# G-Quadratic, LG-Quadratic, and Koszul Quotients of Exterior Algebras

Jason McCullough  Iowa State University, Department of Mathematics, Ames, IA,
USA<EMAIL_ADDRESS>and  Zachary Mere  Iowa State University, Department
of Mathematics, Ames, IA, USA<EMAIL_ADDRESS>
###### Abstract.

This paper introduces the study of LG-quadratic quotients of exterior
algebras, showing that they are Koszul, as in the commutative case. We
construct an example of an LG-quadratic algebra that is not G-quadratic and
another example that is Koszul but not LG-quadratic. This is only the second
known Koszul algebra that is not LG-quadratic and the first that is
noncommutative.

###### Key words and phrases:

Koszul algebra, exterior algebra, G-quadratic, LG-quadratic, edge ideal,
generic quadrics

###### 2010 Mathematics Subject Classification:

Primary: 16S37, 13P10; Secondary: 13C40, 11E04, 05E40

##  1\. Introduction

Let $\mathbb{K}$ be a field and let $E=\bigwedge_{\mathbb{K}}\langle
e_{1},\ldots,e_{n}\rangle$ denote an exterior algebra over $\mathbb{K}$ on $n$
variables. The purpose of this paper is to investigate the Koszul and
G-quadratic properties of quotients of $E$. In the commutative setting, we
have the following implications:

$\text{quadratic GB}\Rightarrow\text{G-quadratic}\Rightarrow\text{LG-
quadratic}\Rightarrow\text{Koszul}\Rightarrow\text{quadratic}.$

Each of these implications is strict; see [9, p. 292]. The third implication
is particularly interesting as there is only one known commutative Koszul
algebra that is not LG-quadratic due to Conca [8, Example 1.20].

Over an exterior algebra, we show that the same implications hold and that all
of them are strict. In particular, we introduce the notion of LG-quadratic
quotients of an exterior algebra and prove that they are Koszul (Theorem 3.2).
We construct an LG-quadratic quotient of an exterior algebra that is not
G-quadratic (Theorem 4.5), thus answering a question of Thieu [34, Example
5.2.2]. We also construct a Koszul quotient of an exterior algebra that is not
LG-quadratic (Theorem 5.4). This is then only the second such algebra and the
first that is noncommutative. It is also non-obstructed, in the language of
Conca [8, Example 1.21], meaning that there are algebras defined by quadratic
monomials with the same Hilbert function. This is one reason for the
difficulty in proving that our construction is indeed not LG-quadratic.

One of the primary motivations for this paper comes from the theory of Orlik-
Solomon algebras of complex hyperplane arrangements, which are quotients of
exterior algebras. It follows from work of Peeva [28] and Björner and Zieglier
[4] that a hyperplane arrangement is supersolvable if and only if its Orlik-
Solomon algebra has a defining ideal with a quadratic Gröbner basis (in a
particular system of coordinates). It is an open question whether there are
Koszul Orlik-Solomon algebras arising from non-supersolvable arrangements; see
[31, Problem 82] and [33, p. 487]. The two notions are equivalent for
hypersolvable arrangements [20], graphic arrangements [32], Dirichlet
arrangements [24], root ideal arrangements [18], and others [36].

The rest of this paper is organized as follows. Section 2 collects the
necessary notation and background results. In Section 3 we show that LG-
quadratic quotients of exterior algebras are Koszul. In Section 4 we construct
an LG-quadratic algebra (which is necessarily Koszul) that is not G-quadratic.
We first consider the depth of edge ideals associated to graphs and construct
our example as a quotient of an algebra defined by quadratic monomials. In
Section 5 we show that six generic quadrics in six exterior variables define a
Koszul algebra that is not LG-quadratic. A final Section 6 collects some
remaining relevant examples.

##  2\. Background

In this section we fix notation and recall some results from the literature.
Throughout this paper, let $\mathbb{K}$ denote a field and let $V$ be a
$\mathbb{K}$-vector space of dimension $n$. Let $e_{1},\ldots,e_{n}$ denote a
fixed basis of $V$ and let $E=\bigwedge(V)=\bigwedge_{\mathbb{K}}\langle
e_{1},\ldots,e_{n}\rangle$ denote the exterior algebra of $V$. We view
$E=\oplus_{i=0}^{n}E_{i}$ as a skew-commutative, graded ring with
$\deg(e_{i})=1$ for $1\leq i\leq n$ and with $E_{i}$ denoting the
$\mathbb{K}$-span of the degree $i$ monomials in $E$. An $E$-module is graded
if there is a $\mathbb{K}$-vector space direct sum decomposition
$M=\oplus_{i}M_{i}$ such that $E_{i}\cdot M_{j}\subseteq M_{i+j}$ for all
integers $i$ and $j$. A monomial in $E$ is an expression of the form
$e_{i_{1}}\wedge e_{i_{2}}\wedge\cdots\wedge e_{i_{t}}$; we usually omit the
wedge products and write $e_{i_{1}}e_{i_{2}}\cdots e_{i_{t}}$ instead. An
ideal generated by monomials is a monomial ideal.

###  2.1. Free resolutions, Hilbert series, Poincare series

Let $R$ be a $\mathbb{Z}$-graded $\mathbb{K}$-algebra, and let $M$ be a
finitely generated, graded $R$-module. The Hilbert function of $M$ is
$\mathop{\mathrm{HF}}\nolimits(M,i)=\dim_{\mathbb{K}}(M_{i})$. Its generating
series is the Hilbert series
$\mathop{\mathrm{HS}}\nolimits_{M}(t)=\sum_{i}HF(M,i)t^{i}$. We recall that
$\mathop{\mathrm{HS}}\nolimits_{E}(t)=(1+t)^{n}$.

The module $M$ has a minimal graded free resolution $\mathbf{F}_{\bullet}$
over $R$, where $F_{i}=\bigoplus_{j}R(-j)^{\beta_{i,j}}$. Here $R(-j)$ denotes
a rank-one graded free $R$-module with $R(-j)_{i}=R_{i-j}$. The numbers
$\beta^{R}_{i,j}(M)$ are the graded Betti numbers of $M$ and are invariants of
$M$. The $i$th total Betti number of $M$ is
$\beta^{R}_{i}(M)=\sum_{j}\beta^{R}_{i,j}(M)$. We may alternatively compute
the Betti numbers of $M$ as
$\beta^{R}_{i,j}(M)=\dim_{\mathbb{K}}\mathop{\mathrm{Tor}}\nolimits_{i}^{R}(M,\mathbb{K})_{j}$
and
$\beta^{R}_{i}(M)=\dim_{\mathbb{K}}\mathop{\mathrm{Tor}}\nolimits_{i}^{R}(M,\mathbb{K})$.
Note that most resolutions over $E$ are infinite. The generating series of the
total Betti numbers is the Poincare series of $M$ denoted
$P^{R}_{M}(t)=\sum_{i\geq 0}\beta^{R}_{i}(M)t^{i}$.

###  2.2. Depth, regular elements, and singular varieties

Let $M$ be a graded $E$-module. An element $\ell\in E_{1}$ is regular on $M$
if

$\\{m\in M\mid\ell m=0\\}=\ell M.$

Equivalently, $\ell\in E_{1}$ is regular if

$M(-1)\xrightarrow{\ell\cdot}M\xrightarrow{\ell\cdot}M(1)$

is an exact sequence of graded $E$-modules.

If $\ell\in E_{1}$ is not $M$-regular, we say that $\ell$ is $M$-singular. The
set of $M$-singular elements in $E_{1}$, denoted $V_{E}(M)$, is called the
singular variety of $M$. The following theorem of Aramova, Avramov, and Herzog
will be useful when attempting to compute singular varieties (termed rank
varieties in [1].)

######  Theorem 2.1 ([1, Theorem 3.1]).

If $\mathbb{K}$ is a field, then the singular varieties of finite graded
$E$-modules $M$ and $N$ satisfy the following properties:

  1. (1)

$V_{E}(M)$ is a union of linear subspaces in $E_{1}$.

  2. (2)

If $M\subseteq N$, then each of $V_{E}(M),V_{E}(N),V_{E}(N/M)$ is contained in
the union of the other two.

A sequence of elements $\ell_{1},\ldots,\ell_{d}$ is a regular sequence on $M$
if $\ell_{i}$ is regular on $M/(\ell_{1},\ldots,\ell_{i-1})M$ for
$i=1,\ldots,d$. The maximal length of a regular sequence on $M$ is the depth
of $M$ denoted $\mathrm{depth}_{E}(M).$ Note that $\mathrm{depth}_{E}(M)>0$ if
and only if $V_{E}(M)\neq E_{1}$. Moreover, $HS_{M}(t)$ is divisible by
$(1+t)^{0pt_{E}(M)}$.

The complexity of an $E$-module $M$ is

$\mathop{\mathrm{cx}}\nolimits_{E}(M)=\inf\\{c\in\mathbb{Z}\,|\,\beta_{i}^{E}(M)\leq\alpha
i^{c-1}\text{ for some }\alpha\in\mathbb{R}\\}.$

There is the following analogue of the Auslander-Buchsbaum theorem connecting
depth and complexity.

######  Theorem 2.2 ([1, Theorem 3.2]).

If $\mathbb{K}$ is an infinite field and $M$ is a finitely generated
$E$-module, then every maximal regular sequence has $0pt_{E}(M)$ elements and

$0pt_{E}(M)+\mathop{\mathrm{cx}}\nolimits_{E}(M)=n.$

###  2.3. Gröbner bases

Fix a monomial order $<$ on $E$ and a graded ideal $I\subseteq E$. Denote the
leading term of $f\in E$ by $LT(f)$ and the set of monomials of $f$ with
nonzero coefficients by $\mathop{\mathrm{Supp}}\nolimits(f)$, called the
support of $f$. The initial ideal of $I$ is $in_{<}(I)=(LT(f):f\in I)$. Then
$I$ has a unique reduced Gröbner basis $g_{1},\ldots,g_{t}\in I$; in other
words, $(LT(g_{1}),\ldots,LT(g_{t}))=in_{<}(I)$, each leading coefficient is
$1$, and no monomial in $\mathop{\mathrm{Supp}}\nolimits(g_{i})$ is divisible
by $LT(g_{j})$ for any $1\leq j\leq t$. There is a Buchberger algorithm for
computing Gröbner bases and we refer the reader to [16] for more details. In
particular, we use that $HS_{E/I}(t)=HS_{E/in_{<}(I)}(t)$ [2, Corollary 1.2].

Just as with initial ideals over a polynomial ring, the graded Betti numbers
of a graded ideal over an exterior algebra can only go up when one passes to
the initial ideal. One can then prove the analogous result for depth.

######  Proposition 2.3.

Let $E$ be an exterior algebra and let $I$ be a graded ideal in $E$. Then
$0pt_{E}(E/in_{<}(I))\leq 0pt_{E}(E/I)$.

###### Proof.

By [2, Proposition 1.8], $\beta_{ij}(E/I)\leq\beta_{ij}(E/in_{<}(I))$ for all
$i,j$. Hence

$\displaystyle\mathop{\mathrm{cx}}\nolimits_{E}(E/I)$
$\displaystyle=\inf\\{c\in\mathbb{Z}\mid\beta_{i}^{E}(E/I)\leq\alpha
i^{c-1}\text{ for some }\alpha\in\mathbb{R}\\}$
$\displaystyle\leq\inf\\{c\in\mathbb{Z}\mid\beta_{i}^{E}(E/in_{<}(I))\leq\alpha
i^{c-1}\text{ for some }\alpha\in\mathbb{R}\\}$
$\displaystyle=\mathop{\mathrm{cx}}\nolimits_{E}(E/in_{<}(I)).$

The claim then follows from Theorem 2.2. ∎

###  2.4. Koszul algebras and regularity

A positively graded $\mathbb{K}$-algebra $R$ is said to be Koszul if its
residue field $\mathbb{K}$ has a linear free resolution as an $R$-module.
Equivalently, $R$ is Koszul if $\beta^{R}_{ij}(\mathbb{K})=0$ for $i\neq j$.
For more details about Koszul algebras, see [14] and [29].

If $R=E/I$ is a quotient of an exterior algebra $E$ by some graded ideal $I$,
$R$ is said to be G-quadratic if $I$ has a Gröbner basis consisting of
quadrics (homogeneous elements of degree 2) with respect to some coordinates
on $E_{1}$ and some monomial ordering on $E$. A result of Fröberg [13] shows
that if $I$ is a monomial ideal, then $R$ is Koszul. It follows from a
standard deformation argument [14] that if $R$ is G-quadratic, then $R$ is
again Koszul.

Lastly, $R$ is LG-quadratic if there exists a G-quadratic algebra $A$ and a
regular sequence of linear forms $\ell_{1},\dots,\ell_{r}$ such that $R\cong
A/(\ell_{1},\dots,\ell_{r})$. LG-quadratic algebras were studied in the
commutative setting by Caviglia [7], Avramov, Conca, and Iyengar [3] and
others. We prove in the following section that LG-quadratic quotients of
exterior algebras are Koszul; it is immediate from the definition that all
G-quadratic algebras are LG-quadratic. That the G-quadratic, LG-quadratic, and
Koszul quotients of exterior algebras are distinct classes constitutes the
bulk of this paper. For related work on Koszul algebras and exterior algebras,
see [34], [35], [22], and [23].

##  3\. LG-quadratic algebras

In this section we show that the Koszul property is preserved when killing a
regular element. As a consequence, we show that LG-quadratic quotients of
exterior algebras are Koszul.

We first recall the following change of rings spectral sequence.

######  Theorem 3.1 ([30, Theorem 10.73]).

Let $R\rightarrow S$ be a graded map of graded $\mathbb{K}$-algebras, let $M$
be a graded $R$-module, and let $N$ be a graded $S$-module. Then there exists
a spectral sequence with second page

$(E_{2})_{p,q}=\mathop{\mathrm{Tor}}\nolimits^{S}_{p}(\mathop{\mathrm{Tor}}\nolimits^{R}_{q}(M,S),N)$

that converges to (as the total homology of a certain double complex)

$\mathop{\mathrm{Tor}}\nolimits^{R}_{p+q}(M,N).$

We show that, just as in the commutative case, LG-quadratic algebras are
Koszul.

######  Theorem 3.2.

Let $E$ be an exterior algebra over $\mathbb{K}$, let $I\subset E$ be an
ideal, and let $R=E/I$. If $\ell\in R_{1}$ is regular, then $R$ is Koszul if
and only if $\overline{R}=R/(\ell)$ is Koszul. In particular, if $R$ is LG-
quadratic, then $R$ is Koszul.

###### Proof.

Apply the change of rings spectral sequence to the canonical quotient map
$R\rightarrow\overline{R}$, where $M=N=\mathbb{K}$.

We can simplify the terms on the second page as follows.

First, we can compute
$\mathop{\mathrm{Tor}}\nolimits^{R}_{q}(\mathbb{K},\overline{R})$ by taking
the $R$-free resolution

$\dots\rightarrow
R(-2)\xrightarrow{l\cdot}R(-1)\xrightarrow{l\cdot}R\rightarrow\overline{R}\rightarrow
0.$

Tensoring with $\mathbb{K}$, the maps become zero, so the $q$th homology is
just $\mathbb{K}(-q)$.

Let $F_{\bullet}\rightarrow\mathbb{K}$ be a minimal graded free resolution of
$\mathbb{K}$ over $\overline{R}$, and let
$F_{i}=\bigoplus_{j\in\mathbb{Z}}\overline{R}(-j)^{\beta_{ij}}$. Twisting
preserves exactness, so $F_{\bullet}(-q)\rightarrow\mathbb{K}(-q)$ is also a
minimal graded free resolution over $\overline{R}$. Tensoring with
$\mathbb{K}$, the maps become zero, so the $p$th homology is

$\mathop{\mathrm{Tor}}\nolimits^{\overline{R}}_{p}(\mathbb{K}(-q),\mathbb{K})=\bigoplus_{j\in\mathbb{Z}}\mathbb{K}(-q-j)^{\beta_{p,j}}.$

Suppose $\overline{R}$ is Koszul. Then the bigraded Betti numbers
$\beta_{i,j}=0$ whenever $i\neq j$, so
$(E_{2})_{p,q}=\mathbb{K}(-p-q)^{\beta_{p,p}}$. The differential goes right
two places and down one, i.e. from $\mathbb{K}(-p-q)$ to $\mathbb{K}(-p-q-1)$.
Since the differentials are degree-preserving, they must therefore be zero.
Thus, the spectral sequence collapses at the second page. Since there is a
filtration of $\mathop{\mathrm{Tor}}\nolimits^{R}_{p+q}(M,N)$ given by the
terms $(E_{\infty})_{p,q}$, we have

$\dim_{\mathbb{K}}\mathop{\mathrm{Tor}}\nolimits^{R}_{p+q}(M,N)_{j}=\sum_{i=0}^{p+q}\dim_{\mathbb{K}}\mathbb{K}(-p-q)^{\beta{p,p}}_{j},$

which is zero whenever $j\neq p+q$. Hence $R$ is Koszul.

Now suppose $\overline{R}$ is not Koszul. Then there exist numbers $p,j\geq 0$
such that $\beta_{p,j}\neq 0$ and $p\neq j$. Since $P_{\bullet}$ is minimal
and graded, we must have $j>p$. Assume $p$ is minimal. Then the differential
on page two of the spectral sequence with target $(E_{2})_{p,0}$ comes from
$\mathbb{K}(-p+1)^{\beta_{p,p}}$. By grading, that differential must be zero.
Likewise, the differential on page two whose source is $(E_{2})_{p,0}$ has
$(E_{2})_{p+2,-1}=0$ as its target, so that map must also be zero. So this
term of the spectral sequence has stabilized, and the
$(E_{\infty})_{p-i,i}$-filtration of
$\mathop{\mathrm{Tor}}\nolimits^{R}_{p}(\mathbb{K},\mathbb{K})$ gives us

$\dim_{\mathbb{K}}\mathop{\mathrm{Tor}}\nolimits^{R}_{p}(\mathbb{K},\mathbb{K})_{j}\geq\dim_{\mathbb{K}}\mathop{\mathrm{Tor}}\nolimits^{\overline{R}}_{p}(\mathbb{K},\mathbb{K})_{j}\geq
1.$

This proves that $R$ is not Koszul either. ∎

A very similar theorem is proved by Conca, De Negri, and Rossi [9, Theorem
3.2] in the commutative setting.

##  4\. Depth of exterior edge ideals and LG-quadratic algebras

Our first task is to construct an LG-quadratic quotient of an exterior algebra
that is not G-quadratic. A proposed example is given by Thieu [34, Example
5.2.2(ii)], however we show in Example 6.1 that while Thieu’s example has no
quadratic Gröbner basis in a fixed system of coordinates, it is a monomial
ideal after a change of coordinates and hence G-quadratic afterall. We
construct a different example as a quotient of an algebra defined by the edge
ideal of a graph over an exterior algebra. We begin with notation on edge
ideals of graphs over the exterior algebra.

Let $G=(V,F)$ denote a finite, simple graph on vertex set $V=\\{1,\ldots,n\\}$
with edge set $F$. Let $E=\bigwedge_{\mathbb{K}}\langle
e_{1},\ldots,e_{n}\rangle$ denote the corresponding exterior algebra and let
$S=\mathbb{K}[x_{1},\ldots,x_{n}]$ the corresponding symmetric algebra. We
define the symmetric and exterior edge ideals associated to $G$ as follows:

$I_{S}(G)=\left(\left\\{x_{i}x_{j}\,|\,\\{i,j\\}\in F\right\\}\right)\subseteq
S,\quad\quad I_{E}(G)=\left(\left\\{e_{i}e_{j}\,|\,\\{i,j\\}\in
F\right\\}\right)\subseteq E.$

Edge ideals over polynomial rings have been extensively studied (see e.g.
[26], [37], or [10]) while those over exterior algebras have received less
attention; see e.g. [17]. More generally, monomial ideals over $E$ were
studied by Aramova, Herzog, and Hibi [2] and by Aramova, Avramov, and Herzog
[1]. For terminology on graphs, we refer the reader to [5].

First we collect two basic lemmas comparing the singular varieties of modules
in extensions of the underlying exterior algebras.

######  Lemma 4.1.

Let $E=\bigwedge_{\mathbb{K}}\langle e_{1},\ldots,e_{n}\rangle$ denote the
exterior algebra on $n$ variables and let $M$ be an $E$-module. If
$E^{\prime}=E\otimes_{\mathbb{K}}\bigwedge_{\mathbb{K}}\langle x\rangle$ and
$M^{\prime}=M\otimes_{E}E^{\prime}$, then
$V_{E^{\prime}}(M^{\prime})=V_{E}(M)$.

###### Proof.

Note that as $E$-modules, $E^{\prime}=E\oplus Ex$ and $M^{\prime}=M\oplus Mx$.
Let $m^{\prime}=(a,bx)\in M^{\prime}$, where $a,b\in M$ are homogeneous with
$\deg(b)=\deg(a)-1$. Let $\ell^{\prime}=(\ell,\alpha x)\in E_{1}^{\prime}$,
where $\ell\in E_{1}$ and $\alpha\in\mathbb{K}$. We must show that
$\ell^{\prime}$ is regular on $M^{\prime}$ if and only if $\ell$ is regular on
$M$ or $\alpha\neq 0$.

First suppose $\ell$ is regular on $M$ and $\ell^{\prime}m^{\prime}=0$. Then

$\ell^{\prime}m^{\prime}=(\ell a,\ell bx+(-1)^{\deg(a)}\alpha ax)=0,$

and both coordinates vanish. Since $\ell$ is $M$-regular and $\ell a=0$, we
have $a=\ell a^{\prime}$ for some $a^{\prime}\in M$ with
$\deg(a^{\prime})=\deg(a)-1$. Since

$0=\ell bx+(-1)^{\deg(a)}\alpha ax=\ell bx+(-1)^{\deg(a)}\alpha\ell
a^{\prime}x=\ell(b+(-1)^{\deg(a)}\alpha a^{\prime})x,$

we have $b+(-1)^{\deg(a)}\alpha a^{\prime}=\ell b^{\prime}$ for some
$b^{\prime}\in M$. Therefore

$\displaystyle\ell^{\prime}(a^{\prime},b^{\prime}x)$ $\displaystyle=(\ell
a^{\prime},\ell b^{\prime}x+\alpha xa^{\prime})$
$\displaystyle=(a,(b+(-1)^{\deg(a)}\alpha
a^{\prime})x+(-1)^{\deg(a^{\prime})}\alpha a^{\prime}x)$
$\displaystyle=(a,bx)=m^{\prime}.$

Therefore $\ell^{\prime}$ is $M^{\prime}$-regular.

Suppose $\ell^{\prime}m^{\prime}=0$ and $\alpha\neq 0$. Then
$a=(-1)^{\deg(a)}\alpha^{-1}\ell b$ and it follows that

$\ell^{\prime}((-1)^{\deg(a)}\alpha^{-1}b,0)=(a,bx).$

Hence $\ell^{\prime}$ is $M^{\prime}$-regular.

Conversely, suppose $\ell^{\prime}=(\ell,\alpha x)$ is $M^{\prime}$-regular.
If $\alpha\neq 0$, we are done. Otherwise, suppose $\alpha=0$ and $\ell a=0$
with $a\in M$. Then $\ell^{\prime}(a,0)=0$ and so $(a,0)=\ell^{\prime}(c,dx)$,
for some $c,d\in M$. In particular, $\ell c=a$, showing that $\ell$ is
$M$-regular. The claim then follows. ∎

######  Lemma 4.2.

Let $E=\bigwedge_{\mathbb{K}}\langle e_{1},\ldots,e_{n}\rangle$ denote the
exterior algebra on $n$ variables and let $M$ be an $E$-module. If
$E^{\prime}=E\otimes_{\mathbb{K}}\bigwedge_{\mathbb{K}}\langle x\rangle$ and
$M^{\prime}=M\otimes_{E}E^{\prime}/(x)$, then
$V_{E^{\prime}}(M^{\prime})=\bigcup_{i}\mathrm{span}_{\mathbb{K}}\langle
A_{i}\cup\\{x\\}\rangle$, where
$V_{E}(M)=\bigcup_{i}\mathrm{span}_{\mathbb{K}}\langle A_{i}\rangle$.

###### Proof.

We use that as $E$-modules, $E^{\prime}=E\oplus Ex$ and $M^{\prime}=M$. Let
$m\in M$ and $\ell^{\prime}=(\ell,\alpha x)\in E^{\prime}_{1}$, where $\ell\in
E_{1}$ and $\alpha\in\mathbb{K}$. It is clear that $\ell^{\prime}m=\ell m$,
and thus $\ell^{\prime}$ is $M$-regular if and only if $\ell$ is $M$-regular.
The claim follows. ∎

Next we compute the depth and singular variety of the edge ideal of a path.
The corresponding result over the polynomial ring is well-known.

######  Proposition 4.3.

Let $P_{n}$ denote the path graph on $n$ vertices, and let
$E=\bigwedge_{\mathbb{K}}\langle e_{1},\ldots,e_{n}\rangle$ be the
corresponding exterior algebra.

  1. (1)

If $n\equiv 1\,(\\!\\!\\!\mod 3)$, then $0pt_{E}(E/I_{E}(P_{n}))=1$ and

$V_{E}(E/I_{E}(P_{n}))=\bigcup_{i=0}^{\lfloor
n/3\rfloor}\mathrm{span}_{\mathbb{K}}\langle
e_{1},\ldots,\widehat{e_{3i+1}},\ldots,e_{n}\rangle,$

where $\widehat{e_{i}}$ denotes that $e_{i}$ is omitted. In particular,
$\sum_{i=0}^{\lfloor n/3\rfloor}e_{3i+1}$ is a regular element on
$E/I_{E}(P_{n})$.

  2. (2)

If $n\not\equiv 1\,(\\!\\!\\!\mod 3)$, then $0pt_{E}(E/I_{E}(P_{n}))=0$ and
$V_{E}(E/I_{E}(P_{n}))=E_{1}$.

###### Proof.

We proceed by induction on $n$. If $n=1$, then $I=(0)$, $E/I\cong
E=\bigwedge_{\mathbb{K}}\langle e_{1}\rangle$, and $V_{E}(E/I)=E_{1}$. If
$n=2$, then $I=(e_{1}e_{2})$ and $E/I=\bigwedge_{\mathbb{K}}\langle
e_{1},e_{2}\rangle/(e_{1}e_{2})$, so that $xy=0$ for all $x,y\in E_{1}$. Hence
there are no regular elements on $E/I$ and $V_{E}(E/I)=E_{1}$. If $n=3$, then
$I=(e_{1}e_{2},e_{2}e_{3})\subseteq\bigwedge_{\mathbb{K}}\langle
e_{1},e_{2},e_{3}\rangle$, and $I:e_{2}=(e_{1},e_{2},e_{3})$. Again there are
no regular elements and $V_{E}(E/I)=E_{1}$.

Now suppose $n\geq 4$ and that the proposition holds for smaller $n$. Let
$I=I(P_{n})=(e_{1}e_{2},e_{2}e_{3},\ldots,e_{n-1}e_{n})\subseteq E$. Let
$J=I_{E}(P_{n-1})E$. One computes that
$J:(e_{n-1}e_{n})=I_{E}(P_{n-2})E+(e_{n-2},e_{n-1},e_{n})$. We have the
following short exact sequence of $E$-modules:

$0\to\frac{E}{J:(e_{n-1}e_{n})}\to\frac{E}{J}\to\frac{E}{I}\to 0.$

Note that $E/J=\frac{\bigwedge_{\mathbb{K}}\langle
e_{1},\ldots,e_{n-1}\rangle}{I_{E}(P_{n-1})}\otimes_{\mathbb{K}}\bigwedge_{\mathbb{K}}\langle
e_{n}\rangle$ and

$E/(J:(e_{n-1}e_{n}))=\frac{\bigwedge_{\mathbb{K}}\langle
e_{1},\ldots,e_{n-3}\rangle}{I_{E}(P_{n-3})}\otimes_{\mathbb{K}}\frac{\bigwedge_{\mathbb{K}}\langle
e_{n-2},e_{n-1},e_{n}\rangle}{(e_{n-2},e_{n-1},e_{n})}.$

If $n\equiv 1\,(\\!\\!\\!\mod 3)$, then by induction and Lemma 4.1, we have
$V_{E}(E/J)=\mathrm{span}_{\mathbb{K}}\langle e_{1},\ldots,e_{n-1}\rangle$.
Similarly, by Lemma 4.2,

$V_{E}(E/(J:(e_{n-1}e_{n})))=\bigcup_{i=0}^{\lfloor
n/3\rfloor-1}\mathrm{span}_{\mathbb{K}}\langle
e_{1},\ldots,\widehat{e_{3i+1}},\ldots,e_{n-3},e_{n-2},e_{n-1},e_{n}\rangle.$

Since none of the components of $V_{E}(E/J)$ and $V_{E}(E/(J:(e_{n-1e_{n}})))$
contain each other, it follows from Theorem 2.1(2) that
$V_{E}(E/I)=V_{E}(E/J)\cup V_{E}(E/(J:(e_{n-1e_{n}})))$ and the conclusion
follows.

If $n\equiv 2\,(\\!\\!\\!\mod 3)$, we use the same short exact sequence as
above, but in this case, Lemmas 4.1 and 4.2 combined with the induction
hypothesis imply that $V_{E}(E/(J:(e_{n-1e_{n}})))=E_{1}$, while
$V_{E}(E/J)\neq E_{1}$. Again, by Theorem 2.1(4), we have
$V_{E}(E/I)=V_{E}(E/J)\cup V_{E}(E/(J:(e_{n-1e_{n}})))=E_{1}$.

Finally, if $n\equiv 0\,(\\!\\!\\!\mod 3)$, we consider the following short
exact sequence of $E$-modules:

$0\to\frac{E}{I:e_{n}}\to\frac{E}{I}\to\frac{E}{I+(e_{n})}\to 0.$

As $E/(I+(e_{n})=\bigwedge_{\mathbb{K}}\langle
e_{1},\ldots,e_{n-1}\rangle/I_{E}(P_{n-1})\otimes_{\mathbb{K}}\bigwedge_{\mathbb{K}}\langle
e_{n}\rangle/(e_{n}),$ we have $V_{E}(E/(I+(e_{n}))=E_{1}$ by induction and
Lemma 4.2. One computes that

$I:e_{n}=(e_{1}e_{2},\ldots,e_{n-3}e_{n-2},e_{n-1},e_{n}),$

and so

$E/(I:e_{n})=\bigwedge_{\mathbb{K}}\langle
e_{1},\ldots,e_{n-2}\rangle/I_{E}(P_{n-2})\otimes_{\mathbb{K}}\bigwedge_{\mathbb{K}}\langle
e_{n-1},e_{n}\rangle/(e_{n-1},e_{n}),$

and thus $V_{E}(E/(I:e_{n}))\neq E_{1}$ by Lemma 4.2 and the inductive
hypothesis. Finally by Theorem 2.1(2), we get $V_{E}(E/I)=E_{1}$. ∎

If we consider the commutative edge ideal $I_{S}(P_{n})$ in a polynoial ring
$S=\mathbb{K}[x_{1},\ldots,x_{n}]$, then
$0pt_{S}(S/I_{S}(P_{n})))=\left\lceil\frac{n}{3}\right\rceil$ (see e.g. [11,
Corollary 5.10]), and we see that $0pt_{S}(S/I_{S}(G))-0pt_{E}(E/I_{E}(G))$
can be arbitrarily large.

The previous result will be used to construct an LG-quadratic ideal that is
not G-quadratic. First, we need the following result on Hilbert functions,
which could be checked by brute force, computing the Hilbert series of
$E/I_{E}(G)$ for each graph with $6$ vertices and $6$ edges. We give a more
conceptual argument.

######  Lemma 4.4.

There is no graph $G$ such that
$\mathop{\mathrm{HS}}\nolimits_{E/I_{E}(G)}(t)=1+6t+9t^{2}+t^{3}$.

###### Proof.

The $n$th coefficient on the Hilbert series of a graph is the number of
independent sets of $G$ of size $n$. So if $G$ is a graph with such a Hilbert
series, then it must have $6$ vertices, $\binom{6}{2}-9=6$ edges, and
precisely one set of three vertices such that no two of them are adjacent.

Suppose we have a graph $G$ on $6$ vertices where $\\{a,b,c\\}$ is the lone
independent set.

abcdef

That leaves $9$ potential edges with one vertex in $\\{a,b,c\\}$ and the other
in $\\{d,e,f\\}$ (colored red), and three with both vertices in $\\{d,e,f\\}$
(colored black). Since we want $G$ to have $6$ edges, we have to eliminate $6$
of the potential edges.

However, we can eliminate no more than three of the red edges; otherwise, two
of the eliminated edges would share a vertex in $\\{d,e,f\\}$, creating
another independent set (the shared vertex and the two from $\\{a,b,c\\}$
corresponding to the other endpoints of those two edges).

But then we would be forced to eliminate all three black edges, making
$\\{d,e,f\\}$ a second independent set. In any case, it is impossible for $G$
to have precisely one independent set of size $3$. ∎

We can now construct our example as a quotient of an edge ideal algebra.

######  Theorem 4.5.

There exist LG-quadratic quotients of exterior algebras that are not
G-quadratic.

###### Proof.

Consider the graph $P_{7}$, a path of length $7$, with corresponding edge
ideal

$I_{E}(P_{7})=(e_{1}e_{2},e_{2}e_{3},e_{3}e_{4},e_{4}e_{5},e_{5}e_{6},e_{6}e_{7}).$

A quick computation shows that

$\mathop{\mathrm{HS}}\nolimits_{E/I_{E}(P_{7})}(t)=1+7t+15t^{2}+10t^{3}+t^{4}=(1+6t+9t^{2}+t^{3})(1+t).$

(The first three terms are clear. That
$\mathop{\mathrm{HF}}\nolimits(E/I_{E}(P_{7}),3)=10$ follows from the fact
that there are 10 independent sets of size $3$ in $P_{7}$ and
$\mathop{\mathrm{HF}}\nolimits(E/I_{E}(P_{7}),4)=1$ since there is one
independent set $\\{x_{1},x_{3},x_{5},x_{7}\\}$ of size $4$.) By Proposition
4.3, $0pt_{E}(E/I_{E}(P_{7}))=1$ and $e_{1}+e_{4}+e_{7}$ is a regular element
on $E/I_{E}(P_{7})$. Thus by definition,

$R=\frac{E}{I_{E}(P_{7})+(e_{1}+e_{4}+e_{7})}\cong\frac{\bigwedge_{\mathbb{K}}\langle
e_{1},e_{2},e_{3},e_{4},e_{5},e_{6}\rangle}{(e_{1}e_{2},e_{2}e_{3},e_{3}e_{4},e_{4}e_{5},e_{5}e_{6},e_{6}(e_{1}+e_{4}))}$

is LG-quadratic and hence Koszul by Theorem 3.2. Moreover,
$\mathop{\mathrm{HS}}\nolimits_{R}(t)=1+6t+9t^{2}+t^{3}.$ If $R$ were
$G$-quadratic, there would be a quadratic monomial ideal $J$ corresponding to
a graph on $6$ vertices with the same Hilbert series, but this is impossible
by Lemma 4.4. Therefore $R$ is not $G$-quadratic. ∎

######  Remark 4.6.

This construction gives a simpler proof that the algebra $B(0)$ of Shelton and
Yuzvinsky from [33, Example 5.2] is Koszul. There the authors define

$B(0)=\bigwedge\\!\phantom{.}_{\mathbb{K}}\langle
x_{1},x_{2},x_{3},x_{4},x_{5},x_{6}\rangle/(x_{1}x_{6},x_{2}x_{4},x_{3}x_{5},x_{1}x_{4},x_{2}x_{5},(x_{4}-x_{3})x_{6})$

as a certain deformation of an Orlik-Solomon algebra of a hyperplane
arrangement. They use a complicated Koszul duality argument to show that it is
Koszul. Here we recover this by recognizing $B(0)$ as $R$ from Theorem 4.5
under the identification:

$e_{1}\leftrightarrow-x_{3},\quad e_{2}\leftrightarrow x_{5},\quad
e_{3}\leftrightarrow x_{2},\quad e_{4}\leftrightarrow x_{4},\quad
e_{5}\leftrightarrow x_{1},\quad e_{6}\leftrightarrow x_{6}.$

##  5\. A Koszul quotient of an exterior algebra which is not LG-quadratic

In the case of commutative algebras, Conca [8] has shown that not every Koszul
algebra is LG-quadratic. As of this writing, this is the lone example of a
Koszul algebra that is not LG-quadratic.

######  Example 5.1 ([8, Example 1.20]).

Consider the polynomial ring $\mathbb{K}[a,b,c,d]$ and the ideal $I=(ac,ad,ab-
bd,a^{2}+bc,b^{2})$. That the quotient ring $R=S/I$ is Koszul follows from a
filtration argument. That $R$ is not LG-quadratic follows by showing that no
quadratic monomial ideal has the same h-polynomial. This example negatively
answered a question of Caviglia [7, Question 1.2.6]. See also [3, Question
6.4] and [9, Question 2.12].

The authors were motivated to answer the following parallel question:

######  Question 5.2.

Is there a Koszul quotient of an exterior algebra that is not LG-quadratic?

The purpose of this section is to answer this question in the affirmative. One
distinction between our construction and that of Conca is that our example is
non-obstructed – that is, there exist quadratic monomial ideals with the same
Hilbert function as our example. Thus it takes significant work to show that
the example we construct is not LG-quadratic.

To prove Koszulness, we take advantage of the following result of Fröberg and
Löfwall:

######  Theorem 5.3 ([15, Theorem 10.2]).

Let $\mathbb{K}$ be a field of characteristic $0$, let
$E=\bigwedge_{\mathbb{K}}\langle e_{1},\ldots,e_{n}\rangle$, and let
$I=(f_{1},\ldots,f_{t})$ be an ideal generated by $t$ generic quadrics. If
$t\geq\binom{n}{2}-\frac{n^{2}}{4}$, then $E/I$ is Koszul.

The primary goal in this section is then to prove the following result.

######  Theorem 5.4.

Let $E=\bigwedge_{\mathbb{C}}\langle e_{1},\ldots,e_{6}\rangle$. Consider an
ideal $I=(f_{1},\ldots,f_{6})\subseteq E$ generated by $6$ generic quadrics.
Then $E/I$ is Koszul but not LG-quadratic.

The proof of this theorem will require several steps, which we now summarize.
That such an ideal is Koszul follows from Theorem 5.3. We show that all
quadrics in such an ideal have rank at least 4 by computing the height of the
ideal of $4\times 4$ minors of a generic alternating matrix. The Hilbert
series for the ideal of $6$ generic quadrics in 6 exterior algebra variables
is $1+6t+9t^{2}=(1+3t)^{2}$. If this ideal defined an LG-quadratic algebra,
then since Hilbert functions are preserved when passing to the initial ideal,
there would be a graph $G$ on $n\geq 6$ vertices with $6$ edges whose edge
ideal would have Hilbert series $(1+3t)^{2}(1+t)^{n-6}$. If there were no such
graphs, we would be done. However, there are three such graphs, ignoring
isolated vertices. Finally we must check that for each graph $G$, if $I$ is an
ideal with initial ideal $in_{<}(I)=I_{E}(G)$, then $I$ must contain a rank
$2$ quadric. This step is long and technical since we cannot make assumptions
about the monomial order itself. Since the rank of a quadric can only go down
after killing a linear form, we conclude that the ideal of $6$ generic
quadrics in $6$ variables cannot be LG-quadratic.

The idea to consider the ranks of the quadrics defining a $G$-quadratic
algebra is inspired by a result of Eisenbud, Reeves, and Totaro [12, Theorem
19]; however, their result gives a bound on rank in terms of the codimension
on a polynomial ideal and the number of variables. We need a stronger
statement in the exterior algebra setting. First we prove a lower bound on the
rank of quadrics in ideals generated by generic quadrics in an exterior
algebra.

######  Proposition 5.5.

Fix positive integers $n,r,t$. Let $I=(f_{1},\ldots,f_{t})\subseteq
E=\bigwedge_{\mathbb{C}}\langle e_{1},\ldots,e_{n}\rangle$ be an ideal
generated by $t$ generic quadrics. Then
$\mathop{\mathrm{rank}}\nolimits(q)\geq 2r$ for all nonzero quadrics $q\in I$
if and only if $t\leq\frac{(n-2r+1)(n-2r+2)}{2}$.

###### Proof.

First we identify a quadric $q=\sum_{i<j}\alpha_{i,j}e_{i}e_{j}$ with the
point $p=[\alpha_{1,2}:\cdots:\alpha_{n-1,n}]$ in the projective space
$\mathbb{P}=\mathbb{P}_{\mathbb{C}}^{\binom{n}{2}-1}$, so that
$q=\underline{e}A\underline{e}^{\mathsf{T}}$, where $A$ is the alternating
$n\times n$ matrix with $(i,j)$ entry $\frac{1}{2}\alpha_{i,j}$ if $i>j$ and
$-\frac{1}{2}\alpha_{j,i}$ if $j>i$, and $0$ along the diagonal. Then
$\mathop{\mathrm{rank}}\nolimits(q)<2r$ if and only if $p\in V(J)$, where $J$
is the ideal of $(2r)\times(2r)$ minors of a generic alternating $n\times n$
matrix $X$. It follows from [6, Corollary 2.6] that
$\sqrt{J}=\mathrm{Pf}_{2r}(X)$, where $\mathrm{Pf}_{2r}(X)$ denotes the ideal
of $(2r)\times(2r)$ Pfaffians of $X$. By [21, Corollary 2.5],
$\mathop{\mathrm{codim}}\nolimits(J)=\mathop{\mathrm{codim}}\nolimits(\mathrm{Pf}_{2t}(X))=\frac{(n-2r+1)(n-2r+2)}{2}$.

A choice of $t$ generic quadrics corresponds to a choice of $t$ generic points
from $\mathbb{P}$, whose span is a linear space $L$ of dimension $t-1$. Thus
if $t\leq\frac{(n-2r+1)(n-2r+2)}{2}$, then $L\cap V(J)=\varnothing$. The
result follows. ∎

######  Remark 5.6.

While the radical of the ideal of $(2r)\times(2r)$ minors of a generic
alternating matrix is an ideal of Pfaffians, it is not primary to the ideal of
Pfaffians. Rather there are often embedded components. It would be interesting
to compute the full primary decomposition of these ideals.

######  Corollary 5.7.

If $I=(f_{1},\ldots,f_{6})\subseteq E=\bigwedge_{\mathbb{C}}\langle
e_{1},\ldots,e_{6}\rangle$ an ideal generated by generic quadrics, then every
nonzero quadric in $I$ has rank at least $4$.

###### Proof.

We apply the previous proposition with $n=6$, $r=2$ and $t=6$. ∎

######  Corollary 5.8.

If $q_{1},q_{2}\in\bigwedge_{\mathbb{C}}\langle
e_{1},e_{2},e_{3},e_{4}\rangle$ are linearly independent quadrics, then there
is a quadric $q\in(q_{1},q_{2})$ with $\mathop{\mathrm{rank}}\nolimits(q)=2$.

###### Proof.

We argue as in the proof of Propostion 5.5. The subvariety of
$\mathbb{P}_{\mathbb{C}}^{5}$ corresponding to the space of rank at most $2$
quadrics corresponds to the hypersurface
$X=V(x_{1,2}x_{3,4}-x_{1,3}x_{2,4}+x_{1,4}x_{2,3})$. Two linearly independent
quadrics correspond to a line $L$ which must intersect $X$ in at least one
point. ∎

Next we consider edge ideals of graphs that have the right Hilbert series to
match an LG-quadratic lift of the ideal of six generic quadrics in six
variables. We will eventually show that none of these can possibly be the
initial ideal of an ideal of quadrics of large rank.

######  Lemma 5.9.

Let $G$ be a graph on $n$ vertices and let $E=\bigwedge_{\mathbb{C}}\langle
e_{1},\ldots,e_{n}\rangle$ denote the corresponding exterior algebra. If
$\mathop{\mathrm{HS}}\nolimits_{E/I_{E}(G)}(t)=(1+3t)^{2}(1+t)^{n-6}$, then
$G$ must be one of the following graphs. The number of isolated vertices in
graphs of class (i) is given by $n-5-i$, represented below by discrete nodes
to the right.

(1)

(2)

(3)

In particular, $G$ has no vertex of degree greater than $2$.

For the proof, we will need the following lemma. To simplify notation, we
write $\mathop{\mathrm{HS}}\nolimits(G)$ for
$\mathop{\mathrm{HS}}\nolimits_{E}(E/I_{E}(G))$, where $G$ is a graph on $n$
vertices and $E=\bigwedge_{\mathbb{C}}\langle e_{1},\ldots,e_{n}\rangle.$

######  Lemma 5.10.

If $G=G_{1}\sqcup G_{2}$, then
$\mathop{\mathrm{HS}}\nolimits(G)=\mathop{\mathrm{HS}}\nolimits(G_{1})\cdot\mathop{\mathrm{HS}}\nolimits(G_{2})$.

###### Proof.

Let $E_{1},E_{2}$ denote the corresponding exterior algebras for $G_{1}$ and
$G_{2}$, respectively so that $E=E_{1}\otimes_{\mathbb{C}}E_{2}$. The
statement follows from the fact that
$E_{1}/I_{E_{1}}(G_{1})\otimes_{\mathbb{C}}E_{2}/I_{E_{2}}(G_{2})=E/I_{E}(G)$.
∎

###### Proof of Lemma 5.9..

Recall that the coefficient on $t^{n}$ in $\mathop{\mathrm{HS}}\nolimits(G)$
is the number independent sets of $G$ of size $n$.

Let $n=6$, so $\mathop{\mathrm{HS}}\nolimits(G)=(1+3t)^{2}$. If $G$ has more
than one connected component, then the Hilbert series of each component must
be $1+3t$, which corresponds to a graph on three vertices with three edges,
i.e. a triangle. Suppose $G$ is connected. It cannot be a tree, since it has
six vertices and six edges, so there must be a cycle. It is easy to eliminate
the six graphs with six edges on six vertices with a cycle of length four or
greater by calculating their Hilbert series, because they all have independent
sets of size $3$. So, there exists a $3$-cycle in $G$. Let’s call it
$a_{1}a_{2}a_{3}$, and label the remaining vertices $b,c,d$. Without loss of
generality, $bc$ is an edge since there must be an edge among $b,c,d$. There
must also be an edge among $a_{i},b,d$ for all $i$ since $\\{a_{i},b,d\\}$ is
not independent, but the only possibility with two remaining edges is $bd$.
The same argument for $a_{i},c,d$ shows that $cd$ must be an edge. This
completes the proof for this case.

Let $n=7$, so $\mathop{\mathrm{HS}}\nolimits(G)=(1+3t)^{2}(1+t)$. If $G$ is
connected, then it is a tree on seven vertices. We can classify these trees by
girth, i.e., the maximal length of a path in $G$. There is only one such tree
of girth $m=2$, obtained by connecting one vertex to all other vertices, but
its Hilbert series has the wrong coefficient on $t^{4}$. For girth at least
three, consider the following pictures.

If $m=3$, then there is an induced path of length three as above and the
remaining edges must be chosen among the dashed lines. There cannot be an edge
among the three vertices in the bottom row since that would produce a path of
length at least four. In any case, the four red square vertices form an
independent set, which means this graph has the wrong Hilbert series.

If $m=4$ and the two remaining vertices are connected, then the above graph is
the only possibility, and its Hilbert series has the wrong coefficient on
$t^{4}$ as shown by the red square vertices.

If $m=4$ and the two remaining vertices are not connected, then the rest of
the edges must be chosen from the dashed lines above, and in any case the four
red square vertices will have no edges between them.

If $m=5$, the remaining edge must be one of the dashed lines, but then the
three red square vertices and one of the green pentagon vertices will form an
independent set of size $4$.

If $m=6$, then $G$ is a simple path on seven vertices $a-b-c-d-e-f-g$, and
$a,c,e,g$ is an independent set.

Thus, $G$ is not connected, i.e. $G=G_{1}\sqcup G_{2}$. It is worth noting
that if any graph has Hilbert series $(1+t)^{k}$, then that graph is $k$
isolated vertices (with no edges). This allows us to perform induction if we
find a connected component with such a Hilbert series. The only way to factor
$(1+3t)^{2}(1+t)$ into two polynomials such that neither is $1+t$ is
$\mathop{\mathrm{HS}}\nolimits(G_{1})=1+3t$ and
$\mathop{\mathrm{HS}}\nolimits(G_{2})=(1+3t)(1+t)$. Then $G_{1}$ is a triangle
as shown above, and $G_{2}$ is a graph on four vertices with three edges. If
we assume $G$ has no isolated vertices (which would reduce to the case $n=6$),
then $G_{2}$ is connected, hence a tree. There are only two trees on four
vertices, and the only one with the right Hilbert series is the simple path
$a-b-c-d$.

Let $n=8$. The Hilbert series indicates that $G$ has eight vertices and six
edges, so it must have at least two connected components $G=G_{1}\sqcup
G_{2}$. The only way to factor
$\mathop{\mathrm{HS}}\nolimits(G)=(1+3t)^{2}(1+t)^{2}$ into two polynomials so
that neither is a power of $1+t$ (hence isolated vertices and reduction to a
case on fewer vertices) is $\mathop{\mathrm{HS}}\nolimits(G_{i})=(1+3t)(1+t)$
for $i=1,2$. But then, as shown above, $G_{i}$ either has isolated vertices or
is the simple path on four vertices.

Let $n\geq 9$. Then $G$ has at least nine vertices and exactly six edges,
which implies at least three connected components. There is no way to factor
$(1+3t)^{2}(1+t)^{n-6}$ into three polynomials such that none is a power of
$1+t$, so this reduces to one of the earlier cases. ∎

As above, to a quadratic form $q\in E=\bigwedge_{\mathbb{C}}\langle
e_{1},\ldots,e_{n}\rangle$, we associate an $n\times n$ alternating matrix $A$
so that

$q=\underline{e}A\underline{e}^{\mathsf{T}},$

where $\underline{e}=(e_{1},\ldots,e_{n})$. The rank of $q$ is then
$\mathop{\mathrm{rank}}\nolimits(q)=\mathop{\mathrm{rank}}\nolimits(A)$. It is
well-known that the rank of an alternating matrix must be even [19, Corollary
1, p. 351], say $\mathop{\mathrm{rank}}\nolimits(q)=2r$, and that after a
change of variables, we can write

$q=\sum_{i=1}^{r}e_{2r-1}e_{2r}.$

As we are interested in Gröbner bases, which are sensitive to changes of
coordinates, we need the following lemma. Note that when $\ell\in E_{1}$, we
write $e_{i}>\ell$ if $e_{i}>e_{j}$ for all variables
$e_{j}\in\mathop{\mathrm{Supp}}\nolimits(\ell)$.

######  Lemma 5.11.

Let $q\in E=\bigwedge_{\mathbb{C}}\langle e_{1},\ldots,e_{n}\rangle$ be a
quadric of rank $2r$ and fix a monomial order $<$ on $E$. Then

$q=\alpha_{1}(e_{i_{1}}+\ell_{1,1})(e_{j_{1}}+\ell_{1,2})+\cdots+\alpha_{r}(e_{i_{r}}+\ell_{r,1})(e_{j_{r}}+\ell_{r,2}),$

where $\alpha_{1},\ldots,\alpha_{r}\in\mathbb{C}^{\ast}$,
$e_{i_{1}},\ldots,e_{i_{r}},e_{j_{1}},\ldots,e_{j_{r}}$ are distinct
variables, $\ell_{s,t}\in E_{1}$ with support disjoint from
$\\{e_{i_{1}},\ldots,e_{i_{s}},e_{j_{1}},\ldots,e_{j_{t}}\\}$ and with
$e_{i_{s}}>\ell_{s,1}$ and $e_{j_{s}}>\ell_{s,2}$ for all $1\leq s\leq r$, and

$e_{i_{1}}e_{j_{1}}>e_{i_{2}}e_{j_{2}}>\cdots>e_{i_{r}}e_{j_{r}}.$

###### Proof.

First note that any such $q$ as above has rank $2r$. We proceed by induction
on $r$. Let $e_{i_{1}}e_{j_{1}}$ be the initial monomial of a rank $2r$
quadric $q$ so that $q=\alpha_{1}e_{i_{1}}e_{j_{1}}+q^{\prime}$, with
$\alpha_{1}\in\mathbb{C}^{\ast}$ and all terms in $q^{\prime}$ smaller than
$e_{i_{1}}e_{j_{1}}$. By grouping terms divisible by either $e_{i_{1}}$ or
$e_{j_{1}}$, we can rewrite this as

$q=\alpha_{1}(e_{i_{1}}+\ell_{1,1})(e_{j_{1}}+\ell_{2,1})+q^{\prime\prime},$

where all variables in the supports of $\ell_{1,1}$ and $\ell_{1,2}$ are
smaller than $e_{i_{1}}$ and $e_{j_{1}}$, respectively. In particular, the
variables $e_{i_{1}}$ and $e_{j_{1}}$ are not in the support of
$q^{\prime\prime}$ and so
$\mathop{\mathrm{rank}}\nolimits(q^{\prime\prime})=\mathop{\mathrm{rank}}\nolimits(q)-2$.
Applying the inductive hypothesis to $q^{\prime\prime}$, the result follows. ∎

The most technical part of the proof is showing that any ideal whose initial
ideal matched the edge ideals from Lemma 5.9 must contain a rank $2$ quadric,
which is precisely the following lemma.

There is one subtle point that deserves mentioning; when writing the rank $4$
quadric $q=e_{1}e_{2}+e_{1}e_{3}+e_{2}e_{4}$ in the form of the previous
lemma, we have $q=(e_{1}-e_{4})(e_{2}+e_{3})-e_{3}e_{4}$, but $e_{3}e_{4}$ is
not in the support of $q$. Care must be taken with the arguments when this
occurs. However, we may sometimes avoid this problem. Given a quadric $q$ as
in the previous Lemma, since $e_{i_{1}}>\ell_{1,1}$, we can make the change of
variables $e_{i_{1}}\mapsto e_{i_{1}}-\ell_{1,1}$ without changing the initial
ideal. Continuing in this way, we may always assume that one quadric in a
Gröbner basis has the following simpler form

$q=\alpha_{1}e_{i_{1}}e_{j_{1}}+\cdots+\alpha_{r}e_{i_{r}}e_{j_{r}}.$

After another change of variables we may assume that $\alpha_{i}=1$ for all
$i$.

######  Lemma 5.12.

Let $G$ be one of the three graphs from Lemma 5.9. If $I\subseteq$ is a graded
ideal and $<$ is a monomial order such that $in_{<}(I)=I_{E}(G)$, then there
is a rank $2$ quadric in $I$.

###### Proof.

We may assume that $I$ has a reduced Gröbner basis of quadrics of rank at
least $4$, or else we are done.

Claim 1: $in_{<}(I)=I_{E}(G)$, where $G$ is the graph from Case (1) from Lemma
5.9.

Let $q$ denote one of the elements of the reduced Gröbner basis with leading
term $wx$. By Lemma 5.11 (and remarks afterward), we can write
$q=wx+yz+\cdots$, where $wx>yz$ are distinct variables,
$\ell_{1},\ldots,\ell_{4}\in E_{1}$ and
$w>\ell_{1},x>\ell_{2},\ldots,z>\ell_{4}$. Multiplying $q$ by $w$ or $x$, we
see that $wyz,xyz\in in_{<}(I)$. As $yz\in\mathop{\mathrm{Supp}}\nolimits(q)$,
we have $yz\notin in_{<}(I)$. Therefore either $wy\in in_{<}(I)$ or $wz\in
in_{<}(I)$. Similarly $xy\in in_{<}(I)$ or $xz\in in_{<}(I)$. We conclude that
each vertex of $G$ with degree at least one actually has degree at least $2$.
The only possibility for $G$ is Case (1) from Lemma 5.9.

At this point we may assume that $in_{<}(I)=(uv,uw,vw,xy,xz,yz)$.

Claim 2: The Gröbner basis of $I$ contains no quadric $q$ with
$\mathop{\mathrm{rank}}\nolimits(q)\geq 6$.

Suppose $I$ has a rank $\geq 6$ quadric $q$ in its reduced Gröbner basis.
After a linear change of variables that does not change the initial ideal, we
may assume that $q$ has the form $q=uv+wa+bc+\cdots$, where $u,v,w,a,b,c$ are
distinct variables. Taking multiples of $q$ we see that $uabc,vabc\in
in_{<}(I)$, which forces $ab\in in_{<}(I)$ or $ac\in in_{<}(I)$. Without loss,
we may assume $q$ has the form $q=uv+wx+yc+\cdots$, where
$c\notin\\{u,v,w,x,y,z\\}$. (If $c\in\\{u,v,w,x,y,z\\}$, the only choice would
be $c=z$, in which case $yz$ would be both the initial monomial of a term in
the reduced Gröbner basis of $I$ and in the support of $q$, a contradiction.)
Replacing the other quadrics by their initial terms does not change the
Hilbert function and so $I^{\prime}=(uw,vw,xy,xz,yz,q)$ has the same Hilbert
function as $I$. Taking the initial ideal of $I^{\prime}$, we see that
$I^{\prime\prime}:=(uv,uw,vw,xy,xz,yz,uxyc,vxyc)\subseteq in_{<}(I^{\prime})$.
A calculation shows that the Hilbert series of
$E^{\prime}/I^{\prime\prime}=1+7t+15t^{2}+8t^{3}$, where
$E^{\prime}=\bigwedge_{\mathbb{K}}\langle c,u,v,w,x,y,z\rangle$, while
$(1+3t)^{2}(1+t)=1+7t+15t^{2}+9t^{3}$. Therefore, the Hilbert Series of
$E/in_{<}(I)$ is too small, and we conclude that the reduced Gröbner basis for
$I$ consists of rank $4$ quadrics.

Claim 3: The reduced Gröbner basis of $I$ contains a quadric of the form
$xy+za$, where $in_{<}(I)=(uv,uw,vw,xy,xz,yz)$ and $a\notin\\{u,v,w,x,y,z\\}$.

Suppose this is not the case. Then each of the six quadrics in the Gröbner
basis of $I$ has the form
$(a+\ell)(b+\ell^{\prime})+(c+\ell^{\prime\prime})(d+\ell^{\prime\prime\prime})$,
where $\\{a,b,c\\}=\\{u,v,w\\}$ and $d\in\\{x,y,z\\}$ or
$\\{a,b,c\\}=\\{x,y,z\\}$ and $d\in\\{u,v,w\\}$. Without loss, we may assume
that $u>v>w$ and $x>y>z$. One of the quadrics then has the form
$(y+\ell)(z+\ell^{\prime})+(x+\ell^{\prime\prime})(d+\ell^{\prime\prime\prime})$,
with $\ell,\ell^{\prime},\ell^{\prime\prime},\ell^{\prime\prime\prime}\in
E_{1}$, $d\in\\{u,v,w\\}$ and $yz>xd$. As $x>y>z$, we must have $z>d\geq w$.
Another quadric has the form
$(v+m)(w+m^{\prime})+(u+m^{\prime\prime})(f+m^{\prime\prime\prime})$, with
$m,m^{\prime},m^{\prime\prime},m^{\prime\prime\prime}\in E_{1}$,
$f\in\\{x,y,z\\}$ and $vw>uf$. This forces $w>f\geq z$, which contradicts that
$z>w$. This finishes the proof of Claim 3.

Claim 4: $I$ contains a rank $2$ quadric.

By Claim 3, $I$ contains a quadric of the form $q=xy+za$, where
$a\notin\\{u,v,w,x,y,z\\}$. A second quadric has the form
$q^{\prime}=(x+k)(z+k^{\prime})+\alpha(y+k^{\prime\prime})(d+k^{\prime\prime\prime})$,
where $k,k^{\prime},k^{\prime\prime},k^{\prime\prime\prime}\in E_{1}$,
$\alpha\in\mathbb{K}^{\ast}$, and $d$ is a variable distinct from $x,y,z$.
Considering $xq-q^{\prime}a$, the possibly nonzero leading terms are
$xk^{\prime}a,kza,yda,yk^{\prime\prime\prime}a,k^{\prime\prime}da$. By our
assumptions no monomial from $in_{<}(I)$ can divide any of these and thus they
are all zero. This forces
$c,k,k^{\prime},k^{\prime\prime},k^{\prime\prime\prime}\in(a)$. It follows
that $q,q^{\prime}$ are linearly independent quadrics in
$\bigwedge_{\mathbb{K}}\langle x,y,z,a\rangle$. By Corollary 5.8, $I$ contains
a rank $2$ quadric. ∎

######  Example 5.13.

The hypotheses of the previous lemma cannot be weakened considerably. Consider
the ideal

$I=(e_{1}e_{2}+e_{3}e_{4},e_{1}e_{3}+e_{2}e_{4},e_{2}e_{3}+e_{1}e_{4},e_{5}e_{6}+e_{7}e_{8},e_{5}e_{7}+e_{6}e_{8},e_{6}e_{7}+e_{5}e_{8}),$

in $E=\bigwedge_{\mathbb{K}}\langle e_{1},\ldots,e_{8}\rangle$. It is clear
that all $6$ generators of $I$ have rank $4$ and that they form a reverse
lexicographic Gröbner basis of $I$. The initial ideal
$in_{revlex}(I)=I_{E}(G)$, where $G$ is the graph from Case (1) in Lemma 5.9.
Note that there is a rank $2$ quadric in $I$:

$(e_{1}e_{2}+e_{3}e_{4})+(e_{1}e_{3}+e_{2}e_{4})=(e_{1}+e_{4})(e_{2}-e_{3})\in
I.$

This is in some sense a limiting case later in the proof.

Now that we have all of the ingredients, we can complete the proof of Theorem
5.4.

###### Proof of Theorem 5.4..

That $E/I$ is Koszul follows from Theorem 5.3. It follows from [15, Theorem
10.5] that the Hilbert series of $E/I$ is $HS_{E/I}(t)=1+6t+9t^{2}$. Suppose
$I$ is LG-quadratic. Then there is an ideal $J\subseteq
E^{\prime}=\bigwedge_{\mathbb{C}}\langle y_{1},\ldots,y_{n}\rangle$ equipped
with a monomial order $<$ such that $J$ has a quadratic Gröbner basis and
there is a regular sequence $\underline{\ell}=\ell_{1},\ldots,\ell_{d}$ on
$E^{\prime}/J$ such that $E^{\prime}/(J+(\underline{\ell}))\cong E/I$. It
follows that $HS_{E^{\prime}/J}(t)=(1+6t+9t^{2})(1+t)^{n-6}$. Then
$in_{<}(J)=I(G)$ for one of the three graphs from Lemma 5.9. By Lemma 5.12,
$J$ contains a rank $2$ quadric. Since rank can only go down when we kill a
regular element, $I$ contains a rank $2$ quadric. This contradicts Proposition
5.7 and the result follows. ∎

######  Remark 5.14.

In light of this example, it is reasonable to ask if one can construct similar
examples of commutative Koszul algebras that are not LG-quadratic. Indeed,
Fröberg and Löfwall proved [15, Theorem 7.1] that if
$A=\mathbb{K}[x_{1},\ldots,x_{n}]/I$, where $I$ is generated by $t$ generic
quadrics, then $A$ is Koszul if and only if $t\leq n$ or $t\geq n^{2}/4+n/2$.
The first case corresponds to generic complete intersections which are known
to be LG-quadratic [8, Remark 1.19]. The second case is similar to the setting
considered here. One could compute a lower bound on the ranks of the quadrics
in $I$ by computing the codimension of the appropriate ideal of minors of a
generic symmetric matrix. However, it is not clear what one can say about the
ranks of quadrics appearing in a quadratic Gröbner basis can be in a
polynomial ring. The results in [12] depend on the number of variables, which
makes arguments relating to the LG-quadratic property difficult.

##  6\. Examples

In this section we collect examples to compare quadratic ideals over exterior
algebras. In particular, we have the following implications for quadratic
quotients of an exterior algebra $E/I$:

$\text{quadratic GB}\Rightarrow\text{G-quadratic}\Rightarrow\text{LG-
quadratic}\Rightarrow\text{Koszul}\Rightarrow\text{quadratic}.$

The first two implications follow from the definitions. The third implication
is a consequence of Theorem 3.2. The last implication is well-known [14].

Theorems 4.5 and 5.4 show that the second and third implications are strict,
respectively. The examples below show that the remaining implications are
strict as well.

######  Example 6.1.

A G-quadratic ideal that has no quadratic Gröbner basis in the given
coordinate system
This example is due to Thieu [34, Example 5.2.2.ii], who shows that it is
Koszul via a filtration argument and asks whether it is G-quadratic. We
include a shorter proof of Koszulness here by showing that it is G-quadratic.
Fix a field $\mathbb{K}$ with $\mathrm{char}(\mathbb{K})\neq 2$. Let
$E=\bigwedge_{\mathbb{K}}\langle e_{1},e_{2},e_{3},e_{4}\rangle$ be a
4-variable exterior algebra and consider
$I=(e_{1}e_{2}-e_{3}e_{4},e_{1}e_{3}-e_{2}e_{4})$. First we show that $I$ has
no quadratic Gröbner basis with respect to any monomial order. Indeed, fix a
monomial order $\prec$ and observe that the initial terms of the generators of
$I$ with respect to $\prec$ are $(e_{i}e_{j},e_{i}e_{k})$ for some distinct
integers $1\leq i,j,k\leq 4$. It is clear that
$I_{3}=(e_{1},e_{2},e_{3},e_{4})^{3}$, while not every degree $3$ monomial in
$E$ is in $(e_{i}e_{j},e_{i}e_{k})$. Therefore $I$ has no quadratic Gröbner
basis. However, since the $\mathrm{char}(\mathbb{K})\neq 2$, we have

$\displaystyle I$
$\displaystyle=((e_{1}e_{2}-e_{3}e_{4})+(e_{1}e_{3}-e_{2}e_{4}),(e_{1}e_{2}-e_{3}e_{4})-(e_{1}e_{3}-e_{2}e_{4}))$
$\displaystyle=((e_{1}+e_{4})(e_{2}+e_{3}),(e_{1}-e_{4})(e_{2}-e_{3})).$

Thus after a linear change of variables, $I$ becomes a quadratic monomial
ideal, which is obviously G-quadratic.

That there are quadratic quotients of an exterior algebra that are not Koszul
is well known. This is true even when the defining ideal is principal.

######  Example 6.2.

A quadratic principal ideal that is not Koszul This example is due to Nguyen
[27] and is derived from [25]. We include a proof here for completeness. Let
$E=\bigwedge_{\mathbb{C}}\langle e_{1},e_{2},e_{3},e_{4}\rangle$ and let
$I=(e_{1}e_{2}+e_{3}e_{4})$. One checks that the Hilbert series of $E/I$ is
$\mathop{\mathrm{HS}}\nolimits_{E/I}(t)=1+4t+5t^{2}$. If $E/I$ were Koszul,
then by [14, Definition-Theorem 1] its Poincare series would be

$P_{E/I}(t)=\frac{1}{\mathop{\mathrm{HS}}\nolimits_{E/I}(-t)}=1+4t+11t^{2}+24t^{3}+41t^{4}+44t^{5}-29t^{6}\cdots,$

which has a negative coefficient. Therefore $E/I$ cannot be Koszul.

## Acknowledgements

The authors thank Aldo Conca and Hal Schenck for useful conversations. The
first author was supported by NSF grant DMS-1900792.

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