6
$\begingroup$

I wonder if some hypersurfaces of multi-degree $(1, ..., 1)$ in a product of projective spaces are toric, with polytope a union of polytopes of toric divisors of the ambient space. This question is motivated by the following examples.

  • The first Hirzebruch surface is isomorphic to a bidegree $(1,1)$ surface in $\mathbb{P}^2 \times \mathbb{P}^1$ (see On a Hirzebruch surface). Its polygon can be seen as the union of a triangle and a square, i.e. as a union of two faces of the polytope of $\mathbb{P}^2 \times \mathbb{P}^1$.
  • Likewise, $\mathbb{P}(\mathcal{O}_{\mathbb{P}^1} \oplus \mathcal{O}_{\mathbb{P}^1} \oplus \mathcal{O}_{\mathbb{P}^1}(1))$ is isomorphic to a hypersurface of bidegree $(1,1)$ in $\mathbb{P}^3 \times \mathbb{P}^1$ (the proof is the same as in the case of the first Hirzebruch surface) and its polytope can be seen as a union of a prism and a simplex.
  • A degree six del Pezzo surface is isomorphic to a tridegree $(1,1,1)$ surface in $(\mathbb{P}^1)^3$ (see the answer to Explicit defining equations for del Pezzo surfaces). Its polygon, a hexagon, can be seen as a union of three paralellograms (i.e. three faces of the polytope of $(\mathbb{P}^1)^3$).

Is there a general statement about this (I have not been able to find any) or is it just a coincidence ?

$\endgroup$
7
  • 1
    $\begingroup$ At the first glance it looks you might want to consider the sums, not unions, of polytopes: the toric variety of the sum of two polytopes is isomorphic to the closure of the "diagonal image" of the torus in the product of the toric varieties of the original polytopes. It is easy to show; I worked out the details in Proposition VI.6 in "How many zeroes?" (arxiv.org/abs/1806.05346). $\endgroup$
    – pinaki
    Commented Dec 6, 2023 at 15:15
  • 1
    $\begingroup$ A smooth hypersurface of degree $(1,1)$ in $\mathbb{P}^2 \times \mathbb{P}^2$ is not toric. $\endgroup$
    – Sasha
    Commented Dec 6, 2023 at 15:18
  • 1
    $\begingroup$ @Sasha Or, in other words, the toric hypersurface of degree $(1,1)$ in $\mathbb P^2\times\mathbb P^2$ is not smooth =) It is the zero set of $x_1y_1-x_2y_2$ which has 7 torus-fixed points of which one is singular. It is indeed given by the Minkowski sum of two triangles. $\endgroup$ Commented Dec 10, 2023 at 21:37
  • $\begingroup$ Thank you everyone for your comments ! $\endgroup$
    – Yromed
    Commented Dec 16, 2023 at 17:49
  • $\begingroup$ I can prove that a smooth complete intersection in $\prod \mathbb{P}^{a_i}$ of multidegree $(1,\ldots,1)$ is toric only if all but one of the $a_i$ is one. It follows pretty simply from combining some known results. Is it interesting for you? $\endgroup$
    – Nick L
    Commented Jun 4 at 19:05

2 Answers 2

3
$\begingroup$

First of all, as mentioned in the comments, multiprojective toric varieties standardly arise from Minkowski sums rather than unions. Let me phrase this in the context of toric varieties of lattice point sets (I'll identify the toric variety of a polytope with that of its lattice point set but cf. last sentence). For a finite set $A\subset\mathbb Z^n$ consisting of $a_0,\dots,a_d$ its toric variety is the closure in $\mathbb P^d$ of the image of $(\mathbb C^*)^n$ under $x\mapsto(x^{a_0}:\dots:x^{a_d})$. Consider finite sets $A_1,\dots,A_k\subset\mathbb Z^n$ with $A_i$ consisting of $a^i_0,\dots,a^i_{d_i}$. Then from the Segre embedding it is easy to see that the toric variety of $A_1+\dots+A_k$ is realized inside $\mathbb P^{d_1}\times\dots\times\mathbb P^{d_k}$ as the closure of the image of $$x\mapsto(x^{a^1_0}:\dots:x^{a^1_{d_1}})\times\dots\times(x^{a^k_0}:\dots:x^{a^k_{d_k}}).$$

Now, modulo replacing unions with Minkowski sums, I believe what you describe is a general phenomenon. Consider the toric hypersurface of multidegree $(1,\dots,1)$ in $\mathbb P^{d_1}\times\dots\times\mathbb P^{d_k}$. It is the zero set of $X^1_0\dots X^k_0-X^1_{d_1}\dots X^k_{d_k}$ where $X^i_0,\dots,X^i_{d_k}$ are the homogeneous coordinates on the $i$th factor. I claim that it is the toric variety of a Minkowski sum of $k$ simplices of dimensions $d_1,\dots,d_k$ defined as follows. Denote $n=d_1+\dots+d_k$ and consider $\mathbb R^n$ with coordinates enumerated by $[0,n-1]$. For $i\in[1,k-1]$ let $\Delta_{i}$ be the unit simplex consisting of points $a$ with $a_i=0$ for $i<d_1+\dots+d_{i-1}$ or $i>d_1+\dots+d_i$ and $\sum_{i=d_1+\dots+d_{i-1}}^{d_1+\dots+d_i} a_i=1$. Let $\Delta_k$ be an analogous unit simplex in the subspace spanned by coordinates $d_1+\dots+d_{k-1},\dots,n-1,0$. One may visualize points in $\mathbb R^n$ as $k$-gons with a number in every vertex and $d_i-1$ more numbers inside the $i$th edge, then $\Delta_i$ consists of points with the $d_i+1$ numbers on the $i$th edge adding up to $1$ and all other numbers $0$.

Next, let $A_i$ be the set of lattice points in $\Delta_i$, specifically, for $j\in[0,d_i]$ let $a^i_j\in A_i$ be the point with coordinate $d_1+\dots+d_{i-1}+j\!\!\mod n$ equal to $1$ and all others $0$. In accordance with the above, the toric variety of $A_1+\dots+A_k$ is cut out in $\mathbb P^{d_1}\times\dots\times\mathbb P^{d_k}$ by the kernel of $X^i_j\mapsto t_i z^{a^i_j}$ for formal variables $z_0,\dots,z_{n-1}$ and $t_1,\dots,t_k$. This kernel contains $X^1_0\dots X^k_0-X^1_{d_1}\dots X^k_{d_k}$ because $$\tag{1}\label{sum}a^1_0+\dots+a^k_0=a^1_{d_1}+\dots+a^k_{d_k}.$$ Moreover, it is not hard to check that the kernel is generated by this binomial. This is the combinatorial fact that any equality of the form $$\sum_{i=1}^k\sum_{j=1}^{r_i}a^i_{\alpha_j}=\sum_{i=1}^k\sum_{j=1}^{r_i}a^i_{\beta_j}$$ follows from \eqref{sum}.

We see that our toric hypersurface is the toric variety of $A_1+\dots+A_k$. To identify this with the toric variety of $P=\Delta_1+\dots+\Delta_k$ one checks the "Minkowski sum property": $P\cap\mathbb Z^n=A_1+\dots+A_k$. I claim that this is also pretty straightforward. If one wishes to (more conventionally) define the toric variety of $P$ as associated with its normal fan rather than its set of lattice points, then one would further need to prove that $P$ is normal or very ample or the like.

$\endgroup$
2
  • $\begingroup$ Thank you very much for taking the time to write such a detailed answer. I was trying to work with unions with the hope of obtaining families of toric hypersurfaces degenerating to a union of toric divisors, but I see now why I rather consider Minkowski sums. $\endgroup$
    – Yromed
    Commented Dec 16, 2023 at 17:47
  • $\begingroup$ @Yromed To be fair, it is no coincidence that these polytopes admit nice subdivisions. A pretty popular theme are the so-called mixed subdivisions of Minkowski sums. For sums of simplices these are just subdivisions with each cell a sum of faces of the summands. I don't know much about this but $P$ should admit many such subdivisions. $\endgroup$ Commented Dec 16, 2023 at 23:20
2
$\begingroup$

Let $X \subset \prod_{i=1}^{n} \mathbb{P}^{a_i}$ be a smooth hypersurface of multidegree $(1,\ldots,1)$. I claim $X$ is toric only if $a_i=1$ for all but one $i$.

For $n=2$, this is Lemma 5 of V. M. Buchstaber and N. Ray, Toric manifolds and complex cobordisms, Uspekhi Mat. Nauk 53 (1998), no. 2(320), 139–140 (Russian); English transl., Russian Math. Surveys 53 (1998), no. 2, 371–373. In fact in this case they are toric $\iff$ $\min\{a_1,a_2\}=1$.

Next, we may apply this to get the general statement. Suppose for a contraction that two $a_i$ are greater than $1$, say $a_{i_1},a_{i_{2}}$. Consider the projection to the "orthogonal complement" of a (generic) factor $\mathbb{P}^{a_{i_1}} \times \mathbb{P}^{a_{i_2}} \times \{p\}$, then some fibre is smooth hypersurface of multidegree $(1,1)$ in $\mathbb{P}^{a_{i_1}} \times \mathbb{P}^{a_{i_2}} \times \{p\}$ hence not toric by the above. To prove the smoothness, apply Bertini to the linear system given by translating the intersections with $X$ with $\mathbb{P}^{a_{i_1}} \times \mathbb{P}^{a_{i_2}} \times \{p\}$ to $\mathbb{P}^{a_{i_1}} \times \mathbb{P}^{a_{i_2}} \times \{p_0\}$ for some fixed $p_0$, if this linear system had fixed part $X$ would contain some product of linear factors, but the restriction to any product of linear factors is a hypersurface with degree $(1,\ldots,1)$. Hence a generic element is smooth.

But this contradictis Proposition 2.7 of https://arxiv.org/pdf/2208.09680, recalling that the fibers of a toric morphism are toric. The condition of Proposition 2.7 is equivalent to fibers being connected in this generality, which is true because they are ample divisors in the orthogonal product of projective spaces.

$\endgroup$

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