k-2} h_{k-i}(1,t,\dots,t^{g-k-2},t^{g-k+2},\dots,t^{2g-2k-1}) \Bigr), \end{aligned} \end{equation}
Empty ranges are omitted, and $h_r=0$ for $r<0$. Each $h_r$ is a sum of monomials with nonnegative integer coefficients. Thus $M_{g,k,i}(t)$ is effective for all $g\ge 2$, $0\le k\le g-2$, and $0\le i\le k+1$.

An exceptional object for the irregular Waldron--Witaszek Fano fourfold

Fri, 25 Sep 2026 00:00:00 +0000

Today's arXiv update includes Joe Waldron and Jakub Witaszek's paper, “An irregular smooth Fano fourfold in positive characteristic”. Here, irregular means that $\mathrm{H}^1(X,\mathcal{O}_X)\neq 0$. This is not possible for Fano varieties in characteristic zero, by Kodaira vanishing. But the authors construct a Fano fourfold over the algebraic closure of $\mathbb{F}_2$ with $\mathrm{H}^1(X,\mathcal{O}_X)\cong\overline{\mathbb{F}}_2$.

Whilst this is very interesting for questions about rationality, I am always thinking about derived categories. And, unlike for Fano varieties in characteristic zero, the isomorphism $\mathrm{H}^1(X,\mathcal{O}_X)\cong\overline{\mathbb{F}}_2$ implies that $\mathcal{O}_X$ (or for that matter, any line bundle!) is not an exceptional object. This is unfortunate because we like to start decomposing derived categories of Fano varieties by considering an exceptional sequence of line bundles and then continuing from there.

So what's up with the derived category of this Fano fourfold? It turns out there is an exceptional object, and the verification is almost done by the authors already! Line bundles are out, but there is a rank-2 vector bundle in their writeup that we can consider. Let's recall their setup a bit, although I will omit the precise construction, as I do not feel qualified to talk about regular foliations in characteristic two.

An exceptional rank-2 bundle

In Proposition 3.4 they consider a finite flat morphism $\pi\colon H\to X$ of degree 2, where $H$ is a complete intersection of a quadric and a cubic in $\mathbb{P}^6$ with 63 singular points. In Lemma 4.2 they construct a short exact sequence \[ 0\to\mathcal{O}_X\to\pi_*\mathcal{O}_H\to K\to 0 \] where $K$ is defined as $\ker(\pi_*\mathcal{O}_H(1)\to\pi_*\mathcal{O}_H(2))$, for a derivation that arises from their construction. The final ingredient for us is Proposition 4.5, which gives the isomorphism $\pi^*K\cong\mathcal{O}_H(1)$. We have now seen the main player for our observation.

Claim $\pi_*\mathcal{O}_H$ is exceptional.

Whilst munching on my lunch, I realized that there is a shorter proof.

New and shorter proof. Let us write $E=\pi_*\mathcal{O}_H$. We have, using that $\pi$ is both finite and flat, that \[ \mathrm{Ext}_X^\bullet(E,E) =\mathrm{Ext}_X^\bullet(E,\pi_*\mathcal{O}_H) \cong\mathrm{Ext}_H^\bullet(\pi^*E,\mathcal{O}_H) \cong\mathrm{Ext}_H^\bullet(\mathcal{O}_H,\pi^*E^\vee). \] The long exact sequence in cohomology attached to the dual of the sequence defining $E$ pulled back to $H$ (again using that $\pi$ is flat), together with $\mathrm{H}^\bullet(H,\mathcal{O}_H)\cong\overline{\mathbb{F}}_2[0]$ and $\mathrm{H}^\bullet(H,\mathcal{O}_H(-1))=0$ (using that $H$ is a complete intersection) gives us the sought-after exceptionality.$\square$

Original proof. Let us write $E=\pi_*\mathcal{O}_H$. Because $E$ has rank two, we have \[ E^\vee\cong E\otimes\det(E)^\vee\cong\pi_*\mathcal{O}_H\otimes K^\vee\cong\pi_*(\mathcal{O}_H\otimes\mathcal{O}_H(-1))\cong\pi_*\mathcal{O}_H(-1) \] by the usual pairing for vector bundles, the projection formula, and the properties recalled above. This gives us \[ \mathrm{H}^\bullet(X,E^\vee)\cong\mathrm{H}^\bullet(H,\mathcal{O}_H(-1))\cong 0 \] because $H$ is a complete intersection, so we can use the Koszul resolution. Note that, as in the proof of Lemma 4.1, this also gives us that \[ \mathrm{H}^\bullet(X,E)\cong\mathrm{H}^\bullet(H,\mathcal{O}_H)\cong\overline{\mathbb{F}}_2[0]. \]

Now, consider the defining short exact sequence for $E$ and tensor it with $E^\vee$ to get \[ 0\to E^\vee\to E^\vee\otimes E\to E^\vee\otimes K\to 0. \] Again, because $E$ has rank two, we have $E^\vee\otimes K\cong E$. We have computed the cohomology of the outer terms, and we thus get \[ \mathrm{H}^\bullet(X,E^\vee\otimes E)\cong\overline{\mathbb{F}}_2[0], \] which proves the claim. $\square$

Hence, we have \[ \mathbf{D}^{\mathrm{b}}(X) = \langle E^\perp,E\rangle \] where $E^\perp=\{F\in\mathbf{D}^{\mathrm{b}}(X)\mid\mathbf{R}\mathrm{Hom}(E,F)=0\}$. Note that, by the acyclicity used above, we have $\mathcal{O}_X\in E^\perp$.

Questions

I need to start preparing my lecture for next week, so I'm leaving you (and myself) with the following questions:

What are the Hodge numbers of this Fano fourfold? Note that using Proposition 5.1, we can actually compute that $\mathrm{h}^0(X,\Omega_X^1)\geq 6$, e.g., using my cohomology tables for complete intersections;
Does the Hochschild–Kostant–Rosenberg decomposition for Hochschild homology hold for this variety? It is not automatic for fourfolds in characteristic 2!
How can we decompose the derived category further? Note that $E^\vee$ is also exceptional, but we cannot fit it in a sequence together with $E$.

Rationality of hypersurfaces: a new website

Fri, 11 Sep 2026 00:00:00 +0000

After seeing John Christian Ottem, Stable irrationality of quartic sixfolds on the arXiv yesterday, I was reminded of my plans a little while ago to make a website like mgnbar.info surveying the state of the art on a precise question. Inspiration for this I found in

Without further ado, see rationality.fanography.info. It is just a single page, with features very much like mgnbar.info. I’m not sure I like the methods overview, maybe I will get rid of it at some point.

A noncommutatively rigid Fano fivefold

Wed, 09 Sep 2026 00:00:00 +0000

Here's a question I have had for a while, ever since reading Noncommutative rigidity of the moduli stack of stable pointed curves by Shinnosuke Okawa and Taro Sano.

They prove the cool result that $\mathrm{HH}^2(\overline{\mathcal{M}}_{g,n})=0$ (except when $(g,n)=(0,5)$, and possibly when $(g,n)=(4, 0), (3, 1), (3, 0), (2, 2), (2, 1), (2, 0), (1, 3), (1, 2)$). Here $\mathrm{HH}^2$ is the second Hochschild cohomology, and it measures the deformations of the category of coherent sheaves. This is the noncommutative rigidity in the title of their paper. Note that they are considering a smooth projective Deligne–Mumford stack whenever $g\geq 1$.

Finding noncommutatively rigid varieties

Because I like Fano varieties so much, I was wondering whether we can find a Fano variety $X$ for which the second Hochschild cohomology vanishes, i.e., does there exist a noncommutatively rigid Fano variety?

Note that it is in fact easy to find a non-Fano example of a noncommutatively rigid smooth projective surface: a fake projective plane is of this shape. But I wanted a Fano variety!

Finding noncommutatively rigid Fano varieties

In dimension 2, we have that $\overline{\mathcal{M}}_{0,5}$ is the del Pezzo surface of degree 5, and whilst it is rigid as a variety, it has a 6-dimensional space of Poisson structures, and thus lots of interesting noncommutative deformations. The same holds for the other rigid del Pezzo surfaces: lots of interesting noncommutative deformations, which are truly at the heart of noncommutative algebraic geometry.

In dimension 3, we can use Fanography, which contains the data from the paper Polyvector fields for Fano 3-folds that I wrote a few years ago with Enrico Fatighenti and Fabio Tanturri, to conclude that there are also no noncommutatively rigid Fano 3-folds.

In an ongoing project with many people on classifying Fano 4-folds we are also computing polyvector cohomology, but none of the cases studied so far seem to satisfy the required vanishings.

The example

Now, to dimension 5! PartialFlagVarieties.jl is a useful first selection mechanism, as it makes it possible to consider many zero loci in partial flag varieties at once when looking for candidates. This is how I found a complete intersection in $\mathbb{P}^2\times\mathbb{P}^2\times\mathbb{P}^2\times\mathbb{P}^3$, of degrees $(1,1,0,0)$, $(0,1,0,1)$, $(0,0,1,1)$, $(0,0,1,1)$, which had the potential of being an example. With a little help from an LLM, I could translate the remainder of the problem into a Macaulay2 computation, which confirmed that it is indeed an example.

I was originally writing this up as a blogpost, but it got longer and longer, so I felt like a pdf was more appropriate: A noncommutatively rigid Fano fivefold. It contains all the Julia and Macaulay2 code needed to verify the claims, so that you can check them yourself, and the code is also available in the repository noncommutatively-rigid-fano-fivefold. For now I have no plans to submit it for publication anywhere, but feel free to convince me otherwise.

Brauer groups of Fano 3-folds

Wed, 02 Sep 2026 00:00:00 +0000

After returning from holidays earlier this summer, I organized all the stuff on my desk a bit, and I ran into various half-finished computations and notes. Whilst some might eventually become (short) papers, most of them will (at least initially) just be fun blog posts.

The first topic I want to tackle is the Brauer group of Fano 3-folds. In the wonderful paper Fano varieties with torsion in the third cohomology group by John Christian Ottem and Jørgen Vold Rennemo, they construct even-dimensional Fano varieties (from dimension 4 onwards) for which $\mathrm{H}^3(X,\mathbb{Z})\cong\mathbb{Z}/2\mathbb{Z}$. For a Fano variety, the torsion subgroup of $\mathrm{H}^3(X,\mathbb{Z})$ is its cohomological Brauer group.

For a del Pezzo surface there is nothing to be said about (torsion in) $\mathrm{H}^3(X,\mathbb{Z})$: we have that $\mathrm{H}^3(X,\mathbb{Z})=0$. But a priori there could be something interesting happening in dimension 3, as $\mathrm{H}^3(X,\mathbb{Z})$ can certainly be non-zero. The introduction of the Ottem–Rennemo paper states the following:

In dimension 3, there are 105 deformation classes of Fano varieties, and direct inspection shows that, in each class, the group $\mathrm{H}^3(X,\mathbb{Z})$ is torsion-free.

When asked about the case of Fano 3-folds during a talk, one of the authors suggested that they had not written down the details of the direct inspection, and certainly the paper does not contain these details. So let's just do it here! After all, to paraphrase Sir Mix-a-Lot: I like Fano 3-folds and I cannot lie.

A first reduction

We will do a case-by-case analysis to show that $\mathrm{H}^3(X,\mathbb{Z})$ is torsion-free. Kodaira vanishing and the exponential sequence give \[ \operatorname{Br}(X) \cong \mathrm{H}^3(X,\mathbb{Z})_{\mathrm{tors}}. \] On the other hand, the universal coefficient theorem gives \[ \mathrm{H}^3(X,\mathbb{Z})_{\mathrm{tors}} \cong \operatorname{Hom}\left( \mathrm{H}_2(X,\mathbb{Z})_{\mathrm{tors}}, \mathbb{Q}/\mathbb{Z} \right). \] In particular, $\operatorname{Br}(X)=0$ if and only if $\mathrm{H}_2(X,\mathbb{Z})$ is torsion-free. I guess it is a matter of taste and preference which of these groups is the most familiar to you. We will freely use these different perspectives.

We will also use that this torsion is constant in a smooth proper family. Indeed, by Ehresmann's theorem such a family is locally trivial in the differentiable category, so its integral cohomology groups, including their torsion subgroups, form a local system. It therefore suffices to check one convenient representative of each deformation family.

Our first goal is to significantly reduce the number of families to consider, using the following facts:

the Brauer group of $\mathbb{P}^3$ is trivial;
the Brauer group is a stable birational invariant of smooth projective varieties.

The birational links in the Iskovskikh–Mori–Mukai classification first reduce the problem from 105 deformation families to the 30 primitive ones. Of these primitive families, 17 are rational. Using the descriptions and rationality results collected in Fanography, we are thus left with the following 13 families:

ID	description
1–1	a double cover of $\mathbb{P}^3$ branched along a smooth sextic surface
1–2	a quartic hypersurface $X_4\subset\mathbb{P}^4$
1–3	a complete intersection $X_{2,3}\subset\mathbb{P}^5$ of a quadric and a cubic
1–4	a complete intersection $X_{2,2,2}\subset\mathbb{P}^6$ of three quadrics
1–5	a Gushel–Mukai 3-fold: a section of $\operatorname{Gr}(2,5)$ by a codimension-2 linear subspace and a quadric
1–7	a codimension-5 linear section of $\operatorname{Gr}(2,6)$ in its Plücker embedding
1–11	the double Veronese cone $V_1=X_6\subset\mathbb{P}(1,1,1,2,3)$
1–12	the quartic double solid: a double cover of $\mathbb{P}^3$ branched along a smooth quartic surface
1–13	a cubic hypersurface $V_3\subset\mathbb{P}^4$
2–2	a double cover of $\mathbb{P}^1\times\mathbb{P}^2$ branched along a smooth divisor of bidegree $(2,4)$
2–6	a Verra 3-fold: a double cover of the $(1,1)$-divisor $W_6\subset\mathbb{P}^2\times\mathbb{P}^2$, branched along a smooth anticanonical divisor
2–8	a double cover of $V_7=\operatorname{Bl}_p\mathbb{P}^3$, branched along a smooth anticanonical divisor meeting the exceptional divisor smoothly
3–1	a double cover of $(\mathbb{P}^1)^3$ branched along a smooth divisor of tridegree $(2,2,2)$

Awesome! Only 13 cases to consider.

Tools for the remaining cases

Let's tackle those remaining cases. Our first tool is the following.

Lefschetz in degree 2 for complete intersections. Let $M$ be a smooth projective variety, and let $X\subset M$ be obtained by successively taking smooth ample divisors, with $\dim X\geq 3$. Then $\mathrm{H}_2(X,\mathbb{Z})\cong\mathrm{H}_2(M,\mathbb{Z})$.

This is the integral Lefschetz hyperplane theorem applied successively; see, for instance, Theorem 3.1.17 in Positivity in Algebraic Geometry I.

Looking at our table, we see that quite a few of the interesting cases admit such a description, where $M$ is a partial flag variety. Such varieties have a Schubert cell decomposition and therefore torsion-free integral homology in every degree. In particular, this applies to 1–2, 1–3, 1–4, 1–5, 1–7, and 1–13.

There are now 7 families of Fano 3-folds we still have to deal with. A significant portion of them can be dealt with using the following variation of Lefschetz.

Lefschetz in degree 2 for cyclic covers. Let $X\to Y$ be a cyclic cover of smooth projective varieties of dimension at least 3, branched along a smooth ample divisor. Then $\mathrm{H}_2(X,\mathbb{Z})\cong\mathrm{H}_2(Y,\mathbb{Z})$.

This variation of the Lefschetz theorem for complete intersections is Proposition 1.11 in Topological properties of cyclic coverings branched along an ample divisor.

Again looking at our table, we see that quite a few of the remaining cases admit such a description. Namely, this tool applies to 1–1, 1–12, 2–2, 2–6, 2–8, and 3–1. We are left only with 1–11.

To deal with this remaining case, we want to have a version of Lefschetz that works for weighted complete intersections. Luckily, there is one!

Lefschetz in degree 2 for weighted complete intersections. Let $X\subset\mathbb{P}(a_0,\ldots,a_N)$ be a strongly smooth weighted complete intersection of dimension at least 3. Then \[ \mathrm{H}_2(X,\mathbb{Z})\cong\mathbb{Z}. \] Here strongly smooth means that $X$ is quasismooth and the greatest common divisor of the weights corresponding to the nonzero coordinates of a point is independent of the point.

This variation of the Lefschetz theorem for complete intersections is Proposition 6(i)–(ii) in Monodromy and Betti numbers of weighted complete intersections.

This applies to 1–11: if one of the weight-one coordinates is non-zero, the relevant greatest common divisor is 1; if they all vanish, the weight-two and weight-three coordinates are both non-zero and their weights are coprime.

Conclusion

We have thus proved that the Brauer group of a Fano threefold vanishes! It was indeed a direct inspection, but one must know which tools to use for the inspection.

New paper: The hyperkähler period-index conjecture is false

Tue, 11 Aug 2026 00:00:00 +0000

This post is about The hyperkähler period-index conjecture is false, joint with James Hotchkiss.

If I were still posting fortnightly links, Alexander Perry: The period-index conjecture is false would have prominently featured, because it disproves (as the title suggests) the period-index conjecture, which is famously a theorem due to de Jong in dimension 2, with a beautiful proof. But Alex found counterexamples in every dimension at least 3, with some assistance from an LLM.

The period-index conjecture

To understand the statement of the period-index conjecture, let us consider a field $K$ of transcendence degree $d$ over an algebraically closed field $k$. We attach two integers to a Brauer class $\alpha\in\mathop{\rm Br}(K)$:

the index $\mathop{\rm ind}(\alpha)$ (which measures the size of the unique division algebra in the Brauer class)
the period $\mathop{\rm per}(\alpha)$ (which is the order of $\alpha$ in the torsion abelian group $\mathop{\rm Br}(K)$)

The period-index conjecture, attributed to Colliot-Thélène, then predicts that \[ \mathop{\rm ind}(\alpha)\mid\mathop{\rm per}(\alpha)^{d-1}. \]

Using a bit of Galois cohomology, one can prove that

$\mathop{\rm per}(\alpha)\mid\mathop{\rm ind}(\alpha)$
the period and the index have the same prime factors

so certainly for every individual $\alpha$ there is some exponent for which the index divides the corresponding power of the period. The conjecture claims there is a uniform bound, which is moreover given in terms of the transcendence degree.

As a special case of this we can consider a smooth projective variety $X$ together with its function field $k(X)$, for which $\mathop{\rm Br}(X)\hookrightarrow\mathop{\rm Br}(k(X))$. Johan de Jong proved that it suffices to prove the period-index conjecture for $k(X)$ for the classes in the image of this inclusion, the unramified classes.

Alex constructed unramified classes in every dimension at least 3 which violate the period-index conjecture.

The hyperkähler period-index conjecture

Hyperkähler varieties (see also hyperkaehler.info) are a special kind of variety. They have rich geometry, strong constraints, and many tools to study them. In particular, their geometry is controlled to a large extent by $\mathrm{H}^2(X,\mathbb{Z})$, together with its Beauville–Bogomolov–Fujiki form.

Because of this, and various pieces of evidence, Huybrechts conjectured the stronger hyperkähler period-index conjecture, predicting that \[ \mathop{\rm ind}(\alpha)\mid\mathop{\rm per}(\alpha)^{(\dim X)/2} \] for unramified Brauer classes on smooth projective hyperkähler varieties.

Given that Alex started from an incorrect LLM-assisted counterexample, James and I independently tried constructing an LLM-assisted counterexample for the hyperkähler period-index conjecture, ending up with examples in different deformation types of hyperkähler fourfolds. Given the similarity in tools used (namely James's Hodge-theoretic index), it felt natural to write a joint paper.

So what are the counterexamples? First, observe that for the fourfolds we consider (whose integral cohomology is torsion-free) it is possible to write every Brauer class in terms of $\mathrm{H}^2(X,\mathbb{Z})$, as $\mathop{\rm Br}(X)\cong(\mathrm{H}^2(X,\mathbb{Z})/\mathop{\rm NS}(X))\otimes\mathbb{Q}/\mathbb{Z}$.

Then it becomes a matter of writing down classes in the lattice, giving us the polarization, the Brauer class, and a way to check the conditions for our two obstruction lemmas. For more details one is referred to the paper.

The results are

a period-2 Brauer class on certain hyperkähler fourfolds of type $\mathrm{Kum}^2$, whose index is divisible by 8
a period-2 Brauer class on certain hyperkähler fourfolds of type $\mathrm{K3}^{[2]}$, whose index is divisible by 8
a period-5 Brauer class on certain hyperkähler fourfolds of type $\mathrm{K3}^{[2]}$, whose index is divisible by 125

In each case the index is one factor too big: the conjecture predicts $\mathop{\rm ind}(\alpha)\mid 4$ resp. $\mathop{\rm ind}(\alpha)\mid 25$. The classical period-index conjecture only predicts $\mathop{\rm ind}(\alpha)\mid\mathop{\rm per}(\alpha)^3$, so these are not, as far as we know, counterexamples to it.

This construction lives just outside the known cases of the hyperkähler period-index conjecture:

Huybrechts proved it for $X$ admitting a Lagrangian fibration, for classes whose period is coprime to an integer depending on $X$.
Huybrechts proved it for $X$ the Hilbert scheme of points on a K3 surface, for all classes, with a variant for generalised Kummer varieties.
Hotchkiss–Maulik–Shen–Yin–Zhang proved it for $X$ of $\mathrm{K3}^{[n]}$-type of Picard rank at least 2, for non-special coprime classes.
Bottini–Huybrechts removed the condition on the Picard rank, and also cover the special coprime classes whose period is squarefree.

Hyperelliptic varieties: quotients of complex tori by finite groups

Tue, 21 Jul 2026 00:00:00 +0000

There is a new website, hyperelliptic.ncag.info, on the classification of hyperelliptic (or generalized hyperelliptic) varieties in complex dimensions 2, 3 and 4. For now it lives as a subdomain of ncag.info.

A hyperelliptic variety is a quotient $X = T/G$ of a complex torus $T$ by a finite group $G$ acting freely and without translations. In dimension 1 these are the elliptic curves, in dimension 2 the seven bielliptic surfaces of Bagnera and De Franchis; the word has nothing to do with hyperelliptic curves. The remarkable thing is that all the numerical invariants (the Hodge diamond, the order of the canonical bundle, the number of moduli, the irregularity, the polyvector fields, the twisted Hodge numbers, the Hochschild cohomology) depend only on the tangent representation $\rho\colon G \to \mathrm{GL}(V)$, and can be read off from its character theory. The website computes them all with OSCAR, for every group in the classifications of Uchida–Yoshihara, Lange and Catanese–Demleitner (dimension 3) and of Demleitner (the 79 groups in dimension 4).

This all started from a collaboration with Andreas Demleitner and Pedro Núñez on the Albanese morphism for these varieties. Working on that paper I learned a great many things about hyperelliptic varieties from Andreas, and this website is in a sense a place to keep all of it: the invariants we computed by hand, and many more, now generated automatically and cross-checked against the literature.

As with my other websites it is a static site built (using LLMs) with Hugo. Feature requests, corrections and contributions are very welcome, on GitHub or by email.

New paper: Fano 4-fold quiver moduli from subspace quivers

Wed, 15 Jul 2026 00:00:00 +0000

Whilst it might look like I am hyperproductive these few weeks, with a new blogpost coming up almost every day about something I have done, I am mostly just putting things online that were finished before, in preparation of the summer holidays.

This post is about Fano 4-fold quiver moduli from subspace quivers, a new paper joint with Markus Reineke. Let's discuss the origin story of this paper in a way that is not appropriate for the paper's introduction, but that is perfect for a blog post.

Origin

It all started from Laurent Manivel's A four-dimensional cousin of the Segre cubic. In this paper, he studies a Fano fourfold, defined as the zero locus of a vector bundle on the product of two Grassmannians. Looking at the properties he finds for this Fano fourfold, my mind was applying the duck test in the following form:

If it looks like a quiver moduli space, swims like a quiver moduli space, and quacks like a quiver moduli space, then it probably is a quiver moduli space.

Quiver moduli are after all very special varieties: rational, Hodge–Tate, rigid. Laurent's Segre cubic cousin had all these properties. Through its relation to the Segre cubic, I also quickly guessed the quiver and dimension vector that should do the trick: the 6-subspace quiver, and $\mathbf{d}=(1,1,1,1,1,2;3)$. However, I couldn't pin down an isomorphism.

Then in February Markus came to Utrecht as the Springer visiting chair. Markus realized there is actually a fun classification problem hiding here: which other subspace quivers give rise to Fano 4-folds? The reason to be interested in subspace quivers is that, besides their moduli spaces being rigid, their moduli spaces also have no infinitesimal automorphisms, making them extra special.

It turns out that up to natural identifications, there are precisely 4 such Fano 4-fold subspace quiver moduli: one being the (expected) Segre cousin, another being the Fano model of $\mathop{\rm Bl}_6\mathbb{P}^4$, which is also $(\mathbb{P}^1)^7//\mathrm{PGL}_2$, the moduli space of 7 points on $\mathbb{P}^1$ (let's not spell out stability).

The other two are also very interesting: one admits a natural map to $\mathbb{P}^2$, exhibiting it as something called an involution surface bundle (the first time I encountered one in the wild), and another which looks a lot like the Segre cubic cousin.

And in the end, we also managed to prove that Manivel's Segre cubic cousin was indeed the subspace quiver moduli space we expected it to be all along. All's well that ends well.

Working with quiver moduli

What I like about this paper is that it is so very explicit with quiver moduli, and applies so many of the tools that have been developed for them. I consider it a bit of an advertisement, and I hope that it gets picked up by people who, a priori, would not work with quiver moduli, but now realize how useful this perspective can be!

Related to this is also the use of QuiverTools, which makes it possible to compute many invariants of quiver moduli. For the 4 cases in this paper, the code is available as FanoFourfoldSubspaceQuiverModuli.jl. And the tools we have used in this paper were also the inspiration for some of the new features for QuiverTools, more about those later.

Note that the duck test for quiver moduli doesn't always work so well! We are also working on a paper that explains how well the duck test works for Fano 3-folds. Stay tuned for that.

Also note that there is another work-in-progress (don't expect either of them before the start of my summer break, though!) where we take the involution surface bundle Fano 4-fold subspace quiver moduli, and take its connection to the extended Dynkin quiver $\widetilde{\mathrm{D}}_4$, to find 5 infinite series of even-dimensional Fano varieties attached to extended Dynkin quivers, with lots of interesting properties. Stay tuned for that too!

Cubic 4-folds: special cubic fourfolds and their K3 surfaces

Mon, 13 Jul 2026 00:00:00 +0000

Since early 2023 there has been a website at cubics.fanography.info on the relationship between cubic fourfolds and K3 surfaces, but I never announced it I think: a first draft version was written in just 2 days and I didn’t know what to do next. I’m not sure it is in any state or form “finished”, but given that I’m on a website spree lately, I might as well get this one out too. So this is that announcement, three years late.

Hassett divisors

A smooth cubic fourfold $X\subseteq\mathbb{P}^5$ is special if it contains a surface not homologous to a complete intersection; these form the Hassett divisors $\mathcal{C}_d$ in the moduli space of cubic fourfolds.

The website is a table with one row for every discriminant $d$, collecting some things that are known about $\mathcal{C}_d$: the various notions of an associated (twisted) K3 surface, rationality, the Kodaira dimension, Fourier–Mukai partners, and which surfaces the generic member contains, with clickable references. It goes back to a chart of Nicolas Addington, whose comments shaped the website.

As with my other websites it is a static site built with Hugo. Feature requests, corrections and contributions are very welcome, on GitHub or by email.

LLMs were used to build this, which made the whole process much faster.