# Reducibility of a family of local Artinian Gorenstein algebras111Keywords:
Artinian Gorenstein, local algebra, symmetric decomposition, components,
deformation, Hilbert function, Jordan type, Lefschetz property, Macaulay dual
generator, parametrization. 2020 Mathematics Subject Classification: Primary:
13H10; Secondary: 13E10, 13M05, 14B07, 14C05

Anthony Iarrobino
Department of Mathematics, Northeastern University, Boston, MA 02115, USA.
Pedro Macias Marques
Departamento de Matemática, Escola de Ciências e Tecnologia, Centro de
Investigação
em Matemática e Aplicações, Instituto de Investigação e Formação Avançada,
Universidade de Évora, Rua Romão Ramalho, 59, P–7000–671 Évora, Portugal

(December 27, 2021)

###### Abstract

The Jordan type of an Artinian algebra is the Jordan block partition
associated to multiplication by a generic element of the maximal ideal. We
study the Jordan type for Artinian Gorenstein (AG) local (non-graded) algebras
$A$, and the interaction of Jordan type with the symmetric decomposition of
the Hilbert function $H(A)$. We give examples of Gorenstein sequences $H$ for
which the family $\mathrm{Gor}(H)$ of AG algebras having Hilbert function $H$
has three irreducible components, each corresponding to a symmetric
decomposition of $H$. The component structure results from the intersection of
two opposing filtrations of the family $\mathrm{Gor}(H)$ of AG algebras: that
by Jordan type satisfies the usual dominance property; the second filtration,
by symmetric decomposition, satisfies a known semicontinuity property. Our
examples are in codimension three – the lowest codimension of such an example,
as $\mathrm{Gor}(H)$ is irreducible in codimension two. We also compare the
Jordan type of the AG algebra $A$ and of its associated graded algebra
$A^{\ast}$, showing that they can be different. We explore a problem of
realizability of partial symmetric decompositions.

###### Contents

  1. 1 Introduction.
    1. 1.1 Basic Notions.
    2. 1.2 Background and Summary.
  2. 2 Preliminary results.
    1. 2.1 Specialization of Jordan type.
    2. 2.2 Specialization of symmetric decomposition.
    3. 2.3 When the graded quotient $Q_{A}(0)$ of $A$ has codimension two.
    4. 2.4 Jordan type and symmetric decomposition.
    5. 2.5 Jordan types of $A$ and of $A^{\ast}$.
    6. 2.6 Several irreducible components of $\mathrm{Gor}(H)$.
    7. 2.7 Exotic summands.
    8. 2.8 Relatively compressed $a$-modifications
  3. 3 Applying Jordan type and symmetric decomposition for a local AG algebra.
    1. 3.1 Examples of AG families $\mathrm{Gor}(H)$ having three irreducible components.
    2. 3.2 Infinite families of $\mathrm{Gor}(H)$ having at least two irreducible components.
    3. 3.3 Realizability.
    4. 3.4 Further questions.

##  1 Introduction.

Let $R={\mathsf{k}}\\{x_{1},\ldots,x_{r}\\}$ be the regular local ring in $r$
variables over an infinite field $\mathsf{k}$; for simplicity we will assume
in this paper that $R$ has standard grading. We assume that $A$ is an Artinian
Gorenstein (AG) local algebra $A=R/I$ with maximal ideal $\mathfrak{m}_{A}$
(or $\mathfrak{m}$, when $A$ is understood) of socle degree $j=j(A)$, such
that $A/\mathfrak{m}=\mathsf{k}$. Recall that the Hilbert function $H(A)$
satisfies $H(A)=(h_{0},h_{1},\ldots,h_{j})$ where
$h_{i}=H(A)_{i}=\dim_{\mathsf{k}}(\mathfrak{m}^{i}/\mathfrak{m}^{i+1})$ and
$h_{j}>0$. We will say that $H=(1,r,\ldots,1_{j})$ is a _Gorenstein sequence_
of codimension $r$ if there is an AG local algebra $A$, quotient of $R$ such
that $H(A)=H$. The algebras we study are in general non-homogeneous and our
Gorenstein sequences $H$ will in general not be symmetric.222In the
literature, a Gorenstein sequence $H$ is often assumed to be symmetric, in
which case there is a homogeneous AG algebra with Hilbert function $H$, as
$H(A)$ symmetric implies $A^{\ast}$ is AG ([Wat, Proposition 9] or [I2,
Proposition 1.7]). For $H=(1,r,\ldots,1)$ a Gorenstein sequence, we will
denote by $\mathrm{Gor}(H)$ the reduced variety whose closed points are the
Gorenstein quotients $A$ of $R$ having Hilbert function $H$. In this paper we
exhibit a new method for showing that certain varieties $\mathrm{Gor}(H)$ for
$H$ of codimension three, have several irreducible components – each
corresponding to a _symmetric decomposition_ of $H$.

###  1.1 Basic Notions.

The first author showed that the associated graded algebra
$A^{\ast}=\mathrm{Gr}_{\mathfrak{m}}(A)$ has a filtration by ideals $C_{A}(a)$
whose successive quotients $Q_{A}(a)$ are reflexive ${\mathsf{k}}$ modules. We
first define these ideals and subquotients of $A^{\ast}$, that we term the
_symmetric decomposition_ for $A$. This is our first basic ingredient.

######  Definition 1.1 (Symmetric Decomposition).

Let $A$ be an AG algebra. We define the ideal $C(a)\subset A^{\ast}$ by

$C(a)_{i}=\mathfrak{m}^{\,i}\cap(0:\mathfrak{m}^{\,j+1-a-i})/\bigl{(}\mathfrak{m}^{\,i+1}\cap(0:\mathfrak{m}^{\,j+1-a-i})\big{)}.$
(1.1)

We let $Q(a)_{i}=C(a)_{i}/C(a+1)_{i}$.

Then

$Q(a)_{i}\cong\frac{\mathfrak{m}^{\,i}\cap(0:\mathfrak{m}^{\,j+1-a-i})}{\mathfrak{m}^{\,i}\cap(0:\mathfrak{m}^{\,j-a-i})+\mathfrak{m}^{\,i+1}\cap(0:\mathfrak{m}^{\,j+1-a-i})}.$
(1.2)

Recall from [I2, Theorem 1.5] or [IM, §1.1],

######  Lemma 1.2 (Symmetric Decomposition of $A^{\ast}$).

Let $A$ be an Artinian Gorenstein algebra, then we have a filtration

$A^{\ast}=C_{A}(0)\supset C_{A}(1)\supset C_{A}(2)\supset\cdots\supset
C_{A}(j-2)\supset C_{A}(j-1)=0,$ (1.3)

whose successive quotients $Q_{A}(a)=C_{A}(a)/C_{A}(a+1)$, satisfy, $Q_{A}(0)$
is a standard-graded Artinian Gorenstein algebra, while each $Q_{A}(a)$,
$a\geq 1$ satisfies
$\mathrm{Hom}_{\mathsf{k}}\bigl{(}Q(a)_{i},\mathsf{k}\bigr{)}\cong
Q(a)_{j-a-i}$, that is, the double dual of $Q(a)$ as graded
$\mathsf{k}$-module is isomorphic to $Q(a)$, after the shift; the center of
symmetry of $Q(a)$ is $(j-a)/2$. 333We will show in Example 1.6 that $Q(a)$ is
not necessarily reflexive as an $A^{\ast}$-module.

Letting $H_{A}(a)$ be the Hilbert function of $Q_{A}(a)$, we define the
_symmetric decomposition_ $\mathcal{D}(A)$ of the Hilbert function $H(A)$ as
the sequence

$\mathcal{D}(A)=\bigl{(}H_{A}(0),H_{A}(1),\ldots,H_{A}(j-2)\bigr{)}.$ (1.4)

This symmetric decomposition has recently been used by several groups,
particularly in the study of short or special Artinian Gorenstein algebras,
and in the study of the scheme lengths of forms: see [BJMR, ER1, ER2, Je] and
the discussion and references in [IM]. Writing $\mathsf{m}$ for the maximal
ideal of $R$, we denote by
$\mathrm{Gor}(H)\subset\mathrm{Grass}(n,R/\mathsf{m}^{n})$, the family (an
algebraic set) of AG algebra quotients $I\to R\to A=R/I$ of $R$ having Hilbert
function $H$: thus, a point $p_{A}$ of $\mathrm{Gor}(H)$ specifies the ideal
$I\subset R$. We denote by $\mathrm{Gor}(\mathcal{D})\subset\mathrm{Gor}(H)$
the subfamily of $\mathrm{Gor}(H)$ parametrizing AG algebras whose associated
graded algebras have symmetric decomposition $\mathcal{D}$.Our next basic
ingredient is the Jordan type $P_{\ell,A}$ for a pair consisting of an
Artinian local algebra $A$ and an element $\ell\in\mathfrak{m}_{A}$. This was
introduced in [H-W, §3.5] and in [IMM1, Definition 2.1, Section 2.2], as a
generalization of the much-studied concept of $A$ being weak or strong
Lefschetz (see [H-W]).

######  Definition 1.3.

[IMM1], [H-W, §3.5] The _Jordan type_ $P_{\ell}$ of an element
$\ell\in\mathfrak{m}$ is the partition giving the Jordan block decomposition
of the multiplication map $m_{\ell}:A\to A$, $m_{\ell}(a)=\ell\cdot a$. Since
$\ell$ is nilpotent, the eigenvalues of $m_{\ell}$ are zero. We will say that
$(A,\ell)$ is a _strong Lefschetz pair_ if the Jordan type
$P_{\ell}=H(A)^{\vee}$, the conjugate partition (switch rows and columns in
the Ferrers graph) of $H(A)$.444To obtain $H(A)^{\vee}$ we reorder the
sequence $H(A)$ to make it non-increasing, and then take the conjugate. We say
that $A$ is strong Lefschetz (in [IMM1] this is termed “strong Lefschetz
Jordan type”) if there exists a strong Lefschetz pair $(A,\ell)$. The _Jordan
type_ $P_{A}$ (the generic Jordan type) of $A$ is the partition $P_{\ell}$ for
$\ell$ a generic element in $\mathfrak{m}_{A}$.

When $P_{A}=(p_{1},\ldots,p_{t})$ (we will assume $p_{1}\geq
p_{2}\geq\cdots\geq p_{t}$) we may write $A=\oplus\mathcal{L}(i)$, where each
$\mathcal{L}(i)\cong{\mathsf{k}}[t]/(t^{p_{i}})$ is a simple $k[t]$-module
under the action $t(a)=\ell\cdot a$. We call such a submodule of $A$ an
_$\ell$ -string_ $\mathcal{L}(i)$ of $A$, and we will call the representatives
of $t^{i},i\in[0,p_{1}-1]$ in $A$ the _beads_ of the string.

######  Example 1.4.

The algebra $A={\mathsf{k}}\\{x,y\\}/(x^{2}+y^{3},\,xy)$ of Hilbert function
$H(A)=(1,2,1,1)$ is strong Lefschetz as $\ell=(x+y)$ satisfies
$\ell^{3}=y^{3}\not=0$ in $A$, but $\ell^{4}=0$, so there is a string $\langle
1,\ell,\ell^{2},\ell^{3}\rangle$. Here $\langle x+y^{2}\rangle$ is a length-
one string as $\ell(x+y^{2})=0$, implying the Jordan type is
$P_{\ell}=(4,1)=H(A)^{\vee}$.

We briefly recall the Macaulay duality for an Artinian Gorenstein ring: the
local ring $R=\mathsf{k}\\{x_{1},\ldots,x_{n}\\}$ acts on the divided power
ring ${S=\mathsf{k}_{DP}[X_{1},\ldots,X_{r}]}$ by contraction: thus
$x_{i}^{\,k}\circ X_{i}^{\,j}=X_{i}^{\,j-k}$ if $j\geq k$ and is $0$
otherwise. For an ideal $I\subset R$ we denote by $I^{\perp}=\\{F\in S\mid
h\circ F=0,\forall h\in I\\}$; for an element $F\in S$ we denote by
$\operatorname{\mathrm{Ann}}F\subset R$ the ideal
$\operatorname{\mathrm{Ann}}F=\\{h\in R\mid h\circ F=0\\}$. Let $H$ be a
length $n$ Gorenstein sequence of codimension $r$, one occurring as the
Hilbert function of an Artinian Gorenstein algebra quotient of $R$ of socle
degree $j$ ($H_{j}\not=0$, $H_{i}=0$ for $i>j$). We denote by $S_{\leq j}$ the
$R$-submodule of $S$ comprised of all elements in $S$ of degree at most $j$,
and by $\mathcal{E}_{j}$ the set of classes $\overline{F}$ of degree-$j$
elements by the contraction action of units in $R$; that is for $F$, $G$ of
degree $j$ in $S_{\leq j}$ the class $\overline{F}=\overline{G}$ if there
exists a unit $u\in R^{\ast}$ such that $G=u\circ F$; we denote by
$\mathcal{E}_{j,H}$ the subset of classes in $\mathcal{E}_{j}$ such that
$H(R\circ F)=H$ (equivalently, so $H(R/\operatorname{\mathrm{Ann}}F)=H$). Then
we have

######  Lemma 1.5.

[Mac2],[I2, Lemma 1.1] The Artinian Gorenstein (AG) algebras $A=R/I$ of socle
degree $j$ correspond 1 to 1 to the set $\mathcal{E}_{j,H}$ of classes
$\overline{F}$ of degree $j$ elements $F\in S_{\leq j}$ $\mod$ the contraction
action by units of $R$. The class $\overline{F}$ determines the submodule
$A^{\vee}=R\circ F\subset S$ of Hilbert function $H$. That is, we define

$\displaystyle\iota$ $\displaystyle:\ \mathrm{Gor}(H)\to\mathcal{E}_{j,H}$
$\displaystyle\iota(A)$ $\displaystyle=\overline{F},\text{ where }R\circ
F=I^{\perp}$ (1.5) $\displaystyle\iota^{\prime}$
$\displaystyle:\mathcal{E}_{j,H}\to\mathrm{Gor}(H),$
$\displaystyle\iota^{\prime}(\overline{F})$ $\displaystyle=A_{F}\text{ where
}A_{F}=R/\operatorname{\mathrm{Ann}}F.$

Then $\iota$ and $\iota^{\prime}=\iota^{-1}$ are inverse isomorphisms.

For $F=F_{j}+F_{j-1}+\cdots$ and $A=R/\operatorname{\mathrm{Ann}}F$, the
maximum length graded Artinian Gorenstein quotient of $A^{\ast}$ is
$Q_{A}(0)=A^{\ast}/C_{A}(1)$, and
$Q_{A}(0)=R/\operatorname{\mathrm{Ann}}F_{j}$.

We term an element $F\in S$ determining an AG algebra $A=R/I$,
$I=\operatorname{\mathrm{Ann}}F$, a Macaulay _dual generator_ of $A$. It is a
way to view and to construct AG algebras, that we will use frequently. For a
more detailed study see [I2, IM, KlKl].

Next, we show by example that, although the modules $Q(a)$ satisfy a nice
reflexive property as $\mathsf{k}$-vector spaces, they are not reflexive in
general as $A^{*}$-modules.

######  Example 1.6.

Let ${R=\mathsf{k}[x,y,z]}$, consider ${F=X^{3}YZ+Y^{4}}$, and take
${A=R/\operatorname{\mathrm{Ann}}F}$. Then ${H(A)=(1,3,5,4,3,1)}$, and its
symmetric decomposition is

$\mathcal{D}=\bigl{(}H(0)=(1,3,4,4,3,1),\,H(1)=(0,0,1,0,0)\bigr{)}.$

We have that
${\operatorname{\mathrm{Ann}}F=(z^{2},\,xy^{2},\,y^{3}-x^{3}z,\,y^{2}z,\,x^{4})}$
and ${A^{*}=R/(\operatorname{\mathrm{Ann}}F)^{*}}$ has a two-dimesional socle,
generated by $y^{2}$ and $x^{3}yz$, since
$(\operatorname{\mathrm{Ann}}F)^{*}=(z^{2},\,xy^{2},\,y^{3},\,y^{2}z,\,x^{4})$.
We can check that $Q(1)=\langle y^{2}\rangle$, and therefore
${\mathrm{Hom}_{A^{*}}\bigl{(}Q(1),A^{*}\bigr{)}=\langle\phi_{1},\phi_{2}\rangle}$,
where $\phi_{1}(y^{2})=y^{2}$ and $\phi_{2}(y^{2})=x^{3}yz$, since the image
of $y^{2}$ must be in the socle of $A^{*}$. But now we see that the double
dual of $Q(1)$ has dimension four as a $\mathsf{k}$-vector space, since a
morphism from $\mathrm{Hom}_{A^{*}}\bigl{(}Q(1),A^{*}\bigr{)}$ to $A^{*}$ is
defined by

$\phi_{1}\mapsto a_{1}y^{2}+b_{1}x^{3}yz,\qquad\phi_{2}\mapsto
a_{2}y^{2}+b_{2}x^{3}yz.$

In particular, we cannot have an isomorphism between $Q(1)$ and its double
dual, so $Q(1)$ is not reflexive as an $A^{*}$-module.

###  1.2 Background and Summary.

We will assume henceforth that the regular local ring
$R={\mathsf{k}}\\{x,y\\}$ or ${\mathsf{k}}\\{x,y,z\\}$, according to context –
unless otherwise stated. Given a Gorenstein sequence $H$ (not necessarily
symmetric) recall that by $\mathrm{Gor}(H)$ we understand the (reduced)
variety whose geometric points are the local Gorenstein quotients $A=R/I$
having Hilbert function $H$. It is known, since J. Briançon’s thesis [Bri] and
the first author’s Memoir [I1] that for $T$ any O-sequence of codimension two
(satisfying Equation (2.6)), and for $\sf k$ algebraically closed, the variety
$Z_{T}$ parametrizing all ideals of Hilbert function $T$ is smooth and
irreducible, and is fibred by affine spaces over the smooth projective variety
$G_{T}$ parametrizing graded ideals of Hilbert function $T$ (see Lemma 2.8
below). For a codimension two Gorenstein sequence $H$ (see Lemma 2.8.A) the
variety $\mathrm{Gor}(H)$ has Zariski closure $Z_{H}$. Also in codimension
two, given $H$, there is a unique symmetric decomposition [I2, Theorem 2.2];
finally, $\mathrm{Gor}(H)$ is smooth of known dimension [Bri, I1], fibred by
opens in an affine space over $\mathrm{GGor}(H)$ [I1] (see Section 2.3 and for
the fibration also [HaHy]). Thus, $\mathrm{Gor}(H)$ is irreducible in
codimension two.

In codimension three S. Diesel showed that the _graded_ Gorenstein algebras
$\mathrm{GGor}(H)$ of Hilbert function $H$ form an irreducible family of known
dimension [Di]: to prove this she used the D. Buchsbaum - D. Eisenbud Pfaffian
structure theorem [BuEi]. She also showed results about the poset of
generator-relation degrees corresponding to a given Hilbert function $H$ (for
a summary, see [IK, Theorem 5.25]). In codimension four, the family
$\mathrm{GGor}(H)$ of graded Gorenstein algebras $A=R/I$ of Hilbert function
$H$ may itself have several irreducible components: this was shown in [IS,
Theorem 4.9] for the infinite class of graded Gorenstein sequences beginning
$H=(1,4,7,h,b,\ldots)$, $8\leq h\leq 10$, $b\leq 3h-17$. A general method here
in codimension four is interpreting the ideal $(I_{\leq 3})$ generated by the
initial portion of $I$ as the ideal of a curve in $\mathbb{P}^{3}$. Then some
of the complicated component structure of the Hilbert scheme of curves on
$\mathbb{P}^{3}$ can be shown to transfer to give irreducible components of
$\mathrm{GGor}(H)$ (see also [Kl2] and [IK, Chapter 6]).

#### Summary.

Section 2 presents tools we will use, as well as examples and questions.
Sections 2.1 and 2.2 report the well-known results of how Jordan types
(Section 2.1) and symmetric decompositions $\mathcal{D}$ (Section 2.2) may
specialize. Section 2.3 gives a brief summary of parametrization results about
graded Gorenstein algebras in codimension two, for the case - normal for us -
when $Q(0)$ has codimension two. Section 2.4 asks about the relation between
Jordan bases for an AG algebra $A$ and $A^{\ast}$, and the symmetric
decomposition of $A^{\ast}$: can we find a Jordan basis, or even a pre-Jordan
basis for $A$ (not requiring each string to end in zero, Definition 2.10) that
induces a basis for each $Q_{A}(a)$. In general we cannot (Example 2.15), and
this is an insurmountable obstruction to directly defining a notion of Jordan
order type for non-graded AG algebras that would generalize the symmetric
Jordan degree type (specify the initial degrees of strings) for graded AG
algebras introduced implicitly in [HW] and explicitly in [IMM1, §2.6] and
[CGo].

In Section 2.5 we study in codimension three the Jordan types of $A$, of its
associated graded algebra $A^{\ast}$ and of the subquotients of $A^{\ast}$
that we term the symmetric decomposition of $A^{\ast}$ (Definition 1.1). There
is, as in the graded case, an inequality between the Jordan type of $A$ and
the conjugate partition of the Hilbert function $H(A)$ (Lemma 2.4 below from
[IMM1, Theorem 2.5]). We give several examples where $P_{A}>P_{A^{\ast}}$,
including Example 2.22 where $H(A)$ is symmetric. We introduce in Definition
2.23 further invariants, a contiguous partition $P_{c}(H)$ associated to $H$,
and related invariants $P_{\mathcal{D}}$ and $P_{c}(\mathcal{D})$ relevent
when $H$ or a component $H(a)$ of $\mathcal{D}$ is non-unimodal; and we give
in Lemma 2.25 an inequality between $P_{c}(H)$ and $P_{A,\ell}$ when $H(A)=H$.

In Section 2.6 our Example 2.27 is our first using the previous tools to show
that $\mathrm{Gor}(H)$ \- here $H=(1,3,4,4,4,2,1)$ \- may have several
irreducible components. There, a single stratum
$\mathrm{Gor}(\mathcal{D}_{2})$ has several irreducible components (Claim
2.28). This example shows also that a concantenated Jordan type associated to
the symmetric decomposition of $A^{\ast}$ may be larger in the dominance
partial order than that of $A^{\ast}$, a surprise to us. In Example 2.31 we
give a simple case illustrating for $H=(1,3,5,4,4,2,1)$ our method of showing
that $\mathrm{Gor}(H)$ may have several irreducible components in codimension
three.

In Section 2.7 we give a concise presentation of the theory of exotic summands
of the Macaulay dual generator $F$, from [BJMR, §3.2]; in Lemma 2.35 we
determine the number of exotic parameters of $\mathrm{Gor}(H)$ when
$H\bigl{(}Q(0)\bigr{)}$ has codimension two, and $H(A)$ has codimension three.
This will help us determine the dimensions of some of the irreducible
components of $\mathrm{Gor}(H)$ in the following sections. The concept of
relatively compressed modification in Section 2.8 from [I2, IM] helps in
giving the structure of some of the components $\mathrm{Gor}(\mathcal{D})$ we
will see later.

In Section 3 we study the irreducible components of several $\mathrm{Gor}(H)$
in codimension three.

In contrast to the case of symmetric $H$ in codimension three, here the
variety $\mathrm{Gor}(H)$ parametrizing all AG algebras of a non-symmetric
Hilbert function $H$ may be reducible. This was shown by the first author for
the Hilbert function $H=(1,3,3,2,1,1)$; the proof used the semicontinuity of
symmetric decomposition (Lemma 2.6) and a dimension calculation [I2, Theorem
4.2]. Our main results in Section 3.1 introduce a new method for showing the
reducibility of certain $\mathrm{Gor}(H)$ in codimension three, by comparing
the Jordan type of $A$ under dominance partial order, and a natural partial
order on symmetric decompositions. We use this method to show that there are
three irreducible components for $H=(1,3,4,4,3,2,1)$, corresponding to the
three occurring symmetric decompositions:

######  Theorem 1.

(Theorem 3.1) The Gorenstein sequence $H=(1,3,4,4,3,2,1)$ has three symmetric
decompositions,

$\displaystyle\mathcal{D}_{1}$
$\displaystyle=\bigl{(}H(0)=(1,2,3,4,3,2,1),\,H(1)=H(2)=0,\,H(3)=(0,1,1,0)\bigr{)}$
$\displaystyle\mathcal{D}_{2}$
$\displaystyle=\bigl{(}H(0)=(1,2,3,3,3,2,1),\,H(1)=0,\,H(2)=(0,1,1,1,0)\bigr{)}$
$\displaystyle\mathcal{D}_{3}$
$\displaystyle=\bigl{(}H(0)=(1,2,2,2,2,2,1),\,H(1)=(0,1,2,2,1,0)\bigr{)}.$

The generic Jordan type of these algebras, is, respectively
$P_{1}=(7,5,3,2,1)$, $P_{2}=(7,5,3,3)$, and $P_{3}=(7,5,4,2)$. These symmetric
decompositions stratify $\mathrm{Gor}(H)$ into three irreducible components of
dimensions, respectively, $29$, $34$ and $28$ for $\mathcal{D}_{1}$,
$\mathcal{D}_{2}$ and $\mathcal{D}_{3}$.

We extend this result to the infinite set of Gorenstein sequences
$H=(1,3,4^{k},3,2,1)$, $k\geq 2$ in Theorem 3.3. In Section 3.2, Example 3.5
and Theorem 3.6, we exhibit a doubly infinite sequence of families
$\mathrm{Gor}(H)$ in codimension three, each $H$ having having at least two
symmetric decompositions, each of which determines an irreducible component of
$\mathrm{Gor}(H)$. Our method of proof uses the dominance partial order on
Jordan type (Lemma 2.3) and the semicontinuity of symmetric decomposition
(Lemma 2.6, [I1, §4]). Perhaps surprisingly, these work in opposite
directions.

There are obvious questions concerning the growth of the number of irreducible
components of $\mathrm{Gor}(H)$ as function of the codimension, socle degree
and Sperner number (maximum height) of $H$, which we plan to address in a
subsequent paper.

In Section 3.3 we discuss whether the natural conditions on a potential
symmetric decomposition of $H$ implied by Lemma 1.2 are sufficient: is any
such potential decomposition _realizable_ , that is, does it actually occur
for some AG algebra $A$? An example with answer “no” was given for
$H=(1,4,3,4,3,1)$,
$\mathcal{D}=\bigl{(}H(0)=(1,3,3,3,3,1),\,H(1)=(0,1,0,1,0)\bigr{)}$ in [I2,
p.99]; S. Masuti and M. Rossi have shown that the answer is “Yes” for socle
degree at most four [MasR]. Very recently S. Masuti has obtained partial
results in socle degree five, suggesting that there are only a very few non-
realizable decompositions [Mas]. We show a second main result (Theorem 3.10):

######  Theorem 2.

Let $R=\mathsf{k}\\{x_{1},\ldots,x_{r}\\}$, $S=\mathsf{k}_{DP}[X_{1},\ldots
X_{r}]$, $R^{\prime}=R\\{w\\}$, $S^{\prime}=S_{DP}[W]$. Let $B$ be a graded
Gorenstein algebra of Hilbert function $H(0)$ of socle degree $j$, and assume
that $B=R/I_{B}$ where $I_{B}=\operatorname{\mathrm{Ann}}f\cap R$ for an
element $f\in S_{j}$. Assume further that $j\geq 5$ and that $I_{B}$ is
generated in degrees less or equal $j-3$.555In effect, $j\geq 8$. Let
$H(1)=(0,1,0,\ldots,0,1_{j-2},0)$, and assume that $H=H(0)+H(1)$ satisfies the
Macaulay conditions. Then $\mathcal{D}=\bigl{(}H(0),H(1)\bigr{)}$ is not
realizable as the (complete) decomposition sequence of an AG algebra quotient
$A$ of $R^{\prime}$ having $Q(0)$ isomorphic to $B$.

The question of non-ubiquity asks whether there is a decomposition
$\mathcal{D}(A):H=\sum_{0}^{k}H(a)$ such that some
$H^{\prime}=\sum_{0}^{k^{\prime}}H(a)$ is not realizable as $\mathcal{D}(B)$
for a value $k^{\prime}<k$. In [IM] we showed a partial non-ubiquity result
where we imposed further conditions on $B$. Our examples here, however, do not
imply a non-ubiquity result, and the general question of non-ubiquity remains
open.

We discuss related questions and pose further problems in Section 3.4.

##  2 Preliminary results.

We set out ingredients we will use to work with deformations of algebras.

###  2.1 Specialization of Jordan type.

We first recall the dominance partial order on partitions of $n$. In Lemma 2.4
(see [IMM1, Theorem 2.5]) we remind that the conjugate $H(A)^{\vee}$
(Definition 1.3) to $H(A)$ is an upper bound for the Jordan type of an element
$\ell\in\mathfrak{m}_{A}$.

######  Definition 2.1.

[Dominance order on partitions of $n$.] Let $P=(p_{1},p_{2},\ldots,p_{s})$,
$p_{1}\geq p_{2}\geq\cdots\geq p_{s}$ and
$P^{\prime}=(p_{1}^{\prime},p_{2}^{\prime},\ldots p_{s^{\prime}}^{\prime})$
with $p^{\prime}_{1}\geq p^{\prime}_{2}\geq\cdots\geq
p^{\prime}_{s^{\prime}}$. Then we say $P\leq P^{\prime}$ in the dominance
partial order if for each $i\in[1,\mathrm{max}\\{s,s^{\prime}\\}]$ we have

$\sum_{k=1}^{i}p_{i}\leq\sum_{k=1}^{i}p^{\prime}_{i}.$ (2.1)

The following results are well-known. We thank Steven Kleiman for the two
references D. Mumford [Mum, Prop. 8, p. 44] (with an unnecessary hypothesis of
$W$ irreducible) and R. Hartshorne [Hart, Exercise 5.8(c), p. 125] for the
first Lemma.

######  Lemma 2.2 (Flat family).

Suppose $\mathsf{k}$ is algebraically closed, that $W$ is a reduced variety
and let $\pi:M\to W$ be an algebraic family, whose fibres are Artinian
schemes. Then $M$ is a flat family if and only if the length of the fibres
$M_{w}=\pi^{-1}(w)$, $w\in W$ is constant.

######  Lemma 2.3.

Suppose $\mathsf{k}$ is algebraically closed. Let $\mathcal{A}(w)$, $w\in W$
be a flat family of Artinian algebras over a reduced, irreducible parameter
space $W$. Then there is a generic Jordan type $P_{W}$: it is the generic
Jordan type of $\mathcal{A}(w)$ that occurs for an open dense set of $w\in W$.
Let $w_{0}\in W$. Then the generic Jordan type $P_{0}$ of
$\mathcal{A}(w_{0})$, satisfies $P_{W}\geq P_{0}$ in the dominance partial
order of Definition 2.1.

###### Proof.

This is straightforward to show from the semicontinuity of the rank of
$\ell(t)^{i}$, see [CM, Thm. 6.2.5] or [IMM1, Corollary 2.44]. $\square$

We have from [IMM1, Theorem 2.5]

######  Lemma 2.4.

Let $A=(A,\mathfrak{m},{\mathsf{k}})$ be a local Artinian algebra over
$\mathsf{k}$ and let $M\subset A^{k}$ be an $A$-module, with Hilbert function
$H(M)$. For any $\ell\in\mathfrak{m}_{A}$, its Jordan type satisfies

$P_{\ell,M}\leq H(M)^{\vee}$ (2.2)

in the dominance partial order.

Note that the associated graded algebra $A^{\ast}$ of an Artinian algebra $A$
is a specialization of $A$, hence $P_{A}\geq P_{A^{\ast}}$.

######  Lemma 2.5 (Dominance).

Suppose $\mathsf{k}$ is algebraically closed, and let $\mathcal{A}(w)$, $w\in
W$ be a flat family of finite AG algebras over a reduced, irreducible
parameter space $W$. Denote by $P^{\ast}_{W}$ the generic Jordan type of
$\mathcal{A}^{\ast}(w)$ for $w$ in an open dense subset of $W$; and by
$P^{\ast}_{0}$ the generic Jordan type of $A_{0}^{\ast}$ where
$A_{0}=\mathcal{A}(w_{0})$. Then $P^{\ast}_{W}\geq P^{\ast}_{0}$ in the
dominance partial order of Definition 2.1.

###### Proof.

By Lemma 2.2 the family $\mathcal{A}^{\ast}(w)$, $w\in W$ is also flat, as the
fibres have constant length. By examining the behavior of
$\mathfrak{m}_{\mathcal{A}(w)}^{\,i}$ we may conclude that the the family
$\mathcal{A}^{\ast}(w)$, $w\in W$ indeed specializes to $A_{0}^{\ast}$.

This also follows from the principle that localization is a flat functor, and
$A\to A^{\ast}$ is a localization. The conclusion now follows from Lemma 2.3.

$\square$

###  2.2 Specialization of symmetric decomposition.

We recall from [I2, §4] that there is a partial order $\leq$ on symmetric
decompositions of a fixed Hilbert function $H$ such that
$\overline{\mathrm{Gor}(\mathcal{D})}\cap\mathrm{Gor}(\mathcal{D}^{\prime})\not=\emptyset\Rightarrow\mathcal{D}\leq\mathcal{D}^{\prime}$,
that is, the symmetric decompositions $\mathcal{D}$ of the Hilbert function
$H$ satisfy a semicontinuity property consistent with the partial order (Lemma
2.6). This result follows from the semicontinuity of
$\dim\bigl{(}\mathfrak{m}^{i}\cap(0:\mathfrak{m}^{b})\bigr{)}$ when the
Hilberrt function $H(A)$ is fixed: fixing the Hilbert function fixes
$\dim_{\sf k}\mathfrak{m}^{i}$ and by duality fixes $\dim_{\sf
k}(0:\mathfrak{m}^{b})$, so the intersection dimension is semicontinuous.

Given a symmetric decomposition $\mathcal{D}$ of a fixed AG Hilbert function
$H$, we write

$H\bigl{(}\mathfrak{m}^{i}/\bigl{(}\mathfrak{m}^{i}\cap(0:\mathfrak{m}^{b})\bigr{)}\bigr{)}=(0,\ldots,0,n_{i,b},n_{i+1,b},\ldots,n_{j,b}).$
(2.3)

We set
$N_{i,b}=\dim_{\mathsf{k}}\bigl{(}\mathfrak{m}^{i}/\bigl{(}\mathfrak{m}^{i}\cap(0:\mathfrak{m}^{b})\bigr{)}\bigr{)}=\sum_{k=i}^{j}n_{k,b}$.
We have that

$N_{i,b}=\sum_{a=0}^{j-b-i}\sum_{k=i}^{j-b-a}H(a)_{k},$ (2.4)

the sum of the entries of $\mathcal{D}$ (as usually arranged) in a triangle
bounded on the left by the leading degree $i$ and on the right by the rising
diagonal ending at $H(0)_{j-b}$.

######  Lemma 2.6.

[I2, Lemmas 1.13, 4.1A,B] Let $\mathcal{A}(w)$, $w\in W$ be a (flat) family of
Artinian Gorenstein algebras of fixed Hilbert function $H$, parametrized by a
scheme $W$, and fix integers $i,b,m$. The condition
$N_{i,b}\bigl{(}\mathcal{A}(w)\bigr{)}\leq m$ is a closed condition on the
family $\mathcal{A}(w)$, $w\in W$.

###### Proof.

Fixing $H$ fixes the vector space dimensions of $\mathfrak{m}^{i}$ and
$(0:\mathfrak{m}^{b})$ (by duality
$\dim_{\mathsf{k}}{(0:\mathfrak{m}^{b})}=\dim_{\mathsf{k}}A-\dim_{\mathsf{k}}\mathfrak{m}^{b}$).
Thus $N_{i,b}\leq m$ expresses the condition that the two subspaces
$\mathfrak{m}_{\mathcal{A}(w)}^{\,i}$ and
${\bigl{(}0:\mathfrak{m}_{\mathcal{A}(w)}^{\,b}\bigr{)}}$ of $\mathcal{A}(w)$,
each having a fixed dimension, intersect in dimension at least
${(\dim_{\mathsf{k}}\mathfrak{m}^{i}-m)}$. This is a closed condition on $w\in
W$. $\square$

See Figure 1, where $N_{2,3}$ for $\mathcal{D}_{1}$ is $8$, for
$\mathcal{D}_{2}$ is $7$ and for $\mathcal{D}_{3}$ is 6, implying there can be
no specializations from $\mathrm{Gor}(\mathcal{D}_{i})$ to
$\mathrm{Gor}({\mathcal{D}_{j}})$ for $i>j$, when $H=(1,3,4,4,3,2,1)$. We
define a partial order on symmetric decompositions,

$\mathcal{D}_{1}\geq\mathcal{D}_{2}\text{ if
}N_{i,b}\bigl{(}\mathcal{D}_{1})\bigr{)}\geq
N_{i,b}\bigl{(}\mathcal{D}_{2})\bigr{)}\text{ for each pair $(i,b)$}.$ (2.5)

######  Corollary 2.7 ($H_{A}(0)$ as obstruction to deformation in
$\mathrm{Gor}(H)$).

Suppose that $\mathcal{D}_{s}$, $\mathcal{D}_{t}$ are two symmetric
decompositions of $H$ such that
$\mathcal{D}_{s}(0)_{i}<\mathcal{D}_{t}(0)_{i}$, Then
$\overline{\mathrm{Gor}({\mathcal{D}_{s}})}\cap\mathrm{Gor}(\mathcal{D}_{t})=\emptyset.$

###### Proof.

Take $b=j-i$ in the Lemma 2.6 $\square$

In the Figure 1, the Corollary for $i=3$ shows again that there is no
specialization from $\mathrm{Gor}(\mathcal{D}_{i})$ to
$\mathrm{Gor}(\mathcal{D}_{j})$ for $i>j$, with $H=(1,3,4,4,3,2,1)$.

$N_{2,3}$:$\begin{array}[]{ccccccc}1&2&\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{3}&\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{4}&3&2&1\\\
&1&\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{1}&&&&\\\
&&\mathcal{D}_{1}&&&\end{array}$
$\begin{array}[]{ccccccc}1&2&\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{3}&\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{3}&3&2&1\\\
&1&\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{1}&1&&&\\\
&&\mathcal{D}_{2}&&&\end{array}$

$\begin{array}[]{ccccccc}1&2&\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{2}&\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{2}&2&2&1\\\
&1&\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{2}&1&&&\\\
&&\mathcal{D}_{3}&&&\end{array}$
We have
$\overline{\mathrm{Gor}(\mathcal{D}_{3})}\cap\bigl{(}\mathrm{Gor}(\mathcal{D}_{2})\cup\mathrm{Gor}(\mathcal{D}_{1})\bigr{)}=\emptyset$,

$\overline{\mathrm{Gor}(\mathcal{D}_{2})}\cap\mathrm{Gor}(\mathcal{D}_{1})=\emptyset$.

Figure 1: Obstruction to specialization of symmetric decompositions,
$H=(1,3,4,4,3,2,1)$ (Lemma 2.6).

As mentioned, there is a unique symmetric decomposition of $H$ when
$H=(1,2,\ldots)$ is a Gorenstein sequence of codimension two (so the Hilbert
function of a complete intersection). Not much is known about the poset
$\mathfrak{D}(H)$ of symmetric decompositions of a given Gorenstein sequence
$H$ of codimension at least three: see Question 3.24. We do not even know
which (non-graded) Gorenstein sequences in codimension three occur for
complete intersections!

###  2.3 When the graded quotient $Q_{A}(0)$ of $A$ has codimension two.

We first record some well-known facts about parametrizing AG algebras of
codimension two (Lemmas 2.8 and 2.9). We have already stated the results we
will need about specialization of Jordan type (Lemma 2.3) and the
semicontinuity of symmetric decompositions (Lemma 2.6). Let

$T=(1,2,\ldots,d,t_{d},t_{d+1},\ldots,t_{j},0),\text{ where }d\geq
t_{d}\geq\cdots\geq t_{j}>0,$ (2.6)

be a sequence possible for the Hilbert function of an Artinian quotient of
$R={\mathsf{k}}\\{x,y\\}$ (called a codimension two O-sequence) and let
$e_{i}=t_{i-1}-t_{i},n=|T|=\sum t_{i}$. Here $d$ is the order of ideals
defining algebras of Hilbert function $T$: $\mathfrak{m}^{d}\supset I$ but
$\mathfrak{m}^{d+1}\not\supset I$. Recall that a codimension two Artinian
Gorenstein algebra is a complete intersection. Then we have (for the case
$Z_{T}$ in (B.) see also [Bri]),

######  Lemma 2.8.

See [I1, Theorem 2.12]

  1. A.

[Mac1] The sequence $T$ of Equation (2.6) can occur for an Artinian Gorenstein
quotient of the local ring $R={\mathsf{k}}\\{x,y\\}$ if and only if $e_{i}\leq
1$ for each $i\geq d$.

  2. B.

Let $Z_{T}$ parametrize the Artinian algebra quotients $A=R/I$ having Hilbert
function $T$, and $G_{T}$ parametrize the graded quotients having Hilbert
function $T$. Then $Z_{T}$, $G_{T}$ are smooth irreducible varieties,
$\pi:Z_{T}\to G_{T}:\pi(A)=\mathrm{Gr}_{\mathfrak{m}}(A)=A^{\ast}$ is a bundle
with affine space fibres over the projective variety $G_{T}$; the natural
inclusion map $G_{T}\to Z_{T}$ is a zero-section of $\pi$. Each of $Z_{T}$,
$G_{T}$ has a cover by opens in affine spaces. We have

$\displaystyle\dim Z_{T}$ $\displaystyle=n-\sum_{i\geq d}e_{i}(e_{i}+1)/2$
$\displaystyle=n-d-\sum e_{i}(e_{i}-1)/2;$ (2.7) $\displaystyle\dim G_{T}$
$\displaystyle=\sum_{i\geq d}(e_{i}+1)(e_{i+1}).$ (2.8)

######  Lemma 2.9 (Codimension two implies strong Lefschetz).

Let $A$ be a local Artinian algebra of codimension two, thus
$A={\mathsf{k}}\\{x,y\\}/I$ and assume that $\mathrm{char}\ \mathsf{k}$ is
either zero or greater than the socle degree of $A$. Then $A$ has a linear
element $\ell$ that is strong Lefschetz.

###### Proof.

This arises from standard basis theory of J. Briançon [Bri] (in characteristic
zero), or its analogue in characteristic $p$ [I1]. See also [IMM1, Lemma 2.14]
or [I2, Theorem 2.9]). $\square$

We can use Lemma 2.9 as part of ruling out strong Lefschetz for AG algebras
$A\in\mathrm{Gor}(\mathcal{D})$ for certain decompositions where $A$ has
codimension three but $A(0)$ has codimension two. See Example 2.15.

###  2.4 Jordan type and symmetric decomposition.

We are concerned here with the interplay between Jordan type and the symmetric
decomposition $\mathcal{D}(A)$ of the Hilbert function $H(A)$. Here, we have
many questions, including whether and how we can define a “Jordan order type”
for non-homogeneous AG algebras. Examples 2.14 and 2.15 illustrate some of the
issues.666In work in progress with Johanna Steinmeyer [IMS] we have concluded
that although there is a notion of sequential Jordan type for non-homogeneous
AG algebras, our several tries at a “Jordan order type” have not succeeded,
see Question 3.20 .

######  Definition 2.10.

A _pre-Jordan basis_ $\mathfrak{B}$ of $A$ with respect to a non-unit $\ell\in
A$, of Jordan type $P_{\ell}=(p_{1},p_{2},\ldots,p_{s})$ is a basis of $A$ as
a vector space over $\mathsf{k}$

$\mathfrak{B}=(\ell^{k}b_{i}:1\leq i\leq s,\,0\leq k<p_{i})$ (2.9)

where $\ell^{p_{i}}b_{i}\in{\mathsf{k}}[\ell]\cdot\langle
b_{1},b_{2},\ldots,b_{i-1}\rangle$. Recall that if for each $i\in[1,s]$,
$\ell^{p_{i}}b_{i}=0$ then $\mathfrak{B}$ is a Jordan basis for $(A,\ell)$. We
will call a sequence $(b_{i},\ell b_{i},\ldots,\ell^{p_{i}-1}b_{i})$ as in
Equation 2.9 a _pre-string_ for the pre-Jordan basis, whose beads are
$\ell^{a}b_{i},a\in[0,p_{i}-1]$.

######  Remark 2.11.

We will have use for the fact, for $A$ an AG algebra (not necessarily graded)
and $\ell,h\in A$,

$\text{If }\ell\in\mathfrak{m}^{a}\text{ and
}h\in\mathfrak{m}^{b}\cap(0:\mathfrak{m}^{c})\text{ then }\ell\cdot
h\in\mathfrak{m}^{a+b}\cap(0:\mathfrak{m}^{c-a}).$ (2.10)

Also if $\ell\in(0:\mathfrak{m}^{e})$ then $\ell h\in(0:m^{e-b})$.

As a consequence, if $h\in A$ represents an element of $Q(a)_{i}$ and $\ell h$
represents an element of $Q(b)_{k}$ where $b<a$ then $k\geq i+2$.

Our next example $A$ has a Jordan basis compatible with the $Q(a)$
decomposition.

######  Example 2.12.

Let $R={\mathsf{k}}\\{x,y\\}$, take $f=X^{[4]}Y+Y^{[4]}$, and let
$A=R/\operatorname{\mathrm{Ann}}f$. Then
$\operatorname{\mathrm{Ann}}f=(xy^{2},\,y^{3}-x^{4})$, the Hilbert function is
$H(A)=(1,2,3,2,2,1)$, and its symmetric decomposition is

$\mathcal{D}=\bigl{(}H(0)=(1,2,2,2,2,1),\,H(1)=(0,0,1,0,0)\bigr{)}.$

Here $P_{A,x}=(5,5,1)$: the following strings give a Jordan basis for $(A,x)$
whose elements can be partitioned into bases for each $Q(a)_{i}$, note that
$x^{5}=xy^{2}=0$ in $A$.

$Q(0)$ | $\begin{matrix}1&x&x^{2}&x^{3}&x^{4}&&\\\ &y&xy&x^{2}y&x^{3}y&x^{4}y\end{matrix}$
---|---
$Q(1)$ | $\begin{matrix}\phantom{1}&\phantom{x}&y^{2}\end{matrix}$

Here $P_{A,\ell}=(6,4,1)$ for $\ell=x+y$. The following is a pre-Jordan basis
for $(A,\ell)$ again whose elements can be partitioned into bases for each
$Q(a)_{i}$

$Q(0)$ | $\begin{matrix}1&\ell&\ell^{2}&\ell^{3}&\ell^{4}&\ell^{5}\\\ &y&\ell y&\ell^{2}y&\ell^{3}y&\end{matrix}$
---|---
$Q(1)$ | $\begin{matrix}\phantom{1}&\phantom{y}&y^{2}\end{matrix}$

It is not a Jordan basis, because ${\ell\cdot\ell^{3}y=x^{4}y\neq 0}$ and
${\ell y^{2}=x^{4}\neq 0}$. However, we can change this basis and get a Jordan
basis that agrees with the $Q(a)$ decompositon of $A^{\ast}$:

$Q(0)$ | $\begin{matrix}1&\ell&\ell^{2}&\ell^{3}&\ell^{4}&\ell^{5}\\\ &y-\tfrac{1}{5}\ell&\ell y-\tfrac{1}{5}\ell^{2}&\ell^{2}y-\tfrac{1}{5}\ell^{3}&\ell^{3}y-\tfrac{1}{5}\ell^{4}&\end{matrix}$
---|---
$Q(1)$ | $\begin{matrix}\phantom{1}&\phantom{y-\tfrac{1}{5}\ell}&y^{2}-(x^{3}-yx^{2})\end{matrix}$

Here $(0:\mathfrak{m}^{3})=(x^{3},x^{2}y,y^{2})$ so
$h=x^{3}-yx^{2}=x^{2}(x-y)\in\mathfrak{m}^{3}\cap(0:\mathfrak{m}^{3})$. Now
$(x+y)(x^{2})(x-y)=x^{2}(x^{2}-y^{2})=x^{4}$, so we have
$\ell\cdot(x^{3}-h)=0$.

It is easy to see that the linear form $y$ has a Jordan basis for
$P_{y}=(5,2,2,2)$ since $xy^{2}=0$. Thus every linear form has a Jordan basis
compatible with the $Q(a)$ decomposition. As we shall see later in Examples
2.14 and 2.15 this in general does not occur.

######  Question 2.13.

How does the symmetric decomposition $\mathcal{D}(A)$ restrict the Jordan type
of $A$? In particular can the strong Lefschetz property depend on the
decomposition $\mathcal{D}(A)$?

Answer. The answer to the second question is “Yes”, see Theorem 3.1 where only
the third decomposition is compatible with strong Lefschetz. We develop
partial answers to the first question in many of the results and examples of
this paper.We have the following counterexample due to Chris McDaniel, of a
non-homogeneous Artinian Gorenstein algebra $A$ for which there is no Jordan
basis consistent with $H(A)$ so, in particular, no Jordan basis consistent
with the $Q(a)$ decomposition.

######  Example 2.14.

(C. McDaniel, private communication) Let $F=XY^{3}+X^{2}Y\in k_{DP}[X,Y]$,
$A={\mathsf{k}}\\{x,y\\}/I$,
$I=\operatorname{\mathrm{Ann}}F=(x^{2}-xy^{2},y^{4})$. Note
$x^{3}=(x+y^{2})(x^{2}-xy^{2})+xy^{4}\in I$. Then $H(A)=(1,2,2,2,1)$ and it
can be readily shown that $A$ has no Jordan basis consistent with $H(A)$. Note
that since $A$ has a symmetric Hilbert function, the $Q(a)$ decomposition of
$A^{\ast}$ has a single component $Q_{A}(0)=A^{\ast}$. Thus, $A$ does not have
a basis consistent with the $Q(a)$ decomposition; but $A^{\ast}$, being
graded, does have a Jordan basis consistent with its Hilbert function.

We also have a result of the second author that every Artinian algebra has a
pre-Jordan basis consistent with the Hilbert function [Mar].

The next Example 2.15 is of a codimension three AG algebra that for a
particular linear form does not have a pre-Jordan basis that is consistent
with the $Q(a)$ decomposition. This is an apparent complication in an attempt
to define a notion of Jordan order type for non-graded AG algebras (Question
3.20).

######  Example 2.15.

Let $R=\mathsf{k}\\{x,y,z\\}$, take $f=X^{4}Y^{4}+X^{3}Y^{3}Z+Z^{5}$ and let
$A=R/\operatorname{\mathrm{Ann}}f$. Then (from Macaulay)
$\operatorname{\mathrm{Ann}}f=(yz^{2},xz^{2},y^{4}z,x^{4}z,y^{5},xy^{4}-y^{3}z,x^{4}y-x^{3}z,x^{5},x^{3}y^{3}-x^{2}y^{2}z-z^{4})$,
$H(A)=(1,3,6,8,7,5,3,2,1)$ of length $36$. Then we have
$Q(0)=R/(x^{5},y^{5},z)$, $Q(2)\cong z\cdot R/(x^{3},y^{3},z^{2})$ and
$Q(3)=\langle z^{2},z^{3}\rangle$.

$\mathcal{D}=\bigl{(}H(0)=(1,2,3,4,5,4,3,2,1),\,H(2)=(0,1,2,3,2,1,0),\,H(3)=(0,0,1,1,0,0)\bigr{)}.$

We wish to show that the multiplication by $\ell=z\in A$ does not admit a pre-
Jordan basis inducing a basis of the $Q(a)$ decomposition. Suppose by way of
contradiction that there is a pre-Jordan basis $B$ that can be partitioned
into subsets inducing a basis for each $Q(a)_{i}$. Since
${z^{5}=x^{4}y^{4}\neq 0}$ in $A$, but ${z^{6}=0}$, there is a string of
length $6$ in $B$. Also, we have ${z^{5}\mathfrak{m}=0}$, so this string must
start in order $0$, and we can scale it such that it starts in ${1+b}$, with
${b\in\mathfrak{m}}$. Its second element is ${z+zb}$ and has order $1$. On the
other hand, ${(z+zb)\circ f=X^{3}Y^{3}+Z^{4}+b\circ(X^{3}Y^{3}+Z^{4})}$. Since
${b\circ(X^{3}Y^{3}+Z^{4})}$ is a polynomial of degree at most $5$, it cannot
cancel the term $X^{3}Y^{3}$, so ${(z+zb)\circ f}$ has degree $6$, which tells
us that ${z+zb\in(0:\mathfrak{m}^{7})\setminus(0:\mathfrak{m}^{6})}$. Thus,
${\mathfrak{m}\cap(0:\mathfrak{m}^{7})}$ is the smallest intersection of
powers of the maximal ideal by Loewy ideals that ${z+zb}$ belongs to: so
$z+zb$ must represent a nonzero class in

$Q(1)_{1}=\frac{\mathfrak{m}\cap(0:\mathfrak{m}^{7})}{\mathfrak{m}^{2}\cap(0:\mathfrak{m}^{7})+\mathfrak{m}\cap(0:\mathfrak{m}^{6})}=\frac{(0:\mathfrak{m}^{7})}{\mathfrak{m}^{2}\cap(0:\mathfrak{m}^{7})+(0:\mathfrak{m}^{6})}.$

Nonetheless, ${z+zb=(z-xy)+(xy+zb)}$, with ${z-xy\in(0:\mathfrak{m}^{6})}$,
because $(z-xy)\circ f=-X^{2}Y^{2}Z+Z^{4}$, and
${xy+zb\in\mathfrak{m}^{2}\cap(0:\mathfrak{m}^{7})}$. So ${z+zb}$ represents
the zero class in $Q(1)_{1}$, contradicting the assumption that $B$ is a pre-
Jordan basis for $A$.

Some comments.

This argument relied on the fact that there is an element that represents a
non-zero class in a module $Q(a)_{i}$ (here ${1+b}$ in $Q(0)_{0}$) but when we
multiply by $\ell=z$ we get an element that represents zero in all modules
$Q(a)_{i}$ where it fits. We know that we can always find a pre-Jordan basis
compatible with the Hilbert function – a weaker condition – and the
corresponding modules do not have this behavior: any non-zero element of $A$
represents a non-zero class in the quotient
$\mathfrak{m}^{i}/\mathfrak{m}^{i+1}$, where $i$ is its order.

Now take $h=z-xy$, representing a non-zero element in $Q(2)_{1}$. Then for
$\ell=z$ we have $\ell\cdot h$ is zero in all $Q(a)_{i}$ where it would fit:
the smallest intersection of powers of the maximal ideal by Loewy ideals that
it belongs to is $\mathfrak{m}^{2}\cap(0:\mathfrak{m}^{5})$, making it a good
candidate for $Q(2)_{2}$. However, we can write $zh$ as
$(zh+x^{2}y^{2})-(x^{2}y^{2})$, with
${zh+x^{2}y^{2}\in\mathfrak{m}^{2}\cap(0:\mathfrak{m}^{4})}$ and
$x^{2}y^{2}\in\mathfrak{m}^{3}\cap(0:\mathfrak{m}^{5})$, showing it represents
the zero class in $Q(2)_{2}$. Nonetheless, $z^{2}h$ represents a non-zero
class in $Q(3)_{3}$ going to a higher $a=3$, and $z^{3}h$ represents a non-
zero class in $Q(2)_{5}$ going back to $a=2$, but jumping by two in order, as
expected.

Note that a generic enough $\ell=x+y+z$ has Jordan type $(9,7,5,5,3,3,2,1,1)$
which is exactly that expected from concatenating the SLJT of the $Q(a)$
decomposition (Macaulay computation). However, $\ell=x+y+z$ is not a SL
element for $A$ as $H^{\vee}=(9,7,6,4,4,3,2,1)$.

######  Definition 2.16.

Given an element $f\in S$, by the _order_ $\mathfrak{o}(g)$ of $g=\varrho\circ
f$ we mean the highest power of the maximum ideal ${\mathsf{m}_{R}}$ of $R$
such that ${g\in{\mathsf{m}_{R}}^{\mathfrak{o}(g)}\circ f}$, but is not in
${{\mathsf{m}_{R}}^{\mathfrak{o}(g)+1}\circ f}$. The _degree_ of $g$ is $k$ if
$g\in S_{\leq k}$ but not in $S_{<k}$.

######  Remark 2.17.

(Exotic summands of $f$) Let $H$ be a Gorenstein sequence of socle degree $j$
and $\mathcal{D}=\bigl{(}H(0),\ldots,\,H(j-2)\bigr{)}$ a symmetric
decomposition of $H$. Let $n_{i}=n_{i}(\mathcal{D})=\sum_{u=0}^{i}H(u)_{1}$.
Following [BJMR, Definition 3] we say that $f\in\mathcal{S}$ determining
$A=R/\operatorname{\mathrm{Ann}}f\in\mathrm{Gor}(\mathcal{D})$ is in _standard
form_ if after a linear change of variables in $\mathcal{S}$ to
$X_{1},\ldots,X_{r}$, we have for each $i$,
$f_{j-i}\in{\mathsf{k}}_{DP}[X_{1},\ldots,X_{n_{i}}]$: otherwise $f$ is said
to have _exotic_ terms. See Section 2.7 below. In the previous Example 2.15
where $f=X^{4}Y^{4}+X^{3}Y^{3}Z+Z^{5}$ the term $X^{3}Y^{3}Z$ is an _exotic
summand_ in this sense. We can see this because the variable $Z$ is a partial
of $f$ of order $\mathfrak{o}(Z)=5$, as we can obtain it by doing
${-x^{2}y^{2}(z-xy)\circ f=Z}$; we have $n_{0}=n_{1}=2$, so $f_{8-0}$ and
$f_{8-1}$ should be written in two variables $X,Y$, thus standard terms
involving $Z$ have degree at most $6$, and any term involving $Z$ of higher
degree is exotic (see [BJMR, Definitions 3 and 4] or [IM, Definition 2.3,
§2.2]). Exotic summands provide examples of unexpected behavior, as happened
here with the element $z$: we have that ${z\circ f=X^{3}Y^{3}+Z^{4}}$, so
${z\in(0:\mathfrak{m}^{7})}$, but the term $X^{3}Y^{3}$ can be cancelled using
a partial of higher order, namely ${xy\circ f=X^{3}Y^{3}+X^{2}Y^{2}Z}$. This
explains why ${z-xy\in(0:\mathfrak{m}^{6})}$, a crucial fact in showing that
${z+zb}$ does not represent any non-zero class in the modules $Q(a_{i})$.

An example of $(A,\ell),\ell$ generic, where $P_{A,\ell}$ does not come from
the $Q(a)$-decomposition is given below (Example 2.27). There, $A$ is not
strong Lefschetz, although $Q(0)$, being of codimension two, is strong
Lefschetz (Lemma 2.9). Quite commonly, however, a local AG algebra $A$ of
codimension three having $Q_{A}(0)$ of codimension two is strong Lefschetz, as
in the next example.

######  Example 2.18.

Let $R={\mathsf{k}}\\{x,y,z\\}$, take $f=X^{[4]}Y+Z^{[3]}$, and let
$A=R/\operatorname{\mathrm{Ann}}f$. Then
$\operatorname{\mathrm{Ann}}f=(y^{2},\,z^{2}-x^{4}y,x^{5})$, the Hilbert
function is $H(A)=(1,3,3,2,2,1)$, and its symmetric decomposition is

$\mathcal{D}=\bigl{(}H(0)=(1,2,2,2,2,1),\,H(2)=(0,1,1,0)\bigr{)}.$

It is straightforward to check that $\ell=x+y+z$ is a SL element for both the
concatenation of the $Q(a)$ decomposition, and for $A$.

######  Definition 2.19.

Let $A$ be an AG algebra of socle degree $j$ and let $\mathfrak{B}$ be a basis
of $A$ as a vector space over $\mathsf{k}$. We say that $\mathfrak{B}$ is
_compatible with the symmetric decomposition_ if, for any integers $a,i\geq
0$, $Q_{A}(a)_{i}$ is generated by the set

$\bigl{\\{}v\in\mathfrak{B}\mid
v\in\bigl{(}\mathfrak{m}^{i}\cap(0:\mathfrak{m}^{j+1-a-i})\bigr{)}\setminus\bigl{(}\mathfrak{m}^{i+1}\cap(0:\mathfrak{m}^{j+1-a-i})+\mathfrak{m}^{i}\cap(0:\mathfrak{m}^{j-a-i})\bigr{)}\bigr{\\}}$

######  Question 2.20.

  1. (i).

Given a pre-Jordan basis for the multiplication by an element
$\ell\in\mathfrak{m}$ in an AG algebra $A$, compatible with the $Q(a)$
decomposition of $A$, can we adjust it by subtracting from each $b_{i}$ a
suitable element in ${\mathsf{k}}[\ell][b_{1},\ldots,b_{i-1}]$ to give a
Jordan basis compatible with the $Q(a)$ decomposition of $A^{\ast}$?

  2. (ii).

Let $A$ be an AG algebra and let $\ell$ be any element in $\mathfrak{m}_{A}$.
Then is there a pre-Jordan basis $\mathfrak{B}=\\{v_{1},\ldots,v_{n}\\}$ for
the multiplication map by $\ell$ in $A$ such that

$\\#\mathfrak{B}\cap\mathfrak{m}_{A}^{i}\cap(0:\mathfrak{m}_{A}^{j+1-a-i})=\sum_{b\geq
a,k\geq i}\dim Q(b)_{k}$

for each pair $a,i\geq 0$?

Answer NO! to each. Example 2.14 is a counterexample to (i); and Example 2.15
is a counterexample for (ii).

######  Example 2.21 (Representative $h\in A$ of an element in $Q(a)$).

Consider $F=X^{4}+X^{2}Y$, where $A=R/I$ with
$I=\operatorname{\mathrm{Ann}}F=(xy-x^{3},y^{2})$, and $H(A)=(1,2,1,1,1)$,

$\mathcal{D}(A)=\bigl{(}H(0)=(1,1,1,1,1),\,H(2)=(0,1,0)\bigr{)}.$

Here $A^{\ast}={\mathsf{k}}[x,y]/(xy,y^{2},x^{5})$ with filtration
$A^{\ast}=C(0)=C(1)\supset C(2)=yA^{\ast}$. Recall that $Q(2)_{1}$ is a
quotient of
$\mathfrak{m}\cap(0:\mathfrak{m}^{4+1-2-1})=\mathfrak{m}\cap(0:\mathfrak{m}^{2})$.
We have that $h=y-x^{2}$ satisfies $xh\circ f=y^{2}h\circ F=0$, so
$h\in\mathfrak{m}\cap(0:\mathfrak{m}^{2})$, and represents the element $y\in
Q(2)\subset A^{\ast}$. Note, however, that $y\notin(0:\mathfrak{m}^{2})$ as
$x^{2}y\circ F\not=0$, so $y\in A$ itself is not a representative of
$Q(2)_{1}$, but, rather, represents an element of $Q(1)_{1}$.

Moral: One cannot pass to $Q_{A}(a)$ in a natural way just from $A^{\ast}$;
that is, the sequence $C(0)\supset C(1)\supset\ldots\supset C(j-2)$ arises
from the additional structure of Gorenstein duality on the algebra $A$. Thus,
in the Example 2.21 one cannot pass first to the initial form $y$ of $y-x^{2}$
to see its image in $Q(2)$. Recall also that there are graded algebras
$A^{\ast}$ with several distinct symmetric decompositions [IM, §1.6].

###  2.5 Jordan types of $A$ and of $A^{\ast}$.

Lemma 2.3 shows that the Jordan type for $P_{A,\ell}$ is greater or equal that
of $P_{A^{\ast},\ell}$ in the dominance partial order. The examples in this
section show that $P_{A}$ may be strictly greater than $P_{A^{\ast}}$, and we
explore the cause. Our first Example 2.22 is one in which the Hilbert function
is symmetric. Then $A$ AG implies that $Q_{A}(0)=A^{\ast}$ and is a graded AG
algebra (Lemma 1.5). We introduce further invariants of $H$ and $\mathcal{D}$
(Definition 2.23), and recall a key inequality $P_{c}\bigl{(}H(M)\bigr{)}\geq
P_{\ell,M}$(Lemma 2.25). We also explain using the Jordan strings in Example
2.26 how $A$ and $A^{\ast}$ may have different Jordan types.

#### Symmetric Hilbert function, $A,A^{\ast}$ of different Jordan type.

We will now see that even when the AG algebra $A$ has symmetric Hilbert
function $H(A):\dim_{\mathsf{k}}A_{i}=\dim_{\mathsf{k}}A_{j-i}$ for $0\leq
i\leq j/2$, in which case $A^{\ast}$ is also Gorenstein, we may have
$P_{A,\ell}>P_{A^{\ast},\ell}.$ This example arose from discussion with
Johanna Steinmeyer.

######  Example 2.22.

Let $R={\mathsf{k}}\\{x,y,z\\}$, $S={\mathsf{k}}_{DP}[X,Y,Z]$ let
$F=X(Y^{4}+Z^{4})+X^{3}+Y^{2}Z^{3}$, let
$A_{F}=R/\operatorname{\mathrm{Ann}}F$, for which $H(A_{F})=(1,3,5,5,3,1)$;
the defining ideal is

$I=\operatorname{\mathrm{Ann}}F=(x^{2}-z^{4},xyz,xz^{2}-y^{2}z,xy^{3}-yz^{3},y^{4}-z^{4}),$

with $7$ generators. We have for Jordan type of $A$ with respect to $x$,
$P_{x,A_{F}}=(4,2^{6},1^{2})$, The associated graded ideal is

$I^{\ast}=\operatorname{\mathrm{Ann}}(X(Y^{4}+Z^{4})+Y^{2}Z^{3})=(x^{2},y^{2}z-xz^{2},xyz,y^{4}-z^{4},xy^{3}-yz^{3}),$

with JT $P_{x,A^{\ast}}=(2^{8},1^{2})$ and
$\mathcal{S}_{x,A^{\ast}}=(2\uparrow_{0}^{4},2\uparrow_{1}^{3},1_{2,3}).$ We
have $P_{A_{F},x}>P_{A^{\ast},x}$ in the dominance partial order. Note that
here both $A$ and $A^{\ast}$ are strong Lefschetz with respect to $\ell=x+y+z$
(calculation in Macaulay2).

#### Jordan type and filtrations.

Recall that the Hilbert function of a finite length graded module $M$ is the
sequence $H(M)=(h_{0},\ldots,h_{j})$ where $h_{i}=\dim_{\mathsf{k}}M_{i}$.
Given a Hilbert function sequence $H$, recall that we denote by $H^{\vee}$ the
conjugate partition (switch rows and columns in the Ferrers graph) of the
partition rearrangement of $H$ (Definition 1.3).

The Example 2.27 in the next section below shows that the Jordan type
associated to $\mathcal{D}(A)$ can be strictly greater than that associated to
$A^{\ast}$, which was a surprise to us (Equation (2.17)). The explanation is
that for some $a\geq 0$, the $A^{\ast}$-module $Q(a)$ may satisfy
$P_{Q(a),\ell}<H(a)^{\vee}$, rather than equality, which would be the analogue
for $Q(a)$ of strong-Lefschetz. To prepare, we first give some notation and
state some results about filtrations.

When $M$ is standard-graded and $H(M)$ is non-unimodular, then $H(M)^{\vee}$
is simply unattainable as a Jordan type. Now we will introduce the concept of
contiguous partition of $H$, from [IMM1, Definition 2.17], as the best we
might do (Lemma 2.25).

######  Definition 2.23 (Conjugate and contiguous partitions from $H$,
relative Lefschetz property).

i. The _contiguous partition_ $P_{c}(H)$ of the Hilbert function $H$ is the
partition whose parts are the lengths of the maximal contiguous row segments
of the bar graph of $H$.

ii. We say a linear form $\ell\in A_{1}$ in a graded Artinian algebra $A$ has
the _Lefschetz property relative to $H$_ if its Jordan type is equal to the
contiguous partition of $H$:

$P_{\ell}=P_{c}(H).$

iii. The _concatenation_ $P_{1}\bigcup P_{2}$ of two partitions $P_{1}$ of
$a$, and $P_{2}$ of $b$, is the partition of $a+b$ obtained by concatenating
(placing alongside) the parts of $P_{1}$ and of $P_{2}$.

Thus, for a sequence $H_{1}=(1,3,5,2,3)$ we have $P_{c}(H_{1})=(5,4,2,1,1,1)$
but for $H_{2}=(1,3,5,3,2)$ we have $P_{c}(H_{2})=(5,4,3,1,1)=H_{2}^{\vee}$.
For $P_{1}=(4,3,1)$, $P_{2}=(5,3,2)$ we have $P_{1}\bigcup
P_{2}=(5,4,3,3,2,1)$.

Given a symmetric decomposition
$\mathcal{D}=\bigl{(}H(0),\,H(1),\ldots\bigr{)}$ of $H$, we denote by
$P_{\mathcal{D}}=\bigl{(}H(0)^{\vee},H(1)^{\vee},\ldots\bigr{)}$ the
concatenation of the conjugate partitions to $(H(0),H(1),\ldots)$. We denote
by $P_{c}(\mathcal{D})$ the concatenation of the set of
$\bigl{\\{}P_{c}\bigl{(}H(a)\bigr{)}\bigr{\\}}$ of contiguous partitions.

$\displaystyle P_{\mathcal{D}}$ $\displaystyle=\bigl{(}H(0)^{\vee}\cup
H(1)^{\vee}\cup\ldots\bigr{)}$ (2.11) $\displaystyle P_{c}(\mathcal{D})$
$\displaystyle=P_{c}\bigl{(}H(0)\bigr{)}\cup
P_{c}\bigl{(}H(1)\bigr{)}\cup\cdots.$ (2.12)

We will need the contiguous partitions $P_{c}(H)$, $P_{c}(\mathcal{D})$
whenever we deal with non-unimodal Hilbert functions $H(A)$ or non-unimodal
component Hilbert functions $H(a)$.

We will write $P_{Q(A),\ell}$ for the concatenation of
$\\{P_{Q_{A}(a),\ell}\\}$, the actual Jordan types of the graded modules
$Q(a)$ with respect to $\ell$. Since $Q_{A}(a)$ is a graded $A^{\ast}$ module,
we have by Lemma 2.25 below

$P_{Q(a),\ell}\leq P_{c}(H_{A}(a))\leq P(H_{A}(a)).$ (2.13)

We say that a filtration $\mathcal{F}:M=M(0)\supset M(1)\supset\cdots\supset
M(c)\supset M(c+1)=0$ of a module $M$ over a standard graded Artin algebra $A$
is _compatible_ with the $\mathfrak{m}$-adic filtration, if $\mathfrak{m}\cdot
M(i)\subset M(i)$ for each $i\in[0,c]$.

######  Lemma 2.24.

Let $A$ be an Artinian standard graded algebra, and $M$ a module over $A$ with
a stratification $\mathcal{F}:M=M(0)\supset M(1)\supset\cdots\supset
M(c)\supset M(c+1)=0$ that is compatible with the $\mathfrak{m}$-adic
filtration. Denote the successive quotient modules by
$Q_{\mathcal{F}}(i)=M(i)/M(i+1)$ for $i\in[0,c]$. Let $\ell\in A_{1}$ be a
linear form. Then we have (the right side is the concatenation),

$P_{M,\ell}\geq\bigcup_{i}\\{P_{Q(i),\ell}\\}.$ (2.14)

Now let $A$ be an AG algebra, and consider the $Q(a)$ subquotients of
$A^{\ast}$ (Definition 1.1). Then

$P_{A^{\ast},\ell}\geq\bigcup_{a}\\{P_{Q(a),\ell}\\}.$ (2.15)

###### Proof of Lemma 2.24.

It suffices to show (2.14), and to show it for $c=1$. Let $P=P_{M,\ell}$ have
$t$ parts, and let $M=\oplus_{1}^{t}L(i)$ be a decomposition of $M$ into $t$
$\ell$-strings (Definition 1.3). For a string
$L(i)=\bigl{(}m(i),\ldots,m(i)^{p_{i}-1}\bigr{)}$ of length $p_{i}$ in $M$,
let $n(i)=m(i)\ell^{s_{i}}$ be its first element in $M(1)$. Then its image in
$Q(0)=M/M(1)$ is the string $m(i),\ell m(i),\ldots,\ell^{s_{i}-1}m(i)$ and its
image in $Q(1)=M(1)$ is the string
$\bigl{(}n(i),\ldots,n(i)^{p_{i}-s_{i}}\bigr{)}$. We will show Claim.
$Q(0)=\oplus L(i)/n(i)L(i)$ is a Jordan decomposition, and $Q(1)=\oplus
n(i)L(i)$ is a Jordan decomposition of $Q(1)=M(1)$.

Proof of Claim: Suppose that $Q(0)$ has Jordan decomposition $P(0)$ and $Q(1)$
has Jordan decomposition $P(1)$ with respect to $\ell$. Then the concatenation
gives a Jordan pre-basis for $M$ (we just do not have $\ell^{p_{i}}m(i)=0$).
The Jordan type $P_{M}$ is the maximum of the partitions associated to Jordan
pre-bases. so $P_{M}\geq P(0)\cup P(1)$. $\square$

It follows that there is equality in Equation (2.14) in the case $c=1$ only if
each $\ell$-string of $M(1)$ does not have a precursor $n^{\prime}(i)$ in $M$
such that $\ell n^{\prime}(i)=n(i)$. We recall [IMM1, Theorem 2.20],

######  Lemma 2.25.

For a finite graded module $M$ over a standard-graded Artinian algebra $A$
with Hilbert function $H(M)$, we have for any linear form $\ell\in A_{1}$

$P_{c}\bigl{(}H(M)\bigr{)}\geq P_{\ell,M}.$

Since our subquotients $Q_{A}(a)$ are standard graded, Lemma 2.25 implies that
$P_{Q(a),\ell}\leq P_{c}\bigl{(}H(Q(a)\bigr{)}$. In the Example 2.31 we have
$P_{A^{\ast}}>P_{c}(\mathcal{D})$.

We give some further examples, here and in the next section, where
$P_{A}>P_{A^{\ast}}$. In the next example $Q(1)$ is non-simple, as
$H(1)=(0,1,0,1,0)$. This is reflected in the Jordan type of $A^{\ast}$ but not
in that of $A$, as $\\{z,\ell z=x^{3}\\}$ is a length-two pre-string
(Definition 2.10) in $A$, but $\ell z=0$ in $A^{\ast}$.

######  Example 2.26 (Cute).

Let $R={\mathsf{k}}\\{x,y,z\\}$, $S={\mathsf{k}}_{DP}\\{X,Y,Z\\}$, consider
${F=X^{[3]}Y^{[2]}+Y^{[3]}Z}$, and let ${A=S/I}$, with
${I=\operatorname{\mathrm{Ann}}F}$. Then $H(A)=(1,3,3,4,2,1)$ and its
symmetric decomposition is

$\mathcal{D}=\bigl{(}H(0)=(1,2,3,3,2,1),\ H(1)=(0,1,0,1,0)\bigr{)},$

while $H(A)^{\vee}=(6,4,3,1),$ We will show that for $\ell=x+y+z$

$H(A)^{\vee}>P_{A,\ell}=(6,4,2,2)>P_{A^{\ast},\ell}=(6,4,2,1,1)=P_{c}(\mathcal{D}).$
(2.16)

The generators of $I$ are given by

$I=(xz,yz-x^{3},z^{2},xy^{3},y^{4}),$

and the associated graded ideal $I^{\ast}$ defining $A^{\ast}$ is

$I^{\ast}=(xz,yz,z^{2},xy^{3},x^{4},y^{4}),$

the annihilator of $\langle X^{[3]}Y^{[2]},Z,Y^{3}\rangle$. A calculation in
Macaulay2 shows that for $\ell=x+y+z$, or a random linear combination that the
Jordan type is $P_{A,\ell}=(6,4,2,2)$, while
$P_{A^{\ast},\ell}=(6,4,2,1,1)=P_{c}(\mathcal{D})$. Since
${H(A)^{\vee}=(6,4,3,1)}$ we see that $A$ is not SL (although $Q(0)$, being of
codimension two, is strong Lefschetz). We now explain the difference between
$P_{A}$ and $P_{A^{\ast}}$.

First, for $A^{\ast}$ taking $\ell=x+y+z$, its action on generators
$1,x,y^{2}$ gives pre-strings of lengths $(6,4,2)$ (the length $6$ pre-string
is a string, since $\ell^{7}=0$). In particular
$\ell^{2}y^{2}=x^{2}y^{2}\in\langle\ell^{4},\ell^{3}x\rangle$ in $A^{\ast}$,
so there is a length-two pre-string $\langle y^{2},y^{2}\ell\rangle$. We may
take $z\in A^{\ast}$ as the generator of another pre-string of $A^{*}$, for
which $\ell\cdot z=0$ in $A^{\ast}$; and the last pre-string of $A^{\ast}$ is
$\\{y^{3}\\}$, also of length one, as $\ell y^{3}=0$ in $A^{\ast}$. Thus,
$P_{A^{\ast},\ell}=(6,4,2,1,1)$.

What changes for $A$ is that $\ell z=xz+yz+z^{2}=x^{3}$, giving a length two
pre-string, and $\ell y^{2}=y^{3}+xy^{2}$ is linearly independent of
$\langle\ell^{3},\ell^{2}x,x^{3}\rangle$ so $\\{y^{2},\ell y^{2}\\}$ remains a
length-two pre-string, yielding $P_{A,\ell}=(6,4,2,2)$.

What is cute here, is that one $\ell$ pre-string of $A$ starts representing an
element, class of $z$ in $Q(1)_{1}$, but has next bead $\ell z=x^{3}$
representing a class in $Q(0)_{3}$, a jumping behavior. This illustrates the
principle that if $z\in Q(a)$ and $\ell z$ represents an element of a lower
$Q(b),b<a$, then $\ell z$ must have a jump in degree (Remark 2.11).

###  2.6 Several irreducible components of $\mathrm{Gor}(H)$.

In this section we give two initial examples of codimension three Gorenstein
sequences $H$ such that $\mathrm{Gor}(H)$ has several irreducible components,
Example 2.27 where $H=(1,3,4,4,4,2,1)$, and Example 2.31, where
$H=(1,3,5,4,4,2,1)$. In between, we discuss a potential further way to
distinguish components of $\mathrm{Gor}(H)$, using the tangent space to the
Hilbert scheme (Remark 2.29).

In the next example the AG algebra $B$ is in an irreducible component of
$\mathrm{Gor}(\mathcal{D}_{2})$ that is comprised of strong Lefschetz AG
algebras, whose associated graded algebras are also strong Lefschetz. The
algebra $A$ is in a different irreducible component of the same
$\mathrm{Gor}(\mathcal{D}_{2})$, comprised of non-strong Lefschetz algebras.
In the symmetric decomposition $\mathrm{Gor}({\mathcal{D}_{1}})$ a general
enough element $C$ has the same Jordan type as $A$; because of the
specialization restrictions on symmetric decompositions, no family of algebras
in $\mathrm{Gor}(\mathcal{D}_{2})$ can specialize to any algebra in
$\mathrm{Gor}(\mathcal{D}_{1})$. Because strong Lefschetz dominates non-strong
Lefschetz, there are no specializations from a family in
$\mathrm{Gor}(\mathcal{D}_{1})$ to an element of
$\mathrm{Gor}(\mathcal{D}_{2})$ that is strong Lefschetz. This contrast
assures that $\mathrm{Gor}(H)$, $H=(1,3,4,4,4,2,1)$ has at least two
irreducible components. But we show that $\mathrm{Gor}(\mathcal{D}_{2})$
itself has two irreducible components, neither in the closure of
$\mathrm{Gor}(\mathcal{D}_{1})$, so that $\mathrm{Gor}(H)$ has three
components.

Recall from Equation (2.11) that, given a symmetric decomposition
$\mathcal{D}$, we define $P(\mathcal{D})=\bigcup_{a=0}^{j-2}H(a)^{\vee}$, and
$P_{c}(\mathcal{D})=\bigcup_{a=0}^{j-2}P_{c}(H(a))$; these can be quite
different from the actual concatenation $P_{Q(A)}$ of
$\\{P_{Q(a)},a=0,1,\ldots,j-2\\}$ with respect to a generic linear form
$\ell$. Here $P_{Q(A)}$ must be dominated by $P_{A^{\ast}}$ since $Q(a)$ are
the quotients arising from a filtration of $A^{\ast}$. We now illustrate this.
We also have in $A,C$ examples of a non-strong-Lefschetz AG algebra such that
$Q(0)$ is strong-Lefschetz - as the codimension of $Q(0)$ is two (Lemma 2.9).

######  Example 2.27 (Jordan type of $A$ and of $A^{\ast}$ different,
$\mathrm{Gor}(\mathcal{D}_{2})$ has two irreducible components,
$\mathrm{Gor}(H)$ has three).

abscked Let $R={\mathsf{k}}[x,y,z]$, $S={\mathsf{k}}_{DP}[X,Y,Z]$, fix
$H=(1,3,4,4,4,2,1)$ of socle degree $6$.

The algebra $C$ and $\mathrm{Gor}(\mathcal{D}_{1})$: Let

$\mathcal{D}_{1}=\bigl{(}H(0)=(1,2,3,4,3,2,1),\,H(1)=(0,1,0,0,1,0),\,H(2)=(0,0,1,0,0)\bigr{)}.$

Let $W=X^{3}Y^{3}+ZY^{4}+Z^{4}$, then
$J=\operatorname{\mathrm{Ann}}W=(xz,yz-x^{3},z^{3}-y^{4})$,
$C=R/J\in\mathrm{Gor}(\mathcal{D}_{1})$. Here
$Q_{C}(0)={\mathsf{k}}[x,y]/(x^{4},y^{4})$ and $Q^{\vee}(1)=\langle
Z,Y^{4}\rangle,Q^{\vee}(2)=\langle Z^{2}\rangle$. Also
$J^{\ast}=(xz,yz,z^{3},y^{4},x^{4}y,x^{5})$. We have
$P(\mathcal{D}_{1})=(7,5,3,2,1^{2}),P_{c}(\mathcal{D}_{1})=(7,5,3,1^{4}),$ and

$H^{\vee}=(7,5,4,3)$. Let $\ell=x+y+z$ we have 777All calculations were in
Macaulay, and we found $\ell=x+y+z$ is general enough in this example.

$H^{\vee}>P_{C,\ell}=(7,5,3,3,1)>P_{C^{\ast},\ell}=(7,5,3,2,2)>P(\mathcal{D}_{1})>P_{c}(\mathcal{D}_{1})=P_{Q(A)}.$

The algebra $A$ and $\mathrm{Gor}(\mathcal{D}_{2})$: Let

$\mathcal{D}_{2}=(H(0)=(1,2,3,3,3,2,1),H(1)=(0,1,1,1,1)).$

Let $F_{A}=X^{2}Y^{4}+X^{3}YZ$, then
$I=\operatorname{\mathrm{Ann}}F=(z^{2},xz-y^{3},y^{2}z,x^{4},x^{3}y^{2})$,
$A=R/I\in\mathrm{Gor}(\mathcal{D}_{2})$. Here
$Q_{A}(0)={\mathsf{k}}[x,y]/(x^{3},y^{5}),Q^{\vee}(1)=\langle
Z,YZ,X^{3},X^{3}Y\rangle$, and
$A^{\ast}=R/(z^{2},xz,y^{2}z,x^{4},y^{5},x^{3}y^{2})$. We have
$H^{\vee}=P(\mathcal{D}_{2})=P_{c}(\mathcal{D}_{2})=(7,5,4,3)$. We have for
$\ell=x+y+z$ (also for random $\ell$),

$P(\mathcal{D}_{2})=P_{c}(\mathcal{D}_{2})=(7,5,4,3)>P_{A,\ell}=(7,5,3,3,1)>P_{A^{\ast},\ell}=(7,5,3,2,2)=P_{Q_{A},\ell}.$
(2.17)

Thus $A,A^{\ast}$, respectively have the same generic Jordan types as,
respectively, $C,C^{\ast}$, and they are not strong Lefschetz. Explanation for
$P(\mathcal{D}_{2})>P_{A^{\ast}}=P_{Q(A)}$: note that $P_{Q(1),\ell}=(2,2)$,
there being the two $\ell$-strings on $Q(1)$ of $\langle YZ,Z\rangle$ and
$\langle X^{3}Y,X^{3}\rangle$. And the component Jordan type $P_{Q(a),\ell}$
may, as here, be smaller than $P_{\mathcal{D}_{1}}$. A choice of strings for
$(A,\ell)$ giving a pre-Jordan basis is (here $z\ell^{3}=0$ while
$z\ell^{2}\in\mathfrak{m}^{4}$)

$A=\langle(1,\ell,\ldots,\ell^{6});(y,y\ell,y\ell^{2},y\ell^{3},y\ell^{4});(z,z\ell,z\ell^{2}),(x^{2},x^{2}\ell,x^{2}\ell^{2}),(x^{3})\rangle.$

We will later use that the Hilbert function
$H(R/I_{A}^{2})=(1,3,6,10,12,12,{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{12}},10,7,3)$,
so
$H(I_{A}/I_{A}^{2})=(0,0,2,6,8,10,{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{11}},10,7,3)$.

The algebra $B$ and $\mathrm{Gor}(\mathcal{D}_{2})$: Let $G=X^{2}Y^{4}+Z^{5}$,
and set
$B=R/J,J=\operatorname{\mathrm{Ann}}G=(xz,yz,x^{3},y^{5},x^{2}y^{4}-z^{5})$
satisfies $B\in\mathrm{Gor}(\mathcal{D}_{2})$ but now $Q(1)=\langle
Z,Z^{2},Z^{3},Z^{4}\rangle$ is cyclic, and $B,B^{\ast}$ are both strong
Lefschetz. That is, for $\ell=x+y+z$

$H^{\vee}=(5,4,3,3)=P_{B,\ell}=P_{B^{\ast},\ell}.$

We will need that the Hilbert function of $R/I_{B}^{2}$ is
$(1,3,6,10,12,12,{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{12}},8,6,4,2)$,
so
$H(I_{B}/I_{B}^{2})=(0,0,2,6,8,10,{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{11}},8,6,4,2)$.

Dimension of $\mathrm{Gor}(\mathcal{D}_{1})$: Choose $\langle
X,Y\rangle\subset S_{1}$ a $\mathbb{P}^{2}$ choice. Let $F$ be a dual
generator of $C\in\mathrm{Gor}(\mathcal{D}_{1})$. By Theorem 2.8, Equation
(2.7) the dimension of the (not necessarily graded) quotients
$Z_{H(0)}=n_{0}-4=12$. By [IM, Lemma 1.40], $H(1)=(0,1,0,0,1)$ now implies
that the $Z$ portion of $F_{5}$ is linear in $Z$. So it has the form
$ZW,W\in{\mathsf{k}}[X,Y]_{4}$ giving a dimension 5 fibering. We have the
space $zR_{1}$ free to act on $F_{4}$, and we must have $\dim(zR_{1})\circ
F_{4}\mod{\mathsf{k}}[X,Y]_{2}=1$, this requires
$F_{4}=L^{4}\mod{\mathsf{k}}[X,Y]_{4}$ (as we have already accounted for the
${\mathsf{k}}[X,Y]_{4}$ portion of $F_{4}$, thus we choose $L=az+bx+cy$, a
three-dimension choice. We may add arbitrary elements from $S_{3},S_{2}$ not
already in $R\circ(F_{6}+F_{5}+F_{4})$ without changing $H$, so we may add
$(r_{2}-4)+(r_{3}-4)=(6-4)+(10-4)$, obtaining a total dimension of
$2+12+5+3+8=30$ for $\mathrm{Gor}(\mathcal{D}_{1})$.Dimension of $A$ component
of $\mathrm{Gor}(\mathcal{D}_{2})$. Choosing $\langle X,Y\rangle$. Let $F$ be
a dual generator of an algebra with the Jordan type of $A$ in
$\mathrm{Gor}(\mathcal{D}_{2})$: then by Equation (2.7), again the dimension
of the not necessarily graded quotients $Z_{H(0)}=n_{0}-3=12$. As for
$\mathcal{D}_{1}$ we are choosing a $ZW,W\in{\mathsf{k}}[X,Y]_{4}$, general
enough, so this is a 5 dimensional choice (we care about the proportionality
coefficient). The degree-4 component of $F$ in $Z$ must be annihilated by
$(I_{2})^{2}=\langle z^{2},z\ell)^{2}\rangle$, a 3-dimensional space and is
viewed mod $R\circ F_{\geq 5}$ of dimension $4$ so gives a $10-4-3=3$
dimensional space. Finally we may add lower degree terms in $Z$: giving
$(r_{2}-4)+(r_{3}-4)=8$; we have for dimension of this component
$2+12+5+3+8=30$, again.

Dimension of $B$ component of $\mathrm{Gor}(\mathcal{D}_{2})$. The choice of
$F_{6}$, as before is fibred over $\mathbb{P}^{2}$ by a 12-dimensional
$Z_{H(0)}$. Now choose the $Z$-related portion of $F_{5}$ as $L^{5}$, a 3
dimensional choice. We may add on elements of $(I^{2})^{\perp}$ to $F_{4}$, as
we must have $I_{2}\circ F_{4}\subset(\operatorname{\mathrm{Ann}}F)_{2}$. Here
$(I^{2})_{4}=(xz,yz)^{2}$ (since, after coordinate change $F_{6}+F_{5}$ is a
connected sum), so, again, as for $A$, the $Z$-portion of $F_{4}$ is a
$10-4-3=3$-dimensional choice. Finally $F_{3}+F_{2}$ adds 8 to the dimension,
giving $2+12+3+3+8=28$ dimension family.

We have now shown

######  Claim 2.28.

$\mathrm{Gor}(\mathcal{D}_{2})$ itself has two irreducible components, one
corresponding to $F_{5}=ZG_{4}\mod{{\mathsf{k}}}[X,Y]$ as for the algebra $A$;
the other component satisfies $F_{5}=L^{5}\mod{{\mathsf{k}}}[X,Y]$, as for the
algebra $B$. The components have respective dimensions 30 for $A$ and 28 for
$B$.

###### Proof of Claim.

An algebra in the family related to $A$ cannot specialize to a general element
related to $B$ by the closure properties of Jordan type (Lemma 2.3). But the
$B$-component has smaller dimension, so a family in the $B$-related component
cannot specialize to a general element in the $A$-component. $\square$

Specialization: proof that $\mathrm{Gor}(H)$ has three components.

Since $H_{C}(0)_{3}=4>H_{B}(0)=3$ we have by Corollary 2.7 that no subfamily
of $\mathrm{Gor}(\mathcal{D}_{2})$ can specialize to an algebra in
$\mathrm{Gor}(\mathcal{D}_{1})$; this and that $P_{B}=(7,5,4,3)$ dominates
$P_{C}=(7,5,3,3,1)$ shows that we have at least two irreducible components of
$\mathrm{Gor}(H),H=(1,3,4,4,4,2,1)$, corresponding to these two symmetric
decompositions; but we have just shown that $\mathrm{Gor}(\mathcal{D}_{2})$
itself has two irreducible components, neither in the closure of
$\mathrm{Gor}(\mathcal{D}_{1})$.

#### Tangent space to the Hilbert scheme $\mathrm{Hilb}^{n}(\mathbb{A}^{2})$
at $A$.

We introduce this mainly as a theme to further explore elsewhere: we do not
use this tangent space in an essential way in this paper.

######  Remark 2.29.

Negative deformations, tangent space. The tangent space to the Hilbert scheme
$\mathrm{Hilb}^{n}(\mathbb{A}^{2})$ at a point corresponding to the length-$n$
Artinian algebra $A=R/I$ is $\mathrm{Hom}(I,A)$; for a graded AG algebra we
would have $\mathrm{Hom}(I,A)\cong I/I^{2}$ with a reversal in degree [IK,
Theorem 3.9, Remark 3.10]; however we are not sure how this might extend to
the non-graded case.

Calculation in Macaulay for the algebras $A$, $B$, $C$ above, all of Hilbert
function $H(A)=(1,3,4,4,4,2,1)$, shows, remarkably, an equality in the lengths
$|H(R/I_{A}^{\,2})|=|H(R/I_{B}^{2}{\,2})|=|H(R/I_{C}^{\,2})|$, and in fact, we
have $|H(I/I^{2})=3\cdot|H(R/I)|$ for each of the ideals
$I=I_{A},I_{B},I_{C}$.888A result of J.O. Kleppe shows that as a consequence
of the D. Buchsbaum-D.Eisenbud Pfaffian structure theorem applied to graded AG
algebras of codimension three, the Hilbert function $H(R/I^{2})$ depends only
on $H(R/I)$, ([Kl1, Proposition 2.5], [IK, Proposition 4.25]). This Hilbert
function result does not generalize to nongraded AG algebras of codimension
three; however in each example we have checked the length
$|H(I/I^{2})|=3|H(R/I)|$ (See Note 3.2 p. 3.2).

That the dimension $\dim_{\mathsf{k}}(I/I^{2})=3\dim(R/I)$, is a consequence
of the smoothability of all codimension three Artinian Gorenstein algebras. We
note:

Principle. For a constant length family $\Xi(t),t\in T$, of Artinian algebras
$\Xi(t)=R/I(t)$, the sum $\sum_{i\geq i_{0}}H(\Xi(t))_{i}$ for a fixed $i_{0}$
is an upper semicontinuous function. Higher values are more special: one
cannot deform from a family with higher sum to one of lower sum. The reason
is, assuming that the socle degrees of $\Xi(t)$ are $j$ or below, so working
in the finite dimensional quotient ring $R^{\prime}=R/\mathfrak{m}^{j+1}$ that
for $\Xi(t)=R^{\prime}/I(t)$ the dimension of the intersection
$I(t)\cap\mathfrak{m}^{i_{0}}\subset R^{\prime}$ is semicontinuous.

This implies

######  Lemma 2.30.

Consider two AG algebras $A$, $B$ of codimension three, and assume
$H(A)=H(B)$. Let $\mathcal{A}=\\{A(t),t\in T\\}$, be a family of algebras
satisfying $H\bigl{(}A(t)\bigr{)}=H(A)$ and
$H\bigl{(}I_{A(t)}/I_{A(t)}^{\,2}\bigr{)}=H(I_{A}/I_{A}^{\,2})$. Assume that
there is an integer $i_{0}$ such that

$\sum_{i\geq
i_{0}}\dim_{\mathsf{k}}\bigl{(}(I_{A(t)})_{i}/(I_{A(t)})_{i}^{\,2}\bigr{)}>\sum_{i\geq
i_{0}}\dim_{\mathsf{k}}\bigl{(}(I_{B})_{i}/(I_{B})_{i}^{\,2}\bigr{)}.$ (2.18)

Then there can be no specialization from the family $\mathcal{A}$ to $B$.

###### Proof.

We have

$H(R/I_{A}^{\,2})=H(I_{A}/I_{A}^{\,2})+H(R/I_{A})=H(I_{A}/I_{A}^{\,2})+H(A),$
(2.19)

so, given $H(A)=H(B)$ and the fact
$\dim_{\mathsf{k}}(I_{A}/I_{A}^{\,2})=3\dim(R/I_{A})$ for height three local
AG algebras, we conclude the result from the Principle above. $\square$

For, example, in codimension two or less, the curvilinear length $n$ local
algebras of Hilbert function $H=(1,1,\ldots,1_{n-1})$ specialize to all other
local algebras of length $n$, but no family of non-curvilinears can specialize
to a curvilinear algebra.999Reason: each colength $n$ ideal $I$ of $R$
contains $\mathfrak{m}^{n}$, so we may work with $I/\mathfrak{m}^{n}\subset
R/\mathfrak{m}^{n}$. The codimension of
$I\cap\mathfrak{m}^{i_{0}}/\mathfrak{m}^{n}$ in
$\mathfrak{m}^{i_{0}}/\mathfrak{m}^{n}$ for a fixed $i_{0}\leq n$ is
$\sum_{i\geq i_{0}}H(R/I)$. So a larger sum greater than $c$ implies a smaller
intersection of $I$ with a fixed ideal $\mathfrak{m}^{i_{0}}$, which is an
open condition on ideals $I/\mathfrak{m}^{n}$ of fixed length in
$R/\mathfrak{m}^{n}$.

We have from calculation in Macaulay2 for the algebras $A$, $B$, $C$ above in
Example 2.27 $\begin{array}[]{c||c|c|}R/I&H(R/I^{2})&H(I/I^{2})\\\ \hline\cr
A&(1,3,6,10,12,12,{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{12}},10,7,3)&(0,0,2,6,8,10,{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{11}},10,7,3)\\\
\hline\cr
B&(1,3,6,10,12,12,{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{12}},8,6,4,2)&(0,0,2,6,8,10,{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{11}},8,6,4,2)\\\
\hline\cr
C&(1,3,6,10,12,12,{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{11}},9,7,4,1)&(0,0,2,6,8,10,{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{10}},9,7,4,1)\\\
\hline\cr\end{array}$ Here, we may compare $H(R/I_{A}^{\,2})$ and
$H(R/I_{B}^{\,2})$ as within $\mathrm{Gor}(H)$ the lengths of $H(A)$ and of
$H(R/I_{A}^{\,2})$ are constant. Thus, it follows from the table above that
there are no specializations from a family like $B$ to $A$ ($i_{0}=8$) nor $C$
to $B$ ($i_{0}=7$) nor $B$ to $C$ ($i_{0}=10$). Warning: we have not checked
the $H(R/I^{\,2})$ for generic algebras of these irreducible components of
$\mathrm{Gor}(H)$.

#### When $H(A)^{\vee}>P_{A}>P_{A^{\ast}}>P_{c}(\mathcal{D})$, and a second
$\mathrm{Gor}(H)$ with two irreducible components.

The following example was discussed in [IM] with reference to the
decomposition $\mathcal{D}_{1}$, but now we consider the Jordan type. We will
see that $A$ is not strong Lefschetz, and that for a generic linear form
$\ell$ (see Definition 2.23), we have
$H(A)^{\vee}>P_{A,\ell}>P_{A^{\ast},\ell}>P_{c}(\mathcal{D}_{1}).$ We also
show that the family $\mathrm{Gor}(H)$ has two irreducible components,
corresponding to the two symmetric decompositions $\mathcal{D}_{1}$ and
$\mathcal{D}_{2}$ of $H$. This is the main theme we will explore in Section 3.

######  Example 2.31.

(i). We recall from [IM, Example 1.31], let
$R={\mathsf{k}}\\{x,y,z\\},S={\mathsf{k}}_{DP}[X,Y,Z]$ and $F=X^{[3]}\cdot
Y^{[3]}+Z(X^{[4]}+Y^{[4]})\in S$. Then $A=R/\operatorname{\mathrm{Ann}}F$ has
Hilbert function $H_{F}=(1,3,5,4,4,2,1)$ and symmetric decomposition

$\mathcal{D}_{1}=\big{(}H_{F}(0)=(1,2,3,4,3,2,1),\ H_{F}(1)=(0,1,0,0,1,0),\
H_{F}(2)=(0,0,2,0,0)\big{)}.$ (2.20)

We will show that, while $H(A)^{\vee}=P_{c}(H(A))=(7,5,4,3,1)$, for a generic
$\ell$,

$P_{A,\ell}=(7,5,3,3,1,1)>P_{A^{\ast},\ell}=(7,5,3,2,2,1)=\mathcal{D}^{\vee}>P_{c}(\mathcal{D}_{1})=(7,5,3,2,1,1,1).$
(2.21)

The ideal
$I=\operatorname{\mathrm{Ann}}F=(z^{2},xyz,x^{2}z-xy^{3},y^{2}z-x^{3}y,x^{4}-y^{4})$;
the associated graded ideal $I^{\ast}$ defining $A^{\ast}$ is
$I^{\ast}=(z^{2},xyz,x^{2}z,y^{2}z,x^{4}-y^{4},xy^{4},yx^{4}).$

Here $Q_{F}^{\vee}(1)=Q_{F}(1)\circ F=\langle x^{4}\circ f,z\circ
f\rangle=\langle Z,X^{[4]}+Y^{[4]}\rangle,\quad Q_{F}^{\vee}(2)=Q_{F}(2)\circ
F=\langle(-zx+y^{3})\circ f,(-zy+x^{3})\circ F\rangle=\langle XZ,YZ\rangle.$

Note, $H(A)^{\vee}=(7,5,4,3,1)$, $\mathcal{D}_{1}^{\vee}=(7,5,3,2,2,1)$ while
$P_{c}(\mathcal{D}_{1})=(7,5,3,2,1,1,1)$. Also, for any
$\ell,P_{Q_{F}(1),\ell}=(1,1)=P_{c}(H(1)$ so each $P_{Q_{F}(a),\ell}$ achieves
its maximum $P_{c}(H(a))$ under Lemma 2.25. Nevertheless, as we shall see,
$H(A^{\ast})>P_{c}(\mathcal{D_{1}})$.

The structure of $Q_{F}(1)$ suggests that the Jordan type of $A^{\ast}$ might
be quite special compared to that of $A$. This is confirmed by the remark,
that for any non-zero linear form $\ell=ax+by+cz$, the homomorphism
$m_{\ell^{2}}:A^{\ast}_{1}\to A^{\ast}_{3}$ has a kernel, as $z\ell^{2}=0\mod
I^{\ast}$. We may use as $\mathsf{k}$-basis

$A=\langle
1;x,y,z;x^{2},xy,y^{2},zx,zy;x^{3},x^{2}y,xy^{2},y^{3};x^{4},x^{3}y,xy^{3},y^{4};x^{3}y^{2},x^{2}y^{3};x^{3},y^{3}\rangle.$

For $\ell=ax+by+cz$ general enough both $A,A^{\ast}$, have linearly
independent strings $L_{1}=\\{1,\ell,\ldots,\ell^{6}\\}$,
$L_{2}=\\{x,x\ell,\ldots,x\ell^{4}\\}$. A pre-Jordan basis for $A$ is pre-
Jordan basis

$L_{1},L_{2},L_{3}=\\{z,z\ell,z\ell^{2}\\},L_{4}=\\{y^{2},y^{2}\ell,y^{2}\ell^{2}\\},L_{5}=\\{x^{2}\\},L_{6}=\\{x^{4}\\}=\\{y^{4}\\}.$

To establish this, note that $z\ell^{3}\equiv
z(x^{3}+y^{3})=x^{3}y^{2}+x^{2}y^{3}\\}$ is in $\langle L_{1}\oplus
L_{2}\rangle$ as $H(A)_{5}=~{}2$. Here $x^{2}\ell$ does not kick off a class
$\langle x^{4}\rangle$, we know that $A_{4}$ is already full. This shows
$P_{A,\ell}=(7,5,3,3,1,1)$. For $A^{\ast}$ we have strings
$L_{1},L_{2},L^{\prime}_{3}=\\{z,z\ell\\}$ (as $z\ell^{2}\in I^{\ast}$). Also
$L^{\prime}_{4}=\\{y^{2},y^{2}\ell,y^{2}\ell^{2}\\}$ (here $A_{5}$ is already
in the span of $L_{1},L_{2}$). We have
$L^{\prime}_{5}=\\{zy,zy\ell\\},L^{\prime}_{6}=\\{xy,xy\ell\\},L^{\prime}_{7}=x^{2}$,
giving $P_{\ell,A^{\ast}}=(7,5,3,2,2,1)$.

Calculation in Macaulay 2 confirms that $P_{A,\ell}=(7,5,3,3,1,1)$ for $\ell$
random or $\ell=(x+y+z)$;
$P_{A^{\ast},\ell}=(7,5,3,2,2,1)=\mathcal{D}_{1}^{\vee}>P_{c}(\mathcal{D}_{1})$
for $\ell$ random and
$P_{A^{\ast},\ell}=(7,5,3,2,1,1,1)=P_{c}(\mathcal{D}_{1})$ for $\ell=(x+y+z)$.

(ii). The algebra $B=R/\operatorname{\mathrm{Ann}}G,G=X^{[2]}\cdot
Y^{[4]}+Z^{[5]}+(Y+Z)^{[3]})\in S$ has the same Hilbert function
$H=(1,3,,5,4,4,2,1)$ and decomposition

$\mathcal{D}_{2}=\mathcal{D}_{B}=\left(H(0)=(1,2,3,3,3,2,1),H(1)=(0,1,1,1,1),H(2)=(0,0,1)\right).$

Since $\mathcal{D}_{2}<\mathcal{D}_{1}$ from $A$ in the order of Lemma 2.6
there can be no specialization from a subfamily of
$\mathrm{Gor}(\mathcal{D}_{2})$ to an element of
$\mathrm{Gor}(\mathcal{D}_{1}).$ But it is easy to check that
$P_{B}=(7,5,4,3,1)$, strictly greater in the dominance partial order than
$P_{A}=(7,5,3,3,1,1)$ (Lemma 2.3) so there is no specialization from a family
in $\mathrm{Gor}(\mathcal{D}_{1})$ to a general element of
$\mathrm{Gor}(\mathcal{D}_{2})$. It is not hard to show that each family is
irreducible. So $\mathrm{Gor}(H)$ has two irreducible components, each
corresponding to a symmetric decomposition of $H$.

Examples 2.22, 2.26, 2.27 and 2.31 are of algebras with $P_{A}>P_{A^{\ast}}$.
Example 2.27 is of a non-SL algebra $A$ such that $Q_{A}(0)$ is SL. Johanna
Steinmeyer asked the following in our MFO (Oberwolfach) discussion of October
2020.

######  Question 2.32.

(a) Can we find a strong Lefschetz AG algebra $A$ such that $A^{\ast}$ is not
SL? (b) Can we show that $A$ SL implies $Q_{A}(0)$ is SL?

Both questions are open.

###  2.7 Exotic summands.

When parametrizing families of AG algebras with given Hilbert function $H$, we
can use both the dual generator $f$ and the ideal
$\operatorname{\mathrm{Ann}}f$. Since we can compute the Hilbert function and
its symmetric decomposition $\mathcal{D}$ from the vector space $R\circ f$, of
partials of $f$, it is crucial to understand how changes in the summands of
$f$ in each degree may affect $H$ and $\mathcal{D}$. In a preprint that led to
[BJMR], A. Bernardi and K. Ranestad noticed that some summands had variables
that were unexpected in high degrees, given the decomposition $\mathcal{D}$.
They termed them _exotic summands_. For convenience we repeat the definition,
given in Remark 2.17.

######  Definition 2.33 (Exotic summand of $f$).

(See [BJMR, Definition 3 and 4], [IM, Definition 2.3, §2.2]). Let $H$ be a
Gorenstein sequence of socle degree $j$ and
$\mathcal{D}=\bigl{(}H(0),\,H(1),\ldots,\,H(j-2)\bigr{)}$ a symmetric
decomposition of $H$. Let $n_{i}=n_{i}(\mathcal{D})=\sum_{u=0}^{i}H(u)_{1}$.
We say that $f\in\mathcal{S}$ determining
$A=R/\operatorname{\mathrm{Ann}}f\in\mathrm{Gor}(\mathcal{D})$ admits a
_standard form_ if, after a change of variables in $\mathcal{S}$ to
$X_{1},\ldots,X_{r}$, for each $i$,
$f_{j-i}\in{\mathsf{k}}_{DP}[X_{1},\ldots,X_{n_{i}}]$. Otherwise $f$ is said
to have _exotic_ terms.101010For $f=(X-Y)^{7}+Y^{5}$, a change of variables
$X_{1}=X-Y$ leads to $f=X_{1}^{7}+Y^{5}$, which is in standard form. When
there is an exotic summand, no linear change of variables leads to $f$ in
standard form.

For examples, see Remark 2.17 earlier and Example 2.34 below. It is known that
up to isomorphism of local algebras we may choose the dual generator to be in
standard form: more precisely, given the AG algebra $A$ we may determine
$B=R/J$, $J=\operatorname{\mathrm{Ann}}G$ such that $B\cong A$ and $G$ is in
standard form. This is shown in [I2, Theorem 5.3], but with more detail in
[IM, Theorem 2.7]).111111This standard form is not obtained by simply omitting
the exotic summands, which would not in general preserve the isomorphism
class.

######  Example 2.34.

Let $R={\mathsf{k}}\\{x,y,z\\}$, $S=\mathsf{k}_{DP}[X,Y,Z]$, and consider
${A=R/\operatorname{\mathrm{Ann}}f}$, with ${f=X^{3}Y^{3}+YZ^{3}}$ in $S$. We
can see that $A$ has Hilbert function ${H=(1,3,5,5,3,2,1)}$, with symmetric
decomposition

$\mathcal{D}=\bigl{(}H(0)=(1,2,3,4,3,2,1),\,H(2)=(0,1,2,1,0)\bigr{)}.$

If we take a close look at $R\circ f$, we see that the action of monomials
$x^{p}y^{q}$, with ${0\leq p,q\leq 3}$ on $f$ explains $H(0)$, while the
partials

$\displaystyle z\circ f$ $\displaystyle=YZ^{2}$ $\displaystyle yz\circ f$
$\displaystyle=Z^{2}$ $\displaystyle yz^{2}\circ f$ $\displaystyle=Z$
$\displaystyle z^{2}\circ f$ $\displaystyle=YZ$

account for $H(2)$. Note that $x\circ f$ and $y\circ f$ are partials of degree
$5$, while $z\circ f$ is a partial of degree $3$, which helps to explain why
$z\circ f$ and its subsequent partials contribute to $H(2)$. This is not
surprising, since $Z$ only occurs in $f$ in a summand of degree $4$. We say
that $Z$ is a partial of order $3$, in the sense that it is obtained by the
action of $yz^{2}$, an element of order $3$, and cannot be obtained by the
action of an element of higher order. (For instance, even if ${z^{3}\circ
f}=Y$, we know that $Y$ is a partial of order $5$, because ${x^{3}y^{2}\circ
f=Y}$.)

Now it would be tempting to think that any dual generator $g$ defining an AG
algebra ${B=R/\operatorname{\mathrm{Ann}}g}$ with the same Hilbert function
$H$ and decomposition $\mathcal{D}$ would have summands in degrees $6$ and $5$
that could be written with only two variables (up to a linear change of
variables, if necessary). This is true for degree $6$, but not for degree $5$.
For example, if we take ${g=X^{3}Y^{3}+X^{2}Y^{2}Z}$ then $B$ has Hilbert
function $H$ with decomposition $\mathcal{D}$. Although $Z$ occurs in degree
$5$, it is a partial of order $3$, since ${(-xyz+x^{2}y^{2})\circ g=Z}$, and
we cannot obtain it with the action of an element of higher order. The reason
for this behaviour is that we can take ${\varphi=xy}$ and write
${g=g_{\mathrm{st}}+g_{\mathrm{ex}}}$, with

$\displaystyle g_{\mathrm{st}}$ $\displaystyle=X^{3}Y^{3}-XYZ^{2}-Z^{3}$
$\displaystyle g_{\mathrm{ex}}$ $\displaystyle=Z(\varphi\circ
X^{3}Y^{3})+Z^{2}(\varphi^{2}\circ X^{3}Y^{3})+Z^{3}(\varphi^{3}\circ
X^{3}Y^{3})$

Note that ${(z-xy)\circ g_{\mathrm{ex}}=0}$. We say that $f$ and
$g_{\mathrm{st}}$ are in the standard form, and we call $X^{2}Y^{2}Z$ an
exotic summand of $g$ (see [BJMR, Definitions 3 and 4] or [IM, Definition 2.3,
§2.2]).

The following result is a consequence of [BJMR, Proposition 6], which implies
that an element $f$ defining an algebra
$A=R/\operatorname{\mathrm{Ann}}f\in\mathrm{Gor}(\mathcal{D})$ can be written
$f=f_{st}+f_{ex}$ where $f_{st}$ is standard, and $f_{ex}$ can be expressed
using $f_{st}$. For any ${a\geq 0}$, let ${n_{a}=\sum_{i\leq a}H(a)_{1}}$.

######  Lemma 2.35 (Exotic count, two to three variables).

Let $H$ be a Gorenstein sequence with $H_{1}=3$; let $a$ be an integer at
least $2$, and let $\mathcal{D}$ be a symmetric decomposition of $H$
satisfying $n_{a-1}=2$ and $n_{a}=3$. Let
$R/\operatorname{\mathrm{Ann}}f\in\mathrm{Gor}(\mathcal{D})$. Then we can
choose a basis $X,Y,Z$ for $S_{1}$ such that $X$ and $Y$ are partials of $f$
of order at least $j-a$ and $Z$ is a partial of order $j-a-1$ – in other
words, ${X,Y\in\mathfrak{m}^{j-a}\circ f}$ and
${Z\in(\mathfrak{m}^{j-a-1}\circ f)\setminus(\mathfrak{m}^{j-a}\circ f)}$.
Furthermore, we may write $f=f_{\mathrm{st}}+f_{\mathrm{ex}}$, where
$f_{\mathrm{st}}$ is standard (i.e., for any degree ${d>j-a}$,
${(f_{\mathrm{st}})_{d}\in\mathsf{k}_{\mathrm{DP}}[X,Y]}$) and

$f_{\mathrm{ex}}=Z\varphi(f_{\mathrm{st}})+Z^{[2]}\varphi^{2}(f_{\mathrm{st}})+Z^{[3]}\varphi^{3}(f_{\mathrm{st}})+\cdots$
(2.22)

where $\varphi\in{\mathsf{k}}\\{x,y\\}_{\geq 2}$ and the degree of
$\varphi(f_{\mathrm{st}})$ is at least ${j-a}$.

In particular, if we fix $f_{\mathrm{st}}$, and let
$A(f_{\mathrm{st}})\subset\mathrm{Gor}(\mathcal{D})$ be

$A(f_{\mathrm{st}})=\\{A=R/\operatorname{\mathrm{Ann}}f\mid
f=f_{0}+f_{ex}\\},$

then $A(f_{\mathrm{st}})$ is an affine space – parametrizing
$\varphi(f_{\mathrm{st}})$ in Equation (2.22); it has dimension

$\dim A(f_{\mathrm{st}})=\sum_{b=0}^{a-1}\sum_{u=j-a}^{j-b-2}H(b)_{u}.$ (2.23)

###### Proof.

The Equation (2.22) results from [BJMR, Proposition 6], where the number of
variables is ${n=3}$, and all variables occur as partials od $f$, since
${n_{a}=3}$. The algebra $A\in A(f_{\mathrm{st}})$ is determined uniquely by
the choice of terms $\varphi(f_{\mathrm{st}})$ of $f_{ex}$ in Equation (2.23).
These are parametrized by their top degree components. $\square$

######  Remark 2.36.

Note that if $n_{0}=2$, we have ${H(b)_{j-b-1}=0}$ for any ${b>0}$ so the sum
in (2.23) becomes ${\sum_{u=j-a}^{j-2}H_{u}}$.

We apply Lemma 2.35 and the Remark in Theorem 3.1, as part of determining the
dimension of the irreducible components of
$\mathrm{Gor}(H),H=(1,3,4,4,3,2,1)$, and similarly in Theorem 3.3 where
$H=(1,3,4^{k},3,2,1),k\geq 1$.

###  2.8 Relatively compressed $a$-modifications

We study the effect of adding generically chosen terms of degree $j-a$ to a
dual generator $F$. We suppose that a dual generator
$F=F_{j}+\cdots+F_{j+1-a}\in S={\mathsf{k}}_{DP}[X_{1},\ldots,X_{r}]$ has been
fixed for an AG algebra $A=R/I,R={\mathsf{k}}\\{x_{1},\ldots,x_{r}\\}$. Let
$F^{\prime}=F+G,G\in S_{\leq j-a}$; then we term
$A^{\prime}=R/\operatorname{\mathrm{Ann}}F^{\prime}$ an $a$-modification of
$A$. Given $F$, we define $M(A,a)$ to be the following sequence, symmetric
about $(j-a)/2$. Recall $r_{i}=\dim_{\mathsf{k}}R_{i}$, and we denote by
$H=(h_{0},\ldots,h_{j}$ the Hilbert function $H=H(A)$.

$M(F,a)=\begin{cases}r_{i}-h_{i}&\text{ for }i\leq(j-a)/2\\\
M(F,a)_{j-a-i}&\text{ for }i>(j-a)/2.\end{cases}$ (2.24)

In words, we symmetrize the difference $(H(R)-H(A))_{\leq{j-a}/2}$. For
example if $R={\mathsf{k}}\\{x,y,z\\}$ and $j=7,a=1$ and
$H(A)=(1,2,2,2,2,2,1)$, then $M(F,1)=(0,1,4,4,1)$.

######  Definition 2.37.

We say that $F^{\prime}=F+G,G\in S_{j-a}$ is a relatively compressed
$a$-modification of $F$ if $H(A^{\prime})=H(A)+M(A,a)$.

The next Lemma is a result of J. Emsalem and the first author [I2, Theorem
3.3], see also [IM, §1.2, and Proposition 1.18].

######  Lemma 2.38.

Let $F^{\prime}$ satisfy $F^{\prime}=F+G,F=F_{j}+\cdots+F_{j+1-a}$ be given
and let $G^{\prime}\in S_{\leq j-a}$, let
$A^{\prime}=R/\operatorname{\mathrm{Ann}}F^{\prime}$. Then

$H(A^{\prime})\leq H(A)+M(A,a).$ (2.25)

Also, if $G_{j-a}\in S_{j-a}$ is general enough we have equality in Equation
(2.25). The family of relatively compressed $a$-modifications of $F$ has
dimension $\sum_{i=0}^{j-a}(r_{i}-H(A)_{i})$, and is parametrized by general
enough choices of elements $G_{\leq j-a}\in S_{\leq
j-a}\mod\mathfrak{m}^{a}\circ F$.

######  Example 2.39.

Let $R={\mathsf{k}}\\{x,y,z\\}$ and
$A=R/I,I=(z,xy,x^{6}-y^{6})=\operatorname{\mathrm{Ann}}(X^{5}+Y^{5})$ of
Hilbert function $H=(1,2,2,2,2,2,1)$. Take $G_{5}=\sum_{1}^{4}L_{i}^{[5]}$
where $L_{1},\ldots,L_{4}$ are four general enough linear forms in $R_{1}$.
Then $A^{\prime}=R/\operatorname{\mathrm{Ann}}(F+G)$ will have Hilbert
function $H^{\prime}=H(A^{\prime})=H+(0,1,4,4,1)=(1,3,6,6,3,2,1)$. The
elements $F^{\prime}=F+G$ with $F_{6}=X^{6}+Y^{6}$ and $G\in S_{\leq 5}$ are
parametrized by an open dense in an affine space of dimension

$(1,3,6,10,15,21)-(1,3,6,6,3,2)=((10_{3}-6_{3})+(15_{4}-3_{4})+(21_{5}-2_{5})=35.$

The AG algebras of HF $H^{\prime}$ with decomposition
$\mathcal{D}=(H(0)=H,H(1)=(0,1,4,4,1)$ are parametrized by a variety of
dimension four (choose $\langle X,Y\rangle$ from $R_{1}$, then pick $X,Y$) for
the graded AG algebra quotient of $R$ having HF $H$. So the AG algebras of
decomposition $\mathcal{D}$ form an irreducible family of dimension $39$. This
is the unique symmetric decomposition for $H^{\prime}$, so
$\mathrm{Gor}(H^{\prime})$ is irreducible of dimension 39.

######  Question 2.40.

Given the Jordan type $P_{A,\ell}$ what are the possible Jordan types
$P_{A^{\prime},\ell}$ of a relatively compressed $a$-modification $A^{\prime}$
of $A$? Does it satisfy an inequality involving the JT of $A$ and the
difference $H(A^{\prime})-H(A)$ where $A^{\prime}$ is the RCM of $A$?

We use relatively compressed modifications in the proof of (iv) of Theorem
3.1, p. 29.

##  3 Applying Jordan type and symmetric decomposition for a local AG algebra.

The main results of the paper introduce a new method of showing that
$\mathrm{Gor}(H)$ can have several irreducible components, corresponding to
the symmetric decompositions of $H$: namely the combination of the dominance
partial order on Jordan types (Lemma 2.3) and the partial order on symmetric
decompositions (Lemma 2.6 and Corollary 2.7). We have already used these to
show the existence of $\mathrm{Gor}(H)$ having two irreducible components, in
Section 2.6, Examples 2.27 and 2.31. The following examples have at least
three irreducible components. A simplifying aspect of Theorem 3.1 is that each
component $H(Q(a))$ is unimodal, leading to equality between the Jordan types
of the algebras and their associated graded algebras, in contrast to Example
2.27 above.

###  3.1 Examples of AG families $\mathrm{Gor}(H)$ having three irreducible
components.

We give next our first example with three irreducible components, which we
describe in detail. We extend the result in Theorem 3.3 to
$H=(1,3,4^{k},3,2,1),k\geq 1$, giving an infinite sequence of such examples.
In the following Theorem the proof that
$\mathrm{Gor}(\mathcal{D}_{1}),\mathrm{Gor}(\mathcal{D}_{2})$ and
$\mathrm{Gor}{\mathcal{D}_{3}}$ must determine at least three irreducible
components of $\mathrm{Gor}(H)$ is immediate from Lemmas 2.3 and 2.6 once we
determine the generic Jordan type for each $\mathrm{Gor}(\mathcal{D}_{i})$;
but we go further to describe the structure of each
$\mathrm{Gor}(\mathcal{D}_{i})$, showing it is irreducible, and specifying
their dimensions.

######  Theorem 3.1 ($Gor(H)$ with three irreducible components, different
Jordan types, one SL).

The Gorenstein sequence $H=(1,3,4,4,3,2,1)$ has three symmetric decompositions
(a),(b),(c).

  1. (i.)

(a.) A generic algebra in $\mathrm{Gor}(\mathcal{D}_{1})$,

$\mathcal{D}_{1}=\bigl{(}H(0)=(1,2,3,4,3,2,1),\,H(1)=H(2)=0,\,H(3)=(0,1,1,0)\bigr{)}$

will have Jordan type $P_{1}=(7,5,3,2,1)$. Take $F=X^{3}Y^{3}+Z^{3}$. Then
$A=R/\operatorname{\mathrm{Ann}}F$,
$\operatorname{\mathrm{Ann}}F=(xz,yz,z^{3}-x^{3}y^{3},x^{4},y^{4})$ is in
$\mathrm{Gor}(\mathcal{D}_{1})$.

Also having this decomposition is
$A^{\prime}=R/\operatorname{\mathrm{Ann}}F^{\prime}$,
$F^{\prime}=X^{6}+Y^{6}+(X+Y)^{[6]}+(X-Y)^{[6]}+Z^{3}$.

Here $H^{\vee}=(7,5,4,2)$,
$A^{\ast}=R/\operatorname{\mathrm{Ann}}(X^{3}Y^{3},Z^{2})$, and for
$\ell=x+y+z$, we have
$P_{A,\ell}=P_{A^{\ast},\ell}=P_{\mathcal{D}_{1}}=P_{1}$.

(b.) A generic algebra in $\mathrm{Gor}(\mathcal{D}_{2})$

$\mathcal{D}_{2}=\bigl{(}H(0)=(1,2,3,3,3,2,1),\,H(1)=0,\,H(2)=(0,1,1,1,0)\bigr{)}$

will have Jordan type $P_{2}=(7,5,3,3)$. Let
$G=X^{6}+Y^{6}+(X+Y)^{[6]}+Z^{4}$, then
$\mathcal{B}=R/\operatorname{\mathrm{Ann}}G$,
$\operatorname{\mathrm{Ann}}G=(zx,zy,xy(x-y),2z^{4}-x^{6},x^{5}+y^{5}-3xy^{4})$
is in $\mathrm{Gor}(\mathcal{D}_{2})$. We have
$\mathcal{B}^{\ast}=R/\operatorname{\mathrm{Ann}}(G-Z^{4},Z^{3})$ and
$P_{\mathcal{B},\ell}=P_{\mathcal{B}^{\ast},\ell}=P_{D_{2}}=P_{2}$.

(c.) Finally, a generic algebra in $\mathrm{Gor}(\mathcal{D}_{3})$,

$\mathcal{D}_{3}=\bigl{(}H(0)=(1,2,2,2,2,2,1),\,H(1)=(0,1,2,2,1,0)\bigr{)},$

will have strong Lefschetz Jordan type $P_{3}=(7,5,4,2)$. Take
$\mathcal{C}=R/\operatorname{\mathrm{Ann}}W$,
$W=X^{6}+Y^{6}+(X+Y)^{[5]}+Z^{5}$.
$\operatorname{\mathrm{Ann}}W=\bigl{(}xz,yz,xy(x-y),z^{5}-x^{6},x^{6}-y^{6}\bigr{)}$.
Then $\mathcal{C}\in\mathrm{Gor}(\mathcal{D}_{3})$. Here
$\mathcal{C}^{\ast}=R/\operatorname{\mathrm{Ann}}(W-Z^{5},Z^{4})$ and with
$\ell=x+y+z$
$H^{\vee}=P_{\mathcal{C},\ell}=P_{C^{\ast},\ell}=P_{\mathcal{D}_{3}}=P_{3}$.

There are no further symmetric decompositions of $H$.

  2. (ii.)

In the partial order of Equation (2.5) we have
$\mathcal{D}_{1}\geq\mathcal{D}_{2}\geq\mathcal{D}_{3}$ (see Figure 1), and

$\overline{\mathrm{Gor}(\mathcal{D}_{3})}\cap\left(\mathrm{Gor}(\mathcal{D}_{2})\cup\mathrm{Gor}(\mathcal{D}_{1})\right)=\emptyset\text{
and
}\quad\overline{\mathrm{Gor}(\mathcal{D}_{2})}\cap\mathrm{Gor}(\mathcal{D}_{1})=\emptyset.$
(3.1)

  3. (iii.)

In the dominance partial order $P_{3}\geq P_{2}\geq P_{1}$. We have that no
subfamily $\mathcal{A}(w)\subset\mathrm{Gor}(\mathcal{D}_{i})$ can specialize
to an element $\mathcal{A}(w_{0})$ of $\mathrm{Gor}({D_{j}})$ having generic
Jordan type $P_{j}$ when $i<j$.

  4. (iv.)

Each $\mathrm{Gor}(\mathcal{D}_{i})$, $i\in[1,3]$ is an irreducible component
of $\mathrm{Gor}(H)$. They have dimensions, respectively $29$, $34$ and $28$
for $\mathcal{D}_{1}$, $\mathcal{D}_{2}$, and $\mathcal{D}_{3}$.

###### Proof.

Note that these three are the only occurring symmetric decompositions of $H$.
Indeed, the three sequences $H(0)$ run through all the height 2 symmetric
Gorenstein sequences below $H$. A height two algebra $C(a)$ cannot have a
Hilbert function jump up of two from $H_{i}(C(a))$ to $H_{i+1}(C(a))$
(Equation (2.6)). This implies that the first non-zero $H(a)$ for $a>0$, must
begin $(0,1,\ldots)$, and it has center $(6-a)/2$; later summands must have
smaller center of symmetry. Inspection of symmetric splittings of $H-H(a)$ now
show these are the only decompositions.

We next determine the generic Jordan types for each decomposition.

i.a. Jordan type for a generic $A\in\mathcal{D}_{1}(H)$.

First let ${A(0)=\mathsf{k}[x,y,z]/\operatorname{\mathrm{Ann}}F}$, with
${F=X^{[3]}Y^{[3]}+Z^{[3]}}$, $F\in\mathcal{D}_{1}$. Then a computation by
Macaulay2 shows that the Jordan type of $\ell=x+y+z$ is
${P_{A(0),\ell}=(7,5,3,2,1)}$. To show that this is the Jordan type of a
general AG algebra $A$ having symmetric decomposition $\mathcal{D}_{1}$, we
should note that by Lemma 2.4 ${P_{A,\ell}}$ is bounded above by
$H^{\vee}=(7,5,4,2)$. The only sequences of integers that are strictly greater
than $P_{A,\ell}$ and dominated by $H^{\vee}$ are $(7,5,3,3)$, $(7,5,4,1,1)$,
and $H^{\vee}$ itself. So if we exclude all these cases, we are done. We first
exclude $(7,5,3,3)$, arguing in detail. We then will exclude the last two, an
easier check.

Let $A\in\mathrm{Gor}(\mathcal{D}_{1})$ and suppose that $\ell$ in
$\mathfrak{m}_{A}$ has Jordan type $P_{A,\ell}=(7,5,3,3)$. Then we can
consider four strings corresponding to a pre-Jordan basis:

$\displaystyle S_{1}$ $\displaystyle=\\{1,\ell,\ldots,\ell^{6}\\}$
$\displaystyle S_{2}$ $\displaystyle=\\{b,\ell b,\ldots,\ell^{4}b\\}$
$\displaystyle S_{3}$ $\displaystyle=\\{c,\ell c,\ell^{2}c\\}$ $\displaystyle
S_{4}$ $\displaystyle=\\{d,\ell d,\ell^{2}d\\}.$

Since $\ell^{6}$ is non-zero, we know that
${\ell\in\mathfrak{m}_{A}\setminus\mathfrak{m}_{A}^{\,2}}$, i.e. $\ell$ has
order one. Also $b$ cannot belong to ${\mathfrak{m}_{A}^{\,2}}$, otherwise
$\ell^{4}b$ would be in ${\mathfrak{m}_{A}^{\,6}=\langle\ell^{6}\rangle}$.
Furthermore, by subtracting scalar-multiples of elements in $S_{1}$ to the
elements of $S_{2}$ if necessary, we may assume that
${b\in\mathfrak{m}_{A}\setminus\mathfrak{m}_{A}^{\,2}}$. Let us now show that
both $\ell$ and $b$ represent non-zero, independent classes in

${Q_{A}(0)_{1}=\frac{\mathfrak{m}_{A}}{\mathfrak{m}_{A}^{\,2}+(0:\mathfrak{m}_{A}^{\,5})}}.$

It is straightforward to see that we cannot have
${\ell\in\mathfrak{m}_{A}^{\,2}+(0:\mathfrak{m}_{A}^{\,5})}$, otherwise
$\ell^{6}=0$. On the other hand, if
${b\in\langle\ell\rangle+\mathfrak{m}_{A}^{\,2}+(0:\mathfrak{m}_{A}^{\,5})}$,
say ${b=\alpha\ell+b_{1}+b_{2}}$, with ${b_{1}\in\mathfrak{m}_{A}^{\,2}}$ and
${b_{2}\in(0:\mathfrak{m}_{A}^{\,5})}$, we would have
${\ell^{4}b=\alpha\ell^{5}+\ell^{4}b_{1}+\ell^{4}b_{2}}$, with $\ell^{4}b_{1}$
and $\ell^{4}b_{2}$ both in ${\mathfrak{m}_{A}^{\,6}=\langle\ell^{6}\rangle}$,
a contradiction. In a similar way, we can show that all the elements in
$S_{1}$ and $S_{2}$ represent non-zero, independent classes in $Q_{A}(0)$.
Arguing as we did for $b$, we can assume that ${c,d\in\mathfrak{m}_{A}}$.
Moreover, since
${\dim_{\mathsf{k}}\mathfrak{m}_{A}/\mathfrak{m}_{A}^{\,2}=3}$, we may assume
that ${d\in\mathfrak{m}_{A}\setminus\mathfrak{m}_{A}^{\,2}}$ and
${c\in\mathfrak{m}_{A}^{\,2}\setminus\mathfrak{m}_{A}^{\,3}}$ (we cannot have
$c$ in $\mathfrak{m}_{A}^{\,3}$, otherwise $\ell^{2}c$ would be in
${\mathfrak{m}_{A}^{\,5}=\langle\ell^{5},\ell^{6},\ell^{4}b\rangle}$). Let us
now show that $c$ represents a non-zero class in

${Q_{A}(0)_{2}=\frac{\mathfrak{m}_{A}^{\,2}}{\mathfrak{m}_{A}^{\,3}+\mathfrak{m}_{A}^{\,2}\cap(0:\mathfrak{m}_{A}^{\,4})}},$

not belonging to the span of the classes of $\ell^{2}$ and $\ell b$. If we had
${c\in\langle\ell^{2},\ell
b\rangle+\mathfrak{m}_{A}^{\,3}+\mathfrak{m}_{A}^{\,2}\cap(0:\mathfrak{m}_{A}^{\,4})}$,
we could write ${c=\alpha\ell^{2}+\beta\ell b+c_{1}+c_{2}}$, with
${c_{1}\in\mathfrak{m}_{A}^{\,3}}$ and
${c_{2}\in\mathfrak{m}_{A}^{\,2}\cap(0:\mathfrak{m}_{A}^{\,4})}$, and we would
have ${\ell^{2}c=\alpha\ell^{4}+\ell^{3}b+\ell^{2}c_{1}+\ell^{2}c_{2}}$, with
$\ell^{2}c_{1}$ in $\mathfrak{m}_{A}^{\,5}$ and $\ell^{2}c_{2}$ in
${\mathfrak{m}_{A}^{\,2}\cap(0:\mathfrak{m}_{A}^{\,4})}$. The symmetric
decomposition $\mathcal{D}_{2}$ tells us that
${\dim_{\mathsf{k}}Q_{A}(0)_{4}=3=\dim_{\mathsf{k}}\mathfrak{m}_{A}^{\,4}-\dim_{\mathsf{k}}\mathfrak{m}_{A}^{\,5}}$.
Since

${Q_{A}(0)_{4}=\frac{\mathfrak{m}_{A}^{\,4}}{\mathfrak{m}_{A}^{\,5}+\mathfrak{m}_{A}^{\,4}\cap(0:\mathfrak{m}_{A}^{\,2})}},$

this implies that
${\mathfrak{m}_{A}^{\,4}\cap(0:\mathfrak{m}_{A}^{\,2})\subseteq\mathfrak{m}_{A}^{\,5}}$.
so both $\ell^{2}c_{1}$ and $\ell^{2}c_{2}$ lie in
${\mathfrak{m}_{A}^{\,5}=\langle\ell^{5},\ell^{6},\ell^{4}b\rangle}$, again a
contraction. So far, we have established that all the elements in $S_{1}$,
$S_{2}$ and $S_{3}$ represent non-zero, independent classes in $Q_{A}(0)$.
Since ${\dim_{\mathsf{k}}Q_{A}(3)_{1}=1}$ and

${Q_{A}(3)_{1}=\frac{(0:\mathfrak{m}_{A}^{\,3})}{\mathfrak{m}_{A}^{\,2}\cap(0:\mathfrak{m}_{A}^{\,3})+(0:\mathfrak{m}_{A}^{\,2})}},$

we see that ${(0:\mathfrak{m}_{A}^{\,3})\setminus\mathfrak{m}_{A}^{\,2}}$ is a
non-empty subset of ${\mathfrak{m}_{A}\setminus\mathfrak{m}_{A}^{\,2}}$. Since
the classes of $\ell$, $b$, and $d$ span
${\mathfrak{m}_{A}/\mathfrak{m}_{A}^{\,2}}$, we can write
${d=d^{\prime}+\alpha\ell+\beta b+d^{\prime\prime}}$, with
${d^{\prime}\in(0:\mathfrak{m}_{A}^{\,3})}$, ${\alpha,\beta\in\mathsf{k}}$,
and ${d^{\prime\prime}\in\mathfrak{m}_{A}^{\,2}}$. So we get
${\ell^{2}d=\ell^{2}d^{\prime}+\alpha\ell^{3}+\beta\ell^{2}b+\ell^{2}d^{\prime\prime}}$,
where $\ell^{2}d^{\prime}$ lies in
${\mathfrak{m}_{A}^{\,6}=\langle\ell^{6}\rangle}$ and
$\ell^{2}d^{\prime\prime}$ lies in
${\mathfrak{m}_{A}^{\,4}=\langle\ell^{4},\ell^{5},\ell^{6},\ell^{3}b,\ell^{4}b,\ell^{2}c\rangle}$,
a final contradiction. This shows that $(7,5,3,3)$ cannot be the Jordan type
of $\ell$.

Suppose now that $P_{A,\ell}=(7,5,4,\ldots)$. Then the set of strings would
include

$\displaystyle S_{1}$ $\displaystyle=\\{1,\ell,\ldots,\ell^{6}\\}$
$\displaystyle S_{2}$ $\displaystyle=\\{b,\ell b,\ldots,\ell^{4}b\\}$
$\displaystyle S_{3}$ $\displaystyle=\\{c,\ell c,\ell^{2}c,\ell^{3}c\\}.$

As in the previous case, we can see that all the elements in $S_{1}$ and
$S_{2}$ represent non-zero, independent classes in $Q_{A}(0)$. Also as we did
before for $d$, since ${\dim_{\mathsf{k}}Q_{A}(3)_{1}=1}$, we can write
${c=c^{\prime}+\alpha\ell+\beta b+c^{\prime\prime}}$, with
${c^{\prime}\in(0:\mathfrak{m}_{A}^{\,3})}$, ${\alpha,\beta\in\mathsf{k}}$,
and ${c^{\prime\prime}\in\mathfrak{m}_{A}^{\,2}}$. But this means that
${\ell^{3}c=\ell^{3}c^{\prime}+\alpha\ell^{4}+\beta\ell^{3}b+\ell^{3}c^{\prime\prime}}$,
where ${\ell^{3}c^{\prime}=0}$ and $\ell^{3}c^{\prime\prime}$ lies in
${\mathfrak{m}_{A}^{\,5}=\langle\ell^{5},\ell^{6},\ell^{4}b\rangle}$, a
contradiction. This finishes the proof that the general Jordan type of
$A\in\mathrm{Gor}(\mathcal{D}_{1})$ is ${(7,5,3,2,1)}$.

i.b. Jordan type for a generic element of $\mathrm{Gor}(\mathcal{D}_{2})$. We
need only establish that the generic Jordan type on
$\mathrm{Gor}(\mathcal{D}_{2})$ is $P_{\mathcal{B}}=(7,5,3,3)$. Since we have
an algebra $B$ of that Jordan type we need only eliminate strong Lefschetz for
algebras in $\mathrm{Gor}(\mathcal{D}_{2})$. This can be done in a similar way
to the argument for $\mathrm{Gor}(\mathcal{D}_{2})$, by supposing that we have
a set of strings form a pre-Jordan basis including

$\displaystyle S_{1}$ $\displaystyle=\\{1,\ell,\ldots,\ell^{6}\\}$
$\displaystyle S_{2}$ $\displaystyle=\\{b,\ell b,\ldots,\ell^{4}b\\}$
$\displaystyle S_{3}$ $\displaystyle=\\{c,\ell c,\ell^{2}c,\ell^{3}c\\}.$

and writing ${c=c^{\prime}+\alpha\ell+\beta b+c^{\prime\prime}}$, with
${c^{\prime}\in(0:\mathfrak{m}_{A}^{\,4})}$, ${\alpha,\beta\in\mathsf{k}}$,
and ${c^{\prime\prime}\in\mathfrak{m}_{A}^{\,2}}$. Then we see that
${\ell^{3}c^{\prime}\in(0:\mathfrak{m}_{A})=\mathfrak{m}_{A}^{\,6}}$ and the
argument follows.

i.c. Jordan type for a generic element of $\mathrm{Gor}(\mathcal{D}_{3})$.
Taking $\ell=x+y+z$ the Jordan type of $(\mathcal{C},\ell)$ is
$P_{\mathcal{C}}=(7,5,4,2)=H(\mathcal{C})^{\vee}$, so $\mathcal{C}$ is SL. By
Lemma 2.4 this is the maximum possible Jordan type in $\mathrm{Gor}(H)$, so it
is the generic Jordan type on $\mathrm{Gor}(\mathcal{D}_{3})$. Proof of
(ii),(iii). Immediate from the above and Lemmas 2.3 and 2.6 (see Figure
1).Proof of (iv): Structure and irreducibility of the symmetric decomposition
loci.

(iv)(a.) $\mathcal{D}_{1}$. By Lemma 2.8, Equation (2.7), for a fixed choice
of $V\subset R_{1}$ of dimension two, say $V$,the dimension of the family
$Z_{H(0)}(V)$, the variety of quotients of the regular local ring
${\mathsf{k}}\\{V\\}$ having Hilbert function $H(0)=(1,2,3,4,3,2,1)$ is
$n-d=16-4=12$; this is a fibre over the projective plane $\mathbb{P}^{2}$
parametrizing $V$. Fix $V$ and choose one such algebra $A(0)$ determined by a
polynomial $F(0)$ in ${\mathsf{k}}_{DP}[X,Y]$: now setting $F=F_{0}+G_{3}$
with $G$ generic in $S_{3}\mod(R\circ F)_{3}$, where
$S={\mathsf{k}}_{DP}[X,Y,Z]$, we obtain what is termed a relatively compressed
$6-3=3$-modification of $A(0)$ (Definition 2.37, Lemma 2.38). Here $G_{3}$ is
parametrized by an open dense in an affine space of dimension $\dim
S_{3}-H_{3}=10-4$; we may add to $G_{3}$ an arbitrary element $G_{2}\in
S_{2}\mod R\circ F$: giving an additional affine space fibre of dimension 2.
We so far have determined a fibred space of dimension $14+8=22$. By Lemma 2.35
and Equation (2.23) the exotic summands involving $Z$ give a fibration of
fibre dimension $H(0)_{3}+H(0)_{4}=4+3=7$ (adding terms in degrees $4,5$ to
$F$ \- multiplying an old partial by $Z$). Thus
$\mathrm{Gor}(\mathcal{D}_{1})$ is irreducible of dimension $29$, with the
successive fibration structure just given.

(iv)(b.) $\mathcal{D}_{2}$. Again by Lemma 2.8, the dimension of the family
$Z_{H(0)}(V)$ for $H(0)=(1,2,3,3,3,2,1)$ is $n-d=15-3=12$; the choice of $V\in
S_{1}$ gives 2 more, so $14$. Given $A(0)$ determined by $F$ we form $F+G_{4}$
where $G_{4}$ satisfies the very restrictive condition
$\dim_{\mathsf{k}}R_{2}\circ G_{4}\mod\langle X^{2},XY,Y^{2}\rangle=1$. Here
$G_{4}$ must be viewed mod $\langle R_{2}\circ F\rangle$, a three dimensional
subspace $W$ of $\langle X^{4},X^{3}Y,X^{2}Y^{2},XY^{3},Y^{4}\rangle$. We have
$R_{2}\circ G_{4}\mod\langle X^{2}XY,Y^{2}\rangle=\langle L^{2}\rangle$ or
$La_{1},a_{1}\in\langle X,Y\rangle$ and we may write
$G_{4}=\sum_{0}^{3}\alpha_{i}L^{4-i}$ with $\alpha_{i}\in{\sf k}[X,Y]_{i}$ we
have $\alpha_{2}=0$ (to assure $\dim_{\sf k}A_{2}=4$, otherwise there is no
restriction on $G_{4}$. Counting dimensions we have two for the choice of
$\langle L\rangle$, $1+2+4$ for choice of $\alpha_{0},\alpha_{1},\alpha_{3}$,
so $9$ (this is a not unexpected codimension three out of the possible choices
for an element of $R_{4}/R_{2}\circ F$). For $G_{3}$ we add $\dim R_{3}-4=6$
and for $G_{2}$ add $\dim R_{2}-4=2$ giving us $14+9+6+2=31$ so far. The
exotic summands by Equation (2.23) add $H(0)_{4}=3$, giving a total dimension
of $34$ for $\mathrm{Gor}(\mathcal{D}_{2})$.

(iv)(c.) $\mathcal{D}_{3}$. First choose $F_{6}+F_{5}$: here $F_{6}$ is the
sum $L_{1}^{[6]}+L_{2}^{[6]}$ or $L_{1}^{[5]}L_{2}$, and
$F_{5}=L_{3}^{[5]}+L_{4}^{[5]}$, or $L_{3}^{[4]}\cdot L_{4}$, where each of
$L_{3},L_{4}\in S_{1}$ are linearly independent from $L_{1},L_{2}$. By
$\mathrm{PGL}(2)$ action we may assume that the linear forms are
$X,Y,Z,X+Y+Z$, essentially choosing 4 points in $\mathbb{P}^{2}$, so an eight
dimensional choice, but we care about the scale, so total of 11 dimensions for
$F_{6}+F_{5}$. Now we add elements $G_{i}\in\langle R_{i},i\in[2,4]\rangle$,
so

$F=X^{6}+Y^{6}+Z^{5}+(X+Y+Z)^{[5]}+G_{4}+G_{3}+G_{2}.$

Note that $V=I_{2}^{\ast}=\langle
X^{2},Y^{2},Z^{2},(X+Y+Z)^{[2]}\rangle^{\perp}=\langle(xy-xz,xy-yz)\rangle$.
We need $I_{2}^{\ast}\circ G_{4}\subset\langle
X^{2},Y^{2},Z^{2},(X+Y+Z)^{[2]}$, which we can test121212This analysis is
similar to one for the dimension of the $A$ component in Example 2.27 p. 2.27.
by $I_{2}^{\ast}\circ\langle I_{2}^{\ast}\circ G_{4}\rangle=0$, that is
$G_{4}\in W^{\perp},W=V^{2}=\langle
x^{2}(y-z)^{2},xy(y-z)(x-z),y^{2}(x-z)^{2}\rangle$. Since $W$ is a vector
space of dimension 3, $W^{\perp}\subset S_{4}$ has dimension $(15-3)=12$, but
we must consider $G_{4}\mod\langle
X^{4},Y^{4},z\circ(Z^{5}+(X+Y+Z)^{[5]}\rangle$, a 3-dimensional subspace of
$W^{\perp}$ so we obtain that $G_{4}$ is parametrized by an affine space
bundle $\mathbb{A}^{9}$ over the choice of $F_{6}+F_{5}$. Now $G_{3},G_{2}$
may be chosen arbitrarily mod powers of $X,Y,Z,(X+Y+Z)$, so from vector spaces
of dimension $(r_{2}-h_{2})+(r_{3}-h_{3})=(2+6)=8$. By Equation (2.23) there
are no exotic terms. We have $\dim\mathrm{Gor}(\mathcal{D}_{3})=(11+9+8)=28$.

(iv) Overview: we have
$\dim\mathrm{Gor}(\mathcal{D}_{2})=34>\dim\mathrm{Gor}(\mathcal{D}_{1})=29>\dim\mathrm{Gor}(\mathcal{D}_{3})=28$
showing that, just from a dimension argument, the closure of
$\mathrm{Gor}(\mathcal{D}_{3})$ cannot be all of
$\mathrm{Gor}(\mathcal{D}_{2})$ or $\mathrm{Gor}(\mathcal{D}_{1})$. The Jordan
types for generic elements of the three $\mathrm{Gor}(\mathcal{D}_{i})$ are
$P_{3}>P_{2}>P_{1}$ by Lemma 2.3; this in contrast with Lemma 2.6 on symmetric
decompositions $\mathcal{D}_{1}\geq\mathcal{D}_{2}\geq\mathcal{D}_{3}$ and the
irreducibility of each $\mathrm{Gor}(\mathcal{D}_{i})$ shows that we have
three irreducible components of $\mathrm{Gor}(H)$. However, the statement from
the semicontinuity of symmetric decompositions is stronger: for example, no
family of AG algebras $A\in\mathrm{Gor}(\mathcal{D}_{2})$ can specialize to an
algebra in $\mathrm{Gor}(\mathcal{D}_{1})$!

This completes the proof of the Theorem. $\square$

######  Note 3.2 (Discussion of $H(R/I^{2})$ after Remark 2.29).

Here $H(R/I)=(1,3,4,4,3,2,1)$ of length $|H(R/I)|=18$.

$\begin{array}[]{c||c|c|}R/I&H(R/I^{2})&H(I/I^{2})\\\
\hline\cr\hline\cr{A}&(1,3,6,10,12,12,{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{9}},9,6,4)&((0,0,2,6,9,10,{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{8}},9,6,4)\\\
\hline\cr{\mathcal{B}}&(1,3,6,10,12,12,{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{10}},8,5,3,2)&(0,0,2,6,9,10,{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{9}},8,5,3,2)\\\
\hline\cr{\mathcal{C}}&(1,3,6,10,12,12,{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{11}},7,5,3,2)&(0,0,2,6,9,10,{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{10}},7,5,3,2)\\\
\hline\cr\end{array}$ The lengths $|H(I/I^{2})|=3\cdot|H(R/I)|=54$ for each of
$I=I_{A},I=I_{B},I=I_{C}$. So according to the discussion of Remark 2.29 we
would have $C$ possibly most general, but $B$, $A$ incomparable. To check: the
Hilbert functions for generic elements $A,B,C$ of the components of
$\mathrm{Gor}(H)$.

The next theorem extends the main results of Theorem 3.1 (case $k=2$) to
arbitrary $H=(1,3,4^{k},3,2,1),k\geq 2.$

######  Theorem 3.3.

Let $k\geq 2$. Consider the Gorenstein sequence $H(k)=(1,3,4^{k},3,2,1)$ of
socle degree $j=k+4$ and length $|H(k)|=4k+10$. It has three symmetric
decompositions (a),(b),(c):

  1. (i.)

(a.) A generic algebra in $\mathrm{Gor}(\mathcal{D}_{1})$,

$\mathcal{D}_{1}=\left(H(0)=(1,2,3,4^{k-1},3,2,1),H(1)=\cdots=H(k)=0,H(k+1)=(0,1,1)\right)$

will have Jordan type $P_{1}=(k+5,k+3,k+1,k-1,2)$ Let
$F(k)=X^{3}Y^{k+1}+Z^{3}$ The algebra
$A(k)=R/\operatorname{\mathrm{Ann}}F(k)$,
$\operatorname{\mathrm{Ann}}F(k)=(xz,yz,z^{3}-x^{3}y^{k+1},x^{4},y^{k+2})$ is
in $\mathrm{Gor}(\mathcal{D}_{1})$.

(b.) A generic algebra in $\mathrm{Gor}(\mathcal{D}_{2})$

$\mathcal{D}_{2}=\left(H(0)=(1,2,3^{k+1},2,1),H(2)=(0,1^{k+1})\right),$

will have Jordan type $P_{2}=(k+5,k+3,k+1.k+1)$. Let
$G=X^{k+4}+Y^{k+4}+(X+Y)^{[k+4]}+Z^{4}$, then
$\mathcal{B}=R/\operatorname{\mathrm{Ann}}G$,
$\operatorname{\mathrm{Ann}}G=(zx,zy,xy(x-y),z^{4}-x^{k+4},x^{k+3}+y^{k+3}-3xy^{k+2})$
is in $\mathrm{Gor}(\mathcal{D}_{2})$.

(c.) Finally, a generic algebra in $\mathrm{Gor}(\mathcal{D}_{3})$,

$\mathcal{D}_{3}=\left(H(0)=(1,2^{k+3},1),H(1)=(0,1,2^{k},1)\right).$

will have strong Lefschetz Jordan type $P_{3}=(k+5,k+3,k+2,k)$. Take
$\mathcal{C}=R/\operatorname{\mathrm{Ann}}W,W=X^{k+4}+Y^{k+4}+(X+Y)^{k+3}+Z^{k+3}$.
$\operatorname{\mathrm{Ann}}W=(xz,yz,xy(x-y),z^{k+3}-x^{k+4},x^{k+4}-y^{k+4})$.
Then $\mathcal{C}\in\mathrm{Gor}(\mathcal{D}_{3})$. Note that, taking
$\ell=x+y+z$ the Jordan type of $(\mathcal{C},\ell)$ is the SL $P_{3}$.

There are no further symmetric decompositions.

  2. (ii.)

In the partial order of Equation (2.5) we have
$\mathcal{D}_{1}\geq\mathcal{D}_{2}\geq\mathcal{D}_{3}$ (see Figure 1), and

$\overline{\mathrm{Gor}(\mathcal{D}_{3})}\cap\left(\mathrm{Gor}(\mathcal{D}_{2})\cup\mathrm{Gor}(\mathcal{D}_{1})\right)=\emptyset\text{
and
}\quad\overline{\mathrm{Gor}(\mathcal{D}_{2})}\cap\mathrm{Gor}(\mathcal{D}_{1})=\emptyset.$
(3.2)

  3. (iii.)

In the dominance partial order $P_{3}\geq P_{2}\geq P_{1}$. We have that no
subfamily $\mathcal{A}(w)\subset\mathrm{Gor}(\mathcal{D}_{i})$ can specialize
to an element $\mathcal{A}(w_{0})$ of $\mathrm{Gor}(\mathcal{D}_{j})$ having
generic Jordan type $P_{j}$ when $i<j$.

  4. (iv.)

Each $\mathrm{Gor}(\mathcal{D}_{i}),i\in[1,3]$ is an irreducible component of
$\mathrm{Gor}(H)$. They have dimensions, respectively, $8k+13,3k+22$ and
$8k+12$ for $\mathcal{D}_{1},\mathcal{D}_{2}$ and $D_{3}$.

###### Proof.

Most of the proof is a straightforward extension of that of Theorem 3.1.
However, we verify the dimensions.

a. We consider the first two decompositions. Here $\dim\mathcal{D}$ has
contributions from

  1. (i)

$\mathbb{P}^{2}$, choice of $\langle x,y\rangle$);

  2. (ii)

$|Q(0)|-\nu(Q(0))$, dimension of family of quotients of ${\mathsf{k}}[x,y]$
having Hilbert function $H(Q(0))$ (Lemma 2.8B);

  3. (iii)

dimension of modification for $Q(1)$, and $8=(10-4)+(6-4)$ for a general
element $G_{3}+G_{2}$ in degree 3,2, outside of $A^{\vee}_{2,3}$ for
$H=(1,3,4,4,\ldots)$;

  4. (iv)

exotic summands (Lemma 2.35 and Remark 2.36).

For $\mathcal{D}_{1}$ this is $2+4(k+2)-4+6+2+(4k-1)=8k+13$. Here the exotics
occur in degrees $4$ to $j-1$ and have dimension
$\sum_{i=3}^{j-2}H(Q(0)_{i}=|Q(0)|-9=4k+8-9=4k-1$ by Equation (2.23).

For $\mathcal{D}_{2}$ we have $2+(3k+9)-3+3+8+H(0)_{4}=3k+22$ (the exotic
contribution is $H(0)_{4}=3$).

(b).We now consider $\mathcal{D}_{3}$.

First choose $F_{j}+F_{j-1}$: here $F_{j}$ is the sum
$L_{1}^{[j]}+L_{2}^{[j]}$ or $L_{1}^{[j-1]}L_{2}$, and
$F_{5}=L_{3}^{[j-1]}+L_{4}^{[j-1]}$, or $L_{3}^{[j-2]}\cdot L_{4}$, where each
of $L_{3},L_{4}\in S_{1}$ are linearly independent from $L_{1},L_{2}$. As in
the case $k=2$ by $\mathrm{PGL}(2)$ action we may assume that the linear forms
are $X,Y,Z,X+Y+Z$, essentially choosing 4 points in $\mathbb{P}^{2}$, so an
eight dimensional choice, but we care about the scale, so total of 11
dimensions for $F_{j}+F_{j-1}$. Now we add elements $G_{i}\in\langle
R_{i},i\in[2,j-2]\rangle$, so
$F=F_{j}+F_{j-1}+\sum_{i=5}^{j-2}G_{i}+(G_{3}+G_{2}).$ Here we must choose
$G_{j-2},\ldots,G_{5}$ so they do not add to the Hilbert function: for
example, a generic $G_{4}$ (relatively compressed modification) would add
$(0,0,2,0)$ giving $H=(1,3,6,4,\ldots)$. Note that $V=I_{2}^{\ast}=\langle
X^{2},Y^{2},Z^{2},(X+Y+Z)^{[2]}\rangle^{\perp}=\langle(xy-xz,xy-yz)\rangle$.
We need $I_{2}^{\ast}\circ G_{4}\subset\langle
X^{2},Y^{2},Z^{2},(X+Y+Z)^{[2]}$, which we can test by
$I_{2}^{\ast}\circ\langle I_{2}^{\ast}\circ G_{4}\rangle=0$, that is $G_{4}\in
W^{\perp},W=V^{2}=\langle x^{2}(y-z)^{2},xy(y-z)(x-z),y^{2}(x-z)^{2}\rangle$.
Since $W$ is a vector space of dimension 3, $W^{\perp}\subset S_{4}$ has
dimension $(15-3)=12$. This corresponds to $W$ defining a scheme of length 12
the order-two vanishing at 4 points of $\mathbb{P}^{2}$
$(1,0,0),(0,1,0),(0,0,1),(1,1,1)$, each local ideal being
$\mathfrak{m}_{p_{i}}^{2}$ of colength 3. But we must consider
$G_{i}\mod\langle R_{j-i}\circ(X^{j},Y^{j},((z,xy)\cap
R_{j-2})\circ(Z^{j-1}+(X+Y+Z)^{[j-1]}\rangle$. This is a 3-dimensional
subspace of $W^{\perp}$ when $i=j-2=k+2$ but 4-dimensional when $4\leq i\leq
j-3$ so we obtain that $G_{4}+\cdots+G_{k+2}$ is parametrized by an affine
space bundle $\mathbb{A}^{8}+\cdots\mathbb{A}^{8}+\mathbb{A}^{9}$ of total
dimension $8(k-2)+9$ over the choice of $F_{j}+F_{j-1}$. And $G_{3},G_{2}$ may
be chosen arbitrarily mod powers of $X,Y,Z,(X+Y+Z)$, so from vector spaces of
dimension $(r_{2}-h_{2})+(r_{3}-h_{3})=(2+6)=8$. By Equation (2.23) there are
no exotic terms. We have
$\dim\mathrm{Gor}(\mathcal{D}_{3})=(11+8(k-2)+9+8)=8k+12$. $\square$

######  Question 3.4.

[Concordance] Is there a codimension three Gorenstein sequence $H$, and two
symmetric decomposition strata $\mathcal{D}_{1},\mathcal{D}_{2}$ such that the
partial orders on
$\mathrm{Gor}(\mathcal{D}_{1}),\mathrm{Gor}(\mathcal{D}_{2})\subset\mathrm{Gor}(H)$
from Jordan type and from semicontinuity for $\mathcal{D}_{1},\mathcal{D}_{2}$
(Lemma 2.6) are concordant? That is, $\mathcal{D}_{1}<\mathcal{D}_{2}$ and
likewise $P_{1}\leq P_{2}$? Or, at least consistent - say
$P_{\mathcal{D}_{1}}=P_{\mathcal{D}_{2}}$ for a generic linear form? See
Question 3.24.

###  3.2 Infinite families of $\mathrm{Gor}(H)$ having at least two
irreducible components.

We give an infinite series of examples of codimension three families
$\mathrm{Gor}(H)$ with several irreducible components - just below and Theorem
3.6, where $H=(1,3,4^{k},3^{s},2,1)$ with $k\geq 1,s\geq 2$. Let
$R={\mathsf{k}}[x,y,z]$. First, recall that a Gorenstein algebra
$A=R/\operatorname{\mathrm{Ann}}F$ of codimension three and socle degree $j$
is a _connected sum_ of a socle-degree $j$algebra
$B={\mathsf{k}}[x.,y]/\operatorname{\mathrm{Ann}}G,G\in{\mathsf{k}}[X,Y]$ of
codimension 2, and $C={\mathsf{k}}[z]/\operatorname{\mathrm{Ann}}Z^{k},k\leq
j$a of codimension $1$ if we can write $F=G+Z^{k}$. In that case, given
$\ell\in\langle x,y\rangle$ the JT of $A$ with respect to $\ell+z$ is obtained
by adjoining the part $(k-1)$ to $P_{B,\ell}$, if $k<j$, and by adjoining
$(k-2)$ to $P_{B,\ell}$ if $k=j$.

######  Example 3.5 (Lefschetz property contributing to component structure of
$Gor(H)$).

We take $H(A)=(1,3,4,4,3,3,3,2,1)$ which permits two symmetric decompositions.
Let $F=X^{[8]}+Y^{[8]}+(Z)^{[7]}+(X+Y+Z)^{[4]}$ (a power sum), and
$A=R/\operatorname{\mathrm{Ann}}F$, of Hilbert function
$H(A)=(1,3,4,4,3,3,3,2,1)$ and symmetric decomposition
$\mathcal{D}_{1}=\mathcal{D}_{A}$,

$\mathcal{D}_{1}=\left(H_{A}(0)=(1,2,2,2,2,2,2,2,1),H_{A}(1)=(0,1,1,1,1,1,1),H_{A}(3)=(0,0,1,1)\right).$

Here even $A^{\ast}$ is SL, as it is readily seen to have Jordan type
$P_{\ell},\ell=(x+y+z)$ satisfying $P_{\ell}=H(A)^{\vee}=(9,7,6,2)$.

Now let $G=X^{[8]}+Y^{[8]}+(X+Y)^{[8]}+Z^{[4]}$, and
$B=R/\operatorname{\mathrm{Ann}}G$. Then $\mathcal{D}_{2}=\mathcal{D}_{B}$
satisfies

$\mathcal{D}_{2}=\left(H_{B}(0)=(1,2,3,3,3,3,3,2,1),H_{B}(4)=(0,1,1,1)\right),$

of the same Hilbert function $H(B)=H(A)$. However, since $G$ is a connected
sum of components of different degrees, it follows that the Jordan type of $B$
is the concatenation of that for $Q(0)=R/\operatorname{\mathrm{Ann}}G_{8}$,
which is $(9,7,5)$ and that for $Q(4)$, which is $3$. So
$P_{B,\ell}=(9,7,5,3)<H_{B}^{\vee}$, and $B$ is not SL.

Specialization: $\mathrm{Gor}(H)$ has two irreducible components: Since no
element of $\mathrm{Gor}(\mathcal{D}_{2})$ is strong Lefschetz, no subfamily
of $\mathrm{Gor}(\mathcal{D}_{2})$ can specialize to a SL element of
$\mathrm{Gor}(\mathcal{D}_{1})$ (Lemma 2.3). But no family of algebras of
decomposition $\mathcal{D}_{1}$ can specialize to one of decomposition
$\mathcal{D}_{2}$, since $H_{0}(B)_{2}>H_{0}(A)_{2}$ (Corollary 2.7).

Here the two-component structure does not follow simply from the
semicontinuity of $\mathcal{D}$ (see Lemma 2.6 and [I2, §4]) but uses the
Jordan type in an essential way.

Parametrization of algebras having decomposition $\mathcal{D}_{1}$. This
should be an irreducible family. For the associated graded algebra, we choose
two variables to work in (a choice parametrized by $\mathbb{P}^{2}$); then for
$F_{8}$ the sum of (eighth powers of) two linear forms, in the two variables,
up to constant linear multiple $(2+1)$, then a general linear form in three
variables (with constant coefficient mattering) for $F_{7}$ and similarly for
$F_{8}$, for a total of $2+3+3+3=11$ dimensions. The dimension of
$Gor(\mathcal{D}_{A})$ may be somewhat larger.

Parametrization of algebras having decomposition $\mathcal{D}_{2}$: It would
appear that the family of AG algebras of decomposition $\mathcal{D}_{B}$
should be irreducible, of dimension at least that of the family of stratified
$B^{\ast}$ (stratified by sequences $C(0)\supset C(1)\supset\cdots$ having the
same Hilbert functions of quotients $H(a)=H(Q(a))$). Here
$Q(0)=R/\operatorname{\mathrm{Ann}}G_{8}$, is parametrized by $G_{8}$ having
Hilbert function $H_{B}(0)$. For $G_{8}$ choose a subspace of two variables
$\langle x,y\rangle$ from three, then take a sum of three eighth powers of
linear forms, with coefficients, mod constant multiple of all: so $G_{8}$ is
parametrized by a bundle of dimension $(3(1)+2)$ over $\mathbb{P}^{2}$, so has
dimension 7. Then one needs to choose a single linear form in three variables
for $G_{4}$, where the constant multiple matters, so $3$ more, so total
dimension 10 for the family of associated graded algebras having decomposition
$\mathcal{D}_{B}$. But the family of AG algebras with decomposition
$\mathcal{D}_{B}$ could be larger, as $G$ may have terms that are hidden.

The following result generalizes the Example 3.5 and has essentially the same
proof.

######  Theorem 3.6.

Let $H=(1,3,4^{k},3^{s},2,1)$ where $k\geq 1,s\geq 2$, of socle degree
$j=k+s+3$ and $F=X^{[j]}+Y^{[j]}+(Z)^{[j-1]}+(X+Y+Z)^{[k+2]}$, determining an
algebra $A=R/\operatorname{\mathrm{Ann}}F$ of symmetric decomposition

$\mathcal{D}_{1}=\left(H(0)=(1,2^{j-1},1),H(1)=(0,1^{j-2}),H(j-k-3)=(0,0,1^{k})\right),$

of Hilbert function $H$. The algebra $A$, hence general enough algebras of
decomposition $\mathcal{D}_{1}$ are strong Lefschetz for $\ell=(x+y+z)$ \-
they have JT $H^{\vee}=(j+1,j-1,j-2,k)$.

Now let $G=X^{[j]}+Y^{[j]}+(X+Y)^{[j]}+Z^{[k+2]}$, and
$B=R/\operatorname{\mathrm{Ann}}G$, where $\mathcal{D}_{B}$ has symmetric
decomposition

$\mathcal{D}_{2}=\left(H_{B}(0)=(1,2,3^{j-3},2,1),H_{B}(j-k-2)=(0,1^{k+1})\right),$

of the same Hilbert function $H(B)=H(A)$. However, since $G$ is a connected
sum of components of different degrees, it follows that the Jordan type of $B$
is the concatenation of that for $Q(0)=R/\operatorname{\mathrm{Ann}}G_{j}$,
which is $(j+1,j-1,j-3)$ and that for $Q(j-k-2)$, which is $(k+1)$. So
$P_{B,\ell}=(j+1,j-1,j-3,k+1)<H^{\vee}$. Thus, general elements of
$\mathrm{Gor}(\mathcal{D}_{2})$ are not weak Lefschetz, but not strong
Lefschetz.

A family of elements of $\mathrm{Gor}(\mathcal{D}_{1})$ cannot specialize to
any element of $\mathrm{Gor}(\mathcal{D}_{2})$ (Corollary 2.7), and a family
of elements of $\mathrm{Gor}(\mathcal{D}_{2})$ cannot specialize to an element
of $\mathrm{Gor}(\mathcal{D}_{1})$ that is strong Lefschetz (Lemma 2.3). These
strata comprise two irreducible components of $\mathrm{Gor}(H)$ (for certain
$H$ there may be more, see Theorem 3.1).

######  Remark 3.7.

It is easy to set up families of codimension three Gorenstein sequences of
higher Sperner number (maximum value of $H$) where similar arguments should
work to show that $\mathrm{Gor}(H)$ has at least two irreducible components.
For example, let $H=(1,3,5^{k},4^{s},3,2,1)$ with $k\geq 2$ of socle degree
$j=k+s+4$ and length $5k+4s+10$, and note that
$H^{\vee}=(k+s+5,k+s+3,k+s+2,k+s,k)$. Let

$\mathcal{D}_{1}=\left(H(0)=(1,2,3^{k+s+1},2,1),H(3)=(0,0,1^{k+s}),H(s+3)=(0,1^{k+1})\right),$

and note that $P_{\mathcal{D}_{1}}=(k+s+5,k+s+3,k+s+1,k+s,k+1)$. Let

$\mathcal{D}_{2}=\left(H(0)=(1,2,3,4^{k+s-1},3,2,1),H(s+3)=(0,1^{k+1},H(k+s)=(0,0,1)\right),$

and note that $P_{\mathcal{D}_{2}}=(k+s+5,k+s+3,k+s+1,k+s-1,k+1,1)$ which is
less than $P_{\mathcal{D}_{1}}$, so again the expected Jordan types under
dominance partial order are opposite to the symmetric decomposition order.

######  Question 3.8.

Can we get an unbounded number of irreducible components of $\mathrm{Gor}(T)$
as a function of socle degree $j$, in the limit as $j\to\infty$? What is the
function $b(j,r)=$ max # components of an AG algebra of socle degree $j$ and
embedding dimension $r$? For $r=2,b(j,2)=1$ for all $j$ (since $T$ determines
$\mathcal{D}$ and $\mathrm{Gor}(T)$ is irreducible in codimension two). Can we
show it goes to $\infty$ for $r=3$ and $j$ large?

######  Question 3.9.

The map $A\to A^{\ast}$ for algebras $A\in\mathrm{Gor}(T)$ induces an
algebraic map of schemes $\pi:\mathrm{Gor}(T)\to G(T)$ parametrizing graded
algebras of Hilbert function $T$. Can we conclude $Gor(\mathcal{D})$ is
irreducible if the image $\pi(Gor(\mathcal{D}))\subset G(T)$ is irreducible
and there is a sufficiently strong conditions on the fibres?

###  3.3 Realizability.

We assume in this section that $A$ is a local Artinian Gorenstein algebra of
socle degree $j$, so $A=R/I$ where $R$ is a standard graded regular local
ring. Recall that the Hilbert function $H(A)$ for such $A$ satisfies three
conditions:

  1. (i).

$H=\sum_{a=0}^{j-2}H(a)$, where each $H(a)$ is symmetric about $(j-a)/2$.

  2. (ii).

For each $a,0\leq a\leq j-2$ we have that $\sum_{u=0}^{a}H(u)$ is an
O-sequence (possible for a Hilbert function of an Artinian algebra).

  3. (iii).

The symmetric sequence $H(0)$ occurs as the Hilbert function of a graded
Gorenstein algebra.

Here the first and third conditions are from Lemma 1.2 and Equation (1.3); the
second condition results from
$H(A^{\ast}/\mathcal{C}(a+1))=\sum_{u=0}^{a}H(u)$. It is natural to ask if
there are further conditions on the symmetric decompositions $\mathcal{D}(A)$
so on the component Hilbert functions $H(A)$ for AG algebras. If it actually
occurs as the complete Hilbert function decomposition for an AG algebra, we
call $\mathcal{D}(A)$ _realizable_. S. Masuti and M. Rossi showed that all
socle degree four potential decomposition sequences $\mathcal{D}$ satisfying
the three conditions are realizable [MasR, §3.6,3.7].

In [IM, p. 6] we introduced the concept of non-ubiquity: a realizable
decomposition with $H(a)\not=0$ is non-ubiquitous if there is some $b<a$ such
that $\mathcal{D}_{\leq b}=\left(H(0),H(1),\ldots,H(b)\right)$ is not
realizable. We could only show partial non-ubiquity, where we imposed an
additional condition on $B$ [IM, §8]. We here exhibit non-realizable sequences
in codimension three, for socle degree $j\geq 8$, satisfying the three
conditions. A realizable complete intersection or Gorenstein algebra $A$ that
is non-ubiquitous cannot arise from our examples. We were led to our non-
realizability results in response to a question asked us by S. Masuti, and
another about the - unknown - Hilbert functions of codimension three complete
intersections, by M. Rossi at a talk in the IIT Bombay Virtual Commutative
Algebra seminar of December 29, 2020. Subsequently to our examples, S. Masuti
has extended the all-decompositions-realizable result of [MasR] to obtain
partial results in socle degree $5$ [Mas]. We now give a general construction
of some non-realizable candidate symmetric decompositions.

######  Theorem 3.10.

Let $R={\mathsf{k}}[x_{1},\ldots,x_{r}],S={\mathsf{k}}_{DP}[X_{1},\ldots
X_{r}$, $R^{\prime}=R[w],S^{\prime}=S[W]$. Let $B$ be a graded Gorenstein
algebra of Hilbert function $H(0)$ of socle degree $j$, and assume that
$B=R/I_{B}$ where $I_{B}=\operatorname{\mathrm{Ann}}f\cap R$ for an element
$f\in S_{j}$. Assume further that $j\geq 5$ and that $I_{B}$ is generated in
degrees less or equal $j-3$.131313In effect, $j\geq 8$. Let
$H(1)=(0,1,0,\ldots,0,1_{j-2},0)$, and assume that $H=H(0)+H(1)$ satisfies the
Macaulay conditions. Then $\mathcal{D}=\left(H(0),H(1)\right)$ is not
realizable as the (complete) decomposition sequence of an AG algebra quotient
$A$ of $R^{\prime}$ having $Q(0)$ isomorphic to $B$. Nor can it occur as part
of a decomposition sequence $\mathcal{D}$ for an AG quotient of $R$ having
$Q(0)$ isomorphic to $B$.

###### Proof.

Assume by way of contradiction that
$A=\operatorname{\mathrm{Ann}}F,F=f+f_{j-1}+\cdots$ has decomposition