# Necessary conditions for the existence of Morita Contexts in the bicategory
of Landau-Ginzburg Models

Yves Baudelaire Fomatati
Department of Mathematics and Statistics, University of Ottawa,
Ottawa, Ontario, Canada K1N 6N5.
<EMAIL_ADDRESS>
> Abstract

We use a matrix approach to study the concept of Morita context in the
bicategory $\mathcal{LG}_{K}$ of Landau-Ginzburg models on a particular class
of objects. In fact, we first use properties of matrix factorizations to state
and prove two necessary conditions to obtain a Morita context between two
objects of $\mathcal{LG}_{K}$. Next, we use a celebrated result (due to Schur)
on determinants of block matrices to show that these necessary conditions are
not sufficient. Finally, we state a trivial sufficient condition.

Keywords. Matrix factorizations, tensor product, Morita equivalence.
Mathematics Subject Classification (2020). 16D90, 15A23, 15A69, 18N10.

In the sequel, $K$ is a commutative ring with unity and $R$ will denote the
power series ring $K[[x_{1},\cdots,x_{n}]]$ or its subring of polynomials
$K[x_{1},\cdots,x_{n}]$. It would always be clear which ring we are referring
to.

##  1 Introduction

A Morita context also called pre-equivalence data [4], is a generalization of
Morita equivalence between categories of modules. Two rings $T$ and $S$ are
called Morita equivalent if the categories of left $T-$Modules ($T-$Mod) and
of left $S-$Modules $(S-$Mod) are equivalent. The prototype of Morita
equivalent rings is provided by a ring $T$ and the ring of $n\times n$
matrices over $T$ (for details, see corollary 22.6 of [2]).
It is evident that if two rings are isomorphic, they are Morita equivalent.
However, the converse is not true in general. There exist rings that are not
isomorphic, yet are Morita equivalent (cf. p. 470 of [22]). In fact, it
suffices to take a ring $T$ and the ring of $n\times n$ matrices over $T$.
However, there is a partial converse which holds. Indeed, if two rings are
Morita equivalent, then their centers are isomorphic. In particular, if the
rings are commutative, then they are isomorphic ([22], p.494). Because of this
result, Morita equivalence is interesting solely in the situation of
noncommutative rings. For more details on Morita equivalence, see [2]. This
notion can be generalized using the notion of Morita contexts.
Morita contexts were first introduced in the bicategory of unitary rings and
bimodules as $6$-tuples $(A,B,M,N,\phi,\psi)$; where $A$ and $B$ are rings,
$M$ is an A-B-bimodule, $N$ is a B-A-bimodule and $\phi:M\otimes N\rightarrow
A$ and $\psi:N\otimes M\rightarrow B$ are homomorphisms satisfying
$\phi\otimes M=M\otimes\psi$ and $N\otimes\phi=\psi\otimes N$. For a
characterization of Morita context in this bicategory, see theorem 5.4 of
[25]. Bass (cf. chap. 2, section 4.4 of [3]) proved that the Morita context
$(A,B,M,N,\phi,\psi)$ is a Morita equivalence if and only if $M$ is both
projective and a generator in the category of $A-$modules. Recall (cf. [22])
that if $A$ is a ring, an $A$-module $P$ is projective if for every surjective
$A$-linear map $f:M\rightarrow N$ and every $A$-linear map $g:P\rightarrow N$
there is a unique $A$-linear map $h:P\rightarrow M$ such that $g=fh$.
Morita contexts were recently studied in many bicategories [25]. But they have
not yet been studied in the bicategory $\mathcal{LG}_{K}$ of Landau-Ginzburg
models (section 2.2 of [6]) over a commutative ring $K$ whose intricate
construction ([8]) is reminiscent of, but more complex than that of the
bicategory of associative algebras and bimodules. In this paper, we study the
concept of Morita context in the bicategory of Landau-Ginzburg models. A
Landau-Ginzburg model is a model in solid state physics for superconductivity.
$\mathcal{LG}_{K}$ possesses adjoints (also called duals, cf. [8]) and this
helps in explaining a certain duality that exists in the setting of Landau-
Ginzburg models in terms of some specified relations (cf. page 1 of [8]). The
objects of $\mathcal{LG}_{K}$ are potentials which are polynomials satisfying
some conditions (definition 2.4 p.8 of [8]). But unlike [8], we do not
restrict ourselves to potentials. It turns out that the authors of [8] used
potentials to suit their purposes because even if we take the objects of
$\mathcal{LG}_{K}$ to be polynomials rather than potentials and then apply the
construction of $\mathcal{LG}_{K}$ given in [8], we obtain virtually the same
bicategory except that we now have more objects. Thus in this paper, the
objects of $\mathcal{LG}_{K}$ are simply polynomials.
There are many reasons for studying the notion of Morita context. The first
reason is that it generalizes the very important notion of Morita equivalence.
Another reason is that it is used to prove some celebrated results. For
example, the Morita context which has been introduced in [23] was used since
to prove Wedderburn theorem on the structure of simple rings [3]. Morita
contexts were also used in [1] to obtain various results: Goldie’s theorem
([16], [17]) on the ring of quotients of semi-prime rings and as a
specialization, Wedderburn’s structure theorems of semi-simple Artinian rings
were obtained. Other applications, though sometimes not stated in an explicit
form, can be found in various places (e.g. [18], p.75]).
In this paper, we use properties of matrix factorizations to give necessary
conditions to obtain a Morita context between two objects of the bicategory
$\mathcal{LG}_{K}$. Thus, our first main result is the following theorem.

Theorem A.
Let

  * •

$(R,f)$ and $(S,g)$ be two objects of $\mathcal{LG}_{K}$.

  * •

$X\in\mathcal{LG}_{K}((R,f),(S,g))=hmf(R\otimes_{K}S,g-f)$ i.e.,
$X:(R,f)\rightarrow(S,g)$ is a finite rank matrix factorization of $g-f$.

  * •

$Y\in\mathcal{LG}_{K}((S,g),(R,f))=hmf(S\otimes_{K}R,f-g)$ i.e.,
$Y:(S,g)\rightarrow(R,f)$ is a finite rank matrix factorization of $f-g$.
such that $X\otimes Y$ and $Y\otimes X$ are finite rank matrix factorizations.

  * •

$\Delta_{f}\in\mathcal{LG}_{K}((R,f),(R,f))=hmf(R\otimes_{K}R,f\otimes id-
id\otimes f)$ i.e., $\Delta_{f}:(R,f)\rightarrow(R,f)$ is a finite rank matrix
factorization of $f\otimes id-id\otimes f$.

  * •

$\Delta_{g}\in\mathcal{LG}_{K}((S,g),(S,g))=hmf(R\otimes_{K}S,g\otimes id-
id\otimes g)$ i.e., $\Delta_{g}:(S,g)\rightarrow(S,g)$ is a finite rank matrix
factorization of $g\otimes id-id\otimes g$.

  * •

$\eta:X\otimes_{R}Y\rightarrow 1_{(R,f)}=\Delta_{f}$ and let $(P,Q)$ and
$(R,T)$ be pairs of matrices representing respectively the finite rank matrix
factorizations $X\otimes Y$ and $\Delta_{f}$.

  * •

$\rho:Y\otimes_{S}X\rightarrow 1_{(S,g)}=\Delta_{g}$ and let
$(P^{\prime},Q^{\prime})$ and $(R^{\prime},T^{\prime})$ be pairs of matrices
representing respectively the finite rank matrix factorizations $Y\otimes X$
and $\Delta_{g}$.

Then: A necessary condition for $\Gamma=(X,Y,\eta,\rho)$ to be a Morita
Context is

$\begin{cases}\eta^{1}=0,\;and\;\eta^{0}Q=0,\\\
\rho^{1}=0,\;and\;\rho^{0}Q^{\prime}=0.\end{cases}$

where for ease of notation, we wrote $\rho^{i}$ and $\eta^{i}$ respectively
for the matrices of $\rho^{i}$ and $\eta^{i}$, $i=0,1$.

Next, thanks to a celebrated result (due to Schur [26], [27]) on determinants
of block matrices, we observe that if $X$ and $X^{\prime}$ are respectively
two matrix factorizations of two arbitrary polynomials, then the determinants
of the four matrices appearing in the Yoshino tensor products
$X\widehat{\otimes}X^{\prime}=(P,Q)$ and
$X^{\prime}\widehat{\otimes}X=(P^{\prime},Q^{\prime})$ are all equal.
Moreover, when we translate this in $\mathcal{LG}_{K}$ where a $1-$morphism is
a matrix factorization of the difference of two polynomials, we find out that
those four determinants are all equal to zero. So our second main result is
stated as follows:

Theorem B.
Let $(R,f)$ and $(S,g)$ be two objects in $\mathcal{LG}_{K}$ and let

$X:(R,f)\rightarrow(S,g)\,\,and\,\,X^{\prime}:(S,g)\rightarrow(R,f)$

be 1-morphisms in $\mathcal{LG}_{K}$. If $X\otimes X^{\prime}=(P,Q)$ and
$X^{\prime}\otimes X=(P^{\prime},Q^{\prime})$, then

$det(P)=det(Q)=det(P^{\prime})=det(Q^{\prime})=0$

Thanks to this result we conclude that the necessary conditions earlier stated
are not sufficient.
This paper is organized as follows: In the next section, we review the notion
of matrix factorization. In section 3, we recall properties of matrix
factorizations. Section 4 is a recall of the definition of the bicategory of
Landau-Ginzburg models. The notion of Morita contexts in $\mathcal{LG}_{K}$ is
discussed in section 5. Finally, we discuss further problems in the last
section.

##  2 Matrix Factorizations

In this section, we first recall the definition of a matrix factorization and
describe the category of matrix factorizations of a power series $f$. Next,
the definition of Yoshino’s tensor product is recalled.

In 1980, Eisenbud came up with an approach of factoring both reducible and
irreducible polynomials in $R$ using matrices. For instance, the polynomial
$f=x^{2}+y^{2}$ is irreducible over the real numbers but can be factorized as
follows:

$\begin{bmatrix}x&-y\\\ y&x\end{bmatrix}\begin{bmatrix}x&y\\\
-y&x\end{bmatrix}=(x^{2}+y^{2})\begin{bmatrix}1&0\\\ 0&1\end{bmatrix}=fI_{2}$

We say that $(\begin{bmatrix}x&-y\\\ y&x\end{bmatrix},\begin{bmatrix}x&y\\\
-y&x\end{bmatrix})$ is a $2\times 2$ matrix factorization of $f$.

######  Definition 2.1.

[29], [10]
An $n\times n$ matrix factorization of a power series $f\in\;R$ is a pair of
$n$ $\times$ $n$ matrices $(P,Q)$ such that $PQ=fI_{n}$, where $I_{n}$ is the
$n\times n$ identity matrix and the coefficients of $P$ and of $Q$ are taken
from $R$.

When $n=1$, we get a $1$ $\times$ $1$ matrix factorization of $f$, i.e.,
$f=[g][h]$ which is simply a factorization of $f$ in the classical sense. But
in case $f$ is not reducible, this is not interesting, that’s why we will
mostly consider $n>1$.
The original definition of a matrix factorization was given by Eisenbud [13]
as follows: a matrix factorization of an element $x$ in a ring $A$ (with
unity) is an ordered pair of maps of free $A-$modules $\phi:F\rightarrow G$
and $\psi:G\rightarrow F$ s.t., $\phi\psi=x\cdot 1_{G}$ and $\psi\phi=x\cdot
1_{F}$. Though this definition is valid for any arbitrary ring (with unity),
in order to effectively study matrix factorizations, it is important to
restrict oneself to specific rings. Working with specific rings makes it
possible to easily give examples and it also allows one to carry out
computations in a well-defined framework. Yoshino [29] restricted himself to
matrix factorizations of power series. In this section, we will restrict
ourselves to matrix factorizations of a polynomial.

######  Example 2.1.

Let $h=xy+xz^{2}+yz^{2}$.
We give a $2\times 2$ matrix factorization of $h$:

$\begin{bmatrix}z^{2}&y\\\ x&-x-y\end{bmatrix}\begin{bmatrix}x+y&y\\\
x&-z^{2}\end{bmatrix}=(xy+xz^{2}+yz^{2})\begin{bmatrix}1&0\\\
0&1\end{bmatrix}=hI_{2}$

Thus;

$(\begin{bmatrix}z^{2}&y\\\ x&-x-y\end{bmatrix},\begin{bmatrix}x+y&y\\\
x&-z^{2}\end{bmatrix})$

is a $2\times 2$ matrix factorization of $h$.

We now propose a simple straightforward algorithm to obtain an $n\times n$
matrix factorization from one that is $m\times m$, where $m<n$.

Simple straightforward algorithm: Let $(P,Q)$ be an $m\times m$ matrix
factorizations of a power series $f$. Suppose we want an $n\times n$ matrix
factorization of $f$, where $n>m$.
Let $(i,j)$ stand for the entry in the $i^{th}$ row and $j^{th}$ column.

  * •

Turn $P$ and $Q$ into $n\times n$ matrices by filling them with zeroes
everywhere except at entries $(i,j)$ where $1\leq i\leq m$ and $1\leq j\leq
m$.
Then for all entries $(k,k)$ with $k>m$, Either:

    1. 1.

In $P$, replace the diagonal elements (which are zeroes) with $f$
And

    2. 2.

In $Q$, replace the diagonal elements (which are zeroes) with 1

Or:
Interchange the roles of $P$ and $Q$ in steps $1.$ and $2.$ above.

It is evident that this simple algorithm works not only for polynomials but
also for any element in a unital ring.

The standard algorithm to factor polynomials using matrices is found in [10]
and for an improved version see [14] and [15].

###  2.1 The category of matrix factorizations of $f\in R$

The category of matrix factorizations of a power series $f\in
R=K[[x]]:=K[[x_{1},\cdots,x_{n}]]$ denoted by $MF(R,f)$ or $MF_{R}(f)$, (or
even $MF(f)$ when there is no risk of confusion) is defined [29] as follows:
$\bullet$ The objects are the matrix factorizations of $f$.
$\bullet$ Given two matrix factorizations of $f$; $(\phi_{1},\psi_{1})$ and
$(\phi_{2},\psi_{2})$ respectively of sizes $n_{1}$ and $n_{2}$, a morphism
from $(\phi_{1},\psi_{1})$ to $(\phi_{2},\psi_{2})$ is a pair of matrices
$(\alpha,\beta)$ each of size $n_{2}\times n_{1}$ which makes the following
diagram commute:

$\textstyle{K[[x]]^{n_{1}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\psi_{1}}$$\scriptstyle{\alpha}$$\textstyle{K[[x]]^{n_{1}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\beta}$$\scriptstyle{\phi_{1}}$$\textstyle{K[[x]]^{n_{1}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha\;\;\;\;\;\;\;\;\;\;(\bigstar)}$$\textstyle{K[[x]]^{n_{2}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\psi_{2}}$$\textstyle{K[[x]]^{n_{2}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi_{2}}$$\textstyle{K[[x]]^{n_{2}}}$

That is,

$\begin{cases}\alpha\phi_{1}=\phi_{2}\beta\\\
\psi_{2}\alpha=\beta\psi_{1}\end{cases}$

For a detailed definition of this category, see [14].

###  2.2 Yoshino’s Tensor Product of Matrix Factorizations and its variants

In this subsection, we recall the definition of the tensor product of matrix
factorizations denoted $\widehat{\otimes}$, constructed by Yoshino [29] using
matrices. This will be useful when we will be describing the notion of Morita
Context in $\mathcal{LG}_{K}$ (cf. section 5). In the sequel, except otherwise
stated $S=K[[y_{1},\cdots,y_{r}]]=K[[y]]$, where $K$ is a commutative ring
with unity and $r\geq 1$.

######  Definition 2.2.

[29] Let $X=(\phi,\psi)$ be an $n\times n$ matrix factorization of a power
series $f\in R$ and $X^{\prime}=(\phi^{\prime},\psi^{\prime})$ an $m\times m$
matrix factorization of $g\in S$. These matrices can be considered as matrices
over $L=K[[x,y]]$ and the tensor product $X\widehat{\otimes}X^{\prime}$ is
given by
($\begin{bmatrix}\phi\otimes 1_{m}&1_{n}\otimes\phi^{\prime}\\\
-1_{n}\otimes\psi^{\prime}&\psi\otimes
1_{m}\end{bmatrix},\begin{bmatrix}\psi\otimes
1_{m}&-1_{n}\otimes\phi^{\prime}\\\ 1_{n}\otimes\psi^{\prime}&\phi\otimes
1_{m}\end{bmatrix}$)
where each component is an endomorphism on $L^{n}\otimes L^{m}$.

It is easy to see [29] that $X\widehat{\otimes}X^{\prime}$ is a matrix
factorization of $f+g$ of size $2nm$.
In the following example, we consider matrix factorizations in one variable.

######  Example 2.2.

Let $X=(x^{2},x^{2})$ and $X^{\prime}=(y^{2},y^{4})$ be $1\times 1$ matrix
factorizations of $f=x^{4}$ and $g=y^{6}$ respectively. Then
$X\widehat{\otimes}X^{\prime}=(\begin{bmatrix}x^{2}\otimes 1&1\otimes y^{2}\\\
-1\otimes y^{4}&x^{2}\otimes 1\end{bmatrix},\begin{bmatrix}x^{2}\otimes
1&-1\otimes y^{2}\\\ 1\otimes y^{4}&x^{2}\otimes
1\end{bmatrix})=(\begin{bmatrix}x^{2}&y^{2}\\\
-y^{4}&x^{2}\end{bmatrix},\begin{bmatrix}x^{2}&-y^{2}\\\
y^{4}&x^{2}\end{bmatrix})$
And

$\begin{bmatrix}x^{2}&y^{2}\\\
-y^{4}&x^{2}\end{bmatrix}\begin{bmatrix}x^{2}&-y^{2}\\\
y^{4}&x^{2}\end{bmatrix}=(x^{2}+y^{6})\begin{bmatrix}1&0\\\ 0&1\end{bmatrix}$

This shows that $X\widehat{\otimes}X^{\prime}$ is a matrix factorization of
$f+g=x^{2}+y^{6}$ and it is of size $2(1)(1)=2$.

In the next example, we consider matrix factorizations in two variables.

######  Example 2.3.

Let

X= ($\begin{bmatrix}x&-y\\\ y&x\end{bmatrix},\begin{bmatrix}x&y\\\
-y&x\end{bmatrix}$), and X’= ($\begin{bmatrix}-x&y\\\
-y&-x\end{bmatrix},\begin{bmatrix}x&y\\\ -y&x\end{bmatrix}$)
be $2\times 2$ matrix factorizations of $f=x^{2}+y^{2}$ and $g=-x^{2}-y^{2}$
respectively. Let

A= $\begin{bmatrix}x&-y\\\ y&x\end{bmatrix}\otimes\begin{bmatrix}1&0\\\
0&1\end{bmatrix}$, B= $\begin{bmatrix}1&0\\\
0&1\end{bmatrix}\otimes\begin{bmatrix}-x&y\\\ -y&-x\end{bmatrix}$, C=
$\begin{bmatrix}-1&0\\\ 0&-1\end{bmatrix}\otimes\begin{bmatrix}x&y\\\
-y&x\end{bmatrix}$,

D= $\begin{bmatrix}x&y\\\ -y&x\end{bmatrix}\otimes\begin{bmatrix}1&0\\\
0&1\end{bmatrix}$, A’= $\begin{bmatrix}x&y\\\
-y&x\end{bmatrix}\otimes\begin{bmatrix}1&0\\\ 0&1\end{bmatrix}$, B’=
$\begin{bmatrix}-1&0\\\ 0&-1\end{bmatrix}\otimes\begin{bmatrix}-x&y\\\
-y&-x\end{bmatrix}$,

C’= $\begin{bmatrix}1&0\\\ 0&1\end{bmatrix}\otimes\begin{bmatrix}x&y\\\
-y&x\end{bmatrix}$, and D’= $\begin{bmatrix}x&-y\\\
y&x\end{bmatrix}\otimes\begin{bmatrix}1&0\\\ 0&1\end{bmatrix}$

Then
$X\widehat{\otimes}X^{\prime}=(\begin{bmatrix}A&B\\\
C&D\end{bmatrix},\begin{bmatrix}A^{\prime}&B^{\prime}\\\
C^{\prime}&D^{\prime}\end{bmatrix})$

$=(\left(\begin{array}[]{cccccccc}x&0&-y&0&-x&y&0&0\\\ 0&x&0&-y&-y&-x&0&0\\\
y&0&x&0&0&0&-x&y\\\ 0&y&0&x&0&0&-y&-x\\\ -x&-y&0&0&x&0&y&0\\\
y&-x&0&0&0&x&0&y\\\ 0&0&-x&-y&-y&0&x&0\\\ 0&0&y&-x&0&-y&0&x\\\
\end{array}\right),\left(\begin{array}[]{cccccccc}x&0&y&0&x&-y&0&0\\\
0&x&0&y&y&x&0&0\\\ -y&0&x&0&0&0&x&-y\\\ 0&-y&0&x&0&0&y&x\\\
x&y&0&0&x&0&-y&0\\\ -y&x&0&0&0&x&0&-y\\\ 0&0&x&y&y&0&x&0\\\
0&0&-y&x&0&y&0&x\\\ \end{array}\right))$

If we call these two $8\times 8$ matrices $P$ and $Q$ respectively, then
$PQ=0\cdot I_{8}=0,$ where $I_{8}$ is the identity matrix of size $8$ and the
last zero is the zero matrix of size $8$. Hence,
$X\widehat{\otimes}X^{\prime}$ is a matrix factorization of
$f+g=x^{2}+y^{2}-x^{2}-y^{2}=0$ of size $2(2)(2)=8$.

Yoshino’s tensor product has three mutually distinct functorial variants as
can be seen in [15]

##  3 Properties of Matrix Factorizations

We will now state and prove some properties of matrix factorizations of
polynomials. Some of them will be used when studying the notion of Morita
Context between two objects in the bicategory $\mathcal{LG}_{K}$.
In this section, except otherwise stated $R=K[x_{1},\cdots,x_{n}]=K[x]$. All
statements and proofs in this section are taken from [10] except for
proposition 3.2. They are reproduced here (sometimes with slight
modifications) for the purposes of completeness.

######  Lemma 3.1.

If $PQ=fI_{n}$, then det($P$) divides $f^{n}$. If in addition, $f$ is
irreducible in $R$, then det($P$) is a power of $f$.

######  Corollary 3.1.

If $0\neq f\in R$ and $PQ=fI_{n}$, then over the field of fractions
$\mathcal{F}$ of $R$, $P$ is invertible.

Since $P$ is invertible over $\mathcal{F}$, the unique $Q$ such that
$PQ=fI_{n}$ is $Q=P^{-1}fI_{n}=\frac{f}{det(P)}adj(P)$, where $adj(P)$ is the
adjoint of the matrix $P$.

Now, $adj(P)$ is a matrix over $R=K[x_{1},\cdots,x_{n}]$. So, if $det(P)$
divides $f$, then $Q$ will also be a matrix over $R$. However, it is possible
for $Q$ not to have entries in $R$, and therefore $P$ will not appear in any
matrix factorization of $f$ over $R$. We now prove that matrices appearing in
a matrix factorization commute with each other.

######  Proposition 3.1.

If $0\neq f\in R$ and $P,\;Q$ are $n\times n$ matrices so that $PQ=fI_{n}$,
then $QP=fI_{n}$.

######  Corollary 3.2.

If $0\neq f\in R$ and $PQ=fI_{n}$, then over the field of fractions
$\mathcal{F}$ of $R$, $P$ and $Q$ are invertible.

######  Remark 3.1.

It is important to note that corollary 3.1 and corollary 3.2 actually say that
matrices that appear in a matrix factorization of a nonzero polynomial are
invertible over the field of fractions $\mathcal{F}$ of $R$. This result will
soon be very useful when describing the notion of Morita Context in
$\mathcal{LG}_{K}$.

We state and prove another property of matrix factorizations thanks to which
we can conclude that an $n\times n$ matrix factorization of a polynomial $f$
is not unique.

######  Proposition 3.2.

If $0\neq f\in R$ and $P,\;Q$ are $n\times n$ matrices such that $PQ=fI_{n}$,
then $Q^{t}P^{t}=fI_{n}$, where $Q^{t}$ (respectively $P^{t}$) stands for the
transpose of $Q$ (respectively $P$).

###### Proof.

Assume $0\neq f\in R$ and $P,\;Q$ are $n\times n$ matrices such that
$PQ=fI_{n}$. then:

$\displaystyle PQ=fI_{n}$ $\displaystyle\Rightarrow(PQ)^{t}=(fI_{n})^{t}$
$\displaystyle\Rightarrow
Q^{t}P^{t}=fI_{n}\;since\;(f)^{t}=f\,viewed\;as\,a\,1\times 1\,matrix$

∎

Before we proceed, it is worthwhile stating some well-known facts in the
literature, see notes on tensor products by Conrad (for points 1. and 2. below
see theorem 4.9, example 4.11 of [9], point 3. simply generalizes point 2.).

######  Lemma 3.2.

  1. 1.

Let $K[x_{1},x_{2},\cdots,x_{n}]$ and
$K[x^{\prime}_{1},x^{\prime}_{2},\cdots,x^{\prime}_{m}]$ be free $K-$modules
with respective bases $\\{e_{i}\\}^{n}_{i=1}$ and
$\\{e^{\prime}_{j}\\}^{m}_{j=1}$. Then $\\{e_{i}\otimes
e^{\prime}_{j}\\}_{i=1,\cdots n;j=1,\cdots m}$ is a basis of
$K[x_{1},x_{2},\cdots,x_{n}]\otimes
K[x^{\prime}_{1},x^{\prime}_{2},\cdots,x^{\prime}_{m}]$.

  2. 2.

The $K-$modules $K[x_{1},x_{2},\cdots,x_{n}]\otimes
K[x^{\prime}_{1},x^{\prime}_{2},\cdots,x^{\prime}_{m}]$ and
$K[x_{1},x_{2},\cdots,x_{n},x^{\prime}_{1},x^{\prime}_{2},\cdots,x^{\prime}_{m}]$
are isomorphic as $K-$modules.

  3. 3.

If we let $x$ stand for $x_{1},x_{2},\cdots,x_{n}$ and $x^{(l)}$ stand for
$x^{(l)}_{1},x^{(l)}_{2},\cdots,x^{(l)}_{n_{l}}$, where
$n,l,n_{l}\in\mathbb{N}$, $n=n_{0}$, and $\\{x^{(l)}\\}$ means $x$ with $l$
primes, e.g $x^{(2)}=x^{\prime\prime},\;x^{(0)}:=x$.
Then more generally, we have:

$K[x,x^{(1)},\cdots,x^{(l)}]\cong\bigotimes_{p=0,\cdots,l}K[x^{(p)}].$

##  4 The bicategory $\mathcal{LG}_{K}$ of Landau-Ginzburg models

We quickly recall the construction of the bicategory $\mathcal{LG}_{K}$ of
Landau-Ginzburg models. In definition 5.5 of [14], a $B-$category is defined
to be a structure having all requirements of a bicategory but without
necessarily satisfying the unit morphisms requirement. In order to easily
recall the construction of $\mathcal{LG}_{K}$, we first construct a
$B-$category which we call $B-Fac$ and then we proceed to the unit
construction. The objects of $B-Fac$ are polynomials $f$ denoted by pairs
$(R,f)$ where $f\in R=K[x]$. Here we do not impose restrictions on the objects
of our bicategory as was originally done in [8], where the authors instead
consider potentials (Definition 2.4, p.8 of [8]). This generalization we do at
the level of the objects does not pose a problem in the construction of
$\mathcal{LG}_{K}$. The end result is simply a bicategory which has more
objects than the original $\mathcal{LG}_{K}$ defined in [8], we still call it
$\mathcal{LG}_{K}$.
We first recall the notion of linear factorizations which is an ingredient for
the construction of $\mathcal{LG}_{K}$.

######  Definition 4.1.

(p.8 of [8]) Linear factorization
A linear factorization of $f\in R=K[x]$ is a $\mathbb{Z}_{2}-$graded
$R-$module $X=X^{0}\oplus X^{1}$ together with an odd (i.e., grade reversing)
$R-$linear endomorphism $d:X\longrightarrow X$ such that $d^{2}=f\cdot
id_{X}$.
$f\cdot id_{X}$ stands for the endomorphism $x\mapsto f\cdot x,\,\,\forall
x\in X$.

Since we are dealing with a $\mathbb{Z}_{2}-$grading and $d$ is odd, we can
also say $d$ is a degree one map. $d$ is called a twisted differential in [7].
$d$ is actually a pair of maps $(d^{0},d^{1})$ that we may depict as follows:

$\textstyle{X^{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d^{0}}$$\textstyle{X^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d^{1}}$$\textstyle{X^{0}}$

and the stated condition on them is: $d^{0}\circ d^{1}=f\cdot id_{X^{1}}$ and
$d^{1}\circ d^{0}=f\cdot id_{X^{0}}$.
If $X$ is a free $R-$module, then the pair $(X,d)$ is called a matrix
factorization and we often refer to it by $X$ without explicitly mentioning
the differential $d$.

######  Remark 4.1.

If $M_{0}$ and $M_{1}$ are respectively matrices of the $R-$linear
endomorphisms $d^{0}$ and $d^{1}$, then the pair $(M_{0},M_{1})$ would be a
matrix factorization of $f$ according to definition 2.1.

######  Definition 4.2.

(p.9 [8]) Morphism of linear factorizations
A morphism of linear factorizations $(X,d_{X})$ and $(Y,d_{Y})$ is an even
(i.e., a grade preserving) $R-$linear map $\phi:X\longrightarrow Y$ such that
$d_{Y}\phi=\phi d_{X}$.

Concretely (see page 19 of [21]), $\phi$ is a pair of maps
$\textstyle{X^{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi^{0}}$$\textstyle{Y^{0}}$
and
$\textstyle{X^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi^{1}}$$\textstyle{Y^{1}}$
such that the following diagram commutes:

$\textstyle{X^{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{{d^{0}_{X}}}$$\scriptstyle{\phi^{0}}$$\textstyle{X^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi^{1}}$$\scriptstyle{{d^{1}_{X}}}$$\textstyle{X^{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi^{0}}$$\textstyle{Y^{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{{d^{0}_{Y}}}$$\textstyle{Y^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{{d^{1}_{Y}}}$$\textstyle{Y^{0}}$

######  Definition 4.3.

(p.9 [8]) homotopic linear factorizations
Let $(X,d_{X})$ and $(Y,d_{Y})$ be linear factorizations. Two morphisms
$\varphi,\psi:X\longrightarrow Y$ are homotopic if there exists an odd
$R-$linear map $\lambda:X\longrightarrow Y$ such that $d_{Y}\lambda+\lambda
d_{X}=\psi-\varphi$.
More precisely, the following diagram commutes:

$\textstyle{X_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d_{X}}$$\scriptstyle{\psi-\phi}$$\textstyle{X_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\lambda_{0}}$$\scriptstyle{\psi-\phi}$$\scriptstyle{d_{X}}$$\textstyle{X_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\lambda_{1}}$$\scriptstyle{\psi-\phi\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\dagger}}$$\textstyle{Y_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d_{Y}}$$\textstyle{Y_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d_{Y}}$$\textstyle{Y_{1}}$

i.e.,

$d_{Y}\circ\lambda_{0}+\lambda_{1}\circ d_{X}=\psi-\phi$

######  Notations 4.1.

We keep the following notations used in [8]:
$\bullet$ We denote by $HF(R,f)$ the category of linear factorizations of
$f\in R$ modulo homotopy.
$\bullet$ We also denote by $HMF(R,f)$ its full subcategory of matrix
factorizations.
$\bullet$ Furthermore, we write $hmf(R,f)$ for the full subcategory of finite
rank matrix factorizations, viz. the matrix factorizations whose underlying
$R$-module is free of finite rank.

######  Remark 4.2.

(p.9 of [8])
$HMF(R,f)$ is idempotent complete ([5], [24]). As earlier stated, we work with
polynomials rather than power series, so $hmf(R,f)$ is not necessarily
idempotent complete [20]. The idempotent closure of $hmf(R,f)$ (denoted by
$hmf(R,f)^{\omega}$) is a full subcategory of $HMF(R,f)$ whose objects are
those matrix factorizations which are direct summands of finite-rank matrix
factorizations in the homotopy category. Moreover, $hmf(R,f)^{\omega}$ is an
idempotent complete category. Taking the idempotent completion is necessary
because the composition of $1$-morphisms in $\mathcal{LG}_{K}$ results in
matrix factorisations which, while not finite-rank, are summands in the
homotopy category of something finite-rank. There are two natural ways to
resolve this: work throughout with power series rings and completed tensor
products, or work with idempotent completions.

Let ($R=K[x],f$) and ($S=K[y],g$) be elements of $B-Fac$. The small category
$B-Fac((R,f),(S,g))$ is defined as follows:

$B-Fac((R,f),(S,g)):=hmf(R\otimes S,1_{R}\otimes g-f\otimes
1_{S})^{\omega}=hmf(K[x,y],g-f)^{\omega}$

viz. a 1-morphism between two polynomials $f$ and $g$ is a matrix
factorization of $g-f$.
Then given two composable $1-$cells $X\in B-Fac((R,f),(S,g))$ and $Y\in
B-Fac((S,g),(T,h))$, we define their composition using Yoshino’s tensor
product as discussed in subsection 2.2. $Y\circ X:=Y\otimes_{S}X$
$\in\,HMF(R\otimes_{K}T,1_{R}\otimes h-f\otimes 1_{T})$ which is a
$\mathbb{Z}_{2}-$graded module, where:

$(Y\otimes_{S}X)^{0}=(Y^{0}\otimes_{S}X^{0})\oplus(Y^{1}\otimes_{S}X^{1})\;\;and\;\;(Y\otimes_{S}X)^{1}=(Y^{0}\otimes_{S}X^{1})\oplus(Y^{1}\otimes_{S}X^{0}),$

the differential ([8]) is

$d_{Y\otimes X}=d_{Y}\otimes 1+1\otimes d_{X},$

where the tensor product is taken over $S$ and $1\otimes d_{X}$ has the usual
Koszul signs when applied to elements. That is;

$d_{Y\otimes X}(y,x)=d_{Y}(y)\otimes x+(-1)^{i}y\otimes
d_{X}(x)\,\,\,\,\,\,\,\,\,$

(see p.28 [21]) where $y\in Y^{i}$.
By remark 2.1.8 on p.29 of [6], $Y\otimes_{S}X$ is a free module of infinite
rank over $R\otimes_{K}T$. However, the argument of Section 12 of [12] shows
that it is naturally isomorphic to a direct summand in the homotopy category
of something finite-rank. So, we may define $Y\circ X:=Y\otimes_{S}X$
$\in\,hmf(R\otimes_{K}T,1_{R}\otimes h-f\otimes
1_{T})^{\omega}=B-Fac((R,f),(T,h))$.

We now define the tensor product of morphisms of matrix factorizations. Let
$X_{1}$, $X_{2}$ be objects of $B-Fac((R,f),(S,g))$ and $Y_{1}$, $Y_{2}$ be
objects of $B-Fac((S,g),(T,h))$. Let $\alpha:X_{1}\rightarrow X_{2}$ and
$\beta:Y_{1}\rightarrow Y_{2}$ be two morphisms, then we define their tensor
product in the obvious way $\beta\otimes\alpha:Y_{1}\otimes X_{1}\rightarrow
Y_{2}\otimes X_{2}$ in $B-Fac((R,f),(T,h))$.
With the above data, the composition (bi-)functor is entirely determined in
our B-category:

$\star_{(R,f),(S,g),(T,h)}:B-Fac((R,f),(S,g))$ x
$B-Fac((S,g),(T,h))\rightarrow B-Fac((R,f),(T,h))$

$(X,Y)\mapsto Y\otimes_{S}X$

The definition of the associativity morphism is easy to state. In fact, for
$X\in B-Fac((R,f),(S,g))$, $Y\in B-Fac((S,g),(T,h))$ and $Z\in
B-Fac((T,h),(P,r))$, the associator is the 2-isomorphism

$a_{Z,Y,X}:Z\otimes(Y\otimes X)\rightarrow(Z\otimes Y)\otimes X$

given by the usual formula

$z\otimes(y\otimes x)\rightarrow(z\otimes y)\otimes x$

where $x\in X$, $y\in Y$ and $z\in Z$.

######  Lemma 4.1.

$B-Fac$ is a $B-$category.

We now discuss the construction of the units in $\mathcal{LG}_{K}$.

###  4.1 Unit $1-$morphisms in $\mathcal{LG}_{K}$

Here, we will construct the identity 1-cells. We let
$R=K[x_{1},x_{2},\cdots,x_{n}]$.

From lemma 3.2, we have in particular that

$R\otimes_{K}R\cong
K[x_{1},x_{2},\cdots,x_{n},x^{\prime}_{1},x^{\prime}_{2},\cdots,x^{\prime}_{n}].$

where $x_{i}=x_{i}\otimes 1$ and $x^{{}^{\prime}}_{i}=1\otimes x_{i}$.
The subscript $"K"$ in $\otimes_{K}$ will be very often omitted for ease of
notation. We need an object $\Delta_{f}:(R,f)\rightarrow(R,f)$ in
$hmf(R\otimes R,f\otimes id-id\otimes f)$ or equivalently,
$hmf(K[x_{1},x_{2},\cdots,x_{n},x^{\prime}_{1},x^{\prime}_{2},\cdots,x^{\prime}_{n}],h(x,x^{\prime})),$
where $h(x,x^{\prime})=f(x)-f(x^{\prime}),$ where $id$ stands for $1_{R}$.
Recall (cf. section 5.5 of [28]): The exterior algebra $\bigwedge(V)$ of a
vector space $V$ over a field K is defined as the quotient algebra of the
tensor algebra; $T(V)=\bigoplus_{i=1}^{\infty}T^{i}(V)=K\bigoplus
V\bigoplus(V\otimes V)\bigoplus(V\otimes V\otimes V)\bigoplus\cdots,$ by the
two-sided ideal $I$ generated by all elements of the form $x\otimes x$ for
$x\in V$. Symbolically, $\bigwedge(V)=T(V)/I$. The exterior product $\wedge$
of two elements of $\bigwedge(V)$ is the product induced by the tensor product
$\otimes$ of $T(V)$. That is, if

$\pi:T(V)\rightarrow\bigwedge(V)=T(V)/I$

is the canonical surjection, and if a and b are in $\bigwedge(V)$, then there
are $\alpha$ and $\beta$ in $T(V)$ such that $a=\pi(\alpha)$ and
$b=\pi(\beta)$ and $a\wedge b=\pi(\alpha\otimes\beta)$. Let
$\theta_{1},\theta_{2},\cdots,\theta_{n}$ be formal symbols111That is, we
declare those symbols to be linearly independent by definition. We consider
the $R\otimes R-$module:

$\Delta_{f}=\bigwedge\\{\bigoplus_{i=1}^{n}(R\otimes R)\theta_{i}\\}$

This is an exterior algebra generated by $n$ anti-commuting variables, the
$\theta_{i}s$ modulo the relations that the $\theta$’s anti-commute, that is
$\theta_{i}\wedge\theta_{j}=-\theta_{j}\wedge\theta_{i}$. Typically, we will
omit the wedge product and write for instance $\theta_{i}\wedge\theta_{j}$
simply as $\theta_{i}\theta_{j}$. Here, the ”$\wedge$” product is taken over K
just like the tensor product. A typical element of $\Delta_{f}$ is $(r\otimes
r^{\prime})\theta_{i_{1}}\theta_{i_{2}}\cdots\theta_{i_{k}}$ or equivalently
$h(x_{1},x_{2},\cdots,x_{n},x^{\prime}_{1},x^{\prime}_{2},\cdots,x^{\prime}_{n})\theta_{i_{1}}\theta_{i_{2}}\cdots\theta_{i_{k}}$
where $i_{1},\cdots,i_{k}\in\\{1,\cdots,n\\}$ and
$h(x_{1},x_{2},\cdots,x_{n},x^{\prime}_{1},x^{\prime}_{2},\cdots,x^{\prime}_{n})$
$\in$
$K[x_{1},x_{2},\cdots,x_{n},x^{\prime}_{1},x^{\prime}_{2},\cdots,x^{\prime}_{n}]$.
$\Delta_{f}$ as an algebra is finitely generated by the set of formal symbols
$\\{\theta_{1},\cdots,\theta_{n}\\}$.
$\Delta_{f}$ as an $R\otimes R-$module is generated by the set containing the
empty list and all products of the form $\theta_{i_{1}}\dots\theta_{i_{k}}$
where $i_{1},\cdots,i_{k}\in\\{1,\cdots,n\\}$. The action of $R\otimes R$ is
the obvious one.

$\Delta_{f}$ is endowed with the $\mathbb{Z}_{2}-$grading given by
$\theta-$degree (where deg$\theta_{i}=1$ for each $i$). Thus
$deg\theta_{i}^{2}=0\;and\;deg\theta_{i}\theta_{j}=0$.
Next, we define the differential as follows:

$d:\Delta_{f}\longrightarrow\Delta_{f}$
$d(-)=\sum_{i=1}^{n}[(x_{i}-x_{i}^{{}^{\prime}})\theta_{i}^{\ast}(-)+\partial_{i}(f)\theta_{i}\wedge(-)]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdots\,\,\,\natural$

Where $\theta_{i}^{\ast}$ is the unique derivation extending the map
$\theta_{i}^{\ast}(\theta_{j})=\delta_{ij}$ and as mentioned in [8], it acts
on an element $\theta_{i_{1}}\theta_{i_{2}}\cdots\theta_{i_{k}}$ of the
exterior algebra by the Leibniz rule with Koszul signs. In fact,

$\theta_{i}^{\ast}(\theta_{j_{1}}\theta_{j_{2}}\cdots\theta_{j_{k}})=\begin{cases}0\,for\,i\neq
j,\,\forall j\in\\{{j_{1}},{j_{2}},\cdots,{j_{k}}\\}\\\
(-1)^{p+1}\theta_{j_{1}}\theta_{j_{2}}\cdots\hat{\theta_{i}}\cdots\theta_{j_{k}}\,\,otherwise\end{cases}$

where $\hat{\theta_{i}}$ signifies that $\theta_{i}$ has been removed, and $p$
is the position of $\theta_{i}$ in
$\theta_{j_{1}}\theta_{j_{2}}\cdots\theta_{i}\cdots\theta_{j_{k}}$

And:

$\partial_{i}:k[x_{1},\cdots,x_{n},x_{1}^{{}^{\prime}},\cdots,x_{n}^{{}^{\prime}}]\longrightarrow
k[x_{1},\cdots,x_{n},x_{1}^{{}^{\prime}},\cdots,x_{n}^{{}^{\prime}}]$

is defined by,

$\partial_{i}(h)=\frac{h(x_{1}^{{}^{\prime}},\cdots,x_{i-1}^{{}^{\prime}},x_{i},\cdots,x_{n},x_{1}^{{}^{\prime}},\cdots,x_{n}^{{}^{\prime}})-h(x_{1}^{{}^{\prime}},\cdots,x_{i}^{{}^{\prime}},x_{i+1},\cdots,x_{n},x_{1}^{{}^{\prime}},\cdots,x_{n}^{{}^{\prime}})}{x_{i}-x_{i}^{{}^{\prime}}}$

where for ease of notation we wrote $h$ as argument of $\partial_{i}$ instead
of the more cumbersome notation
$\partial_{i}(h(x_{1},\cdots,x_{n},x_{1}^{{}^{\prime}},\cdots,x_{n}^{{}^{\prime}}))$.
The following lemma will be useful in the definition of the left and the right
units of $\mathcal{LG}_{K}$.

######  Lemma 4.2.

There is a canonical map of factorizations
$\pi:\Delta_{f}\longrightarrow R$ given by $\pi[(r\otimes
r^{\prime})\theta_{i_{1}}\theta_{i_{2}}\cdots\theta_{i_{k}}]=\delta_{k,0}rr^{\prime}$.
$\pi$ is in fact the composition of the projection
$\pi^{\ast}:\Delta_{f}\longrightarrow R\otimes R$ to $\theta-$degree $0$,
followed by the multiplication map $m:R\otimes R\longrightarrow R=R_{0}\oplus
R_{1}$, where we endow $R$ with the trivial grading i.e., $R_{0}=R$ and
$R_{1}=\\{0\\}$.

###  4.2 The left and the right units of $\mathcal{LG}_{K}$

In this subsection, we recall the definition of the left and right identities
of the bicategory $\mathcal{LG}_{K}$. We will denote the right (respectively
left) unit by $\rho$ (respectively $\lambda$).
Consider a $1$-morphism $X\in$ $hmf(R\otimes S,1_{R}\otimes g-f\otimes
1_{S})^{\omega}=hmf(R\otimes S,id\otimes g-f\otimes id)^{\omega}$.
Thus, $X$ is a matrix factorization of $id\otimes g-f\otimes id$ and is also
an $R\otimes S-$module.
Let $1_{X}:X\longrightarrow X$ be the identity map and $\pi$ be the projection
defined in lemma 4.2.

######  Remark 4.3.

Observe that any $S-$module $N$ can be considered as an $R-$module by letting
$rn:=f(r)n$ where $f:R\longrightarrow S$ is a homomorphism of commutative
rings. It is easy to see that the $R\otimes S-$module $X$ can be considered as
an $R-$module via the following $K-$homomorphism of commutative unitary rings
$f:R\longrightarrow R\otimes S$ defined by $f(r)=r\otimes 1_{S}$, and hence
one can also see $X$ as an $R\otimes R-$module by means of the following
(multiplicative map which is a) $K-$homomorphism of commutative unitary rings
$m:R\otimes R\longrightarrow R$ defined by $m(r\otimes
r^{\prime})=rr^{\prime}$. It is not difficult to see that the $R\otimes
R-$module $\Delta_{f}$ can be considered as an $R-$module via the following
homomorphism of commutative unitary rings $f:R\longrightarrow R\otimes R$
defined by $f(r)=r\otimes 1_{R}$.

Thanks to this remark, it makes sense to form the following tensor product
over $R$: $X\otimes_{R}R$ and $X\otimes_{R}\Delta_{f}$ since $X$ and
$\Delta_{f}$ can be viewed as $R-$modules. Consequently, we will simply write
$X\otimes R$ and $X\otimes\Delta_{f}$ for ease of notation.
Similarly, since the $R\otimes S-$module $X$ and the $S\otimes S-$module
$\Delta_{g}$ can be viewed as $S-$modules, we can form the module
$\Delta_{g}\otimes_{S}X$ that we simply write as $\Delta_{g}\otimes X$.
Also consider the map $u:X\otimes R\longrightarrow X$ defined by $u(x\otimes
r)=xr$. This definition makes sense since $X$ can be viewed as an $R-$module.
$u$ is an isomorphism (See example 1 page 363 of [11]).

Now, define $\rho_{X}:X\otimes\Delta_{f}\longrightarrow X$ by
$\rho_{X}:=u\circ(1_{X}\otimes\pi)$ and $\lambda_{X}:\Delta_{g}\otimes
X\longrightarrow X$ by $\lambda_{X}:=u\circ(\pi\otimes 1_{X})$.
$\rho_{X}$ and $\lambda_{X}$ are clearly morphisms in $hmf(R\otimes
S,id\otimes g-f\otimes id)^{\omega}$.
$\rho$ is natural w.r.t. $2-$morphisms in the variable $X$ and there is no
direct inverse for $\rho_{X}$, for each $X$ (See section 5.2 of [14]).

##  5 Morita contexts in $\mathcal{LG}_{K}$

Before discussing the notion of Morita context in $\mathcal{LG}_{K}$, we
define what it is in an arbitrary bicategory.

######  Definition 5.1.

[25]
Let $\mathcal{B}$ be a bicategory with natural isomorphisms a, r and l. Given
two 0-cells A and B, we define a Morita Context between A and B as a four-
tuple $\Gamma=(f,g,\eta,\rho)$ consisting of two 1-cells $f\in
Hom_{\mathcal{B}}(A,B)$ and $g\in Hom_{\mathcal{B}}(B,A)$, and two 2-cells
$\eta:fg\rightarrow 1_{A}$ and $\rho:gf\rightarrow 1_{B}$ such that the
following diagrams commute

|
---|---
$\textstyle{(gf)g\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{a}$$\scriptstyle{\rho 1}$$\textstyle{g(fg)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1\eta}$$\textstyle{1_{B}g\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{l}$$\textstyle{g1_{A}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{r}$$\textstyle{g}$ |
---|---
$\textstyle{(fg)f\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{a}$$\scriptstyle{\eta
1}$$\textstyle{f(gf)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1\rho}$$\textstyle{1_{A}f\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{l}$$\textstyle{f1_{B}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{r}$$\textstyle{f}$

Equationally, we have:

  1. 1.

$r\circ 1\eta\circ a=l\circ\rho 1$

  2. 2.

$r\circ 1\rho\circ a=l\circ\eta 1$

######  Remark 5.1.

[25]

  * •

Observe that any adjunction $<\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
6.75pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
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13.65971pt\raise 6.1111pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.75pt\hbox{$\scriptstyle{f}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 30.75pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
30.75pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{B}$}}}}}}}\ignorespaces}}}}\ignorespaces,\lx@xy@svg{\hbox{\raise
0.0pt\hbox{\kern
7.0434pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
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0.0pt\hbox{$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
14.24844pt\raise 5.1875pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-0.8264pt\hbox{$\scriptstyle{g}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 31.0434pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
31.0434pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{A}$}}}}}}}\ignorespaces}}}}\ignorespaces,\lx@xy@svg{\hbox{\raise
0.0pt\hbox{\kern
7.6pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{1_{A}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
14.9679pt\raise 4.50694pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
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3.0pt}}}}}}\ignorespaces{\hbox{\kern 31.6pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
31.6pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{fg}$}}}}}}}\ignorespaces}}}}\ignorespaces,\lx@xy@svg{\hbox{\raise
0.0pt\hbox{\kern
8.55035pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{gf\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
15.8125pt\raise 5.1875pt\hbox{{}\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\hbox{\hbox{\kern 0.0pt\raise-0.8264pt\hbox{$\scriptstyle{\eta}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 32.55035pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
32.55035pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{1_{B}}$}}}}}}}\ignorespaces}}}}\ignorespaces>$ becomes
a Morita context as soon as its unit $\varepsilon$ is invertible.

  * •

A Morita context is strict if both $\eta$ and $\rho$ in the foregoing
definition are isomorphisms. Strict Morita contexts and adjoint equivalences
are basically the same. One can switch between them by inverting the unit
$\varepsilon$ of the adjunction.

Nota Bene: It is perhaps good to mention that what is called Morita context in
this paper is instead called wide right Morita context from $B$ to $A$ in
[19]. In [25], it is also called an abstract bridge. Both authors declare that
the notion of left Morita context is defined by reversing 2-cells. We will not
deal with left Morita context in our work.

In the sequel, we will write $K[x]$ (respectively $K[y]$) for
$K[x_{1},x_{2},\cdots,x_{n}]$ ($K[y_{1},y_{2},\cdots,y_{m}]$), where $x_{i}$
and $y_{j}$ are indeterminates for $i,j\in\\{1,2,\cdots,max(n,m)\\}$.

Description of Morita context in $\mathcal{LG}_{K}$
Let $(R,f)$ and $(S,g)$ be two objects of $\mathcal{LG}_{K}$, that is
polynomials such that $f\in R=K[x]$ and $g\in S=K[y]$. In all of this section,
except otherwise stated, we want to keep the following remark and assumption
in mind:

######  Remark 5.2.

Let $X\in\mathcal{LG}_{K}((R,f),(S,g))$ and
$Y\in\mathcal{LG}_{K}((S,g),(R,f))$ be $1-$morphisms of $\mathcal{LG}_{K}$, we
normally have:

  1. 1.

$X\in\mathcal{LG}_{K}((R,f),(S,g))=hmf(R\otimes_{K}S,g-f)^{\omega}$ i.e.,
$X:(R,f)\rightarrow(S,g)$ is a matrix factorization which is a direct summand
of a finite rank matrix factorization of $g-f$.

  2. 2.

$Y\in\mathcal{LG}_{K}((S,g),(R,f))=hmf(S\otimes_{K}R,f-g)^{\omega}$ i.e.,
$Y:(S,g)\rightarrow(R,f)$ is a matrix factorization which is a direct summand
of a finite rank matrix factorization of $f-g$.

  3. 3.

$\Delta_{f}\in\mathcal{LG}_{K}((R,f),(R,f))=hmf(R\otimes_{K}R,f\otimes id-
id\otimes f)$ i.e., $\Delta_{f}:(R,f)\rightarrow(R,f)$ is a finite rank matrix
factorization of $f\otimes id-id\otimes f$.

  4. 4.

$\Delta_{g}\in\mathcal{LG}_{K}((S,g),(S,g))=hmf(S\otimes_{K}S,g\otimes id-
id\otimes g)$ i.e., $\Delta_{g}:(S,g)\rightarrow(S,g)$ is a finite rank matrix
factorization of $g\otimes id-id\otimes g$.

Observe that the $1-$morphism $X$ as defined in the previous remark can also
be a finite rank matrix factorization i.e., an object of
$hmf(R\otimes_{K}S,g-f)$ which is a subcategory of
$hmf(R\otimes_{K}S,g-f)^{\omega}$. A similar observation holds for $Y$.
We would like to use determinant of matrices to discuss the notion of Morita
context in $\mathcal{LG}_{K}$, therefore, it is important for us to deal with
entities that are of finite rank. That is why we need the following
assumption.
Assumption: We restrict our study of Morita context in $\mathcal{LG}_{K}$ to
those objects $(R,f)$ and $(S,g)$ which are such that the following
$1-$morphisms are all finite rank matrix factorizations: $X$, $Y$, $X\otimes
Y$ and $Y\otimes X$.
Since we will only be dealing with finite rank matrix factorizations, we will
sometimes intentionally omit the phrase ”finite rank” in front of the phrase
”matrix factorization”,.

A Morita Context between objects $(R,f)$ and $(S,g)$ of $\mathcal{LG}_{K}$ is
a four-tuple $\Gamma=(X,Y,\eta,\rho)$ where:

  * •

$X:(R,f)\rightarrow(S,g)$ is a matrix factorization of $g-f$.

  * •

$Y:(S,g)\rightarrow(R,f)$ is a matrix factorization of $f-g$.

  * •

$\eta:X\otimes_{R}Y\rightarrow 1_{(R,f)}=\Delta_{f}$ where $\Delta_{f}$ is the
identity on $(R,f)$ in $\mathcal{LG}_{K}$ i.e., a matrix factorization of
$f\otimes id-id\otimes f$ as seen in the previous section.

  * •

$\rho:Y\otimes_{S}X\rightarrow 1_{(S,g)}=\Delta_{g}$

such that the following diagrams (thereafter referred to as the $M.C.LG_{K}$
diagrams) commute up to homotopy:

|
---|---
$\textstyle{(Y\otimes_{S}X)\otimes_{R}Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{a}$$\scriptstyle{\rho\otimes 1_{Y}}$$\textstyle{Y\otimes_{S}(X\otimes_{R}Y)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1_{Y}\otimes\eta}$$\textstyle{\Delta_{g}\otimes_{S}Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{l_{Y}}$$\textstyle{Y\otimes_{R}\Delta_{f}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{r_{Y}}$$\textstyle{Y}$ |
---|---
$\textstyle{(X\otimes_{R}Y)\otimes_{S}X\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{a}$$\scriptstyle{\eta\otimes
1_{X}}$$\textstyle{X\otimes_{R}(Y\otimes_{S}X)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{1_{X}\otimes\rho}$$\textstyle{\Delta_{f}\otimes_{R}X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{l_{X}}$$\textstyle{X\otimes_{S}\Delta_{g}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{r_{X}}$$\textstyle{X}$

That is, the following two conditions hold:

    1. 1.

$r_{Y}\circ 1_{Y}\otimes\eta\circ a\approx l_{Y}\circ\rho\otimes 1_{Y}$

    2. 2.

$r_{X}\circ 1_{X}\otimes\rho\circ a\approx l_{X}\circ\eta\otimes 1_{X}$

where $\approx$ stands for the homotopy relation.

equivalently:

  1. 1.

$\exists\lambda:Z\rightarrow Y$ s.t. $d_{Y}\lambda+\lambda d_{Z}=\psi-\phi$,
where $Z=(Y\otimes_{S}X)\otimes_{R}Y$, $\psi=r_{Y}\circ 1_{Y}\otimes\eta\circ
a$, and $\phi=l_{Y}\circ\rho\otimes 1_{Y}$

  2. 2.

$\exists\xi:Z^{\prime}\rightarrow X$ s.t. $d_{X}\xi+\xi
d_{Z^{\prime}}=\psi^{\prime}-\phi^{\prime}$, where
$Z^{\prime}=(X\otimes_{R}Y)\otimes_{S}X$, $\psi^{\prime}=r_{X}\circ
1_{X}\otimes\rho\circ a$, and $\phi^{\prime}=l_{X}\circ\eta\otimes 1_{X}$

We would now like to find necessary conditions on $\eta$ and $\rho$ to obtain
a Morita context in $\mathcal{LG}_{K}$. We will use a matrix approach because
it is easier to use matrices than linear transformations in our setting. Thus,
we will have recourse to one of the properties of matrix factorizations we
studied earlier (cf. corollaries 3.1 and 3.2); namely that matrices that
appear in a matrix factorization of a nonzero polynomial are invertible.
In the sequel, except otherwise stated, matrices that appear in a matrix
factorization are of a fixed size $n\in\mathbb{N}$. So we will not bother to
mention the sizes of pair of matrices constituting a matrix factorization.
Let $(P,Q)$ and $(R,T)$ be pairs of matrices representing respectively the
matrix factorizations $X\otimes_{R}Y$ and $\Delta_{f}$. We assume $f$ is not a
constant polynomial, thus $\Delta_{f}$ is not the zero polynomial and so, by
corollaries 3.1 and 3.2, we have that $R$ (respectively $T$) is invertible
over the field of fractions of $R$ (respectively over the field of fractions
of $T$).
We know that
$\eta=(\eta^{0},\eta^{1}):X\otimes Y\rightarrow 1_{(R,f)}=\Delta_{f}$
is a morphism of matrix factorizations if the following diagram commutes:
$\textstyle{(X\otimes
Y)^{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{{d^{0}_{X\otimes
Y}}}$$\scriptstyle{\eta^{0}}$$\textstyle{(X\otimes
Y)^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\eta^{1}}$$\scriptstyle{{d^{1}_{X\otimes
Y}}}$$\textstyle{(X\otimes
Y)^{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\eta^{0}}$$\textstyle{(\Delta_{f})^{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{{d^{0}_{\Delta_{f}}}}$$\textstyle{(\Delta_{f})^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{{d^{1}_{\Delta_{f}}}}$$\textstyle{(\Delta_{f})^{0}}$.

That is:

$\begin{cases}\eta^{0}d^{1}_{X\otimes Y}=d^{1}_{\Delta_{f}}\eta^{1}\\\
\eta^{1}d^{0}_{X\otimes Y}=d^{0}_{\Delta_{f}}\eta^{0}\end{cases}$

In matrix form:

$\sharp\begin{cases}\eta^{0}Q=T\eta^{1}...i)\\\
\eta^{1}P=R\eta^{0}...ii)\end{cases}$

where for ease of notation we wrote $\eta^{j}$ for the matrix of $\eta^{j}$,
$j=0,1$.
Now, $X$ (respectively $Y$) being a matrix factorization of $g(y)-f(x)$
(respectively $f(x)-g(y)$) we obtain from the Yoshino’s tensor product of
matrix factorizations (cf. definition 1.2 of [29]) that, $X\otimes Y$ is a
factorization of $(g(y)-f(x))+(f(x)-g(y))=0$. This means that $PQ=0$.

from $ii)$, since $R$ is invertible, we have:

$\displaystyle\eta^{1}P=R\eta^{0}$ $\displaystyle\Rightarrow
R^{-1}\eta^{1}P=\eta^{0}\cdots\bigstar$ $\displaystyle\Rightarrow
R^{-1}\eta^{1}PQ=\eta^{0}Q$
$\displaystyle\Rightarrow\eta^{0}Q=0\cdots\bigstar\bigstar$

Putting this in $i)$, since $T$ is invertible, we obtain:

$\displaystyle T\eta^{1}=\eta^{0}Q=0$
$\displaystyle\Rightarrow\eta^{1}=T^{-1}0=0$
$\displaystyle\Rightarrow\eta^{1}=0$

We would now like to give a necessary condition on $\rho$. The process to
obtain this necessary condition is completely analogous to what we did in the
case of $\eta$. But we will present it here for the sake of clarity.

Let $(P^{\prime},Q^{\prime})$ and $(R^{\prime},T^{\prime})$ be pair of
matrices representing respectively the matrix factorizations $Y\otimes_{S}X$
and $\Delta_{g}$. We assume $g$ is not a constant polynomial, thus
$\Delta_{g}$ is not the zero polynomial and so, by corollaries 3.1 and 3.2, we
have that $R^{\prime}$ and $T^{\prime}$ are invertible.
We know that
$\rho=(\rho^{0},\rho^{1}):Y\otimes X\rightarrow 1_{(S,g)}=\Delta_{g}$
is a morphism of matrix factorizations if the following diagram commutes:
$\textstyle{(Y\otimes
X)^{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{{d^{0}_{Y\otimes
X}}}$$\scriptstyle{\rho^{0}}$$\textstyle{(Y\otimes
X)^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho^{1}}$$\scriptstyle{{d^{1}_{Y\otimes
X}}}$$\textstyle{(Y\otimes
X)^{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\rho^{0}}$$\textstyle{(\Delta_{g})^{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{{d^{0}_{\Delta_{g}}}}$$\textstyle{(\Delta_{g})^{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{{d^{1}_{\Delta_{g}}}}$$\textstyle{(\Delta_{g})^{0}}$

That is:

$\begin{cases}\rho^{0}d^{1}_{Y\otimes X}=d^{1}_{\Delta_{g}}\rho^{1}\\\
\rho^{1}d^{0}_{Y\otimes X}=d^{0}_{\Delta_{g}}\rho^{0}\end{cases}$

In matrix form:

${\dagger}\begin{cases}\rho^{0}Q^{\prime}=T^{\prime}\rho^{1}...i)^{\prime}\\\
\rho^{1}P^{\prime}=R^{\prime}\rho^{0}...ii)^{\prime}\end{cases}$

where for ease of notation we wrote $\rho^{i}$ for the matrix of $\rho^{i}$,
$i=0,1$.
Now, $X$ (respectively $Y$) being a matrix factorization of $g(y)-f(x)$
(respectively $f(x)-g(y)$) we obtain from the Yoshino’s tensor product of
matrix factorization (cf. definition 1.2 of [29]) that $Y\otimes X$ is a
factorization of $(f(x)-g(y))+(g(y)-f(x))=0$. This means that
$P^{\prime}Q^{\prime}=0$.

from $ii)^{\prime}$, since $R^{\prime}$ is invertible, we have:

$\displaystyle\rho^{1}P^{\prime}=R^{\prime}\rho^{0}$ $\displaystyle\Rightarrow
R^{\prime-1}\rho^{1}P^{\prime}=\rho^{0}$ $\displaystyle\Rightarrow
R^{\prime-1}\rho^{1}P^{\prime}Q^{\prime}=\rho^{0}Q^{\prime}$
$\displaystyle\Rightarrow\rho^{0}Q^{\prime}=0$

Putting this in $i)^{\prime}$, since $T^{\prime}$ is invertible, we get:

$\displaystyle T^{\prime}\rho^{1}=\rho^{0}Q^{\prime}=0$
$\displaystyle\Rightarrow\rho^{1}=T^{\prime-1}0=0$
$\displaystyle\Rightarrow\rho^{1}=0$

We now gather our results in the following theorem.

######  Theorem 5.1.

Let

  * •

$(R,f)$ and $(S,g)$ be two objects of $\mathcal{LG}_{K}$.

  * •

$X\in\mathcal{LG}_{K}((R,f),(S,g))=hmf(R\otimes_{K}S,g-f)$ i.e.,
$X:(R,f)\rightarrow(S,g)$ is a finite rank matrix factorization of $g-f$.

  * •

$Y\in\mathcal{LG}_{K}((S,g),(R,f))=hmf(S\otimes_{K}R,f-g)$ i.e.,
$Y:(S,g)\rightarrow(R,f)$ is a finite rank matrix factorization of $f-g$.
such that $X\otimes Y$ and $Y\otimes X$ are finite rank matrix factorizations.

  * •

$\Delta_{f}\in\mathcal{LG}_{K}((R,f),(R,f))=hmf(R\otimes_{K}R,f\otimes id-
id\otimes f)$ i.e., $\Delta_{f}:(R,f)\rightarrow(R,f)$ is a finite rank matrix
factorization of $f\otimes id-id\otimes f$.

  * •

$\Delta_{g}\in\mathcal{LG}_{K}((S,g),(S,g))=hmf(R\otimes_{K}S,g\otimes id-
id\otimes g)$ i.e., $\Delta_{g}:(S,g)\rightarrow(S,g)$ is a finite rank matrix
factorization of $g\otimes id-id\otimes g$.

  * •

$\eta:X\otimes_{R}Y\rightarrow 1_{(R,f)}=\Delta_{f}$ and let $(P,Q)$ and
$(R,T)$ be pairs of matrices representing respectively the finite rank matrix
factorizations $X\otimes Y$ and $\Delta_{f}$.

  * •

$\rho:Y\otimes_{S}X\rightarrow 1_{(S,g)}=\Delta_{g}$ and let
$(P^{\prime},Q^{\prime})$ and $(R^{\prime},T^{\prime})$ be pairs of matrices
representing respectively the finite rank matrix factorizations $Y\otimes X$
and $\Delta_{g}$.

Then: A necessary condition for $\Gamma=(X,Y,\eta,\rho)$ to be a Morita
Context is

$\begin{cases}\eta^{1}=0,\;and\;\eta^{0}Q=0,\\\
\rho^{1}=0,\;and\;\rho^{0}Q^{\prime}=0.\end{cases}$

We will now prove that these necessary conditions are not sufficient thanks to
an auxiliary lemma (lemma 5.1) which is proved using a celebrated result (Fact
5.1) on determinant of block matrices due to Schur (e.g [26], [27] (theorem
3)).

######  Fact 5.1.

If $A,B,C$ and $D$ are $n\times n$ matrices, and

$M=\begin{bmatrix}A&B\\\ C&D\end{bmatrix},$

Then the following hold:

  1. 1.

$det(M)=det(AD-CB)$ if $AC=CA$

  2. 2.

$det(M)=det(AD-BC)$ if $CD=DC$

  3. 3.

$det(M)=det(DA-BC)$ if $BD=DB$

  4. 4.

$det(M)=det(DA-CB)$ if $AB=BA$

The following auxiliary lemma states that the determinants of the matrices in
the matrix factorizations $X\otimes X^{\prime}$ and $X^{\prime}\otimes X$ are
all equal.

######  Lemma 5.1.

Let $X=(\phi,\psi)$ be an $n\times n$ matrix factorization of $f\in R$ and
$X^{\prime}=(\phi^{\prime},\psi^{\prime})$ be an $m\times m$ matrix
factorizations of $g\in S$ where $\phi$ and $\psi$ (respectively
$\phi^{\prime}$ and $\psi^{\prime}$) are matrices over $K[x]$ (respectively
$K[y]$). These matrices can be considered as matrices over $L=K[x,y]$ and let
$X\widehat{\otimes}X^{\prime}=(P,Q)=(\begin{bmatrix}\phi\otimes
1_{m}&1_{n}\otimes\phi^{\prime}\\\ -1_{n}\otimes\psi^{\prime}&\psi\otimes
1_{m}\end{bmatrix},\begin{bmatrix}\psi\otimes
1_{m}&-1_{n}\otimes\phi^{\prime}\\\ 1_{n}\otimes\psi^{\prime}&\phi\otimes
1_{m}\end{bmatrix})$ and

$X^{\prime}\widehat{\otimes}X=(P^{\prime},Q^{\prime})=(\begin{bmatrix}\phi^{\prime}\otimes
1_{n}&1_{m}\otimes\phi\\\ -1_{m}\otimes\psi&\psi^{\prime}\otimes
1_{n}\end{bmatrix},\begin{bmatrix}\psi^{\prime}\otimes
1_{n}&-1_{m}\otimes\phi\\\ 1_{m}\otimes\psi&\phi^{\prime}\otimes
1_{n}\end{bmatrix})$

where each component is an endomorphism on $L^{n}\otimes L^{m}$.
Then

$\begin{cases}det(P)=det(P^{\prime}),\\\ det(Q)=det(Q^{\prime})\end{cases}$

Furthermore, all the four determinants are equal.

###### Proof.

We will make use of fact 5.1(2), to compute the determinants of $P$,
$P^{\prime}$, $Q$ and $Q^{\prime}$ which are block matrices. Looking at $P$,
in order to apply fact 5.1(2), we first have to observe that
$(-1_{n}\otimes\psi^{\prime})(\psi\otimes
1_{m})=-1_{n}\psi\otimes\psi^{\prime}1_{m}=-\psi\otimes\psi^{\prime}=\psi(-1_{n})\otimes
1_{m}\psi^{\prime}=(\psi\otimes 1_{m})(-1_{n}\otimes\psi^{\prime})$ and
looking at $P^{\prime}$ we equally observe that
$(-1_{m}\otimes\psi)(\psi^{\prime}\otimes
1_{n})=-1_{m}\psi^{\prime}\otimes\psi
1_{n}=-\psi^{\prime}\otimes\psi=\psi^{\prime}(-1_{m})\otimes
1_{n}\psi=(\psi^{\prime}\otimes 1_{n})(-1_{m}\otimes\psi)$, we can compute the
determinants of $P$ and $P^{\prime}$ as follows:

$\displaystyle det(P)$ $\displaystyle=det[(\phi\otimes 1_{m})(\psi\otimes
1_{m})-(1_{n}\otimes\phi^{\prime})(-1_{n}\otimes\psi^{\prime})]$
$\displaystyle=det[\phi\psi\otimes
1_{m}+1_{n}\otimes\phi^{\prime}\psi^{\prime}]$
$\displaystyle=det[f1_{n}\otimes 1_{m}+1_{n}\otimes
g1_{m}]\,since\,\phi\psi=f1_{n}\,and\,\phi^{\prime}\psi^{\prime}=g1_{m}$
$\displaystyle=det[f(1_{n}\otimes 1_{m})+g(1_{n}\otimes
1_{m})]\,by\,properties\,of\,\otimes$
$\displaystyle=det[(f+g)1_{nm}]\,since\;1_{n}\otimes 1_{m}=1_{nm}$
$\displaystyle=(f+g)^{nm}det(1_{nm})\;by\,properties\;of\;determinants$
$\displaystyle=(f+g)^{nm}$ $\displaystyle det(P^{\prime})$
$\displaystyle=det[(\phi^{\prime}\otimes 1_{n})(\psi^{\prime}\otimes
1_{n})-(1_{m}\otimes\phi)(-1_{m}\otimes\psi)]$
$\displaystyle=det[\phi^{\prime}\psi^{\prime}\otimes
1_{n}+1_{m}\otimes\phi\psi]$ $\displaystyle=det[g1_{m}\otimes
1_{n}+1_{m}\otimes
f1_{n}]\,since\,\phi\psi=f1_{n}\,and\,\phi^{\prime}\psi^{\prime}=g1_{m}$
$\displaystyle=det[g(1_{m}\otimes 1_{n})+f(1_{m}\otimes 1_{n})]$
$\displaystyle=det[(g+f)1_{mn}]\,since\;1_{m}\otimes 1_{n}=1_{mn}$
$\displaystyle=(g+f)^{mn}det(1_{mn})\;by\,properties\;of\;determinants$
$\displaystyle=(f+g)^{mn}\;by\;commutativity\;in\;K[x,y]$
$\displaystyle=det(P)\,as\;desired.$

Likewise, in order to use fact 5.1(2) to compute $det(Q)$, from $Q$, we
observe that $(1_{n}\otimes\psi^{\prime})(\phi\otimes
1_{m})=1_{n}\phi\otimes\psi^{\prime}1_{m}=\phi\otimes\psi^{\prime}=\phi
1_{n}\otimes 1_{m}\psi^{\prime}=(\phi\otimes
1_{m})(1_{n}\otimes\psi^{\prime})$ and looking at $Q^{\prime}$ we equally
observe that $(1_{m}\otimes\psi)(\phi^{\prime}\otimes
1_{n})=1_{m}\phi^{\prime}\otimes\psi
1_{n}=\phi^{\prime}\otimes\psi=\phi^{\prime}1_{m}\otimes
1_{n}\psi=(\phi^{\prime}\otimes 1_{n})(1_{m}\otimes\psi)$ we can compute the
determinants of $Q$ and $Q^{\prime}$ as follows:

$\displaystyle det(Q)$ $\displaystyle=det[(\psi\otimes 1_{m})(\phi\otimes
1_{m})-(-1_{n}\otimes\phi^{\prime})(1_{n}\otimes\psi^{\prime})]$
$\displaystyle=det[\psi\phi\otimes
1_{m}+1_{n}\otimes\phi^{\prime}\psi^{\prime}]$
$\displaystyle=det[f1_{n}\otimes 1_{m}+1_{n}\otimes
g1_{m}]\,since\,\psi\phi=f1_{n}\,and\,\phi^{\prime}\psi^{\prime}=g1_{m}$
$\displaystyle=det[f(1_{n}\otimes 1_{m})+g(1_{n}\otimes
1_{m})]\,by\,properties\,of\,\otimes$
$\displaystyle=det[(f+g)1_{nm}]\,since\;1_{n}\otimes 1_{m}=1_{nm}$
$\displaystyle=(f+g)^{nm}det(1_{nm})\;by\,properties\;of\;determinants$
$\displaystyle=(f+g)^{nm}$ $\displaystyle det(Q^{\prime})$
$\displaystyle=det[(\psi^{\prime}\otimes 1_{n})(\phi^{\prime}\otimes
1_{n})-(-1_{m}\otimes\phi)(1_{m}\otimes\psi)]$
$\displaystyle=det[\psi^{\prime}\phi^{\prime}\otimes
1_{n}+1_{m}\otimes\phi\psi]$ $\displaystyle=det[g1_{m}\otimes
1_{n}+1_{m}\otimes
f1_{n}]\,since\,\psi\phi=f1_{n}\,and\,\phi^{\prime}\psi^{\prime}=g1_{m}$
$\displaystyle=det[g(1_{m}\otimes 1_{n})+f(1_{m}\otimes 1_{n})]$
$\displaystyle=det[(g+f)1_{mn}]\,since\;1_{m}\otimes 1_{n}=1_{mn}$
$\displaystyle=(g+f)^{mn}det(1_{mn})\;by\,properties\;of\;determinants$
$\displaystyle=(f+g)^{mn}\;by\;commutativity\;in\;K[x,y]$
$\displaystyle=det(Q)\,as\;desired.$

Clearly, all the four determinants are equal. ∎

######  Remark 5.3.

It is good to observe that in the case of a Morita Context in
$\mathcal{LG}_{K}$, the $f$ and the $g$ we have in the above auxiliary lemma
are actually additive inverses of each other by definition of 1-morphisms in
$\mathcal{LG}_{K}$. In fact, if $X$ is a morphism from the polynomial $f_{1}$
to the polynomial $f_{2}$, then $X$ is a matrix factorization of
$f_{2}-f_{1}=f$. And if $Y$ is a morphism from the polynomial $f_{2}$ to the
polynomial $f_{1}$, then $Y$ is a matrix factorization of $f_{1}-f_{2}=g$. We
see that $f=-g$.

Thus, we have the following result which is actually a consequence of lemma
5.1.

######  Theorem 5.2.

Let $(R,f)$ and $(S,g)$ be two objects in $\mathcal{LG}_{K}$ and let
$X:(R,f)\rightarrow(S,g)$ and $X^{\prime}:(S,g)\rightarrow(R,f)$ be
1-morphisms in $\mathcal{LG}_{K}$. If $X\otimes X^{\prime}=(P,Q)$ and
$X^{\prime}\otimes X=(P^{\prime},Q^{\prime})$, then
$det(P)=det(Q)=det(P^{\prime})=det(Q^{\prime})=0$

###### Proof.

The proof follows immediately from the remark and lemma 5.1. ∎

It follows from this theorem that $Q$ in theorem 5.1 is not invertible and so,
the equality $\eta^{0}Q=0$ will never have a unique solution for $\eta^{0}$.
There would be several solutions from this equality, among which the one(s)
that will be sufficient to obtain a Morita context.

######  Remark 5.4.

The fact that $Q$ is not invertible helps to see that the necessary condition
given for $\eta$ in theorem 5.1 is not sufficient. In fact, in the discussion
that precedes the statement of that theorem, we cannot reverse the direction
of the implication symbol from $\bigstar\bigstar$ to $\bigstar$. Indeed,
suppose $\eta^{0}Q=0$ then since $PQ=0$, we have $\eta^{0}Q=R^{-1}\eta^{1}PQ$
implying $R\eta^{0}Q=\eta^{1}PQ$ which implies $(R\eta^{0}-\eta^{1}P)Q=0$.
Now, $Q$ being noninvertible, we cannot obtain from here that
$(R\eta^{0}-\eta^{1}P)=0$ which is $\bigstar$. So, that necessary condition is
not a sufficient one.
Similarly, since it also follows from theorem 5.2 that $Q^{\prime}$ in theorem
5.1 is not invertible, the necessary condition given for $\rho$ in theorem 5.1
is not sufficient.

Though the necessary conditions we found are not sufficient, there is a
trivial sufficient condition. In fact, it is easy to see that two equal
morphisms of linear factorizations are homotopic; since it would suffice to
take $\lambda=0$ in definition 4.3. We immediately have the following remark
which gives a (trivial) sufficient condition to obtain a Morita context in
$\mathcal{LG}_{K}$.

######  Remark 5.5.

Let $(R,f)$ and $(S,g)$ be two objects of $\mathcal{LG}_{K}$. Let
$X:(R,f)\rightarrow(S,g)$ be a matrix factorization of $g-f$ and
$Y:(S,g)\rightarrow(R,f)$ a matrix factorization of $f-g$. Then
$\Gamma=(X,Y,0,0)$ is a Morita Context in $\mathcal{LG}_{K}$. That is,
provided we have $X$ and $Y$, it suffices to take $\eta=0$ and $\rho=0$ in
definition 5.1 to obtain a Morita Context in $\mathcal{LG}_{K}$ .

This follows from the fact that the maps $r$ and $l$ are morphisms of matrix
factorizations and so, they are linear. Consequently, the image of zero under
these morphisms is zero.
In fact:
If $\eta=0$ and $\rho=0$, then:

  1. 1.

$\psi=r_{Y}\circ 1_{Y}\otimes\eta\circ a=0$ and $\phi=l_{Y}\circ\rho\otimes
1_{Y}=0$

  2. 2.

$\psi^{\prime}=r_{X}\circ 1_{X}\otimes\rho\circ a=0$, and
$\phi^{\prime}=l_{X}\circ\eta\otimes 1_{X}=0$

Hence, the $M.C.LG_{K}$ diagrams commute up to homotopy if:
$\exists\lambda:Z\rightarrow Y$ s.t. $d_{Y}\lambda+\lambda d_{Z}=\psi-\phi=0$,
where $Z=(Y\otimes_{S}X)\otimes_{R}Y$ and $\exists\xi:Z^{\prime}\rightarrow X$
s.t. $d_{X}\xi+\xi d_{Z^{\prime}}=\psi^{\prime}-\phi^{\prime}=0$, where
$Z^{\prime}=(X\otimes_{R}Y)\otimes_{S}X$

Hence, it now suffices to choose $\lambda=0$ and $\xi=0$ to see that
$\Gamma=(X,Y,0,0)$ is a Morita Context in $\mathcal{LG}_{K}$.
A straightforward consequence of theorem 5.1 and remark 5.5 is the following:

######  Corollary 5.1.

A necessary and sufficient condition on $\eta^{1}$ and $\rho^{1}$ for
$\Gamma=(X,Y,\eta,\rho)$ to be a Morita Context is $\eta^{1}=0=\rho^{1}$.

######  Example 5.1.

Consider $h=-x^{2}+1\in\mathbb{R}[x]=R$ and $g=y^{2}+1\in\mathbb{R}[y]=S$. A
Morita Context between $(R,h)$ and $(S,g)$ is a quadruple
$\Gamma=(X,X^{\prime},\eta,\rho)$ where:

  * •

$X:(R,h)\rightarrow(S,g)$ is a matrix factorization of
$g-h=y^{2}+x^{2}\in\mathbb{R}[x,y]$. We take
$X=(\begin{bmatrix}x&-y\\\ y&x\end{bmatrix},\begin{bmatrix}x&y\\\
-y&x\end{bmatrix})$ since $\begin{bmatrix}x&-y\\\
y&x\end{bmatrix}\begin{bmatrix}x&y\\\ -y&x\end{bmatrix}=(g-h)\cdot I_{2}$

  * •

$X^{\prime}:(S,g)\rightarrow(R,h)$ is a matrix factorization of
$h-g=-y^{2}-x^{2}\in\mathbb{R}[x,y]$. We take
$X^{\prime}=(\begin{bmatrix}-x&y\\\ -y&-x\end{bmatrix},\begin{bmatrix}x&y\\\
-y&x\end{bmatrix})$ since $\begin{bmatrix}-x&y\\\
-y&-x\end{bmatrix}\begin{bmatrix}x&y\\\ -y&x\end{bmatrix}=(h-g)\cdot I_{2}$

  * •

$\eta=X\otimes X^{\prime}\rightarrow\Delta_{h}$, $x\otimes x^{\prime}\mapsto
0$ viz. $\eta$ is the zero map.

  * •

$\rho=X^{\prime}\otimes X\rightarrow\Delta_{g}$, $x^{\prime}\otimes x\mapsto
0$ viz. $\rho$ is the zero map.

So,

$\Gamma=[(\begin{bmatrix}x&-y\\\ y&x\end{bmatrix},\begin{bmatrix}x&y\\\
-y&x\end{bmatrix}),(\begin{bmatrix}-x&y\\\
-y&-x\end{bmatrix},\begin{bmatrix}x&y\\\ -y&x\end{bmatrix}),0,0]$

is a Morita context between $(R,h)$ and $(S,g)$.

######  Remark 5.6.

  1. 1.

It is good to mention that in such a setting (remark 5.5) not all Morita
Contexts between two objects are identical. In fact, they differ at the level
of the matrix factorizations $X$ and $Y$.

  2. 2.

Intuitively, Morita contexts being pre-equivalences, it is not interesting to
study cases where the two polynomials $f$ and $g$ are equal.

##  6 FURTHER PROBLEMS

In this paper, we gave necessary conditions on $\eta$ and $\rho$ to obtain a
Morita context in $\mathcal{LG}_{K}$. But the only sufficient condition we
were able to give on a $4$-tuple $(X,Y,\eta,\rho)$ to be a Morita context was
the trivial one, i.e., $\eta=\rho=0$. An interesting question would be to find
nontrivial sufficient conditions on $\eta$ and $\rho$.

> Acknowledgments

Part of this work was carried out during my Ph.D. studies in mathematics at
the University of Ottawa in Canada. I am grateful to Prof. Dr. Richard Blute
who was my Ph.D. supervisor for all the fruitful interactions.
I gratefully acknowledge the financial support of the Queen Elizabeth Diamond
Jubilee scholarship during my Ph.D. studies.

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