Cyclically presented groups are finitely presented groups whose relations are invariant under cyclic permutation of generators. Since their introduction, they have provided a setting uniting combinatorial group theory, number theory, and low-dimensional topology. A key theme is expressing algebraic invariants in terms of arithmetic data from the presentation.

\medskip
A particularly important family is given by the Fibonacci-type cyclically presented
groups
\begin{equation}\label{eq:Gnmk}
G_n(m,k)
=
\langle x_0,\dots,x_{n-1}
\mid
x_i x_{i+m}=x_{i+k}\ (0\le i<n)
\rangle,
\qquad (\text{indices mod } n).
\end{equation}

These groups were first discussed in \cite{MR1634446,MR387419} (see also the survey \cite{MR3010816}). They include the classical Fibonacci groups $F(2,n)=G_n(1,2)$\cite{Conway1965}, the Sieradski groups $S(2,n)=G_n(2,1)$ \cite{MR830041}, and the Gilbert-Howie groups $H(n,m)=G_n(m,1)$ \cite{MR1332862}, as well as many generalisations. These groups have been studied from various perspectives, including finiteness and asphericity \cite{MR2003626,MR2464807,MR1622427,MR1298588,MR1332862,MR2488144}, hyperbolicity \cite{MR4223853,MR4243359,MR4424974}, perfectness \cite{MR1700021,MR2661287} (see also \cite{MR4315549}), and topology \cite{MR1332862}.

\medskip
A basic invariant of a finitely presented group is its abelianization.
For cyclically presented groups, this invariant is especially accessible: abelianization gives an integer circulant relation matrix, so its structure is determined by the Smith normal form of this circulant matrix.
Equivalently, the \(\mathbb{Z}\)-rank of the abelianization, that is, the minimal number of generators of its torsion-free part, is determined by the greatest common divisor of the \emph{representer} polynomial (see Section~\ref{prelim}) and $t^n-1$. For the Fibonacci-type presentation \eqref{eq:Gnmk}, the representer polynomial is
\[
1+t^m-t^k.
\]
The \(\mathbb{Z}\)-rank of $G_n(m,k)^{\mathrm{ab}}$ is then equal to
the degree of $\gcd(1+t^m-t^k,t^n-1)$, that is, the number of common roots of
$1+t^m-t^k$ and $t^n-1$, counted with multiplicity.

\medskip

For a generic choice of parameters $n$ and $m$, the abelianization of $H(n,m)$ is finite or of $\mathbb{Z}$-rank 1. In particular, the cases with $\mathbb{Z}$-rank 2 are exceptional. When the abelianization has $\mathbb{Z}$-rank 2, it is necessary that $m \equiv 2 \pmod n$. This places these groups in close relation with the Sieradski groups, for which $m=2$. This motivates the problem of determining exactly when the abelianization is free and has $\mathbb{Z}$-rank 2. Guided by extensive computations and structural results, Noferini and Williams proposed the following conjecture which characterizes exactly when this phenomenon occurs.

\label{prelim}

This section lays the foundation for the remainder of the paper by presenting the core ideas and setting the notation that will be used consistently throughout. All notations introduce

\begin{abstract}
\noindent
The Gilbert–Howie groups $H(n, m)$ form a notable subclass within the broader family of Fibonacci-type cyclically presented groups $G_n(m, k)$. Noferini and Williams conjectured that the abelianization $H(n, m)^{ab}$ is torsion-free with \(\mathbb{Z}\)-rank $2$ if and only if $n \equiv 0 \pmod{6}$ and $m \equiv 2 \pmod{n}$. We confirm this conjecture by proving that \[ \operatorname{Res}(F, G) = \pm 1, \] where \[ F = \frac{1 + t^m - t}{\Phi_6} \quad \text{and} \quad G = \frac{t^n - 1}{\Phi_6}, \] with $\Phi_6$ denoting the sixth cyclotomic polynomial. The proof uses a minimality argument, reducing the general problem to three cases: $m = 2 + \frac{n}{3}$, $m = 2 + \frac{n}{2}$, and $m = 2 + \frac{2n}{3}$.
These cases are then handled using polynomial resultant analysis and field-theoretic methods.
 As a consequence, we complete the classification of all $G_n(m, k)$ that arise as labelled oriented graph groups.

For a generic choice of parameters $n$ and $m$, the abelianization of $H(n,m)$ is finite or of $\mathbb{Z}$-rank 1. In particular, the cases with $\mathbb{Z}$-rank 2 are exceptional. When the abelianization has $\mathbb{Z}$-rank 2, it is necessary that $m \equiv 2 \pmod n$. This places these groups in close relation with the Sieradski groups, for which $m=2$. This motivates the problem of determining exactly when the abelianization is free and has $\mathbb{Z}$-rank 2. Guided by extensive computations and structural results, Noferini and Williams proposed the following conjecture which characterizes exactly when this phenomenon occurs.

The forward implication follows from cyclotomic roots of order six for the trinomial $1+t^m-t$ under the stated congruences (see \cite[Lemma 4.2]{MR4418964}).
The converse is more difficult as one must rule out all other possible cyclotomic coincidences uniformly in $n$ and $m$. This brings us to the purpose of this paper, which is to prove Conjecture~\ref{conj}.

\begin{proof}
Suppose for contradiction that $\operatorname{Res}(F,G)=1$ and that $v\nmid u$. Then $v>1$. By \cite[Lemma~4.3]{MR4418964}, one of the following holds:
\[
v\mid m-1,\quad v\mid m-2,\quad v\mid 1.
\]
Since $\gcd(v,m-2)=1$ and $v>1$, the only possibility is $v\mid m-1$. Write $m=1+\alpha v$ for some $\alpha\in\mathbb{N}$. Since $6\mid m-2$, it follows that $6\mid\alpha v- 1$. Set $\alpha_0 = n/v$. Since $\alpha_0$ is even, an argument similar to that used in the proof of Lemma~\ref{lucas} shows that $t^v + 1$ divides $t^n - 1$. Since $\gcd(6, v) = 1$, we have
\[
\gcd(t^v+1,\Phi_6(t))=1.
\]
Hence $t^v+1$ divides $G(t)$, and so for some $S(t)\in\mathbb{Z}[t]$, we may write
\[
G(t)=(t^v+1)S(t).
\]
By \eqref{eq:res-mult}, we have
\[
\operatorname{Res}(F,G)
=\operatorname{Res}(F,t^v+1)\operatorname{Res}(F,S).
\]
Since $\operatorname{Res}(F,G)=1$, it follows that
\begin{equation}\label{eq:resultant-F}
\operatorname{Res}(F,t^v+1)=1.
\end{equation}

\begin{corollary}\label{special}
Let $n=2\cdot 3^s$ with $s\geq 2$. Then $H(n,2\pm n/3)^{ab}\not\cong \mathbb{Z}^2.$
\end{corollary}
\begin{proof}
This is exactly the situation where $f(t)=t^m-t+1$ with $m=2\pm n/3\pmod n$. Set $K=\pm n/3\pmod n$ and apply Lemma~\ref{lem:v1_reduction} and Lemma~\ref{grow}.
\end{proof}

\begin{lemma}\label{minimality}
Under Assumptions~\ref{std}, let $u=\tau v$ for some $\tau\ge 1$. If $a\ge b+2$ or $s\ge r+2$ or $(a=b+1$ and $s=r+1)$, then there exists an integer $n_1$ with $n_1\mid n$ and $n_1<n$, and an integer
$m_1$ such that $2<m_1<n_1$, $m_1\equiv m\pmod{n_1}$, and $m_1\equiv2\pmod6$.
\end{lemma}

In this section, we give a proof of Conjecture~1.1. To begin, let us recall the setting. Our aim is to show that $H(n,m)^{ab}$ is isomorphic to $\mathbb{Z}^2$ if and only if $n \equiv 0 \pmod{6}$ and $m \equiv 2 \pmod{n}$. By~ \cite[Lemma 4.2]{MR4418964}, a necessary and sufficient condition for \(\mathbb{Z}\)-rank to be~$2$ is $m \equiv 2 \pmod{6}$ and $n \equiv 0 \pmod{6}$. Hence, we write
\[
n = 2^{a} 3^{s} v \quad \text{and} \quad m = 2^{b} 3^{r} u,
\]
where $\gcd(uv,6)=1$ and $a,s,b,r \geq 1$.

\medskip
A direct computation shows that
\[
H(18,14)^{\mathrm{ab}} \cong H(18,8)^{\mathrm{ab}} \cong \mathbb{Z}^2 \oplus \mathbb{Z}/19\mathbb{Z}, \qquad
 H(12,8)^{\mathrm{ab}} \cong \mathbb{Z}^2 \oplus \mathbb{Z}/5\mathbb{Z}.
\]
In particular, all three groups have \(\mathbb{Z}\)-rank~2 and nontrivial torsion.
Hence, while the conditions \(n \equiv 0 \pmod{6}\) and \(m \equiv 2 \pmod{6}\) are sufficient for the abelianization to have \(\mathbb{Z}\)-rank~2, they are not sufficient to ensure that it is torsion-free.

\begin{conjecture}[Noferini--Williams \cite{MR4418964}]\label{conj}
For $n\ge1$ and $0\le m<n$, $H(n,m)^{\mathrm{ab}}\cong\mathbb{Z}^2$
if and only if
$n\equiv0\pmod6$ and
$m\equiv2\pmod n.$
\end{conjecture}

The forward implication follows from cyclotomic roots of order six for the trinomial $1+t^m-t$ under the stated congruences (see \cite[Lemma 4.2]{MR4418964}).
The converse is more difficult as one must rule out all other possible cyclotomic coincidences uniformly in $n$ and $m$. This brings us to the purpose of this paper, which is to prove Conjecture~\ref{conj}.

\medskip
Our method is arithmetic: the \(\mathbb{Z}\)-rank~2, torsion-free condition reduces to $\operatorname{Res}(F,G)=1,$ where
\[
F=\frac{1+t^m-t}{\Phi_6(t)},\quad
G=\frac{t^n-1}{\Phi_6(t)}.
\]
The proof proceeds by a minimality argument. In Section~\ref{tech}, we show that the conjecture holds for the groups $H(n, 2 \pm n/3)$ and $H(n, 2 + n/2)$. We also derive strong necessary conditions on $n$ and $m$ that any minimal counterexample must satisfy. In Section \ref{sec:mainproof}, we show that any such minimal counterexample must in fact belong to one of these families. Consequently, no minimal counterexample exists, and the conjecture follows.

The introduction outlines a proof of Conjecture~\ref{conj} as follows. The forward implication is obtained by exhibiting “cyclotomic roots of order six for the trinomial $1+t^m-t$ under the stated congruences.” For the converse, the paper uses an arithmetic method: the “$\mathbb{Z}$-rank~2, torsion-free condition reduces to $\operatorname{Res}(F,G)=1$,” with
\[
F=\frac{1+t^m-t}{\Phi_6(t)},\quad
G=\frac{t^n-1}{\Phi_6(t)}.
\]
The rest of the argument is “a minimality argument.” First (Section~\ref{tech}) the conjecture is proved for the families $H(n, 2 \pm n/3)$ and $H(n, 2 + n/2)$, and “strong necessary conditions on $n$ and $m$ that any minimal counterexample must satisfy” are derived. Then (Section~\ref{sec:mainproof}) it is shown that any minimal counterexample “must in fact belong to one of these families,” so “no minimal counterexample exists,” yielding the conjecture.

The introduction outlines a proof of
\begin{conjecture}[Noferini--Williams \cite{MR4418964}]\label{conj}
For $n\ge1$ and $0\le m<n$, $H(n,m)^{\mathrm{ab}}\cong\mathbb{Z}^2$
if and only if
$n\equiv0\pmod6$ and
$m\equiv2\pmod n.$
\end{conjecture}
as follows. The forward implication is obtained by exhibiting “cyclotomic roots of order six for the trinomial $1+t^m-t$ under the stated congruences.” For the converse, the paper uses an arithmetic method: the “$\mathbb{Z}$-rank~2, torsion-free condition reduces to $\operatorname{Res}(F,G)=1$,” with
\[
F=\frac{1+t^m-t}{\Phi_6(t)},\quad
G=\frac{t^n-1}{\Phi_6(t)}.
\]
The rest of the argument is “a minimality argument.” First, we use the technical results underpinning the proof of the main theorem:
\label{tech}
In this section, we present the technical results that underpin the proof of Conjecture~\ref{conj}. All notation is as in Section~\ref{std}.

\medskip
Our first result extends \cite[Corol
In this part the conjecture is proved for the families $H(n, 2 \pm n/3)$ and $H(n, 2 + n/2)$, and “strong necessary conditions on $n$ and $m$ that any minimal counterexample must satisfy” are derived. Then (Section~\ref{sec:mainproof}) it is shown that any minimal counterexample “must in fact belong to one of these families,” so “no minimal counterexample exists,” yielding the conjecture.

Question:
For integers \(n\ge 1\) and \(0\le m<n\), let
\[
H(n,m)=\langle x_0,\dots,x_{n-1}\mid x_i x_{i+m}=x_{i+1}\ (0\le i<n)\rangle,
\]
with indices taken modulo \(n\). Which of the following statements is equivalent to the abelianization \(H(n,m)^{\mathrm{ab}}\) being isomorphic to \(\mathbb{Z}^2\)?

\(n\equiv 0\pmod{6}\) and \(m\equiv 2\pmod n\).

Choices:
B. \(n\equiv 0\pmod{6}\) and \(m\equiv 2\pmod{6}\).
C. \(n\equiv 0\pmod{6}\) and \(m\equiv 2\pmod{6}\).
D. \(n\equiv 0\pmod{6}\) and either \(m\equiv 2\pmod n\) or \(m\equiv 2-\frac{n}{3}\pmod n\).
E. \(n\equiv 0\pmod{6}\) and \(m\equiv 2\pmod{\gcd(n,6)}\).

replace mod-n congruence by mod-6 rank-only condition

drop torsion-free strengthening from the exact mod-n condition to the weaker rank-2 congruence condition

promote special minimal-counterexample families to actual solutions

weaken dependence on n from congruence modulo n to congruence modulo gcd(n,6)


\label{lem:kqexist} Let $\mathbb{F}$ be a field.
 Then there exists a uniform Kronecker quotient over $\mathbb{F}$.

Let $\mathbb{F}_m$ denote the vector space of $m\times m$ matrices
over the field $\mathbb{F}$ with characteristic $\chr(\mathbb{F})$, where $m\in\mathbb{N}$. We will denote
by $\mathbb{F}_m\otimes\mathbb{F}_n=\mathbb{F}_{mn}$ the tensor
product of $\mathbb{F}_m$ and $\mathbb{F}_n$. By the matrix $E_{ij}$
we mean a matrix in $\mathbb{F}_m$ with a 1 in row $i$ and column $j$ and
0 in every other entry. In general, the order of the matrix $E_{ij}$ will
be clear from the context. The trace $\tr(A)$ of a matrix $A\in\mathbb{F}_m$
is the sum of the main diagonal entries of $A$.

\subsection{Kronecker products and Kronecker quotients}

Let $A\in\mathbb{F}_m$ and $B\in\mathbb{F}_s$. The Kronecker product
\cite{graham81a,horn91,kronecker3}
$A\otimes B\in\mathbb{F}_{ms}$ is the $ms\times ms$ matrix over $F$ with entries (here $:=$ denotes a definition)
\begin{equation}
 \label{eq:kronent}
 (A\otimes B)_{(i-1)s+p,(j-1)s+q} := (A)_{i,j}(B)_{p,q},
\end{equation}
where
 $i,j\in\{1,\ldots,m\}$ and
 $p,q\in\{1,\ldots,s\}$.
In other words, we can write in block matrix form
\begin{equation*}
A\otimes B=\begin{bmatrix}
                    (A)_{1,1}B & (A)_{1,2}B & \ldots & (A)_{1,n} B \\
                    (A)_{2,1}B & (A)_{2,2}B & \ldots & (A)_{2,n} B \\
                    \vdots     & \vdots     & \ddots & \vdots      \\
                    (A)_{m,1}B & (A)_{m,2}B & \ldots & (A)_{m,n} B
           \end{bmatrix}.
\end{equation*}
We note that the Kronecker product may be viewed as a family of operations
on vector spaces of square matrices (indexed by the orders of the matrices involved),
or as an operation on the graded vector space of all square matrices. In other words,
the Kronecker product is the family of operations
$\otimes:\mathbb{F}_m\times\mathbb{F}_n\to\mathbb{F}_{mn}$ indexed by $m,n\in\mathbb{N}$.

In the same way that a Kronecker product is a family
of operations on various matrix vector spaces (or on the graded vector space of
all square matrices), a Kronecker quotient may be viewed as a family of operations
on various matrix vector spaces (or as a partially defined operation on the graded
vector space of all square matrices). In other words, we consider the family
$\oslash:\mathbb{F}_{mn}\times\mathbb{F}_n\to\mathbb{F}_m$ indexed by $m,n\in\mathbb{N}$
satisfying the following definition.

\begin{definition}[Kronecker quotient \cite{kronquotient,leopardi05a}]
 \label{def:kquotient} A (right) Kronecker quotient is an operation $\oslash$ satisfying
 \begin{equation*}
  (A\otimes B)\oslash B = A
 \end{equation*}
 for all matrices $A\in\mathbb{F}_{m}$ and $B\in\mathbb{F}_{n}$
 ($B$ non-zero) and $m,n\in\mathbb{N}$.
\end{definition}

In general, $(M\oslash A)\otimes A\neq M$ for $mn\times mn$ matrices $M$
and $n\times n$ non-zero matrices $A$. A detailed study of Kronecker quotients
appears in \cite{kronquotient}. Throughout this article, we will consider
only right Kronecker quotients and differences, analogous results hold for
left Kronecker quotients and differences in an obvious way. Uniform Kronecker
quotients are of particular interest, since the family of operations are
related to each other via algebraic equations. More precisely, there is a generating
operation for the entire family in the algebraically obvious way:

\begin{definition}
 \label{def:kquniform} A Kronecker quotient $\oslash$ is a uniform Kronecker quotient if
 \begin{equation*}
  (A\otimes C)\oslash B = A\otimes (C\oslash B)
 \end{equation*}
 for all matrices $A\in\mathbb{F}_{m}$, $B\in\mathbb{F}_n$, $C\in\mathbb{F}_p\otimes\mathbb{F}_n$ and $m,n,p\in\mathbb{N}$,
 and
 \begin{equation*}
  (A+B)\oslash C = (A\oslash C)+(B\oslash C), \qquad
  (kA)\oslash C = k(A\oslash C)
 \end{equation*}
 for all matrices $A$, $B$ and $C$, and scalars $k\in\mathbb{F}$ whenever the operations are defined.
\end{definition}

\begin{remark}
 An equivalent definition for uniform Kronecker quotients
 is as follows, by simply setting $p=1$ in Definition \ref{def:kquniform}:
 A Kronecker quotient $\oslash$ is a uniform Kronecker quotient if
 \begin{math}
  (A\otimes C)\oslash B = A\otimes (C\oslash B)
 \end{math}
 for all matrices $A\in\mathbb{F}_{m}$, $B,C\in\mathbb{F}_n$ and $m,n\in\mathbb{N}$,
 and
 \begin{math}
  (A+B)\oslash C = (A\oslash C)+(B\oslash C),
 \end{math}
 \begin{math}
  (kA)\oslash C = k(A\oslash C)
 \end{math}
 for all matrices $A$, $B$ and $C$, and scalars $k\in\mathbb{F}$ whenever the operations are defined.
\end{remark}

In \cite{kronquotient}, examples of uniform Kronecker quotients over the real
and complex numbers are given. The following lemma establishes the existence
of a uniform Kronecker quotient over an arbitrary field $\mathbb{F}$.

Let $\mathbb{F}_m$ denote the vector space of $m\times m$ matrices
over the field $\mathbb{F}$ with characteristic $\chr(\mathbb{F})$, where $m\in\mathbb{N}$. We will denote
by $\mathbb{F}_m\otimes\mathbb{F}_n=\mathbb{F}_{mn}$ the tensor
product of $\mathbb{F}_m$ and $\mathbb{F}_n$. By the matrix $E_{ij}$
we mean a matrix in $\mathbb{F}_m$ with a 1 in row $i$ and column $j$ and
0 in every other entry. In general, the order of the matrix $E_{ij}$ will
be clear from the context. The trace $\tr(A)$ of a matrix $A\in\mathbb{F}_m$
is the sum of the main diagonal entries of $A$.

Let $A\in\mathbb{F}_m$ and $B\in\mathbb{F}_s$. The Kronecker product
\cite{graham81a,horn91,kronecker3}
$A\otimes B\in\mathbb{F}_{ms}$ is the $ms\times ms$ matrix over $F$ with entries (here $:=$ denotes a definition)
\begin{equation}
 \label{eq:kronent}
 (A\otimes B)_{(i-1)s+p,(j-1)s+q} := (A)_{i,j}(B)_{p,q},
\end{equation}
where
 $i,j\in\{1,\ldots,m\}$ and
 $p,q\in\{1,\ldots,s\}$.
In other words, we can write in block matrix form
\begin{equation*}
A\otimes B=\begin{bmatrix}
 (A)_{1,1}B & (A)_{1,2}B & \ldots & (A)_{1,n} B \\
 (A)_{2,1}B & (A)_{2,2}B & \ldots & (A)_{2,n} B \\
 \vdots & \vdots & \ddots & \vdots \\
 (A)_{m,1}B & (A)_{m,2}B & \ldots & (A)_{m,n} B
 \end{bmatrix}.
\end{equation*}
We note that the Kronecker product may be viewed as a family of operations
on vector spaces of square matrices (indexed by the orders of the matrices involved),
or as an operation on the graded vector space of all square matrices. In other words,
the Kronecker product is the family of operations
$\otimes:\mathbb{F}_m\times\mathbb{F}_n\to\mathbb{F}_{mn}$ indexed by $m,n\in\mathbb{N}$.

\begin{definition}[Kronecker quotient \cite{kronquotient,leopardi05a}]
 \label{def:kquotient} A (right) Kronecker quotient is an operation $\oslash$ satisfying
 \begin{equation*}
 (A\otimes B)\oslash B = A
 \end{equation*}
 for all matrices $A\in\mathbb{F}_{m}$ and $B\in\mathbb{F}_{n}$
 ($B$ non-zero) and $m,n\in\mathbb{N}$.
\end{definition}

\begin{definition}
 \label{def:kquniform} A Kronecker quotient $\oslash$ is a uniform Kronecker quotient if
 \begin{equation*}
 (A\otimes C)\oslash B = A\otimes (C\oslash B)
 \end{equation*}
 for all matrices $A\in\mathbb{F}_{m}$, $B\in\mathbb{F}_n$, $C\in\mathbb{F}_p\otimes\mathbb{F}_n$ and $m,n,p\in\mathbb{N}$,
 and
 \begin{equation*}
 (A+B)\oslash C = (A\oslash C)+(B\oslash C), \qquad
 (kA)\oslash C = k(A\oslash C)
 \end{equation*}
 for all matrices $A$, $B$ and $C$, and scalars $k\in\mathbb{F}$ whenever the operations are defined.
\end{definition}

\begin{remark}
 An equivalent definition for uniform Kronecker quotients
 is as follows, by simply setting $p=1$ in Definition \ref{def:kquniform}:
 A Kronecker quotient $\oslash$ is a uniform Kronecker quotient if
 \begin{math}
 (A\otimes C)\oslash B = A\otimes (C\oslash B)
 \end{math}
 for all matrices $A\in\mathbb{F}_{m}$, $B,C\in\mathbb{F}_n$ and $m,n\in\mathbb{N}$,
 and
 \begin{math}
 (A+B)\oslash C = (A\oslash C)+(B\oslash C),
 \end{math}
 \begin{math}
 (kA)\oslash C = k(A\oslash C)
 \end{math}
 for all matrices $A$, $B$ and $C$, and scalars $k\in\mathbb{F}$ whenever the operations are defined.
\end{remark}

In \cite{kronquotient}, examples of uniform Kronecker quotients over the real
and complex numbers are given. The following lemma establishes the existence
of a uniform Kronecker quotient over an arbitrary field $\mathbb{F}$.

\begin{definition}
 \label{def:kquniform} A Kronecker quotient $\oslash$ is a uniform Kronecker quotient if
 \begin{equation*}
  (A\otimes C)\oslash B = A\otimes (C\oslash B)
 \end{equation*}
 for all matrices $A\in\mathbb{F}_{m}$, $B\in\mathbb{F}_n$, $C\in\mathbb{F}_p\otimes\mathbb{F}_n$ and $m,n,p\in\mathbb{N}$,
 and
 \begin{equation*}
  (A+B)\oslash C = (A\oslash C)+(B\oslash C), \qquad
  (kA)\oslash C = k(A\oslash C)
 \end{equation*}
 for all matrices $A$, $B$ and $C$, and scalars $k\in\mathbb{F}$ whenever the operations are defined.
\end{definition}

\begin{proof}
 Let $B,C\in\mathbb{F}_m$ with $C\neq 0$. Then there exists
 an non-zero entry $(C)_{i,j}\neq 0$ in $C$. Now let $f:\mathbb{F}_m\to\mathbb{N}\times\mathbb{N}$
 be defined by $f(C)=(i,j)$ and $f(0_m)=(1,1)$, i.e. $f$ chooses a non-zero entry of $C$, unless
 $C$ is the zero matrix. Define $\text{---}\oslash C$
 as the linear extension of
 \begin{equation*}
 (A\otimes B)\oslash C := A\dfrac{(B)_{f(C)}}{(C)_{f(C)}}
 \end{equation*}
 for all $A\in\mathbb{F}_s$, $s\in\mathbb{N}$. Then
 $\oslash$ is a uniform Kronecker quotient by construction.
\end{proof}

\begin{definition}
 \label{def:kqdefineskd} Let $\oslash$ be a Kronecker quotient. The Kronecker difference $\ominus$ induced by $\oslash$ is the
 Kronecker difference given by
 \begin{equation*}
 M\ominus B := (M-I_m\otimes B)\oslash I_n
 \end{equation*}
 for all $M\in\mathbb{F}_{mn}$, $B\in\mathbb{F}_n$ and $m,n\in\mathbb{N}$.
\end{definition}

\begin{proposition}
 \label{prop:induceddiffprop} Let $\oslash$ be a Kronecker quotient and let $\ominus$
 be the induced Kronecker difference
 \begin{equation*}
 X\ominus B:=(X-I_m\otimes B)\oslash I_n
 \end{equation*}
 for all $X\in \mathbb{F}_{mn}$ and $B\in\mathbb{F}_n$.
 \begin{enumerate}[label=(\alph*)]
 \item
 If $(A\oslash B)^T = A^T\oslash B^T$
 then $(A\ominus B)^T = A^T\ominus B^T$.\phantom{$\dfrac1n$}
 \item
 If $\tr(A\oslash B) = \dfrac{\tr(A)}{\tr(B)}$
 then $\tr(A\ominus B) = \dfrac1n(\tr(A)-m\tr(B))$.
 \item
 If $\oslash$ is uniform then $\ominus$ is uniform.\phantom{$\dfrac1n$}
 \item
 If $(kA)\oslash B = k(A\oslash B)$
 then $(kA)\ominus(kB) = k(A\ominus B)$.\phantom{$\dfrac1n$}
 \item
 If $(A+C)\oslash B = A\oslash B+C\oslash B$
 then $(A+C)\ominus(B+D) = A\ominus B+C\ominus D$.\phantom{$\dfrac1n$}
 \end{enumerate}
\end{proposition}
\begin{proof}
 The proof of (c) appears in Proposition \ref{prop:ukqdefinesukd}, which also proves (d) and (e). The proofs of $(a)$ and $(b)$ each follow
 by straightforward calculation:
 \begin{multline*}
 (A\ominus B)^T = ((A-I_m\otimes B)\oslash I_n)^T = (A-I_m\otimes B)^T\oslash I_n \\
 = (A^T-I_m\otimes B^T)\oslash I_n = A^T\ominus B^T,
 \end{multline*}
 \begin{multline*}
 \tr(A\ominus B) = \tr((A-I_m\otimes B)\oslash I_n) \\
 = \dfrac{\tr(A-I_m\otimes B)}n
 = \dfrac1n(\tr(A)-m\tr(B)).
 \qedhere
 \end{multline*}
\end{proof}

\begin{lemma}
 \label{lem:unicanon} Let $\upsilon_n\in\mathbb{F}_n$, $n\in\mathbb{N}$ be a sequence of matrices with
 $\tr(\upsilon_n)=1$. Then the Kronecker difference defined below is uniform, where
 \begin{equation*}
 A\ominus B = \tr_{12}(\alpha_{m,n}^T(A\otimes I_m-I_m\otimes B\otimes I_m)),\qquad
 \alpha_{m,n}:= \sum_{i,j=1}^m E_{ij}\otimes\upsilon_n\otimes E_{ji},
 \end{equation*}
 for all $A\in\mathbb{F}_m\otimes\mathbb{F}_n$ and $B\in\mathbb{F}_n$, $m,n\in\mathbb{N}$.
\end{lemma}

\begin{theorem}
 \label{thm:unicanon2} Let $\ominus$ be a uniform Kronecker difference. Then there exists a sequence of matrices
 $\upsilon_n\in\mathbb{F}_n$, $n\in\mathbb{N}$, with $\tr(\upsilon_n)=1$ and matrices
 $\gamma_{m,n}\in\mathbb{F}_{m^2n}$, $m,n\in\mathbb{N}$, with $\tr_{1}(\gamma_{m,n})=0$ and $\tr_{2}(\gamma_{m,n})=0$ and
 \begin{equation*}
 A\ominus B = \tr_{12}(\alpha_{m,n}^T(A\otimes I_m-I_m\otimes B\otimes I_m)),\qquad
 \alpha_{m,n}:= \sum_{i,j=1}^m E_{ij}\otimes\upsilon_n\otimes E_{ji} + \gamma_{m,n},
 \end{equation*}
 for all $A\in\mathbb{F}_m\otimes\mathbb{F}_n$ and $B\in\mathbb{F}_n$, $m,n\in\mathbb{N}$.
\end{theorem}

\begin{proof}
 By Remark \ref{rmk:trivial}, for each $n\in\mathbb{N}$ there exists a matrix $\upsilon_n\in\mathbb{F}_n$
 such that $\tr(\upsilon_n)=1$ and $A\ominus B=\tr(\upsilon_n^T(A-B))$ for all $A,B\in\mathbb{F}_n$.
 Following the proof of Theorem \ref{thm:canon2} (the choice of $E_{11}$ was arbitrary -- we
 required a matrix with unit trace), there exists $\gamma_{m,n}\in\mathbb{F}_{m^2n}$ (dependent on $\upsilon_n$) such that
 $\tr_2(\gamma_{m,n})=0$, and
 \begin{equation*}
 A\ominus B = \tr_{12}(\alpha_{m,n}^T(A\otimes I_m-I_m\otimes B\otimes I_m)),\qquad
 \alpha_{m,n}:= \sum_{i,j=1}^m E_{ij}\otimes\upsilon_n\otimes E_{ji} + \gamma_{m,n}
 \end{equation*}
 for all $A\in\mathbb{F}_m\otimes\mathbb{F}_n$ and $B\in\mathbb{F}_n$, $m,n\in\mathbb{N}$.
 It remains to
 show that $\tr_1(\gamma_{m,n})=0$.
 First let us consider $B,C\in\mathbb{F}_n$.
 Since $\ominus$ is uniform, and following the proof of Lemma \ref{lem:unicanon},
 $(A\oplus C)\ominus B=A\oplus (C\ominus B)$ yields that (recalling that $\tr_2(\gamma_{m,n})=0$ and by use of Lemma \ref{lem:blockpartial})
 \begin{align*}
 \lefteqn{\tr_{12}(\gamma_{m,n}^T(A\otimes I_n\otimes I_m+I_m\otimes C\otimes I_m-I_m\otimes B\otimes I_m))}\qquad & \\
 &= \Btr(\tr_2(\gamma_{m,n})^T(A\otimes I_m))
 + \Btr(\tr_{1}(\gamma_{m,n})^T ((C-B)\otimes I_m)) \\
 &= \Btr(\tr_{1}(\gamma_{m,n})^T ((C-B)\otimes I_m)) \\
 &= 0,
 \end{align*}
 for all $A\in\mathbb{F}_m$, $B,C\in\mathbb{F}_n$ and $m,n\in\mathbb{N}$. Writing $\tr_{1}(\gamma_{m,n})$
 in the form
 \begin{equation*}
 \tr_{1}(\gamma_{m,n}) = \sum_{i,j=1}^n E_{ij}\otimes G_{ij}
 \end{equation*}
 and setting $C-B=E_{kl}$ yields that
 \begin{equation*}
 \sum_{i,j=1}^n \Btr((E_{ji}\otimes G_{ij}^T)(E_{kl}\otimes I_m))
 = \sum_{j=1}^n \Btr(E_{jl}\otimes G_{kj}^T)
 = G_{kl}^T = 0.
 \end{equation*}
 Since $j$ and $k$ were arbitrary, $\tr_{1}(\gamma_{m,n}) = 0$ for all $m,n\in\mathbb{N}.$
 \def\ignore{\color{red}
 We require that
 \begin{equation*}
 (A\oplus C)\ominus B= A\oplus(C\ominus B)=A\oplus 0_p + 0_m\oplus(C\ominus B)
 \end{equation*}
 for all $m,n,p\in\mathbb{N}$, $A\in\mathbb{F}_m$, $C\in\mathbb{F}_{p}\otimes\mathbb{F}_n$ and $B\in\mathbb{F}_n$.
 By Lemma \ref{lem:unicanon} we need only satisfy
 \begin{gather*}
 \tr_{12}(\gamma_{mp,n}^T(A\otimes I_{pn}\otimes I_{mp}+I_{m}\otimes C\otimes I_{mp}-I_{mp}\otimes B\otimes I_{mp})) \\
 = 0_m\oplus(\tr_{12}(\gamma_{p,n}^T(C\otimes I_p-I_p\otimes B\otimes I_p)))
 \end{gather*}
 for all $A\in\mathbb{F}_m$, $B\in\mathbb{F}_n$ and
 $C\in\mathbb{F}_p\otimes\mathbb{F}_n$. Equivalently, since $\tr_2(\gamma_{mp,n})=0$ (over $\mathbb{F}_{n}$) and $\tr_1(\gamma_{m,n})=0$ (over $\mathbb{F}_m$),
 \begin{equation}
 \label{eq:unigamma}
 \tr_{12}(\gamma_{mp,n}^T(I_{m}\otimes C-I_{mp}\otimes B)\otimes I_{mp})
 = I_m\otimes(\tr_{12}(\gamma_{p,n}^T(C\otimes I_p)))
 \end{equation}
 for all $B\in\mathbb{F}_n$, and $C\in\mathbb{F}_p\otimes\mathbb{F}_n$.
 If $C=0$, we obtain for all $B\in\mathbb{F}_n$
 \begin{equation}
 \label{eq:unigammaB}
 \tr_{12}(\gamma_{mp,n}^T(I_{mp}\otimes B\otimes I_{mp})) = 0_{mp}.
 \end{equation}
 Hence,
 \begin{equation}
 \label{eq:unigammaC}
 \tr_{12}(\gamma_{mp,n}^T(I_{m}\otimes C\otimes I_{mp}))
 = I_m\otimes(\tr_{12}(\gamma_{p,n}^T(C\otimes I_p)))
 \end{equation}
 for all $C\in\mathbb{F}_p\otimes\mathbb{F}_n$.
 First, let us write $\gamma_{mp,n}$ and $\gamma_{p,n}$ in block form, namely
 \begin{equation*}
 \gamma_{p,n} = \sum_{k,l=1}^{pn} E_{kl}\otimes H_{kl},\qquad
 \gamma_{mp,n} = \sum_{i,j=1}^{m}\sum_{k,l=1}^{pn}E_{ij}\otimes E_{kl}\otimes H_{ij;kl}
 \end{equation*}
 where each $H_{kl}\in\mathbb{F}_p$ and each $H_{ij;kl}\in\mathbb{F}_{m}\otimes\mathbb{F}_p$.
 It follows from \eqref{eq:unigammaC} that
 \begin{equation*}
 \sum_{i,j=1}^m\sum_{k,l=1}^{pn}
 \tr(E_{ji})\tr(E_{lk}C)H_{ij;kl}^T
 =
 \sum_{k,l=1}^{pn}\tr(E_{lk}C)I_m\otimes H_{kl}^T
 \end{equation*}
 for all $C\in\mathbb{F}_p\otimes\mathbb{F}_n$.
 If $C=E_{kl}\in\mathbb{F}_{pn}$, then
 \begin{equation*}
 \sum_{i=1}^m H_{ii;kl}^T= I_m\otimes H_{kl}^T.
 \end{equation*}
 In particular, if $p=1$ we obtain
 \begin{equation*}
 \tr_1(\gamma_{m,n}) = \gamma_{1,n}\otimes I_m.
 \end{equation*}
 Now \eqref{eq:unigammaB} becomes
 \begin{equation*}
 \tr(\gamma_{1,n}^TB)\otimes I_{mp} = 0_{mp}
 \end{equation*}
 for all $B\in\mathbb{F}_n$. Hence, $\gamma_{1,n}=0$ and $\tr_1(\gamma_{m,n})=0$ for all $m,n\in\mathbb{N}$.
 }
\end{proof}

\begin{proof}
 Let $B,C\in\mathbb{F}_m$ with $C\neq 0$. Then there exists
 an non-zero entry $(C)_{i,j}\neq 0$ in $C$. Now let $f:\mathbb{F}_m\to\mathbb{N}\times\mathbb{N}$
 be defined by $f(C)=(i,j)$ and $f(0_m)=(1,1)$, i.e. $f$ chooses a non-zero entry of $C$, unless
 $C$ is the zero matrix. Define $\text{---}\oslash C$
 as the linear extension of
 \begin{equation*}
  (A\otimes B)\oslash C := A\dfrac{(B)_{f(C)}}{(C)_{f(C)}}
 \end{equation*}
 for all $A\in\mathbb{F}_s$, $s\in\mathbb{N}$. Then
 $\oslash$ is a uniform Kronecker quotient by construction.
\end{proof}

\subsection{Kronecker sums and Kronecker differences}

The Kronecker sum $A\oplus B$ of an $m\times m$ matrix $A$ and an
$n\times n$ matrix $B$ is given by \cite[pp. 268]{horn91}
\begin{equation*}
 A\oplus B = A\otimes I_n + I_m\otimes B.
\end{equation*}
Over the real and complex numbers, we have
\begin{equation*}
 \exp(A\oplus B) = \exp(A)\otimes\exp(B).
\end{equation*}
The Kronecker sum uniquely satisfies the above equation. The Kronecker sum
also arises in the study of matrix equations of the form
\begin{equation*}
 BX+XA^T = Y
\end{equation*}
where $X$ is the subject of the equation. Taking the vectorization (vec)
on both sides of the equation, and using the fact that $\vc(ABC)=(C^T\otimes A)\vc(B)$,
yields \cite[pp. 268]{horn91}
\begin{equation*}
 (A\oplus B) \vc(X) = \vc(Y).
\end{equation*}

This article defines and studies Kronecker differences and their connection with Kronecker
quotients \cite{kronquotient}.
Similar to the definition of Kronecker quotients, we consider the family $\ominus:\mathbb{F}_{mn}\times\mathbb{F}_n\to\mathbb{F}_m$
indexed by $m,n\in\mathbb{N}$ satisfying the following definition:

\begin{definition}
 \label{def:kqdifference} A Kronecker difference is an operation $\ominus$ satisfying
 \begin{equation*}
  (A\oplus B)\ominus B = A
 \end{equation*}
 for all matrices $A\in\mathbb{F}_{m}$ and $B\in\mathbb{F}_{n}$,
 $m,n\in\mathbb{N}$.
\end{definition}

Every Kronecker quotient induces a Kronecker difference in the following way.

\begin{definition}
 \label{def:kqdefineskd} Let $\oslash$ be a Kronecker quotient. The Kronecker difference $\ominus$ induced by $\oslash$ is the
 Kronecker difference given by
 \begin{equation*}
  M\ominus B := (M-I_m\otimes B)\oslash I_n
 \end{equation*}
 for all $M\in\mathbb{F}_{mn}$, $B\in\mathbb{F}_n$ and $m,n\in\mathbb{N}$.
\end{definition}

\begin{remark}
 \label{rem:kqdefineskd} It is easily verified that this expression satisfies the condition for a
 Kronecker difference, i.e. for every $A\in\mathbb{F}_m$, $B\in\mathbb{F}_n$ and $m,n\in\mathbb{N}$,
 \begin{align*}
  (A\oplus B)\ominus B
   &= (A\otimes I_n + I_m\otimes B)\ominus B \\
   &= (A\otimes I_n + I_m\otimes B - I_m\otimes B)\oslash I_n \\
   &= (A\otimes I_n)\oslash I_n \\
   &= A.
 \end{align*}
\end{remark}

\begin{remark}[Duality 1]
 \label{rem:kddefineskq} Definition \ref{def:kqdefineskd} has a weak dual, namely: Let $\oslash$ be a Kronecker quotient such that
 \begin{equation*}
  M\oslash B = (M(0_m\oplus B)^{-1})\ominus 0_n
 \end{equation*}
 for all $M\in\mathbb{F}_{mn}$, invertible $B\in\mathbb{F}_n$ and $m,n\in\mathbb{N}$,
 where $\ominus$ is a Kronecker difference. Indeed, for invertible $B$,
 $(0_m\oplus B)^{-1} = (I_m\otimes B)^{-1} = I_m\otimes B^{-1}$ and
 \begin{align*}
  (A\otimes B)\oslash B
   &= ((A\otimes I_n)(I_m\otimes B))\oslash B \\
   &= ((A\otimes I_n)(I_m\otimes B)(I_m\otimes B)^{-1})\ominus 0_n \\
   &= (A\oplus 0_n)\ominus 0_n \\
   &= A.
 \end{align*}
\end{remark}

The notion of uniformity in Kronecker quotients has an analogue for Kronecker differences.
\begin{definition}
 \label{def:kduniform} A Kronecker difference $\ominus$ is a uniform Kronecker difference if
 \begin{equation*}
  (A\oplus C)\ominus B = A\oplus (C\ominus B)
 \end{equation*}
 for all matrices $A\in\mathbb{F}_{m}$, $B\in\mathbb{F}_n$, $C\in\mathbb{F}_p\otimes\mathbb{F}_n$ and $m,n,p\in\mathbb{N}$,
 and
 \begin{equation*}
  (A+B)\ominus (C+D) = (A\ominus C)+(B\ominus D), \qquad
  (kA)\ominus (kC) = k(A\ominus C)
 \end{equation*}
 for all matrices $A$, $B$, $C$ and $D$, and scalars $k\in\mathbb{F}$ whenever the operations are defined.
\end{definition}
Once again, we immediately have a connection between uniform Kronecker quotients and uniform Kronecker differences.

\begin{proposition}
 \label{prop:ukqdefinesukd} Let $\oslash$ be a uniform Kronecker quotient.
 Then the induced Kronecker difference $\ominus$ is uniform.
\end{proposition}

\begin{proof}
 Following the definition in Definition \ref{def:kqdefineskd},  \begin{equation*}
  M\ominus B := (M-I_m\otimes B)\oslash I_n
 \end{equation*}
 for all $M\in\mathbb{F}_{mn}$, $B\in\mathbb{F}_n$ and $m,n\in\mathbb{N}$,
 where $\oslash$ is a uniform Kronecker quotient.
 In Remark \ref{rem:kqdefineskd} it was shown that $\ominus$ is a Kronecker difference,
 it remains to show that this Kronecker difference is uniform.
 We have for all $A\in\mathbb{F}_{m}$, $B\in\mathbb{F}_n$, and $C\in\mathbb{F}_p\otimes\mathbb{F}_n$,
 \begin{align*}
  (A\oplus C)\ominus B
   &= (A\otimes I_p\otimes I_n + I_m\otimes C - I_m\otimes I_p\otimes B) \oslash I_n \\
   &= (A\otimes I_p\otimes I_n)\oslash I_n + (I_m\otimes C)\oslash I_n - (I_m\otimes I_p\otimes B) \oslash I_n \\
   &= A\otimes I_p + I_m\otimes(C\oslash I_n) - I_m\otimes I_p\otimes(B\oslash I_n) \\
   &= A\otimes I_p + I_m\otimes[(C\oslash I_n) - I_p\otimes (B\oslash I_n)] \\
   &= A\otimes I_p + I_m\otimes [(C-I_p\otimes B)\oslash I_n] \\
   &= A\otimes I_p + I_m\otimes (C\ominus B) \\
   &= A\oplus (C\ominus B).
 \end{align*}
 Finally, for all $A,B\in\mathbb{F}_{mn}$ and $C,D\in\mathbb{F}_n$,
 \begin{align*}
  (A+B)\ominus (C+D)
   &= (A+B-I_m\otimes C-I_m\otimes D)\oslash I_n \\
   &= (A-I_m\otimes C+ B-I_m\otimes D)\oslash I_n \\
   &= (A-I_m\otimes C)\oslash I_n + (B-I_m\otimes D)\oslash I_n \\
   &= A\ominus C + B\ominus D
 \end{align*}
 and for any scalar $k\in\mathbb{F}$,
 \begin{align*}
  (kA)\ominus (kC)
   &= (kA-I_m\otimes(kC))\oslash I_n \\
   &= [k(A-I_m\otimes C)]\oslash I_n \\
   &= k[(A-I_m\otimes C)\oslash I_n] \\
   &= k(A\ominus C). \qedhere
 \end{align*}
\end{proof}

Before studying Kronecker differences in more detail, we will need some additional definitions
and technical results.
The presentations of these definitions and technicalities are left for Appendix \ref{sec:orthogonality} and Appendix \ref{sec:lemmata}.

Let $B,C\in\mathbb{F}_m$ with $C\neq 0$. Choose a non-zero entry $(C)_{i,j}\neq 0$ and define a selection map $f:\mathbb{F}_m\to\mathbb{N}\times\mathbb{N}$ by $f(C)=(i,j)$ (and $f(0_m)=(1,1)$). Define $\text{---}\oslash C$ as the linear extension of
\[
(A\otimes B)\oslash C := A\dfrac{(B)_{f(C)}}{(C)_{f(C)}}\quad\text{for all }A\in\mathbb{F}_s.
\]
Then “$\oslash$ is a uniform Kronecker quotient by construction.”

Let $B,C\in\mathbb{F}_m$ with $C\neq 0$. Choose a non-zero entry $(C)_{i,j}\neq 0$ and define a selection map $f:\mathbb{F}_m\to\mathbb{N}\times\mathbb{N}$ by $f(C)=(i,j)$ (and $f(0_m)=(1,1)$). Define $\text{---}\oslash C$ as the linear extension of
\[
(A\otimes B)\oslash C := A\dfrac{(B)_{f(C)}}{(C)_{f(C)}}\quad\text{for all }A\in\mathbb{F}_s.
\]
Then $\oslash$ is a uniform Kronecker quotient by construction.

Question:
Let \(\mathbb F\) be a field, and for each \(m\in\mathbb N\) let \(\mathbb F_m\) denote the vector space of \(m\times m\) matrices over \(\mathbb F\). Let \(\otimes:\mathbb F_m\times\mathbb F_n\to\mathbb F_{mn}\) be the usual Kronecker product. A (right) Kronecker quotient is an operation \(\oslash\) on square matrices over \(\mathbb F\) such that for every \(m,n\in\mathbb N\), every \(A\in\mathbb F_m\), and every nonzero \(B\in\mathbb F_n\), one has \((A\otimes B)\oslash B=A\). It is called a uniform Kronecker quotient if, in addition, for all compatible sizes, all \(A\in\mathbb F_m\), \(B\in\mathbb F_n\), \(C\in\mathbb F_p\otimes\mathbb F_n=\mathbb F_{pn}\), and all scalars \(k\in\mathbb F\),
\[
(A\otimes C)\oslash B = A\otimes (C\oslash B),
\]
and whenever defined,
\[
(A_1+A_2)\oslash C=(A_1\oslash C)+(A_2\oslash C),\qquad (kA)\oslash C=k(A\oslash C).
\]
Under these assumptions, which existence statement holds?

There exists a uniform Kronecker quotient over \(\mathbb F\); equivalently, there exists an operation \(\oslash\) on square matrices over \(\mathbb F\) satisfying \((A\otimes B)\oslash B=A\) for every \(A\in\mathbb F_m\) and every nonzero \(B\in\mathbb F_n\), and also satisfying the uniformity and linearity conditions above for all compatible matrix sizes.

Choices:
B. There exists a Kronecker quotient \(\oslash\) on square matrices over \(\mathbb F\) such that for every \(m,n\in\mathbb N\), every \(A\in\mathbb F_m\), and every nonzero \(B\in\mathbb F_n\), one has \((A\otimes B)\oslash B=A\); moreover, one may choose \(\oslash\) so that for each fixed size \(n\) there is a single pair of indices \((i_n,j_n)\) with the property that for every nonzero \(C\in\mathbb F_n\), the quotient is obtained by reading the \((i_n,j_n)\)-entry of \(C\) and extending linearly, while still satisfying the stated uniformity and linearity conditions for all compatible sizes.
C. There exists a Kronecker quotient \(\oslash\) over \(\mathbb F\); that is, there exists an operation \(\oslash\) on square matrices over \(\mathbb F\) such that \((A\otimes B)\oslash B=A\) for every \(A\in\mathbb F_m\) and every nonzero \(B\in\mathbb F_n\).
D. There exists an operation \(\oslash\) on square matrices over \(\mathbb F\) such that for every \(m,n\in\mathbb N\), every \(A\in\mathbb F_m\), and every nonzero \(B\in\mathbb F_n\), one has \((A\otimes B)\oslash B=A\), and such that the same operation is defined and satisfies the uniformity and linearity identities for all matrices \(B\in\mathbb F_n\), including \(B=0\).
E. There exists a uniform Kronecker quotient over \(\mathbb F\), and one may choose it so that for every nonzero \(C\in\mathbb F_n\) the map \(M\mapsto M\oslash C\) is given by a formula depending only on \(C\) and not on any auxiliary choice, namely by selecting a canonically determined nonzero entry of \(C\) and extending linearly, while satisfying \((A\otimes B)\oslash B=A\) for every \(A\in\mathbb F_m\) and every nonzero \(B\in\mathbb F_n\).

dependence of the selected index on the matrix C itself

nonzero restriction on the divisor matrix

arbitrary choice function for a nonzero entry replaced by canonical choice


\begin{example}\label{ex:origra}
Consider the homogeneous manifold
\[
Gr^{+}(2,n)=SO(n)/SO(n-2)\times SO(2)=Q_{n},
\]
with the natural effective $T$-action.
Here, if $n=2r$ then $T=T^{r}/{\pm 1}$, and if $n=2r+1$ then $T=T^{r}$.
This oriented Grassmanian is a connected double cover of the real Grassmanian $Gr(2,n)$.
This implies $Gr^{+}(2,3)=S^3$.
It is not difficult to show that $Gr^+(2,4)$ is homeomorphic to $S^2\times S^2$.
(Furthermore, $Gr^+(2,6)$ is homeomorphic to the complex Grassmanian $Gr_{\C}(2,4)$.)
The orbit space is homeomorphic to $\Sigma^{\lfloor n/2\rfloor} \C P^{\lfloor n/2\rfloor-2}$ by~\cite{S19}.
In particular, the orbit space is ${\lfloor n/2\rfloor}$-connected (where the suspension of an empty set is by definition a point).
In particular, it is contractible for $n=3,4$.
This homogeneous manifold has type $D_{r}/D_{r-1}\times T^{1}$ for $n=2r$ and $B_{r}/B_{r-1}\times T^1$ for $n=2r+1$, where $r\geq 2$.
Here, $B_1=A_1$, $D_1=T^{1}$, and $D_2=(A_1)^2$.
Since $Gr^{+}(2,2r)=Q_{2r}$ and $Gr^{+}(2,2r+1)=Q_{2r+1}$,
It is known that $H^{odd}(Q_{n};\mathbb{Z})=0$.
Therefore, Theorem~\ref{thm:AM} implies that the orbit space is $4$-acyclic for $n=2r$ and $3$-acyclic for $n=2r+1$.
Therefore, our result agrees with previous known results, in particular, in dimensions $n=6,7,8$.
\end{example}

\begin{example}\label{ex:comgra}
Consider the homogeneous manifold
\[
Gr(2,5)=SU(5)/S(U(2)\times U(3)),
\]
with the natural effective $T^{4}$-action.
This is the complex Grassmanian of two-planes in $\C^5$.
The fixed-point data at the identity is a root system of type $A_{4}/A_{1}\times A_{2}\times T^{1}$.
By Theorems~\ref{main} and Corollary~\ref{cor:acyc},
this GKM-manifold has $4$-acyclic orbit space $Gr(2,5)/T^4$.
On the other hand, by~\cite{BT} or~\cite{S21},
\[
\widetilde{H}_{*}(Gr(2,5)/T^4;\Z)\simeq
\begin{cases}
\Z/2\Z,\ n=5,\\
\Z,\ n=8,\\
0,\ \mbox{else}.
\end{cases}
\]
This agrees with the above acyclicity claim.
\end{example}

\begin{corollary}\label{cor:acyc}
If $G/H$ is a symmetric GKM$_{3}$ space (see Theorem~\ref{thm:cart} and (1), (2) of Theorem~\ref{main})
with the following coefficient $\Bbbk$ for the reduced homology, then $\widetilde{H}_{i}(T\backslash G/H;\Bbbk)=0$ for $i\le 4$.

If $\Bbbk=\mathbb{Z}$, the the list is as follows:
\begin{center}
\begin{tabular}{|l|l|l|l|l|} \hline
label & pair & $G$ & $H$ &  \\ \hline
   {\rm AIII} & $(A_{n},A_{i}\times A_{n-1-i}\times T^{1})$ & $SU(n+1)$ & $S(U(i+1)\times U(n-i))$ & $0\le i\le n-1$ \\
   {\rm BDI}    & $(D_{n}, D_{n-1}\times T^{1})$ & $SO(2n)$ & $SO(2n-2)\times SO(2)$ &    \\
    {\rm DIII} & $(D_{n}, A_{n-1}\times T^{1})$ & $SO(2n)$ & $U(n)$ & \\
    {\rm CII} & $(C_{n},C_{i}\times C_{n-i})$ & $Sp(n)$ & $Sp(i)\times Sp(n-i)$ & $1\le i\le n-1$ \\
    {\rm EIII} & $(E_{6}, D_{5}\times T^{1})$ & $E_{6}$ & $(Spin(10)\times T^{1})/\mathbb{Z}_{4}$ & \\
    {\rm EVII} & $(E_{7},E_{6}\times T^{1})$ & $E_{7}$ & $(E_{6}\times T^{1})/\mathbb{Z}_{3}$ & \\
   {\rm FII} & $(F_{4}, B_{4})$ & $F_{4}$ & $Spin(9)$ & \\
   \hline
 \end{tabular}
\end{center}

If $\Bbbk=\mathbb{Z}[\frac{1}{2}]$, the the list is as follows:
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
label & pair  & $G$  & $H$  &  \\ \hline
  {\rm BDI}  & $(D_{n}, D_{i}\times D_{n-i})$ & $SO(2n)$ & $SO(2i)\times SO(2n-2i)$ & $i>2$,   $n-i>2$  \\
    {\rm EII} & $(E_{6}, A_{5}\times A_{1})$ & $E_{6}$ & $(SU(6)\times SU(2))/\mathbb{Z}_{2}$ & \\
      {\rm EV} & $(E_{7},A_{7})$ & $E_{7}$ & $SU(8)/\mathbb{Z}_{2}$ & \\
      {\rm EVI} & $(E_{7},D_{6}\times A_{1})$ & $E_{7}$ & $(Spin(12)\times SU(2))/\mathbb{Z}_{2}$ & \\
    {\rm EIX} & $(E_{8},E_{7}\times A_{1})$ & $E_{8}$ & $(E_{7}\times SU(2))/\mathbb{Z}_{2}$ & \\
   \hline
 \end{tabular}
 \end{center}
\end{corollary}

\begin{lemma}\label{lm:23dep}
Let $G$ be a simple Lie group of type $A$--$D$, and $H$ be
its proper maximal connected (closed) maximal rank subgroup.
Then one has the following:
\begin{enumerate}
\item $k(G/H)=2$ if and only if the weighted signed graph $\Gamma(G,H)$ has a subgraph (with weighted vertices) appeared in Figure~\ref{example_weighted_vertex}, where $i<j<k$.
\begin{figure}[H]
\begin{tikzpicture}
\begin{scope}[xscale=0.5, yscale=0.5]

ll(-2,-2)circle (5pt);
ll(-2,2)circle (5pt);
ll(2,-2)circle (5pt);
\draw[thick](-2,-2)--(-2,2);
\node[above] at (-2,2) {$w_{V}(i)$};
\draw[thick](2,-2)--(-2,2);
\node[below] at (-2,-2) {$w_{V}(j)$};
\draw[thick](-2,-2)--(2,-2);
\node[below] at (2,-2) {$w_{V}(k)$};

ll(5,2)circle (5pt);
ll(5,-2)circle (5pt);
ll(9,-2)circle (5pt);
\draw[dashed](5,-2)--(5,2);
\node[above] at (5,2) {$w_{V}(i)$};
\draw[dashed](9,-2)--(5,2);
\node[below] at (5,-2) {$w_{V}(j)$};
\draw[thick](9,-2)--(5,-2);
\node[below] at (9,-2) {$w_{V}(k)$};

ll(14,2)circle (5pt);
ll(14,-2)circle (5pt);
\draw[thick](14,-2)to[out=75,in=-75](14,2);
\draw[dashed](14,-2)to[out=105,in=-105](14,2);
\draw[thick](14,2.7) circle [x radius=0.7cm, y radius=0.7cm];
\node[above] at (14,3.4) {$w_{V}(i)$};
\node[below] at (14,-2) {$w_{V}(j)$};

ll(19,2)circle (5pt);
ll(19,-2)circle (5pt);
\draw[dashed](19,-2)to[out=105,in=-105](19,2);
\draw[thick](19,2.7) circle [x radius=0.7cm, y radius=0.7cm];
\node[above] at (19,3.4) {$w_{V}(i)$};
\draw[thick](19,-2.7) circle [x radius=0.7cm, y radius=0.7cm];
\node[below] at (19,-3.4) {$w_{V}(j)$};

ll(23,2)circle (5pt);
ll(23,-2)circle (5pt);
\draw[thick](23,-2)to[out=75,in=-75](23,2);
\draw[thick](23,2.7) circle [x radius=0.7cm, y radius=0.7cm];
\node[above] at (23,3.4) {$w_{V}(i)$};
\draw[thick](23,-2.7) circle [x radius=0.7cm, y radius=0.7cm];
\node[below] at (23,-3.4) {$w_{V}(j)$};

\end{scope}
\end{tikzpicture}
\caption{$2$-independent subgraphs.}
\label{example_weighted_vertex}
\end{figure}
\item $k(G/H)=3$ if the weighted signed graph $\Gamma(G,H)$ has neither of the subgraphs (with weighted vertices) depicted in Figure~\ref{example_weighted_vertex} but has a weighted subgraph appeared in  Figure~\ref{example_weighted_vertex2}, where $i<j<k<l$.
\begin{figure}[H]
\begin{tikzpicture}
\begin{scope}[xscale=0.5, yscale=0.5]

ll(-2,-2)circle (5pt);
ll(-2,2)circle (5pt);
ll(2,-2)circle (5pt);
ll(2,2)circle (5pt);
\draw[thick](-2,-2)--(-2,2)--(2,2)--(2,-2)--cycle;
\node[above] at (-2,2) {$w_{V}(i)$};
\node[below] at (-2,-2) {$w_{V}(j)$};
\node[below] at (2,-2) {$w_{V}(k)$};
\node[above] at (2,2) {$w_{V}(l)$};

\coordinate (A) at (5,2);
\coordinate (B) at (5,-2);
\coordinate (C) at (9,2);
ll(A)circle (5pt);
ll(B)circle (5pt);
ll(C)circle (5pt);
\draw[thick](A)to[out=240,in=120](B);
\node[above] at (A) {$w_{V}(i)$};
\draw[dashed](A)to[out=300,in=60](B);
\node[below] at (B) {$w_{V}(j)$};
\draw[thick](A)to[out=330,in=210](C);
\node[above] at (C) {$w_{V}(k)$};
\draw[dashed](A)to[out=30,in=150](C);

\coordinate (a) at (12,2);
\coordinate (b) at (12,-2);
\coordinate (c) at (16,-2);
\coordinate (d) at (16,2);
ll(a)circle (5pt);
ll(b)circle (5pt);
ll(c)circle (5pt);
ll(d)circle (5pt);

\draw[dashed] (a)--(b)--(d)--(c)--(a);

\node[above] at (a) {$w_{V}(i)$};
\node[below] at (b) {$w_{V}(j)$};
\node[below] at (c) {$w_{V}(k)$};
\node[above] at (d) {$w_{V}(l)$};

\end{scope}
\end{tikzpicture}
\caption{$3$-independent subgraphs.}
\label{example_weighted_vertex2}
\end{figure}
\end{enumerate}
\end{lemma}

\begin{theoremM}\label{thmM:mp}
Let $G/H$ be any simply connected, homogeneous space such that $T\subset H\subset G$ is a maximal torus, ${\rm rank}\ T=n$, and $G$ is a compact, connected, simple Lie group.
Then one has
\[
k(G/H)=2,3\ \text{or}\ n(=\frac{1}{2}\dim G/H).
\]
Furthermore,
$k(G/H) = 3$ or $n$ if and only if $G/H$ is a maximal rank compact symmetric space with rank $>2$ other than $B_{n}/D_{i}\times B_{n-i}$, $C_{n}/A_{n-1}\times T^{1}$, $F_{4}/C_{3}\times A_{1}$.
\end{theoremM}

\label{sect:1}

\subsection{Motivation and Main theorem}
\label{sect:1.1}

Let $M$ be a $2m$-dimensional, smooth $T$-manifold with a nonempty set of isolated fixed points $M^{T}$, where $T\cong (S^{1})^{n}$ and we assume that the $T$-action is {\it almost effective}, i.e., the kernel of the $T$-action is finite.
There is the following irreducible decomposition
\begin{align}
\label{cpx_on_p}
T_{p}M\cong \bigoplus_{i=1}^{m}V_{\alpha_{p,i}},
\end{align}
of the isotropy representation at any fixed point $p\in M^{T}$, where $V_{\alpha_{p,i}}$ is the (complex) irreducible representation space of $T$, and $\alpha_{p,i}\in \mathfrak{t}_{\mathbb{Z}}^{*}/\pm 1$ for $i=1,\ldots, m$ is determined up to a sign.
Here, $\mathfrak{t}_{\mathbb{Z}}^{*}\cong \Z^{n}$ denotes the character lattice of
the dual of the Lie algebra $\mathfrak{t}^{*}$ of $T$.
Set $S_{p}:=\{\alpha_{p,i}\ |\ i=1,\ldots, m\}\subset \mathbb{Z}^{n}$ and define the following numbers:
\begin{align}
\label{independency}
k(p):=\max\{q\ |\ \text{every $q$ elements in $S_{p}$ are linearly independent}\},\
k(M):=\min\limits_{p\in M^{T}} k(p).
\end{align}
If $2\le k(M)$, then this manifold is called a \textit{GKM manifold}~\cite{GKM},~\cite{GZ}.
A GKM manifold is called a \textit{$j$-independent} or \textit{GKM$_{j}$) manifold} if $k(M)\geq j$ holds (see \cite{AMS, GS}. In \cite{AM}, this is called a {\it $j$-generality}).
If $k(M)=m$ then $n=m$; in such case this manifold is called a {\it torus manifold}~\cite{MP},~\cite{HaMa}.
Mikiya Masuda has asked the following question to the first named author:
\begin{probM}
\label{Masuda-problem}
Let $M$ be an equivariantly formal (i.e., $H^{odd}(M)=0$) $2m$-dimensional GKM$_4$ manifold with a $T$-action.
Is $M$ homeomorphic to a torus manifold, and is the given $T$-action obtained as the restriction of a half-dimensional torus $T^{m}$-action to a subtorus $T$?
\end{probM}

The main result of the present paper is given as follows (see Theorem \ref{main}), where by a {\it maximal rank compact symmetric space} we mean a compact symmetric space $G/H$ that is a homogeneous space of maximal rank (also referred to as a {\it class 1 symmetric space of inner type}).

\begin{theorem}
\label{main}
Assume that $G$ is simply connected, simple Lie group and $H$ be its maximal rank
subgroup such that ${\rm rank}(G)={\rm rank}(H)=\dim T=n$.
Then, $k(G/H)=2,3,n$ and the following holds:
\begin{enumerate}
\item $k(G/H)=n={\rm rank}(G)={\rm rank}(H)$ if and only if $(G,H)$ is one of the following types:
\begin{description}
\item[Type $A_{n}$] $(A_{n},A_{n-1}\times T^{1})$;
\item[Type $B_{n}$] $(B_{n}, D_{n})$.
\end{description}
\item $k(G/H)=3(\not=n)$ if and only if
$(G,H)$ is one of the pairs appered in Theorem \ref{thm:cart} except for the above two cases, $(B_{n},D_{i}\times B_{n-i})$, $(C_{n},A_{n-1}\times T^{1})$ and $(F_{4}, C_{3}\times A_{1})$.
\item $k(G/H)=2$, otherwise (all of the other cases).
\end{enumerate}
\end{theorem}

\begin{abstract}
Let $G/H$ be a simply connected homogeneous space of maximal rank.
Then the maximal torus $T$-action on $G/H$ is a GKM manifold.
We call the $T$-action $j$-independent if any $i(\leq j)$ pairwise distinct isotropy weights at a fixed point are linearly independent.
Using weighted graphs, we show that the maximal independence of $G/H$ is $2$, $3$ or $n=\dim T$, and that the cases of $3$ or $n=\dim T$ correspond to some symmetric spaces of rank $>2$.
As a corollary,
using the results of Ayzenberg and Masuda, the lower-degree reduced homology groups (with appropriate coefficients) of the orbit space $T\backslash G/H$ vanish.
\end{abstract}

Let $M$ be a $2m$-dimensional, smooth $T$-manifold with a nonempty set of isolated fixed points $M^{T}$, where $T\cong (S^{1})^{n}$ and we assume that the $T$-action is {\it almost effective}, i.e., the kernel of the $T$-action is finite.
There is the following irreducible decomposition
\begin{align}
\label{cpx_on_p}
T_{p}M\cong \bigoplus_{i=1}^{m}V_{\alpha_{p,i}},
\end{align}
of the isotropy representation at any fixed point $p\in M^{T}$, where $V_{\alpha_{p,i}}$ is the (complex) irreducible representation space of $T$, and $\alpha_{p,i}\in \mathfrak{t}_{\mathbb{Z}}^{*}/\pm 1$ for $i=1,\ldots, m$ is determined up to a sign.
Here, $\mathfrak{t}_{\mathbb{Z}}^{*}\cong \Z^{n}$ denotes the character lattice of
the dual of the Lie algebra $\mathfrak{t}^{*}$ of $T$.
Set $S_{p}:=\{\alpha_{p,i}\ |\ i=1,\ldots, m\}\subset \mathbb{Z}^{n}$ and define the following numbers:
\begin{align}
\label{independency}
k(p):=\max\{q\ |\ \text{every $q$ elements in $S_{p}$ are linearly independent}\},\
k(M):=\min\limits_{p\in M^{T}} k(p).
\end{align}
If $2\le k(M)$, then this manifold is called a \textit{GKM manifold}~\cite{GKM},~\cite{GZ}.
A GKM manifold is called a \textit{$j$-independent} or \textit{GKM$_{j}$) manifold} if $k(M)\geq j$ holds (see \cite{AMS, GS}. In \cite{AM}, this is called a {\it $j$-generality}).
If $k(M)=m$ then $n=m$; in such case this manifold is called a {\it torus manifold}~\cite{MP},~\cite{HaMa}.
Mikiya Masuda has asked the following question to the first named author:
\begin{probM}
\label{Masuda-problem}
Let $M$ be an equivariantly formal (i.e., $H^{odd}(M)=0$) $2m$-dimensional GKM$_4$ manifold with a $T$-action.
Is $M$ homeomorphic to a torus manifold, and is the given $T$-action obtained as the restriction of a half-dimensional torus $T^{m}$-action to a subtorus $T$?
\end{probM}

As an application of Theorem~\ref{thmM:mp}, together with the acyclicity results of~\cite{AMS},~\cite{AM}, we obtain vanishing (in degrees $\leq 4$) for reduced homology with an appropriate coefficients of the orbit spaces for homogeneous GKM manifolds, see Corollary \ref{cor:acyc}.
Our results are consistent with those in~\cite{S19},~\cite{S21},~\cite{BT}, while providing new insights, particularly for several symmetric spaces.

\subsection{The torus action on the homogeneous space $G/H$}
\label{sect:2.3}
Let $G$ be a compact, connected, semi-simple Lie group, i.e., $\pi_{1}(G)$ is finite, and
$T$ be its maximal torus, and $H$ be a closed, connected subgroup of $G$ such that $T\subseteq H\subset G$, i.e., a {\it maximal rank subgroup} of $G$.
It is known in this case that
the homogeneous space $G/H$ is a $2m$-dimensional, simply connected manifold satisfying $H^{odd}(G/H;\mathbb{Q})=0$, see e.g.~\cite{MiTo, Bu} and~\cite[Theorem 2.4]{GHZ}.
Below we recall the description of the tangential weights of the $T$-action on $G/H$ by following these references.

\begin{proposition}
\label{k-independence}
Let $G$ be a compact, connected semi-simple Lie group, $H$ be its maximal rank subgroup and $T\subseteq H\subset G$ be a maximal torus.
Then $k(G/H)=k(gH)$ for any $gH\in (G/H)^{T}$.
\end{proposition}
This proposition shows that if we compute the $k(G/H)$-independence on $\Delta_{G}^{+}\setminus \Delta_{H}^{+}\simeq \Delta_{G,H}/\{\pm 1\}$, then we have the $k(G/H)$-independence of all tangent spaces on the fixed points of $G/H$
(Cf.~Example~\ref{ex_depends_k(p)}).
In this paper, we determine the $k$-independence for all maximal rank homogeneous spaces $G/H$.

\begin{theorem}
\label{thm:cart}
Let $G/H$ be a maximal rank compact symmetric spaces with rank $>2$.
Then, the pair $(G,H)$ corresponds to one of the following list.
\begin{center}
\begin{tabular}{|l|l|l|l|l|} \hline
 label & pair & $G$ & $H$ & \\ \hline
 {\rm AIII} & $(A_{n},A_{i}\times A_{n-1-i}\times T^{1})$ & $SU(n+1)$ & $S(U(i+1)\times U(n-i))$ & $0\le i\le n-1$ \\
 {\rm BDI} & $(B_{n}, D_{i}\times B_{n-i})$ & $SO(2n+1)$ & $SO(2i)\times SO(2n-2i+1)$ & $1\le i\le n$ \\
 & $(D_{n}, D_{i}\times D_{n-i})$ & $SO(2n)$ & $SO(2i)\times SO(2n-2i)$ & $1\le i\le n-1$ \\
 {\rm DIII} & $(D_{n}, A_{n-1}\times T^{1})$ & $SO(2n)$ & $U(n)$ & \\
 {\rm CI} & $(C_{n}, A_{n-1}\times T^{1})$ & $Sp(n)$ & $U(n)$ & \\
 {\rm CII} & $(C_{n},C_{i}\times C_{n-i})$ & $Sp(n)$ & $Sp(i)\times Sp(n-i)$ & $1\le i\le n-1$ \\
 {\rm EII} & $(E_{6}, A_{5}\times A_{1})$ & $E_{6}$ & $(SU(6)\times SU(2))/\mathbb{Z}_{2}$ & \\
 {\rm EIII} & $(E_{6}, D_{5}\times T^{1})$ & $E_{6}$ & $(Spin(10)\times T^{1})/\mathbb{Z}_{4}$ & \\
 {\rm EV} & $(E_{7},A_{7})$ & $E_{7}$ & $SU(8)/\mathbb{Z}_{2}$ & \\
 {\rm EVI} & $(E_{7},D_{6}\times A_{1})$ & $E_{7}$ & $(Spin(12)\times SU(2))/\mathbb{Z}_{2}$ & \\
 {\rm EVII} & $(E_{7},E_{6}\times T^{1})$ & $E_{7}$ & $(E_{6}\times T^{1})/\mathbb{Z}_{3}$ & \\
 {\rm EVIII} & $(E_{8},D_{8})$ & $E_{8}$ & $Ss(16)$ & \\
 {\rm EIX} & $(E_{8},E_{7}\times A_{1})$ & $E_{8}$ & $(E_{7}\times SU(2))/\mathbb{Z}_{2}$ & \\
 {\rm FI} & $(F_{4}, C_{3}\times A_{1})$ & $F_{4}$ & $(Sp(3)\times Sp(1))/\mathbb{Z}_{2}$ & \\
 {\rm FII} & $(F_{4}, B_{4})$ & $F_{4}$ & $Spin(9)$ & \\
 \hline
 \end{tabular}
\end{center}

\subsection{Main theorem}
\label{sect:3.1}
If we assume $G$ is simply connected, then the standard $T$-action on the simply connected homogeneous space $G/H$ is equivariantly diffeomorphic to the following product of the homogeneous spaces (see e.g.~\cite[Section 2.2]{Ku10}):
\begin{align}
\label{splitting}
G/H\cong G_{1}/H_{1}\times \cdots \times G_{r}/H_{r},
\end{align}
where $G_{i}$ is a compact, simply connected simple Lie group, $H_{i}$ be its maximal rank subgroup and there is an isomorphism $T\cong T_{1}\times \cdots \times T_{r}$ such that $T_{i}\subset H_{i}\subset G_{i}$ is a maximal torus for $i=1,\ldots, r$.
Moreover, we can easily show the following proposition. (Recall that a torus manifold is equipped with a half-dimensional torus action.)
\begin{proposition}
\label{split_computation}
Assume that $G/H$ is not a torus manifold.
Then, there is the following equality for the splitting \eqref{splitting}:
\begin{align*}
k(G/H)=\min \{k(G_{i}/H_{i})\ |\ 1\le i\le r \}.
\end{align*}
\end{proposition}

\begin{theorem}
\label{main}
Assume that $G$ is simply connected, simple Lie group and $H$ be its maximal rank
subgroup such that ${\rm rank}(G)={\rm rank}(H)=\dim T=n$.
Then, $k(G/H)=2,3,n$ and the following holds:
\begin{enumerate}
\item $k(G/H)=n={\rm rank}(G)={\rm rank}(H)$ if and only if $(G,H)$ is one of the following types:
\begin{description}
\item[Type $A_{n}$] $(A_{n},A_{n-1}\times T^{1})$;
\item[Type $B_{n}$] $(B_{n}, D_{n})$.
\end{description}
\item $k(G/H)=3(\not=n)$ if and only if
$(G,H)$ is one of the pairs appered in Theorem \ref{thm:cart} except for the above two cases, $(B_{n},D_{i}\times B_{n-i})$, $(C_{n},A_{n-1}\times T^{1})$ and $(F_{4}, C_{3}\times A_{1})$.
\item $k(G/H)=2$, otherwise (all of the other cases).
\end{enumerate}
\end{theorem}

Note that by the classification of the homogeneous torus manifolds in \cite{Ku09}, $k(G/H) =n$ if and only if $G/H$ is equivariantly diffeomorphic to $A_{n}/A_{n-1}\times T^{1}(\cong \mathbb{C}P^{n})$ or $B_{n}/D_{n}(\cong S^{2n})$.
Moreover, the independence of $M_{1}\times \cdots \times M_{r}$ coincides with
the minimal independence of $M_{i}$'s (if it is not a torus manifold, see Proposition~\ref{split_computation}), and the independence of any connected homogeneous space $G/H$ is equal to that of its universal covering space $G_{1}/H_{1}\times \cdots \times G_{r}/H_{r}$.
Therefore, together with these facts, Theorem~\ref{thmM:mp} answers Problem~\ref{Masuda-problem} positively for the class of homogeneous GKM manifolds.

As an application of Theorem~\ref{thmM:mp}, together with the acyclicity results of~\cite{AMS},~\cite{AM}, we obtain vanishing (in degrees $\leq 4$) for reduced homology with an appropriate coefficients of the orbit spaces for homogeneous GKM manifolds, see Corollary \ref{cor:acyc}.
Our results are consistent with those in~\cite{S19},~\cite{S21},~\cite{BT}, while providing new insights, particularly for several symmetric spaces.

\subsection{Sketch of the proof}
\label{sect:1.2}

The proof of Theorem~\ref{thmM:mp} follows by computing $k(G/H)$ for every homogeneous GKM manifold $G/H$.
Here, $k(G/H)=k(gH)$ for every fixed point $gH\in (G/H)^{T}$ (see~\cite{GHZ}).
We prove that $k(G/H)\geq 3$ implies that $G/H$ is a maximal rank compact symmetric space with rank $>2$ using the well known characterization of symmetric spaces in terms of the Lie algebra $T_{eH} G/H$ (see Lemma~\ref{auxiliary lemma}).
For maximal rank compact symmetric space $G/H$ such that $G$ is a classical Lie group (i.e., type $A$--$D$), the problem of finding maximal independence for $G/H$ reduces to the study of the matroid formed by the complement $\Delta^{+}_{G,H}:=\Delta^{+}_{G}\setminus \Delta^{+}_{H}$ of positive root systems.
The analysis of circuits for such a matroid is possible using {\it weighted signed graphs}, which were introduced in~\cite{Ha} and studied in~\cite{Za} (also see \cite{FGLP} and \cite{FGGP}).
We apply this technique together with Borel-de Siebenthal theory~\cite{BoSi} and Cartan's classification of symmetric spaces \cite{C27} to compute $k(G/H)$ for all pairs $(G,H)$ except for $G$ being of type $F_{4},E_{6},E_{7},E_{8}$.
(We remark that signed graphs under the name of Levi graphs or crystallographs were also used in~\cite{Re} to study root subsystems.
The difference of our application for signed graphs is that we study \textit{complements} of root subsystems.)
In the remaining cases, i.e. for $G$ being of type $F_{4},E_{6},E_{7},E_{8}$, the proof is a simple case study.

\subsection{Structure of the paper}
\label{sect:1.3}

We briefly outline the structure of the present paper.
In Section~\ref{sec:prep}, we recall the isotropy representation for homogeneous GKM manifolds, Borel-de Siebenthal theory, Cartan's classification of symmetric spaces and homology vanishing of the orbit space for independent GKM manifolds.
One of the main contributions of this paper is to clarify the relationship between symmetric spaces and GKM manifolds from the viewpoint of independence.
The value of $k(G/H)$ for each homogeneous space $G/H$ is listed in Theorem~\ref{main}.
This result implies $\widetilde{H}_{i}(T\backslash G/H;\Bbbk)=0$ for all $i\leq k(G/H)+1$, where $\Bbbk$ is additionally specified (see Corollary~\ref{cor:acyc} for symmetric GKM$_{3}$ cases).
This homology vanishing is compared with several previously known results in Examples~\ref{ex:origra} and~\ref{ex:comgra}.
The remainder of the paper is the proof of Theorem~\ref{main}.
In Section~\ref{sect:4}, linearly dependent triples are described in several cases in terms of weighted signed graphs (see Lemma~\ref{lm:23dep}).
In Sections~\ref{sect:5}--~\ref{sect:7}, we analyse linear dependencies for root subsystems of classical types, using Borel-de Siebenthal theory and weighted signed graphs.
In Sections~\ref{sect:8}--~\ref{sect:9},
the remaining cases (when $G$ is of type $F_{4},E_{6},E_{7},E_{8}$) are studied.

\

\noindent {\bf Acknowledgements:} The authors gratefully acknowledge funding of the Deutsche Forschungsgemeinschaft
(DFG, German Research Foundation): Project number 561158824 (the second named author's Walter Benjamin Fellowship).
We thank M.~Masuda for many valuable discussions on toric manifolds without whom we would not be able to conduct this project, and to O.~Goertsches for the interest to our paper.
We also thank M.~Nakagawa for giving us a useful information on symmetric spaces.
The first author was partially supported by JSPS KAKENHI Grant Number 21K03262.
He would like to thank the Philipps University of Marburg for providing him with an excellent environment for research during his stay in August 2025.
The second named author is grateful to Okayama University of Science  and Osaka Metropolitan University for a very comfortable work atmosphere, where the initial part of the present work was performed during his research visits in 2023.

The proof of Theorem~\ref{thmM:mp} follows by computing $k(G/H)$ for every homogeneous GKM manifold $G/H$. Here, $k(G/H)=k(gH)$ for every fixed point $gH\in (G/H)^{T}$ (see~\cite{GHZ}).

One shows that $k(G/H)\geq 3$ implies that $G/H$ is a maximal rank compact symmetric space with rank $>2$, using the well known characterization of symmetric spaces in terms of the Lie algebra $T_{eH}G/H$ (see Lemma~\ref{auxiliary lemma}).

For maximal rank compact symmetric spaces $G/H$ with $G$ a classical Lie group (type $A$--$D$), finding maximal independence reduces to studying the matroid formed by the complement of positive root systems $\Delta^{+}_{G,H}:=\Delta^{+}_{G}\setminus \Delta^{+}_{H}$. The analysis of circuits in this matroid is carried out using \textit{weighted signed graphs} (as in~\cite{Ha,Za}, also \cite{FGLP,FGGP}). Applying this technique together with Borel-de Siebenthal theory~\cite{BoSi} and Cartan's classification of symmetric spaces~\cite{C27} yields the computation of $k(G/H)$ for all pairs $(G,H)$ except when $G$ is of type $F_{4},E_{6},E_{7},E_{8}$.

In the remaining cases ($G$ of type $F_{4},E_{6},E_{7},E_{8}$), the proof is described as a simple case study.

The proof of Theorem~\ref{thmM:mp} follows by computing $k(G/H)$ for every homogeneous GKM manifold $G/H$. Here, $k(G/H)=k(gH)$ for every fixed point $gH\in (G/H)^{T}$ (see GHZ).

One shows that $k(G/H)\geq 3$ implies that $G/H$ is a maximal rank compact symmetric space with rank $>2$, using the well known characterization of symmetric spaces in terms of the Lie algebra $T_{eH}G/H$. We first prove the following lemma.
\begin{lemma}\label{auxiliary lemma}
Under the hypothesis in Theorem~\ref{main},
if $k(G/H)\ge 3$, then  $G/H$ is a symmetric space given in Theorem~\ref{thm:cart}.
\end{lemma}

For maximal rank compact symmetric spaces $G/H$ with $G$ a classical Lie group (type $A$--$D$), finding maximal independence reduces to studying the matroid formed by the complement of positive root systems $\Delta^{+}_{G,H}:=\Delta^{+}_{G}\setminus \Delta^{+}_{H}$. The analysis of circuits in this matroid is carried out using \textit{weighted signed graphs} (as in Ha and Za, also FGLP and FGGP). Applying this technique together with Borel-de Siebenthal theory BoSi and Cartan's classification of symmetric spaces C27 yields the computation of $k(G/H)$ for all pairs $(G,H)$ except when $G$ is of type $F_{4},E_{6},E_{7},E_{8}$.

In the remaining cases ($G$ of type $F_{4},E_{6},E_{7},E_{8}$), the proof is described as a simple case study.,

\label{thmM:mp}
Let $G/H$ be any simply connected, homogeneous space such that $T\subset H\subset G$ is a maximal torus, ${\rm rank}\ T=n$, and $G$ is a compact, connected, simple Lie group.
Then one has
\[
k(G/H)=2,3\ \text{or}\ n(=\frac{1}{2}\dim G/H).
\]
Furthermore,
$k(G/H) = 3$ or $n$ if and only if $G/H$ is a maximal rank compact symmetric space with rank $>2$ other than $B_{n}/D_{i}\times B_{n-i}$, $C_{n}/A_{n-1}\times T^{1}$, $F_{4}/C_{3}\times A_{1}$.,

Classification or Bijection
Biconditional or Equivalence

Question: Let $G/H$ be a simply connected homogeneous space on which a maximal torus $T\subset H\subset G$ acts, where $G$ is a compact, connected, simple Lie group and ${\rm rank}\,T=n$. For a $T$-fixed point $p$, let $S_p$ be the set of isotropy weights of the $T$-representation on $T_p(G/H)$, define $k(p)=\max\{q\mid \text{every } q \text{ elements of } S_p \text{ are linearly independent}\}$, and set $k(G/H)=\min_{p\in (G/H)^T} k(p)$. Under these assumptions, $k(G/H)$ can only be $2$, $3$, or $n$. Which of the following statements is equivalent to the condition that $k(G/H)=3$ or $k(G/H)=n$?

$G/H$ is a maximal rank compact symmetric space, meaning a compact symmetric space with ${\rm rank}(G)={\rm rank}(H)=n>2$, and $G/H$ is not one of the spaces $B_n/(D_i\times B_{n-i})$, $C_n/(A_{n-1}\times T^1)$, or $F_4/(C_3\times A_1)$.

Choices:
B. $G/H$ is a maximal rank compact symmetric space, meaning a compact symmetric space with ${\rm rank}(G)={\rm rank}(H)=n$, and $G/H$ is not one of the spaces $B_n/(D_i\times B_{n-i})$, $C_n/(A_{n-1}\times T^1)$, or $F_4/(C_3\times A_1)$.
C. $G/H$ is a maximal rank compact symmetric space with ${\rm rank}(G)={\rm rank}(H)=n>2$.
D. $G/H$ is a compact symmetric space with ${\rm rank}(G)={\rm rank}(H)=n>2$, and $k(G/H)=n$ unless $G/H$ is one of the spaces $B_n/(D_i\times B_{n-i})$, $C_n/(A_{n-1}\times T^1)$, or $F_4/(C_3\times A_1)$, in which case $k(G/H)=3$.
E. $G/H$ is a maximal rank compact symmetric space, meaning a compact symmetric space with ${\rm rank}(G)={\rm rank}(H)=n>2$, and $G/H$ is not one of the spaces $B_n/(D_i\times B_{n-i})$ or $C_n/(A_{n-1}\times T^1)$.


\label{lem:L2Green}
Let $(M,g)$ be a compact connected $2$-dimensional Riemannian manifold without boundary, and let $G$ be the symmetric mean-zero Green function of $-\Delta$.
It then holds that $G\in L^2(M\times M, dx\otimes dx)$, and so
\[
\sigma^2 \coloneqq \int_{M}\int_{M} G(x,y)^2\,dx\,dy <\infty.
\]

Let $(M,g)$ be a compact connected two-dimensional Riemannian manifold without boundary, $\mathrm{d}_g$ is the induced distance, and let $G(x,y)$ denote the symmetric mean-zero Green function of the Laplacian (Definition \ref{def:green}). We work with the normalized volume measure $dx=\mathrm{vol}(M)^{-1}\,\mathrm{d}\mathrm{vol}$. Whenever we integrate in the variable $y$, we write $dy$ for the same measure. For points $x_1,\dots,x_n\in M$, we define the empirical measure:
\[
\mu_n \coloneqq \frac1n\sum_{i=1}^n \delta_{x_i}.
\]
Steinerberger \cite{Ste21} showed that for $d\ge 3$, the following Green--Wasserstein inequality holds:
\[
W_2(\mu_n,dx)\ \lesssim_{M}\ n^{-1/d} + \frac1n\left|\sum_{i\neq j} G(x_i,x_j)\right|^{1/2}.
\]
In \cite[Problem 53]{SteOpen}, he gives the following estimate:
\begin{equation}\label{eq:Steinerberger2D}
W_2\!\left(\frac1n\sum_{k=1}^n \delta_{x_k},\,dx\right)
\lesssim_M
\frac1n\left|\sum_{i\neq j} G(x_i,x_j)\right|^{1/2}
+
\begin{cases}
\sqrt{\frac{\log n}{n}} & \text{if } d=2,\\
n^{-1/d} & \text{if } d\ge 3.
\end{cases}
\end{equation}
In this note, we use
\[
\sum_{i\neq j}(\cdots)\ \coloneqq\ \sum_{\substack{1\le i,j\le n\\ i\neq j}}(\cdots)
\]
to denote the ordered sum over distinct indices. Notice that, by symmetry of $G$, it is twice the sum over $i<j$. Steinerberger asked whether the factor $\sqrt{\log n}$ in \eqref{eq:Steinerberger2D} can be removed while keeping the same Green-energy term. The precise question is as follows.

\begin{problem}[Steinerberger {\cite[Problem 53]{SteOpen}}]\label{prob:Steinerberger}
In dimension $d=2$, can one replace the remainder $\sqrt{\frac{\log n}{n}}$ in \eqref{eq:Steinerberger2D}
by $\frac{1}{\sqrt n}$ while keeping the same off-diagonal Green term? In other terms, does there exist
$C_M>0$ such that for all $n\in\mathbb N$ and all $x_1,\dots,x_n\in M$,
\[
W_2\!\left(\frac1n\sum_{k=1}^n \delta_{x_k},\,dx\right)
\le C_M\left(\frac1{\sqrt n}
+\frac1n\left|\sum_{i\neq j} G(x_i,x_j)\right|^{1/2}\right)?
\]
\end{problem}
\noindent Our main result, Theorem \ref{thm:main}, gives a negative answer to Problem \ref{prob:Steinerberger}.

\begin{remark}[The diagonal]\label{rem:diagonal}
We note that the Green function $G$ is smooth on $(M\times M)\setminus \{(x,x):x\inM\}$ with a logarithmic singularity along the diagonal.
In a deterministic context, if $x_i=x_j$ for some $i\neq j$, then the term $G(x_i,x_j)$ is not finite. It is then natural to interpret the Green term in \eqref{eq:Steinerberger2D} and Problem \ref{prob:Steinerberger} as $+\infty$, in which case the inequality holds trivially.
In the random settings that we use in this note, collisions occur with probability $0$ because $dx$ is non-atomic.
\end{remark}

We now recall some definitions that will be useful in the main argument.
For Borel probability measures $\mu,\nu$ on $M$, let us denote by $\Gamma(\mu,\nu)$ the set of couplings of $\mu$ and $\nu$ on $M\times M$.
The quadratic Wasserstein distance is
\[
W_2(\mu,\nu)\coloneqq \left(\inf_{\gamma\in \Gamma(\mu,\nu)} \int_{M\times M} \mathrm{d}_g(x,y)^2\,\,\mathrm{d}\gamma(x,y)\right)^{1/2}.
\]
Given that $M$ is compact, it follows that $W_2(\mu,\nu)$ is always finite and satisfies $W_2(\mu,\nu)\le \mathrm{diam}(M)$. Let $\Delta$ be the Laplace--Beltrami operator on $(M,g)$, which is realized here as a self-adjoint operator on $L^2(M,dx)$ with domain $H^2(M)$. Again, since $M$ is compact and connected, $\ker(-\Delta)$ consists of the constant functions.

\begin{definition}[Mean-zero Green function]\label{def:green}
A mean-zero Green function for $-\Delta$ is a measurable function
\[
G:M\times M \to \mathbb{R}\cup\{+\infty\}
\]
with the following properties.
\begin{enumerate}
\item For each $x\in M$, $y\mapsto G(x,y)$ is locally integrable on $M$ and satisfies the normalization:
\begin{equation}\label{eq:GreenMeanZero}
\int_{M} G(x,y)\,dy=0.
\end{equation}
\item For each $f\in C^\infty(M)$ with $\int_{M} f\,dx=0$, the function
\[
u(x)\coloneqq \int_{M} G(x,y)\, f(y)\,dy
\]
is a weak solution of $-\Delta u = f$ with $\int_{M}u\,dx=0$.
\end{enumerate}
\end{definition}

Recall that the uniqueness and existence of the mean-zero Green function, as a distribution kernel, are guaranteed. We can also choose it to be symmetric (for example, see \cite{Aubin98} or \cite{Rosenberg97}). We now fix a symmetric mean-zero Green function $G$, so that $G(x,y)=G(y,x)$ holds.
By symmetry and \eqref{eq:GreenMeanZero}, we obtain:
\[
\int_{M}G(y,x)\,dy=0\qquad\text{for every }x\in M.
\]

\begin{remark}[Local singularity]\label{rem:logsing}
In dimension $2$, the following classical local expansion (in geodesic normal coordinates) is true:
\[
G(x,y)= -\frac{1}{2\pi}\log \mathrm{d}_g(x,y) + H(x,y).
\]
Note that $H$ extends continuously to the diagonal and is smooth off the diagonal.
In particular, we have $G(x,\cdot)\in L^1(M,dx)$ for each fixed $x$.
\end{remark}

Let $(M,g)$ be a compact connected two-dimensional Riemannian manifold without boundary, $\mathrm{d}_g$ is the induced distance, and let $G(x,y)$ denote the symmetric mean-zero Green function of the Laplacian (Definition \ref{def:green}). We work with the normalized volume measure $dx=\mathrm{vol}(M)^{-1}\,\mathrm{d}\mathrm{vol}$. Whenever we integrate in the variable $y$, we write $dy$ for the same measure. For points $x_1,\dots,x_n\in M$, we define the empirical measure:
\[
\mu_n \coloneqq \frac1n\sum_{i=1}^n \delta_{x_i}.
\]
Steinerberger \cite{Ste21} showed that for $d\ge 3$, the following Green--Wasserstein inequality holds:
\[
W_2(\mu_n,dx)\ \lesssim_{M}\ n^{-1/d} + \frac1n\left|\sum_{i\neq j} G(x_i,x_j)\right|^{1/2}.
\]
In \cite[Problem 53]{SteOpen}, he gives the following estimate:
\begin{equation}\label{eq:Steinerberger2D}
W_2\!\left(\frac1n\sum_{k=1}^n \delta_{x_k},\,dx\right)
\lesssim_M
\frac1n\left|\sum_{i\neq j} G(x_i,x_j)\right|^{1/2}
+
\begin{cases}
\sqrt{\frac{\log n}{n}} & \text{if } d=2,\\
n^{-1/d} & \text{if } d\ge 3.
\end{cases}
\end{equation}
In this note, we use
\[
\sum_{i\neq j}(\cdots)\ \coloneqq\ \sum_{\substack{1\le i,j\le n\\ i\neq j}}(\cdots)
\]
to denote the ordered sum over distinct indices. Notice that, by symmetry of $G$, it is twice the sum over $i<j$. Steinerberger asked whether the factor $\sqrt{\log n}$ in \eqref{eq:Steinerberger2D} can be removed while keeping the same Green-energy term. The precise question is as follows.

\begin{remark}[The diagonal]\label{rem:diagonal}
We note that the Green function $G$ is smooth on $(M\times M)\setminus \{(x,x):x\inM\}$ with a logarithmic singularity along the diagonal.
In a deterministic context, if $x_i=x_j$ for some $i\neq j$, then the term $G(x_i,x_j)$ is not finite. It is then natural to interpret the Green term in \eqref{eq:Steinerberger2D} and Problem \ref{prob:Steinerberger} as $+\infty$, in which case the inequality holds trivially.
In the random settings that we use in this note, collisions occur with probability $0$ because $dx$ is non-atomic.
\end{remark}

We now recall some definitions that will be useful in the main argument.
For Borel probability measures $\mu,\nu$ on $M$, let us denote by $\Gamma(\mu,\nu)$ the set of couplings of $\mu$ and $\nu$ on $M\times M$.
The quadratic Wasserstein distance is
\[
W_2(\mu,\nu)\coloneqq \left(\inf_{\gamma\in \Gamma(\mu,\nu)} \int_{M\times M} \mathrm{d}_g(x,y)^2\,\,\mathrm{d}\gamma(x,y)\right)^{1/2}.
\]
Given that $M$ is compact, it follows that $W_2(\mu,\nu)$ is always finite and satisfies $W_2(\mu,\nu)\le \mathrm{diam}(M)$. Let $\Delta$ be the Laplace--Beltrami operator on $(M,g)$, which is realized here as a self-adjoint operator on $L^2(M,dx)$ with domain $H^2(M)$. Again, since $M$ is compact and connected, $\ker(-\Delta)$ consists of the constant functions.

\begin{definition}[Mean-zero Green function]\label{def:green}
A mean-zero Green function for $-\Delta$ is a measurable function
\[
G:M\times M \to \mathbb{R}\cup\{+\infty\}
\]
with the following properties.
\begin{enumerate}
\item For each $x\in M$, $y\mapsto G(x,y)$ is locally integrable on $M$ and satisfies the normalization:
\begin{equation}\label{eq:GreenMeanZero}
\int_{M} G(x,y)\,dy=0.
\end{equation}
\item For each $f\in C^\infty(M)$ with $\int_{M} f\,dx=0$, the function
\[
u(x)\coloneqq \int_{M} G(x,y)\, f(y)\,dy
\]
is a weak solution of $-\Delta u = f$ with $\int_{M}u\,dx=0$.
\end{enumerate}
\end{definition}

Recall that the uniqueness and existence of the mean-zero Green function, as a distribution kernel, are guaranteed. We can also choose it to be symmetric (for example, see \cite{Aubin98} or \cite{Rosenberg97}). We now fix a symmetric mean-zero Green function $G$, so that $G(x,y)=G(y,x)$ holds.
By symmetry and \eqref{eq:GreenMeanZero}, we obtain:
\[
\int_{M}G(y,x)\,dy=0\qquad\text{for every }x\in M.
\]

\begin{remark}[Local singularity]\label{rem:logsing}
In dimension $2$, the following classical local expansion (in geodesic normal coordinates) is true:
\[
G(x,y)= -\frac{1}{2\pi}\log \mathrm{d}_g(x,y) + H(x,y).
\]
Note that $H$ extends continuously to the diagonal and is smooth off the diagonal.
In particular, we have $G(x,\cdot)\in L^1(M,dx)$ for each fixed $x$.
\end{remark}

\begin{definition}[Mean-zero Green function]\label{def:green}
A mean-zero Green function for $-\Delta$ is a measurable function
\[
G:\M\times \M \to \mathbb{R}\cup\{+\infty\}
\]
with the following properties.
\begin{enumerate}
\item For each $x\in \M$, $y\mapsto G(x,y)$ is locally integrable on $\M$ and satisfies the normalization:
\begin{equation}\label{eq:GreenMeanZero}
\int_{\M} G(x,y)\,dy=0.
\end{equation}
\item For each $f\in C^\infty(\M)$ with $\int_{\M} f\,dx=0$, the function
\[
u(x)\coloneqq \int_{\M} G(x,y)\, f(y)\,dy
\]
is a weak solution of $-\Delta u = f$ with $\int_{\M}u\,dx=0$.
\end{enumerate}
\end{definition}

\begin{equation}\label{eq:GreenMeanZero}
\int_{\M} G(x,y)\,dy=0.
\end{equation}

\begin{equation}\label{eq:Steinerberger2D}
\W\!\left(\frac1n\sum_{k=1}^n \delta_{x_k},\,dx\right)
\lesssim_\M
\frac1n\left|\sum_{i\neq j} G(x_i,x_j)\right|^{1/2}
+
\begin{cases}
\sqrt{\frac{\log n}{n}} & \text{if } d=2,\\
n^{-1/d} & \text{if } d\ge 3.
\end{cases}
\end{equation}

\begin{problem}[Steinerberger {\cite[Problem 53]{SteOpen}}]\label{prob:Steinerberger}
In dimension $d=2$, can one replace the remainder $\sqrt{\frac{\log n}{n}}$ in \eqref{eq:Steinerberger2D}
by $\frac{1}{\sqrt n}$ while keeping the same off-diagonal Green term? In other terms, does there exist
$C_\M>0$ such that for all $n\in\mathbb N$ and all $x_1,\dots,x_n\in \M$,
\[
\W\!\left(\frac1n\sum_{k=1}^n \delta_{x_k},\,dx\right)
\le C_\M\left(\frac1{\sqrt n}
+\frac1n\left|\sum_{i\neq j} G(x_i,x_j)\right|^{1/2}\right)?
\]
\end{problem}

Let $(\M,g)$ be a compact connected two-dimensional Riemannian manifold without boundary, $\dist$ is the induced distance, and let $G(x,y)$ denote the symmetric mean-zero Green function of the Laplacian (Definition \ref{def:green}). We work with the normalized volume measure $dx=\mathrm{vol}(\M)^{-1}\dd\mathrm{vol}$. Whenever we integrate in the variable $y$, we write $dy$ for the same measure. For points $x_1,\dots,x_n\in \M$, we define the empirical measure:
\[
\mu_n \coloneqq \frac1n\sum_{i=1}^n \delta_{x_i}.
\]
Steinerberger \cite{Ste21} showed that for $d\ge 3$, the following Green--Wasserstein inequality holds:
\[
\W(\mu_n,dx)\ \lesssim_{\M}\ n^{-1/d} + \frac1n\left|\sum_{i\neq j} G(x_i,x_j)\right|^{1/2}.
\]
In \cite[Problem 53]{SteOpen}, he gives the following estimate:
\begin{equation}\label{eq:Steinerberger2D}
\W\!\left(\frac1n\sum_{k=1}^n \delta_{x_k},\,dx\right)
\lesssim_\M
\frac1n\left|\sum_{i\neq j} G(x_i,x_j)\right|^{1/2}
+
\begin{cases}
\sqrt{\frac{\log n}{n}} & \text{if } d=2,\\
n^{-1/d} & \text{if } d\ge 3.
\end{cases}
\end{equation}
In this note, we use
\[
\sum_{i\neq j}(\cdots)\ \coloneqq\ \sum_{\substack{1\le i,j\le n\\ i\neq j}}(\cdots)
\]
to denote the ordered sum over distinct indices. Notice that, by symmetry of $G$, it is twice the sum over $i<j$. Steinerberger asked whether the factor $\sqrt{\log n}$ in \eqref{eq:Steinerberger2D} can be removed while keeping the same Green-energy term. The precise question is as follows.

\begin{remark}[The diagonal]\label{rem:diagonal}
We note that the Green function $G$ is smooth on $(\M\times \M)\setminus \{(x,x):x\in\M\}$ with a logarithmic singularity along the diagonal.
In a deterministic context, if $x_i=x_j$ for some $i\neq j$, then the term $G(x_i,x_j)$ is not finite. It is then natural to interpret the Green term in \eqref{eq:Steinerberger2D} and Problem \ref{prob:Steinerberger} as $+\infty$, in which case the inequality holds trivially.
In the random settings that we use in this note, collisions occur with probability $0$ because $dx$ is non-atomic.
\end{remark}

We now recall some definitions that will be useful in the main argument.
For Borel probability measures $\mu,\nu$ on $\M$, let us denote by $\Gamma(\mu,\nu)$ the set of couplings of $\mu$ and $\nu$ on $\M\times \M$.
The quadratic Wasserstein distance is
\[
\W(\mu,\nu)\coloneqq \left(\inf_{\gamma\in \Gamma(\mu,\nu)} \int_{\M\times \M} \dist(x,y)^2\,\dd\gamma(x,y)\right)^{1/2}.
\]
Given that $\M$ is compact, it follows that $\W(\mu,\nu)$ is always finite and satisfies $\W(\mu,\nu)\le \mathrm{diam}(\M)$. Let $\Delta$ be the Laplace--Beltrami operator on $(\M,g)$, which is realized here as a self-adjoint operator on $L^2(\M,dx)$ with domain $H^2(\M)$. Again, since $\M$ is compact and connected, $\ker(-\Delta)$ consists of the constant functions.

\begin{definition}[Mean-zero Green function]\label{def:green}
A mean-zero Green function for $-\Delta$ is a measurable function
\[
G:\M\times \M \to \mathbb{R}\cup\{+\infty\}
\]
with the following properties.
\begin{enumerate}
\item For each $x\in \M$, $y\mapsto G(x,y)$ is locally integrable on $\M$ and satisfies the normalization:
\begin{equation}\label{eq:GreenMeanZero}
\int_{\M} G(x,y)\,dy=0.
\end{equation}
\item For each $f\in C^\infty(\M)$ with $\int_{\M} f\,dx=0$, the function
\[
u(x)\coloneqq \int_{\M} G(x,y)\, f(y)\,dy
\]
is a weak solution of $-\Delta u = f$ with $\int_{\M}u\,dx=0$.
\end{enumerate}
\end{definition}

\begin{remark}[Local singularity]\label{rem:logsing}
In dimension $2$, the following classical local expansion (in geodesic normal coordinates) is true:
\[
G(x,y)= -\frac{1}{2\pi}\log \dist(x,y) + H(x,y).
\]
Note that $H$ extends continuously to the diagonal and is smooth off the diagonal.
In particular, we have $G(x,\cdot)\in L^1(\M,dx)$ for each fixed $x$.
\end{remark}

\begin{proof}
Recall that $\mathrm{vol}$ denotes the Riemannian volume measure on $(\M,g)$ and $dx = \mathrm{vol}(\M)^{-1}\dd\mathrm{vol}$. Let $\mathrm{inj}(\M)>0$ be the injectivity radius of $(\M,g)$ and set $r_0\coloneqq \mathrm{inj}(\M)/4$. We now apply standard results on the singularity structure of the Green function in dimension $2$ (for instance, see \cite{Aubin98}). In particular, the function
\[
H(x,y)\coloneqq G(x,y)+\frac{1}{2\pi}\log \dist(x,y),
\qquad x\neq y,
\]
extends continuously to the set $\{(x,y)\in \M\times\M: \dist(x,y)\le 2r_0\}$ (this also holds across the diagonal). Observe that this set is compact, so there exists $C_0>0$ such that
\[
|H(x,y)|\le C_0\qquad\text{whenever }\dist(x,y)\le 2r_0.
\]
As a result, for all $x\neq y$ with $\dist(x,y)<2r_0$, it follows that
\begin{equation}\label{eq:GreenLogBound}
|G(x,y)| \le \frac{1}{2\pi}|\log \dist(x,y)|+C_0 \le C\bigl(1+|\log \dist(x,y)|\bigr)
\end{equation}
for some constant $C>0$.

We now split $\M\times\M$ into the near-diagonal region
\[
D\coloneqq \{(x,y)\in\M\times\M: \dist(x,y)<r_0\}
\]
and its complement $D^c$. Notice that $D^c$ is compact and is always a positive distance away from the diagonal, so $G$ is smooth on $D^c$ and therefore bounded there. We obtain:
\[
\int_{D^c} G(x,y)^2\,dx\,dy<\infty.
\]
Let us now analyze $D$. By \eqref{eq:GreenLogBound}, we set
\[
I(x)\coloneqq \int_{B_{r_0}(x)} \bigl(1+|\log \dist(x,y)|\bigr)^2\,dy,
\]
so that
\[
\int_{D} G(x,y)^2\,dx\,dy
\le C^2\int_{\M} I(x)\,dx.
\]
We also fix $x\in\M$. Note that on the ball $B_{r_0}(x)$, we may use geodesic polar coordinates centered at $x$: $y=\exp_x(\rho\theta)$ with $\rho\in(0,r_0)$ and $\theta\in\mathbb{S}^1$. In these coordinates, the volume form has the following form:
\[
\dd\mathrm{vol}(y)=J_x(\rho,\theta)\,\dd\rho\,\dd\theta,
\qquad
J_x(\rho,\theta)=\rho\,a_x(\rho,\theta),
\]
where $a_x$ is smooth and satisfies $a_x(0,\theta)=1$.
In particular, by compactness, there exists $C_1>0$ with:
\begin{equation}\label{eq:JacobianBound}
J_x(\rho,\theta)\le C_1\rho\qquad\text{for all }x\in\M,\ \theta\in\mathbb{S}^1,\ \rho\in(0,r_0)
\end{equation}
(see, for instance, \cite{Chavel84}).
We now apply the normalization of $dx$ and obtain:
\begin{align*}
\int_{B_{r_0}(x)} \bigl(1+|\log \dist(x,y)|\bigr)^2\,dy
&=\frac{1}{\mathrm{vol}(\M)}\int_{\mathbb{S}^1}\!\!\int_0^{r_0} \bigl(1+|\log \rho|\bigr)^2 J_x(\rho,\theta)\,\dd\rho\,\dd\theta\\
&\le \frac{C_1}{\mathrm{vol}(\M)}\int_{\mathbb{S}^1}\!\!\int_0^{r_0} \bigl(1+|\log \rho|\bigr)^2 \rho\,\dd\rho\,\dd\theta.
\end{align*}
The right-hand side is finite since near $0$ the function $\rho\mapsto \rho(\log \rho)^2$ is integrable. We apply the substitution $t=-\log\rho$ (so $\rho=e^{-t}$ and $\rho\,\dd\rho=-e^{-2t}\,\dd t$) and get:
\[
\int_0^{r_0} \bigl(1+|\log \rho|\bigr)^2\rho\,\dd\rho
=\int_{\infty}^{-\log r_0} \bigl(1+|t|\bigr)^2(-e^{-2t})\,\dd t
=\int_{-\log r_0}^{\infty} \bigl(1+|t|\bigr)^2 e^{-2t}\,\dd t <\infty.
\]
It follows that $\int_{B_{r_0}(x)} (1+|\log \dist(x,y)|)^2\,dy$ is bounded uniformly in $x$.
If we integrate over $x$, we arrive at $\int_D G^2\,dx\,dy<\infty$. Finally, we combine the estimates on $D$ and $D^c$. This finishes the proof of $\sigma^2<\infty$.
\end{proof}

\begin{theorem}[The $\sqrt{\log n}$ remainder]\label{thm:main}
Let $G$ be the symmetric mean-zero Green function of $-\Delta$.
There does not exist a constant $C_{\M}>0$ such that for all $n\in\mathbb{N}$ and all $x_1,\dots,x_n\in \M$,
\begin{equation}\label{eq:NoUniversalO1sqrtN}
\W\!\left(\frac1n\sum_{i=1}^n\delta_{x_i},\, dx\right)
\le
C_{\M}\left(\frac{1}{\sqrt{n}}+\frac1n\left|\sum_{i\neq j} G(x_i,x_j)\right|^{1/2}\right).
\end{equation}
In particular, the $\sqrt{\log n}$ factor in the two-dimensional inequality \eqref{eq:Steinerberger2D} cannot be removed if the unrenormalized Green-energy term were to be preserved.
\end{theorem}

\begin{proof}
Recall that $\mathrm{vol}$ denotes the Riemannian volume measure on $(M,g)$ and $dx = \mathrm{vol}(M)^{-1}\,\mathrm{d}\mathrm{vol}$. Let $\mathrm{inj}(M)>0$ be the injectivity radius of $(M,g)$ and set $r_0\coloneqq \mathrm{inj}(M)/4$. We now apply standard results on the singularity structure of the Green function in dimension $2$ (for instance, see \cite{Aubin98}). In particular, the function
\[
H(x,y)\coloneqq G(x,y)+\frac{1}{2\pi}\log \mathrm{d}_g(x,y),
\qquad x\neq y,
\]
extends continuously to the set $\{(x,y)\in M\timesM: \mathrm{d}_g(x,y)\le 2r_0\}$ (this also holds across the diagonal). Observe that this set is compact, so there exists $C_0>0$ such that
\[
|H(x,y)|\le C_0\qquad\text{whenever }\mathrm{d}_g(x,y)\le 2r_0.
\]
As a result, for all $x\neq y$ with $\mathrm{d}_g(x,y)<2r_0$, it follows that
\begin{equation}\label{eq:GreenLogBound}
|G(x,y)| \le \frac{1}{2\pi}|\log \mathrm{d}_g(x,y)|+C_0 \le C\bigl(1+|\log \mathrm{d}_g(x,y)|\bigr)
\end{equation}
for some constant $C>0$.

We now split $M\timesM$ into the near-diagonal region
\[
D\coloneqq \{(x,y)\inM\timesM: \mathrm{d}_g(x,y)<r_0\}
\]
and its complement $D^c$. Notice that $D^c$ is compact and is always a positive distance away from the diagonal, so $G$ is smooth on $D^c$ and therefore bounded there. We obtain:
\[
\int_{D^c} G(x,y)^2\,dx\,dy<\infty.
\]
Let us now analyze $D$. By \eqref{eq:GreenLogBound}, we set
\[
I(x)\coloneqq \int_{B_{r_0}(x)} \bigl(1+|\log \mathrm{d}_g(x,y)|\bigr)^2\,dy,
\]
so that
\[
\int_{D} G(x,y)^2\,dx\,dy
\le C^2\int_{M} I(x)\,dx.
\]
We also fix $x\inM$. Note that on the ball $B_{r_0}(x)$, we may use geodesic polar coordinates centered at $x$: $y=\exp_x(\rho\theta)$ with $\rho\in(0,r_0)$ and $\theta\in\mathbb{S}^1$. In these coordinates, the volume form has the following form:
\[
\,\mathrm{d}\mathrm{vol}(y)=J_x(\rho,\theta)\,\,\mathrm{d}\rho\,\,\mathrm{d}\theta,
\qquad
J_x(\rho,\theta)=\rho\,a_x(\rho,\theta),
\]
where $a_x$ is smooth and satisfies $a_x(0,\theta)=1$.
In particular, by compactness, there exists $C_1>0$ with:
\begin{equation}\label{eq:JacobianBound}
J_x(\rho,\theta)\le C_1\rho\qquad\text{for all }x\inM,\ \theta\in\mathbb{S}^1,\ \rho\in(0,r_0)
\end{equation}
(see, for instance, \cite{Chavel84}).
We now apply the normalization of $dx$ and obtain:
\begin{align*}
\int_{B_{r_0}(x)} \bigl(1+|\log \mathrm{d}_g(x,y)|\bigr)^2\,dy
&=\frac{1}{\mathrm{vol}(M)}\int_{\mathbb{S}^1}\!\!\int_0^{r_0} \bigl(1+|\log \rho|\bigr)^2 J_x(\rho,\theta)\,\,\mathrm{d}\rho\,\,\mathrm{d}\theta\\
&\le \frac{C_1}{\mathrm{vol}(M)}\int_{\mathbb{S}^1}\!\!\int_0^{r_0} \bigl(1+|\log \rho|\bigr)^2 \rho\,\,\mathrm{d}\rho\,\,\mathrm{d}\theta.
\end{align*}
The right-hand side is finite since near $0$ the function $\rho\mapsto \rho(\log \rho)^2$ is integrable. We apply the substitution $t=-\log\rho$ (so $\rho=e^{-t}$ and $\rho\,\,\mathrm{d}\rho=-e^{-2t}\,\,\mathrm{d} t$) and get:
\[
\int_0^{r_0} \bigl(1+|\log \rho|\bigr)^2\rho\,\,\mathrm{d}\rho
=\int_{\infty}^{-\log r_0} \bigl(1+|t|\bigr)^2(-e^{-2t})\,\,\mathrm{d} t
=\int_{-\log r_0}^{\infty} \bigl(1+|t|\bigr)^2 e^{-2t}\,\,\mathrm{d} t <\infty.
\]
It follows that $\int_{B_{r_0}(x)} (1+|\log \mathrm{d}_g(x,y)|)^2\,dy$ is bounded uniformly in $x$.
If we integrate over $x$, we arrive at $\int_D G^2\,dx\,dy<\infty$. Finally, we combine the estimates on $D$ and $D^c$. This finishes the proof of $\sigma^2<\infty$.
\end{proof}

Using the 2D Green function singularity, write
\[
H(x,y)\coloneqq G(x,y)+\frac{1}{2\pi}\log \mathrm{d}_g(x,y)
\]
which “extends continuously” to a compact neighborhood of the diagonal, hence is bounded there. This gives the near-diagonal logarithmic bound
\[
|G(x,y)|\le C\bigl(1+|\log \mathrm{d}_g(x,y)|\bigr)\qquad (\mathrm{d}_g(x,y)<2r_0).
\]
Split \(M\times M\) into the near-diagonal region \(D=\{\mathrm{d}_g(x,y)<r_0\}\) and its complement \(D^c\). On \(D^c\), “\(G\) is smooth … and therefore bounded,” so \(\int_{D^c}G^2<\infty\).

On \(D\), use the bound above and reduce to showing finiteness of
\[
I(x)\coloneqq \int_{B_{r_0}(x)}\bigl(1+|\log \mathrm{d}_g(x,y)|\bigr)^2\,dy.
\]
In geodesic polar coordinates \(y=\exp_x(\rho\theta)\), the Jacobian satisfies \(J_x(\rho,\theta)=\rho a_x(\rho,\theta)\le C_1\rho\), hence
\[
I(x)\le \frac{C_1}{\mathrm{vol}(M)}\int_{\mathbb{S}^1}\!\!\int_0^{r_0}(1+|\log\rho|)^2\rho\,d\rho\,d\theta.
\]
The radial integral is finite because “near \(0\) the function \(\rho\mapsto \rho(\log\rho)^2\) is integrable,” shown via the substitution \(t=-\log\rho\) giving an integrand \((1+|t|)^2e^{-2t}\). This yields a uniform bound on \(I(x)\), so integrating over \(x\) gives \(\int_D G^2<\infty\). Combining \(D\) and \(D^c\) proves \(G\in L^2(M\times M)\) and hence \(\sigma^2<\infty\).

Using the 2D Green function singularity, write
\[
H(x,y)\coloneqq G(x,y)+\frac{1}{2\pi}\log \mathrm{d}_g(x,y)
\]
which extends continuously to a compact neighborhood of the diagonal, hence is bounded there. This gives the near-diagonal logarithmic bound
\[
|G(x,y)|\le C\bigl(1+|\log \mathrm{d}_g(x,y)|\bigr)\qquad (\mathrm{d}_g(x,y)<2r_0).
\]
Split \(M\times M\) into the near-diagonal region \(D=\{\mathrm{d}_g(x,y)<r_0\}\) and its complement \(D^c\). On \(D^c\), \(G\) is smooth and therefore bounded, so \(\int_{D^c}G^2<\infty\).

On \(D\), use the bound above and reduce to showing finiteness of
\[
I(x)\coloneqq \int_{B_{r_0}(x)}\bigl(1+|\log \mathrm{d}_g(x,y)|\bigr)^2\,dy.
\]
In geodesic polar coordinates \(y=\exp_x(\rho\theta)\), the Jacobian satisfies \(J_x(\rho,\theta)=\rho a_x(\rho,\theta)\le C_1\rho\), hence
\[
I(x)\le \frac{C_1}{\mathrm{vol}(M)}\int_{\mathbb{S}^1}\!\!\int_0^{r_0}(1+|\log\rho|)^2\rho\,d\rho\,d\theta.
\]
The radial integral is finite because near \(0\) the function \(\rho\mapsto \rho(\log\rho)^2\) is integrable; via the substitution \(t=-\log\rho\) the integrand becomes \((1+|t|)^2e^{-2t}\). This yields a uniform bound on \(I(x)\), so integrating over \(x\) gives \(\int_D G^2<\infty\). Combining \(D\) and \(D^c\) proves \(G\in L^2(M\times M)\) and hence \(\sigma^2<\infty\).

Question: Let $(M,g)$ be a compact connected $2$-dimensional Riemannian manifold without boundary, let $dx=\operatorname{vol}(M)^{-1}\,d\operatorname{vol}$ be the normalized volume measure, and let $G:M\times M\to \mathbb{R}\cup\{+\infty\}$ be a symmetric mean-zero Green function for $-\Delta$, meaning that $G(x,y)=G(y,x)$, for each fixed $x\in M$ the map $y\mapsto G(x,y)$ is locally integrable and satisfies $\int_M G(x,y)\,dy=0$, and for each $f\in C^\infty(M)$ with $\int_M f\,dx=0$, the function $u(x)=\int_M G(x,y)f(y)\,dy$ is a weak solution of $-\Delta u=f$ with $\int_M u\,dx=0$. Which statement holds for every such $(M,g)$ and $G$?

The Green function is square-integrable on $M\times M$ with respect to the product measure $dx\otimes dx$; equivalently, $G\in L^2(M\times M,dx\otimes dx)$ and therefore
\[
\sigma^2\coloneqq \int_M\int_M G(x,y)^2\,dx\,dy<\infty.
\]

Choices:
B.
The Green function is essentially bounded on $M\times M$ with respect to the product measure $dx\otimes dx$; equivalently, $G\in L^\infty(M\times M,dx\otimes dx)$ and therefore
\[
\int_M\int_M G(x,y)^2\,dx\,dy<\infty.
\]
C.
For each fixed $x\in M$, the section $y\mapsto G(x,y)$ belongs to $L^2(M,dy)$, and in particular
\[
\int_M G(x,y)^2\,dy<\infty\qquad\text{for every }x\in M.
\]
D.
The Green function is square-integrable off the diagonal but not on all of $M\times M$; more precisely, for every $\varepsilon>0$ one has
\[
G\in L^2\bigl(\{(x,y)\in M\times M:\ d_g(x,y)\ge \varepsilon\},dx\otimes dx\bigr),
\]
whereas
\[
\int_M\int_M G(x,y)^2\,dx\,dy=+\infty.
\]
E.
For each fixed $x\in M$, the map $y\mapsto G(x,y)$ belongs to $L^p(M,dy)$ for every $p>2$, and consequently
\[
G\in L^p(M\times M,dx\otimes dx)\qquad\text{for every }p>2.
\]

logarithmic singularity near the diagonal is bounded only after subtracting the log term, not in absolute size

joint $L^2(M\times M)$ conclusion weakened to fiberwise $L^2$ in the $y$-variable for each fixed $x$

near-diagonal integrability of $\rho(\log\rho)^2$

critical exponent from the logarithmic singularity: $L^2$ holds but no gain to all $p>2$


Question: Let $(M,g)$ be a compact connected $2$-dimensional Riemannian manifold without boundary, and let $dx=\operatorname{vol}(M)^{-1}\,d\operatorname{vol}$ be the normalized volume measure on $M$. Let $G:M\times M\to \mathbb{R}\cup\{+\infty\}$ be a symmetric mean-zero Green function for the Laplace--Beltrami operator $-\Delta$, meaning that $G(x,y)=G(y,x)$, for each fixed $x\in M$ the function $y\mapsto G(x,y)$ is locally integrable and satisfies $\int_M G(x,y)\,dy=0$, and for every $f\in C^\infty(M)$ with $\int_M f\,dx=0$, the function $u(x)=\int_M G(x,y)f(y)\,dy$ is a weak solution of $-\Delta u=f$ with $\int_M u\,dx=0$. Which statement holds for such a Green function?

The Green function satisfies $G\in L^2(M\times M,dx\otimes dx)$. Consequently, the quantity \[\sigma^2\coloneqq \int_M\int_M G(x,y)^2\,dx\,dy\] is finite.

Choices:
B. The Green function satisfies $G\in L^p(M\times M,dx\otimes dx)$ for every $p\in[1,\infty]$. Consequently, in particular, $G$ is essentially bounded on $M\times M$ and \[\int_M\int_M G(x,y)^2\,dx\,dy<\infty.\]
C. The Green function satisfies $G\in L^1(M\times M,dx\otimes dx)$. Consequently, the quantity \[\int_M\int_M |G(x,y)|\,dx\,dy\] is finite.
D. For every compact connected $2$-dimensional Riemannian manifold without boundary, there exists a constant $C_M>0$ such that for every symmetric mean-zero Green function $G$ of $-\Delta$ one has \[\sup_{x\in M}\int_M G(x,y)^2\,dy\le C_M.\] In particular, $G\in L^2(M\times M,dx\otimes dx)$.
E. The Green function satisfies $G\in L^2\big((M\times M)\setminus\{(x,x):x\in M\},dx\otimes dx\big)$, but \[\int_M\int_M G(x,y)^2\,dx\,dy=+\infty\] because the logarithmic singularity along the diagonal is not square-integrable in dimension $2$.

logarithmic singularity only yields finite L2, not boundedness or all Lp

drop square-integrability and retain only absolute integrability

uniform-in-x L1 bound on local model does not imply uniform L2_y section bound

misclassifies the 2D logarithmic singularity as non-square-integrable near the diagonal