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\begin{document}


\title[Veronese surfaces]{Veronese surfaces and line arrangements}

\author[Tom Graber and Kristian Ranestad]{Tom Graber and Kristian Ranestad$^{*}$}
\thanks{$^{*}$ Supported by the Research Council of Norway and by MSRI}





\address{Department of Mathematics,
University of California, Berkeley, CA 94720} \email{graber@math.berkeley.edu}

\address{Matematisk Institutt, Universitetet i Oslo P.O.Box 1053, Blindern,
N-0316 Oslo,  NORWAY } \email{ranestad@math.uio.no}

\date{\today}



\subjclass{Primary 14J; Secondary 14D}

\begin{abstract}
We formulate a generalization of Castelnuovo`s lemma on rational normal curves to d-uple Veronese surfaces and apply it to define a Hilbert scheme compactification of line arrangements in the Hilbert scheme of  Veronese surfaces.  In the case of arrangements of five lines, we find the compactification to be isomorphic to a Del Pezzo surface of degree 5.  We thus recover previous results of Kapranov and Hacking.  
\end{abstract}



\maketitle


Paul Hacking recently studied compact moduli of hyperplane arrangements using
stable pairs of a possibly reducible variety and a divisor on it \cite{PH}.  His compactification  coincides with a Hilbert scheme compactification of so called special Veronese varieties due to Kapranov \cite{Kap}.  
 
In this note we consider line arrangements in the plane and define what coincides with Kapranov's special Veronese varieties of dimension 2.   In particular, we recover an analogue of Castelnuovo's lemma on rational normal curves for $d-$uple
Veronese surfaces.  This sets up a natural map of Hacking's moduli space of
line arrangements of degree $d+3$ to the Hilbert scheme of $d-$uple
Veronese surfaces.   We give a different and direct proof that the corresponding Hilbert scheme compactification of
the  moduli of stable $5$-line
arrangements is naturally isomorphic to a smooth Del Pezzo surface of degree
$5$.  Our enumeration of singular surfaces in the family, recovers Hacking's classification. 

\section{A Castelnuovo type lemma}

We start with an analogue of

\begin{lemma}\label{Cast}[Castelnuovo] Through $d+3$ points in linear
general position in $\PP^{d}$ there is a unique rational normal curve.
\end{lemma}
\noindent due to Kapranov:
\begin{lemma}\label{Cav}\cite{Kap}  Given $d+4$ general $\PP^{d}$'s in $P^N$  with
$N=(d+2)(d+1)/2-1$, such that any two of the $\PP^{d}$'s meet in exactly a
point, and any set of $d+1$ intersection points span a $\PP^{d}$. Then there
is a unique $d$-uple Veronese surface that contains the
$(d+4)(d+3)/2$ intersection points.
\end{lemma}
\begin{remark}
Note that we need the first ``general'', since our proof is not
constructive.
\end{remark}
\proof  First we assume that $V$ is a $d-$uple Veronese surface through
all the intersection points.  By Castelnuovo's lemma, in each $\PP^{d}$
there is a unique rational normal curve of degree $d$ through the $d+3$
intersection points.
We show that $V$ actually must intersect each $\PP^{d}$ in the unique
rational normal curve given by Castelnuovo's lemma.  For this, fix a 
$\PP^d$ and let $\Gamma$ be the $d+3$ intersection points in it.  Let $C$ be a
general curve in the linear system $|(d+3)L|$ on $V$ that contains
$\Gamma$, where $L$ is the class of a line in $V\cong \PP^{2}$. Since
$\Gamma $ is smooth, $C$ is smooth.  In fact $d+3$ lines through the $d+3$
points generate a linear system without base points outside $\Gamma$, so by
Bertini $C$ is smooth. Since $C\subset V$ is the canonical embedding,
$\Gamma$ moves in a net on $C$.  But $C$ has only one
$g_{d+3}^2$, so $\Gamma$ must therefore coincide with the $L\cap C$
for some rational normal curve $L_0\in |L|$ of degree $d$ on $V$.  Clearly $L_0$
coincides with the unique rational normal curve given by Castelnuovo's
lemma. Thus $V$ contains $d+4$ rational normal curves of degree $d$.

On the other hand any set of $d+4$ general members of the linear system
$|L|$ on $V$ gives rise to $d+4$ pairwise intersecting $\PP^d$'s in 
$\PP^{N}$.

Now, consider the family of marked Veronese surfaces, where the marking is
a line arrangement of degree  $d+4$, i.e. a collection of $d+4$ lines in
the plane.  Clearly this family is irreducible.  We compute the dimension by adding the dimension of the orbit of Veronese surfaces to the dimension of the family of line arrangements and get 
$$(d+2)^2(d+1)^2/4-9+2(d+4)=d(d+4)(d^2+2d+5)/4$$
Compare this with the family of sets of $d+4$ pairwise intersecting 
$\PP^d$'s.  This family is clearly also irreducible.  We compute its dimension as a configuration of $d=4$ points in the Grassmannian satisfying the relevant Schubert condition and get
$$(d+4)(d+1)((d+2)(d+1)/2-(d+1))-((d+2)(d+1)/2-2(d+1)+1)(1+2+...+d+3)$$
$$=d(d+4)((d+1)^2/2-(d-1)(d+3)/4)=d(d+4)(d^2+2d+5)/4.$$
  Since the two
families have the same dimension, the lemma follows as soon as the
Veronese surface is unique for a general configuration of $d$-planes.
For this, let $C$ be the
line arrangement of degree $d+4$.    The $d$-th
Veronese embedding is given by $K_C-L_C$, where $L_C$ is the divisor of
the embedding in
$\PP^{2}$. The divisor $L_C$ is unique (given by the divisors determined by
its d+4 line
components), so $K_C-L_C$ is unique, and therefore also the Veronese
surface.
\QED




\section{Compactifying the muduli of stable line arrangements}

Consider the family $Y_d$ of $d-$uple Veronese surfaces that
intersect $d+3$ pairwise intersection $\PP^d$'s in a rational normal
curve.  Now, the family $X_d$ of $\PP^d$'s that intersect each of the
$d+3$ fixed $\PP^d$'s has dimension $2d$ in the Grassmannian.

Note that, by Lemma \ref{Cav}, there is a rational map $X_d\to Y_d$  whose
general fiber is a (dual) Veronese surface.
Thus

\begin{proposition}  The family $Y_{d}$ of $d-$uple Veronese surfaces that
intersect $d+3$ pairwise intersecting $\PP^d$'s in a rational normal
curve, has dimension $2d-2$.
\end{proposition}


\begin{definition}  A line arrangement of degree $d>3$ is called stable, if no
three lines are concurrent.  \end{definition}
The family  of stable line arrangements of degree
$d+3$ modulo automorphisms has dimension $2d-2$.
Now, clearly any Veronese surface in $Y_{d}$ defines a stable line 
arrangement of degree $d+3$. 
\begin{lemma} Two Veronese surfaces in $Y_d$ determine distinct stable line arrangements of degree $d+3$.\end{lemma}
\proof  The embedding of a line arrangement of degree $d+3$ on a $d$-uple Veronese surface  is given by the canonical linear series, so the Veronese surface is unique.\QED

 Therefore the moduli of stable line arrangements has a compactification in the Hilbert scheme of $d$-uple Veronese surfaces.   More 
precisely, we define $Y_{d}^{H}$ to be the reduced closure of $Y_{d}$ 
in the Hilbert scheme of Veronese surfaces.


Here we treat the case of $d=2$. So we consider the family of Veronese
surfaces in $\PP^5$, defined according to the lemma by fixing five 
$\PP^2$'s and letting a sixth plane $P$ move in the family of planes that meet
all the first five.  This family of planes form a subvariety $X$ in 
${\bf Gr}(3,6)$ of dimension $4$, since it is the proper intersection of five Schubert
hyperplane sections, while the corresponding family $Y_{2}$ of Veronese
surfaces has dimension $2$.

Let us show directly how a stable line arrangement of degree $5$ gives 
a Veronese surface intersecting five given pairwise 
intersecting  planes in conics. So, consider a 
stable line arrangement given by linear forms $l_{0},l_{1},\ldots 
l_{4}$ on $\PP ^2$. In terms of coordinates in $\PP ^2$ we may assume that
$l_1=y_0$, $l_2=y_1$, $l_3=y_2$, $l_4=y_0+y_1+y_2$ and $l_5=y_0+e_1y_1+e_2y_2$, where $e_1e_2\not=0, e_1\not=1, e_2\not=1$ and $e_1\not= e_2$.  Given five planes $P_i$ in $\PP^5$ that pairwise intersect, we may choose coordinates such that the first four planes 
$P_{1},\ldots,P_{4}$ are defined by 
$x_{0}=x_{1}=x_{2}=0$, $x_{2}=x_{3}=x_{4}=0$, $x_{0}=x_{4}=x_{5}=0$ 
and $x_{1}=x_{3}=x_{5}=0$, and that the fifth plane is defined by
$$x_{0}+a_{3}x_{3}+a_{4}x_{4}+a_{5}x_{5}=0,$$
$$x_{1}+b_{3}x_{3}+b_{4}x_{4}+b_{5}x_{5}=0,$$
$$x_{2}+c_{3}x_{3}+c_{4}x_{4}+c_{5}x_{5}=0,$$
where $${\rm det}\left(\begin{matrix} a_{3}&a_{4}&a_{5}\cr  b_{3}&b_{4}&b_{5}\cr 
c_{3}&c_{4}&c_{5}\cr\end{matrix}\right)=0,
c_{5}=0, b_{4}=0 \quad{\rm and}\quad a_{3}=0,$$ to make the fifth plane intersect the 
first four.  This reduces to the condition that the equations of the fifth 
plane $P_{5}$ are
$$x_{0}+a_{4}x_{4}+a_{5}x_{5}=0,$$
$$x_{1}+b_{3}x_{3}+b_{5}x_{5}=0,$$
$$x_{2}+c_{3}x_{3}+c_{4}x_{4}=0,$$
and 
$${\rm det}\left(\begin{matrix}0&a_{4}&a_{5}\cr  b_{3}&0&b_{5}\cr 
c_{3}&c_{4}&0\cr\end{matrix}\right)=0.$$
The last condition may be reformulated to say that there exist  nonzero coefficients $s_1,s_2,s_3$ such that $-s_2b_3+s_3c_3=s_1a_4+s_3c_4=s_1a_5-s_2b_5=0$.  Thus we may give the equations of $P_5$ the form:
$$s_1x_{0}+Bx_{4}+Cx_{5}=0,$$
$$s_2x_{1}+Ax_{3}+Cx_{5}=0,$$
$$s_3x_{2}+Ax_{3}-Bx_{4}=0.$$
We now define the Veronese map by a choice of basis for the quadrics 
on $\PP^2$: 
$$d_0x_{0}=l_{1}l_{3}, d_1x_{1}=l_{1}l_{4}, d_2x_{2}=l_{1}l_{2}, 
d_3x_{3}=l_{2}l_{4}, d_4x_{4}=l_{2}l_{3}, d_5x_{5}=l_{3}l_{4}$$
where the coefficients $d_0,\ldots,d_5$ are all nonzero and to be determined.
In coordinates $y_i$ we get
$$d_0x_{0}=y_{0}y_{2}, d_1x_{1}=y_{0}(y_{0}+y_{1}+y_{2}), 
d_2x_{2}=y_{0}y_{1},$$
$$
d_3x_{3}=y_{1}(y_{0}+y_{1}+y_{2}), d_4x_{4}=y_{1}y_{2}, 
d_5x_{5}=y_{2}(y_{0}+y_{1}+y_{2}).$$
The image of $l_{5}=0$ lies in the plane defined by the three linear 
forms
$$e_{2}d_0x_{0}+(d_1x_{1}-d_0x_{0}-d_2x_{2})+e_{1}d_2x_{2},$$
$$d_2x_{2}+e_{2}d_4x_{4}+e_{1}(d_3x_{3}-d_2x_{2}-d_4x_{4}),$$
and
$$d_0x_{0}+e_{1}d_4x_{4}+e_{2}(d_5x_{5}-d_0x_{0}-d_4x_{4})$$
or after reordering and reduction
$$(1-e_{2})d_0x_{0}+(e_{1}-e_2)d_4x_{4}+e_{2}d_5x_{5}$$

$$d_1x_{1}+e_{1}d_3x_{3}+e_{2}d_5x_{5},$$
and
$$(1-e_{1})d_2x_{2}+e_{1}d_3x_{3}+(e_{2}-e_{1})d_4x_{4}.$$
So this plane coincides with $P_{5}$  when 

$(1-e_{2})d_0=s_1, d_1=s_2, (1-e_1)d_2=s_3, e_{1}d_3=A, (e_{1}-e_2)d_4=B, e_{2}d_5=C$, which determines the coefficients $d_0,\ldots, d_5$ as promised.  
An immediate consequence of this computation is
\begin{proposition} The surface $Y$ is rational. 
\end{proposition}
\par
We shall now describe a compactification $Y$ of $Y_{2}$, that we shall 
eventually show coincides with the Hilbert scheme compactification $Y_{2}^{H}$.
For each Veronese surface the union of secant lines form a 
singular cubic hypersurface. The cubic hypersurface is defined by the 
determinant of a symmetric $3\times 3$ matrix of linear forms, and it 
is singular precisely along the Veronese surfaces.  Therefore $Y_{2}$ 
as a reduced variety has a natural embedding in the space of cubic hypersurfaces.
We take $Y$ to be the reduced closure of $Y_{2}$ in this embedding.
Our main result in this section is the description of $Y$.  

\begin{theorem}  The compactification $Y$ of the family $Y_{2}$ of Veronese surfaces 
in $\PP^5$ that meet five
pairwise intersecting planes in conics, form a smooth Del Pezzo surface of degree
$5$.  The lines on the Del Pezzo surface form the boundary 
$Y\setminus Y_{2}$ and parameterizes degenerate
Veronese surfaces.  A  point that lies on one line corresponds to a
surface with a smooth cubic scroll component and a plane through the 
directrix of the scroll, while the intersection point between two 
lines corresponds to a surface with a smooth quadric component and two 
plane components meeting each other in a point and the quadric in a line, one 
from each ruling.
\end{theorem}

\begin{remark}  Kapranov \cite{Kap} proves that by association, the moduli space $Y_2$ is isomorphic to the moduli space $M_{0,5}$ of $5$-pointed $\PP^1$ which is isomorphic to a Del Pezzo surface of degree $5$.\end{remark}

\proof   Consider the family of cubic hypersurfaces singular along a Veronese surface of the family $Y_2$. 
Their linear span is contained in the vector space $V$ of cubic hypersurfaces 
that are singular at the ten points of intersection of the
five planes $P_i$.  This space $V$ has dimension at least $6$:  The space of
cubics that contain the five pairwise intersecting planes is $16$-dimensional.  A singularity at
an intersection point imposes one linear condition, so altogether the 
singularities imposes at most ten conditions,
which leaves at least a $6$-dimensional vector space for the $10$-nodal cubics.

\begin{lemma}\label{sing}  No cubic
in $V$ is singular along one of the five planes $P_{i}$.
\end{lemma}
\proof  Let $F$ be a cubic singular along the plane $P_{1}$, say. If so let $P'$ be a $3-$space
spanned by $P_{1}$ and the point of intersection between two of the other
planes.  Since $F$ is singular in the intersection point, the span $P'$ 
must be contained in $F$.   Consider the hyperplane $H_{12}$ spanned 
by $P_{1}$ and $P_{2}$.   By the previous argument the restriction of 
$F$ to $H_{12}$ decomposes into three $\PP^3$'s that contain $P_{1}$, one for each 
intersection point of $P_{2}$ with the remaining three planes.  On the 
other hand $F$ contains $P_{2}$, so $H_{12}$ must be a component of 
$F$.  Similarly the hyperplanes spanned by $P_{1}$ and $P_{3}$, 
$P_{4}$ and $P_{5}$ are all components of $F$. But $F$ is a cubic, so 
this is absurd.\QED


Consequently, there is precisely a net of cubics in $V$
that are singular at some point in $P_{i}$ outside the original four.  In fact
since the cubics already contain $P_{i}$, it is at most $3$ conditions to be
singular at some point in the plane, and if it is less than $6$ 
conditions to be singular at two general points outside the original 
four, then there would be a cubic hypersurface singular along the 
whole plane, contradicting Lemma \ref{sing}.  
In particular
\begin{corollary}\label{cub} The space $\PP (V)$ of cubic hypersurfaces that are 
singular in the ten intersection points of the five planes $P_{i}$ is 
$5-$dimensional and the cubic hypersurfaces that are singular at some point 
in $P_{i}$ outside the original four form a $\PP^2$ and are all singular along the unique conic 
passing through the five points.\end{corollary}

Let $V_{i}\subset \PP(V)$ be the subvariety of cubic hypersurfaces 
that are singular along some conic in $P_{i}$.

\begin{lemma} The variety $V_{i}$ is a rational cubic scroll, the 
    Segre embedding of $\PP^1\times\PP^2$. \end{lemma}
    \proof 
By Corollary \ref{cub} 
there is a plane in $V_{i}$ of cubic hypersurfaces for each conic.  Since the conics 
moves in a pencil, the variety $V_{i}$ is the union of a pencil of 
planes.  Furthermore, no two planes in this pencil have 
a common point, because the corresponding cubic hypersurface would 
be singular along $P_{i}$ contradicting  Lemma \ref{sing}. But the only pencils of planes with this 
property are those of the Segre embedding of $\PP^1\times\PP^2$, so 
$V_{i}$ must be such a cubic scroll.\QED

The five cubic scrolls $V_{i}$ are distinct: Let $H_{ij}$ be 
the hyperplane spanned by $P_{i}$ an $P_{j}$, let $p_{ij}$ be the intersection point $P_{i}\cap P_{j}$, and let $H_{i}(jk,lm)$ 
be the hyperplane spanned by $P_{i}$ and the points $p_{jk}$ and 
$p_{lm}$.  Then $H_{25}\cup H_{34}\cup H_{5}(12,34)$ and $H_{12}\cup 
H_{34}\cup H_{1}(25,34)$ form cubic hypersurfaces that belong to 
$\PP (V)$. They are both singular along the conic 
$(H_{25}\cup H_{34})\cap P_{1}$ in $P_{1}$, but they are singular at 
the distinct conics  $(H_{25}\cup H_{13})\cap P_{4}$ and  $(H_{35}\cup 
H_{12})\cap P_{4}$ in $P_{4}$.   Since the planes in $V_{i}$ 
correspond to conics in $P_{i}$, the planes in $V_{1}$ and $V_{4}$ 
say, cannot coincide.  Therefore, the $V_{i}$ are also distinct.

Now, a cubic scroll is defined by the $2\times 2$ minors of a $2\times 
3$ matrix of linear forms, i.e. be three quadrics.   Since the 
intersection of two of these quadrics is a threefold of degree $4$,  two distinct cubic 
scrolls have at most one defining quadrics in common.  
Therefore the ideal of $Y$ contains at least $5$ quadrics, and $Y$ is 
contained in five distinct cubic scrolls.  In particular, $Y$ has 
degree at most $5$. 


Next, we will enumerate degenerations of Veronese surfaces in 
$V_{2}$.  So we consider limit schemes $T$ in the Hilbert scheme of the Veronese surfaces in our 
family. A limit scheme $T$ must clearly satisfy the following two conditions:

1. The Hilbert polynomial of $T$ coincides with that of a smooth Veronese
surface. In particular it has a surface component of degree $4$ whose
reduced part is connected in codimension $1$.

2. The scheme theoretic intersection with any one of the five planes is at
least a plane conic that passes through the four intersection points of
this plane with the other planes.

First, we need some notation. As above we denote by $p_{ij}$ the intersection point $P_i\cap P_j$, and by
$H_{ij}$ the hyperplane spanned by $P_{i}$ and $P_{j}$.   Denote by 
$P_{ijk}$ the plane spanned by the interesection points of $P_{i}$, 
$P_{j}$ and $P_{k}$, and let $L_{ijk}$ be the line of intersection  
$P_{ijk}\cap H_{lm}$.

To enumerate the possible degenerate Veronese surfaces,
we first exclude two cases.  First, one of the planes $P_i$ cannot be a
component of a degenerate Veronese surface: If so, the residual surface $S$
has degree $3$ and intersect the four remaining planes in at least a
conic. Any two of these four planes span a hyperplane that intersect $S$
in a curve of degree at least $4$, so $S$ has a component in each of these
$6$ hyperplanes. But any three of these hyperplanes intersect in a plane,
therefore $S$ has at least $4$ components.  This is a contradiction.
Secondly, any quadric surface component cannot intersect one of the five
planes, say $P_5$ in a conic:  If so this component intersects the
remaining four planes in at most two lines, in which case these lines meet
at an intersection point of two planes, say $p_{34}$.  The residual
surface $S$ has degree $2$ and intersect $P_1$ and $P_2$  in at least a
conic, and $P_3$ and $P_4$ in at least a line.  Again, this means that the
hyperplanes spanned by $H_{12}$, $H_{13}$, $H_{14}$, $H_{23}$ and
$H_{24}$ all contain a component of $S$.  This is possible only if $S$
consists  of the two planes $P_{123}$ and $P_{124}$.  But then the intersection of $S$ and the
quadric is $0$-dimensional, against condition 1 above.

Now, any component of a degenerate Veronese surface meets a plane $P_i$ in a
conic section, a line, a point or not at all.
Consider components of reducible degenerate Veronese surfaces.  By the first
condition, they are planes, quadrics or cubic scrolls.
Since at most two planes $P_i$ lie in a $\PP^4$ any cubic scroll
intersects the union of the $P_i$ in a curve of degree at most $7$. In
that case it intersect two planes in a conic, and the remaining three in a
line.  Furthermore these three lines are disjoint, so the scroll is 
smooth.
Since a $\PP^3$ intersect at most $4$ planes in a line, and no quadric
component can intersect a plane in a conic, the intersection between a
quadric component and the five planes is at most $4$ lines.  In that case
they form a  quadrangle defined by the pairwise intersection of two pairs
of planes  and the quadric surface component is smooth.
Finally, any plane intersect the five planes in at most three lines, in
which case it is the plane spanned by the intersection points of three
planes.  

\begin{lemma}  If $T$ is a singular Veronese surface in the family $Y$, then $T$
is the reducible union of a plane and a smooth cubic scroll intersecting
the plane along the directrix, or the reducible union of two planes and a
smooth quadric surface intersecting the planes along a line one from each
ruling.
\end{lemma}

\proof  First, an irreducible singular degeneration of a Veronese surface
is a cone over a rational normal curve.  Since the intersection with each
plane is a conic, the vertex of the cone is in each plane, absurd. A
nonreduced component is a multiple structure on a plane or a double
structure on a quadric. The bound of the intersection of the reduced
component with the five planes excludes this possibility also.  Therefore a
singular Veronese surface in the family is reduced, and the components are
planes, smooth quadrics or cubic scrolls.  Furthermore, by degree reasons
alone, any plane component must meet the union of the five planes in $3$
lines, any quadric surface component is smooth and must meet the five
planes in $4$ lines.  Furthermore if the singular Veronese surface has a
quadric surface component, the residual components must be two planes,
that meet the quadric in a line, one from each ruling. Finally, any cubic
scroll intersect the five planes in a curve of degree $7$ with two conic
components and $3$ disjoint lines.  In particular the scroll is smooth,
and the residual component is a plane through the directrix.\QED

We may now enumerate the degenerate Veronese surfaces in the above 
notation:


1. Any plane component  coincides with the plane spanned by the
intersection points of three planes $P_{ijk}=<p_{ij},p_{jk},p_{ik}>$.

2. Any quadric component is smooth and coincides with  the quadric through
say $L_{123}$ and $L_{345}$ which contain the points $p_{14},p_{24},
p_{15}, p_{25}$.
This quadric lies in the $3$-space $<P_1,P_2>\cap <P_4,P_5>$.

3. Any cubic scroll component is smooth and coincides with a scroll
through three lines and meeting two planes in conics through their
intersection point.  For each pair of planes there is a pencil of them, i.e. for each plane
component there is a pencil of them.  For the plane $P_{ijk}$ this pencil
of cubic scrolls have the line $L_{ijk}$ as a common directrix.

Altogether we find ten plane components in degenerations, fifteen 
smooth quadric surface components of degenerations, and ten pencils 
of smooth cubic scroll components.


For each of these degenerate Veronese surfaces the union of secant lines form a cubic
fourfold:
If the Veronese surface is smooth, then the cubic is the determinant of a
symmetric matrix.
If the surface is the union of a plane and a cubic scroll, then the cubic
is the union of the $\PP^4$ of the cubic scroll and the unique rank $3$ quadric with
vertex the plane which contains the cubic scroll.
If the surface is the union of two planes and a quadric surface, then the
cubic is the union of three $\PP^4$'s.  Notice furthermore that in 
each case the union of the secant lines form the unique 
cubic hypersurface that is singular along the degenerate Veronese 
surface.


Therefore there is a curve consisting of ten lines in $\PP (V)$ 
parameterising singular cubics of degenerate Veronese surfaces.

\begin{lemma} The surface $Y\subset \PP (V)$ contains at most ten lines.\end{lemma}

\proof Assume that a line in $Y$ contains the cubic of a smooth 
Veronese surface, and consider the base locus of the pencil.  Since 
every cubic of the pencil is singular along some Veronese surface, by 
Bertini, the union of the Veronese surfaces must be contained in the 
base locus of the pencil.  In at least one of the planes $P_{i}$ the pencil 
of Veronese surfaces restricts to the pencil of conics through the 
four intersection points, since otherwise the pencil of Veronese 
surfaces have five conics in common, which is impossible.  
Corresponding to the singular conics in the pencils, the Veronese 
must be degenerate, so the cubic is decompose in a hyperplane 
and a rank $3$ quadric with vertex a plane $P_{ijk}$, or in three hyperplanes.  If the base locus 
contains a Veronese surface, this must be contained in the rank $3$ 
quadric and the vertex plane is a tangent plane to the Veronese 
surface.  For each singular conic there is a rank $3$ quadric 
containing the Veronese surface, such that the vertex plane is tangent 
to the Veronese surface. But two of these three quadrics must have a 
vertex plane that passes through the same intersection point $p_{ij}$ 
on $P_{i}$.  On the other hand this is a point on the Veronese surface, so in fact it 
must be the point of tangency, and the two vertex planes must coincide, which 
is absurd.  
Therefore all cubics of the pencil are cubics of degenerate 
Veronese surfaces, and the ten lines of degenerate
Veronese surfaces are the only possible lines on $Y$.
\QED

Clearly, both $Y$ and the ten lines of degenerate Veronese 
surfaces lie in the five cubic scrolls $V_{i}$.  Recall that $Y$ has 
degree at most $5$.
   On the other hand $Y$ spans $\PP(V)$: Otherwise it is a 
possibly reducible cubic scroll, but $Y$ contains contains at most ten lines.  
If $Y$ has degree $4$, it is again a scroll or a possibly degenerate 
Veronese surface.  Only in the case of a Veronese surface are there 
at most ten lines.  But a Veronese surface is not contained in any 
smooth cubic threefold scroll, so we are left with the sole possibility that $Y$ 
has degree $5$ and is contained in $5$ quadrics. 
This is possible only if the ideal of $Y$ is generated by five quadrics, 
the Pfaffians of a $5\times 5$ matrix of linear forms.  Thus $Y$ is 
the scheme theoretic intersection of the cubic scrolls, the ten lines 
all lie on $Y$ and $Y$ is a 
smooth Del Pezzo surface.
\QED

\begin{corollary} $Y$ coincides with the Hilbert scheme compactification $Y_2^H$ of $Y_2$.
\end{corollary}
\proof  The family $Y$ of surfaces is clearly flat, so there is a morphism $f: Y\to Y_2^H$.  On the other hand the singular cubic hypersurfaces associated to each surface in $Y$ sets up a morphism $g: Y_2^H\to Y$  inverse to $f$, so $f$ must be an isomorphism.\QED
  \vskip 3mm
\noindent {\bf Acknowledgement.} We thank Paul Hacking for pointing out Kapranov's work on special Veronese varieties, and Andreas Holmsen for posing a question on line transversals to certain sets of planes in $\PP^4$ that inspired us to consider these families of Veronese surfaces.
 \bibliography{ver}
\begin{thebibliography}{10}

\bibitem{PH} P.~Hacking: Compact moduli of hyperplane arrangements, preprint, math.AG/0310479


\bibitem{Kap} Kapranov, M. M. Veronese curves and Grothendieck-Knudsen moduli space $\overline M\sb {0,n}$.  J. Algebraic Geom.  2  (1993),  no. 2, 239--262.




\end{thebibliography}
\end{document}



























Notice that this scroll is 
defined by three quadrics, and that the only lines in it are those of 
planes, and those mapped to a point in the projection to $\PP^2$.

 Consider a pencil of cubic hypersurfaces in the space $V$ distinct from the ten lines, 
such that at least two of them $F_{0:1}$ and $F_{1:0}$ are singular along a 
Veronese surface or a degenerate Veronese surface. Fix a plane 
$P_{i}$ and consider the restriction of the cubics to this plane.  In the plane
the cubics of $Y$ are singular along a conic, while the general cubic hypersurfaces 
of $V$ are
singular only in four points.  Let $x=y=z=0$ be the equations for the plane $P_i$,
and assume that
$F_{0:1}$ and $F_{1:0}$ are singular along the conics $q_1$ and $q_2$,
respectively.  Then the equation for a cubic hypersurface in the pencil
generated by $F_{0:1}$ and $F_{1:0}$ has the form $F_{s:t}=sq_1L_1+tq_2L_2 +
g(s,t)$, where the $L_1$ and $L_2$ are linear forms in $x,y,z$ and
$g(s,t)$ is quadratic in $x,y,z$.  If $q_1$ and $q_2$ are not proportional or have a common line component,
then $F_{s:t}$ is singular along a a conic in $P_{i}$ only if $st=0$ 
or $L_{1}$ and $L_{2}$ are proportional.  In the latter case each 
cubic in the pencil is singular along a conic in $P_{i}$.  Consider 
the restriction of the pencil of cubics to the hyperplane $H_{1}$ defined by the 
linear form $L_{1}$.  These cubics are all singular along the plane 
$P_{i}$.  The intersection of $H_{1}$ with the four remaining planes are 
lines.   The pencil of cubics have the form:  $$ax^2+bxy+cy^2$$ where 
$a,b,c$ are linear in $u,v,w$ and in the parameters $s,t$ of the 
pencil.  Any such cubic threefold is the union of a one 
parameter family of planes. 


, the general cubic of the pencil has only four singular points in the plane $P_i$. 
In that case the linear forms $L_1$ and $L_2$ are not proportional and the cubic hypersurface 
$V(F_{s:t})$ is singular along a conic in $P_i$ if and only if $st=0$.
Therefore a third cubic hypersurface in $V$ could be singular along a conic in $P_i$ only if $q_1$ and $q_2$ are proportional, i.e. the two original, possibly degenerate, Veronese surfaces intersect along a conic section. 
Now, this argument may be applied at all five planes. So unless the 
pencil contains only two cubic hypersurfaces belonging to $Y$, we may assume that
the  two possibly degenerate Veronese surfaces intersect
along $5$ conics.  But each quadric hypersurface that contains the five conics contains any possibly degenerate Veronese surface that they are contained in, so the Veronese surfaces must coincide, contrary to assumption.   
\QED



Next, we fix a plane, say $P_1$ and consider the pencil of conic intersections with
Veronese surfaces in the family $Y$. Since each
Veronese surface intersects the plane in a unique conic in the pencil of conics through the four intersection points, the incidence between Veronese
surfaces and conics defines a morphism $\pi_1:Y\to\PP^1$. There is 
precisely at net of cubic hypersurfaces in $V$ that are singular 
along a conic in the plane $P_{1}$, so the
fibers of the map $\pi_1$ are all plane curves. Furthermore the fibers must be conics, since there are only finitely many lines on $Y$ and no trisecants. We may describe the
singular fibers:  The fiber over a singular conic consists of degenerate Veronese surfaces:
Over the conic $L_{23}L_{45}$ there is the pencil with fixed plane
$P_{123}$ and the pencil with fixed plane $P_{145}$, and the two pencils
meet in the degenerate Veronese surface containing both of these planes.
Thus three fibers of the above map are singular conics.  Moreover,
since we have accounted for the lines, there are only these singular
members of the pencil.  The five  pencils  of plane conics  first
implies that the surface $S$ is rational. The only possible singular
points are the three component degenerations.  Thus, $Y$ is an 
irreducible rational surface with at most ten singular points.
Consider its minimal resolution of singularities $Y'$, and the 
pullback $H$
of a hyperplane in ${\bf P} V$. Since $Y$ has five pencils of conic 
curves and only finitely many lines, the adjunction map 
must be constant, i.e.  $Y$ is a Del Pezzo surface.
Since it contains exactly ten lines and has no trisecants, it must 
have degree $5$ and span ${\bf P} V$.















Consider now the family of cubic hypersurfaces in $\Pn 5$ that are
singular along a Veronese surface, and let $S$ be the reduced closure
of this
family.  First of all, $S$ is a surface, since it parametrizes
Veronese surfaces through the intersection points of $5$ planes.
Secondly, it contains a curve of degree 10 with ten linear components
whose dual graph
is a Petersen graph.  Thirdly, we may define five distint pencils of
conics on the surface $S$.  For each of the five planes, consider the
map that associates a Veronese surface to its conic intersection
in that plane.  Since the conic passes through the four intersection
points with the other planes, it belongs to the pencil of conics
through these points.  Therefore the map defines a pencil of curves
on $S$.  Now every Veronese surface intersect the plane in a unique
conic, so the map is in fact a morphism, and the pencil of curves has
no basepoints.  The fibers over the singular conics are easy to
describe.  They are the degenerate Veronese surfaces associated to
planes $P_{ijk}$ meeting our plane in two of the four points.
In particular these fibers consist settheretically of two intersecting
lines in $S$.
On the other hand each fiber is contained in a plane: On the one hand
to cubic in the family is singular along a plane. For a cubic
containg a plane, and singular in the four points of intersection with
the other planes, it is 3 conditions to be singular in a further point
in the plane (if less then some would be singular in the plane),
i.e. there is precisely a plane of cubic hypersurfaces
singular along a conic in one of the five planes.
The three  singular fibers are therefore plane conics.  Moreover,
since we have accounted for the lines, there is only these singular
members of the pencil.  These five  pencils  of plane conics  first
implies that the surface $S$ is rational. The only possible singular
points are the three component degenerations.  Furthermore it has no
proper trisecant lines.  The adjunction map H+K defines a constant
map, so $S$ is the Del Pezzo of degree $5$.








Fix a plane, say $P_1$ and consider the pencil of conic intersections with
Veronese surfaces in the family $Y$. The incidence between Veronese
surfaces and conics defines a morphism $\pi_1:Y->\Pn 1$, since each
Veronese surface intersect in a unique conic. By the above remark the
fibers are all plane curves. By the trisecant lemma they are conics, since
there are only finitely many lines on $Y$. In fact we may describe the
singular fibers more precisely.


We start by describing some degenerations of line 
arrangements of degree $5$.  Consider a $\PP^{2}\times A_1$ with four lines fixed
in $\PP^{2}$ and a fifth moving with $t\in A_1$.  Assume that the five lines
$C_t$ are distinct and no three are concurrent when $t\notequal 0$, while
$C_0$ has one triple point, two triple points or a double line.  In the
first two cases we blow up the product in the triple points of $C_0$ and
get either one exceptional plane and an $F_1$, or two exceptional planes
and a plane blown up in two points as the central fiber.  In the third
case, blow up the double line. The central fiber in this case is a plane
and a scroll $F_1$. We will see that all these degenerations occur in our 
family $Y$.

But first, we look at some divisors in $X$: There are (at least) three kinds of
boundary divisors on $X$:

 1.$D_{ij}$: All planes through the point of intersection $P_{i} \cap P_{j}$ and
meeting the
three remaining ones.  ( this divisor is isomorphic to a projection 
of $\Pn 1x\Pn 1x\Pn 1$
from a point in its Segre embedding)
2.$D^{ij}$: All planes in the span of $P_{i} \cup P_{j}$ that intersect the three
other
ones.  This divisor is isomorphic to the previous one.
3. $D_{ijk}$: All planes that meet the plane spanned by the three
intersection points $P_{i}\cap P_{j}$, $P_{j} \cap P_{k}$ and $P_{k} \cap 
P_{i}$ in a line, and meet the two
remaining
planes. This divisor is a $\Pn 1$-bundle(scroll) over $\PP^{2}$.

The general plane in the first divisor gives a smooth Veronese
surface, the general one in the two other divisors give the union of a
cubic scroll and a plane through its directrix.

The family of planes of a given Veronese surface is a $3-$uple Veronese
surface ( this is the embedding of the dual plane).  In the degenerate
case of a plane and a cubic scroll, this surface degenerates to the union
of a $2-$uple Veronese and a rational scroll $(S(2,3))$ of degree $5$.
[Intersection of boundary divisors:
$D_{123}\cap D_{345}$: Planes through a point (in a $P_3$) intersecting
$P_{123}$ and $P_{345}$ in a line. These planes form a quadric surface on
$X$.

$D_{123}\cap D_{234}$  form a conic section in $X$.]







This curve has  degree $10$ and arithmetic genus $6$:  Each pencil
form a line where the corresponding hypersurfaces have a fixed hyperplane
component. The curve, if it spans $\Pn 5$ is a singular canonical curve
which is the interesction of a Del Pezzo surface of degree $5$ and a
quadric hypersurface.  Our aim is to prove the $Y$ coincides with this Del
Pezzo surface.


We analyze further the space $V$ of cubic hypersurfaces.  First, no cubic
in $V$ is singular along a plane $P$:  If so let $P'$ be a $3-$space
spanned by $P$ and a point of intersection between two of the other
planes.  Since the cubic is singular in the intersection point, its
restriction to $P'$ must be.  But the restriction must be the union of a
double plane at $P$ and a plane through the intersection point, a
contradiction.  Consequently, there is precisely a net of cubics in $V$
that are singular at some point in $P$ outside the original four.  In fact
since the cubics already contain $P$, it is at most $3$ conditions to be
singular at some point in the plane, and if it is $2$ or less conditions
at some point, it would be at most $5$ conditions to be singular in $6$
points not on a conic, so there would be a cubic in $V$ singular along the
plane.




Moving to another of the five planes $P_i$, at least one of the singular
fibers of $\pi_1$ defines a section of the map $\pi_i$.   To show that the
general fiber is smooth, assume they are all singular.  Then there must be
a pencil of cubic hypersurfaces, all singular along some Veronese surface.
But this is impossible, by Bertini. So in fact the only lines in fibers
are those of degenerate Veronese surfaces, which we already accounted for.
In conclusion we have five  pencils of conics on the surface, just like
the Del Pezzo.


Now, we consider the surface$S$ and the five pencils defned..



Therefore they have at least finite scheme of a degree 12 in common.
Thus if more than $2$ cubics in the pencil are singular along a Veronese
surface, then they all
If this is to happen for all five planes, we have to be in the case where
$L_1$ and $L_2$ are proportional in at least 4? of the cases.




 The general fiber is therefore smooth or is another singular conic.
Now, for each plane $P_i$ the intersection with a Veronese surface
belonging

Consider the self map
$\Pn 5\to \Pn 5$ defined by these cubics. This map is generically $3:1$.
The restriction to a Veronese surface maps the surface isomorphically to a
plane.
In the dual space to the target: are these planes the tangent planes to  a
Del Pezzo surface?







For each of these Veronese surfaces the union of secant lines form a
cubic
fourfold:
If the Veronese surface is smooth, then the cubic is the determinant of
a
symmetric matrix.
If the surface is the union of a plane and a cubic scroll, then the
cubic
is the union of the $\Pn 4$ of the cubic scroll and the quadric cone
with
vertex the plane over the base conic of the cubic scroll.
If the surface is the union of two planes and a quadric surface, then
the
cubic is the union of three $\Pn 4$'s.

The cubic hypersurfaces singular at the ten points of intersection of
the
five planes $P_i$ form $6$-dimensional vector space:  The space of
cubics
that contain the five planes is $16$-dimensional.  A singularity at an
intersection point imposes one linear condition, so altogether ten which
leaves a $6$-dimensional vector space for the $10$-nodal cubics.

The surface $Y$ has a natural embedding in the corresponding $\Pn 5$ of
cubic hyper surfaces, since any two degenerate Veronese surfaces
determine
distinct cubic hypersurfaces.

The cubic hypersurfaces of the degenerate Veronese surfaces enumerated
above form a curve of degree $10$ and arithmetic genus $6$:  Each pencil
form a line where the corresponding hypersurfaces have a fixed
hyperplane
component. The curve, if it spans $\Pn 5$ is a singular canonical curve
which is the interesction of a Del Pezzo surface of degree $5$ and a
quadric hypersurface.  Our aim is to prove the $Y$ coincides with this
Del
Pezzo surface.

Fix a plane, say $P_1$ and consider the pencil of conic intersections
with
Veronese surfaces in the family $Y$. The incidence between Veronese
surfaces and conics defines a morphism $\pi_1:Y->\Pn 1$, since each
Veronese surface intersect in a unique conic. The fiber over a singular
conic consists of degenerate Veronese surfaces: Over the conic
$L_{23}L_{45}$ there is the pencil with fixed plane $P_{123}$ and the
pencil with fixed plane $P_{145}$, and the two pencils meet in the
degenerate Veronese surface containing both of these planes.  Thus three
fibers of the above map is a singular conic.
Moving to another of the five planes $P_i$, at least one of the singular
fibers of $\pi_1$ defines a section of the map $\pi_i$.   To show that
the
general fiber is smooth, assume they are all singular.  Then there must
be
a pencil of cubic hypersurfaces, all singular along some Veronese
surface.
But this is impossible, by Bertini. So in fact the only lines in fibers
are those of degenerate Veronese surfaces, which we already accounted
for.
In conclusion we have five  pencils of conics on the surface, just like
the DelPezzo.





  The general fiber is therefore smooth or is another singular conic.
Now, for each plane $P_i$ the intersection with a Veronese surface
belonging

Consider the self map
$\Pn 5\to \Pn 5$ defined by these cubics. This map is generically $3:1$.
The restriction to a Veronese surface maps the surface isomorphically
to a
plane.
In the dual space to the target: are these planes the tangent planes to
  a
Del Pezzo surface?


Consider two Veronese surfaces and their respective singular cubic
threefolds.
If the pencil generated by the two cubics contains another singular
cubic,
then every member is a singular cubic, and they have a fixed component.
In fact look at the restriction of the cubics to a plane.  In the plane
the singular cubics are singular along a conic, while the general is
singular only along the four points.   At most two cubics in a pencil
can
be singular in a conic, unless they have a common component and they all
are:  Let $x=y=z=0$ be the equations for the plane $P_1$, and assume
that
$F_1$ and $F_2$ are singular along the conics $q_1$ and $q_2$,
respectively.  Then the equation for a cubic hypersurface in the pencil
generated by $F_1$ and $F_2$ has the form $F_{s:t}=[sq_1L_1+tq_2L_2 +
g(s,t)$, where the $L_1$ and $L_2$ are linear forms in $x,y,z$ and
$g(s,t)$ is quadratic in $x,y,z$.  The cubic hypersurface $V(F_{s:t}$ is
singular along a conic in $P_1$ if and only if $st=0$, $L_1$ and $L_2$
are
proportional or $q_1$ and $q_2$ are proportional.
In the two latter cases all cubic hypersurfaces in the pencil are
singular
along a conic in the plane.  In the first, the conic moves in the
pencil, while the the special hyperplane tangent along the conic is
fixed. In the latter, the conic is fixed, while the tangent hyperplane
moves in a pencil.  What remains is to show that this cannot happen at
all
five planes.  First of all, two Veronese surfaces in this family has
at most one common conic, since if they have two, then 6 quadrics
restrict to conics through
3 points.  Which means that the two Veroneses lie in three quadrics.
But then the yhree extra planes are also in these quadrics, so they
define a threefold, but this would be a cubic scroll, but this does
not have three pairwise intersecting planes.

Thus if more than $2$ cubics in the pencil are singular along a Veronese
surface, then they all have a common tangent hyperplane at four planes.

If this is to happen for all five planes, we have to be in the case
where
$L_1$ and $L_2$ are proportional in at least 4? of the cases.


A constructive existence proof: Fix three pairwise intersecting planes.
Consider the nets of hyperplanes through each of these.  An
isomorphism between the three nets with no fixed points defines the
planes of a Veronese surface.  Isomorphisms are fixed by three
additional pairwise intersecting planes.  Can we prove that the union
of the family of planes form a cubic hypersurface?





Consider two Veronese surfaces and their respective singular cubic
threefolds.
If the pencil generated by the two cubics contains another singular
cubic,
then every member is a singular cubic, and they have a fixed component.
In fact look at the restriction of the cubics to a plane.  In the plane
the singular cubics are singular along a conic, while the general is
singular only along the four points.   At most two cubics in a pencil
can
be singular in a conic, unless they have a common component and they all
are:  Let $x=y=z=0$ be the equations for the plane $P_1$, and assume
that
$F_1$ and $F_2$ are singular along the conics $q_1$ and $q_2$,
respectively.  Then the equation for a cubic hypersurface in the pencil
generated by $F_1$ and $F_2$ has the form $F_{s:t}=[sq_1L_1+tq_2L_2 +
g(s,t)$, where the $L_1$ and $L_2$ are linear forms in $x,y,z$ and
$g(s,t)$ is quadratic in $x,y,z$.  The cubic hypersurface $V(F_{s:t}$ is
singular along a conic in $P_1$ if and only if $st=0$, $L_1$ and $L_2$
are
proportional or $q_1$ and $q_2$ are proportional.
In the two latter cases all cubic hypersurfaces in the pencil are
singular
along a conic in the plane.
The same argument may be applied at all five planes. So we may assume
that
every cubic in the pencil is singular along a conic in each of the five
planes. For each plane there are two possibilieties: either the conic is
fixed, or the linear form $L$ is fixed.  If the former is the case for
more than one plane, then we would have two Veronese surfaces intersect
along $d$ conics and along $5-d\choose 2$ points, with $d>1$.  Thus the
two Veronese surfaces are in at least $d+1$ quadrics.  This is easily
shown to be impossible. Therefore two veronese surfaces have at most one
conic in common, and in the pencil of cubics it is the linear form $L$
that is fixed for the pencil at at least four planes.
Consider now the dual Veronese surfaces.  The linear form is a point on
the dual Veronese, while the hyperplanes throuh the plane of a conic
define a tangent plane to the dual Veronese.  Thus the dual Veronese
surfaces have at least four points in common where even the tangent
planes
coincides. Dual to the fifth plane, they have a common tangent plane,
but
they do not intersect.  Their union would lie in a net of quadrics, so
the
quadrics define a cubic scroll or a complete intersection.  In the first
case the intersection curve is an elliptic sextic curve. But this curve
cannot have four singular points, so the inersection of the quadrics
must
be a cubic scroll.  On the scroll, the veronese surface muct intersect
each plane in a conic and the scroll must be a cone, so the two veronese
surfaces intersect in a  curve of degree $4$.  But this violates the
previous condition on intersections.