q_{1},q_{2}\rangle$ and some quadric $q_{6}\in\langle q_{3},q_{4}\rangle$. We
may assume $q_{5}=q_{1}$ and $q_{6}=q_{3}$. Now $J=\langle
q_{1},q_{2},q_{3}\rangle=\langle x_{0}l_{1},x_{0}l_{2},x_{1}l_{3}\rangle$.
Because there is a linear syzygy involving $x_{0}l_{1}$ and $x_{1}l_{3}$,
either $l_{1}=x_{1}$ or $l_{3}=x_{0}$. If $l_{3}=x_{0}$, then $S/J$ has Betti
table

$\begin{matrix}1&-&-&-\cr-&3&3&1\cr\end{matrix}.$

If $l_{1}=x_{1}$, then $J=\langle x_{0}x_{1},x_{0}l_{2},x_{1}l_{3}\rangle$. If
$l_{2}=l_{3}$, then $S/J$ has Betti table

$\begin{matrix}1&-&-&-\cr-&3&3&1\cr\end{matrix},$

otherwise $S/J$ has Betti table

$\begin{matrix}1&-&-&\cr-&3&2&\cr\end{matrix}.$

∎

###  4.2. A case by case characterization of the quadratic component

We now examine the cases $b_{12}(S/Q_{F})\in\\{2,\ldots 6\\}.$ As noted
earlier, when $b_{12}\in\\{0,1\\}$ there is nothing to analyze.

####  4.2.1. Case $b_{12}=2$

######  Proposition 4.8.

Let $F$ be a quaternary quartic with $b_{12}(A_{F})=2$. Then $Q_{F}$ is the
ideal of an apolar scheme $Z$, which differs depending on the two possible
Betti tables below.

  1. (1)

If $F$ is of type $[200]$, then $S/Q_{F}$ has Betti table

$\begin{matrix}1&-&-&\cr-&2&-&\cr-&-&1&\cr\end{matrix}$

and $Q_{F}$ is a $(2,2)$ complete intersection.

  2. (2)

If $F$ is of type $[210]$, then $S/Q_{F}$ has Betti table

$\begin{matrix}1&-&-&\cr-&2&1&\cr\end{matrix},$

and $Q_{F}$ is the ideal of the union of a plane and a, possibly embedded,
line.

The set of ideals $Q_{F}$ in the Hilbert scheme, when $F$ is of type $[200]$
or $[210]$, is irreducible.

###### Proof.

If $Q_{F}$ is not a complete intersection, then it has a linear syzygy, so we
are in the case of Proposition 4.3 case $(1)$. Complete intersections form an
open set in the Grassmannian of pencils in the space of quadrics, while the
set of unions of a plane and a line is parameterized by an open set in the
product of the space of planes and the space of lines in ${\mathbb{P}}^{3}$.
The irreducibility for the set of ideals $Q_{F}$ follows. ∎

####  4.2.2. Case $b_{12}=3$

######  Proposition 4.9.

Let $F$ be a quaternary quartic of type $[300]$. Then either

  1. (1)

$S/Q_{F}$ has Betti table $[300a]$ or $[300b]$

$\begin{matrix}1&-&-&-&\cr-&3&-&-&\cr-&-&3&-&\cr-&-&-&1&\cr\end{matrix}$

and $Q_{F}$ is a complete intersection $(2,2,2)$, or

  2. (2)

$S/Q_{F}$ has Betti table $[300c]$

$\begin{matrix}1&-&-&-&\cr-&3&-&-&\cr-&-&4&2&\cr\end{matrix}$

and $Q_{F}$ is the ideal of a union of a line and a (possibly embedded) length
$4$ scheme.

In either case, the set of ideals $Q_{F}$ in the Hilbert scheme is
irreducible.

###### Proof.

The two cases follow from Lemma 4.11 below. The set of $(2,2,2)$ complete
intersections coincides with an open set in the Grassmannian of nets of
quadrics in ${\mathbb{P}}^{3}$, so it is irreducible. The set of ideals of
type $[300c]$ maps dominantly to the Grassmannian of lines, and to the Hilbert
scheme of length $4$ subschemes. The latter two are irreducible, so the set of
ideals of type $[300c]$ is also irreducible. ∎

######  Proposition 4.10.

Let $F$ be a quaternary quartic with $b_{12}(A_{F})=3$, and assume that $F$ is
not of type $[300]$. Then $Q_{F}$ is the ideal of an apolar scheme $Z$,
varying with the Betti table as described below.

  1. (1)

If $F$ is of type $[310]$, then $S/Q_{F}$ has Betti table

$\begin{matrix}1&-&-&-&\cr-&3&1&-&\cr-&-&2&1&\cr\end{matrix}$

and $Q_{F}$ is the ideal of the union $Z$ of a plane conic and a (possibly
embedded) scheme of length $2$.

  2. (2)

If $F$ is of type $[320]$, then $S/Q_{F}$ has Betti table

$\begin{matrix}1&-&-&-&\cr-&3&2&-&\cr\end{matrix}$

and $Q_{F}$ is the ideal a curve $Z$ of degree $3$ defined by a determinantal
net of quadrics.

  3. (3)

If $F$ is of type $[331]$, then $S/Q_{F}$ has Betti table

$\begin{matrix}1&-&-&-&\cr-&3&3&1&\cr\end{matrix}$

and $Q_{F}$ is the ideal of the union $Z$ of a plane and a (possibly embedded)
point.

The set of ideals $Q_{F}$ when $F$ is of type $[310]$, type $[320]$, or type
$[331]$, form irreducible components of the corresponding Hilbert schemes.

###### Proof.

The first two cases follows from lemmas 4.12 and 4.13 below, while in the
third case $Q_{F}$ is the syzygy ideal of the second order linear syzygy
described in Proposition 4.3 case (4). In each case, the set of schemes $Z$
form irreducible components of the Hilbert scheme, so the proposition follows.
∎

######  Lemma 4.11.

If the Betti table of $S/Q_{F}$ has second row $(3,0,0)$, then the Betti table
of $S/Q_{F}$ is either

$\begin{matrix}1&-&-&-&\cr-&3&-&-&\cr-&-&4&2&\cr\end{matrix}$

and ${\mathbb{V}}(Q_{F})$ is the union of a line and a possibly embedded
scheme of length $4$, or

$\begin{matrix}1&-&-&-&\cr-&3&-&-&\cr-&-&3&-&\cr-&-&-&1&\cr\end{matrix}$

and ${\mathbb{V}}(Q_{F})$ is a complete intersection.

###### Proof.

Since there is no linear syzygy, no two of the forms share a common factor,
hence $\operatorname{codim}(Q_{F})=2$. So $Q_{F}$ contains a subideal
$J=\langle q_{1},q_{2}\rangle$ which is a $(2,2)$ complete intersection, with
$Q_{F}=J+\langle q_{3}\rangle$. As a complete intersection $S/J$ is Cohen-
Macaulay of degree $4$, and we have the short exact sequence

(25) $0\longrightarrow S(-2)/(J:q_{3})\stackrel{{\scriptstyle\cdot
q_{3}}}{{\longrightarrow}}S/J\longrightarrow S/Q_{F}\longrightarrow 0.$

It follows from [CAwvtAG]*Lemma 3.6 that the associated primes of $(J:q_{3})$
are all codimension $2$. Hence $\deg(J:q_{3})\in\\{1,\ldots,4\\}$. If
$\deg(J:q_{3})=1$, then $(J:q_{3})$ must be reduced, which would imply that
$(J:q_{3})$ has a linear generator. This generator would yield a linear first
syzygy in the mapping cone resolution of $Q_{F}$, contradicting the hypothesis
that there is no linear first syzygy on $Q_{F}$. So
$\deg(J:q_{3})\in\\{2,3,4\\}$, and all associated primes are of codimension
$2$. Note also that from the long exact sequence in $\operatorname{Ext}$ of
the sequence in Equation 25,
$\operatorname{Ext}^{3}(S/(J:q_{3}),S)\simeq\operatorname{Ext}^{4}(S/Q_{F},S)$,
so $(J:q_{3})$ is Cohen-Macaulay iff $Q_{F}$ is saturated.

The degree of $(J:q_{3})=4$:

In this case $q_{3}$ is a nonzero divisor on $J$ and therefore $Q_{F}$ is a
$(2,2,2)$ complete intersection, with Betti table of $S/Q_{F}$ of the form

$\begin{matrix}1&-&-&-&\cr-&3&-&-&\cr-&-&3&-&\cr-&-&-&1&\cr\end{matrix}.$

The degree of $(J:q_{3})=3$:

In this case, we claim the Betti table of $S/(J:q_{3})$ is

$\begin{matrix}1&-&-&\cr-&3&2&\cr\end{matrix}.$

Since $\deg(Q_{F})=1$, the ideal $Q_{F}$ is contained, after a change of
variables, in the ideal $\langle x_{0},x_{1}\rangle$ of a line $L$. As
$J\subset Q_{F}\subset\langle x_{0},x_{1}\rangle$, we infer $J$ can be
generated by the pair of $2\times 2$ minors involving the last column of the
matrix

$\begin{pmatrix}l_{1}&l_{2}&x_{0}\cr l_{3}&l_{4}&x_{1}\cr\end{pmatrix},$

for an appropriate choice of linear forms $l_{i}$. The three $2\times 2$
minors of the matrix generate the ideal $I$ of a cubic curve, residual to the
line $L$ in the complete intersection ${\mathbb{V}}(J)$. So both
$(l_{1}l_{4}-l_{2}l_{3})x_{0}$ and $(l_{1}l_{4}-l_{2}l_{3})x_{1}$ lie in $J$.
But $q_{3}=l_{5}x_{0}+l_{6}x_{1}$ for some linear forms $l_{5},l_{6}$, so
$(l_{1}l_{4}-l_{2}l_{3})q_{3}\in J$, which means
$(l_{1}l_{4}-l_{2}l_{3})\in(J:q_{3})$ and therefore $I\subset(J:q_{3})$. Both
$I$ and $(J:q_{3})$ have degree $3$, and since $I$ generates the ideal of a
curve, equality holds.

With the resolution of $(J:q_{3})$ in hand, it follows that $S/Q_{F}$ has
Betti table

$\begin{matrix}1&-&-&-&\cr-&3&-&-&\cr-&-&4&2&\cr\end{matrix}.$

In this case ${\mathbb{V}}(q_{3})$ intersects ${\mathbb{V}}(J:q_{3})$ in a
scheme of length $6$, and a subscheme of length $2$ is on $L$. Therefore,
${\mathbb{V}}(Q_{F})$ is the union of a line and zero-scheme of degree $4$.

The degree of $(J:q_{3})=2$:

In this case, $(J:q_{3})$ is the ideal of two skew lines or a double structure
on a line. The set of double structures on a codimension $2$ ideal of degree
$1$ is determined in [Engheta]. After a change of variables, the ideal is
either $\langle x_{0},x_{1}^{2}\rangle$ or $\langle
x_{0}^{2},x_{0}x_{1},x_{1}^{2},x_{0}l_{0}+x_{1}l_{1}\rangle$, with $l_{i}$
homogeneous and $\operatorname{codim}(x_{0},x_{1},l_{0},l_{1})=4$. From this,
it follows that for both possibilities the resolutions have Betti table

$\begin{matrix}1&-&-&-\cr-&4&4&1\cr\end{matrix}.$

Hence from the mapping cone construction, $S/Q_{F}$ would have Betti table

$\begin{matrix}1&-&-&-&-\cr-&3&-&-&-\cr-&-&5&4&1\cr\end{matrix}.$

By Lemma 4.2 this cannot occur.

In particular, the only two Betti tables possible for second row $(3,0,0)$ are
as claimed. ∎

######  Lemma 4.12.

If the Betti table of $S/Q_{F}$ has second row $(3,1,0)$, then $S/Q_{F}$ has
Betti table

$\begin{matrix}1&-&-&-&\cr-&3&1&-&\cr-&-&2&1&\cr\end{matrix}.$

In this case $Q_{F}$ is the ideal of the union of a plane conic and a
(possibly embedded) scheme of length $2$.

###### Proof.

If the first syzygy has rank $3$, then by Proposition 4.3, $Q_{F}$ has a
subideal $J$ generated by $2\times 2$ minors of a $2\times 3$ matrix, which
has $b_{23}=2$, contradiction. Therefore, the rank of the first syzygy is $2$.
This means $Q_{F}=J+\langle q_{3}\rangle$, where $J=\langle
l_{1}l_{3},l_{2}l_{3}\rangle$ and $(J:q_{3})=J$. Hence the Betti table of
$S/Q_{F}$ is

$\begin{matrix}1&-&-&-&\cr-&3&1&-&\cr-&-&2&1&\cr\end{matrix}.$

In this case the scheme defined by $Q_{F}$ is either a plane conic and two
points, for example $\langle x_{0}x_{1},x_{0}x_{2},q\rangle$, or a plane conic
with two embedded points, for example $\langle x_{0}^{2},x_{0}x_{1},q\rangle$.
∎

######  Lemma 4.13.

If the Betti table of $S/Q_{F}$ has second row $(3,2,0)$, then $S/Q_{F}$ has
Betti table

$\begin{matrix}1&-&-&\cr-&3&2&\cr\end{matrix}.$

In this case, $Q_{F}$ is generated by $2\times 2$ minors of a $2\times 3$
matrix of linear forms and is the ideal of a curve of degree $3$ that spans
${\mathbb{P}}^{3}$.

###### Proof.

Since there are only two linear syzygies, there is by Lemma 4.6 at least one
first linear syzygy of $Q_{F}$ having rank $3$. Therefore, by Proposition 4.3,
$Q_{F}$ is generated by the $2\times 2$ minors of a $2\times 3$ matrix and is
the ideal of a curve with Hilbert polynomial $3t+1$. ∎

####  4.2.3. Case $b_{12}=4$

When $F^{\perp}$ is a complete intersection $(2,2,2,2)$, then
$Q_{F}=F^{\perp}$.

######  Proposition 4.14.

Let $F$ be a quaternary quartic, then the following are equivalent

  1. (1)

$Q_{F}=F^{\perp}$

  2. (2)

$S/Q_{F}$ is Artinian.

  3. (3)

$Q_{F}$ contains a complete intersection $(2,2,2,2)$

Furthermore, the set of forms $F$ satisfying these conditions is irreducible.

###### Proof.

Clearly $(1)$ implies $(2)$. Furthermore, if $S/Q_{F}$ is Artinian, then a
general subset of four quadrics in $Q_{F}$ forms a complete intersection, so
$(2)$ implies $(3)$. Finally, if $Q_{F}$ contains a complete intersection $I$,
then $I(4)\subset S_{4}$ has codimension $1$, so coincides with $F^{\perp}$
for a quartic form $F$. Therefore $Q_{F}\subset I=F^{\perp}$.

The complete intersections form a dense open set in the Grassmannian
$Gr(4,S_{2})$, so an irreducible set. ∎

######  Proposition 4.15.

Let $F$ be a quaternary quartic with $b_{12}(A_{F})=4$, and assume $F$ is not
of type $[400]$. Then $Q_{F}$ is saturated and is the ideal of an apolar
scheme $Z$.

  1. (1)

If $F$ is of type $[420]$, then $S/Q_{F}$ has Betti table

$\begin{matrix}&&&&\cr 1&-&-&-&\cr-&4&2&-&\cr-&-&3&2&\cr\end{matrix}$

and $Q_{F}$ is the ideal of a scheme $Z$ of length $6$, the intersection of a
curve of degree $3$ defined by a determinantal net of quadrics, and another
quadric.

  2. (2)

If $F$ is of type $[430]$, then $S/Q_{F}$ has Betti table

$\begin{matrix}&&&&\cr 1&-&-&-&\cr-&4&3&-&\cr-&-&1&1&\cr\end{matrix}$

and $Q_{F}$ is the ideal of the union $Z$ of a line and a scheme of length $3$
that span ${\mathbb{P}}^{3}$.

  3. (3)

If $F$ is of type $[441]$, then $S/Q_{F}$ has Betti table

$\begin{matrix}&&&&\cr 1&-&-&-&\cr-&4&4&1&\cr\end{matrix}.$

If the second order syzygy has rank $3$, then $Q_{F}$ is the ideal of the
union of a plane conic and a (possibly embedded) point. Designate this as type
$[441a]$.

If the second order syzygy has rank $4$, then $Q_{F}$ is the ideal of the
union $Z$ of two skew lines or a double line on a smooth quadric surface.
Designate this as type $[441b]$.

The set of ideals $Q_{F}$, when $F$ is of type $[420]$ or $[430]$, form
irreducible components of the Hilbert scheme, while when $F$ is of type
$[441]$ the corresponding ideals $Q_{F}$ form two irreducible components of
the Hilbert scheme.

###### Proof.

The three cases follow from lemmas 4.16, 4.17 and 4.19 below. In the first two
cases a general form $F$ apolar to $Z$ is of type $[420]$ and type $[430]$
respectively, and the schemes $Z$ form irreducible components of the Hilbert
scheme. In the third case, the two subcases for $Z$ each correspond to
irreducible components of the Hilbert scheme, so the proposition follows. ∎

We say that a quaternary quartic $F$ is of type $[441a]$ when the second order
syzygy of $Q_{F}$ has rank $3$, and that a quaternary quartic $F$ is of type
$[441b]$ when the second order syzygy of $Q_{F}$ has rank $4$.

Note that if the Betti table of $S/Q_{F}$ has second row $(4,0,0)$, then
$Q_{F}$ is a $(2,2,2,2)$ complete intersection.

######  Lemma 4.16.

If the Betti table of $S/Q_{F}$ has second row $(4,2,0)$, then $S/Q_{F}$ has
Betti table

$\begin{matrix}1&-&-&-&\cr-&4&2&-&\cr-&-&3&2&\cr\end{matrix}$

and the scheme defined by $Q_{F}$ is a scheme of length $6$, the intersection
of a determinantal curve of degree $3$ and another quadric.

###### Proof.

By Lemma 4.5 not all first syzygies of $Q_{F}$ have rank $4$. If there is a
linear first syzygy of rank $3$, then by Proposition 4.3, there is a subideal
$J\subset Q_{F}$ such that $J$ is $2\times 2$ minors of a $2\times 3$ matrix
of linear forms, and $Q_{F}=J+\langle q_{4}\rangle$. Since $q_{4}$ is a non-
zero divisor on $J$ iff $(J:q_{4})=J$, the result is immediate in this case,
so suppose $q_{4}$ is a zero divisor, and consider the short exact sequence of
Equation 25, with $q_{3}$ replaced by $q_{4}$. Note that $(J:q_{4})$ has no
linear generator, for such a generator would yield an extra linear syzygy in
the resolution for $Q_{F}$, and the two linear syzygies from $J$ preclude
this. Since $\deg(J)=3$, $\deg(J:q_{4})\in\\{1,2\\}$. Since the associated
primes of $(J:q_{4})$ are a subset of the associated primes of $J$, all
associated primes of $(J:q_{4})$ are of codimension $2$. In particular, if
$\deg(J:q_{4})$ is $1$, then it contains a linear form, a contradiction, and
if $\deg(J:q_{4})$ is $2$ and $(J:q_{4})$ is prime, it is generated by a
linear form and irreducible quadric, again a contradiction. Finally, if
$\deg(J:q_{4})$ is $2$ but it is not prime, it is either a pair of reduced
lines, or a double structure on a line. In the first case, the lines cannot
meet (or there is a linear form in the idea), so the lines are skew, with
resolution below.

$\begin{matrix}1&-&-&-\cr-&4&4&1\cr\end{matrix}.$

If $(J:q_{4})$ is a double structure on a line, then as in the proof of Lemma
4.11, $(J:q_{4})$ is either $\langle x_{0},x_{1}^{2}\rangle$ or $\langle
x_{0}^{2},x_{0}x_{1},x_{1}^{2},x_{0}l_{0}+x_{1}l_{1}\rangle$. The first case
is impossible due to the existence of a linear generator. The second case has
resolution as above. Both possibilities yield a mapping cone resolution for
$S/Q_{F}$ of the form

$\begin{matrix}1&-&-&-&-\cr-&4&2&-&-\cr-&-&4&4&1\cr\end{matrix},$

which is precluded by Lemma 4.2. So a rank 3 linear syzygy implies that
$S/Q_{F}$ has Betti table

$\begin{matrix}1&-&-&-&\cr-&4&2&-&\cr-&-&3&2&\cr\end{matrix}.$

If there is no linear first syzygy of rank $3$, then $Q_{F}$ has two rank $2$
linear syzygies involving different pairs of generators, so is of the form
$\langle x_{0}l_{1},x_{1}l_{1},x_{2}l_{2},x_{3}l_{2}\rangle$, which has Betti
table

$\begin{matrix}1&-&-&-&-\cr-&4&2&-&-\cr-&-&4&4&1\cr\end{matrix}.$

This is precluded by Lemma 4.2. ∎

######  Lemma 4.17.

If the Betti table of $S/Q_{F}$ has second row $(4,3,0)$, then $S/Q_{F}$ has
Betti table

$\begin{matrix}1&-&-&-&\cr-&4&3&-&\cr-&-&1&1&\cr\end{matrix}.$

In this case $Q_{F}$ is the ideal of a a line and a (possibly embedded) scheme
of length $3$.

###### Proof.

We analyze by the rank of linear first syzygies. By Lemma 4.5, not all first
linear syzygies of $Q_{F}$ have rank $4$.

There is a rank 3 linear first syzygy on $Q_{F}$:

Then by Proposition 4.3, there must be a subideal $J\subseteq Q_{F}$ such that
$S/J$ has Betti table

$\begin{matrix}1&-&-&\cr-&3&2&\cr\end{matrix}.$

So $J$ is Cohen-Macaulay of degree $3$ and codimension $2$. Assume
$Q_{F}=J+\langle q_{4}\rangle$. Then $Q_{F}$ has a linear syzygy involving
$q_{4}$, and $(J:q_{4})$ is the ideal of a curve with one linear generator. If
$\deg(J:q_{4})=3$, then $(J:q_{4})=J$, so it cannot have a linear generator,
contradiction. If $\deg(J:q_{4})=1$, then $(J:q_{4})$ has two linear
generators, again a contradiction. So $\deg(J:q_{4})=2$ and $(J:q_{4})$ is a
$(1,2)$ complete intersection. Therefore $S/(J:q_{4})$ has Betti table

$\begin{matrix}1&1&-&\cr-&1&1&\cr\end{matrix},$

and, from the mapping cone, $S/Q_{F}$ has Betti table

$\begin{matrix}1&-&-&-&\cr-&4&3&-&\cr-&-&1&1&\cr\end{matrix}.$

The quadric $q_{4}$ intersects ${\mathbb{V}}(J)$ in a line, and the complete
intersection $(J:q_{4})$ in a scheme of length $4$. But the line meets the
complete intersection in a point, so $Q_{F}$ is the ideal of the union $Z$ of
a line and a possibly embedded scheme of length $3$.

There is no linear first syzygy of rank $\geq 3$ on $Q_{F}$:

By Lemma 4.7, $Q_{F}$ contains either a subideal $J_{1}$ with Betti table of
$S/J_{1}$

$\begin{matrix}1&-&-&\cr-&3&2&\cr\end{matrix}$

or a subideal $J_{2}$ with Betti table of $S/J_{2}$

$\begin{matrix}1&-&-&-&\cr-&3&3&1&\cr\end{matrix}.$

If $J_{1}\subseteq Q_{F}$, then by the same reasoning as the case with rank 3
syzygy above, $S/Q_{F}$ has Betti table

$\begin{matrix}1&-&-&-&\cr-&4&3&-&\cr-&-&1&1&\cr\end{matrix}.$

On the other hand, $J_{2}$ cannot be in $Q_{F}$, because the linear second
syzygy on $J_{2}$ would persist in the resolution of $Q_{F}$.

Therefore, the resolution of $Q_{F}$ with top row $(4,3,0)$ is as claimed. ∎

######  Lemma 4.18.

If $Q_{F}$ admits a second linear syzygy $f$ of order 3, then $I_{f}$ is the
ideal of a plane and a (possibly embedded) point $P$, such that all elements
of $Q_{F}$ vanish at $P$.

###### Proof.

The description of $I_{f}$ follows from Proposition 4.3. For the last
statement, change coordinates so that $P=(1:0:0:0)$ and $I_{f}$ is one of the
following:

  1. (1)

$I=\langle x_{0}x_{1},x_{0}x_{2},x_{0}x_{3}\rangle$

  2. (2)

$I=\langle x_{1}^{2},x_{1}x_{2},x_{1}x_{3}\rangle$

Let $q$ be a quadric $\in Q_{F}\setminus I$, and suppose $q$ does not vanish
at $P$. Then after modifying $q$ using elements from $I$ we can write $q$ as

$q=x_{0}^{2}+f(x_{1},x_{2},x_{3}),\mbox{ with }f\mbox{ quadratic}.$

Now if $I=\langle x_{0}x_{1},x_{0}x_{2},x_{0}x_{3}\rangle$ we see that
$x_{0}g\in I\subset Q_{F}$ for any polynomial $g$ in the variables
$x_{1},x_{2},x_{3}$, and therefore $x_{0}^{3}=qx_{0}-fx_{0}\in Q_{F}$.

It follows that all cubics and in consequence all quartics vanishing on the
plane $x_{0}=0$ are in $Q_{F}$. But this is impossible if $F$ is non-
degenerate. Similarly, if $I=\langle x_{1}^{2},x_{1}x_{2},x_{1}x_{3}\rangle$,
then after reducing modulo $I$, we may assume that
$q=x_{0}^{2}+ax_{0}x_{1}+x_{0}(x_{2},x_{3})$. Then

$x_{0}^{2}x_{1}=qx_{1}-ax_{0}x_{1}^{2}+x_{0}x_{1}l(x_{2},x_{3})\in Q_{F},$

but $\langle x_{0}^{2}x_{1}\rangle+I$ generates all cubics vanishing on the
plane $x_{1}=0$, and again we have a contradiction. ∎

######  Lemma 4.19.

If the Betti table of $S/Q_{F}$ has second row $(4,4,1)$, then the Betti table
of $S/Q_{F}$ is

$\begin{matrix}1&-&-&-&\cr-&4&4&1&\cr\end{matrix}.$

If the second order syzygy has rank $4$, then $Q_{F}$ is the ideal of two skew
lines or a double line in a smooth quadric surface. If the second order syzygy
has rank $3$, then $Q_{F}$ is the ideal of a plane conic and a, possibly
embedded, point.

###### Proof.

Let $f$ denote the second linear syzygy of $Q_{F}$. By Proposition 4.3 we have
the following two possibilities:

$f$ is rank $4$:

Then $I_{f}$ is the ideal of either two skew lines, or of a double line in a
smooth quadric surface, such that $S/I_{f}$ in both cases has Betti table

$\begin{matrix}1&-&-&-&\cr-&4&4&1&\cr\end{matrix}.$

In that case we must have $Q_{F}=I_{f}$.

$f$ is rank $3$:

Then $I_{f}$ is the ideal of a plane and a (possibly embedded) point $P$, and
$S/I_{f}$ has Betti table

$\begin{matrix}1&-&-&-&\cr-&3&3&1&\cr\end{matrix}.$

In that case $Q_{F}=I_{f}+\langle q\rangle$ for some quadric $q$ vanishing at
$P$ as seen in Lemma 4.18. It follows that $Q_{F}$ is the ideal of a plane
conic and a (possibly embedded) point.

Thus the possibilities are as claimed. ∎

####  4.2.4. Case $b_{12}=5$

######  Proposition 4.20.

Let $F$ be a quaternary quartic with $b_{12}(A_{F})=5$. Then the quadratic
ideal $Q_{F}$ is saturated and is the ideal of an apolar scheme $Z$.

  1. (1)

If $F$ is of type $[550]$, then $S/Q_{F}$ has Betti table

$\begin{matrix}&&&&\cr 1&-&-&-&\cr-&5&5&-&\cr-&-&-&1&\cr\end{matrix}$

and $Q_{F}$ is the ideal of a scheme $Z$ of length $5$ with no subscheme of
length $4$ in a plane.

  2. (2)

If $F$ is of type $[551]$, then $S/Q_{F}$ has the Betti table

$\begin{matrix}&&&&\cr 1&-&-&-&\cr-&5&5&1&\cr-&-&1&1&\cr\end{matrix}$

and $Q_{F}$ is the ideal of a scheme $Z$ of length $5$ with a subscheme of
length $4$ in a plane and no subscheme of length $3$ in a line.

  3. (3)

If $F$ is of type $[562]$, then $S/Q_{F}$ has the Betti table

$\begin{matrix}&&&&\cr 1&-&-&-&\cr-&5&6&2&\cr\end{matrix},$

and $Q_{F}$ is the ideal of the union $Z$ of a line and a (possibly embedded)
scheme of length $2$ such that that $Z$ spans ${\mathbb{P}}^{3}$.

For each of the three types, the corresponding set of ideals $Q_{F}$ form an
irreducible component of the Hilbert scheme.

###### Proof.

The three cases follow from the lemmas 4.21 and 4.22 below. In each case, the
schemes $Z$ corresponding to $Q_{F}$ form an irreducible component of the
Hilbert scheme. ∎

We analyze $Q_{F}$ in terms of its second linear syzygies.

######  Lemma 4.21.

If $Q_{F}$ has $b_{12}=5$ and admits a second linear syzygy then $Q_{F}$ is
one of the following:

  1. (1)

$Q_{F}$ is the ideal of a scheme $Z$ of length $5$, with a complete
intersection subscheme of length $4$ in a plane. Then $S/Q_{F}$ has Betti
table

$\begin{matrix}&&&&\cr 1&-&-&-&\cr-&5&5&1&\cr-&-&1&1&\cr\end{matrix}.$

Furthermore, no subscheme of length 3 in $Z$ is contained in a line.

  2. (2)

$Q_{F}$ is the ideal of a line and a (possibly embedded) subscheme of length
$2$, and $S/Q_{F}$ has Betti table

$\begin{matrix}1&-&-&-&\cr-&5&6&2&\cr\end{matrix}.$

###### Proof.

By Proposition 4.3 and Lemma 4.5 we have two possibilities. $b_{34}=1$ and the
second linear syzygy $f$ of $Q_{F}$ has rank $3$ or $4$. If $f$ has rank $3$,
then $I_{f}$ is a subideal of $Q_{F}$ which is the defining ideal of a plane
and a point $P$. By Lemma 4.18 the remaining quadrics must pass through the
point $P$ and when restricted to the plane they vanish on either a complete
intersection scheme of length 4 or on a line and a point. In both cases the
ideal will be saturated and as in the assertion. Note that in the case where
$Q_{F}$ defines a finite scheme $Z$, since it is defined by quadrics, this
finite scheme cannot contain any subscheme of length $3$ contained in a line.
If $b_{34}=1$ and the unique (up to scaling) second linear syzygy $f$ is of
rank 4, then by Proposition 4.3, the ideal $Q_{F}$ contains the ideal of two
skew lines. Adding one more quadric we obtain either an ideal defining four
points or the ideal of a line and two points.

In the first case, consider a subideal $Q^{\prime}$ of $Q_{F}$ consisting of
quadrics passing through two general additional points coplanar with three of
the four points. The subideal $Q^{\prime}$ is generated by three quadrics and
vanishes on a plane and an additional point. But $Q^{\prime}$ has a second
linear syzygy of rank $3$ which is a contradiction with $b_{34}=1$ and $f$ of
rank 4.

In the second case the Betti table of $S/Q_{F}$ is

$\begin{matrix}1&-&-&-&\cr-&5&6&2&\cr\end{matrix}.$

This contradicts $b_{34}=1$, hence the case $b_{34}=1$ with the second linear
syzygy of rank 4 never occurs. ∎

######  Lemma 4.22.

If $F$ is a quaternary quartic such that $A_{F}$ is of type $[550]$, then
$Q_{F}$ is the ideal of a scheme of length $5$, with no subscheme of length
$4$ lying in a plane.

###### Proof.

The Betti table of $A_{F}$ is

$\begin{matrix}1&-&-&-&-&\cr-&5&5&-&-&\cr-&1&-&1&-&\cr-&-&5&5&-&\cr-&-&-&-&1&\cr\end{matrix}.$

Hence $F^{\perp}$ is generated by $Q_{F}$ and a cubic form. Notice that the
second syzygy of degree 5 must be a syzygy only on the quadrics. Furthermore,
the first syzygies of degree $5$ are all Koszul syzygies between the cubic
form and the quadrics. Therefore $S/Q_{F}$ has Betti table

$\begin{matrix}1&-&-&-\cr-&5&5&-\cr-&-&-&1\cr\end{matrix}$

and $I$ is the ideal of a finite scheme $Z$ of length 5.

If $Z$ contains a subscheme of length $4$ in a plane, then at least a
$3$-dimensional space of quadrics in $Q_{F}$ would vanish on the plane of this
subscheme. But then these quadrics have a second order linear syzygy,
contradicting the fact that $Q_{F}$ does not. ∎

####  4.2.5. Case $b_{12}=6$

######  Proposition 4.23.

Let $F$ be a quaternary quartic with $b_{12}(A_{F})=6$. Then $F$ is of type
$[683]$ and the quadratic ideal $Q_{F}$ is saturated. The Betti table of
$S/Q_{F}$ is

$\begin{matrix}1&-&-&-\cr-&6&8&3\cr\end{matrix}$

and $Q_{F}$ is the ideal of a finite apolar scheme $Z$ of length $4$. When $F$
is of type $[683]$, the corresponding set of ideals $Q_{F}$ form an
irreducible component of the Hilbert scheme.

###### Proof.

It follows from the proof of Theorem 2.2 of [SSY] that when $b_{12}=6$ the top
two rows of the Betti table of $S/Q_{F}$ are

$\begin{matrix}1&-&-&-&-\cr-&6&8&3&-\cr\end{matrix}.$

We wish to show that this is the entire Betti table for $S/Q_{F}$. There are a
number of ways to do this. Here we use a generic initial ideal argument. After
a general change of coordinates, in graded reverse lex order the initial
ideal, $\operatorname{gin}(Q_{F})$, will be a strongly stable ideal with 6
quadratic generators (see Appendix B for the terminology and basic results
used here). There are two possibilities:

  1. (1)

The quadradic part of $\operatorname{gin}(Q_{F})$ is
$(x_{0},x_{1},x_{2})^{2}$.

  2. (2)

$\operatorname{gin}(Q_{F})$ contains a monomial $x_{i}x_{3}$.

In the second case, the saturation of $Q_{F}$ is $(Q_{F}:x_{3})$, which will
contain a linear form $L$, and therefore $Q_{F}$ contains
$(x_{0},x_{1},x_{2},x_{3})\cdot L$, which contradicts the known quadratic
linear strand of the Betti table of $S/Q_{F}$ above.

Therefore, we must be in the setting of the first case, where the degree $2$
component of $\operatorname{gin}(Q_{F})$ is $(x_{0},x_{1},x_{2})^{2}$. To
compute a Gröbner basis for $Q_{F}$, notice that there are eight minimal
S-pairs on the six generators of $Q_{F}$ (in these general coordinates), which
all occur in degree $3$. Each of these must reduce to 0, since there are eight
linear syzygies on the generators of $Q_{F}$. Therefore the six generators of
$Q_{F}$ form a Gröbner basis. As the Betti table of $S/Q_{F}$ is pointwise at
most the Betti table of $S/\operatorname{gin}(Q_{F})$, this forces the Betti
table of $S/Q_{F}$ to be the above table.

Therefore, $Q_{F}$ defines a degree 4 zero scheme in ${\mathbb{P}}^{3}$, and
this scheme spans ${\mathbb{P}}^{3}$. Schemes of length $4$ that span
${\mathbb{P}}^{3}$ form an open dense subset of the Hilbert scheme
$\operatorname{Hilb}^{4}({\mathbb{P}}^{3})$, and the proposition follows. ∎

##  5\. Rank and powersum presentations of quartics

In this section we determine the rank $r=\operatorname{r}(F)$ and the variety
of powersum presentations $VSP(F,r)$ of the general form $F$ of each Betti
table type.

Theorem 2.2 of [SSY] proves that a graded Artin Gorenstein algebra $A$ of
regularity and codimenson $4$ has one of $16$ Betti tables; these Betti tables
are reproduced as Table 6 and Table 7 in Appendix A. Recall from Definition
1.1 that ${\mathcal{F}}_{B}\subset{\mathbb{P}}^{34}$ is the set of forms $F$
for which $A_{F}$ has Betti table $B$, labelled as in the Appendix A. We will
analyze the sets ${\mathcal{F}}_{B}$ via the set of quadratic ideals
$\\{Q_{F}|F\in{\mathcal{F}}_{B}\\}$. These sets of quadratic ideals were
identified in Section 4 via their Betti tables, and by Definition 1.2,
${\mathcal{G}}_{B}$ denotes the set of ideals $Q_{F}$ for quaternary quartics
$F$ such that $S/Q_{F}$ has Betti table $B$.

In Section 4 we showed that the sets ${\mathcal{G}}_{B}$ are irreducible,
except for one Betti table. The set ${\mathcal{G}}_{[441]}$ has two components
that we denote ${\mathcal{G}}_{[441a]}$ and ${\mathcal{G}}_{[441b]}$, which
are distinguished by the rank of a second linear syzygy. With these
distinctions the quadratic ideals form $16$ different irreducible components.
We begin by discussing the relation between the sets ${\mathcal{F}}_{B}$ and
${\mathcal{G}}_{B^{\prime}}$, when the latter is an irreducible component of
the set of quadratic ideals that contains an ideal
$Q_{F}\in{\mathcal{G}}_{B^{\prime}}$ for some quartic form $F$ for which
$A_{F}$ has Betti table $B$.

For two sets ${\mathcal{F}}_{B}$ we distinguish some subsets. Recall that the
cactus rank $\operatorname{cr}(F)$ of a form $F$ is the shortest length of a
subscheme apolar to $F$.

######  Definition 5.1.

We set

${\mathcal{F}}_{[300a]}=\\{F|Q_{F}\in{\mathcal{G}}_{[300ab]},\operatorname{cr}(F)=8\\},{\mathcal{F}}_{[300b]}=\\{F|Q_{F}\in{\mathcal{G}}_{[300ab]},\operatorname{cr}(F)=7\\},$
${\rm and}\;\;{\mathcal{F}}_{[300c]}=\\{F|Q_{F}\in{\mathcal{G}}_{[300c]}\\},$

and we set

${\mathcal{F}}_{[441a]}=\\{F|Q_{F}\in{\mathcal{G}}_{[441a]}\\}\;\;{\rm
and}\;\;{\mathcal{F}}_{[441b]}=\\{F|Q_{F}\in{\mathcal{G}}_{[441b]}\\}.$

######  Lemma 5.2.

${\mathcal{F}}_{[300]}={\mathcal{F}}_{[300a]}\sqcup{\mathcal{F}}_{[300b]}\sqcup{\mathcal{F}}_{[300c]}$

and

${\mathcal{F}}_{[441]}={\mathcal{F}}_{[441a]}\sqcup{\mathcal{F}}_{[441b]}.$

###### Proof.

The last part follows from the distinction of the sets of quadratic ideals
${\mathcal{G}}_{[441]}={\mathcal{G}}_{[441a]}\sqcup{\mathcal{G}}_{[441b]}$.
For the first part, the ideals in ${\mathcal{G}}_{[300ab]}$ are the complete
intersections $(2,2,2)$. The forms $F$ with $Q_{F}$ a complete intersection
$(2,2,2)$ and cactus rank $7$, are, by apolarity, precisely those containing
an ideal $J\subset F^{\perp}$ with $J\in\operatorname{Hilb}_{7}$ and no ideal
$J^{\prime}\subset F^{\perp}$ with
$J^{\prime}\in\operatorname{Hilb}_{k},\;\;k<7$. Any ideal
$J^{\prime}\in\operatorname{Hilb}_{k},\;\;k<7$ would have ${\rm
dim}J^{\prime}(2)\geq 4$, so $J^{\prime}\subset F^{\perp}$ would imply
$b_{12}(A_{F})\geq 4$ against the assumption. So

${\mathcal{F}}_{[300b]}=\\{F|Q_{F}\in{\mathcal{G}}_{[300ab]},F^{\perp}\supset
J\;{\rm with}\;J\in\operatorname{Hilb}_{7}\\}.$

Clearly, any form $F$ in the complement of ${\mathcal{F}}_{[300b]}$ in the set

$\\{F|Q_{F}\in{\mathcal{G}}_{[300ab]}\\}$

has cactus rank $8$. In fact, $Q_{F}\in\operatorname{Hilb}_{8}$, and any
$J^{\prime}\subset F^{\perp}$ with
$J^{\prime}\in\operatorname{Hilb}_{k},\;\;k<8$ would contain the complete
intersection $Q_{F}$. Therefore, this complement coincides with
${\mathcal{F}}_{[300a]}$ and the lemma follows. ∎

######  Proposition 5.3.

The following $19$ sets of quaternary quartics are all irreducible:

  1. (1)

The $14$ sets ${\mathcal{F}}_{B}$ for a Betti table $B$ different from $[300]$
and $[441]$.

  2. (2)

The three sets
${\mathcal{F}}_{[300a]},{\mathcal{F}}_{[300b]},{\mathcal{F}}_{[300c]}$.

  3. (3)

The two sets ${\mathcal{F}}_{[441a]},{\mathcal{F}}_{[441b]}$.

###### Proof.

We first set up the natural algebraic incidence between ideals $Q_{F}$ and
forms $F$.

######  Lemma 5.4.

Let $k>0$ and consider the Grassmannian $\mathbf{Gr}(k,S_{2})$ of $k$
dimensional spaces of quadrics in $S$. The set
$X=\\{(Q,[F])\in\mathbf{Gr}(k,S_{2})\times{\mathbb{P}}^{34}\mid q\circ
F=0\mbox{ for all }q\in
Q\\}\subset\mathbf{Gr}(k,S_{2})\times{\mathbb{P}}^{34}$ is closed.

###### Proof.

Let $\Gamma\subseteq S_{2}$ be a $(10-k)$-subspace of $S_{2}$. Assume
$U_{\Gamma}\subset\mathbf{Gr}(k,S_{2})$ is the subset of $k$-planes which only
intersect $\Gamma$ at $0$. Then $U_{\Gamma}\simeq{\mathbb{C}}^{k(n-k)}$ is an
open subset in $\mathbf{Gr}(k,S_{2})$. Assume the local coordinates of
$(U_{\Gamma})$ is $\\{a_{ij}\\}_{1\leq i\leq k,1\leq j\leq n-k}$, that is, a
$k$-plane in $U_{\Gamma}$ can be written as the row space of

(26) $\begin{bmatrix}1&&&a_{1,1}&\dots&a_{1,n-k}&\\\
&\ddots&&\vdots&&\vdots&\\\ &&1&a_{k,1}&\dots&a_{k,n-k}&\end{bmatrix}.$

Therefore $X\cap(U_{\Gamma}\times{\mathbb{P}}^{34})$ is given by zero locus of
polynomials in $a_{i,j}$ and the coordinates of ${\mathbb{P}}^{34}$, hence
closed on $(U_{\Gamma}\times{\mathbb{P}}^{34})$. Note that this argument works
on every patch of an open affine cover $\\{U_{\Gamma}\\}$ of
$\mathbf{Gr}(k,S_{2})$, so $X$ is closed. ∎

In the notation of Lemma 5.4 we may consider the projection maps
$\pi_{1}:X\to\mathbf{Gr}(k,S_{2})$ and $\pi_{2}:X\to{\mathbb{P}}^{34}$. Then
$\pi_{2}(X)=\\{F\in{\mathbb{P}}^{34}\mid b_{12}(S/Q_{F})\geq k\\}$. Assume
$F\in\pi_{2}(X)$. Then the fiber of $\pi_{2}$ over $F$ is
$\\{Q\in\mathbf{Gr}(k,S_{2})\mid Q\subseteq Q_{F}\\}$. In particular, when
$b_{12}(S/Q_{F})=k$, the fiber has only one point. On the other hand, the
fiber of $\pi_{1}$ over a $Q\in\mathbf{Gr}(k,S_{2})$ is the set
$\\{[F^{\prime}]\in{\mathbb{P}}^{34}\mid Q\subseteq Q_{F^{\prime}}\\}$. Thus
there is, for each $b_{12}=b_{12}(S/F^{\perp})=b_{12}(S/Q_{F})$ a regular map

$\varphi:\\{F\in{\mathbb{P}}^{34}|b_{12}(S/F^{\perp})=k\\}\to
G(k,S_{2}),\quad[F]\mapsto[Q_{F}].$

We now let ${\mathcal{G}}$ be an irreducible set of saturated quadratic ideals
in $S$, all with the same Betti table. Let $I\in{\mathcal{G}}$ and let
$B^{\prime}$ be the Betti table of $S/I$. Let $I_{4}\subset S_{4}$ be the
space of quartic forms in $I$. Each quaternary quartic form $F\in R_{4}$
defines a hyperplane $F^{\perp}\subset S_{4}$. If $I_{4}\subset F^{\perp}$,
i.e. $F\in(I_{4})^{\perp}\subset S_{4}^{*}=R_{4}$, then $I\subset Q_{F}$ since
both are saturated. If the latter inclusion is strict, the inclusion
$I_{4}\subset(Q_{F})_{4}$ is strict, therefore $I=Q_{F}$ for the general
$F\in(I_{4})^{\perp}$. In particular this is an irreducible set of dimension
depending only on the common Betti table $B^{\prime}$ of $S/I$ for ideals
$I\in{\mathcal{G}}$. The restriction of $\varphi$ is therefore a map

$\\{F|Q_{F}\in\mathcal{G}_{B^{\prime}}\\}\to\mathcal{G}_{B^{\prime}}\quad
F\mapsto Q_{F}$

with irreducible fibers of the same dimension and the proposition follows.

Finally, the ideals in ${\mathcal{G}}_{[300ab]}$ are the complete
intersections $(2,2,2)$ and form an irreducible set. Therefore the set of
forms $F$ with $Q_{F}$ a complete intersection is also irreducible. Now, the
set of ideals $J$ of subschemes of length $7$ is irreducible and those ideals
$J$ that additionally contain a complete intersection $(2,2,2)$ form an dense
subset ${\mathcal{G}^{\prime}}\subset\operatorname{Hilb}_{7}$. In particular,
as in the above proof for irreducible sets of quadratic ideals, the set of
forms ${\mathcal{F}}_{[300b]}\subset{\mathbb{P}}^{34}$, i.e. the set of forms
$F$ with an ideal $J\subset F^{\perp}$ and $J\in{\mathcal{G}^{\prime}}$, is
irreducible.

By Lemma 5.2, the set ${\mathcal{F}}_{[300a]}$ is the complement of
${\mathcal{F}}_{[300b]}$ in the set of forms $F$ with $Q_{F}$ a complete
intersection $(2,2,2)$. The latter set is irreducible, while
${\mathcal{F}}_{[300a]}$ clearly is a dense subset, so
${\mathcal{F}}_{[300a]}$ is also irreducible. ∎

For each of the irreducible subsets of ${\mathbb{P}}^{34}$ of Proposition 5.3,
we compute the dimension of ${\mathcal{F}}_{B}$, and for a general quartic
$F\in{\mathcal{F}}_{B}$ we determine the rank $\operatorname{r}(F)$ and
$VSP(F,r)$, and characterize as in Section 3, the general set of points
computing $\operatorname{r}(F)$.

######  Proposition 5.5.

Let $F$ be a quaternary quartic that is not a cone. Then $F$ belongs to one of
$19$ disjoint irreducible sets listed in tables 6 and 7 with the given
dimension. For each set the table gives, for a general form $F$ in the set,
the rank $r=r(F)$, the variety $VSP(F,r)$, and the Betti table for a general
set $\Gamma$ of $r$ points apolar to $F$. In each case, the set of points
$\Gamma$ impose independant conditions on quartics.

###### Proof.

By Theorem 2.2 of [SSY] we know that any quartic $F$ which is not a cone has
one of $16$ Betti tables, and hence belong to one of the $19$ sets of
Proposition 5.3.

The rank of a form $F$ has the following lower bound:

######  Lemma 5.6.

Let $b_{12}(F)$ be the dimension of the space of quadratic forms in
$F^{\perp}$, i.e. the space of generators of $Q_{F}$, and let $r(F)$ be the
rank of $F$. Then $r(F)\geq 10-b_{12}(F)$.

###### Proof.

If $\Gamma$ is a set of $r$ points apolar to $F$, then the ideal $I_{\Gamma}$
of $\Gamma$ is contained in $F^{\perp}$. In particular the quadratic part of
$I_{\Gamma}$ must be contained in $Q_{F}$. But $r$ points impose at most $r$
conditions on quadrics, so $10-r\leq{\rm dim}I_{\Gamma}(2)\leq b_{12}(F)$,
which implies

$r\geq 10-b_{12}(F).$

∎

For each of the $19$ irreducible sets, we now argue one by one and start with
forms with Betti table in Table 6 of Section A.

Betti table B | $\operatorname{r}(F)$ | VSP(F,r) |  $\Gamma$ from Table 1 | $\dim({\mathcal{F}}_{B})$
---|---|---|---|---
[683] | $4$ | one point | four points, not in a plane | $15$
[550] | $5$ | one point | five points, no four in a plane | $19$
[400] | $8$ | ${\mathbb{P}}^{3}$ | a $(2,2,2)$ CI | $24$
[320] | $7$ | ${\mathbb{P}}^{1}$ | seven points on a twisted cubic | $24$
[300a] | $8$ | one point | a $(2,2,2)$ CI | $28$
[300b] | $7$ | one point | seven points in a $(2,2,2)$ CI | $27$
[300c] | $7$ | ${\mathbb{P}}^{1}$ | seven points, three on a line | $24$
[200] | $8$ | two points | eight points in a $(2,2)$ CI | $31$
[100] | $9$ | K3 surface | nine points in a quadric | $33$
[000] | $10$ | Fivefold | ten points, not on quadric | $34$
Table 3. VSP for Betti tables from Table 6

Type $[683]$:

The quadratic ideal $Q_{F}$ is, by Proposition 4.23, the ideal of a scheme of
length $4$, i.e. four points for a general $F$. So the rank of $F$ is at most
$4$ and therefore by Lemma 5.6 equal to $4$. Furthermore, the ideal of any
four points computing the rank of $F$ contains a $6$-dimensional space of
quadrics in $F^{\bot}$, i.e. all of $Q_{F}$, so $VSP(F,4)$ is one point. The
dimension of the family of $4$-tuples of points in ${\mathbb{P}}^{3}$ is
$4\times 3=12$ and each $4$-tuple of points span a ${\mathbb{P}}^{3}$ in
${\mathbb{P}}^{34}$, so there is an irreducible $12+3=15$-dimensional family
of forms $F$ with this Betti table in ${\mathbb{P}}^{34}$.

Type $[550]$:

The quadratic ideal $Q_{F}$ is, by Proposition 4.20, the ideal of a scheme of
length $5$, i.e. five points for a general $F$. So $r(F)\leq 5$ and the ideal
generated by these quadrics is the only ideal of five points contained in
$F^{\bot}.$

By Lemma 5.6 the rank is least $5$, so $r(F)=5$ and $VSP(F,5)$ is a point. The
dimension of the family of $5$-tuples of points in ${\mathbb{P}}^{3}$ is
$5\times 3=15$, each $5$-tuple of points span a ${\mathbb{P}}^{4}$ in
${\mathbb{P}}^{34}$, so there is an irreducible $15+4=19$-dimensional family
of forms $F$ with this Betti table in ${\mathbb{P}}^{34}$.

Type $[400]$:

$F^{\bot}$ is a complete intersection of quadrics $(2,2,2,2)$. Any
$3$-dimensional subspace of quadrics in $F^{\bot}$ generate the ideal of a
scheme of length $8$, while no ideal of fewer points are contained in the
complete intersection. Therefore $F$ has rank $8$ and
$VSP(F,8)={\mathbb{P}}^{3}$.

The Grassmannian $G(4,10)$ of $4$-dimensional spaces of quadrics in
${\mathbb{P}}^{3}$ has dimension $4\times 6=24$ and parameterizes $(2,2,2,2)$
complete interesections in ${\mathbb{P}}^{3}$, yielding an irreducible
$24$-dimensional family of forms $F$ with this Betti table in
${\mathbb{P}}^{34}$.

Type $[320]$:

The quadratic ideal $Q_{F}$ is, by Proposition 4.10, a determinantal net, i.e.
for a general $F\in{\mathcal{F}}_{[320]}$, the ideal a twisted cubic curve $C$
that is apolar to $F$. By Lemma 3.1 there is a pencil of subschemes of length
$7$ on $C$ that is apolar to $F$. The ideal of any seven points contain a
$3$-dimensional space of quadrics, so if the ideal is contained $F^{\bot}$ the
seven points are on the twisted cubic curve. By Lemma 5.6, $F$ has rank $7$
and $VSP(F,7)={\mathbb{P}}^{1}$. The dimension of the family of twisted cubic
curves in ${\mathbb{P}}^{3}$ is $12$. Each twisted cubic $C$ is mapped by
$v_{4}$ to a rational normal curve $v_{4}(C)$ of degree $12$ that spans a
${\mathbb{P}}^{12}$ in ${\mathbb{P}}^{34}$, so there is an irreducible
$12+12=24$-dimensional family of forms $F$ with this Betti table in
${\mathbb{P}}^{34}$. In Theorem 2.7 of [Boij], Boij computes the dimension of
components of $\operatorname{Gor}(H)$ determined by $s$ points on a rational
normal curve in ${\mathbb{P}}^{n}$, but types $[320]$ fall outside the range
where his result applies.

Type $[300a]$:

The quadratic ideal $Q_{F}$ define, by Proposition 4.10, a complete
intersection $Z$ of length $8$. When $Z$ consists of eight distinct points and
no cubic in $F^{\bot}$ vanishes in any seven points, but not all eight in $Z$,
then $r=8$ and $Z$ is the only set of eight points apolar to $F$, so
$VSP(F,8)=\\{1\;point\\}$. The dimension of the family of complete
intersections $(2,2,2)$ in ${\mathbb{P}}^{3}$ is $3\times 7=21$, which is the
dimension of $G(3,10)$ the Grassmannian of nets of quadrics in
${\mathbb{P}}^{3}$. Each complete intersection spans a ${\mathbb{P}}^{7}$ in
${\mathbb{P}}^{34}$, so there is an irreducible $21+7=28$-dimensional family
of forms $F$ of rank $8$ with this Betti table in ${\mathbb{P}}^{34}$.

Type $[300b]$:

The quadratic ideal $Q_{F}$ define, by Proposition 4.10, a complete
intersection $Z$ of length $8$. One of the cubic generators in $F^{\bot}$
vanishes on a subscheme $Z^{\prime}$ of $Z$ of length $7$, but not on $Z$.
When $Z^{\prime}$ consists of seven distinct points, $r(F)=7$ and
$VSP(F,7)=\\{\mbox{one point}\\}$. In fact, the rank is at most $7$, while, by
Lemma 5.6, the rank is at least $7$, so equality follows. If $F$ is in the
span of two subset of seven among the eight points $Z$, then it lies in the
span of the common six points of the two subset and the rank is at most $6$.
So there is only one point in $VSP(F,7)$.

The dimension of the family of $7$-tuples of points in ${\mathbb{P}}^{3}$ is
$7\times 3=21$ and each $7$-tuple of points span a ${\mathbb{P}}^{6}$ in
${\mathbb{P}}^{34}$, so there is an irreducible $21+6=27$-dimensional family
of forms $F$ of rank $7$ with this Betti table in ${\mathbb{P}}^{34}$.

Type $[300c]$:

Included in boundary of the family $[300b]$ is the family $[300c]$ of quartic
forms of rank $7$ that are apolar to the set of three collinear points and
four points in general position, see Proposition 4.10. In this case $Q_{F}$ is
a not a complete intersection; it is the ideal of a line and a scheme of
length $4$, i.e. for a general $F\in{\mathcal{F}}_{[300c]}$ a line $L$ and a
set $Z$ of four general points. Forms $F$ of this type has a unique
decomposition $F=F_{1}+F_{2}$, where $F_{1}$ is apolar to $L$ while $F_{2}$ is
apolar to $Z$:

$[F]\in\langle v_{4}(L\cup Z)\rangle\subset{\mathbb{P}}^{34},$

so there are unique forms

$[F_{1}]\in\langle v_{4}(L)\rangle\;\;{\rm and}\;\;[F_{2}]\in\langle
v_{4}(Z)\rangle$

such that

$[F]\in\langle[F_{1}],[F_{2}]\rangle.$

We may assume that $F_{1}$ is a general form apolar to $L$, so by Lemma 3.1
$F_{1}$ has rank $3$ and $VSP(F_{1},3)={\mathbb{P}}^{1}$, while $F_{2}$ is of
type $[683]$ so $VSP(F_{2},4)=\\{point\\}$. Therefore also
$VSP(F,7)={\mathbb{P}}^{1}$. The dimension of the family of lines is $4$, so
the set of $5$-tuples (line + four points) is $4+3+3+3+3=16$-dimensional. Each
$5$-tuple spans a ${\mathbb{P}}^{8}$ in ${\mathbb{P}}^{34}$, so there is an
irreducible $16+8=24$-dimensional family of forms $F$ with this Betti table in
${\mathbb{P}}^{34}$.

Type $[200]$:

The quadratic ideal $Q_{F}$ is, by Proposition 4.8, a complete intersection
$2,2)$, so for a general $F\in{\mathcal{F}}_{[200]}$ it is the ideal of an
elliptic quartic curve $C$ that is apolar to $F$. By Lemma 3.1, $F$ is apolar
to two sets of eight points on $C$. By Lemma 5.6, the rank of $r(F)=8$ and
$VSP(F,8)=\\{\mbox{two points}\\}$. The family of elliptic quartics in
${\mathbb{P}}^{3}$ has dimension $16$. Each elliptic quartic curve $C$ is
mapped by $v_{4}$ to an elliptic normal curve $v_{4}(C)$ of degree $16$
spanning a ${\mathbb{P}}^{15}$ in ${\mathbb{P}}^{34}$, yielding an irreducible
$16+15=31$-dimensional family of forms $F$ with this Betti table in
${\mathbb{P}}^{34}$.

Type $[100]$:

Let $Q$ be the unique quadric in $F^{\bot}$. By Lemma 5.6 the rank of $F$ is
at least $9$. Now, any set of nine points apolar to $F$ is necessarily
contained in $Q$. The $4$-uple embedding $v_{4}(Q)$ of $Q$ spans a
${\mathbb{P}}^{24}\subset{\mathbb{P}}^{34}$. There is a $18$-dimensional
family of $9$-tuples of points on the quadric that each span a
${\mathbb{P}}^{8}$ in ${\mathbb{P}}^{24}$, so one expects that $F$ is apolar
to a surface of $9$-tuples of points. We show that $VSP(F,9)$ is a K3 surface
for a general form $F$ of type $[100]$, in Section 5.1.

Since $F$ is apolar to a quadric, the middle catalecticant matrix has rank
$9$. A rank $1$ catalecticant matrix is the catalecticant of a rank $1$
quartic form, so the determinant of the catalecticant defines the closure of
the set of forms $F$ of rank $9$, i.e. forms with a Betti table of type
$[100]$. Since the determinant is irreducible we conclude that there is an
irreducible $33$-dimensional family of forms with this Betti table in
${\mathbb{P}}^{34}$.

Type $[000]$:

If $F^{\bot}$ contains no quadric, the rank is at least $10$ by Lemma 5.6. The
family of forms of rank $9$ is dense in a hypersurface, so the general
$F\in{\mathbb{P}}^{34}$ must have rank $10$. By dimension count $VSP(F,10)$ is
a fivefold for a general $F$, and the rank is computed by ten points that are
not in any quadric surface. One can prove that a double quadric have rank
$10$, the following question arise.

######  Question 5.7.

Describe VSP(2Q,10) with $Q\subset\mathbb{P}^{3}$ a quadric.

Note that in this case there is an open orbit of OG(4). Recall that the case
of variety of sums of powers of the double quadric curve give rise to the
Mukai-Umemura example.

Next, we consider the Betti tables in [SSY]*Table 2, (see Table 7 in Appendix
A).

Betti table B | $r$ | VSP(F,r) |  $\Gamma$ from Table 1 | $\dim({\mathcal{F}}_{B})$
---|---|---|---|---
[562] | $5$ | ${\mathbb{P}}^{1}$ | five points, three collinear | $16$
[551] | $5$ | one point |  [1pt]five points, four coplanarand no three collinear | $18$
[441a] | $6$ | ${\mathbb{P}}^{1}$ | six points, five coplanar | $20$
[441b] | $6$ | ${\mathbb{P}}^{1}\times{\mathbb{P}}^{1}$ |  [1pt]six points, three oneach of two skew lines | $17$
[430] | $6$ | ${\mathbb{P}}^{1}$ | six points, three collinear | $20$
[420] | $6$ | one point | six points, no five coplanar | $23$
[331] | $7$ | Fano threefold | seven points, six coplanar | $21$
[310] | $7$ | ${\mathbb{P}}^{1}$ | seven points, five coplanar | $24$
[210] | $8$ | Fano threefold $\cup\;{\mathbb{P}}^{1}\times{\mathbb{P}}^{1}$ |  [1pt]eight points, six coplanar orfive coplanar, three collinear | $25$
Table 4. VSP for Betti tables from Table 7

Type $[562]$:

For a general $F\in{\mathcal{F}}_{[562]}$ the quadratic ideal $Q_{F}$ is, by
Proposition 4.20, the ideal of $\Gamma$: the union of a line $L$ and two
points $p_{1},p_{2}$ that altogether span ${\mathbb{P}}^{3}$. So this $\Gamma$
is apolar to $F$. In the $4$-uple embedding of
$v_{4}:{\mathbb{P}}^{3}\to{\mathbb{P}}^{34}$, this $\Gamma$ is reembedded as
the union of a rational normal quartic curve $v_{4}(L)$ and two points
$v_{4}(p_{1}),v_{4}(p_{2})$, and by apolarity $[F]$ is a point in the span of
this union.

In fact, it lies in the span of the two points and a unique point $[F_{1}]$ in
the span of the quartic curve $v_{4}(L)$. The form $F_{1}$ is apolar to the
line $L$, so by Lemma 3.1, $F_{1}$ has rank $2$, or it has rank $3$ and
$VSP(F_{1},3)={\mathbb{P}}^{1}$. If $F_{1}$ has rank $2$, then $F$ would have
rank $4$, which is impossible by Lemma 5.6, so $F_{1}$ has rank $3$. The ideal
of any five points apolar to $F$ is contained in at least five quadrics, i.e.
in $\Gamma$, so we conclude that also $VSP(F,5)={\mathbb{P}}^{1}$. Now, the
image $v_{4}(\Gamma)$ in ${\mathbb{P}}^{34}$ spans a ${\mathbb{P}}^{6}$, so
for each choice of a line and two points in ${\mathbb{P}}^{3}$ there is a
$6$-dimensional family of forms $F$ of this type.

Notice the $\Gamma$ is unique for a general $F$ apolar to it. Counting the
dimension of the family of schemes $\Gamma$ we get $4$ for the line, and $3$
for each of the points, yielding an irreducible $6+4+3+3=16$ dimensional
family of forms $F$ of this type in ${\mathbb{P}}^{34}$.

Type $[551]$:

For a general $F\in{\mathcal{F}}_{[551]}$ the quadratic ideal $Q_{F}$ is, by
Proposition 4.20, the ideal of five points, of which four lie in a plane. In
the embedding $v_{4}:{\mathbb{P}}^{3}\to{\mathbb{P}}^{34}$ these five points
span a ${\mathbb{P}}^{4}$ that contain the point $[F]$. A general form
$[F^{\prime}]$ in that ${\mathbb{P}}^{4}$ will have the same Betti table for
its apolar ring $A_{F^{\prime}}$. The ideal of any set of five points apolar
to $F$ is generated by at least five quadrics, i.e. by the five quadrics in
$F^{\bot}$, so $VSP(F,5)=\\{\mbox{one point}\\}$. Counting the dimension of
the family of $5$-tuples of points with four in a plane, we get $3$ for the
plane, $8$ for the four points in the plane, and $3$ for the point outside, so
altogether there is an irreducible $4+3+8+3=18$-dimensional family of forms
$F$ of this type in ${\mathbb{P}}^{34}$.

Type $[441a]$:

For a general $F\in{\mathcal{F}}_{[441a]}$ the quadratic ideal $Q_{F}$ is, by
Proposition 4.15, the ideal of $\Gamma$: The union of a plane conic curve $C$
and a point $p$ not in the plane of the conic. The image $v_{4}(C)$ in
${\mathbb{P}}^{34}$ is a rational normal curve of degree $8$. By apolarity
$[F]$ lies in the span of the image of the point $p$ and a unique point
$[F_{1}]$ in the span of $v_{4}(C)$. The quartic form $F_{1}$ is apolar to
$C$, so by Lemma 3.1, it has rank $5$, and $VSP(F_{1},5)={\mathbb{P}}^{1}$.

The ideal of any six points apolar to $F$ is generated by at least four
quadrics, so is contained in $\Gamma$, so also $VSP(F,5)={\mathbb{P}}^{1}$.
Counting the dimension of the family of pairs of a conic and a point, we get
$8$ for the conic and $3$ for the point. The span of their image in
${\mathbb{P}}^{34}$ is $9$-dimensional, so there is an irreducible
$8+3+9=20$-dimensional family of forms $F$ of this type in
${\mathbb{P}}^{34}$.

Type $[441b]$:

For a general $F\in{\mathcal{F}}_{[441b]}$ the quadratic ideal $Q_{F}$ is, by
Proposition 4.15, the ideal of $\Gamma$: The union of two skew lines $L_{1}$
and $L_{2}$. The form $F$ has a unique decomposition $F=F_{1}+F_{2}$, where
$F_{i}$ is apolar to the line $L_{i}$ for $i=1,2$. By Lemma 3.1, each $F_{i}$
has rank $3$ and $VSP(F_{i},3)={\mathbb{P}}^{1}$. The ideal of any six points
apolar to $F$ is generated by at least four quadrics, so is contained in
$L_{1}\cup L_{2}$ and consists of two sets of three points, apolar to $F_{1}$
and $F_{2}$ respectively, so
$VSP(F,6)={\mathbb{P}}^{1}\times{\mathbb{P}}^{1}$. Counting the dimension of
the family of pairs of lines, we get $8$. The span of their image in
${\mathbb{P}}^{34}$ is $9$-dimensional, so there is an irreducible
$8+9=17$-dimensional family of forms $F$ of this type in ${\mathbb{P}}^{34}$.

Type $[430]$:

For a general $F\in{\mathcal{F}}_{[310]}$ the quadratic ideal $Q_{F}$ is, by
Proposition 4.15, the ideal of a line and three points. The ideal of any set
of six points apolar to $F$ has at least four quadric generators, i.e. the
quadrics in in $F^{\bot}$ and a cubic generator. So $F$ is a sum of three
fixed rank $1$ quartics and a unique quartic $F_{1}$ that is apolar to a line.
If $F_{1}$ is apolar to two points on the line, $F$ would be apolar to five
points and $F^{\bot}$ would contain more quadrics, so $F_{1}$ is not apolar to
two points on the line. Therefore by Lemma 3.1, the form $F_{1}$ has rank $3$
and $VSP(F_{1},3)={\mathbb{P}}^{1}$, and thus also
$VSP(F,6)={\mathbb{P}}^{1}$. The dimension of the family of triples of points
is $9$, and the family of lines is $4$-dimensional. The image of the line and
the three independent points in ${\mathbb{P}}^{34}$ span a $7$-space, so there
is an irreducible $9+4+7=20$-dimensional family of forms $F$ of this type in
${\mathbb{P}}^{34}$.

Type $[420]$:

For a general $F\in{\mathcal{F}}_{[420]}$ the quadratic ideal $Q_{F}$ is, by
Proposition 4.15, the ideal of six points. By Castelnuovo’s Lemma [EH]*Theorem
1, there is a unique twisted cubic curve through six points in
${\mathbb{P}}^{3}$ as soon as no subset of four lie in a plane, so the six
points is the intersection of the curve with a quadric. The ideal of any set
of six points is generated by at least four quadrics, i.e. by the quadrics in
$F^{\bot}$. Therefore $VSP(F,6)=\\{1\;point\\}$. The dimension of the family
of $6$-tuples of points is $18$, and $6$ independent points in
${\mathbb{P}}^{34}$ span a $5$-space, so there is a $18+5=23$-dimensional
family of forms $F$ of this type in ${\mathbb{P}}^{34}$.

Type $[331]$:

For a general $F\in{\mathcal{F}}_{[331]}$ the quadratic ideal $Q_{F}$ is, by
Proposition 4.10, the ideal of a plane $P$ and a point outside the plane. So
in the space ${\mathbb{P}}^{34}$ of quartics the point $[F]$ lies in the span
of the point and a unique point $[F_{1}]$ in the span of the image $v_{4}(P)$
of the plane. So $F_{1}$ is a quartic form apolar to the plane. Since there is
a $3$-dimensional space of quadrics in $F^{\bot}$, the rank of $F$ is at least
$7$, so $F_{1}$ has rank at least $6$. Therefore $F_{1}$ is not apolar to any
quadric in the plane, its rank is $6$ and $VSP(F_{1},6)$ is a Fano threefold
by [Mukai]. The ideal of any apolar set of seven points contains a
$3$-dimensional space of quadrics, i.e. all the quadrics in $F^{\bot}$. Hence
any apolar set of seven points contains the point outside the plane, and
$VSP(F,7)=VSP(F_{1},6)$ i.e. a Fano threefold. Counting dimensions we get $3$
for the point and $3$ for the plane, while the span of the point and the image
of the plane in ${\mathbb{P}}^{34}$ is $15$ -dimensional, yielding an
irreducible $3+3+15=21$ dimensional family of forms $F$ of this type in
${\mathbb{P}}^{34}$.

Type $[310]$:

For a general $F\in{\mathcal{F}}_{[310]}$ the quadratic ideal $Q_{F}$ is, by
Proposition 4.10, is the ideal of the union of a conic $C$ and two points
$p_{1},p_{2}$ outside the plane of the conic. The ideal of any set of seven
points apolar to $F$ has at least $3$ quadric generators, i.e. the quadrics in
$F^{\bot}$. So any such set consists of the two points outside the conic and
five points on the conic $C$. The image $v_{4}(C)$ in ${\mathbb{P}}^{34}$ is a
rational normal curve of degree $8$. By apolarity $[F]$ lies in the span of
the image of the two points $p_{1},p_{2}$ and $v_{4}(C)$, so $F$ has a unique
decomposition $F=F_{1}+F_{2}$ where $F_{1}$ is apolar to the two points
$p_{1},p_{2}$ and has rank $2$, while $F_{2}$ is apolar to the conic $C$ and
has rank $5$. By Lemma 3.1, $VSP(F_{2},5)={\mathbb{P}}^{1}$, therefore also
$VSP(F,7)={\mathbb{P}}^{1}$. The dimension of the family of conics in a plane
in ${\mathbb{P}}^{3}$ is $3+5=8$, so the set of triples (conic + two points)
is $8+3+3=14$-dimensional. Each triple spans a ${\mathbb{P}}^{10}$ in
${\mathbb{P}}^{34}$, so there is an irreducible $14+10=24$-dimensional family
of forms $F$ of this type in ${\mathbb{P}}^{34}$.

Type $[210]$:

For a general $F\in{\mathcal{F}}_{[210]}$ the quadratic ideal $Q_{F}$ is, by
Proposition 4.8, the ideal of a plane $P$ and a line $L$ in
${\mathbb{P}}^{3}$. In the space ${\mathbb{P}}^{34}$ of quartics, the span of
$v_{4}(P)$ and $F$ intersects the span of $v_{4}(L)$ in a line that passes
through the intersection point $v_{4}(P\cap L)$. So there is a linear pencil
of lines through $F$ that intersect both the span of $v_{4}(P)$ and the span
of $v_{4}(L)$. The line through $F$ and the point $v_{4}(P\cap L)$ is in this
pencil, so there is a unique line in the pencil that intersects the span of
$v_{4}(P)$ in a form of rank $5$, and similarly a unique line in the pencil
that intersect the span of $v_{4}(L)$ in a form of rank $2$. The first line
defines a decomposition $F=F_{1}+F_{2}$, where $F_{1}$ is apolar to the plane
and has rank $5$, while $F_{2}$ is apolar to the line and has rank $3$. The
second line defines a decomposition $F=F^{\prime}_{1}+F^{\prime}_{2}$, where
$F^{\prime}_{1}$ is apolar to the plane and has rank $6$ and $F^{\prime}_{2}$
is apolar to the line and has rank $2$.

The ideal of any set of eight points contains at least two quadratic
generators, so any such set that is apolar to $F$ must be contained the union
of the line and the plane. And so any set of eight points defines a
decomposition $F=F_{1}+F_{2}$, where $F_{1}$ is apolar to the line and $F_{2}$
is apolar to the plane.

By the above, there are exactly two such decompositions where the sum of the
ranks of the two summands is $8$. When $F_{1}$ has rank $3$, then $F_{2}$ must
have rank $5$ and is apolar to a conic in the plane. By Lemma 3.1,
$VSP(F_{1},3)={\mathbb{P}}^{1}$ and $VSP(F_{2},5)={\mathbb{P}}^{1}$, so
$VSP(F,8)$ contains a component ${\mathbb{P}}^{1}\times{\mathbb{P}}^{1}$ from
the first decomposition.

In the second decomposition $F^{\prime}_{1}$ has rank $2$, then
$F^{\prime}_{2}$ is apolar to the plane and have rank $6$.
$VSP(F^{\prime}_{1},2)=\\{\mbox{one point}\\}$ and $VSP(F^{\prime}_{2},6)$ is
a Fano threefold, [Mukai]. So $VSP(F,8)$ contains the Fano threefold
$VSP(F^{\prime}_{2},6)$ from the second decomposition. We conclude that
$VSP(F,8)$ is the disjoint union of a ${\mathbb{P}}^{1}\times{\mathbb{P}}^{1}$
and a Fano threefold.

In ${\mathbb{P}}^{34}$ the span of $v_{4}(P\cup L)$ is a ${\mathbb{P}}^{18}$,
and the family of pairs of planes $P$ and lines $L$ defined by $Q_{F}$ is
$3+4=7$-dimensional, so there is an irreducible $18+7=25$-dimensional family
of forms $F$ with Betti table of this type in ${\mathbb{P}}^{34}$.

∎

###  5.1. $VSP(F,9)$ for a general quartic form of rank $9$

If $F$ has rank $9$ then $F^{\perp}$ contains a unique quadric form. Here we
assume that the quadric $Q$ defined by this form is smooth, which is certainly
the case for a general form of rank $9$. Since any set $\Gamma$ of nine points
in ${\mathbb{P}}^{3}$ lies on a quadric, any set $\Gamma$ that is apolar to
$F$ is contained in $Q$. In particular the variety $VSP(F,9)$ is contained in
the Hilbert scheme $\operatorname{Hilb}_{9}(Q)$ of schemes of length $9$ on
$Q$. Therefore we restrict our attention to $Q$.

####  5.1.1. Apolarity on $Q$.

On $Q={\mathbb{P}}^{1}\times{\mathbb{P}}^{1}$, the divisors are defined by
bihomogeneous forms. Let

$S_{Q}={\mathbb{C}}[u_{0},u_{1},v_{0},v_{1}]=\oplus_{(a,b)}H^{0}({\mathcal{O}}_{Q}(a,b)$

be the bigraded ring of forms on $Q$.

Then there is a natural restriction map $S\to S_{Q}$, with kernel generated by
the homogeneous ideal $I_{Q}$ of $Q$. Given $F\in R_{4}$, let $(F^{\bot})_{Q}$
be the ideal generated by the image of $F^{\bot}$. When $I_{Q}\subset
F^{\bot}\subset S$, then $(F^{\bot})_{Q}(4)$ has codimension $1$ in
$S_{Q}(4,4)$, so is defined by a linear form $F_{Q}\in S_{Q}(4,4)^{*}$. We
call $F_{Q}$ the restriction of $F$ to $Q$.

Conversely, the natural map

$S_{Q}(4,4)^{*}\to S_{4}^{*}=R_{4}$

is injective, and associates a unique form $F\in R_{4}$ to a form $F_{Q}\in
S_{Q}(4,4)^{*}$.

Now, apolarity extends to bihomogeneous forms. Any form $G$ of bidegree
$(a,b)$ in $S_{Q}$ may be interpreted as a differential operator on $F_{Q}$:

$G(F_{Q})\in S_{Q}(4-a,4-b)^{*};\quad G(F_{Q}):G^{\prime}\mapsto F_{Q}(G\cdot
G^{\prime}).$

We define the apolar ideal $F_{Q}^{\bot}\subset S_{Q}$ as the ideal generated
by the saturation

$F_{Q}^{\bot}=\langle\cup_{(a,b)}\\{G\in S_{Q}(a,b)|S_{Q}(4-a,4-b)\cdot
G\subset{\rm ker}F_{Q}\\}\rangle.$

Then $F^{\bot}|_{Q}\subset F_{Q}^{\bot}$ and $F^{\perp}$ can be recovered from
$F_{Q}^{\bot}$ as the biggest ideal in $S$ whose restriction $F^{\bot}|_{Q}$
to $Q$ is contained in $F_{Q}^{\bot}$.

A scheme $\Gamma$ of length $9$ is apolar to $F$ if $I_{\Gamma}\subset
F^{\perp}$. Any such $\Gamma$ lies in $Q$, so $I_{\Gamma}|_{Q}\subset
F^{\perp}|_{Q}\subset F_{Q}^{\perp}$. Since the ideal $I_{\Gamma}$ is
saturated in $S$, and $F_{Q}^{\perp}$ is saturated in $S_{Q}$, the saturation
of $I_{\Gamma}|_{Q}$ in $S_{Q}$ is contained in $F_{Q}^{\perp}$, which means
that $\Gamma\in VSP(F_{Q},9)\subset\operatorname{Hilb}_{9}(Q)$.

Conversely, let $\Gamma\in VSP(F_{Q},9)$ with $I_{\Gamma}\subset
F_{Q}^{\perp},$ and let $I^{\prime}_{\Gamma}\subset S$ be the ideal of
$\Gamma\subset{\mathbb{P}}^{3}$. Then $I^{\prime}_{\Gamma}\subset
F_{Q}^{\perp}$, and so $\Gamma\in VSP(F,9).$ Since
$\operatorname{Hilb}_{9}(Q)$ is closed in the Hilbert scheme of length $9$
subschemes in ${\mathbb{P}}^{3}$, we have shown

######  Lemma 5.8.

Let $F$ be a quaternary quartic of rank $9$, apolar to a smooth quadric $Q$,
and let $F_{Q}$ be the restriction of $F$ to $Q$. Then

$VSP(F,9)\cong VSP(F_{Q},9).$

######  Lemma 5.9.

Let $\Gamma\subset Q$ be a scheme of length $9$ with
$h^{0}(I_{\Gamma}(4,4))\geq 17$. Then the forms in $H^{0}(I_{\Gamma}(4,4))$
have a fixed component of bidegree $(0,1)$ or $(1,0)$ that contains a
subscheme of length at least $6$ on $\Gamma$.

Furthermore, if $F_{Q}$ is a $(4,4)$-form and the minimal length of a scheme
apolar to $F_{Q}$ is $9$, then $h^{0}(I_{\Gamma}(4,4))=16$ for any scheme
$\Gamma$ of length $9$ that is apolar to $F_{Q}$.

###### Proof.

Let $\Gamma$ be a scheme of length $9$ that impose dependent conditions on
$(4,4)$-forms, i.e. $h^{0}(I_{\Gamma}(4,4))\geq 17$. Then the restriction of
the forms in $I_{\Gamma}(4,4)$ to any irreducible curve $C$ defined by a form
in $I_{\Gamma}(4,4))$ would be special. But any such curve $C$ has geometric
genus at most $9$, so has no special linear system of dimension $\geq 16$. The
forms in $I_{\Gamma}(4,4)$, must therefore have a fixed component, for
dimension reasons, of bidegree $(0,1)$ or $(1,0)$. Furthermore, if this
component has bidegree $(1,0)$ and contains a subscheme $\Gamma^{\prime}$ of
$\Gamma$, and $\Gamma^{\prime\prime}=\Gamma\setminus\Gamma$ is the residual
scheme, then $h^{0}(I_{\Gamma^{\prime\prime}}(3,4))\geq 17$, which means the
$\Gamma^{\prime\prime}$ has length at most $3$. Hence, $\Gamma^{\prime}$ has
length at least $6$.

By Lemma 3.1, any scheme of minimal length that is apolar to a $(4,4)$-form
contains a subscheme of length at most $3$ on a curve of bidegree $(0,1)$ or
$(1,0)$. So the moving part of the forms in $I_{\Gamma}(4,4)$ are $(3,4)$ or
$(4,3)$ forms in the ideal of a scheme $\Gamma_{0}$ of length at least $6$ and
defines a linear system of dimension at least $17$. Again a general such form
is irreducible and defines a curve $C$ of genus at most $6$. But the
restriction of the linear system to $C$ would be special, which is impossible
when the dimension is at least $16$. ∎

######  Proposition 5.10.

. Let $F_{Q}$ be a general $(4,4)$-form on a smooth quadric surface $Q$. Then
$VSP(F_{Q},9)\subset\operatorname{Hilb}_{9}(Q)$ is an irreducible surface.

###### Proof.

Consider the incidence

$I_{9}=\\{(F_{Q},I_{\Gamma}|I_{\Gamma}\subset
F_{Q}^{\perp}\\}\subset{\mathbb{P}}(T(4,4))\times\operatorname{Hilb}_{9}(Q).$

Let $H_{0}\subset\operatorname{Hilb}_{9}(Q)$ be the locus of ideals $J$, with
${\rm dim}J(4,4)=16$, it is a dense subset, so its inverse image $I_{9,0}$ in
$I_{9}$ is irreducible. Hence, the general fiber of the projection
$I_{9,0}\to{\mathbb{P}}(T(4,4))$ is also irreducible. Let $F_{Q}$ be a general
$(4,4)$-form. Then, by Lemma 5.9, the variety $VSP(F_{Q},9)$ is the closure in
$\operatorname{Hilb}_{9}(Q)$ of the fiber over $F_{q}$ in $I_{9,0}$, and the
lemma follows. ∎

Now, let $\Gamma\subset Q$ be a set of nine points on $Q$. Then, by
Proposition 3.2, the space of cubic generators in the ideal of $\Gamma$ modulo
$Q$ is $7$-dimensional, i.e. if $I_{\Gamma}\subset S_{Q}$ is the ideal of
$\Gamma$ on $Q$, then $h^{0}(I_{\Gamma}(3,3))=7$. The forms of bidegree
$(3,3)$ in $I_{\Gamma}$ are, however, not minimal generators. In fact,
$h^{0}(I_{\Gamma}(3,2))=h^{0}(I_{\Gamma}(2,3))=3$ for a general $\Gamma$. On
the other hand, ${\rm dim}_{k}F_{Q}^{\bot}(2,3)={\rm
dim}_{k}F_{Q}^{\bot}(3,2)=6$, so if a set $\Gamma$ of nine points is apolar to
$F$, then $H^{0}(I_{\Gamma}(3,2))\subset F_{Q}^{\bot}(3,2)$ and
$H^{0}(I_{\Gamma}(2,3))\subset F_{Q}^{\bot}(2,3)$.

For a general set $\Gamma$ of nine points on $Q$ the vanishing locus of both
the forms of bidegree $(2,3)$ and those of bidegree $(3,2)$ in the ideal is
$\Gamma$, so we may conclude that $VSP(F_{Q},9)$ has a birational model in the
two Grassmannians $G(3,F_{Q}^{\bot}(2,3))$ and $G(3,F_{Q}^{\bot}(3,2))$. In
particular, by semicontinuity, ${\rm dim}H^{0}(I_{\Gamma}(2,3))\cap
F_{Q}^{\bot}(2,3)\geq 3$ and

${\rm dim}H^{0}(I_{\Gamma}(3,2))\cap F_{Q}^{\bot}(3,2)\geq 3$

for any $\Gamma\in VSP(F_{Q},9)$.

We now characterize the nets of $(2,3)$-forms that vanish on nine general
points on $Q$.

######  Lemma 5.11.

Let $\langle G_{1},G_{2},G_{3}\rangle$ be a net of $(2,3)$-forms on $Q$ whose
vanishing locus is a scheme $\Gamma$ of length $9$, and assume that the ideal
$I_{\Gamma}$ contains no form of bidegree $(1,3)$. Then the net has a syzygy
of bidegree $(1,1)$ and generates the space of $2\times 2$ minors of a matrix

$\begin{pmatrix}A_{1},A_{2},A_{3}\\\
B_{1},B_{2},B_{3}\end{pmatrix},\qquad\mbox{ with deg }A_{i}=(1,1),\mbox{ deg
}B_{i}=(1,2).$

In particular $I_{\Gamma}(2,3)=\langle G_{1},G_{2},G_{3}\rangle$.

###### Proof.

Given the scheme $\Gamma$ of length $9$ and a general pencil of forms of
bidegree $(2,3)$ in $I_{\Gamma}$. Then $\Gamma$ is linked in this pencil to a
scheme $\Gamma_{0}$ on $Q$ of length $3$. Then $\Gamma_{0}$ spans a line in
$Q$ or a plane. If $\Gamma_{0}$ spans a line $L$, this line must have bidegree
$(1,0)$, and one of the forms in the pencil must vanish on $L$. But then that
form has a factor of bidegree $(1,3)$ that vanishes on $\Gamma$, contrary to
the assumption. So $\Gamma_{0}$ spans a plane. It is then a complete
intersection of forms of bidegree $(1,1)$ and $(1,2)$. The ideal of $\Gamma$
therefore contains the $2$-minors of a $2\times 3$ matrix as stated in the
lemma. The first row is then a syzygy among the $2$-minors.

It remains to show that the space of minors coincide with original net of
$(2,3)$ forms. If not, $H^{0}(I_{\Gamma}(2,3))$ is at least $4$-dimensional,
or equivalently, $h^{1}(I_{\Gamma}(2,3))>0$. Let $C\subset Q$ be a curve
defined by a general form $G\in I_{\Gamma}(2,3)$. Then $C$ is irreducible, and
of genus $\leq 2$. From the cohomology of the exact sequence

$0\longrightarrow{\mathcal{O}}_{Q}\stackrel{{\scriptstyle\cdot
G}}{{\longrightarrow}}I_{\Gamma}(2,3)\longrightarrow
I_{C,\Gamma}(2,3)\longrightarrow 0$

we get $h^{1}(I_{C,\Gamma}(2,3))>0$ and $h^{0}(I_{C,\Gamma}(2,3))>2$. So the
linear system $H^{0}(I_{C,\Gamma}(2,3))$ on $C$ is special. But no curve of
genus $\leq 2$ has a special linear system of dimension $>2$. ∎

We let $V_{F_{Q}}(2,3)\subset G(3,F_{Q}^{\bot}(2,3))$ be the locus of nets of
$(2,3)$-forms in $F_{Q}^{\bot}(2,3))$ with a syzygy of bidegree $(1,1)$. The
variety $V_{F_{Q}}(3,2)\subset G(3,F_{Q}^{\bot}(3,2))$ is defined similarly.

We will show that for a general $(4,4)$-form $F_{Q}$ the loci $V_{F_{Q}}(2,3)$
and $V_{F_{Q}}(3,2)$ are surfaces with two smooth components, one birational
to $VSP(F_{Q},9)$ and the other consisting of nets of forms whose common zeros
is a scheme of length $8$, in fact a complete intersection $((1,3),(2,2))$ and
$((3,1),(2,2))$ respectively.

To make this statement precise we need some lemmas on generality assumptions.

For the $(4,4)$-form $F_{Q}$, consider the space of partials of order $(2,1)$:

$P_{2,3}=\\{G(F_{Q})|G\in H^{0}({\mathcal{O}}_{Q}(2,1))\\}\subset
H^{0}({\mathcal{O}}_{Q}(2,3)).$

Then $F_{Q}^{\bot}(2,3)=P_{2,3}^{\bot}$ defines a rational map

$\begin{matrix}&&&{\mathbb{P}}(H^{0}({\mathcal{O}}_{Q}(2,3))\\\
&v_{2,3}\hskip-11.38092pt&\nearrow&\downarrow\\\
q_{2,3}:&Q&\to&{\mathbb{P}}(F_{Q}^{\bot}(2,3)^{*})\\\ \end{matrix},$

which may be interpreted as the composition of the map
$v_{2,3}:Q\to{\mathbb{P}}(H^{0}({\mathcal{O}}_{Q}(2,3))$ defined by
$H^{0}({\mathcal{O}}_{Q}(2,3)^{*}$ and the projection from the space of
partials $P_{2,3}\subset{\mathbb{P}}(H^{0}({\mathcal{O}}_{Q}(2,3))$. In this
interpretation, the following lemma is a consequence:

######  Lemma 5.12.

The map $q_{2,3}$ is a morphism if and only if $F_{Q}$ has no partials of
order $(2,1)$ and rank $\leq 1$. The map $q_{2,3}$ is an embedding if and only
if $F_{Q}$ has no partials of order $(2,1)$ and cactus rank $\leq 2$.

###### Proof.

A point $p=[G(F)]$ in the intersection $P_{2,3}\cap v_{2,3}(Q)$ is a form
whose ideal $I_{p}\in S_{Q}$ is contained in $G(F)^{\perp}$. Thus, $G(F)$ is
apolar to a point $p$, for some partial $G\in S_{Q}$ of order $(2,1)$ if and
only if $q_{2,3}$ is not defined at $p$. Similarly, a point $s=G(F)$ lies in
the span of a length $2$ scheme $Z\subset Q$, if and only if $Z$ is apolar to
$G(F)$, i.e. $I_{Z}\subset P(F)^{\perp}$. Thus, $G(F)$ is apolar to a scheme
$Z$ of length $2$, for some partial $G\in S_{Q}$ of order $(2,1)$ if and only
if $q_{2,3}$ is not an embedding. ∎

Of course, there is an analogous lemma for $q_{3,2}$.

######  Lemma 5.13.

In the space of $(4,4)$-forms, the following loci are at most divisors:

  1. (1)

the locus of forms apolar to the union of $(0,1)$-curve or a $(1,0)$-curve and
a scheme of length at most $6$.

  2. (2)

the locus of forms apolar to some $(1,3)$ or $(2,2)$-form.

  3. (3)

the locus of forms with a partial of order $(2,1)$ or $(1,2)$ that is apolar
to a scheme of length at most $2$.

  4. (4)

the locus of forms that is apolar to some scheme $\Gamma$ of length $9$ that
impose dependent conditions on $(4,4)$ -forms.

###### Proof.

For the first case, let $Z$ be the union of a $(0,1)$-curve $L$ and a finite
scheme $Z_{0}$ of length at most $5$ disjoint from $L$. If a $(4,4)$-form
$F_{Q}$ is apolar to $Z$, then $F_{Q}$ decomposes into
$F_{Q}=F_{Q,L}+F_{Q,0}$, where $F_{Q,L}$ is apolar to $L$ and $F_{Q,0}$ is
apolar to $Z_{0}$. Now, since $v_{4,4}(L)$ spans a ${\mathbb{P}}^{4}$ and
there is a pencil of $(0,1)$ curves, the forms apolar to a $(0,1)$-curve has
dimension $4+1=5.$ Similarly the set of forms apolar to $Z_{0}$ is
$5$-dimensional and the Hilbert scheme $\operatorname{Hilb}_{k}(Q)$ is
$2k$-dimensional, so the set of forms apolar to $Z_{0}$ is at most
$17$-dimensional. The set of forms $F_{Q}$ with a decomposition
$F_{Q}=F_{Q,L}+F_{Q,0}$ is therefore at most $23$-dimensional, i.e. at most a
divisor in the space of $(4,4)$-forms.

Similarly the set of $(4,4)$-forms apolar to a fixed $(2,2)$ curve or a
$(1,3)$-curve is $15$-dimensional, respectively $16$-dimensional, while the
families of such curves are $8$, respectively $7$\- dimensional. So the set of
forms apolar to some $(2,2)$-curve or some $(1,3)$-curves is at most
$23$-dimensional, i.e. at most a divisor in the space of $(4,4)$-forms.

The ideal of a general scheme $Z$ of length $2$ on $Q$ is generated by two
$(1,1)$-forms say $A_{1}$ and $A_{2}$. A partial $G(F_{Q})$ of degree $(2,3)$
is apolar to $Z$ only if $A_{1}(G(F_{Q}))=A_{2}(G(F_{Q}))=0$, i.e. if both
$A_{1}(F_{Q})$ and $A_{2}(F_{Q})$ are annihilated by the $(2,1)$ form $G$.

Any $(1,2)$ form $G\in S_{Q}(1,2)$ defines a linear map differentiation

$Cat_{G}:S_{Q}(3,3)^{*}\to S_{Q}(1,2)^{*};\quad H\mapsto G(H).$

Consider its restriction to the space of degree $(3,3)$-partials

$P_{3,3}(F)=\\{A(F_{Q})|A\in S_{Q}(1,1)\\}\subset S_{Q}(3,3)^{*}.$

Now, $G\in H^{\perp}$ only if $Cat_{G}$ has corank at least $1$. Similarly,
$G$ annihilates a pencil of forms in $P_{3,3}(F)$ only if $Cat_{G}$ restricted
to $P_{3,3}(F)$ has corank at least $2$. Since ${\rm dim}P_{3,3}(F)=4$ and
${\rm dim}S_{Q}(1,2)^{*}=6$, the corank $2$ locus has expected dimension $8$,
i.e. the set of $[G]\in{\mathbb{P}}(S_{Q}(2,1))$, such that $Cat_{G}$ is
expected to be empty for general $F$. This is also easily confirmed with the
example

$F_{Q}=x_{0}^{4}y_{0}^{4}+x_{1}^{4}y_{0}^{4}+x_{0}^{3}x_{1}y_{0}y_{1}^{3}+x_{0}^{2}x_{1}^{2}y_{0}^{2}y_{1}^{2}+x_{1}^{4}y_{1}^{4}.$

So the set of $(4,4)$-forms $F_{Q}$ for which a partial $G(F_{Q})$ of degree
$(2,3)$ is apolar to length $2$ subscheme $Z$, defined by two $(1,1)$ forms,
is at most a divisor in the space of $(4,4)$-forms. If a length $2$ scheme $Z$
is contained in a $(0,1)$-curve or a $(1,0)$-curve the similar argument
applies with the same conclusion.

By Lemma 5.9, a $(4,4)$-form $F_{Q}$ of cactus rank $9$, is not part of the
fourth locus, so this latter locus is at most a divisor. ∎

We say that a $(4,4)$-form is sufficiently general if it is not in any of the
loci of Lemma 5.13. Note that any apolar scheme of length at most $8$ has a
$(2,2)$-form in its ideal, so a sufficiently general $(4,4)$ -form is not
apolar to any scheme of length $\leq 8$.

######  Proposition 5.14.

Let $F_{Q}$ be a sufficiently general $(4,4)$ form on a quadric $Q$, and let
$\langle G_{1},G_{2},G_{3}\rangle$ be a net of forms in $F_{Q}^{\bot}(2,3)$
with a syzygy of bidegree $(1,1)$, i.e. a net that belongs to
$V_{F_{Q}}(2,3)$. Let

$\Gamma={\mathbb{V}}(G_{1},G_{2},G_{3}).$

Then $\Gamma$ is a subscheme of length $9$ apolar to $F_{Q}$, i.e. $\Gamma\in
VSP(F_{Q},9)$, or $\Gamma$ is a complete intersection $((1,3),(2,2))$ of
length $8$. In the latter case the $(1,1)$-forms of the syzygy vanish in a
common point on $Q$.

###### Proof.

Assume $\langle G_{1},G_{2},G_{3}\rangle$ is a net of $(2,3)$-forms that are
apolar to $F_{Q}$ and have a syzygy $A_{1}G_{1}+A_{2}G_{2}+A_{3}G_{3}=0$ where
the $A_{i}$ have bidegree $(1,1)$. We analyze case by case:

  1. (1)

${\rm rank}\langle A_{1},A_{2},A_{3}\rangle=2$

  2. (2)

${\rm rank}\langle A_{1},A_{2},A_{3}\rangle=3$ and the forms
$G_{1},G_{2},G_{3}$ have a common component.

  3. (3)

${\rm rank}\langle A_{1},A_{2},A_{3}\rangle=3$ and the vanishing locus
${\mathbb{V}}(G_{1},G_{2},G_{3})$ is finite.

In case $(1)$ we may assume $A_{3}=0$ and $A_{1}G_{1}=A_{2}G_{2}$. If $A_{1}$
or $A_{2}$ is irreducible, then there is a $(1,2)$-form $G_{0}$ such that
$G_{1}=A_{2}G_{0}$ and $G_{2}=A_{1}G_{0}$. The form $G_{0}(F_{Q})$ has
bidegree $(3,2)$, and $A_{1}(G_{0}(F_{Q}))=A_{2}(G_{0}(F_{Q}))=0$. But
$(A_{1},A_{2})$ generates the ideal of a scheme of length $2$, so
$G_{0}(F_{Q})$ is apolar to a subscheme of length $2$. This is excluded by the
generality assumptions of $F_{Q}$. If $A_{1}$ and $A_{2}$ are reducible with a
common factor of bidegree $(0,1)$, then we may assume that
$B_{1}G_{1}=B_{2}G_{2}$ for $(1,0)$-forms $B_{i}$ such that $G_{1}=B_{2}G_{0}$
and $G_{2}=B_{1}G_{0}$ . Then $G_{0}$ is a $(1,3)$-form and $G_{0}(F_{Q})$ is
a $(3,1)$-form that is annihilated by every $(0,1)$ form. But that means
$G_{0}$ is apolar to $F_{Q}$, again contrary to the assumption on $F_{Q}$. If
$A_{1}$ and $A_{2}$ are reducible with a common factor of bidegree $(1,0)$,
then a similar argument implies that there is a $(2,2)$ form apolar to
$F_{Q}$, against the generality assumption.

For cases $(2)$ and $(3)$, we first notice that when $F_{Q}$ is sufficiently
general, then, by Lemmas 5.12 and 5.13, the map
$\phi_{2,3}:Q\to{\mathbb{P}}^{5}$ defined by $F_{Q}^{\bot}(2,3)$ is an
embedding. Therefore, the only curves on $Q$ that are mapped into a plane by
$\phi_{2,3}$ are the curves of bidegree $(0,1)$.

In case $(2)$ we therefore may assume that the common factor $B$ of the
$G_{i}$ has bidegree $(0,1)$. So $G_{i}=BH_{i}$ where the $H_{i}$ have
bidegree $(2,2)$ and $A_{1}H_{1}+A_{2}H_{2}+A_{3}H_{3}=0$. The $A_{i}$ have at
most one common zero, otherwise we would be in case $(1)$ again. If the
$A_{i}$ have one common zero, then the $H_{i}$ vanish on a common scheme $Z$
of length $5$, which would be the complete intersection of a $(2,1)$-form and
a $(1,2)$-form. The three $H_{i}$ would span a net of forms in the
$4$-dimensional space of $(2,2)$-forms in the ideal of $Z$. Still, one may
compute that the three $(2,3)$ forms $BH_{i}$ generate the space of
$(4,4)$-forms in the ideal of $Z\cup\\{B=0\\}$. This means that if the
$BH_{i}$ lie in $F_{Q}^{\perp}$, then $Z\cup\\{B=0\\}$ is apolar to $F_{Q}$,
contrary to the generality assumption.

If the $(1,1)$-forms have no common zeros, then the $(2,2)$-forms $H_{i}$
generate the ideal of a scheme $Z$ of length $6$, and the union of this scheme
and the curve $\\{B=0\\}$ is apolar to $F$, again contrary to the generality
assumption on $F_{Q}$.

In case $(3)$ we may assume that the $G_{i}$ are all irreducible. Since
$A_{1}G_{1}+A_{2}G_{2}+A_{3}G_{3}=0$, the $A_{i}G_{i}$ form a pencil. In the
baselocus $\Gamma_{P}$ of this pencil of degree $24$, each $A_{i}$ contains a
length $7$ subscheme, and the intersection of any two $A_{i}$ has length $2$.
If the three $A_{i}$ have no common zeros the intersection of the three
$G_{i}$ is a subscheme of length $24-3\times 7+3\times 2=9$ of the baselocus
$\Gamma_{P}$. In this case the vanishing locus $\\{G_{1}=G_{2}=G_{3}=0\\}$ is
a scheme $\Gamma$ of length $9$.

If the common zeros of the three $A_{i}$ is a reduced point, then the
intersection of the three $G_{i}$ is a subscheme of length $24-3\times
7+3\times 2-1=8$ of the baselocus $\Gamma_{P}$. In this case the vanishing
locus $\\{G_{1}=G_{2}=G_{3}=0\\}$ is a scheme $\Gamma$ of length $8$.

If the common zeros of the three $A_{i}$ is scheme of length $2$, then the
$A_{i}$ form a pencil, which is case (1).

Assume, first that the vanishing locus $\\{G_{1}=G_{2}=G_{3}=0\\}$ is a scheme
$\Gamma$ of length $9$, and consider a general pencil in the net. In this
pencil $\Gamma$ is linked to a scheme $\Gamma_{0}$ of length $3$. Then
$\Gamma_{0}$ is a complete intersection $(1,0),(1,3)$ or $(1,1),(1,2)$.
Correspondingly the net is the space of $2\times 2$-minors of a matrix

$\begin{pmatrix}A_{1},A_{2},A_{3}\\\ B_{1},B_{2},B_{3}\end{pmatrix},$

where ${\rm deg}A_{i}=(1,0),{\rm deg}B_{i}=(1,3)$ or ${\rm
deg}A_{i}=(1,1),{\rm deg}B_{i}=(1,2)$ respectively. In the first case there is
a syzygy of bidegree $(1,0)$ which generates syzygies of bidegree $(1,1)$ of
rank $2$. This is excluded in case $(1)$ above. So only the second case is
possible, as stated in the lemma.

To show that $\Gamma$ is apolar to $F_{Q}$ we need to show that
$I_{\Gamma}(a,b)\subset F_{Q}^{\bot}(a,b)$ for any $(a,b)$. By Lemma 5.11 we
know that this is the case for $(a,b)=(2,3)$. The space of forms of degree
$(4,4)$ generated by $\langle G_{1},G_{2},G_{3}\rangle$ has a dimension that
may be computed from the matrix representation: The $(2,3)$-forms generate
$3\cdot 6=18$ form of bidegree $(4,4)$, with two syzygies generated by the
syzygy of degree $(1,1)$, so $16$ linearly independent $(4,4)$-forms. Since
the vanishing locus of the $(2,3)$-forms is $\Gamma$, the general $(4,4)$-form
in this space defines an irreducible curve $C$ of genus at most $9$. The
restriction of the $16$-dimensional space of $(4,4)$-forms to $C$ is
$15$-dimensional, so that space defines nonspecial linear system on $C$. We
conclude that the $16$-dimensional space of $(4,4)$-forms coincides with
$I_{\Gamma}(4,4)$. In particular, $I_{\Gamma}(4,4)\subset F_{Q}^{\bot}(4,4)$.
By apolarity we infer: For any form $G\in I_{\Gamma}$ of bidegree $(a,b)$,
with $a\leq 4,b\leq 4$, the form $G(F_{Q})$ has bidegree $(4-a,4-b)$ and is
annihilated by any form of bidegree $(4-a,4-b)$. Therefore $G(F_{Q})=0$, and
$G\in F_{Q}^{\bot}(a,b)$. We conclude that $I_{\Gamma}(a,b)\subset
F_{Q}^{\bot}(a,b)$ for any $(a,b)$.

Assume, next, that the vanishing locus $\\{G_{1}=G_{2}=G_{3}=0\\}$ is a scheme
$\Gamma$ of length $8$, and consider a general pencil in the net. In this
pencil $\Gamma$ is linked to a scheme $\Gamma_{0}$ of length $4$. In the
relation $A_{1}G_{1}+A_{2}G_{2}+A_{3}G_{3}=0$, when the common zeros of the
$A_{i}$ is a reduced point, the three forms $A_{1},G_{2},G_{3}$ must vanish in
scheme of length $4$. So the common zeros $\Gamma$ of the $G_{i}$ is linked in
$(G_{2},G_{3})$ to a scheme that is the common zeros of $A_{1}$ and a
$(2,2)$-form $B_{1}$, or $A_{1}$ and a $(1,3)$-form $B_{2}$. From the first
case there is a $(1,3)$-form $H_{1}$ in the ideal of $\Gamma$, while from the
second case there is a $(2,2)$-form $H_{2}$ in the ideal of $\Gamma$. For
degree reasons, the ideal of $\Gamma$ is a complete intersection
$(H_{1},H_{2})$.

The multiples of $H_{1}$ and $H_{2}$ generate the $4$-dimensional space
$I_{Z}(2,3)$ of forms of bidegree $(2,3)$, with a $4$-dimensional space of
syzygies of bidegree $(1,1)$. In fact, any $3$-dimensional subspace in
$I_{Z}(2,3)$ has a syzygy of bidegree $(1,1)$, such that the $(1,1)$ forms
have a common zero on $Q$. This can be seen explicitly: As generators of a net
in $I_{Z}(2,3)$ we may choose one form $A_{1}H_{1}$, $B_{1}H_{2}$ and
$A_{2}H_{1}+B_{2}H_{2}$, the $A_{1}$ and $A_{2}$ are $(1,0$-forms, while
$B_{1}$ and $B_{2}$ are $(0,1)$-forms. Then the syzygy of bidegree $(1,1)$
among the three is

$A_{2}B_{1}\cdot
A_{1}H_{1}-A_{1}B_{1}\cdot(A_{2}H_{1}+B_{2}H_{2})+A_{1}B_{2}\cdot
B_{1}H_{2}=0.$

Notice that the $(1,1)$ forms vanish at the point
$p={\mathbb{V}}(A_{1},B_{1})$. When $A_{2}H_{1}+B_{2}H_{2}$ vanishes at this
point, then the three $(2,3)$ forms in the ideal of $Z$ actually vanishes in a
scheme of length $9$. In this case $Z\cup p$ is apolar to $F_{Q}^{\perp}$. But
when $A_{2}H_{1}+B_{2}H_{2}$ does not vanish at $p$, then the three
$(2,3)$-forms in $I_{Z}(2,3)$ does not vanish on an apolar scheme, even if the
net lies in $F_{Q}^{\perp}$.

Thus the net of $(2,3)$ forms in $F_{Q}^{\perp}(2,3)$ with a syzygy in
bidegree $(1,1)$ consists both of nets that define apolar schemes of length
$9$ to $F$, and schemes of length $8$ that are not apolar to $F$.

∎

There is, of course, a completely analogous proposition for $V_{F_{Q}}(3,2)$.

Let $V^{8}_{F_{Q}}(2,3)$ and $V^{9}_{F_{Q}}(2,3)$ be the sets of nets $\langle
G_{1},G_{2},G_{3}\rangle\in G(3,F^{\perp}(2,3))$ with a $(1,1)$-syzygy, such
that ${\mathbb{V}}(G_{1},G_{2},G_{3})$ has length $8$ and length $9$,
respectively. And similarly $V^{8}_{F_{Q}}(3,1)$ and $V^{9}_{F_{Q}}(3,2)$ in
$G(3,F^{\perp}(3,2))$.

Clearly, both $V^{8}_{F_{Q}}(2,3)$ and $V^{9}_{F_{Q}}(2,3)$ are closed,
$V^{8}_{F_{Q}}(2,3)$ since it is the locus where the $(1,1)$-syzygy vanishes
on $Q$, and $V^{9}_{F_{Q}}(2,3)$ since it the locus where the net of $(2,3)$
curves vanish on a scheme of length $9$. Also, since any net in closed variety
$V^{9}_{F_{Q}}(2,3)$ defines an apolar scheme of length $9$ to $F$, the
natural map

$V^{9}_{F_{Q}}(2,3)\to VSP(F_{Q},9)$

is by Proposition 5.14 both surjective and injective. In particular every
point in $VSP(F_{Q},9)$ is apolar to $F$.

######  Proposition 5.15.

When $F_{Q}$ is a general $(4,4)$-form, then $VSP(F_{Q},9)$ is a smooth
surface isomorphic to $V^{9}_{F_{Q}}(2,3)$.

###### Proof.

The last isomorphism follows from the above bijection, as soon as
$VSP(F_{Q},9)$ is smooth.

Consider the subset $U\subset\operatorname{Hilb}_{9}Q$ of schemes $\Gamma$
such that $I_{\Gamma}(4,4)$ is $16$-dimensional. By Lemma 5.9, $U$ is an open
set, that, by Proposition 5.14, contains $VSP(F_{Q},9)$. Let $E_{U}$ be the
rank $16$ vector bundle on $U$ whose fiber over $\Gamma\in U$ is
$I_{\Gamma}(4,4)$, then any $(4,4)$-form $F_{Q}$ defines a section of
$E_{U}^{*}$. The vector bundle is clearly generated by these sections, and the
zeros of a section is the set of $\Gamma$ such that $I_{\Gamma}(4,4)$ is in
the kernel of $F_{Q}$. In particular this zero locus is $VSP(F_{Q},9)\cap U$.
Finally, $U$ is smooth (cf. [Fogarty]), and $VSP(F_{Q},9)\subset U$ for a
general $F_{Q}$, so by the Bertini theorem, $VSP(F_{Q},9)$ is also smooth. ∎

######  Theorem 5.16.

Let $F$ be a general quaternary quartic form of rank $9$, then the variety
$VSP(F,9)$ is a K3 surface isomorphic to a smooth quartic surface in
$\mathbb{P}^{3}$.

###### Proof.

We may assume that $F$ is apolar to a smooth quadric $Q$, and that the
restriction $F_{Q}$ is a sufficiently general $(4,4)$-form as described by
Lemma 5.13. Then by Lemma 5.11 and the proof of Proposition 5.14 to any
$\Gamma\in VSP(F_{Q},9)$ we can associate a unique net of $(1,1)$-forms
representing the $(1,1)$-syzygy among the generators of $I_{\Gamma}(2,3)$. We
hence have a map $\kappa:VSP(F_{Q},9)\to\mathbb{P}(S_{Q}(1,1)^{\vee})$.
Clearly the image of $\kappa$ is contained in the following locus:

$D=\\{<l_{1},l_{2},l_{3}>\in\mathbb{P}(S_{Q}(1,1)^{\vee})\quad|\quad\exists\;q_{1},q_{2},q_{3}\in
F^{\perp}(2,3)\text{ such that }q_{1}l_{1}+q_{2}l_{2}+q_{3}l_{3}=0\\}.$

On the other hand $D$ is described as the rank $\leq 17$ locus of the
composition

$\rho\colon\mathcal{U}_{\mathbb{P}(S_{Q}(1,1)^{\vee})}\otimes
F^{\perp}(2,3)\to S_{Q}(1,1)\otimes
F^{\perp}(2,3)\otimes\mathcal{O}_{\mathbb{P}(S_{Q}(1,1)^{\vee})}\to
F^{\perp}(3,4),$

where $\mathcal{U}_{\mathbb{P}(S_{Q}(1,1)^{\vee})}$ is the universal bundle of
3-spaces in $S_{Q}(1,1)$, the first map comes from the universal bundle
sequence, whereas the second is given by multiplication in $S_{Q}$. Since
$-c_{1}(\mathcal{U}_{\mathbb{P}(S_{Q}(1,1)^{\vee})})$ is the class of the
hyperplane we conclude that $D$ is a hypersurface of degree 6 in
$\mathbb{P}(S_{Q}(1,1)^{\vee})$. Furthermore observe that for $F$ sufficiently
general we have $Q\subset D$. Indeed, if $<l_{1},l_{2},l_{3}>\subset
S_{Q}(1,1)$ are the space of $(1,1)$ forms that vanish at a point $p$ in $Q$,
then the image of the map $\rho$ is contained in $F^{\perp}(3,4)\cap I_{p}$
which has dimension 17 and hence $\rho$ is degenerate over that point. Note
however that the quadric $Q$ is not the image of $VSP(F_{Q},9)$ by $\kappa$.

It remains to see on an example computed in Macaulay2, cf. [M2]*Package
QuaternaryQuartics, that for suitably general $F$ the divisor $D$ is in
general a union of the quadric $Q$ with a smooth quartic surface $X$ meeting
$Q$ transversally and the rank $16$-locus of $\rho$ is empty:

For the $(4,4)$-form

$F_{Q}=x_{0}^{4}y_{0}^{4}+x_{1}^{4}y_{0}^{4}+x_{0}^{3}x_{1}y_{0}y_{1}^{3}+x_{0}^{2}x_{1}^{2}y_{0}^{2}y_{1}^{2}+x_{1}^{4}y_{1}^{4}.$

the $VSP(F_{Q},9)$ is isomorphic to the following nonsingular quartic (of type
$[000]$):

$\displaystyle X(F_{Q})=$ $\displaystyle
949968z_{1}^{4}-6713280z_{2}^{4}-540372420z_{2}^{3}z_{3}+481656096z_{2}^{2}z_{3}^{2}$
$\displaystyle-214742080z_{2}z_{3}^{3}+47738880z_{3}^{4}-100117184265z_{1}z_{2}^{2}z_{4}$
$\displaystyle+4449690134z_{1}z_{2}z_{3}z_{4}-82880z_{1}z_{3}^{2}z_{4}-34412740z_{1}^{2}z_{4}^{2}+949968z_{4}^{4}.$

Here $(z_{1},z_{2},z_{3},z_{4})$ and
$(x_{0}y_{0},x_{0}y_{1},x_{1}y_{0},x_{1}y_{1})$ are dual basis.

We conclude that for a general quartic $F$ of type $[100]$ the $VSP(F,9)$ is a
smooth quartic of type $[000]$, in particular it is a $K3$-surface. ∎

##  6\. Stratifications of the space of quaternary quartics

The open set $U_{34}\subset{\mathbb{P}}^{34}$ of forms that are non-
degenerate, that is, those forms $F$ with no linear form in $F^{\perp}$, is
the disjoint union of the $19$ irreducible sets of Proposition 5.3 and
Proposition 5.5. In this section we show the closure relations between these
sets.

We first we show how they fit in the natural stratification of
${\mathbb{P}}^{34}$ in terms of border rank. The border rank (or secant)
stratification of ${\mathbb{P}}^{34}$ is given by the secant varieties to the
variety $V_{4}=v_{4}({\mathbb{P}}^{3})$ consisting of forms of rank $1$, the
image of the Veronese map of Equation 4. The secant varieties to $V_{4}$,
appearing in Equation 5, form a sequence of closed subvarieties

$V_{4}\subset\operatorname{Sec}_{2}(V_{4})\subset...\subset\operatorname{Sec}_{9}(V_{4})\subset{\mathbb{P}}^{34}.$

The relative complements
$\operatorname{Sec}_{k}(V_{4})\setminus\operatorname{Sec}_{k-1}(V_{4})$, for
$k=2,...,10$ form a stratification of ${\mathbb{P}}^{34}$. Consider the
$10\times 10$ catalecticant matrix $\operatorname{Cat}(F)$ of a quaternary
quartic $F$, which appears in Example 2.10. Let

$\operatorname{Cat}_{k}=\\{[F]|\operatorname{rk}(\operatorname{Cat}(F))\leq
k\\}\subset{\mathbb{P}}^{34}.$

Thus

$\operatorname{Cat}_{k}\cap U_{34}=\cup_{b_{12}\geq
10-k}{\mathcal{F}}_{{[b_{12}b_{23}b_{34}]}}.$

The sets $\operatorname{Cat}_{k}\setminus\operatorname{Cat}_{k-1},k=2,...,10$
form the catalecticant stratification of the space ${\mathbb{P}}^{34}$ of
quaternary quartics, with the obvious closure relation
$\operatorname{Cat}_{k}\supset\operatorname{Cat}_{k-1}$. The relation between
subsets ${\mathcal{F}}_{{[400]}}$ and ${\mathcal{F}}_{{[300a]}}$ explains the
difference between the secant stratification and the catalecticant
stratification:

######  Proposition 6.1.

Let $\operatorname{Sec}_{k}(V_{4})\subset{\mathbb{P}}^{34}$ be the $k$-th
secant variety of $V_{4}$, i.e. the closure of the union of spans of $k$
points on $V_{4}$. Then

  1. (1)

$\operatorname{Sec}_{k}(V_{4})=\operatorname{Cat}_{k}$ if $k\leq 5$,

  2. (2)

$\operatorname{Sec}_{6}(V_{4})=\operatorname{Cat}_{6}\setminus{\mathcal{F}}_{{[400]}}$,

  3. (3)

$\operatorname{Sec}_{7}(V_{4})=\operatorname{Cat}_{7}\setminus({\mathcal{F}}_{{[300a]}}\cup{\mathcal{F}}_{{[400]}})$,

  4. (4)

$\operatorname{Sec}_{8}(V_{4})=\operatorname{Cat}_{8}$,

  5. (5)

$\operatorname{Sec}_{9}(V_{4})=\operatorname{Cat}_{9}$.

###### Proof.

We consider the quadratic part $Q_{F}$ of the apolar ideal $F^{\perp}$ of $F$.
As we observed following Example 2.10,

$\operatorname{rk}(\operatorname{Cat}(F))=10-{\dim}(F^{\bot})_{2}.$

The secant varieties ${\rm Sec}_{k}(V_{4})$ are irreducible, so for each Betti
table $B$, by comparing the rank of a general $F\in{\mathcal{F}}_{B}$ with
$\operatorname{rk}(\operatorname{Cat}(F))$, we get that any quaternary quartic
form $F$ of border rank different from $6$ and $7$ is the limit of forms
$F_{t}$ of rank $r(F_{t})=\operatorname{rk}(\operatorname{Cat}(F))$.

To complete the proof we consider the cases when the quadrics in
$F^{\bot}_{2}$ form a complete intersection $(2,2,2)$ or $(2,2,2,2)$. The
forms $F$ with a complete intersection $(2,2,2,2)$ in $F^{\bot}_{2}$ are those
of type $[400]$, and form an open set in $\operatorname{Cat}_{6}$, whose
general member has rank $8$, so
${\mathcal{F}}_{{[400]}}\subset\operatorname{Sec}_{8}(V_{4})\setminus\operatorname{Sec}_{7}(V_{4})$.

The forms $F$ such that $F^{\bot}_{2}$ is a complete intersection $(2,2,2)$
are those of type $[300]$, and form an open set in $\operatorname{Cat}_{7}$.
The general member is of type $[300b]$ and has rank $8$, so

${\mathcal{F}}_{{[300b]}}\subset\operatorname{Sec}_{8}(V_{4})\setminus\operatorname{Sec}_{7}(V_{4}).$

The forms of rank $7$ with a complete intersection $(2,2,2)$ in $F^{\bot}_{2}$
are those of type $[300a]$, and form an irreducible closed subset. So

${\mathcal{F}}_{{[300a]}}\subset\operatorname{Sec}_{7}(V_{4})\setminus\operatorname{Sec}_{6}(V_{4}),$

as we needed to show. ∎

###  6.1. The Betti stratification

We now show the closure relations between the $19$ irreducible sets of
Proposition 5.5 whose disjoint union is $U_{34}$, the set of nondegenerate
quartics. We call the $19$ sets the Betti strata of $U_{34}$.

For the closure relation between the Betti strata, we start with the following
simple lemma:

######  Lemma 6.2.

Let $\mathcal{F}_{B}\subset{\mathbb{P}}(R_{4})={\mathbb{P}}^{34}$ be one of
the $19$ Betti strata and assume $F\in\mathcal{F}_{B}$. If $Q_{F}\subset
Q_{F^{\prime}}$, then $F^{\prime}\in\overline{\mathcal{F}_{B}}$.

###### Proof.

If $Q_{F}\subset Q_{F^{\prime}}$, then
$(Q_{F^{\prime}})_{4}^{\perp}\subset(Q_{F})_{4}^{\perp}\subset R_{4}$, and so
the lemma follows from the inclusion
${\mathbb{P}}((Q_{F})_{4}^{\perp})\subset\overline{\mathcal{F}_{B}}\subset{\mathbb{P}}(R_{4})$.
∎

First we treat two closure relations between sets $\mathcal{F}_{B}$ with an ad
hoc argument:

######  Proposition 6.3.

${\mathcal{F}}_{[300c]}\subset\overline{{\mathcal{F}}_{[300b]}}\subset\overline{{\mathcal{F}}_{[300a]}}.$

###### Proof.

The set of forms $F$ with $Q_{F}$ a $(2,2,2)$ complete intersection is
irreducible, and by Proposition 5.3, the set ${\mathcal{F}}_{[300a]}$ is a
dense subset with complement ${\mathcal{F}}_{[300b]}$, so
${\mathcal{F}}_{[300b]}\subset\overline{{\mathcal{F}}_{[300a]}}$

Let $Z$ be a scheme of length $7$ whose ideal $J$ contains a complete
intersection $(2,2,2)$. Then $Z$ contains no subscheme of length $3$ contained
in a line. But $Z$ may degenerate in a flat family to a scheme $Z^{\prime}$
with a subscheme of length $3$ in a line. On the other hand, if $Q_{F}=I$ is
the ideal of the union of a line and a subscheme of length $4$, then $F$ has a
decomposition $F_{1}+F_{2}$, where $F_{1}$ is apolar to the line, and $F_{2}$
is apolar to the scheme of length $4$. The form $F_{1}$ is then apolar to a
pencil of subschemes of length $3$ on the line. Then $F$ is apolar to a pencil
of schemes of length $7$ (cf Type $[300c]$: ), with a subscheme of length $3$
in a line. Any such scheme deforms flatly to a complete intersection.
Therefore ${\mathcal{F}}_{[300c]}\subset\overline{{\mathcal{F}}_{[300b]}}$. ∎

We further deduce closure relations between sets $\mathcal{F}_{B}$ from their
relation to closure relations between sets $\mathcal{G}_{B}$ (recall
Definition 1.2).

For every Betti table $B$ for quadratic ideals there is an embedding

$\mathcal{G}_{B}\to\mathbf{Gr}(b_{12},S_{2}),\quad I\mapsto[I_{2}].$

This embedding provides one of two natural closure relations between
irreducible sets of quadratic ideals.

######  Definition 6.4.

We write ${\mathcal{G}}\prec{\mathcal{G}^{\prime}}$ if
${\mathcal{G}}\subset\overline{{\mathcal{G}^{\prime}}}$ and $b_{12}(S/I)$ is
the same for all $I$ in either set. We write
${\mathcal{G}}\sqsubset{\mathcal{G}^{\prime}}$ if a general ideal
$J\in{\mathcal{G}}$ contains some ideal $I\in{\mathcal{G}^{\prime}}$.

######  Proposition 6.5.

The irreducible sets of quadratic ideals $Q_{F}$ for quaternary quartics
satisfy the following relations of the kind ($\prec$), where
$b_{12}=b_{12}(S/Q_{F})$:

  1. (1)

($b_{12}=5$) : $\mathcal{G}_{[562]}$ (a line and two points) $\prec$
$\mathcal{G}_{[551]}$ (five points, four in a plane) $\prec$
$\mathcal{G}_{[550]}$ (five general points).

  2. (2)

($b_{12}=4$) : $\mathcal{G}_{[441a]}$ (a conic and a point) $\prec$
$\mathcal{G}_{[420]}$ (six general),

  3. (3)

($b_{12}=4$): $\mathcal{G}_{[441b]}$ (two skew lines) $\prec$
$\mathcal{G}_{[430]}$ (a line and $3$ points)$\prec$ $\mathcal{G}_{[420]}$
(six general points).

  4. (4)

($b_{12}=3$) : $\mathcal{G}_{[320]}$ (a twisted cubic)$\prec$
$\mathcal{G}_{[300ab]}$ (CI $(2,2,2)$),

  5. (5)

($b_{12}=3$) : $\mathcal{G}_{[300c]}$ (a line and four points)$\prec$
$\mathcal{G}_{[300ab]}$ (CI $(2,2,2)$),

  6. (6)

($b_{12}=3$): $\mathcal{G}_{[331]}$ (a plane and a point) $\prec$
$\mathcal{G}_{[310]}$ (a conic and $2$ points) $\prec$$\mathcal{G}_{[300ab]}$
(CI $(2,2,2)$).

  7. (7)

($b_{12}=2$) : $\mathcal{G}_{[210]}$ (a plane and a line) $\prec$
$\mathcal{G}_{[200]}$ (CI $(2,2)$),

###### Proof.

Interpreted in $\mathbf{Gr}(b_{12},S_{2})$ the closure relation is
straightforward in each of the seven items in the proposition. ∎

######  Proposition 6.6.

The irreducible sets of quadratic ideals $Q_{F}$ for quaternary quartics
satisfy the following relations of the kind $\sqsubset$:

  1. (1)

$\mathcal{G}_{[683]}$ $\sqsubset$ $\mathcal{G}_{[550]}$ $\sqsubset$
$\mathcal{G}_{[420]}$ $\sqsubset$ $\mathcal{G}_{[300ab]}$ $\sqsubset$
$\mathcal{G}_{[200]}$ $\sqsubset$ $\mathcal{G}_{[100]}$.

  2. (2)

$\mathcal{G}_{[683]}$(four general) $\sqsubset$ $\mathcal{G}_{[551]}$ (five
points, four on a plane)

  3. (3)

$\mathcal{G}_{[550]}$(five general) $\sqsubset$ $\mathcal{G}_{[430]}$ (a line
and three points)

  4. (4)

$\mathcal{G}_{[683]}$(four general) $\sqsubset$ $\mathcal{G}_{[562]}$ (a line
and two points) $\sqsubset$ $\mathcal{G}_{[430]}$ (a line and three points)
$\sqsubset$ $\mathcal{G}_{[300c]}$ (a line and four points)

  5. (5)

$\mathcal{G}_{[562]}$ (a line and two points) $\sqsubset$
$\mathcal{G}_{[441b]}$ (two skew lines)

  6. (6)

$\mathcal{G}_{[551]}$ (five points, four on a plane) $\sqsubset$
$\mathcal{G}_{[441a]}$ (a conic and a point) $\sqsubset$ $\mathcal{G}_{[310]}$
(a conic and two points) $\sqsubset$ $\mathcal{G}_{[210]}$ (a plane and a
line)

  7. (7)

$\mathcal{G}_{[441a]}$ (a conic and a point) $\sqsubset$ $\mathcal{G}_{331]}$
(a plane and a point) $\sqsubset$ $\mathcal{G}_{[210]}$ (a plane and a line)

###### Proof.

For each Betti table $B$ and each component
${\mathcal{G}}\subset{\mathcal{G}}_{B}$, an ideal $I\in{\mathcal{G}}$ is the
ideal of a scheme $Z$. By choosing a point $p\in{\mathbb{P}}^{3}$
appropriately, the ideal $J$ of $Z\cup\\{p\\}$ belongs to
${\mathcal{G}}_{B^{\prime}}$. In case (1) the point $p$ is general. In case
(2) it lies in the plane spanned by a subscheme of length $3$ in $Z$. The
remaining cases are similar. ∎

######  Lemma 6.7.

Assume ${\mathcal{G}}_{i}$ is an irreducible component of the locus of
quadratic ideals $Q_{F}$ for forms $F\in{\mathcal{F}}_{B_{i}}$, where
$i\in\\{1,2\\}$. If ${\mathcal{G}}_{1}\sqsubset{\mathcal{G}}_{2}$ or
${\mathcal{G}}_{1}\prec{\mathcal{G}}_{2}$, then
${\mathcal{F}}_{B_{1}}\subset\overline{{\mathcal{F}}_{B_{2}}}$.

###### Proof.

Assume first that ${\mathcal{G}}_{1}\sqsubset{\mathcal{G}}_{2}$. Recall that
$\mathcal{F}_{B_{1}}$ is an irreducible constructible subset in
$\mathbb{P}^{34}$. It follows that it is enough to prove that for a general
$F\in\mathcal{F}_{B_{1}}$ we have $F\in\overline{\mathcal{F}_{B_{2}}}$. Let us
hence fix $F_{1}\in\mathcal{F}_{B_{1}}$ general. Since the map
$\mathcal{F}_{B_{1}}\to{\mathcal{G}}_{1}$ given by $F\mapsto Q_{F}$ is
algebraic $Q_{F_{1}}$ is general in ${\mathcal{G}}_{1}$ and hence by
assumption there exists $F_{2}\in\mathcal{F}_{B_{2}}$ such that
$Q_{F_{2}}\subset Q_{F_{1}}$. Let $L\subset\mathbb{P}^{34}$ be the space of
polynomials of degree 4 which are annihilated by $Q_{F_{2}}$. Clearly $L$ is a
linear subspace such that $F_{1}\in L$. Furthermore, by Lemma 6.2 we have
$L\subset\overline{\mathcal{F}_{B_{2}}}$. In particular
$F_{1}\in\overline{\mathcal{F}_{B_{2}}}$.

Let us now consider the relation ${\mathcal{G}}_{1}\prec{\mathcal{G}}_{2}$.
This part is slightly more delicate. Note that all the relations of type
${\mathcal{G}}_{1}\prec{\mathcal{G}}_{2}$ are listed in Proposition 6.5.
Recall that we have already proven that
$\mathcal{F}_{[300b]}\subset\overline{\mathcal{F}_{[300a]}}$. Furthermore, in
Proposition 5.5 we have a description of all strata $\mathcal{F}_{B}$ in terms
of descriptions of configurations of points computing the rank of a general
element of $\mathcal{F}_{B}$. Observe that for each relation
${\mathcal{G}}_{1}\prec{\mathcal{G}}_{2}$ listed in Proposition 6.5 and not
involving $\mathcal{G}_{[300ab]}$ the ranks of the general elements of
${\mathcal{F}}_{B_{1}}$ and ${\mathcal{F}}_{B_{2}}$ are equal. We verify that,
in these cases, the configurations of points computing the rank of elements of
${\mathcal{F}}_{B_{1}}$ are specialisations of configurations of points
computing ranks of elements of ${\mathcal{F}}_{B_{2}}$.

More precisely, let $F\in{\mathcal{F}}_{B_{1}}$ be general and let
$\Gamma_{F}$ be a configuration of points computing its rank. Let
$\langle\Gamma_{F}\rangle_{4}\subset{\mathbb{P}}^{34}$ be the linear span of
$\Gamma_{F}$ in its embedding in the space of quartics. Then there exists a
sequence of configurations of points $(\Gamma_{n})_{n\in\mathbb{N}}$ such that
their limit is $\Gamma_{F}$ and $\Gamma_{n}$ are configurations of points
computing the ranks of some forms $F_{n}\in{\mathcal{F}}_{B_{2}}$. In fact, in
this case, for each $n$ the general quartic apolar to $\Gamma_{n}$ is in
${\mathcal{F}}_{B_{2}}$. It follows that
$\langle\Gamma_{n}\rangle_{4}\subset\overline{{\mathcal{F}}_{B_{2}}}$.

It is now enough to observe that since all considered configurations of points
give independent conditions on quartics and consist of the same number of
points then we have
$\dim(\langle\Gamma_{n}\rangle_{4})=\dim(\langle\Gamma_{F}\rangle_{4})$. We
infer that $F$ is a limit of some
$F_{n}\in\langle\Gamma_{n}\rangle_{4}\subset\overline{{\mathcal{F}}_{B_{2}}}$,
and therefore $F\in\overline{{\mathcal{F}}_{B_{2}}}$. ∎

We combine Propositions 6.5 and 6.6 with Lemma 6.7 to obtain closure relations
between the Betti strata ${\mathcal{F}}_{B}$. Type $[000]$ is the general
case, hence the closure of $\mathcal{F}_{[000]}$ is ${\mathbb{P}}^{34}$.

######  Proposition 6.8.

The irreducible Betti strata for quaternary quartics satisfy the closure
relations below. In each case, we describe the locus for a general apolar set
$\Gamma$.

  1. (1)

${\mathcal{F}}_{[562]}$ (five points, three on a line)
$\subset\overline{\mathcal{F}_{[551]}}$ (five points, four in a plane)
$\subset\overline{\mathcal{F}_{[550]}}$ (five general points).

  2. (2)

${\mathcal{F}}_{[441a]}$ (six points, five in a plane)
$\subset\overline{\mathcal{F}_{[420]}}$ (six general points),

  3. (3)

${\mathcal{F}}_{[441b]}$ (six points, three on each of two skew lines)
$\subset\overline{\mathcal{F}_{[430]}}$ (six points, three in a line)
$\subset\overline{\mathcal{F}_{[420]}}$ (six general points).

  4. (4)

$\mathcal{F}_{[320]}$ (seven points on twisted cubic)
$\subset\overline{\mathcal{F}_{[300b]}}$ (seven general points),

  5. (5)

$\mathcal{F}_{[300c]}$ (seven points, three in a line)
$\subset\overline{\mathcal{F}_{[300b]}}$ (seven general points),

  6. (6)

$\mathcal{F}_{[331]}$ (seven points, six in a plane)
$\subset\overline{\mathcal{F}_{[310]}}$ (seven points, five in a plane)
$\subset\overline{\mathcal{F}_{[300b]}}$ (seven points in a CI $(2,2,2)$).

  7. (7)

$\mathcal{F}_{[210]}$ (eight points, six in a plane)
$\subset\overline{\mathcal{F}_{[200]}}$ (eight general points),

  8. (8)

$\mathcal{F}_{[683]}$ $\subset\overline{\mathcal{F}_{[550]}}$
$\subset\overline{\mathcal{F}_{[420]}}$
$\subset\overline{\mathcal{F}_{[300b]}}$
$\subset\overline{\mathcal{F}_{[300a]}}$
$\subset\overline{\mathcal{F}_{[200]}}$
$\subset\overline{\mathcal{F}_{[100]}}$.

  9. (9)

$\mathcal{F}_{[683]}$(four general points)
$\subset\overline{\mathcal{F}_{[551]}}$ (five points, four on a plane)

  10. (10)

$\mathcal{F}_{[550]}$(five general points)
$\subset\overline{\mathcal{F}_{[430]}}$ (six points, three in a line)

  11. (11)

$\mathcal{F}_{[683]}$(four general points)
$\subset\overline{\mathcal{F}_{[562]}}$ (five points, three on a line)
$\subset\overline{\mathcal{F}_{[430]}}$ (six points, three on a line)
$\subset\overline{\mathcal{F}_{[300c]}}$ (seven points, three on a line)

  12. (12)

$\mathcal{F}_{[562]}$ (five points, three on a line)
$\subset\overline{\mathcal{F}_{[441b]}}$ (six points, three on each of two
skew lines)

  13. (13)

$\mathcal{F}_{[552]}$ (five points, four on a plane)
$\subset\overline{\mathcal{F}_{[441a]}}$ (six points, five in a plane)
$\subset\overline{\mathcal{F}_{[310]}}$ (seven points, five in a
plane)$\subset\overline{\mathcal{F}_{[021]}}$ (eight points, six in a plane)

  14. (14)

${\mathcal{F}_{[441a]}}$ (six points, five in a plane)
$\subset\overline{\mathcal{F}_{[331]}}$ (seven points, six in a plane)
$\subset\overline{\mathcal{F}_{[210]}}$ (eight points, six in a plane)

The complete intersection case of type $[400]$ requires additional
investigation.

######  Proposition 6.9.

$\mathcal{F}_{[420]}$ (six general points)
$\subset\overline{\mathcal{F}_{[400]}}\subset\overline{\mathcal{F}_{[300a]}}$
(eight points, a complete intersection), while $\mathcal{F}_{[400]}$ does not
lie in the closure of $\mathcal{F}_{[300b]}$ (seven points in a complete
intersection).

###### Proof.

Assume first that a form $F_{0}$ with Betti table $\mathcal{F}_{[400]}$ is a
limit of forms $F_{t}$ in $\mathcal{F}_{[300b]}$, then for each $t\not=0$
there is a net $P_{t}$ of quadrics in $F_{t}^{\perp}$. Let $P_{0}$ be the
limit net. Then $P_{0}$ is a net inside a web $L_{0}$, the web of quadrics in
$F_{0}^{\perp}$. Notice that both $P_{0}$ and $P_{t}$ define a complete
intersection; i.e. a scheme $Z_{t}$ of length $8$. For each $t\not=0$, the
quartic $F_{t}$ is apolar to a distinguished subscheme $Z^{\prime}_{t}\subset
Z_{t}$ of length $7$, which is linked to a point in $Z_{t}$. In the limit,
$Z_{0}$ has a distinguished subscheme $Z^{\prime}_{0}$ of length $7$. The
image $v_{4}(Z^{\prime}_{t})$ in ${\mathbb{P}}^{34}$ spans a
${\mathbb{P}}^{6}$ for each $t$. By apolarity, for each $t\not=0$ the quartic
form $F_{t}$ lies in the span of $v_{4}(Z^{\prime}_{t})$, therefore $F_{0}$
also lies in the span of $v_{4}(Z^{\prime}_{0})$, i.e. $F_{0}$ is apolar to a
scheme of length $7$. But $F_{0}$ is in $\mathcal{F}_{[400]}$, so the rank is
$8$, a contradiction.

Next, assume $F_{0}$ is of type $[420]$, and consider a net $P_{0}$ of
quadrics in the ideal $F_{0}^{\perp}$ that defines a complete intersection: A
general net in the web $L_{0}$ of quadrics in $F_{0}^{\perp}$ suffices. Next,
let $L_{t}$ be a family of webs of quadrics that all contain $P_{0}$, such
that $L_{t}$ defines a complete intersection when $t\not=0$. Then $L_{t}$
generates $F_{t}^{\perp}$ for a quartic $F_{t}$ in $\mathcal{F}_{[400]}$ when
$t\not=0$, and $F_{0}$ is the limit of $F_{t}$. Hence, $\mathcal{F}_{[420]}$
lies in the closure of $\mathcal{F}_{[400]}$.

Finally, ${\mathcal{G}}_{[400]}\sqsubset{\mathcal{G}}_{[300ab]}$, so by Lemma
6.7 and Proposition 5.3,
$\mathcal{F}_{[400]}\subset\overline{\mathcal{F}_{[300a]}}.$ ∎

Now we go in the opposite direction, and prove that certain containments
cannot occur. For this we need upper semi-continuity of some graded Betti
numbers:

######  Lemma 6.10.

Fix a Betti table $B=(\beta_{ij})$, and suppose $I$ is a nondegenerate ideal
such that $I\in\overline{\mathcal{G}_{B}}$. Then the Betti table of $S/I$
satisfies the inequality $b_{i,i+1}(S/I)\geq\beta_{i,i+1}$.

###### Proof.

Note that $b_{i,i+1}=\dim H_{i}((S/I)\otimes\mathcal{K}_{\bullet})_{i+1}$,
where $\mathcal{K}_{\bullet}$ is the Koszul complex. Let
$V=\mathfrak{m}/\mathfrak{m}^{2}$ where $\mathfrak{m}$ is the irrelevant ideal
of $S$. Consider the complex of vector spaces

$(S/I)_{0}\otimes_{{\mathbb{K}}}\wedge^{i+1}V\xrightarrow{\delta_{i+1,1}}(S/I)_{1}\otimes_{{\mathbb{K}}}\wedge^{i}V\xrightarrow{\delta_{i,1}}(S/I)_{2}\otimes_{{\mathbb{K}}}\wedge^{i-1}V$

It is clear that

$b_{i,i+1}=\dim\ker\delta_{i,1}-\dim\mathrm{im}~{}\delta_{i+1,1}.$

Since $I$ contains no linear form, $\delta_{i+1,1}$ does not depend on $I$, so
$\dim\mathrm{im}~{}\delta_{i+1,1}$ is constant. On the other hand,
$\dim\ker\delta_{i,1}=\dim(S/F^{\perp})_{1}\otimes_{{\mathbb{K}}}\wedge^{i}V-\operatorname{rk}\delta_{i,1}$.
As $\operatorname{rk}\delta_{i,1}$ is lower semi-continuous over the parameter
space $G(\beta_{1,2},S_{2})$, we conclude that $b_{i,i+1}$ is upper semi-
continuous. ∎

######  Proposition 6.11.

We have the following non-containments

  1. (1)

$\mathcal{F}_{[441b]}\not\subset\overline{\mathcal{F}_{[550]}}$

  2. (2)

$\mathcal{F}_{[550]}\not\subset\overline{\mathcal{F}_{[441a]}}$

  3. (3)

$\mathcal{F}_{[331]}\not\subset\overline{\mathcal{F}_{[420]}}$

  4. (4)

$\mathcal{F}_{[210]}\not\subset\overline{\mathcal{F}_{[300a]}}$

  5. (5)

$\mathcal{F}_{[331]}\not\subset\overline{\mathcal{F}_{[320]}}$

###### Proof.

The map

${\mathcal{F}}_{B}\to{\mathcal{G}}_{B^{\prime}},\quad F\to Q_{F}$

is surjective for each irreducible stratum ${\mathcal{F}}_{B}$, when
$B^{\prime}$ is the Betti table of $S/Q_{F}$ for some $F\in{\mathcal{F}}_{B}$.
Therefore the first four cases follow immediately from Lemma 6.10.

The last case is more delicate. Let $F_{0}=w^{4}+G_{0}(x,y,z)$. We would like
to show $F_{0}\not\in\overline{\mathcal{F}_{[320]}}$.
Assume $F_{0}$ is a limit of $F_{t}$ in $\mathcal{F}_{[320]}$. For each $t\neq
0$, there is a unique ideal $C_{t}\subseteq F_{t}^{\perp}$ such that $C_{t}$
is a twisted cubic. Therefore, $C_{t}$ is in
$\operatorname{Hilb}^{3m+1}({\mathbb{P}}^{3})$. The limit $C_{0}$ must be a
subideal of $F_{0}^{\perp}$. By [PS85],
$\operatorname{Hilb}^{3m+1}({\mathbb{P}}^{3})$ has two components $H$ and
$H^{\prime}$ where $H$ is the closure of $H_{0}$, the space of smooth twisted
cubics, and $H^{\prime}$ is the closure of

$H_{0}^{\prime}=\\{C^{\prime}\mid~{}C^{\prime}=\mbox{a plane, smooth cubic
curve in ${\mathbb{P}}^{3}$ union a point in ${\mathbb{P}}^{3}$ not on the
curve}\\}.$

All $C_{t}$ with $t\neq 0$ is in $H$, therefore $C_{0}$ must also be in $H$.
On the other hand, note that the second row of of type $[331]$ is $(3,3,1)$,
so $F_{0}^{\perp}$ does not have a subideal in $H\setminus H\cap H^{\prime}$.
Hence $C_{0}\in H\cap H^{\prime}$, which means it is projectively equivalent
to $\langle xz,yz,z^{2},c(x,y,w)\rangle$ for some $c(x,y,w)\in S_{3}$. This is
a contradiction because $\langle xz,yz,z^{2}\rangle$ is not projectively
equivalent to $\langle xw,yw,zw\rangle=F_{0}^{\perp}(2)$. Therefore, the
hypothesis that $F_{0}$ is a limit of $F_{t}$ is false. ∎

Putting everything together yields the stratification appearing in Figure 1 on
the next page. We include in each box the name of the Betti table $B$, the
dimension of the locus $\mathcal{F}_{B}\subseteq{\mathbb{P}}^{34}$, and a set
of points $\Gamma$ such that $\Gamma$ is a minimal set of points apolar to a
general $F\in\mathcal{F}_{B}$.

######  Remark 6.12.

In the next sections we will consider liftings of the Artinian rings $A_{F}$
to AG varieties $X$. The Hilbert polynomial of $X$ then depend only on the
dimension and on the Hilbert function of $A_{F}$. Therefore, if $X$ and
$X^{\prime}$ are liftings of quaternary quartic forms and have the same
Hilbert polynomial, then $X$ deforms to $X^{\prime}$ only if the same holds
for Artinian reductions of $X$ and $X^{\prime}$. And the latter can be read
off of the Betti stratification in Figure 1, since over a locus $\mathcal{H}$
for which the apolar ideals have a fixed Hilbert function the family
$\\{S/F^{\perp}\\}_{F\in\mathcal{H}}$ is flat.

######  Remark 6.13.

Note that in this section we do NOT claim we have a full stratification of
$U_{34}$ by Betti tables. Denote by $P$ a set of subsets of $U_{34}$. By full
stratification we mean that $P$ satisfies the following properties:

  1. (1)

$U_{34}=\coprod_{\mathcal{F}\in P}\mathcal{F}$,

  2. (2)

$\overline{\mathcal{F}}=\coprod_{\mathcal{F^{\prime}}\subseteq\overline{\mathcal{F}}}\mathcal{F}^{\prime}$.

The second property above is equivalent to the so-called _frontier property_ ,
which says if $f\in\mathcal{F}$ and $f\in\overline{\mathcal{F}^{\prime}}$,
then $\mathcal{F}\subseteq\overline{\mathcal{F}^{\prime}}$. Iarrobino and
Kanev provide in [IK]*Section 5 a full stratification of the punctual Hilbert
scheme $\operatorname{Hilb}^{s}({\mathbb{P}}^{n})$. In particular, that
stratification satisfies the frontier property.

The frontier property does not hold in our case. For example, since twisted
cubics degenerate to a union of a line and a plane conic,
$\mathcal{F}_{[320]}$ contains elements with apolar sets in which five of the
seven points are on a plane conic. These elements are in the closure of
$\mathcal{F}_{[310]}$, while the general element in $\mathcal{F}_{[320]}$ is
not, so $\mathcal{F}_{[320]}\cap\overline{\mathcal{F}_{[310]}}\not=\emptyset$
and $\mathcal{F}_{[320]}\not\subset\overline{\mathcal{F}_{[310]}}$.
We may obtain a stratification satisfying the frontier property by applying
the following procedure: Take
$\mathcal{F}_{ij}=\mathcal{F}_{i}\cap\overline{\mathcal{F}_{j}}$ if
$\mathcal{F}_{i}\cap\overline{\mathcal{F}_{j}}$ is neither $\mathcal{F}_{i}$
nor $\emptyset$. Then take
$\mathcal{F}_{i}^{\prime}=\mathcal{F}_{i}-\cup_{j}\mathcal{F}_{ij}$ and
consider the new collection
$P^{\prime}=\\{\mathcal{F}_{i}^{\prime}\\}\cup\\{\mathcal{F}_{ij}\\}$. By
irreducibility of $\mathcal{F}_{i}$,
$\overline{\mathcal{F}_{i}^{\prime}}=\overline{\mathcal{F}_{i}}$. We claim
$P^{\prime}$ satisfies the frontier property: Assume
$f\in\overline{\mathcal{F}_{i}}$ and $f\not\in\mathcal{F}_{i}^{\prime}$. Then
either $f\in\mathcal{F}_{i}$, in which case $f\in\mathcal{F}_{ij}$ for some
$j$, or $f\not\in\mathcal{F}_{i}$, in which case $f\in\mathcal{F}_{ki}$ for
some $k$.

[000] $\dim 34$
---
ten points
[100] $\dim 33$
---
nine points
[200] $\dim 31$
---
eight points
[300a] $\dim 28$
---
eight points, CI
[300b] $\dim 27$
---
seven points
[210] $\dim 25$
---
eight points, six on ${\mathbb{P}}^{2}$
[400] $\dim 24$
---
eight points, CI
[320] $\dim 24$
---
seven points, on a twisted cubic
Type [310] $\dim 24$
---
seven points, five on ${\mathbb{P}}^{2}$
[300c] $\dim 24$
---
seven points, three on ${\mathbb{P}}^{1}$
[420] $\dim 23$
---
six points
[331] $\dim 21$
---
seven points, six on ${\mathbb{P}}^{2}$
[430] $\dim 20$
---
six points, three on ${\mathbb{P}}^{1}$
[441a] $\dim 20$
---
six points, five on ${\mathbb{P}}^{2}$
[550] $\dim 19$
---
five points
[441b] $\dim 17$
---
six points, $3+3$ on ${\mathbb{P}}^{1}$
[551] $\dim 18$
---
five points, four on ${\mathbb{P}}^{2}$
[562] $\dim 16$
---
five points, three on ${\mathbb{P}}^{1}$
[683] $\dim 15$
---
four points

FIGURE 1. Betti table stratification

###  6.2. Relations to Noether-Lefschetz varieties

Each Betti stratum admit the Fermat quartics in its closure, therefore the
general quartic in each stratum is smooth, i.e. defines a smooth K3 surface.
Furthermore, each stratum is invariant with respect to the action of $PGL(3)$,
so the smooth quartics in each stratum form a subset in the moduli space of K3
surfaces. We therefore compare the Betti strata with the Noether-Lefschetz
locus in the moduli space of K3 surfaces.

We consider, for each of the Betti strata, the set of smooth quartics in the
moduli space $\mathcal{M}_{4}$ of K3 surfaces with polarisation of degree 4.
In the latter moduli space we distinguish the Noether-Lefschetz locus of K3
surfaces with Picard number greater than 1. It is a union of an infinite
number of divisors $D_{K}$ called Noether-Lefschetz divisors, indexed by
primitive rank 2 sublattices of the K3 lattice containing the polarisation of
degree 4.

For such a lattice $K$ the divisor $D_{K}$ is the closure in the moduli space
of the locus of all K3 surfaces having $K$ as Picard lattice.

######  Definition 6.14.

For primitive sublattices $K$ of rank $r>2$, $V_{K}$ is a Noether-Lefschetz
variety of codimension $r-1$ in the moduli space $\mathcal{M}_{4}$.

We have the following results.

######  Proposition 6.15.

The following Betti strata of quaternary quartics form Noether-Lefschetz
varieties.

  * •

Quartics of type $[683]$ are Fermat surfaces of Picard number 20 and form a
Noether-Lefschetz variety consisting of a point.

  * •

Quartics of type $[441b]$ are double covers of Kummer surfaces of the product
of two elliptic curves and have Picard rank $18$ and form a $2$-dimensional
Noether-Lefschetz variety.

###### Proof.

For Fermat quartics the result is classical, see for example [Shioda]. H.
Inose [Inose] gave the double cover description of quartics of type $[441b]$,
and their Picard number were computed by Shioda-Inose [Shioda-Inose], see also
[Kuwata]. ∎

######  Corollary 6.16.

The Betti strata of quaternary quartics of types

(27) $\\{[562],[550],[551],[441a],[331],[310],[210]\\}$

do not form Noether-Lefschetz varieties.

###### Proof.

By [Kuwata]*Thm. 1.3, a general quartic of type $[562]$ has Picard number 18.
Indeed, a quartic of type $[562]$ is a Kummer surface associated to a product
of elliptic curves, one of which is the double cover of $\mathbb{P}^{1}$
branched in the points $(1:\zeta_{i})$ with $\zeta_{i}$ being 4-th roots of
unity.

Since the other elliptic curve can be chosen randomly, these are not isogenous
and hence the associated Kummer surface has Picard number 18, by [Kuwata]*Thm.
1.3. Now, since the stratum of type $[441b]$ is irreducible, the stratum of
quartics of type $[562]$ cannot fill a Noether-Lefschetz variety. Consequently
any strata containing $\mathcal{F}_{[562]}$ but not $\mathcal{F}_{[441b]}$
cannot form a Noether-Lefschetz variety. This includes the strata
$\mathcal{F}_{B}$ of quartics of the types in Equation 27. ∎

The proposition above shows in particular that type $[331]$ is not a Noether-
Lefschetz variety. However, in this case we can be more precise. For instance,
the periods of the associated K3 surfaces for type $[331]$ are well
understood: quartics of this type are Galois quadruple covers of
$\mathbb{P}^{2}$ branched along a quartic curve. These were carefully studied
by Kondo in [Kondogenus3, Kondoplanequartics].

######  Lemma 6.17.

The Betti stratum $\mathcal{F}_{[331]}$ of quaternary quartics of type $[331]$
is contained in the family of quartics that are double covers of del Pezzo
surfaces of degree 2. The latter form a Noether-Lefschetz variety $V_{L}$
corresponding to the Picard Lattice $L:=\langle
2\rangle+\langle-2\rangle^{7}$. Furthermore, if $N$ is a Noether-Lefschetz
variety such that $\mathcal{F}_{[331]}\subseteq N$ then $V_{L}\subseteq N$.

###### Proof.

S. Kondo, loc. cit., gives a precise description of the periods of the
associated K3 surfaces. In particular, a general K3 surface of type $[331]$,
which is a general Galois quadruple cover of $\mathbb{P}^{2}$, has Picard
Lattice $L=\langle 2\rangle+\langle-2\rangle^{7}$. We thus have a
$6$-dimensional family of K3 surfaces contained in the $12$-dimensional
Noether-Lefschetz variety $V_{L}$ in the moduli space of polarised K3 surfaces
of degree 4. These do not fill the whole Noether-Lefschetz variety $V_{L}$,
but also are not contained in any smaller Noether-Lefschetz variety. ∎

We can now deduce information about all Betti strata containing the stratum of
type $[331]$.

######  Corollary 6.18.

A Betti stratum $\overline{\mathcal{F}_{B}}$ of quaternary quartics containing
$\mathcal{F}_{[331]}$ cannot be a Noether-Lefschetz variety. In particular,
the Betti stratum $\overline{\mathcal{F}_{[100]}}$ is not a Noether-Lefschetz
divisor. In other words, the K3 surface defined by a very general quartic of
rank $9$ has Picard number $1$.

###### Proof.

Let $q$ be general quartic of the form

$(x^{2}+f_{2}(y,z,t))^{2}+f_{4}(y,z,t),$

for $f_{2}$, $f_{4}$ polynomials of degree 2 and 4 respectively depending only
on the variables $y,z,t$ in $\mathbb{P}(x,y,z,t)$. Then $q$ defines an element
in $V_{L}$ from the proof of Lemma 6.17 and its apolar ring can be checked
with Macaulay2, [M2]*Package QuaternaryQuartics, on the example:

$(x^{2}+y^{2}+z^{2}+t^{2})^{2}+y^{4}+z^{4}+t^{4}$

to have Betti table of type $[000]$. But by Lemma 6.17 any Noether-Lefschetz
variety containing the stratum type $[331]$ must contain the whole variety
$V_{L}$. ∎

######  Remark 6.19.

The fact that the locus of quartics of rank 9 is not a Noether-Lefschetz
divisor can be proven alternatively in the following way. Observe that each
Noether-Lefschetz divisor contains a non-empty locus of hyperelliptic K3’s
that are covers of a smooth quadric. It follows that we can find a flat
1-parameter family of quartics in each Noether-Lefschetz divisor converging to
a double quadric (see [Shah-Trans]). It remains to observe (compute the Betti
table of the apolar ring with Macaulay2, [M2]*Package QuaternaryQuartics )
that a double quadric is of type $[000]$.

######  Remark 6.20.

Let us discuss the types of some other known K3 surfaces. The Vinberg most
singular K3 surface $X_{4}$ given by the equation

$x^{4}=-yzt(y+z+t)$

is of type $[331]$. A general element of the Dwork pencil

$x^{4}+y^{4}+z^{4}+t^{4}-4txyzt=0$

is of type [000]. The K3 quartics $S_{t}\subset\mathbb{P}^{5}$ with
coordinates $x_{1},\dots,x_{5}$ having icosahedral symmetry given by equations

$x_{1}^{4}+\dots+x_{5}^{4}-t(x_{1}^{2}+\dots+x_{5}^{2})^{2}=x_{1}+\dots+x_{5}=0$

for general $t$ are of type [000], however special elements in the pencil are
in smaller strata. For example $S_{0}$ is of type $[550]$.

##  7\. Codimension three varieties in quadrics

We now tackle the question of lifting quartic forms to positive dimensional
AG-varieties. Our goal in this section is to describe AG varieties $X$ of
codimension $3$ in smooth quadric hypersurfaces, in particular those whose
homogeneous coordinate ring has regularity $4$. According to [EPW], AG-
varieties of codimension $3$ in a smooth quadric hypersurface that satisfies
certain cohomological conditions appear as Lagrangian degeneracy loci of a map
between two Lagrangian subbundles of an orthogonal bundle. Furthermore, the
Lagrangian degeneracy loci may have a simpler description as Pfaffian loci. We
start with a description of the Artinian Gorenstein rings.

###  7.1. The Artinian case

Most of the considered Artinian Gorenstein rings up to degree 19 can be seen
explicitly as quotients of the homogeneous coordinate ring of a smooth quadric
surface $Q\subset\mathbb{P}^{3}$. More precisely let $I$ be a homogeneous
ideal in the coordinate ring $S=\mathbb{C}[x_{0},x_{1},x_{2},x_{3}]$ of
$\mathbb{P}^{3}$. Assume that in the ideal there is a smooth quadric $Q$ that
we interpret as $\mathbb{P}^{1}\times\mathbb{P}^{1}$. We can refine the
grading on $S$ so that $\deg x_{0}=\deg x_{1}=(1,0)$ and $\deg x_{2}=\deg
x_{3}=(0,1)$, and assume $Q$ has bidegree $(1,1)$. Let $T=\oplus_{k,l}T(k,l)$
be the bigraded homogeneous coordinate ring of $Q$. We have an exact sequence:

$0\rightarrow S(-1,-1)\stackrel{{\scriptstyle\cdot
Q}}{{\rightarrow}}S\xrightarrow{r}T,$

with the latter map being the restriction map. The image of $r$ is the subring
$\oplus_{k}T(k,k)$.

######  Remark 7.1.

For a fixed ideal Gorenstein ideal $I\subset S$ there might be many different
ideals $J\subset T$ such that $r^{-1}(J)=I$. In particular, if $F$ is the dual
socle generator of $I$, so that $I=F^{\bot}$ and $F_{Q}$ the restriction of
$F$ to $Q$, then $r^{-1}(F_{Q}^{\bot})=F^{\bot}$. The ideal $F_{Q}^{\bot}$ is
the biggest ideal with the property $r^{-1}(F_{Q}^{\bot})=F^{\bot}$. In fact,
there might be smaller ideals $J\subset F_{Q}^{\bot}$ such that
$r^{-1}(J)=F^{\bot}$. From the point of view of apolarity on $T$, note that
the condition $r^{-1}(J)=F^{\bot}$ implies $J(4,4)=F_{Q}^{\perp}(4,4)$,
however the latter condition is weaker.

Remark 7.1 justifies the introduction of the following definition that is
needed in our investigation.

######  Definition 7.2.

Let $F_{Q}$ be a form of degree $(a,b)$ in $T^{\vee}$. We say that an ideal
$J\subset T$ is fully apolar to $F_{Q}$ if $J(a,b)=F_{Q}^{\perp}(a,b)$.

Our aim in this section is to relate the Artinian rings $S/F^{\bot}$ to grade
3 Gorenstein rings obtained as quotients of $T$. The quotient ring
$T/F_{Q}^{\perp}$ is a ring of grade 4. Since the restriction map $r:S\to T\to
T/F_{Q}^{\perp}$ is not surjective, there may still be ideals $J$ of grade 3
such that $r^{-1}(J)=F^{\bot}$. However, this is not always the case. For that
reason we consider the weaker condition of full apolarity to $F_{Q}$. Before
we formulate the proposition let us recall some further terminology.

A Pfaffian ideal $J\subset T$ is an ideal with a resolution of the form

$0\to T(k,l)\to E^{\vee}\xrightarrow{M}E\to J\to 0,$

where $E$ is a free $T$-module and $M$ is an $(2n+1)\times(2n+1)$-dimensional
skew symmetric matrix of bihomogeneous forms. The map $E\to J$ is defined by
the $2n\times 2n$ Pfaffians of $M$. If $n=3$, $J$ is the complete intersection
ideal generated by the three nonzero entries of $M$.

######  Proposition 7.3.

Let $F$ be a quaternary quartic apolar to s smooth quadric $Q$, and let
$F_{Q}$ be the restriction of $F$ to $Q$. Assume $F$ is general in its type.
Then there is an ideal $J$ fully apolar to $F_{Q}$ with a Pfaffian minimal
resolution associated to a skew-symmetric map $\varphi:E^{\vee}\to E$. The
bundle $E$ and the zero blocks of $\varphi$ are listed in the table below, for
each possible type.

Type | $E$ |  Zero blocks of $\varphi$
---|---|---
$[683]$ | $T(2,1)\oplus T(2,1)\oplus T(-1,1)$ |
$[550]$ | $T(2,1)\oplus T(1,2)\oplus T$ |
$[400]$ | $T(1,1)\oplus T(1,1)\oplus T(1,1)$ |
$[300a]$ | $T(1,0)\oplus T(0,1)\oplus 2T(1,1)\oplus T$ |
$[300b]$ | $T(1,0)\oplus T(0,1)\oplus 2T(1,1)\oplus T$ | $T(0,1)\oplus T$
$[300c]$ | $T(1,0)\oplus T(0,1)\oplus 2T(1,1)\oplus T$ |  $T(0,1)\oplus T$, $T(1,0)\oplus T(0,1)$
$[310]$ | $T(1,0)\oplus T(0,1)\oplus 2T(1,1)\oplus T$ |  $T(0,1)\oplus T$, $T(1,0)\oplus T$
$[320]$ | $T(1,0)\oplus 2T(0,1)\oplus T(2,1)\oplus T$ |
$[200]$ | $2T(1,0)\oplus 2T(0,1)\oplus T(1,1)$ |
$[100]$ | $3T(1,0)\oplus 3T(0,1)\oplus T$ |
$[420]$ | $3T(1,1)\oplus 2T$ |
$[430]$ | $T(1,1)\oplus T(2,1)\oplus T(0,1)\oplus 2T$ | $T\oplus T(0,1)$
$[441a]$ | $T(1,2)\oplus T(2,1)\oplus T(0,1)\oplus T(1,0)\oplus T(-1,-1)$ |
$[441b]$ | $T(1,0)\oplus T(1,0)\oplus T(1,3)$ |
$[551]$ | $T(1,2)\oplus T(2,1)\oplus T(1,1)\oplus T(-1,-1)\oplus T$ |
$[562]$ | $T(1,2)\oplus T(2,1)\oplus T(0,1)\oplus T(0,-1)\oplus T$ |
Table 5. The bundle $E$

######  Remark 7.4.

Note that in Proposition 7.3 we consider minimal resolutions of the considered
shape. This in particular implies that the matrix associated to the map
$\varphi$ has no constant nonzero entries. So in all the cases in which, for
degree reasons, constant entries may appear, these constant entries are $0$.

######  Remark 7.5.

Note that the same quartic form $F$ may admit several fully apolar ideals $J$
on a quadric $Q$ apolar to $F$. This includes possible ideals $J$ whose
resolution is different from those listed in Proposition 7.3. Furthermore,
notice that in all cases treated above only type $[100]$ are contained in a
unique quadric. In the remaining cases we have a choice of quadric and the
description depends on the choice. This means that on some quadrics, possibly
even general ones, the restriction $F_{Q}$ might not be fully apolar to a $J$
of the form listed in Proposition 7.3.

###### Proof.

Observe that from the Pfaffian resolution of the ideal $J$ we obtain that the
space $J(4,4)$ has codimension $1$ in $T(4,4)$. It follows that
$J(4,4)=F\perp_{Q}(4,4)$ for some $(4,4)$-form $F_{Q}$. In particular, $J$ is