Veneroni Maps

Abstract

Veneroni maps are a class of birational transformations of projective spaces. This class contains the classical Cremona transformation of the plane, the cubo-cubic transformation of the space and the quarto-quartic transformation of $\mathbb P^4$. Their common feature is that they are determined by linear systems of forms of degree n vanishing along $n+1$ general flats of codimension 2 in $\mathbb P^n$. They have appeared recently in a work devoted to the so called unexpected hypersurfaces. The purpose of this work is to refresh the collective memory of the mathematical community about these somewhat forgotten transformations and to provide an elementary description of their basic properties given from a modern point of view.

1 Introduction

The aim of this note is to give a detailed description of Veneroni’s Cremona transformations in $\mathbb P^n$. They were first described by Veneroni in [13], and then discussed for $n=4$ by Todd in [12] and by Blanch in [2] and for $n\ge 3$ by Snyder and Rusk in [11] (with a focus on $n=5$) and by Blanch again in [1]. The base loci of the Veneroni transformations involve certain varieties swept by lines that were considered for $n=4$ by Segre in [10] and for $n\ge 3$ by Eisland in [6]. Evolution in terminology and rigor can make it a challenge to study classical papers. Our purpose here is to bring this work together in one place, in a form accessible to a modern audience. In order to use Bertini’s Theorem, we assume the ground field $\mathbb K$ has characteristic 0.

Consider $n+1$ distinct linear subspaces $\varPi _0,\dots ,\varPi _{n}\subset \mathbb P^n$ of codimension 2. Let $\mathscr {L}_n$ be the linear system of hypersurfaces in $\mathbb P^n$ of degree n containing $\varPi _0\cup \dots \cup \varPi _{n}$ and let $N+1$ be the vector space dimension of $\mathscr {L}_n$ (we will see that $N=n$ when the $\varPi _j$ are general, hence by semi-continuity we have $N\ge n$). We denote by $v_n:\mathbb P^n\dashrightarrow \mathbb P^N$ the rational map given by $\mathscr {L}_n$. If $N=n$ and if in addition $v_n$ is birational, we refer to $v_n$ as a Veneroni transformation. (Here we raise an interesting question: is $v_n$ birational to its image if and only if $N=n$? When $n=2$, it is not hard to check that $N=n$ always holds and that $v_n$ is always birational.)

When the $\varPi _j$ are general, we will see that $v_n$ is a Veneroni transformation whose inverse is also given by a linear system of forms of degree n vanishing on $n+1$ codimension 2 linear subspaces of $\mathbb P^n$. In this situation, $v_2$ is the standard quadratic Cremona transformation of $\mathbb {P}^2$, $v_3$ is a cubo-cubic Cremona transformation of $\mathbb {P}^3$ (see [4, Example 3.4.3]) and $v_4$ is a quarto-quartic Cremona transformation of $\mathbb {P}^4$ (see [12]). In [8] the quarto-quartic Cremona transformation was used to produce some unexpected hypersurfaces.

The paper is organized as follows: we start in Sect. 2 with characterizing degree $n-1$ hypersurfaces in $\mathbb P^n$, containing n general linear subspaces of codimension 2.

In Sect. 3 we investigate the linear system giving the Veneroni transformation. When the spaces $\varPi _i$ are general, we prove that the dimension of $\mathscr {L}_n$ is $n+1$, we describe the base locus of this system, and prove that $v_n$ is birational.

In Sect. 4 we give the inverse $u_n$ of $v_n$ explicitly and show that $u_n$ is given by a possibly linear subsystem of the linear system of forms of degree n vanishing on a certain set of $n+1$ codimension 2 linear subspaces.

The last section, Sect. 5, is devoted to the additional description of the intersection of two hypersurfaces of the type described in Sect. 2.

2 Codimension 2 Linear Subspaces

Given linear subspaces $\Lambda _1, \dots , \Lambda _s$ of $\mathbb P^n$, a line intersecting them all is called a transversal (for $\Lambda _1, \dots , \Lambda _s$).

Proposition 2.1

Let $\varPi _1,\dots ,\varPi _{n-1}$ be general codimension 2 linear subspaces of $\mathbb P^n$. For every point $p\in \mathbb P^n$, there is a transversal for $\varPi _1,\dots ,\varPi _{n-1}$ through p. If p is general, then there is a unique transversal, which we denote $t_p$, and it meets $\varPi _1\cup \dots \cup \varPi _{n-1}$ in $n-1$ distinct points. If however there are at least two transversals through p, then p lies on a subspace $T_p$ (of dimension $d_p>1$) intersecting each $\varPi _j$ along a subspace of dimension $d_p-1$, $j=1,\dots ,n-1$, and $T_p$ is the union of all transversals for $\varPi _1,\dots ,\varPi _{n-1}$ through p.

Proof

Let H be a general hyperplane in $\mathbb P^n$ and consider the projection $\pi _p:\mathbb P^n\dashrightarrow H$ from $p\in \mathbb P^n$. If $p\not \in \varPi _j$, let $\varPi _j'=\pi _p(\varPi _j)$ and define

$$\varPi '=\bigcap _{{1\le j<n}\atop {p\not \in \varPi _j}}\varPi _j'.$$

The intersection $\varPi '$ is not empty, since each $\varPi _j'$ is a hyperplane in H and $\varPi '$ is the intersection of at most $n-1$ hyperplanes in H. Let $q\in \varPi '$. Then the line $L_{pq}$ is transversal to all $\varPi _i$ (because either $q\in \varPi _i'$, and hence $L_{pq}$ intersects $\varPi _i$, or $p\in \varPi _i$). Conversely, a transversal from p intersects $\varPi '$. Observe that for a general p, the points $\pi ^{-1}(q)|_{\varPi _j}$ are different, so the transversal meets $\varPi _j$ in different points.

Consequently, for a general p there is a unique transversal. If dim $\varPi '=k>0$, then we have a subspace $T_p$ of the transversals of dimension $k+1$. This subspace is a cone over $\varPi '$ and over $\varPi _j\cap T_p$ as well, hence dim $\varPi _j\cap T_p=k$. $\square$

Example 2.2

For 3 general codimension 2 linear subspaces $\varPi _1,\varPi _2,\varPi _3$ of $\mathbb P^4$, the pairwise intersections $\varPi _{ij}=\varPi _i\cap \varPi _j$, $i\ne j$, are points. These three points span a plane T which intersects each $\varPi _i$ in a line. (For $\varPi _1$ this line is the line $L_{23}$ through $\varPi _{12}$ and $\varPi _{13}$, and similarly for $\varPi _2$ and $\varPi _3$.) The lines $L_{12}, L_{13}, L_{23}$ all lie in T, hence every point $p\in T$ has a pencil of transversals, namely the lines in T through p.

Remark 2.3

In the preceding example, not every transversal is in T; this follows from Proposition 2.1. What is more, even if a point p has a pencil of transversals, it need not be true that $p\in T$. Take, for example, a general point $p\in \varPi _1$. The cone on $\varPi _2$ with vertex p intersects $\varPi _3$ in a line L. Every line through p in the plane spanned by p and L is a transversal, so the general point $p\in \varPi _1$ has a pencil of transversals.

Remark 2.4

We will eventually be interested in $n+1$ general codimension 2 subspaces $\varPi _0,\dots ,\varPi _n$ of $\mathbb P^n$. They are defined by $2(n+1)$ general linear forms $f_{j1},f_{j2}$, $j=0,\dots ,n$, where $I_{\varPi _j}=(f_{j1},f_{j_2})$. After a change of coordinates we may assume that $f_{j1}=x_j$ and that $f_{j2}=a_{j0}x_0+\dots +a_{jn}x_n$ with $a_{ji}=0$ if and only if $i=j$. Here the homogeneous coordinate ring R of $\mathbb P^n$ is the polynomial ring $R=\mathbb K[\mathbb P^n]=\mathbb K[x_0,\dots ,x_n]$.

Now, we establish existence and uniqueness of a hypersurface Q of degree $n-1$ containing n general codimension 2 linear subspaces in $\mathbb P^n$ for $n\ge 2$.

Proposition 2.5

Let $\varPi _1,\dots ,\varPi _{n}$ be general codimension 2 linear subspaces of $\mathbb P^n$. Then there exists a unique hypersurface Q of degree $n-1$ containing $\varPi _j$ for $j=1,\dots ,n$. Moreover, Q is reduced and irreducible, it is the union of the transversals for $\varPi _1,\dots ,\varPi _{n}$, and for each point $q\in Q$ we have ${\text {mult}}_qQ\ge r$, where r is the number of indices i such that $q\in \varPi _i$. If q is a general point of Q, then there is a unique transversal for $\varPi _1,\dots ,\varPi _{n}$ through q.

Proof

Let $\Delta$ be the determinantal variety in $(\mathbb P^n)^{n+1}$ of all $(n+1)\times (n+1)$ matrices M of rank at most 2 whose entries are the variables $x_{ij}$. It is known that $\Delta$ is reduced and irreducible of dimension $3n-1$, see [9]. It consists of the locus of points $(p_1,\dots ,p_{n+1})$ whose span in $\mathbb P^n$ is contained in a line.

Let $\pi _i:(\mathbb P^n)^{n+1}\rightarrow \mathbb P^n$ be projection to the ith factor (so $1\le i\le n+1$). Now, for $1\le i\le n$, let $\varPi _i'=\pi _i^{-1}(\varPi _i)$. Then $D=\Delta \cap \bigcap _{1\le i\le n}\varPi _i'$ has dimension $(3n-1)-2n=n-1$. Indeed, $\Delta$ is irreducible, thus intersection with a divisor (preimage of a form by $\pi _j$) drops the dimension by one ($\Delta$ does not lie in one summand, hence cannot lie in the preimage). By Bertini we can do this again and again (2n times, the dimension drops by 2 for every $\varPi _j$). We see that D is reduced and irreducible. Since $\varPi _1\cap \dots \cap \varPi _n=\varnothing$, we see that D is the locus of all points $(p_1,\dots ,p_{n+1})$ such that the span $\langle p_1,\dots ,p_n\rangle$ is a line with $p_i\in \varPi _i$ for $1\le i\le n$ and $p_{n+1}$ being on that line. Thus $\overline{D}=\pi _{n+1}(D)$ is irreducible, properly contains $\varPi _1\cup \dots \cup \varPi _{n}$ and is the union of all transversals for $\varPi _1,\dots ,\varPi _{n}$. (To get $\varPi _j$ in the image of the last projection, take a point p in $\varPi _j$, take a general line $\ell$ through p, and $(\ell \cap \varPi _1, \ell \cap \varPi _2,\dots , \ell \cap \varPi _j=p, \ell \cap \varPi _{j+1}, \dots , \ell \cap \varPi _{n}, p)$ lies in D and projects to $p\in \overline{D}$).

In particular, $\overline{D}$ has dimension $n-1$, and since by Proposition 2.1 there is a line through a general point meeting $n-1$ of the spaces $\varPi _i$ in distinct points, we see that $\deg \overline{D}\ge n-1$.

Below we will check that there is a hypersurface Q of degree $n-1$ containing $\varPi _1\cup \dots \cup \varPi _{n}$. Since any such hypersurface must by Bezout contain all transversals for $\varPi _1,\dots ,\varPi _{n}$, we see that $\deg \overline{D} = n-1$ and $Q=\overline{D}$ and thus that there is a unique hypersurface of degree $n-1$ containing $\varPi _1\cup \dots \cup \varPi _{n}$, and it is irreducible.

To show existence of Q we follow [2]. As mentioned in Remark 2.4, we may assume that the ideal of $\varPi _k$ is

$$I_k=(x_k, f_k=a_{k,0}x_0+\dots +a_{k,k-1}x_{k-1}+a_{k,k+1}x_{k+1}+\dots +a_{k,n}x_n),$$

where we write $f_k$ instead of $f_{k,2}$. By generality we may assume that $a_{i,j}\ne 0$ for all $i\ne j$.

Now consider the $n\times n$ matrix

$$A=\left( \begin{array}{ccccc} -f_1 &{} \dots &{} a_{1,k}x_k &{} \dots &{} a_{1,n}x_n \\ \vdots &{} &{} \vdots &{} &{} \vdots \\ a_{k,1}x_1 &{} \dots &{} -f_k &{} \dots &{} a_{k,n}x_n \\ \vdots &{} &{} \vdots &{} &{} \vdots \\ a_{n,1}x_1 &{} \dots &{} a_{n,k}x_k &{}\dots &{} -f_n \end{array}\right) $$

and let $F=\det (A)$. Note that F is not identically 0 (since its value at the point $(1,0,\dots ,0)$ is not 0) so $\deg F=n$. It is clear, developing $\det (A)$ with respect to the k-th column, that $F\in I_k$ for every $k=1,\dots ,n$. For each k, adding to the kth column of A all of the other columns of A gives a matrix $A_k$ whose entries in the kth column are nonzero scalar multiples of $x_0$; in particular,

$$A_k=\left( \begin{array}{ccccccc} -f_1 &{} \dots &{} a_{1,k-1}x_{k-1} &{}-a_{1,0}x_0 &{} a_{1,k+1}x_{k+1} &{} \dots &{} a_{1,n}x_n \\ \vdots &{} &{} \vdots &{} \vdots &{} \vdots &{} &{} \vdots \\ a_{k-1,1}x_1 &{} \dots &{} -f_{k-1} &{}-a_{k-1,0}x_0 &{} a_{k-1,k+1}x_{k+1} &{} \dots &{} a_{k-1,n}x_n \\ a_{k,1}x_1 &{} \dots &{} a_{k,k-1}x_{k-1} &{}-a_{k,0}x_0 &{} a_{k,k+1}x_{k+1} &{} \dots &{} a_{k,n}x_n \\ a_{k+1,1}x_1 &{} \dots &{} a_{k+1,k-1}x_{k-1} &{}-a_{k+1,0}x_0 &{} -f_{k+1} &{} \dots &{} a_{k+1,n}x_n \\ \vdots &{} &{} \vdots &{} \vdots &{} \vdots &{} &{} \vdots \\ a_{n,1}x_1 &{} \dots &{} a_{n,k-1}x_{k-1} &{}-a_{n,0}x_0 &{} a_{n,k+1}x_{k+1} &{} \dots &{} -f_n \end{array}\right) .$$

so

$$A_1=\left( \begin{array}{cccccc} -a_{1,0}x_0 &{} a_{1,2}x_2 &{}\dots &{} a_{1,k}x_k &{} \dots &{} a_{1,n}x_n \\ -a_{2,0}x_0 &{} -f_2 &{}\dots &{} a_{2,k}x_k &{} \dots &{} a_{2,n}x_n \\ \vdots &{} \vdots &{} &{} \vdots &{} &{} \vdots \\ -a_{k,0}x_0 &{} a_{k,2}x_2 &{} \dots &{} -f_k &{} \dots &{} a_{k,n}x_n \\ \vdots &{} \vdots &{} &{} \vdots &{} &{} \vdots \\ -a_{n,0}x_0 &{} a_{n,2}x_2 &{} \dots &{} a_{n,k}x_k &{}\dots &{} -f_n \end{array}\right) .$$

Thus $F=\det (A)=\det (A_k)=x_0\cdot G$ for some polynomial G. Since $x_0$ is not an element of any $I_k$, it follows that $G\in I_k$ for $k=1,\dots ,n$, hence G vanishes on each of $\varPi _1,\dots ,\varPi _n$. Since $\deg F=n$, we have $\deg (G)=n-1$. Thus G defines a hypersurface Q of degree $n-1$ containing each $\varPi _i$.

Now consider a point $q\in Q$. The matrix $A_k$ will have r columns which vanish at q, where r is the number of indices i such that $q\in \varPi _i$. In particular, each entry in each such column is in the ideal $I_q$. Thus $G=\det (A_k)/x_0\in I_q^r$ so ${\text {mult}}_qQ\ge r$.

Finally assume p is a general point of $\varPi _n$. Since $\varPi _n$ is general, p is a general point of $\mathbb P^n$, hence by Proposition 2.1 there is a unique transversal $t_p$ for $\varPi _1,\dots ,\varPi _{n-1}$ through p, hence $t_p$ is also the unique transversal for $\varPi _1,\dots ,\varPi _{n}$ through p. Thus there is an open neighborhood U of p of points q through each of which there is a unique transversal $t_q$ for $\varPi _1,\dots ,\varPi _{n-1}$, and for those points q of $U\cap Q$, $t_q$ also meets $\varPi _n$, hence for a general point $q\in Q$ there is a unique transversal $t_q$ for $\varPi _1,\dots ,\varPi _{n}$. $\square$

Remark 2.6

Let $p_0,\dots ,p_n$ be the coordinate vertices of $\mathbb P^n$ with respect to the variables $x_0,\dots ,x_n$, so $p_0=(1,0,\dots ,0)$, $\dots$, $p_n=(0,\dots ,0,1)$. We saw in the proof of Proposition 2.5 that $p_0\not \in Q$ (since $F\ne 0$ at $p_0$). Let $A_k'$ be the matrix from the proof of Proposition 2.5 arising after dividing $x_0$ from column k of $A_k$. Then Q is defined by $\det (A_k')=0$ but $A_k'$ at $p_k$ is a matrix which, except for column k, is a diagonal matrix with nonzero entries on the diagonal, and whose kth column has no zero entries. Thus $\det (A_k')\ne 0$ at $p_k$ so $p_k\not \in Q$. In particular, none of the coordinate vertices is on Q.

3 The System $\mathscr {L}_n$

Let us start with some notation. Assume $\varPi _0,\dots , \varPi _n\subset \mathbb P^n$ are general linear subspaces of codimension 2. From the previous section it follows that for each subset $\varPi _0,\dots ,\varPi _{j-1},\varPi _{j+1},\dots ,\varPi _{n}$ of n of them there is a unique hypersurface $Q_j$ of degree $n-1$ containing them. Depending on the context, we may also denote by $Q_j$ the form defining this hypersurface. We may assume $I_{\varPi _i}=(x_i,f_i)$ where $f_j$ is as given in Remark 2.4. In this case we have the $(n+1)\times (n+1)$ matrix

$$B=\left( \begin{array}{cccccc} -f_0 &{} a_{0,1}x_1 &{} \dots &{} a_{0,k}x_k &{} \dots &{} a_{0,n}x_n \\ a_{1,0}x_0 &{}-f_1 &{} \dots &{} a_{1,k}x_k &{} \dots &{} a_{1,n}x_n \\ \vdots &{}\vdots &{} &{} \vdots &{} &{} \vdots \\ a_{k,0}x_0 &{}a_{k,1}x_1 &{} \dots &{} -f_k &{} \dots &{} a_{k,n}x_n \\ \vdots &{} \vdots &{} &{} \vdots &{} &{} \vdots \\ a_{n,0}x_0 &{}a_{n,1}x_1 &{} \dots &{} a_{n,k}x_k &{}\dots &{} -f_n \end{array}\right) .$$

Let $B_i$ be the $n\times n$ submatrix obtained by deleting row i and column i of B (where we have i run from 0 to n). The matrix A in the proof of Proposition 2.5 is thus $B_0$, and we have $\det (B_i)=x_iQ_i$. The next result shows that $v_n$ is the map given by $(x_0,\dots ,x_n)\mapsto (x_0Q_0,\dots ,x_nQ_n)$.

Proposition 3.1

The polynomials $x_iQ_i$, $i=0,\dots ,n$, give a basis for $\mathscr {L}_n$, hence $\dim \mathscr {L}_n=n+1$, so $v_n$ is a rational map to $\mathbb P^n$ whose image is not contained in a hyperplane.

Proof

By Remark 2.6, no coordinate vertex $p_j$ is in $Q_i$ for any i. But $x_iQ_i\in \mathscr {L}_n$ for every i, and $(x_iQ_i)(p_j)\ne 0$ if and only if $i=j$. Thus the polynomials $x_iQ_i$ span a vector space of dimension at least $n+1$.

To show that these sections in fact give a basis, we show that $\dim \mathscr {L}_n = n+1$. We proceed by induction (the proof that $\mathscr {L}_2$ has three independent sections is clear, since three general points impose independent conditions on forms of degree 2 on $\mathbb P^2$). Let A be a fixed hyperplane that contains $\varPi _1$. There is, by Proposition 2.5, a unique section of $\mathscr {L}_n$ containing A, namely $AQ_1$. Moreover, the restrictions to A of sections $s_n$ of $\mathscr {L}_n$ which do not contain A give divisors $s_n\cap A$ of degree n, containing $\varPi _1$, and containing $A\cap \varPi _j, j>1$. So on A, the linear system of restrictions residual to $\varPi _1$ has degree $n-1$ and contains the n general subspaces $\varPi _i\cap A$, $i>1$, of codimension 2. From the inductive assumption this has dimension n, so $\dim \mathscr {L}_n = n+1$.

We may also see the result from the exact sequence

$$0\rightarrow \mathscr {L}_n(-A)\rightarrow \mathscr {L}_n\rightarrow \mathscr {L}_n|_A\rightarrow 0,$$

where A is as above and $\mathscr {L}_n(-A)$ is the linear subsystem of all elements of $\mathscr {L}_n$ containing A. Then, from the inductive assumption, the dimension of $\mathscr {L}_n|_A$ is n, and from Proposition 2.5 the dimension of $\mathscr {L}_n(-A)$ (which is of degree $n-1$ passing through n codimension 2 subspaces in A) is 1. $\square$

Remark 3.2

If the hyperplanes $H_j\supset \varPi _j, j=0,\dots ,n$ are such that any n of them intersect in a point outside all $Q_i$ and $\bigcap _jH_j=\varnothing$, then $H_jQ_j$ are linearly independent.

Proof

If this is not the case, then one of them is linearly dependent of others, let it be $H_0Q_0$. Thus, if $H_jQ_j$ vanish in some point p for $j=1,\dots ,n$, then $H_0Q_0$ also does. Let then $p=\bigcap _{j=1}^nH_j$. Thus, $H_0Q_0$ vanishes on p, but $p\notin H_0$, so $p\in Q_0$, a contradiction. $\square$

Remark 3.3

Observe also, that up to an isomorphism of (the target) $\mathbb P^n$, the map $v_n$ may be defined by any set of $n+1$ linearly independent elements of $\mathscr {L}_n$.

Let $T_n$ be the closure of the union of all lines transversal to $\varPi _0,\dots , \varPi _n$, and let $R_n =Q_0\cap \dots \cap Q_n$ and let $B_n$ be the base locus of $\mathscr {L}_n$ (i.e., the locus where $v_n : \mathbb P^n \dashrightarrow \mathbb P^n$ is not defined). We note that $T_n \subseteq R_n$, by Proposition 2.5.

Proposition 3.4

We have $B_n=\varPi _0\cup \dots \cup \varPi _{n}\cup R_n$.

Proof

Since $v_n$ is given by $(x_0,\dots ,x_n)\mapsto (x_0Q_0,\dots ,x_nQ_n)$, the base locus consists of the common zeros of the $x_iQ_i$. Clearly each $Q_i$ (and hence each $x_iQ_i$) vanishes on $R_n$ (as $R_n$ is the intersection of all $Q_i$.) But $Q_i$ vanishes on $\varPi _j$ for $j\ne i$ and $x_i$ vanishes on $\varPi _i$, so each $x_iQ_i$ vanishes on $\varPi _0\cup \dots \cup \varPi _{n}$. Thus $\varPi _0\cup \dots \cup \varPi _{n}\cup R_n\subseteq B_n$.

Conversely, let p be a point in $B_n$ not in $\varPi _0\cup \dots \cup \varPi _n$. By Remarks 3.2 and 3.3, $v_n$ may be defined by the forms $H_iQ_i$ for sufficiently general $H_i$. Since $H_i$ does not vanish on p, $Q_i$ does for all i. Thus $p\in R_n$, so $B_n\subseteq \varPi _0\cup \dots \cup \varPi _{n}\cup R_n$. $\square$

Proposition 3.5

We have $\dim T_n=n-2$ for $n\ge 3$, and $T_n$ is irreducible for $n>3$.

Proof

Consider the Grassmannian V of lines in $\mathbb P^n$ and the incidence variety $W=\{(v,p)\in V\times \mathbb P^n: p\in L_v\}$, where $L_v$ is the line corresponding to a point $v\in V$. We also have the two projections $\pi _1:W\rightarrow V$ and $\pi _2:W\rightarrow \mathbb P^n$. Then V is an irreducible variety of dimension $2(n-1)$ and degree $\frac{(2(n-1))!}{n!(n-1)!}$ embedded in $\mathbb P^N, N=\left( {\begin{array}{c}n+1\\ 2\end{array}}\right) -1$, see [7], Chap. 1, Sect. 5. The condition of being incident to a codimension 2 linear space is given by a hyperplane in $\mathbb P^N$ (see p. 128 in [3]), so the intersections of V with $n+1$ general hyperplanes gives the locus $\rho _n$ in V parametrizing the lines comprising $T_n$; notice that $\pi _2(\pi _1^{-1}(\rho _n))=T_n$. Thus $\dim \rho _n=2(n-1)-(n+1)=n-3$, so $\dim \pi _1^{-1}(\rho _n)=n-2$, and by Proposition 2.1 the projection $\pi _2$ is generically injective on $\pi _1^{-1}(\rho _n)$ so we have $\dim T_n=n-2$. Moreover, by Bertini’s Theorem, $\rho _n$ (and hence $T_n$) is irreducible when $\dim \rho _n>0$. $\square$

Proposition 3.6

With the notation as above we have $T_n=R_n$ in $\mathbb P^n$.

Proof

Let us start with the following fact. Let $L_0,\dots ,L_k,L$ be lines through a common point p. Let L belong to the space spanned by $L_0,\dots ,L_k,$ let $\mathscr {P}$ be a linear subspace, such that p does not lie on $\mathscr {P}$. Let $L_j$ intersect $\mathscr {P}$ at a point $l_j, j=0,\dots ,k$. Then L intersects $\mathscr {P}$, as the linear combination of a projection of some vectors is a projection of the combination.

Now we can show that the intersection of all $Q_j$ lies in $T_n$, the union of all transversals. Observe, that the opposite inclusion is obvious.

Take a point p in all $Q_j$, but not in any $\varPi _j$. So for each j, there is $L_j$ through p, transversal to all $Q_i$ except $Q_j$. We have $n+1$ such lines, but they must span a space of dimension less than $n+1$ (being in $\mathbb P^n$).

Without loss of generality, let $L_0$ belong to the space spanned by the others. Then using the fact we started with, for $\mathscr{P}=\varPi _0$, we get that $L_0$ intersects $\varPi _0$ (since $L_1,\dots ,L_n$ intersect $\varPi _0$), which finishes the proof.

If $p \in \varPi _j$ for some j, the proof is trivial. $\square$

Proposition 3.7

The Veneroni transformation $v_n: \mathbb P^n\dashrightarrow \mathbb P^n$ is injective off $Q_0\cup \dots \cup Q_n$, hence it is a Cremona transformation.

Proof

Let p, q be two different points off $Q_0\cup \dots \cup Q_n$. Let $H_j$ denote the unique hyperplane through p and $\varPi _j$. Then $\bigcap _{j=0}^nH_j=\{p\}$ as if the intersection of all such $H_i$ is not exactly p, then the intersection $H_0\cap \dots \cap H_n$ is a positive dimensional linear space, and any line through p in this space intersects each $\varPi _i$ and hence is a transversal for $\varPi _0,\dots ,\varPi _n$, and so p, being on a transversal, is in $T_n\subseteq R_n\subseteq B_n$. Take $j_0$ such that $q\notin H_{j_0}$. Then $H_{j_0}Q_{j_0}$ is a non-zero section of $\mathscr {L}_n$ and may be extended to a basis of $\mathscr {L}_n$. Then $v_n$ defined by the sections of this basis separates p and q. Thus $v_n$ is injective off $Q_0\cup \dots \cup Q_n$. $\square$

4 An Inverse for $v_n$

It is of interest to determine an inverse for $v_n$, and to observe that the inverse is again given by forms of degree n vanishing on $n+1$ codimension 2 linear subspaces. We explicitly define such a map $u_n$ and then check that it is an inverse for $v_n:\mathbb P^n\dashrightarrow \mathbb P^n$. If we regard $x_0,\dots ,x_n$ as homogeneous coordinates on the source $\mathbb P^n$ and $y_0,\dots ,y_n$ as homogeneous coordinates on the target $\mathbb P^n$, then $v_n$ is defined by the homomorphism h on homogeneous coordinate rings given by $h(x_0,\dots ,x_n)= (y_0,\dots ,y_n)$, where $y_i= x_iQ_i=\det (B_i)$, as we saw in Sect. 3.

To define $u_n$, we slightly modify matrix B from Sect. 3 by replacing the diagonal entries $-f_i$ in B by $-g_i$ (defined below) and by replacing each entry $a_{i,j}x_j$ in B by $a_{i,j}y_j$ to obtain a new matrix

$$C=\left( \begin{array}{cccccc} -g_0 &{} a_{0,1}y_1 &{} \dots &{} a_{0,k}y_k &{} \dots &{} a_{0,n}y_n \\ a_{1,0}y_0 &{}-g_1 &{} \dots &{} a_{1,k}y_k &{} \dots &{} a_{1,n}y_n \\ \vdots &{}\vdots &{} &{} \vdots &{} &{} \vdots \\ a_{k,0}y_0 &{}a_{k,1}y_1 &{} \dots &{} -g_k &{} \dots &{} a_{k,n}y_n \\ \vdots &{} \vdots &{} &{} \vdots &{} &{} \vdots \\ a_{n,0}y_0 &{}a_{n,1}y_1 &{} \dots &{} a_{n,k}y_k &{}\dots &{} -g_n \end{array}\right) .$$

To define $g_i$, recall that since $f_iQ_i\in \mathscr {L}_n$ for each i and the forms $x_jQ_j$ give a basis for $\mathscr {L}_n$, we can for each i and appropriate scalars $b_{i,j}$ write

$$f_iQ_i=b_{i,0}x_0Q_0+\dots +b_{i,n}x_nQ_n.$$

We define $g_i$ to be $g_i=b_{i,0}y_0+\dots +b_{i,n}y_n$, so we see that $h(g_i)=f_iQ_i$.

As an aside we also note that $b_{i,j}=0$ if and only if $i=j$. (To see this, recall by Remark 2.6 that no $Q_j$ vanishes at any coordinate vertex $p_k$, but $f_i$ vanishes at the coordinate vertex $p_j$ if and only if $i=j$. Thus, evaluating $f_iQ_i=b_{i,0}x_0Q_0+\dots +b_{i,n}x_nQ_n$ at $p_i$ gives $0=b_{i,i}Q_i$, hence $b_{i,i}=0$, while evaluating at $p_j$ for $j\ne i$ gives $0\ne b_{i,j}Q_j$, hence $b_{i,j}\ne 0$.)

Let $C_i$ be the matrix obtained from C by deleting row i and column i. Define a homomorphism $\lambda : \mathbb K[x_0,\dots ,x_n] \rightarrow \mathbb K[y_0,\dots ,y_n]$ by $\lambda (x_i)=\det (C_i)$.

The next result gives an inverse for $v_n$.

Proposition 4.1

The homomorphism $\lambda$ defines a birational map $u_n:\mathbb P^n\dashrightarrow \mathbb P^n$ which is inverse to $v_n$.

Proof

Note that applying h to the entries of C gives the matrix obtained from BD, where D is the diagonal matrix whose diagonal entries are $Q_0,\dots ,Q_n$, from which it is easy to see that $h(\det (C_i))=\det (B_i)Q_0\cdots Q_{i-1}Q_{i+1}\cdots Q_n=x_iQ_0\cdots Q_n$.

We now have $h(\lambda (x_i))=h(\det (C_i))=x_iQ_0\cdots Q_n$, so $u_nv_n=id_U$, where U is the complement of $Q_0\cdots Q_n=0$. Since $v_n$ is a Cremona transformation, so is $u_n$ and thus $u_n$ is the inverse of $v_n$. $\square$

Remark 4.2

We now confirm that the forms $\det (C_i)$ defining $u_n$ have degree n and vanish on $n+1$ codimension 2 linear subspaces $\varPi _i^*\subset \mathbb P^n$. That $\deg (\det (C_i))=n$ is clear, since $C_i$ is an $n\times n$ matrix of linear forms.

Consider the codimension two linear spaces defined by the ideals $J_k=(y_k,g_k)=b_{k,0}y_0+\dots +b_{k,n}y_n$. Since the entries of column k of C are in the ideal $J_k$, it follows that $\det (C_i)$ vanishes on $\varPi _j^*$ for $j\ne i$. It remains to check that $\det (C_i)$ vanishes on $\varPi _i^*$. But let $q\in Q_i$ be a point where $v_n$ is defined. Note that $y_i(v_n(q))=h(y_i)(q)=x_iQ_i(q)=0$ and that $g_i(v_n(q))=h(g_i)(q)=f_iQ_i(q)=0$. Thus $v_n|_{Q_i}$ gives a rational map to $\varPi _i^*$ whose image is in the zero locus of $\det (C_i)$ since $\det (C_i)(v_n(q))=(h(\det (C_i)))(q)=h\lambda (x_i)(q)=(x_iQ_0\cdots Q_n)(q)=0$. Thus $\det (C_i)$ vanishing on $\varPi _i^*$ will follow if we show that $v_n|_{Q_i}$ gives a dominant rational map to $\varPi _i^*$. This in turn will follow if we show for a general $q\in Q_i$ that the fiber over $v_n(q)$ has dimension 1 (since $Q_i$ as dimension $n-1$ and $\varPi _i^*$ has dimension $n-2$). But the space of forms in $\mathscr {L}_n$ vanishing on q is spanned by forms of the form $H_jQ_j$ where $H_j$ is a hyperplane containing q and $\varPi _j$. For a general point q, since the $\varPi _j$ are general, the intersection of any $n-1$ of the $H_j$ with $j\ne i$ has dimension 1. Since the $\varPi _j$ are general, the same is true for a general point $q\in Q_i$ except now, since there is a transversal $t_q$ through q for $\varPi _j$, $j\ne i$, we see that $\cap _{j\ne i}H_j$ still has dimension 1 and is thus exactly $t_q$. Hence the locus of points on which the forms in $\mathscr {L}_n$ vanishing at q vanish is exactly $t_q$. Thus the fiber over $v_n(q)$ has dimension 1, as we wanted to show.

It is still unclear to us whether $u_n$ is itself a Veneroni transformation whenever $v_n$ is. If we denote by $\mathscr {L}_n^*$ the forms in $\mathbb K[y_0,\dots ,y_n]$ of degree n vanishing on $\varPi _0^*\cup \dots \cup \varPi _n^*$, what we saw above is that $u_n$ is defined by an $n+1$ dimensional linear system contained in $\mathscr {L}_n^*$; the issue is whether the linear system is all of $\mathscr {L}_n^*$ (i.e., whether $\dim \mathscr {L}_n^*=n+1$).

In any case, when $\varPi _0,\dots ,\varPi _n$ are general, we now see that $v_n$ gives a birational map $\mathbb P^n\dashrightarrow \mathbb P^n$ whose restriction to $Q_i$ gives a rational map to $\varPi _i^*$ for $i=0,\dots ,n$ and the fiber of $Q_i$ over $\varPi _i^*$ generically has dimension 1. It is convenient to denote the linear system of divisors of degree n vanishing on $\varPi _0\cup \dots \cup \varPi _n$ by $nH-\varPi _0-\dots -\varPi _n$. Similarly, the linear system of divisors of degree $n-1$ vanishing on $\varPi _j$ for $j\ne i$ is represented by $(n-1)H-\varPi _0-\dots -\varPi _n+\varPi _i$. Thus, if $H^*$ is the linear system of divisors of degree 1 on the target $\mathbb P^n$ for $v_n$, then $v_n$ pulls $H^*$ back to $nH-\varPi _0-\dots -\varPi _n$, and it pulls $\varPi _i^*$ back to $Q_i$, represented by $(n-1)H-\varPi _0-\dots -\varPi _n+\varPi _i$. We can represent the pullback by a matrix map $M_n:{\mathbb Z}^{n+1}\rightarrow {\mathbb Z}^{n+1}$ where

$$ M_n=\begin{pmatrix} n&{}n-1&{}n-1&{}\dots &{}n-1\\ -1 &{} 0 &{}-1 &{}\dots &{}-1\\ -1&{}-1&{}0&{}\dots &{}-1\\ \vdots &{}\vdots &{}\vdots &{}&{}\vdots \\ -1&{}-1&{}-1&{}\dots &{}0 \end{pmatrix}. $$

If in fact the spaces $\varPi _i^*$ can be taken to be sufficiently general, then $\dim \mathscr {L}_n^*=n+1$, and $u_n$ pulls H back to $nH^*-\varPi _0^*-\dots -\varPi _n^*$, and it pulls $\varPi _i$ back to $(n-1)H^*-\varPi _0^*-\dots -\varPi _n^*+\varPi _i^*$, and hence is represented by the same matrix $M_n$. Since $M_n^2$ corresponds to the pullback map for $u_nv_n$ and $u_nv_n$ is the identity (where defined), we would expect that $M^2_n=I_n$, which is indeed the case.

5 Intersection of $Q_i$ and $Q_j$

This section is devoted to investigating the intersections of $Q_i$ and $Q_j$, assuming that $\varPi _0,\dots ,\varPi _n$ are general linear subspaces of codimension 2. These intersections were already treated in [11] and in more detail than here, but here we use more modern language.

Without loss of generality assume that $i=0,j=1$, so take $Q_0\cap Q_1$. From the considerations above (Proposition 3.4) we may write

$$Q_0\cap Q_1=T_n\cup \varPi _2\cup \dots \cup \varPi _{n}\cup M_n$$

where $M_n$ is the closure of the complement of $T_n\cup \varPi _2\cup \dots \cup \varPi _{n}$ in $Q_0\cap Q_1$.

Proposition 5.1

The complement of $T_n\cup \varPi _2\cup \dots \cup \varPi _{n}$ in $Q_0\cap Q_1$ is the set of all points $q\in Q_0\cap Q_1$ through which there is no transversal for $\varPi _0,\dots ,\varPi _n$, (in which case there is more than one transversal through q for $\varPi _2,\dots ,\varPi _{n}$).

Proof

For $n=2$ it is easy to check that $Q_0\cap Q_1=\varPi _2$ and that $T_n=M_n=\varnothing$. For $n=3$, keeping in mind that $Q_0=\mathbb P^1\times \mathbb P^1$, $Q_0\cap Q_1$ is a divisor on $Q_0$ of multi-degree (2, 2), consisting of the lines $\varPi _2$ and $\varPi _3$ together with the two transversals for $\varPi _0,\dots ,\varPi _3$ (these two transversals give $T_n$); again $M_n$ is empty. (See, for example, the description of the cubo-cubic Cremona transformation from [4] or [5].)

So now assume that $n\ge 4$. Take a point q from $Q_0\cap Q_1$. Suppose q is not in $\varPi _2\cup \dots \cup \varPi _n$. Since $q\in Q_0$, by Proposition 2.5 there is at least one transversal through q for $\varPi _1,\dots ,\varPi _n$ and since $q\in Q_1$ there is similarly at least one transversal through q for $\varPi _0,\varPi _2\dots ,\varPi _n$. If one of the transversals coming from $q\in Q_0$ is also a transversal coming from $q\in Q_1$, then it follows that the transversal goes through all $\varPi _j$, so the transversal (and hence q) is contained in $T_n$. Otherwise, $q\not \in T_n$, hence there are two lines through q transversal for $\varPi _2,\dots ,\varPi _n$. $\square$

Example 5.2

We close by showing for $n=4$ that the complement of $T_4\cup \varPi _2\cup \dots \cup \varPi _{4}$ in $Q_0\cap Q_1$ is nonempty.

Take three points $p_{ij}$, where $p_{ij}=\varPi _i\cap \varPi _j$, for $j=2,3,4, i\ne j$. Let $\pi$ be the plane spanned by the three points. Take a general point q on $\pi$. From the fact that all $\varPi _j$ are general, we have that q, $p_0:=\pi \cap \varPi _0$ and $p_1:=\pi \cap \varPi _1$ are not on a line. Then the line through q and $p_0$ is a transversal to $\varPi _0,\varPi _2,\varPi _3,\varPi _4$, so it is in $Q_0$ (and in $\pi$ of course). In the same way, the line through q and $p_1$ is a transversal to $\varPi _1,\varPi _2,\varPi _3,\varPi _4$, so it is in $Q_1$, thus q is in $Q_0\cap Q_1$.

To prove that $M_4\not \subset T_4$, take a point r not in $\varPi _2,\varPi _3,\varPi _4$, and consider a projection from r to a general hyperplane. Then the intersection of the images of $\varPi _2,\varPi _3,\varPi _4$ is either a point—and then there is only one transversal to $\varPi _2,\varPi _3,\varPi _4$ through this point—or this intersection is a line, and then we have a plane of transversals from our point r. From this construction it follows that we may have at most a plane of transversals to $\varPi _2,\varPi _3,\varPi _4$. As $\varPi _1, \varPi _0$ are general, the generic transversal on $\pi$ is not transversal to $\varPi _1, \varPi _0$.

Remark 5.3

Snyder and Rusk in [11] assert that $\deg (R_n)=\frac{(n+1)(n-2)}{2}$ and that $\deg (M_n)=\frac{(n-2)(n-3)}{2}$. We plan a future paper explaining these results and showing also precisely that the inverse of a Veneroni transformation is always a Veneroni.