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\title[Geometric vertex decomposition and liaison]{Geometric vertex decomposition and liaison}

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\author{Patricia Klein\thanks{\href{mailto:klein847@umn.edu}{klein847@umn.edu}.}\addressmark{1} \and Jenna Rajchgot \thanks{\href{mailto:rajchgot@math.mcmaster.ca}{rajchgot@math.mcmaster.ca} JR was partially supported by by NSERC grant RGPIN-2017-05732.} \addressmark{2}}

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\address{\addressmark{1}School of Mathematics, University of Minnesota, Minneapolis, MN, USA \\ \addressmark{2}Department of Mathematics and Statistics, McMaster University, Hamilton, ON, Canada}

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\abstract{ Geometric vertex decomposition and liaison are two frameworks that have been used to produce similar results about similar families of algebraic varieties. We establish an explicit connection between these   approaches.
 In particular, we show that each geometrically vertex decomposable ideal is linked by a sequence of elementary $G$-biliaisons of height 1 to an ideal of indeterminates and, conversely, that every $G$-biliaison of a certain type gives rise to a geometric vertex decomposition. 
 
 As a consequence, we can immediately conclude that several well-known  families of ideals are  glicci, including Schubert determinantal ideals, defining ideals of varieties of complexes, and defining ideals of graded lower bound cluster algebras.  We also use the structure of Knutson, Miller, and Yong's geometric vertex decomposition to provide a streamlined implementation of Gorla, Nagel, and Migliore's liaison-theoretic approach to establishing Gr\"obner bases.}

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\keywords{Vertex decomposition, liaison, Gr\"obner bases}

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\section{Introduction}

Determinantal ideals and their generalizations have been explored extensively both in the context of commutative algebra and also in the study of Schubert varieties in flag varieties. This overlap is to be expected because, for example, each ideal generated by the $k\times k$ minors of a generic matrix is the defining ideal of an open patch of a Schubert variety in a Grassmannian; each one-sided ladder determinantal ideal is a Schubert determinantal ideal for a vexillary (i.e., 2143-avoiding) permutation (see eg. \cite{KMY09}); each two sided mixed ladder determinantal ideal is a type $A$ Kazhdan--Lusztig ideal; each ideal generated by the $k\times k$ minors of a generic symmetric matrix is the  defining ideal of an open patch of a Schubert variety in a Lagrangian Grassmannian; and each defining ideal of a variety of complexes is a type $A$ Kazhdan--Lusztig ideal, up to some extra indeterminate generators (see eg. \cite[Ch. 17]{MillerSturmfels}).  

While similar results on the above-mentioned families of ideals appear in the Schubert variety and commutative algebra literatures, it is often different techniques that are used to obtain them.  For example, in \cite{KMY09}, A. Knutson, E. Miller, and A. Yong introduced \emph{geometric vertex decomposition}, a degeneration technique, and used this to study Gr\"obner geometry of Schubert determinantal ideals for vexillary permutations. Independently, \emph{liaison-theoretic} methods were used by E. Gorla in \cite{Gor07} and E. Gorla, J. Migliore, and U. Nagel in \cite{GMN13} to obtain Gr\"obner bases for  various classes of ladder determinantal ideals (including one sided ladder determinantal ideals,
also known as Schubert determinantal ideals for vexillary permutations).   In this extended abstract, we establish an explicit connection between  \emph{geometric vertex decomposition} and \emph{liaison}, and we study implications of this connection. We have three main goals, which we now outline. 



 
Our first goal is to show that it is no coincidence that geometric vertex decomposition and liaison can be used to obtain similar results. Indeed, we prove the following explicit connection between the two techniques (see Corollary \ref{cor:gvdToLia} and Theorem \ref{thm:linkToGVD}):

\begin{mainthm}\label{thm:main1}
Under mild hypotheses, every geometric vertex decomposition gives rise to an elementary $G$-biliason of height $1$.  Every sufficiently ``nice" elementary $G$-biliaison of height $1$ gives rise to a geometric vertex decomposition.
\end{mainthm}

The second motivation for our work comes from a long-standing open question in liaison theory, which asks whether subschemes of $\mathbb{P}^n$ are arithmetically Cohen--Macaulay if and only if they are in the \emph{Gorenstein liaison class of a complete intersection}  (or \emph{glicci}).  It is a standard homological argument that every glicci subscheme of $\mathbb{P}^n$ is arithmetically Cohen--Macaulay.  Hence, the question may be phrased as follows:

\begin{quest}[\protect{\cite[Question 1.6]{KM+01}}]\label{motivation2}
Is every arithmetically Cohen--Macaulay subscheme of $\mathbb{P}^n$ glicci?
\end{quest}

%\noindent For more background on why this question emerges naturally from the history of liaison and for a summary of partial results already in the literature, see Section \ref{sect:liaison}.


By combining our main theorem with some straightforward consequences of geometric vertex decomposition, we give a corollary (stated precisely as Corollary \ref{cor:AutomaticallyGlicci}) from which one can quickly deduce that certain well-known classes of varieties are glicci:

\begin{coro}\label{cor:mainCor}
Let $I$ be a homogenous ideal in a polynomial ring. 
If the Lex-initial ideal of $I$ is the Stanley--Reisner ideal of a vertex decomposable simplicial complex and the vertex decomposition is compatible with the order of the variables, then $I$ is glicci. 
\end{coro}

We discuss three such classes in Section \ref{sect:applications}: matrix Schubert varieties, varieties of complexes, and varieties of graded lower bound cluster algebras.  We present this as evidence in favor of Question \ref{motivation2}, at least in combinatorially-natural settings.  Using the first half of our main theorem, we recover a result of U. Nagel and T. R\"omer from \cite{NR08}, namely that the Stanley--Reisner ideal of a vertex decomposable simplicial complex is glicci.  We show the following (appearing later as Theorem \ref{thm:glicci}):

\begin{thm}
Geometrically vertex decomposable ideals are glicci.
\end{thm}


In  \cite[Lemma 1.12]{GMN13}, it is shown that one can use liaison to compare Hilbert functions when the degrees of the isomorphisms of the $G$-biliaisons involved in an inductive argument are known.  This approach is employed in many of the determinantal cases treated in the literature (\cite{Gor07, Gor08, GMN13}). It is worth noticing that the isomorphisms employed in these papers all have a similar form. 
We explain via geometric vertex decomposition in Theorem \ref{thm:onestep} why this similarity is not a coincidence but, rather, is to be expected.  In that theorem, we associate an explicit isomorphism of degree $1$ to a geometric vertex decomposition. 


In addition to the expository work of describing a unifying structure underlying examples already in the literature, Theorem \ref{thm:onestep} also provides a candidate isomorphism in the style of $G$-biliaison that, in good cases, allows one to use the framework of \cite{GMN13} to prove that a conjectured Gr\"obner basis is, indeed, a Gr\"obner basis.  This corollary has been used in the study of diagonal degenerations of matrix Schubert varieties \cite{Kle}.  Some consequences of Theorem \ref{thm:onestep} on Gr\"obner bases and degenerations appear in Section \ref{sect:GVDGlicci}.




\emph{The structure of this extended abstract.} In Section \ref{sect:GVD}, we review definitions and key lemmas from \cite{KMY09} on geometric vertex decomposition in the unmixed case and record some additional observations about the structure of a geometrically vertex decomposable ideal.  We then briefly review background material on Gorenstein liaison that is necessary for this work.  In Sections \ref{sect:GVDGlicci} and \ref{sect:glicciGVD}, we describe our main theorem (stated above), and related results and examples. In Section \ref{sect:applications}, we prove that certain well-known classes of combinatorially-defined ideals are glicci, via the material in Section \ref{sect:GVDGlicci}.  %Finally, we devote Section \ref{sect:nonpure} to the not necessarily unmixed case, which we relate to vertex decomposition in the not necessarily pure case.

\emph{Notational conventions.} Throughout the extended abstract, we let $\kappa$ be a field, which can be chosen arbitrarily except in Sections \ref{sect:GVDGlicci} and \ref{sect:applications}, where we require that $\kappa$ be infinite.


%\subsection*{Acknowledgements}
%We thank Elisa Gorla, Adam Van Tuyl, and Anna Weigandt for helpful conversations. Part of this work was completed at the Banff International Research Station (BIRS) during the Women in Commutative Algebra workshop in October 2019. We are grateful for the hospitality of the Banff Centre.  The second author was partially supported by NSERC grant RGPIN-2017-05732.








\section{Preliminaries}\label{sect:GVD}

In this section we discuss geometric vertex decomposition, introduced by A. Knutson, E. Miller, and A. Yong in \cite{KMY09}. We first recall the basics of vertex decomposition of simplicial complexes and Stanley--Reisner ideals. Next, we move beyond the monomial ideal case and recall the basics of \emph{geometric} vertex decomposition from \cite{KMY09}. Then we define and study \emph{geometrically vertex decomposable ideals}.  Finally, we recall the information of Gorenstein liaison essential for the purposes of this extended abstract.

\emph{Vertex decomposition and Stanley--Reisner ideals.}
Let $\Delta$ be a simplicial complex on vertex set $[n] = \{1,2,\dots, n\}$ (without an insistence that every $v\in [n]$ necessarily be a face of $\Delta$). 
Given a vertex $v\in \Delta$, define the following three subcomplexes: the \textbf{star} of $v$ is the set $\text{star}_\Delta(v):= \{F\in \Delta\mid F\cup \{v\}\in \Delta\}$,  the \textbf{link} of $v$ is the set $\text{lk}_{\Delta}(v) := \{F\in \Delta\mid F\cup\{v\}\in \Delta, F\cap \{v\}=\emptyset\}$, and the \textbf{deletion} of $v$ is the set $\text{del}_{\Delta}(v) := \{F\in \Delta\mid F\cap \{v\} = \emptyset\}.$  Recall that the \textbf{cone} from $v$ on a simplicial complex $\Delta$ is the smallest simplicial complex that contains the set $\{F\cup \{v\}\mid F\in \Delta\}$. 
Then $\text{star}_\Delta(v)$ is the cone from $v$ on $\text{lk}_\Delta(v)$  and $\Delta = \text{star}_{\Delta}(v)\cup \text{del}_\Delta(v)$. The decomposition of $\Delta$ above is called a \textbf{vertex decomposition}.

A simplicial complex is called \textbf{pure} if all of its facets (i.e., maximal faces) are of the same dimension. 
A simplicial complex $\Delta$ is \textbf{vertex decomposable} if it is pure and if $\Delta = \emptyset$, or $\Delta$ is a simplex, or there is a vertex $v\in \Delta$ such that $\text{lk}_\Delta(v)$ and $\text{del}_{\Delta}(v)$ are vertex decomposable.  Given a simplicial complex $\Delta$ on vertex set $[n]$, one defines the \textbf{Stanley--Reisner ideal} $I_{\Delta}\subseteq \kappa[x_1,\dots, x_n]$ associated to $\Delta$ as  $I_\Delta:= \langle \textbf{x}_F\mid F\subseteq [n], F\notin \Delta\rangle$, where $\textbf{x}_F := \prod_{i\in F}x_i$.  
The association $\Delta\mapsto I_\Delta$ determines a bijection between simplicial complexes on $[n]$ and squarefree monomial ideals in $\kappa[x_1,\dots, x_n]$. We write $\Delta(I)$ for the simplicial complex associated to a squarefree monomial ideal $I$. 

Notice that if $\Delta = \Delta_1\cup \Delta_2$ is a union of simplicial complexes on $[n]$, then $F$ is a non-face of $\Delta$ if and only if it is a non-face of both $\Delta_1$ and $\Delta_2$. Thus, $I_\Delta = I_{\Delta_1}\cap I_{\Delta_2}$. In particular, if $v$ is a vertex of $\Delta$, we may decompose $\Delta$  to get $I_\Delta = I_{\text{star}_\Delta(v)}\cap I_{\text{del}_\Delta(v)}$.  

\begin{lemma}\label{lem:starDelLink}
Let $v\in [n]$ be a vertex of $\Delta$. Writing $I_{\Delta} = \langle x_v^{d_i}q_i\mid 1\leq i\leq m\rangle$ where $q_i$ is a squarefree monomial that is not divisible by $x_v$ and $d_i = 0$ or $1$. Then $I_{\text{star}_\Delta(v)} = \langle q_i \mid 1\leq i\leq m\rangle$,  $I_{\text{lk}_\Delta(v)} = I_{\text{star}_\Delta(v)}+\langle x_v\rangle$, and $I_{\text{del}_\Delta(v)} = \langle q_i \mid d_i = 0\rangle+\langle x_v\rangle$.
\end{lemma}



\emph{Geometric vertex decomposition}
We now discuss \emph{geometric vertex decomposition}, introduced by A. Knutson, E. Miller, and A. Yong in \cite{KMY09}.  Let $R = \kappa[x_1,\dots, x_n]$ be a polynomial ring in $n$ indeterminates and let $y = x_j$ for some $1 \leq j \leq n$.  Define the \textbf{initial $y$-form} $\text{in}_y f$ of a polynomial $f\in R$ to be the sum of all terms of $f$ having the highest power of $y$. That is, if $f = \sum_{i = 0}^n \alpha_i y^i$, where each $\alpha_i\in \kappa[x_1,\dots,\widehat{x}_j,\dots x_n]$ and $\alpha_n\neq 0$, define $\text{in}_y f :=\alpha_n y^n$, which is usually not a monomial. Given an ideal $I\subseteq R$, define $\text{in}_y I: = \langle \text{in}_y f \mid f \in I \rangle.$ 
We say that a monomial order $<$ on $R$ is \textbf{$y$-compatible} if 
$\text{in}_< f = \text{in}_<(\text{in}_y f)$ 
for every $f\in R$, in which case $\text{in}_<(\text{in}_y I) = \text{in}_< I$ for any ideal $I\subseteq R$. 



Let $I\subseteq R$ be an ideal and $<$ a $y$-compatible monomial order.  
With respect to $<$, let $\mathcal{G} := \{y^{d_i}q_i+r_i\mid 1\leq i\leq m\}$ be a Gr\"obner basis of $I$ where $y$ does not divide any $q_i$ and $\text{in}_{y}(y^{d_i}q_i+r_i) = y^{d_i}q_i$. One easily checks that the ideal $\text{in}_y I$ is generated by $\text{in}_y \mathcal{G}:= \{y^{d_i}q_i\mid 1\leq i\leq m\}$  (\cite[Theorem 2.1(a)]{KMY09}). That is, $\text{in}_{y} I = \langle y^{d_i}q_i\mid 1\leq i\leq m\rangle$. 


\begin{definition}[\protect{\cite[Section 2.1]{KMY09}}]\label{def:gvdKMS}
As above, let let $\mathcal{G} := \{y^{d_i}q_i+r_i\mid 1\leq i\leq m\}$ be a Gr\"obner basis of the ideal $I$ with respect to the $y$-compatible term order $<$.  Define $C_{y,I} := \langle q_i\mid 1\leq i\leq m\rangle$ and $
N_{y,I} = \langle q_i\mid d_i = 0\rangle.$ 
When $\text{in}_y I = C_{y,I} \cap (N_{y,I}+\langle y \rangle)$, this decomposition is called a \textbf{geometric vertex decomposition of $I$ with respect to $y$}.  
\end{definition}

The ideals $C_{y,I}$ and $N_{y,I}$ do not depend on the choice of Gr\"obner basis and, in particular, do not depend on the choice of $y$-compatible term order $<$.  
This follows from the facts that $C_{y,I} = (\text{in}_y I :y^\infty)$ by \cite[Theorem 2.1(d)]{KMY09} and that $N_{y,I}+\langle y\rangle = \text{in}_y I+\langle y\rangle$ by \cite[Theorem 2.1 (a)]{KMY09}, together with the observation that $y$ does not appear in the generators of $N_{y,I}$ given in its definition. 

 We say that a geometric vertex decomposition is \textbf{degenerate} if $\sqrt{C_{y,I}} = \sqrt{N_{y,I}}$ or if $C_{y,I} = \langle 1 \rangle$ and \textbf{nondegenerate} otherwise.    As we will see through Lemma \ref{lem:form}, if $C_{y,I} = \langle 1 \rangle$, then some polynomial whose initial $y$-form is a unit multiple of $y$ is an element of $I$, in which case $R/I \cong R/(N_{y,I}+\langle y \rangle)$.   If  $\sqrt{C_{y,I}} = \sqrt{N_{y,I}}$, then $\sqrt{\text{in}_y I} = \sqrt{C_{y,I}} \cap \sqrt{N_{y,I}+\langle y \rangle} = \sqrt{C_{y,I}}$, in which case $\text{in}_y I$, $C_{y,I}$, and $N_{y,I}$ all determine the same variety.  In both of these cases, we may often prefer to study $N_{y,I}$ in the smaller polynomial ring that omits $y$.  This is especially true when $I$ is radical for the following reason:
 
 \begin{proposition}\label{prop:degen-rad}
 If $I$ is radical and has a degenerate geometric vertex decomposition $\text{in}_y I = C_{y,I} \cap (N_{y,I}+\langle y \rangle)$ with $\sqrt{N_{y,I}} = \sqrt{C_{y,I}}$, then the reduced Gr\"obner basis of $I$ does not involve $y$ and $I = \text{in}_y I = C_{y,I} = N_{y,I}$.
 \end{proposition}


%\begin{remark}\label{rem:GVDsimplicial}
%Let $\Delta$ be a simplicial complex on vertex set $[n]$, and let $v$ be a vertex of $\Delta$. 
%The geometric vertex decomposition of $I_\Delta\subseteq R$ with respect to variable $x_v$ agrees with the decomposition $I_{\Delta} = I_{\text{star}_\Delta(v)}\cap I_{\text{del}_\Delta(v)}.$ Indeed, $\text{in}_{x_v}I_\Delta = I_\Delta$, $I_{\text{star}_\Delta(v)} = C_{x_v,I_\Delta}$, and $I_{\text{del}_\Delta(v)} = N_{x_v,I_{\Delta}}+\langle x_v\rangle$ (see the end of Section \ref{sect:vertDecomp}). 
%Observe that since $v\in \Delta$, we have $C_{x_v,I_\Delta} \neq \langle 1\rangle$. Thus, the geometric vertex decomposition is degenerate if and only if $\Delta$ is a cone from $v$ on $\text{lk}_\Delta(v)$.
%\end{remark}


If an ideal $I\subseteq R$ has a generating set $\mathcal{G}$ in which $y^2$ does not divide any term of $g$ for any $g \in \mathcal{G}$, then we say that $I$ is \textbf{squarefree in $y$}. 
It is easy to see (for example, by considering $S$-pair reductions) that every ideal that is squarefree in $y$ has a Gr\"obner basis, with respect to any $y$-compatible term order, such that $y^2$ does not divide any term of any element of the Gr\"obner basis. 

\begin{lemma}\label{lem:form}
If $I \subseteq R$ possesses a geometric vertex decomposition with respect to  a variable $y = x_j$ of $R$, then $I$ is squarefree in $y$, and the reduced Gr\"obner basis of $I$ with respect to any $y$-compatible term order has the form $\{yq_1+r_1,\dots, yq_k+r_k,h_1,\dots, h_\ell\}$ where $y$ does not divide any term of any $q_i$ or $r_i$ for any $ 1 \leq i \leq k$ nor any $h_j$ for any $1 \leq j \leq \ell$.
\end{lemma}

%\begin{proof}
%Fix a $y$-compatible term order $<$, and let $\mathcal{G} = \{y^{d_1}q_1+r_1,\dots, y^{d_m}q_m+r_m\}$ be the reduced Gr\"obner basis of $I$ with respect to $<$, where $y^{d_i}q_i = \text{in}_y(y^{d_i}q_i+r_i)$ and $y$ does not divide any term of $q_i$ for any $1 \leq i \leq m$.   Observe that $yq_i\in C_{y,I} \cap (N_{y,I}+\langle y \rangle)$ for each $1 \leq i \leq m$, so $\text{in}_y I = C_{y,I} \cap (N_{y,I}+\langle y \rangle)$ implies $yq_i\in \text{in}_y I$ for each $1 \leq i \leq m$. Hence, we may assume $d_i \leq 1$ for each $1 \leq i \leq m$.  
%The remaining statements now follow easily.
%\end{proof}




\emph{Geometrically vertex decomposable ideals.} A geometric vertex decomposition of an ideal is analogous to a vertex decomposition of a simplicial complex into a deletion and star. In this subsection, we extend this analogy by considering \emph{geometrically vertex decomposable ideals}, which are analogous to vertex decomposable simplicial complexes. We again let $R = \kappa[x_1,\dots, x_n]$ throughout this subsection. 
Recall that an ideal $P$ of $R$ is an \textbf{associated prime} of the ideal $I$ if $R/P$ is isomorphism to a submodule of $R/I$ and that an ideal $I\subseteq R$ is \textbf{unmixed} if $\dim(R/P) = \dim(R/I)$ for all associated primes $P$ of $I$ .


\begin{definition}\label{def:geometricallyVertexDec}
An ideal $I\subseteq R$ is \textbf{geometrically vertex decomposable} if $I$ is unmixed and if (1)  $I = \langle 1\rangle$ or $I$ is generated by indeterminates in $R$, or (2)  for some variable $y = x_j$ of $R$,  $\text{in}_y I =  C_{y,I}  \cap (N_{y,I}+\langle y\rangle)$ is a geometric vertex decomposition and the contractions of $N_{y,I}$ and $C_{y,I}$ to $\kappa[x_1,\dots,\widehat{y},\dots, x_n]$ are geometrically vertex decomposable.
\end{definition}

We take case $(1)$ to include the zero ideal, whose (empty) generating set vacuously consists only of indeterminates. 
We will soon need observations about the relative heights of the ideals $I$, $C_{y,I}$, and $N_{y,I} $ in the circumstances of condition $(2)$.  The degenerate cases are clear: if $C_{y,I} = \langle 1 \rangle$, then $\hgt(I) = \hgt(N_{y,I} )+1$ and, if $\sqrt{C_{y,I}} = \sqrt{N_{y,I}}$, then $\hgt(I) = \hgt(\text{in}_y I) = \hgt(C_{y,I})=\hgt(N_{y,I} )$. The nondegenerate case is handled by the lemma below.


%We say that the ring $R/I$ is \emph{equidimensional} if $\dim(R/P) = \dim(R/I)$ for all minimal primes $P$ of $I$ or, equivalently, if all irreducible components of the variety of $I$ have the same dimension.  Equidimensionality does not preclude the possibility that $I$ might have embedded primes and so is weaker than unmixedness.

\begin{lemma}\label{lem:height}
If $I\subseteq R$ is an ideal so that $R/I$ is equidimensional and $\text{in}_y I =  C_{y,I}  \cap (N_{y,I} + \langle y \rangle)$ is a nondegenerate geometric vertex decomposition with respect to some variable $y = x_j$ of $R$, then $\hgt(C_{y,I}) = \hgt(I) = \hgt(N_{y,I} )+1$. %Moreover, $R/C_{y,I}$ is equidimensional.
\end{lemma}


%\begin{proof}
%By Lemma \ref{lem:form}, $I$ has a reduced Gr\"obner basis $\mathcal{G}$ of the form $\{yq_1+r_1, \ldots, yq_k+r_k, h_1, \ldots, h_\ell\}$ where $y$ does not divide any term of any $q_i$ or $r_i$, $1 \leq i \leq k$, nor any or $h_j$, $1 \leq j \leq \ell$.  Because $I$ is unmixed, $\hgt(I) = \hgt(\text{in}_y (I))$.  Because Gr\"obner degenerations are flat, $\text{in}_y (I)$ is also unmixed.  Hence, if $C_{y,I}$ and $N_{y,I}+\langle y \rangle $ are incomparable with respect to the partial order by inclusion, then $ \hgt(C_{y,I}) = \hgt(I) = \hgt(N_{y,I} )+1$.

%So, suppose that $N_{y,I} + \langle y \rangle \subseteq C_{y,I}$.  Then, in particular, $y \in C_{y,I}$ and so some element of the Gr\"obner basis $\{q_1, \ldots, q_k, h_1, \ldots, h_\ell\}$ of $C_{y,I}$ has a leading term dividing $y$, which is to say that some $q_i$ is a unit for some $1 \leq i \leq k$ because no term of any $q_i$ or $h_j$ involves $y$ for any $1 \leq i \leq k$ or $1 \leq j \leq \ell$.  But then $C_{y,I} = \langle 1 \rangle$, contradicting non-degeneracy.  

%Suppose that $C_{y,I} \subseteq N_{y,I} +\langle y \rangle$. Because $C_{y,I} \neq N_{y,I}$ by nondegeneracy, $y$ divides some term of some element of $\mathcal{G}$, so $k \geq 1$ and $q_1 \in N_{y,I} +\langle y \rangle$.  Then some term of the Gr\"obner basis $\{h_1, \ldots, h_\ell, y\}$ of $N+\langle y \rangle$ has a leading term dividing the leading term of $q_1$.  But $q_1$ does not involve $y$, and no leading term of $h_i$ for any $1 \leq i \leq \ell$ can divide any term of $q_1$ because $\mathcal{G}$ is the reduced Gr\"obner basis of $I$.
%\end{proof}

Note that if $\Delta$ be a simplicial complex on vertex set $[n]$, then its Stanley--Reisner ideal $I_\Delta \subseteq R$ is geometrically vertex decomposable if and only if $\Delta$ is vertex decomposable.  We next discuss some properties of geometrically vertex decomposable ideals and further connections to vertex decomposable simplicial complexes.

%\begin{proof}
%We proceed by induction on $n$. When $n = 0$, the result is trivial.  So suppose $n \geq 1$ is arbitrary.  Suppose $\Delta$ is vertex decomposable. If $\Delta$ is a simplex, there is nothing to show.  Otherwise, there is some $v\in \Delta$ such that $\text{lk}_\Delta(v)$ and $\text{del}_\Delta(v)$ are vertex decomposable.  As discussed in Remark \ref{rem:GVDsimplicial}, the decomposition $I_\Delta = I_{\text{star}_\Delta(v)}\cap I_{\text{del}_\Delta(v)}$ is a geometric vertex decomposition (where $I_{\text{star}_\Delta(v)} = C_{x_v,I_\Delta}$, and $I_{\text{del}_\Delta(v)} = N_{x_v,I_{\Delta}}+\langle x_v\rangle$).   Furthermore, thinking of $\text{lk}_{\Delta}(v)$ and $\text{del}_\Delta(v)$ as simplicial complexes on vertex set $\{1,\dots, \widehat{v},\dots, n\}$, the Stanley--Reisner ideals of $\text{lk}_{\Delta}(v)$ and $\text{del}_\Delta(v)$ are the contractions of $C_{x_v, I_\Delta}$ and $N_{x_v, I_\Delta}$ to the ring $\kappa[x_1,\dots,\widehat{x}_v,\dots, x_n]$. Hence, by induction, these contracted ideals are geometrically vertex decomposable. It follows that $I_\Delta$ is geometrically vertex decomposable. 

%Conversely, if $I_\Delta$ is geometrically vertex decomposable, then either $I_\Delta = \langle 1 \rangle$ or $I$ is generated by indeterminates, in which case $\Delta = \emptyset$ or $\Delta$ is a simplex, or there is some variable $x_v$ of $R$ so that $I_\Delta = I_{\text{star}_\Delta(v)}\cap I_{\text{del}_\Delta(v)}$ is a geometric vertex decomposition.  If $v \notin \Delta$, then $I_{\text{star}_\Delta(v)} = \langle 1 \rangle$ and $I_\Delta = I_{\text{del}_\Delta(v)}$.  In this case, we may think of $\Delta$ as a simplicial complex on vertex set $\{1,\dots, \widehat{v},\dots, n\}$ corresponding to the geometrically vertex decomposable ideal $N_{x_v,I_{\Delta}} \cap \kappa[x_1,\dots,\widehat{x}_v,\dots, x_n]$, and so the result follows by induction.  If $v \in \Delta$, then the contractions of $C_{x_v,I_\Delta}$ and $N_{x_v,I_{\Delta}}$ to $\kappa[x_1,\dots,\widehat{x}_v,\dots, x_n]$ are geometrically vertex decomposable proper ideals.  By induction, then, $\text{lk}_{\Delta}(v)$ and $\text{del}_\Delta(v)$ are vertex decomposable simplicial complexes on vertex set $\{1,\dots, \widehat{v},\dots, n\}$ and so are also vertex decomposable simplicial complexes when thought of on vertex set $[n]$.
%\end{proof}



\begin{comment}
\begin{proposition}
Let $\Delta$ be a vertex decomposable simplicial complex on vertex set $[n]$. Its Stanley--Reisner ideal $I_\Delta\subseteq R$ is geometrically vertex decomposable.
\end{proposition}

\begin{proof}
We proceed by induction on the number of vertices $m$ in $\Delta$. When $m = 0, 1$, the result is clear. 
So suppose $2\leq m\leq n$ is arbitrary.
If $\Delta$ is a simplex, there is nothing to show. 
Otherwise, there is some $v\in \Delta$ such that $\text{lk}_\Delta(v)$ and $\text{del}_\Delta(v)$ are vertex decomposable. 
As discussed in Remark \ref{rem:GVDsimplicial}, the decomposition $I_\Delta = I_{\text{star}_\Delta(v)}\cap I_{\text{del}_\Delta(v)}$ is a geometric vertex decomposition (where $I_{\text{star}_\Delta(v)} = C_{x_v,I_\Delta}$, and $I_{\text{del}_\Delta(v)} = N_{x_v,I_{\Delta}}+\langle x_v\rangle$).  
Furthermore, thinking of $\text{lk}_{\Delta}(v)$ and $\text{del}_\Delta(v)$ as simplicial complexes on vertex set $\{1,\dots, \widehat{v},\dots, n\}$, the Stanley--Reisner ideals of $\text{lk}_{\Delta}(v)$ and $\text{del}_\Delta(v)$ are the contractions of $C_{x_v, I_\Delta}$ and $N_{x_v, I_\Delta}$ to the ring $\kappa[x_1,\dots,\widehat{x}_v,\dots, x_n]$. Hence, by induction, these contracted ideals are geometrically vertex decomposable. It follows that $I$ is geometrically vertex decomposable. 
\end{proof}
\end{comment}








\begin{proposition}\label{prop:radical}
A geometrically vertex decomposable ideal is radical.
\end{proposition}

%\begin{proof}
%Let $I \subseteq R$ be a geometrically vertex decomposable ideal.  We proceed by induction on $n = \dim(R)$.  We note first that if $I = \langle 0\rangle$, $I = \langle 1\rangle$, or $I$ is generated by indeterminates, the result is immediate.  
%Otherwise, there exists some variable $y = x_i$ such that $\text{in}_y I = C_{y,I}  \cap (N_{y,I}+\langle y\rangle)$ is a geometric vertex decomposition and the contractions of  $C_{y,I}$ and $N_{y,I}$ to $\kappa[x_1,\dots, \widehat{y},\dots, x_n]$ are geometrically vertex decomposable. These contracted ideals are radical by induction, thus so are $C_{y,I}$ and $N_{y,I}$. Hence, $\text{in}_y I = C_{y,I}  \cap (N_{y,I}+\langle y\rangle)$ is radical. Finally, since there is a flat degeneration from $I$ to $\text{in}_y I$, we see that $I$ is radical.
%\end{proof}

%We remark briefly that Proposition \ref{prop:radical} does not require the unmixedness assumptions on $I$, $C_{y,I}$, or $N_{y,I}$. 

We next consider geometrically vertex decomposable ideals that have a certain compatibility with a given lexicographic monomial order. The main result in our discussion of these ideals is Theorem \ref{prop:simplicial}, which we will need in Section \ref{sect:applications} on applications.  
\begin{definition} 
Fix a lexicographic monomial order $<$ on $R$. 
We say that an ideal $I\subseteq R$ is \textbf{$<$-compatibly geometrically vertex decomposable} if $I$ satisfies Definition \ref{def:geometricallyVertexDec} upon replacing item (2) with (2*) for the $<$-largest variable $y$ in $R$, $\text{in}_{y} I = C_{y,I} \cap (N_{y,I}+\langle y\rangle)$ is a   geometric vertex decomposition and the contractions of $N_{y,I}$ and $C_{y,I}$ to $\kappa[x_1,\dots,\widehat{y},\dots, x_n]$ are $<$-compatibly geometrically vertex decomposable  for the naturally induced monomial order on $\kappa[x_1,\dots, \widehat{y},\dots, x_n]$ (which we also call $<$).
\end{definition}


Let $\Delta$ be a simplicial complex on a vertex set $[n]$,  
and let $<$ be a total order on $[n]$. 
We say that a simplicial complex $\Delta$ is \textbf{$<$-compatibly vertex decomposable} if either $\Delta = \emptyset$ or $\Delta$ is a simplex or, for the $<$-largest vertex $v\in \Delta$, $\text{del}_{\Delta}(v)$ and $\text{lk}_{\Delta}(v)$ are $<$-compatibly vertex decomposable.  The following is an easy consequence of \cite[Theorem 2.1]{KMY09}. 
\begin{lemma}\label{lem:compatible}
Suppose that $I\subseteq R$ 
is squarefree in $y = x_j$, and suppose that $<$ is a $y$-compatible monomial order on $R$. Then $\text{in}_< I = \text{in}_< C_{y,I}\cap (\text{in}_< N_{y, I}+\langle y\rangle)$. 
\end{lemma}

%\begin{proof}

%Since $I$ is squarefree in $y$, $I$ has a Gr\"obner basis $\{yq_1+r_1,\dots, yq_k+r_k,h_1,\dots, h_\ell\}$ where $y$ does not divide any term of $q_i$, $r_i$, $1 \leq i \leq k$, nor $h_j$, $1 \leq j \leq \ell$. 
% Let $m_i = \text{in}_<q_i$, $1\leq i\leq k$, and $m_{k+i} = \text{in}_< h_{i}$, $1\leq i\leq \ell$. 
%By \cite[Theorem 2.1(a)]{KMY09}, $\{q_1,\dots, q_{k}, h_1,\dots, h_{\ell}\}$ and $\{h_1,\dots, h_\ell\}$ are Gr\"obner bases for $C_{y,I}$ and $N_{y,I}$ respectively, so $\text{in}_< C_{y,I} = \langle m_i\mid 1\leq i\leq {k+\ell}\rangle$ and $\text{in}_< N_{y,I} = \langle m_{k+i}\mid 1\leq i\leq \ell\rangle$. It is then straightforward to check 

%\begin{equation}\label{eq:initialDecomp}
%\text{in}_< I = \langle ym_1,\dots, ym_k, m_{k+1},\dots, m_{k+\ell}\rangle = \text{in}_< C_{y,I}\cap (\text{in}_< N_{y,I}+\langle y\rangle). \qedhere
%\end{equation}
%\end{proof}


We now state the main result of this section:  

\begin{theorem}\label{prop:simplicial}

An ideal $I\subseteq R$ is $<$-compatibly geometrically vertex decomposable for the lexicographic monomial order $x_1>x_2>\cdots>x_n$ if and only if $\text{in}_< I$ is the Stanley--Reisner ideal of a $<$-compatibly vertex decomposable simplicial complex on $[n]$ for the vertex order $1>2>\cdots>n$. 
\end{theorem}

%\begin{proof}


%If $I = \langle 1\rangle$ or $\langle 0\rangle$, there is nothing to show. So suppose that $I$ is nontrivial and proceed by induction on n = \text{dim}(R). The base case $n = 1$ is straightforward. 

%Suppose that $n\geq 2$ is arbitrary. First assume that $I$ is $<$-compatibly geometrically vertex decomposable and let $y = x_1$. 
%If $I$ is generated by indeterminates, there is nothing to show. Otherwise, we have that $\text{in}_{y} I = C_{y, I}\cap (N_{y, I}+\langle y\rangle)$ is a geometric vertex decomposition, and the contractions $N^c$ and $C^c$ of $N_{y, I}$ and $C_{y, I}$ to $\kappa[x_2,\dots, x_n]$ are $<$-compatibly geometrically vertex decomposable. 
%There are two cases.

%The first case is when $C_{y,I} = \langle 1 \rangle$, which implies that $\text{in}_< I = \text{in}_<N_{y,I}+\langle y\rangle$. By induction, $\text{in}_<N^c$ is the Stanley--Reisner ideal of a $<$-compatibly vertex decomposable simplicial complex. 
%Thus, $\text{in}_<I$, which is equal to $\text{in}_<N_{y,I}+\langle y\rangle$, is too. Indeed, the complexes $\Delta(\text{in}_<N^c)$ and $\Delta(\text{in}_<N_{y,I}+\langle y\rangle)$ are the same (though on different ambient vertex sets).


 %Now assume $C_{y, I} \neq \langle 1 \rangle$.  By induction, $\text{in}_< N^c$ and $\text{in}_< C^c$ are the Stanley--Reisner ideals of $<$-compatibly vertex decomposable simplicial complexes, thus so are $\text{in}_< N_{y, I}+\langle y\rangle$ and $\text{in}_< C_{y, I}+ \langle y\rangle$. 
%\begin{equation}\label{eq:squarefreeDecompCN}
%\text{in}_< I = \text{in}_<C_{y,I}\cap(\text{in}_<N_{y,I}+\langle y\rangle). 
%\end{equation}
%Thus, $I$ is a squarefree monomial ideal. 
%Let $\Delta := \Delta(\text{in}_<I)$. 
%Equation \eqref{eq:initialDecomp} and Lemma \ref{lem:starDelLink} imply that $\text{in}_<C_{y,I}$ and $\text{in}_<N_{y,I}+\langle y\rangle$ are the Stanley--Reisner ideals of $\text{star}_\Delta(1)$ and $\text{del}_\Delta(1)$.
%Thus $\text{in}_<C_{y,I}+\langle y\rangle$ and $\text{in}_<N_{y,I}+\langle y\rangle$ are the Stanley--Reisner ideals of $\text{lk}_\Delta(1)$ and $\text{del}_\Delta(1)$. 
% Hence $\text{lk}_\Delta(1)$ and $\text{del}_\Delta(1)$ are $<$-compatibly vertex decomposable. Thus, $\Delta$ is too. 


%For the converse, assume that $\Delta = \Delta(\text{in}_< I)$ is $<$-compatibly vertex decomposable. 

%Since $y = x_1$ is $<$-largest, and $\text{in}_< I$ is a squarefree monomial ideal by assumption, the reduced Gr\"obner basis of $I$ with respect to $<$ is squarefree in $y$, and so has the form $\mathcal{G} = \{yq_1+r_1,\dots, yq_k+r_k,h_1,\dots, h_\ell\}$, with  $y$ not dividing any term of any $q_i, r_i, h_j$, $1\leq i\leq k$, $1\leq j\leq \ell$. So, by \cite[Theorem 2.1 (b)]{KMY09}, $\text{in}_{y}I = C_{y,I}\cap (N_{y,I}+\langle y\rangle)$, is a geometric vertex decomposition. Again, we have two cases.

%If $C_{y,I} = \langle 1 \rangle$, then $\text{in}_<I = \text{in}_< N_{y,I}+\langle y\rangle$. The complex $\Delta(N_{y,I}+\langle y\rangle)$ is the same as $\Delta(N^c)$. Thus, $\Delta(N^c)$ is $<$-compatibly vertex decomposable. So, by induction, $N^c$ is $<$-compatibly geometrically vertex decomposable. Hence, so too is $I$.

% If $C_{y,I} \neq \langle 1 \rangle$, then we have that $1$ is a vertex of $\Delta$ and, as discussed above, $\text{in}_<C_{y,I}+\langle y\rangle$ and $\text{in}_<N_{y,I}+\langle y\rangle$ are the Stanley--Reisner ideals of $\text{lk}_\Delta(1)$ and $\text{del}_\Delta(1)$. 
%Since $\Delta$ is $<$-compatibly vertex decomposable, so are $\text{lk}_\Delta(1)$ and $\text{del}_\Delta(1)$. Thinking of these complexes as complexes on vertex set $\{2,3,\dots, n\}$, their Stanley--Reisner ideals are $\text{in}_<C^c$ and $\text{in}_<N^c$. So, by induction, $C^c$ and $N^c$ are $<$-compatibly geometrically vertex decomposable. Hence, so too is $I$.
%\end{proof}


The next example shows that there exist geometrically vertex decomposable ideals that are \emph{not} geometrically vertex decomposable compatible with any lexicographic monomial order. 

\begin{example}\label{ex:nolex}
Let $I = \langle y(zs-x^2), ywr, wr(z^2+zx+wr+s^2) \rangle \subseteq \kappa[x,y,z,w,r,s]$. 
Observe that $I$ is squarefree in $y$, and we have a geometric vertex decomposition with $C_{y,I} = \langle zs-x^2,wr \rangle$ and $N_{y,I} = \langle (wr)(zx+s^2+z^2+wr) \rangle$. 
Furthermore, the contractions of $C_{y,I}$ and $N_{y,I}$ to $\kappa[x,z,w,r,s]$ are geometrically vertex decomposable. (To see this, let $C^c$ and $N^c$ denote these contracted ideals. Then $C^c$ and $N^c$ are squarefree in $s$ and $x$, respectively, and $\text{in}_s C^c = \langle zs, wr\rangle$ and $\text{in}_x N^c = \langle wrzx\rangle$.) Hence $I$ is geometrically vertex decomposable.  Yet one readily checks that $I$ has no squarefree initial ideals, so cannot be $<$-compatibly geometrically vertex decomposable for any order $<$ by Theorem \ref{prop:simplicial}. 

\end{example}



\emph{Gorenstein Liaison.} Here we review standard definitions and lemmas in Gorenstein liaison that we will need in this extended abstract. We restrict to an understanding of Gorenstein liaison as it arises in the context of Gr\"obner bases and as it is required to understand Question \ref{motivation2} for our study in combinatorial settings.  For a more thorough introduction, see \cite{MN01}.  
We follow definitions and some notation from \cite{GMN13}, which provides a careful discussion of how liaison theory can be used to make inferences about Gr\"obner bases. Throughout this subsection, we let $R = \kappa[x_0, x_1,\dots, x_n]$ with the standard grading.

\begin{definition}
Let $V_1, V_2, X \subseteq \mathbb{P}^{n}$ be subschemes defined by saturated ideals $I_{V_1}$, $I_{V_2}$, and $I_{X}$ of $R$, respectively, and assume that $X$ is arithmetically Gorenstein.  If $I_X \subseteq I_{V_1} \cap I_{V_2}$ and if $[I_X:I_{V_1}] = I_{V_2}$ and $[I_X: I_{V_2}] = I_{V_1}$, then $V_1$ and $V_2$ are \textbf{directly algebraically $G$-linked} by $X$, and we write $I_{V_1} \sim I_{V_2}$.
\end{definition}

One may generate an equivalence relation using these direct links.

\begin{definition} If there is a sequence of links $V_1 \sim \cdots \sim V_k$ for some $k \geq 2$, then we say that $V_1$ and $V_k$ are in the same \textbf{$G$-liaison class (or Gorenstein liaison class)} and that they are \textbf{$G$-linked} in $k-1$ steps.  Of particular interest is the case in which $V_k$ is a complete intersection, in which case we say that $V_1$ is \textbf{in the Gorenstein liaison class of a complete intersection (abbreviated glicci)}.
\end{definition}


We will say that a homogeneous, saturated, unmixed ideal of $R$ is glicci if it defines a glicci subscheme of $\mathbb{P}^{n}$.  It is because liaison was developed to study subschemes of projective space that the restriction to homogeneous, saturated ideals is natural.  Throughout this extended abstract, we will be interested in $G$-links coming from \emph{elementary $G$-biliaisons}. Indeed, it is through elementary $G$-biliaisons that we connect geometric vertex decomposition to liaison theory.  Let $S$ be a ring.  If $S_P$ is Gorenstein for all prime ideals $P$ of height $0$, then we say that $S$ is $\mathbf{G_0}$.  

\begin{definition}\label{def:Gbiliaison}
Let $I$ and $C$ be homogeneous, saturated, unmixed ideals of $R $ with $\hgt(I) = \hgt(C)$.  Suppose there exist $\ell \in \mathbb{Z} $, a homogeneous Cohen--Macaulay ideal $N \subseteq I \cap C$ of height $\hgt(I)-1$, and an isomorphism $I/N \cong [C/N](-\ell)$ as graded $R/N$-modules.  If $N$ is $G_0$, then we say that $I$ is obtained from $C$ by an \textbf{elementary $G$-biliaison of height $\ell$}.  
\end{definition}

\begin{theorem}[\protect{\cite[Theorem 3.5]{Har07}}]
Let $I$ and $C$ be homogeneous, saturated, unmixed ideals defining subschemes $V_I$ and $V_C$, respectively, of $\mathbb{P}^{n}$.  If $I$ is obtained from $C$ by an elementary $G$-biliaison, then $V_I$ is $G$-linked to $V_C$ in two steps.
\end{theorem}


%\begin{remark}
%\textbf{Even $G$-liaison classes} are equivalence classes of subschemes of $\mathbb{P}^{n}$ of a fixed codimension that are $G$-linked to one another in an even number of steps.  
%Two subschemes in the same \emph{even} $G$-liaison class are more closely related to one another than are two subschemes that can be linked to one another but only in an odd number of steps (see \cite[Section 3]{Nag98}). 
%This provides some intuition for why various classes of generalized determinantal varieties can be linked to one another in an even number of steps, via elementary $G$-biliaisons, yet there is no reason to expect that the intermediate links share a similar form, or are at all easy to describe.
%\end{remark}


%\subsection{Further context on a question in liaison theory}

%The purpose of this subsection is to recall the motivation for the following question:  
%\begin{quest}\cite[Question 1.6]{KM+01}
%Is every arithmetically Cohen--Macaulay subscheme of $\mathbb{P}^n$ glicci?
%\end{quest} 

%Because all complete intersections of a fixed codimension are in the same liaison class, an equivalent formulation of the question is 

%\begin{quest}
%For each codimension, is there exactly one Gorenstein liaison class containing any (or all) Cohen--Macaulay subschemes of $\mathbb{P}^n$?
%\end{quest}

% This question arises by analogy to the special case of complete intersection liaison in codimension 2, where all arithmetically Cohen--Macaulay subschemes are in the (complete intersection and so also Gorenstein) liaison class of a complete intersection \cite[Theorem 3.2]{PS74}.  The same is not true in higher codimensions, however.  In fact, in higher codimensions there are infinitely many complete intersection liaison classes containing arithmetically Cohen--Macaulay schemes \cite[Chapter 7]{KM+01}.  Complete intersection liaison is a well-understood and very satisfying theory in codimension 2, and this failure to generalize to higher codimensions suggests that it is worth searching for a theory that reduces to complete intersection liaison in codimension 2 and also preserves many of its desirable properties in higher codimension.  This is one of the motivations for studying Gorenstein liaison, where better control of the liaison classes containing arithmetically Cohen--Macaulay schemes may still be hoped for in all codimensions.  In particular, an affirmative answer to Question \ref{motivation2} would serve, at least in the eyes of some, as an endorsement of the structure of Gorenstein liaison.

%There are partial results in the direction of an affirmative answer to Question \ref{motivation2}, including the results that standard determinantal schemes \cite[Theorem 1.1]{KM+01}, mixed ladder determinantal schemes from two-sided ladders \cite[Corollary 2.2]{Gor07}, schemes of pfaffians \cite[Theorem 2.3]{DG09}, wide classes of arithmetically Cohen--Macaulay curves in $\mathbb{P}^4$ \cite{CMR00, CMR01}, and arithmetically Cohen-Macaulay schemes defined by Borel-fixed monomial ideals \cite[Theorem 3.5]{MN02} are all glicci.  For more results, see \cite{Har02, Cas03, CDH05, HSS08, Gor08, MN13}.  

%However, the general question remains open.  By connecting geometric vertex decomposition and liaison, we provide more evidence in favor of an affirmative answer to this question and give a framework for assessing membership in the Gorenstein liaison class of a complete intersection for some arithmetically Cohen--Macaulay schemes arising naturally from combinatorial data.


\section{Geometrically vertex decomposable ideals are glicci}\label{sect:GVDGlicci}

Throughout this section, we assume that the field $\kappa$ is infinite, and we let $R$ denote the standard graded polynomial ring $\kappa[x_1, \ldots, x_n]$. 

\emph{An elementary $G$-biliaison arising from a geometric vertex decomposition.}  Our first result uses a geometric vertex decomposition to construct an isomorphism that will constitute an elementary $G$-biliaison when the setting is appropriate.


\begin{theorem}\label{thm:onestep}
Suppose that $I \subseteq R$ is an unmixed ideal possessing a nondegenerate geometric vertex decomposition with respect to some variable $y = x_j$ of $R$.  If $N_{y,I} $ is unmixed, then there is an isomorphism $I/N_{y,I} \cong C_{y,I}/N_{y,I} $ as $R/N_{y,I}$-modules.  If $N_{y,I}$, $C_{y,I}$, and $I$ are homogeneous, then the same map is an isomorphism $I/N_{y,I} \cong [C_{y,I}/N_{y,I} ](-1)$ in the category of graded $R/N_{y,I}$-modules.
\end{theorem}
%\begin{proof}
%Fix a $y$-compatible term order $<$.  From Lemma \ref{lem:form}, we know that the reduced Gr\"obner basis of $I$ has the form $\mathcal{G} = \{yq_1+r_1,\dots, yq_k+r_k,h_1,\dots, h_\ell\}$ where $y$ does not divide any term of $q_i$ or of $r_i$ for any $1 \leq i \leq k$ nor any $h_j$ for $1 \leq j \leq \ell$.   Let $C = C_{y,I} = \langle q_1,\dots, q_k, h_1,\dots, h_\ell\rangle$ and $N = N_{y,I} = \langle h_1,\dots, h_\ell \rangle$. 

%We first observe that $N\subseteq I \cap C$.  To build the desired isomorphism, we will need to find regular elements of $R/N$.  Towards that end, we claim that $\langle q_1, \ldots, q_k \rangle \not\subseteq Q$ for any minimal prime $Q$ of $N$.  If it were, then we would also have $C \subseteq Q$, in violation of Lemma \ref{lem:height}.  Similarly, $\langle yq_1+r_1, \ldots, yq_k+r_k \rangle \not\subseteq Q'$ for any minimal prime $Q'$ of $N$ since, if it were, then we would have $I \subseteq Q'$, in violation of Lemma \ref{lem:height}.  Because $\kappa$ is infinite, we may choose scalars $a_1, \ldots, a_k \in \kappa$ so that neither $u:=a_1q_1+\cdots+a_kq_k$ nor $v:=a_1(yq_1+r_1)+ \ldots+ a_k(yq_k+r_k)$ is an element of any minimal prime of $N$.  Because $\min(N) = \ass(N)$, neither $u$ nor $v$ is a zero-divisor on $R/N$.  

%We may now define a map $\phi_u: C \rightarrow I/N[1/u]$ given by $f \mapsto \dfrac{v}{u}\cdot \dfrac{f}{1}$.  We claim that the image of $\phi_u$ is identically $I/N = I/N[1/u] \cap R/N$, which will allow us to view $\phi_u$ as a map $\phi: C \rightarrow I/N$.  To see that the image of $\phi_u$ is $I/N$, we will show that $\phi_u(q_i) = \dfrac{yq_i+r_i}{1} \in I/N[1/u]$ for all $1 \leq i \leq k$.  It is sufficient to show that each $(yq_i+r_i)u-q_iv \in N$ as elements of $R$.  Indeed, \begin{align*}
%(yq_i+r_i)u-q_iv = &(yq_i+r_i)(a_1q_1+\cdots+a_kq_k) - q_i(a_1(yq_1+r_1)+\cdots+a_k(yq_k+r_k)) \\
%= &r_i(a_1q_1+\cdots+a_kq_k)-q_i(a_1r_1+\cdots+a_kr_k).
%\end{align*}
%Because $yq_i+r_i \in I$ and $v \in I$, we have $r_i(a_1q_1+\cdots+a_kq_k)-q_i(a_1r_1+\cdots+a_kr_k) \in I$.  But $y$ does not divide any term of any $q_j$ or any $r_j$ for $1 \leq j \leq k$, and so the leading term of $r_i(a_1q_1+\cdots+a_kq_k)-q_i(a_1r_1+\cdots+a_kr_k)$ is not divisible by the leading term of any $yq_i+r_i$.  By the assumptions that $\mathcal{G}$ is a Gr\"obner basis of $I$ and that $<$ is $y$-compatible, it must be that $r_i(a_1q_1+\cdots+a_kq_k)-q_i(a_1r_1+\cdots+a_kr_k)$ has a Gr\"obner basis reduction using only the elements of $\mathcal{G}$ not involving $y$, i.e., using only $h_1,\dots, h_\ell $, which implies that $r_i(a_1q_1+\cdots+a_kq_k)-q_i(a_1r_1+\cdots+a_kr_k) \in N$.  Hence, $\phi_u(q_i) = \dfrac{yq_i+r_i}{1} \in I/N[1/u]$ for each $1 \leq i \leq k$.

%A similar computation will show that $\phi_u(h_i) = \dfrac{yh_i}{1} \in I/N[1/u]$ for each $1 \leq i \leq \ell$.  As before, it suffices to show that $yh_iu-h_iv \in N$ as elements of $R$ for every $1 \leq i \leq \ell$: \begin{align*}
%(yh_i)u-h_iv = &(yh_i)(a_1q_1+\cdots+a_kq_k) - h_i(a_1(yq_1+r_1)+\cdots+a_k(yq_k+r_k)) \\
%= &-h_i(a_1r_1+\cdots+a_kr_k).  
%\end{align*}  Again, $h_i, v \in I$ implies $-h_i(a_1r_1+\cdots+a_kr_k) \in I$, and then because no term of $-h_i(a_1r_1+\cdots+a_kr_k)$ is divisible by $y$, we have $-h_i(a_1r_1+\cdots+a_kr_k) \in N$, as desired.

%Because $\mathcal{G}$ is a generating set for $I$ and $\{h_1,\dots, h_\ell\}$ is a generating set for $N$, we see that $\{yq_1+r_1,\dots, yq_k+r_k\}$ generates $I$ over $N$, and so we may let $\phi:C \rightarrow I/N$ be the map given by $\phi(c) = \phi_u(c)$ for every $c \in C$, i.e., $\phi$ is given by multiplication by $\dfrac{u}{v}$, which our computations have shown to be a surjection.  Having established that $\phi(h_i) = 0 \in I/N$, we have $N \subseteq \ker(\phi)$.  And $\ker(\phi) \subseteq N$ because $u$ is a non-zero-divisor on $R/N$.  Therefore, $\phi$ induces an isomorphism $\overline{\phi}: C/N \rightarrow I/N$.  

%It is clear that whenever $N$ is homogeneous so that discussion of degrees makes sense, $\overline{\phi}$ increases degree by $1$ and so  $\overline{\phi}: [C/N](-1) \rightarrow I/N$ is an isomorphism of graded $R/N$-modules.  \end{proof}

%Notice that if, in the proof above, one already knows $q_1$ and $yq_1+r_1$ to be non-zero-divisors on $R/N$, for example if $R/N$ is a domain, one may choose $a_1 = 1$ and $a_i = 0$ for $1<i \leq k$.  In this case, the map $\phi$ will be of the same form used in \cite{Gor07}, \cite{Gor08}, and \cite{GMN13}.

%As indicated above, the primary use of Theorem \ref{thm:onestep} is in the setting of liaison theory (Corollary \ref{cor:gvdToLia}). We will need the following straightforward fact about saturation:

%\begin{lemma}\label{lem:saturated}
%If $I \subseteq R$ is homogeneous and unmixed, then $\sqrt{I}$ is the homogeneous maximal ideal $m$ or $I$ is saturated.  
%\end{lemma}
%\begin{proof}

%Observe that $m$ is an associated prime of $I$ if and only if it is a minimal prime of $I$ if and only if $\sqrt{I} = m$.  
%\end{proof}

\begin{corollary}\label{cor:gvdToLia}
Let $I$ be a homogeneous, saturated, unmixed ideal of $R$ and $\text{in}_y I = C_{y,I} \cap (N_{y,I}+\langle y \rangle)$ a nondegenerate geometric vertex decomposition with respect to some variable $y = x_j$ of $R$.  Assume that $N_{y,I}$ is Cohen--Macaulay and $G_0$ and that $C_{y,I}$ is also unmixed.  Then $I$ is obtained from $C_{y,I}$ by an elementary $G$-biliaison of height $1$.
\end{corollary}

%\begin{proof}
%The height conditions required by the definition of elementary $G$-biliaison are given by Lemma \ref{lem:height}, saturation follows from Lemma \ref{lem:saturated}, and the required isomorphism is constructed in Theorem \ref{thm:onestep}.
%\end{proof}


\begin{comment}

One may sometimes wish that it were true that the homogenization of a Gr\"obner basis of an ideal $I$ in general gave the Gr\"obner basis of the homogenization of $I$.  That equality does not hold in general.  However, we will use Corollary \ref{cor:Gb(I)} to show that in context of geometric vertex decomposition, we do have that convenience.  

Fix a variable $y = x_j$ of $R$, and consider the weight vector $w$ of length $n+1$ whose $i^{th}$ and $n+1^{st}$ coordinates are $1$ and whose other coordinates are $0$.  Understand this weight vector to be assigning $y$ and a new variable $t$ of $R[t]$ weights $1$ and each $x_j$ a weight of $0$ when $i \neq j$.  For an element $f \in R$, let $f^h$ denote its homogenization with respect to $w$ and $I^h = \langle f^h \mid f \in I \rangle$ the ideal generated by the homogenizations of elements of $I$.  We first notice that homogenizations of elements of a Gr\"obner basis of $I$ are generators of $I^h$: Fix a $y$-compatible term order $<$, and let $\mathcal{G}$ denote the reduced Gr\"obner basis of $I$ with respect to $<$, and let $\mathcal{G}^h$ denote the set of homogenizations of elements of $\mathcal{G}$.  Because $I^h$ is generated by the homogenizations of elements of $I$, it is sufficient to show that $f^h \in \langle \mathcal{G}^h \rangle$ for every $f \in I$.  According to Lemma \ref{lem:form}, $\mathcal{G} = \{yq_1+r_1, \ldots, yq_k+r_k, h_1, \ldots, h_k\}$ with $y$ not dividing any term of any $q_i$ or $r_i$ for any $1 \leq i \leq k$ nor any $h_j$ for $1 \leq j \leq \ell$.  Using the deterministic division algorithm (see \cite[Section 15.4]{Eis95}), we may express $f = (\sum_{i = 1}^k \alpha_i(yq_i+r_i))+(\sum_{j = 1}^\ell(\beta_j h_j))$ for some $\alpha_i, \beta_j \in R$ for $1 \leq i \leq k$ and $1 \leq j \leq \ell$ where $\text{in}_<(\beta_j h_j) \notin \langle \{ \text{in}_< \alpha_i(yq_i+r_i) \}_{1 \leq i \leq k} \rangle$ for each $1 \leq j \leq \ell$.  In particular, no term of any $\beta_j h_j$ is divisible by $y$.  Now write each $\alpha_i = \sum_{p = 0}^\infty\gamma_{ip}$ where $\gamma_{ip}$ is divisible by $y^p$ and is not divisible by $y^{p+1}$.  (Of course, all but finitely many of the $\gamma_{ip}$ are $0$, and each such sum is actually finite.)  Then \begin{align*}
f^h &= \left(\sum_{i = 1}^k \left( \sum_{p = 0}^\infty \gamma_{ip}\right) \left(yq_i+r_i\right)\right)^h+\left(\sum_{j = 1}^\ell \left(\beta_j h_j\right)\right)\\
& = \left(\sum_{p = 0}^\infty \sum_{i = 1}^k \gamma_{ip} \left(yq_i+r_i\right)\right)^h+\left(\sum_{j = 1}^\ell \left(\beta_j h_j\right)\right)\\
& = \left(\sum_{p = 0}^\infty \sum_{i = 1}^k \gamma_{ip} \left(yq_i+tr_i\right)\right)+\left(\sum_{j = 1}^\ell \left(\beta_j h_j\right)\right) \in \langle \mathcal{G}^h \rangle
\end{align*} because each $yq_i+tr_i$ and each $h_j$ is an element of $\mathcal{G}^h$ for each $1 \leq i \leq k$ and $1 \leq j \leq \ell$.

Now that we see that $\mathcal{G}^h$ is at least a generating set for $I^h$, we are now ready to use Corollary \ref{cor:Gb(I)} to show that if $I$ is a homogeneous ideal of $R$ (in the standard grading), then $\mathcal{G}^h$ is a Gr\"obner basis of $I^h$.

\begin{corollary}\label{cor:homogGB}
Let $I$ be an unmixed homogeneous ideal of $R$ that admits a geometric vertex decomposition with respect to some variable $y$ of $R$, and let $\mathcal{G}$ be the reduced Gr\"obner basis of $I$ wth respect to the $y$-compatible term order $<$.  Assume that $N_{y,I}$ has no embedded primes.  Then $\mathcal{G}^h = \{g^h \mid g \in \mathcal{G} \}$ is a Gr\"obner basis of $I^h$ under any $y$-compatible term order on $R[t]$ refining $<$.
\end{corollary}
{\color{purple} We're not actually using unmixedness here.  It's just that the height lemma in the mixed case comes later.  We could either move that lemma up, cite it here even though it comes later in the paper, or just leave things how they are (if we want to keep the corollary at all, that is).}
\begin{proof}
The degenerate cases are clear, and so we assume that the geometric vertex decomposition of $I$ with respect to $y$ is nondegenerate.  For convenience, we consider $I$ as an ideal of $R[t]$ and will refer to the term order of $R[t]$ that refines $<$ also as $<$.  By considering $s$-pair reductions, it is clear that $\mathcal{G}$ remains a Gr\"obner basis of $I$ in $R[t]$.  By Lemma \ref{lem:form}, we now $\mathcal{G} = \{yq_1+r_1, \ldots, yq_k+r_k, h_1,\ldots,h_\ell\}$, where $y$ does not divide $q_i, r_i$ for any $1 \leq i \leq k$ nor any $h_j$ for any $1 \leq j \leq \ell$.  By Lemma \ref{lem:height} $\hgt(I), \hgt(C_{y,I})>\hgt(N_{y,I})$.  Now $I^h = \langle yq_1+tr_1, \ldots, yq_k+tr_k, h_1, \ldots, h_\ell \rangle$.  Then because $C_{y,I} =\langle yq_1, \ldots, yq_k , h_1, \cdots h_\ell \rangle  =C_{y,I^h}$ and $N_{y,I} = \langle h_1, \ldots, h_\ell \rangle = N_{y,I^h}$, and the given generating sets of both ideals are Gr\"obner bases by \cite[Theorem 2.1(a)]{KMY09}, we are in the setting of Corollary \ref{cor:Gb(I)}.

Let $M = \begin{pmatrix} 
q_1 & \cdots & q_k\\
tr_1 & \cdots & tr_k
\end{pmatrix}$.  It remains to show that $I_2(M) \subseteq N_{y,I}$.  If $1 \leq i<j \leq k$, then \[
 q_i(tr_j)-tr_i(q_j) = t(q_j(yq_i+r_i)-q_i(yq_j+r_j)) \in I
\] because $yq_i+r_i$ and $yq_j+r_j$ are elements of $I$.  But $y$ does not divide any term of $q_itr_j-tr_iq_j$, and so, by the assumption that $\mathcal{G}$ is a Gr\"obner basis of $I$, $q_itr_j-tr_iq_j$ has a reduction in terms of the given generators of $N_{y,I}$, and so $I_2(M) \subseteq N_{y,I}$.  It now follows from Corollary \ref{cor:Gb(I)} that $\mathcal{G}^h$ is a Gr\"obner basis of $I^h$.  
\end{proof}

Notice that the above result shows that $\text{in}_y I = \text{in}_y I^h$ and that $\text{in}_y I +\langle t \rangle = I^h + \langle t \rangle$, from which it follows that $t$ is a non-zero-divisor on $I^h$.  

\end{comment}

\begin{comment}\begin{theorem}
If the homogeneous ideal $I$ of the standard graded polynomial ring $R = \kappa[x_1,\ldots,x_n,y]$ admits a geometric vertex decomposition with respect to $y$, then $I$ is Cohen--Macaulay if and only if $\text{in}_y I$ is Cohen--Macaulay.
\end{theorem}

\begin{proof}
Let $<$ denote the lexicographic order with respect to $y>x_1>\ldots>x_n$.  It is always the case that $\text{in}_y I$ Cohen--Macaulay implies $I$ Cohen--Macaulay, so we only need to show the other implication, and so we assume that $I$ is Cohen--Macaulay.  If $m$ is the homogeneous maximal ideal of $R$, because $I$ and so $\text{in}_y I$ are homogeneous, it suffices to show that $(R/\text{in}_y I)_m$ is Cohen--Macaulay.  If  $\dim(R/I) = 0$, the result is clear, and so we may fix take $d = \hgt(I)<n+1$ to be arbitrary and proceed by induction on $n+1-d$.  Note also that the result is clear if $n = 0$, and so we will assume throughout that $n \geq 1$.

Let $S = R[t]$, and let $\mu$ denote the maximal ideal $\langle x_1, \ldots, x_n, t \rangle$ of $S$, and let $\prec$ denote the lexicographic order on $S$ with respect to $y>t>x_1>\ldots>x_n$.  By Lemma \ref{lem:form}, the reduced Gr\"obner basis of $I$ with respect to $<$ (and so also $\prec$) has the form $\mathcal{G}_I = \{yq_1+r_1, \ldots, yq_k+r_k, h_1, \ldots, h_\ell\}$.  According to Corollary \ref{cor:homogGB}, $\mathcal{G}_{I^h} = \{yq_1+tr_1, \ldots, yq_k+tr_k, h_1, \ldots, h_\ell \}$ is a Gr\"obner basis of $I^h$ with respect to $\prec$.  Notice that $\text{in}_y I = \text{in}_y I^h$ and so that $\hgt(I) = \hgt(\text{in}_y I) = \hgt(\text{in}_y I^h) = \hgt(I^h)$.  Notice also that $I^h+\langle t \rangle = \text{in}_y I + \langle t \rangle$.  Because $t$ does not divide any term of any elements of the reduced Gr\"obner basis of $\text{in}_y I$, we see \[
\hgt(I^h+\langle t \rangle) = \hgt(\text{in}_y I +\langle t \rangle) = \hgt(\text{in}_y I) + 1 = \hgt(I^h)+1.
\] Hence, $t$ is not in any minimal prime of $I^h$.  Because $I$ is Cohen--Macaulay and so unmixed, $\text{in}_y I = \text{in}_y I^h$ is unmixed, and so $I^h$ is unmixed, and $\min(I^h) = \ass(I^h)$ and $t$ is not a zero-divisor on $I^h$.  It follows that $(S/I^h)_\mu$ is Cohen--Macaulay if and only if $(S/I^h+\langle t \rangle)_\mu \cong (R/\text{in}_y I)_m$ is Cohen--Macaulay.  

We will now show that $(S/I^h)_\mu$ is Cohen--Macaulay.  From now on, we will consider $I$, $I^h$, and $\text{in}_y I$ all as ideals of $S$.  Because both $I$, $I^h$, and $\text{in}_y I$ are unmixed and of height $d<n+1$, the ideal $(x_1, \ldots, x_n,t)$ is not associated to either $I$, $I^h$, or $\text{in}_y I$. Hence, using that $|\kappa| = \infty$, for a sufficiently general choice of $(\alpha_1, \ldots, \alpha_n, \beta) \in \kappa^{n+1}$, $f = \alpha_1 x_1+\cdots+\alpha_n x_n + \beta t$ avoids the minimal primes of $I$, $I^h$, and $\text{in}_y I$ and so is a non-zero-divisor on $(S/I)_n$, $(S/I^h)_n$, and $(S/\text{in}_y I)_n$.  Because $\beta \neq 0$ is an open condition, we may also assume $\beta \neq 0$.  Set $J = I^h+\langle f \rangle$.  We claim that $\mathcal{G}_J = \{ yq_1+tr_1, \ldots, yq_k+tr_k, h_1, \ldots, h_\ell, f \}$ is a Gr\"obner basis of $J$.  Because $\{ yq_1+tr_1, \ldots, yq_k+tr_k, h_1, \ldots, h_\ell\}$ is a Gr\"obner basis of $I^h$, it suffices to observe that because the leading term of $f$ is $\beta t$, the leading terms of $yq_i+tr_i$ and $f$ have no variables in common for any $1 \leq i \leq k$ and the leading terms of $h_j$ and $f$ have no variables in common for any $1 \leq j \leq \ell$.  Similarly, $\{yq_1+r_1, \ldots, yq_k+r_k, h_1, \ldots, h_\ell, f\}$ is a Gr\"obner basis of $I+\langle f \rangle$.  Hence, $\text{in}_y J = \langle yq_1, \ldots, yq_k, h_1, \ldots, h_\ell, f \rangle = \text{in}_y (I+\langle f \rangle)$, all of which are equal to $\text{in}_y I+\langle f \rangle$.  

Because $(S/I)_n$ is Cohen--Macaulay and $f \in n$ is a non-zero-divisor on $(S/I+\langle f \rangle)_n$ is Cohen--Macaulay.  Using homogeneity of $I+\langle f \rangle$, we have that $I+\langle f \rangle$ is a Cohen--Macaulay ideal of $S$ of height $d+1$.  By induction, $\text{in}_y (I+\langle f \rangle)$ is also Cohen--Macaulay, and so 
\end{proof}
\end{comment}

\begin{theorem}\label{thm:glicci}
If $I = I_0 \subseteq R$ is a homogeneous, geometrically vertex decomposable proper ideal, then there is a finite sequence of homogeneous, saturated, unmixed ideals $I_1, \ldots, I_t$ so that $I_{j-1}$ is obtained from $I_{j}$ by an elementary $G$-biliaison of height $1$ for every $1 \leq j \leq t$ and $I_t$ is a complete intersection.  In particular, $I$ is glicci and, hence, Cohen--Macaulay.
\end{theorem}

%\begin{proof} 
%Clearly, it suffices to prove the first claim.  We will proceed by induction on $n = \dim(R)$, noting that the case of a dimension $0$ polynomial ring is trivial.

%We now take $n \geq 1$ to be arbitrary and assume the result for all proper homogeneous ideals $I$ in polynomial rings of dimension $<n$.  If $I$ is a complete intersection, then there is nothing to prove. Otherwise, there exists some variable $y = x_i$ of $R$ for which $\text{in}_y I =  C_{y,I}  \cap (N_{y,I} +\langle y\rangle)$ is a geometric vertex decomposition with the contraction of $N_{y,I} $ and $C_{y,I}$ to $T = \kappa[x_1,\ldots, \hat{y}, \ldots, x_n]$ geometrically vertex decomposable.  

%Suppose first that $ C_{y,I}  = \langle 1 \rangle$, in which case $I = N_{y,I} +\langle y\rangle$.  By induction, with $\tilde{I}_0 = N_{y,I} \cap T$, there is a sequence of ideals $\tilde{I}_1, \ldots, \tilde{I}_t$ of $T$ so that $\tilde{I}_{j-1}$ is obtained from $\tilde{I}_{j}$ by an elementary $G$-biliaison of height $1$ for every $1 \leq j \leq t$ and $\tilde{I}_t$ is a complete intersection.  Setting $I_j = \widetilde{I}_jR+\langle y \rangle$ for every $1 \leq j \leq t$, the result follows from Observation \eqref{Observation2}, above.  Similarly, the result is essentially immediate from the inductive hypothesis together with Observation \eqref{Observation1} in the case $I = C_{y,I} = N_{y,I}$.

%Finally, assume the geometric vertex decomposition with respect to $I$ is nondegenerate, in which case we may apply the inductive hypothesis to $\tilde{I}_1 = C_{y,I} \cap T$.  By induction and in parallel with the previous case, there is a finite sequence of homogeneous, saturated, unmixed ideals $\widetilde{I}_2, \ldots, \widetilde{I}_t$ of $T$ so that $\widetilde{I}_{j-1}$ is obtained from $\widetilde{I}_j$ by an elementary $G$-biliaison of height $1$ in $T$ for every $2 \leq j \leq t$ and $\widetilde{I}_t$ is a complete intersection.  Let $I_j = \widetilde{I}_jR$ for every $2 \leq j \leq t$.  Then with $I_1 = C_{y,I}$, by Observation \eqref{Observation1} above, $I_{j-1}$ is obtained from $I_{j}$ by an elementary $G$-biliaison of height $1$ in $R$ for every $2 \leq j \leq t$ and $I_t$ is a complete intersection.  Hence, it suffices to show that $I$ is obtained from $C_{y,I}$ by an elementary $G$-biliaison of height $1$, but this is Corollary \ref{cor:gvdToLia}. Note that the hypotheses of Corollary \ref{cor:gvdToLia} hold since $C_{y,I}\cap T$ and $N_{y,I}\cap T$ are geometrically vertex decomposable (hence unmixed and radical) and glicci (hence Cohen-Macaulay) by induction.
%\end{proof}

%\begin{corollary}\label{cor:CM}
%If $I \subseteq R$ is a homogeneous, geometrically vertex decomposable proper ideal, then $I$ is Cohen--Macaulay.
%\end{corollary}
%\begin{proof}
%Since glicci implies Cohen--Macaulay, this is immediate from Theorem \ref{thm:glicci}.
%\end{proof}


\emph{Applications to Gr\"obner bases and degenerations} One cannot in general transfer the Cohen--Macaulay property from an ideal to its initial ideal or from one component of a variety to the whole variety.  However, in the context of geometric vertex decomposition, we can use the combination of Cohen--Macaulayness of a homogeneous ideal $I$ and of the component $N_{y,I}+\langle y \rangle$ (equivalently, of $N_{y,I}$) to infer the same about $\text{in}_y I$.  


\begin{corollary}
Suppose that $\text{in}_y I = C_{y,I} \cap (N_{y,I}+\langle y \rangle)$ is a nondegenerate geometric vertex decomposition of the homogeneous ideal $I\subseteq R$ and that both $N_{y,I}$ and $I$ are Cohen--Macaulay. Then, $C_{y,I}$ and $\text{in}_y I$ are Cohen--Macaulay as well.  
\end{corollary}

%\begin{proof}
%For convenience, write $N = N_{y,I}$ and $C = C_{y,I}$.  Because $I$ is Cohen--Macaulay, it is unmixed, and because $N_{y,I}$ is Cohen--Macaulay, it has no embedded primes.  Hence, we may apply Theorem \ref{thm:onestep} to see that $I/N \cong C/N$.  It is easy to see that $C/N \xrightarrow{y} \text{in}_y I/N$ is also an isomorphism, and so $I/N \cong \text{in}_y I/N$.  Let $m$ denote the homogeneous maximal ideal of $R$, and let $H^i_m(M)$ denote the $i^{th}$ local cohomology module of the $R$-module $M$ with support in $m$.  Because $I$ homogeneous implies $\text{in}_y I$ homogeneous, it is sufficient to check Cohen--Macaulayness at $m$.  Let $d = \dim(R/I) = \dim(R/N)-1$.  The short exact sequence $0 \rightarrow I/N \rightarrow R/N \rightarrow R/I \rightarrow 0$ tells us that $H^i_m(I/N) \cong H^{i-1}_m(R/I) = 0$ for all $i \leq d$ because $H^i_m(R/N) = 0$ for all $i \leq d$.  Then from the short exact sequence $0 \rightarrow \text{in}_y I/N \rightarrow R/N \rightarrow R/\text{in}_y I \rightarrow 0$ together with the fact that $\text{in}_y I/N \cong C/N$, we have \[
%H^{i-1}_m(R/\text{in}_y I) \cong H^i_m(\text{in}_y I/N) \cong H^i_m(I/N) \cong H^{i-1}_m(R/I) = 0
%\] for all $i-1<d = \dim(R/\text{in}_y(I))$, and so $R/\text{in}_y(I)$ is Cohen--Macaulay.  

%The argument in the case of $C$ follows the same line using the short exact sequence $0 \rightarrow C/N \rightarrow R/N \rightarrow R/C \rightarrow 0$.  
%\end{proof}


We will now describe conditions that allow one to use the map constructed in Theorem \ref{thm:onestep} in order to conclude that a known set of generators for $I$ forms a Gr\"obner basis when Gr\"obner bases for $C_{y,I}$ and $N_{y,I}$ are known.  The result complements the framework of \cite{KMY09}, in which one begins with a Gr\"obner basis of $I$ and concludes that the resultant generating sets of $C_{y,I}$ and $N_{y,I}$ are also Gr\"obner bases.  The corollary below gives a way to implement the approach of \cite[Lemma 1.12]{GMN13}.

\begin{corollary}\label{cor:Gb(I)}
Let $I = \langle yq_1+r_1,\dots, yq_k+r_k,h_1,\dots, h_\ell \rangle$ be a homogenous ideal of  $R$ with $y = x_j$ some variable of $R$ and $y$ not dividing any term of any $q_i$ for $1 \leq i \leq k$ nor of any $h_j$ for $1 \leq j \leq \ell$. Fix a term order $<$, and suppose that $\mathcal{G}_C = \{q_1,\dots, q_k,h_1,\dots, h_\ell\}$ and $\mathcal{G}_N = \{h_1,\dots, h_\ell\}$ are Gr\"obner bases for the ideals they generate, which we call $C$ and $N$, respectively.  Assume also that the leading term of each $yq_i+r_i$ is $yq_i$ for all $1 \leq i \leq k$, that $\hgt(I)$, $\hgt(C)>\hgt(N)$, and that $N$ is unmixed.  Let $M = \begin{pmatrix}
q_1& \cdots & q_k\\
r_1& \cdots & r_k
\end{pmatrix}.$ If the ideal of $2$-minors of $M$ is contained in $N$, then the given generators of $I$ are a Gr\"obner basis with respect to $<$.
\end{corollary}
%\begin{proof}
%Following the proof of Theorem \ref{thm:onestep}, the conditions that $\hgt(I)$, $\hgt(C)>\hgt(N)$ and that $N$ have no embedded primes imply that the elements $u$ and $v$ of Theorem \ref{thm:onestep} are non-zero-divisors on $R/N$.  The condition that the ideal of $2$-minors of $M$ be contained in $N$ implies that \[
%r_i(a_1q_1 +\cdots+a_kq_k)-q_i(a_1r_1 +\cdots+a_kr_k)  = a_1(r_iq_1-r_1q_i)+\cdots+a_k(r_iq_k-r_kq_i)\in N
%\] for every $1 \leq i \leq k$.  The remainder of the argument from Theorem \ref{thm:onestep} that $\overline{\varphi}: [C/N](-1) \rightarrow I/N$ is an isomorphism remains intact in this setting.

%Set $\tilde{I} = \langle yq_1, \ldots, yq_k, h_1, \ldots, h_k \rangle$.  Because $\mathcal{G}_C$ and $\mathcal{G}_N$ are Gr\"obner bases, we know  that $\text{in}_< C = \langle \text{in}_<(q_1), \ldots, \text{in}_<(q_k), \text{in}_<(h_1), \ldots, \text{in}_<(h_\ell) \rangle$ and $\text{in}_< N = \langle \text{in}_<(h_1), \ldots, \text{in}_<(h_\ell)\rangle $. Now because $<$ is $y$-compatible, $\text{in}_< (yq_i+r_i) = y \cdot \text{in}_<(q_i)$ for each $1 \leq i \leq k$, and so $[\text{in}_<C/\text{in}_<N](-1)\xrightarrow{y} \tilde{I}/\text{in}_<N$ is also an isomorphism.  It follows from Lemma \ref{GVLmodified} that $\tilde{I} = \text{in}_y I$.
%\end{proof}


\begin{example}[The Veronese Embedding]
Using Corollary \ref{cor:Gb(I)}, one can give a concise inductive proof that the usual set of homogeneous equations defining the image of the $d^{\rm{th}}$ Veronese $\nu_d:\mathbb{P}^1 \rightarrow \mathbb{P}^d$ forms a Gr\"obner basis for any $d\geq 1$.  With homogeneous coordinates $[s:t]$ on $\mathbb{P}^1$ and $[x_0: \cdots :x_d]$ on $\mathbb{P}^d$, recall that the $d^{th}$ Veronese is the map $[s:t] \mapsto [s^d:s^{d-1}t:\cdots:st^{d-1}:t^d]$. 
Let $M_d = \begin{pmatrix}
x_0 & x_1 &\cdots & x_{d-1}\\
x_1 & x_2 &\cdots & x_d
\end{pmatrix}$, let $\mathcal{G}_d$ denote the set of $2\times 2$ minors of $M_d$, and let $I = \langle \mathcal{G}_d\rangle$ be the ideal generated by $\mathcal{G}
_d$.
The image of the $\nu_d$ is defined by $I$, 
which is to say that there is a ring isomorphism $\dfrac{\kappa[x_0, \ldots, x_d]}{I} \rightarrow \kappa[s^d, s^{d-1}t, \ldots, st^{d-1}, t^d]\subseteq \kappa[s,t]$ given by $x_i \mapsto s^{d-i}t^i$ for $0 \leq i \leq d$.  One may use Corollary \ref{cor:Gb(I)} in an inductive argument to see that $\mathcal{G}_d$ is a Gr\"obner basis of $I$ with respect to the lexicographic monomial order with $x_d>x_{d-1}>.\cdots>x_1>x_0$ by observing that the ideal generated by the $2\times 2$ minors of $\begin{pmatrix}
x_0& x_1 &\cdots & x_{d-2}\\
x_1x_{d-1}& x_2 x_{d-1} &\cdots & x_{d-1}^2
\end{pmatrix}$ is equal to  $x_{d-1} \cdot N$ and so is contained in $N$.  \qedhere

\end{example}


\section{Some well-known families of ideals are glicci}\label{sect:applications}


Many well-known classes of ideals Gr\"obner degenerate to Stanley--Reisner ideals of vertex decomposable complexes. In this section, we recall a few of these classes and deduce that they are glicci, providing evidence for an affirmative answer to the question of whether every homogeneous Cohen--Macaulay ideal is glicci \cite[Question 1.6]{KM+01}, at least in combintorially-natural settings. As in Section \ref{sect:GVDGlicci}, we will assume throughout this section that the field $\kappa$ is infinite.

The main result we need for our applications is as follows. It is immediately obtained by combining Theorem \ref{prop:simplicial} with Theorem \ref{thm:glicci}. 

\begin{corollary}\label{cor:AutomaticallyGlicci}
Let $I\subseteq \kappa[x_1,\dots, x_n]$ be a homogeneous ideal, and let $<$ denote the lexicographic order with $x_1>x_2>\cdots >x_n$. If $\text{in}_<I$ is the Stanley--Reisner ideal of a $<$-compatibly vertex decomposable simplicial complex on $[n]$ for the vertex order $1>2>\cdots >n$, then $I$ is glicci.
\end{corollary}


We now discuss three classes of ideals which satisfy the hypotheses of Corollary \ref{cor:AutomaticallyGlicci}. We omit many definitions of the ideals in question and instead provide references. 

\emph{Schubert determinantal ideals.} Let $X = (x_{ij})$ be an $n\times n$ matrix of variables and let $R = \kappa[x_{ij}]$ be the polynomial ring in the matrix entries of $X$.   Given a permutation $w\in S_n$, there is an associated generalized determinantal ideal $I_w\subseteq R$, called a \emph{Schubert determinantal ideal}. Schubert determinantal ideals and their corresponding \emph{matrix Schubert varieties} were introduced by Fulton in \cite{Fulton}.

Fix the lexicographical monomial order $<$ on $R$ defined by $x_{ij}>x_{kl}$ if $i<k$ or $i = k$ and $j>l$. This monomial order is \emph{antidiagonal}, that is, the initial term of the determinant of a submatrix $Y$ of $X$ is the product of the entries along the antidiagonal of $Y$. For this monomial order, $\text{in}_<I_w$ is the Stanley--Reisner ideal of a simplicial complex, called a \emph{subword complex}, which is $<$-compatibly vertex decomposable (see \cite{KnutsonMiller} or \cite[Ch. 16.5]{MillerSturmfels}). Corollary \ref{cor:AutomaticallyGlicci} thus immediately implies: 

\begin{proposition}\label{prop:SchubGlicci}
Schubert determinantal ideals are glicci.
\end{proposition}


\emph{Graded lower bound cluster algebras.} Cluster algebras are a class of combinatorially-defined commutative algebras that were introduced by S. Fomin and A. Zelevinsky at the turn of the century \cite{FZI}. \emph{Lower bound algebras}, introduced in \cite{BFZ} are related objects:  each lower bound algebra is contained in an associated cluster algebra, and this containment is equality in certain cases (i.e. in the \emph{acyclic} setting, see \cite[Theorem 1.20]{BFZ}). 


Each (skew-symmetric) lower bound algebra is defined from a quiver. Indeed, given a quiver $Q$, there is an associated polynomial ring $R_Q = \kappa[x_1,\dots, x_n,y_1,\dots, y_n]$ and ideal $K_Q\subseteq R_Q$ such that the lower bound algebra $\mathcal{L}_Q$ associated to $Q$ can be expressed as $\mathcal{L}_Q = R_Q/K_Q$. Fix the lexicographical monomial order with $y_1>\cdots>y_n>x_1>\cdots >x_n$. By \cite[Theorem 1.7]{MRZ} and the proof of \cite[Theorem 3.3]{MRZ}, $\text{in}_<K_Q$ is the Stanley--Reisner ideal of a simplicial complex $\Delta$ on vertex set $\{y_1,\dots, y_n,x_1,\dots, x_n\}$, which has vertex decomposition compatible with $<$. Consequently, by Theorem \ref{prop:simplicial}, we have the following:

\begin{proposition}
The ideal $K_Q$ is geometrically vertex decomposable. When $K_Q$ is homogeneous, it is glicci.
\end{proposition}



\begin{remark}
It follows from \cite[Theorem 1.7]{MRZ} that $K_Q$ is homogeneous if and only if $Q$ has no \emph{frozen vertices} and $Q$ has exactly two arrows entering each vertex and two arrows exiting each vertex.
\end{remark}


\emph{Ideals defining equioriented type $A$ quiver loci.} Let $d_0, d_1, \dots, d_n$ be a sequence of positive integers and consider the product of matrix spaces $\texttt{Mat}$, and product of general linear group $\texttt{GL}$ defined as follows:
\[
\texttt{Hom}:=\oplus_{i = 1}^n \text{Mat}_{d_{i-1}\times d_i}(\kappa), \quad \texttt{GL}:=\oplus_{i = 0}^n \text{GL}_{d_i}(\kappa).
\]
The group $\texttt{GL}$ acts on $\texttt{Hom}$ on the right by conjugation: $(M_i)_{i=1}^n\bullet (g_i)_{i=0}^n = (g_{i-1}^{-1}M_i g_i)_{i=1}^n$. Closures of $\texttt{GL}$-orbits are called \textbf{equioriented type $A$ quiver loci}. \emph{Buchsbaum-Eisenbud varieties of complexes} are special cases of these quiver loci. An introduction to equioriented type $A$ quiver loci and related combinatorics can be found in \cite[Ch. 17]{MillerSturmfels}. 

\begin{proposition}
Equioriented type $A$ quiver loci are glicci. In particular, varieties of complexes are glicci. 
\end{proposition}

%\begin{proof}
This follows because, up to adding some additional indeterminate generators, the ideal of such a quiver locus is a type $A$ Kazhdan--Lusztig ideal. As shown in \cite{WooYongGrobner}, each Kazhdan--Lusztig ideal Gr\"obner degenerates to the Stanley--Reisner ideal of a subword complex, and this degeneration is compatible with the vertex decomposition of the complex.
%\end{proof}



%\begin{proof} 
%Let $\Omega\subseteq \texttt{Hom}$ be an equioriented type $A$ quiver locus, and let $I(\Omega)$ be its (homogeneous and prime) defining ideal in the polynomial ring $\kappa[\texttt{Hom}]$. It follows from results of A. Zelevinsky \cite{Zelevinsky} and V. Lakshmibai and P. Magyar \cite{LakshmibaiMagyar} that there
%is a polynomial ring $R$ with $\kappa[\texttt{Hom}]\subseteq R$, a \emph{Kazhdan--Lusztig ideal} $J\subseteq R$, and an ideal $L$ generated by the indeterminates in $R\setminus \kappa[\texttt{Hom}]$ such that $J = I(\Omega)R+L$. (Here $I(\Omega)R$ denotes the extension of the ideal $I(\Omega)$ to $R$.) As shown in \cite{WooYongGrobner}, each Kazhdan--Lusztig ideal Gr\"obner degenerates to the Stanley--Reisner ideal of a subword complex, and this degeneration is compatible with the vertex decomposition of the complex. Consequently, $J$ is geometrically vertex decomposable. Thus $I(\Omega)$ is geometrically vertex decomposable, hence glicci. 
%\end{proof}


\section{From G-biliaisons to geometric vertex decompositions}\label{sect:glicciGVD}
We now state something of a converse to Theorem \ref{thm:onestep}. 

\begin{theorem}\label{thm:linkToGVD}
Let $I$, $C$, and $N \subseteq I \cap C$ be ideals of $R$, and let $<$ be a $y$-compatible term order.  Suppose that $I$ is squarefree in $y$ and that no term of any element of the reduced Gr\"obner basis of $N$ is divisible by $y$.  Suppose further that there exists an isomorphism $\phi: C/N \xrightarrow{f/g} I/N$ of $R/N$-modules for some $ f, g \in R$, and $\text{in}_y(f)/g = y$.  Then $\text{in}_y I = C \cap ( N+\langle y \rangle )$ is a geometric vertex decomposition of $I$.  
\end{theorem}

%\begin{proof}
%Recall that $I$ must have a Gr\"obner basis of the form $\{yq_1+r_1, \ldots, yq_k+r_k, h_1, \ldots, h_\ell\}$ where $y$ does not divide any term of any $q_i$ or $r_i$ for any $1 \leq i \leq k$ nor any $h_j$ for any $1 \leq j \leq \ell$ because $I$ is squarefree in $y$.  Hence, $\text{in}_y I = \langle yq_1, \ldots, yq_k, h_1, \ldots, h_\ell \rangle$, and this generating set is a Gr\"obner basis of $\text{in}_y I$ by \cite[Theorem 2.1(a)] {KMY09}.

%We claim first that $N = \langle h_1, \ldots, h_\ell \rangle$.  Because no term of any element of the reduced Gr\"obner basis of $N \subseteq I$ is divisible by $y$, each such element must be a polynomial in the $h_i$ for $1 \leq i \leq \ell$, and so $N \subseteq \langle h_1, \ldots, h_\ell \rangle$, from which it follows that $\text{in}_y(N) = N$ and that $y$ is not a zero-divisor on $R/N$.  Conversely, suppose there is some $h_i \in I \setminus N$ for some $1 \leq i \leq \ell$.  Then there exists some $c \in C \setminus N$ and $n \in N$ such that $c \cdot \frac{f}{g} = h_i+n$, where $c$ has been chosen to have the smallest possible $d$ for which $y^d$ divides $\text{in}_y(c)$.  Because $c \cdot \frac{f}{g} = h_i+n$, we have $cf = g(h_i+n)$. Taking initial $y$-forms yields
%\[\text{in}_y(c)\cdot \text{in}_y(f) = \text{in}_y(g)\cdot \text{in}_y(h_i+n) = g\cdot \text{in}_y(h_i+n),\]
%where $\text{in}_y g = g$ by the assumption that $g$ divides $\text{in}_y f$. Dividing both sides by $g$ yields $y\cdot \text{in}_y(c) = \text{in}_y(h_i+n)$. As no term of $h_i$ is divisible by $y$, we see that $y\cdot \text{in}_y(c) = \text{in}_y(n)$. Hence, $y\cdot \text{in}_y(c)\in N$ (as $\text{in}_y N = N$) and so $\text{in}_y(c)\in N$ (as $N:\langle y\rangle = N$).

%Set $c' = c-\text{in}_y(c) \in C \setminus N$, which has an initial $y$-form not divisible by $y^d$.  Now \[
%c' \cdot \frac{f}{g} = (c-\text{in}_y(c)) \frac{f}{g} = h_i+n-\frac{f}{g}\text{in}_y(c).
%\]  But $n-\frac{f}{g}\text{in}_y(c) \in N$, contradicting minimality of $d$.  Hence, $N = \langle h_1, \ldots, h_\ell \rangle$.  

%Next, we claim that $C = \langle q_1, \ldots, q_k, h_1, \ldots, h_\ell \rangle$.  By assumption, $N \subseteq C$, and so it suffices to show that the $q_i$ for $1 \leq i \leq k$ generate $C$ over $N$.  In order to establish this, we will show that $\psi: C/N \xrightarrow{y} \text{in}_y(I)/N$ is an isomorphism.  For each $1 \leq i \leq k$, let $c_i \in C \setminus N$ be a representative of the preimage under $\phi$ of the class of $yq_i+r_i$ in $I/N$ with the smallest possible $d_i$ so that $\text{in}_y(c_i) \notin \langle y^{d_i} \rangle$.  We will show that the image under $\psi$ of the class of $c_i$ is the class of $yq_i$.

%First, we will show that $d_i = 1$ for all $1 \leq i \leq k$.  Suppose, for contradiction, that some $d_i > 1$, i.e., that some $\text{in}_y(c_i) \in \langle y \rangle$.  Then because $\frac{f}{g}c_i = yq_i+r_i+n_i$ for some $n_i \in N$, we may do a similar computation to above to conclude that $\text{in}_y(yq_i+r_i+n_i) = y \cdot \text{in}_y(c_i)\in \langle y^2\rangle$.
%have \[
% \text{in}_y(yq_i+r_i+n_i) = \text{in}_y\left(\frac{f}{g}c_i\right) = y \cdot \text{in}_y(c_i) \in \langle y^2 \rangle.
%\] 
%Because neither $q_i$ nor $r_i$ is divisible by $y$, we must have \[
%y \cdot \text{in}_y(c_i) = \text{in}_y\left(yq_i+r_i+n_i\right) = \text{in}_y(n_i) \in \text{in}_y (N) = N,
%\] and so $\text{in}_y(c_i) \in N$.  Then $c_i' = c_i-\text{in}_y(c_i) \in C\setminus N$ and $\text{in}_y(c_i') \notin \langle y^{d_i} \rangle$.  But still \[
%\frac{f}{g}c_i' = yq_i+r_i+n_i-\frac{f}{g}\text{in}_y(c_i),
%\] and so $c_i'$ also represents a preimage of $yq_i+r_i$ because $n_i-\frac{f}{g}\text{in}_y(c_i) \in N$, contradicting minimality of $d_i$.  Hence, $\text{in}_y(c_i) \notin \langle y \rangle$, which establishes both that $\text{in}_y(c_i) = c_i$ and also that $\text{in}_y(n_i) \notin \langle y^2 \rangle$ for each $1 \leq i \leq k$.

%From the former fact and the relationship $y \cdot \text{in}_y(c_i) = \text{in}_y(yq_i+r_i+n_i)$, we have $yc_i = \text{in}_y(yq_i+r_i+n_i)$.  From the latter, we have $\text{in}_y(yq_i+r_i+n_i) = yq_i$ if no term of $n_i$ is divisible by $y$, and $\text{in}_y(yq_i+r_i+n_i) = yq_i+\text{in}_y(n_i)$ if $n_i$ has some term divisibly by $y$.  But $\text{in}_y(n_i) \in \text{in}_y(N) = N$, and so in either case, the $\psi(c_i)$ is the class of $yq_i$ in $\init_y(I)/N$, which is to say that $\psi$ is surjective.  Also, $\psi$ is injective because $y$ is not a zero-divisor on $R/N$.  Now because the $yq_i$ for $1 \leq i \leq k$ generate $\init_y(I)$ over $N$ and $\psi$ is an isomorphism under which the preimage of the class of $yq_i$ is the class of $q_i$ for each $1 \leq i \leq k$, it must be that $C$ is generated over $N$ by $\{q_1, \ldots, q_k\}$ and that $C = \langle q_1, \ldots, q_k, h_1, \ldots, h_\ell \rangle$.

%By \cite[Theorem 2.1(a)] {KMY09}, the specified generating sets for $\init_y(I)$, $N$, and $C$ are all Gr\"obner bases for them, and so it follows from \cite[Theorem 2.1(b)]{KMY09} that $\text{in}_y(I) = C \cap (N+\langle y \rangle)$ is a geometric vertex decomposition of $I$.  
%\end{proof}

\begin{example}\label{ex:standard}
To illustrate this correspondence between elementary $G$-biliaisons and geometric vertex decomposition, we consider a classical example.  If $I$ is the ideal of $2$-minors of the matrix $M = \begin{pmatrix}
x_{11} & x_{12} & x_{13} \\
x_{21} & x_{22} & x_{23} 
\end{pmatrix}$, $C = \langle x_{11}, x_{12} \rangle$, $N = \langle x_{22}x_{11}-x_{21}x_{12} \rangle$, $f = x_{23}x_{12}-x_{22}x_{13}$, and $g = x_{12}$ in $\kappa[x_{11}, \ldots, x_{23}]$, then the multiplication by $f/g$  map $[C/N](-1) \xrightarrow{f/g} I/N$ gives an elementary $G$-biliaison.  Using any lexicographic order with $x_{23}$ largest, we take $C = C_{x_{23},I}$ and $N = N_{x_{23},I}$, and then $\text{in}_{x_{23}}(I) = C \cap (N+\langle x_{23} \rangle)$ is a geometric vertex decomposition.
\end{example}


\acknowledgements{We thank Elisa Gorla, Adam Van Tuyl, and Anna Weigandt for helpful conversations. Part of this work was completed at the Banff International Research Station (BIRS) during the Women in Commutative Algebra workshop in October 2019. We are grateful for the hospitality of the Banff Centre.  We are also grateful to the anonymous referees, both of this FPSAC extended abstract and also of the corresponding article, all of whose comments improved this work.}

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