# Equidimensional quiver representations and their $U$-invariants

A.N. Panov

###### Abstract

For an arbitrary equidimensional quiver representation, we proposed the method
of construction of a system of free generators of the field of $U$-invariants.
The construction of the section and system of generators depends on the choice
of a map that assign to each vertex one of the arrows incident to it.

Mechanical and Mathematical Department, Samara National Research University,
Samara, Russia

<EMAIL_ADDRESS>
##  1 Introduction

A quiver is a directed graph. A quiver is given by a pair $Q=(V,A)$, where $V$
is the set of vertices and $A$ is the set of arrows. Each arrow $\alpha$
starts at its source $s(\alpha)$ and terminates at its target $t(\alpha)$.
Multiple arrows and loops are allowed.

Let $K$ be a field. A representation of the quiver $Q$ assigns to each vertex
$v\in V$ a linear space $\\{W_{v},~{}v\in V\\}$ defined over the field $K$,
and to each arrow $\alpha$ a linear map $W_{s(\alpha)}\to W_{t(\alpha)}$. Each
representation defines the dimensional vector $n_{Q}=(n_{v})$, where
$n_{v}=\dim W_{v}$.

We identify each linear space $W_{v}$ with the coordinate space $K^{n_{v}}$
and the space of linear maps ${\mathcal{H}}_{\alpha}$ from $W_{s(\alpha)}$ to
$W_{t(\alpha)}$ with the space of matrices
${\mathrm{Mat}}(n_{t(\alpha)},n_{s(\alpha)},K)$ of size $n_{t(\alpha)}\times
n_{s(\alpha)}$.

For each vertex $v\in V$, we consider the group
${\mathrm{GL}}_{v}={\mathrm{GL}}(n_{v})$ with entries in the field $K$, and
its unitriangular group $U_{v}={\mathrm{UT}}(n_{v})$ that consists of all
upper triangular $(n_{v}\times n_{v})$-matrices with ones on the diagonal.
Consider the direct product ${\mathrm{GL}}_{Q}=\prod_{v\in
V}{\mathrm{GL}}_{v}$, its subgroup $U_{Q}=\prod_{v\in V}U_{v}$, and the linear
space

${\mathcal{H}}={\mathcal{H}}_{Q}=\oplus_{\alpha\in A}{\mathcal{H}}_{\alpha}.$

The group ${\mathrm{GL}}_{Q}$ acts on ${\mathcal{H}}$ by the formula

$g.h=(g_{t(\alpha)}X_{\alpha}g_{s(\alpha)}^{-1})_{\alpha\in A},\quad
g=(g_{v})\in{\mathrm{GL}}_{Q},\quad h=(X_{\alpha})\in{\mathcal{H}}.$

The action of ${\mathrm{GL}}_{Q}$ on ${\mathcal{H}}$ determines the
representation $\rho_{g}$ of the group ${\mathrm{GL}}_{Q}$ in the space of
regular functions $K[{\mathcal{H}}]$ by the formula

$\rho_{g}f(h)=f(g^{-1}.h).$

This representation is extended to the action of ${\mathrm{GL}}_{Q}$ on the
field of rational functions $K({\mathcal{H}})$. For any subgroup $G$ of
${\mathrm{GL}}_{Q}$ one can define the algebra of invariants
$K[{\mathcal{H}}]^{G}$ and the field of invariants $K({\mathcal{H}})^{G}$.

The partial case is the action of the group ${\mathrm{GL}}(n)$ on the systems
of matrices $g.(X_{1},\ldots,X_{n})=(gX_{1}g^{-1},\ldots,gX_{n}g^{-1})$. This
representation is associated with the quiver with the set of loops
$\alpha_{1},\ldots,\alpha_{n}$ with the common vertex. The problem of
description of the algebra of ${\mathrm{GL}}(n)$-invariants is solved in the
framework of classical invariant theory in tensors (see [1, 2, 4]). The
algebra of ${\mathrm{GL}}(n)$-invariants is generated by the system of
polynomials $\sigma_{t}(M)$, where $M$ is an arbitrary monomial of
$X_{1},\ldots,X_{n}$, and $\\{\sigma_{t}:~{}~{}1\leqslant t\leqslant n\\}$ are
coefficients of the characteristic polynomial.

For an arbitrary quiver $Q$ and a representation of an arbitrary dimension
$(n_{v})_{v\in V}$, the algebra of ${\mathrm{GL}}_{Q}$-invariants is generated
by the polynomials $\sigma_{t}(X_{\beta_{1}}\cdots X_{\beta_{s}})$, where
$\beta_{1},\ldots,\beta_{s}$ is an arbitrary directed closed path of the
quiver. This theorem in the case of zero characteristic was solved in 1990 in
the paper [3] of L. Le Bruyn, C. Procesi, and in the case of arbitrary
characteristic in 1994 in [6] of S. Donkin.

In the present paper, we consider the action of the subgroup $U_{Q}$ on
${\mathcal{H}}={\mathcal{H}}_{Q}$. The problem of construction of a system of
generators with their relations is considered extremely difficult even for the
case of coadjoint action of ${\mathrm{UT}}(n)$ on the space of
${\mathrm{Mat}}(n)$ (this is associated with the quiver of one loop and one
vertex). We consider the action of $U_{Q}$ on the field $K({\mathcal{H}})$. We
refer to the field of invariants of this action as the field of $U$-invariants
and denote $K({\mathcal{H}})^{U}$. According to the K. Miyata theorem [5], the
field of invariants with respect to an action of a group of inipotent
transformations is rational. Therefore, the field $K({\mathcal{H}})^{U}$ is
rational, i.e. it is a pure transcendental extension of the main field $K$.
Definition. We say that a quiver representation is equidimensional if
$n_{v}=n$ for each $v\in V$.

In the case of equidimensional representation, the linear space
${\mathcal{H}}$ is decomposed into a direct sum of the subspaces
${\mathcal{H}}_{\alpha}$, $\alpha\in A$, which is identified with the space of
$(n\times n)$-matrices ${\mathrm{Mat}}(n)$ defined over the field $K$.

We aim to construct the system of free generators of the field of
$U$-invariants of the equidimensional representation ${\mathcal{H}}$,
$n_{v}=n$, for an arbitrary quiver $Q$. The main result (see Theorem 3) is
formulated in terms of a fixed map $\psi:V\to A$ that assign to each vertex
$v\in V$ one of the arrows $\psi(v)\in A$ incident to it. The constructed
section ${\mathcal{S}}$ and the system of free generators
${\mathbb{P}}_{\alpha}$, $\alpha\in A$, depend on the choice of the map
$\psi$. Earlier in the papers [7, 8, 9], the author constructed such systems
for the adjoint representation and representations on the matrix tuples.

We introduce some notations. For positive integers $a$ and $b$, we denote by
$[a,b]$ the set of all integers $a\leqslant i\leqslant b$. For any integer
$1\leqslant i\leqslant n$, let $i^{\prime}$ stand for the symmetric number to
$i$ with respect to the center of the segment $[1,n]$. We have
$i^{\prime}=n-i+1$.

Consider the $(n\times n)$-matrix $X=(x_{ij})_{i,j=1}^{n}$ with variables
$\\{x_{ij}\\}$. For each $1\leqslant k\leqslant n$, we denote by $D_{k}(X)$
the left lower corner minor of order $k^{\prime}$ for $X$. In particular,
$D_{1}(X)=\det X$.

We denote by ${\mathcal{F}}_{\alpha}$ the subfield of
$K({\mathcal{H}}_{\alpha})$ generated by $D_{k}(X_{\alpha})$, $1\leqslant
k\leqslant n$. The polynomial $D_{k}(X)$ is invariant with respect to the left
and right multiplication by ${\mathrm{UT}}(n)$. Therefore for each $\alpha\in
A$, the polynomials $\\{D_{k}(X_{\alpha})\\}$ are $U$-invariants, i.e.
$\\{D_{k}(X_{\alpha})\\}$ belongs to $K[{\mathcal{H}}]^{U}$. Consider the
subfield ${\mathcal{F}}$ of $K({\mathcal{H}}_{\alpha})^{U}$ generated by
${\mathcal{F}}_{\alpha}$, $\alpha\in A$.

We say that a matrix is anti-diagonal if its entries above and below the anti-
diagonal are zeros. We denote by $\Lambda(n)$ the subspace of all anti-
diagonal $(n\times n)$-matrices over the field $K$. We consider the subspace
of upper anti-triangular matrices ${\mathcal{S}}^{+}(n)$ over the field $K$
(their entries below the anti-diagonal are zeros) and the subspace of lower
anti-triangular matrices ${\mathcal{S}}^{-}(n)$ (their entries above the anti-
diagonal are zeros).

We introduce a linear order on the set of all pairs $\\{(a,b)\\}$, $1\leqslant
a,b\leqslant n$, as follows: $(a,b)\prec(i,k)$, if $b<k$ or $b=k$ and $a>i$.
Then $(n,1)\prec(n-1,1)\prec\ldots\prec(1,1)\prec(n,2)\prec\ldots\prec(1,n)$.
Definition. Let $X=(x_{ij})_{i,j=1}^{n}$ be the matrix with variables
$\\{x_{ij}\\}$.
1) We say that a polynomial $f(X)$ is obtained by a triangular transformation
from the variable $x_{ik}$ if $f(X)=f_{0}(X)x_{ik}+f_{1}(X)$, where $f_{0}(X)$
and $f_{1}(X)$ are polynomials in variables $x_{ab}$, $(a,b)\prec(i,k)$
defined over the same field as $f(X)$, and $f_{0}(X)\neq 0$.
2) We say that the system of polynomials $\\{f_{ik}(X):~{}1\leqslant
i,k\leqslant n\\}$, is obtained by a triangular transformation from the system
$\\{x_{ik}:~{}1\leqslant i,k\leqslant n\\}$, if each $f_{ik}(X)$ is obtained
by a triangular transformation from $x_{ik}$.
Remark. If $\\{f_{ik}(X):~{}1\leqslant i,k\leqslant n\\}$ is obtained by a a
triangular transformation from $\\{x_{ik}:~{}1\leqslant i,k\leqslant n\\}$,
then the first system of polynomials (and the second also) freely generate the
field of rational functions $K(X)$.

##  2 Construction of section

In this section, we construct a section for the action of the subgroup $U_{Q}$
on ${\mathcal{H}}$. Fix a map $\psi:V\to A$ such that for each $v\in V$ the
arrow $\psi(v)$ is incident to $v$. Then for any $\alpha$ from
${\mathrm{Im}}\,\psi$ we have $\alpha=\psi({s(\alpha)})$ or
$\alpha=\psi({t(\alpha)})$. It is possible that
$\alpha=\psi({t(\alpha)})=\psi({s(\alpha)})$. The construction of the section
and the system of free generators depends on the choice of the map $\psi$.

We consider the subspace ${\mathcal{S}}=\oplus_{\alpha\in
A}{\mathcal{S}}_{\alpha}$, where the subspace ${\mathcal{S}}_{\alpha}$ in
${\mathcal{H}}_{\alpha}$ is defined as follows. If $\alpha$ connects different
vertices (i.e.
$\textstyle{t(\alpha)}$$\textstyle{{s(\alpha)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha}$
and ${s(\alpha)}\neq{t(\alpha)}$), we define

${\mathcal{S}}_{\alpha}=\left\\{\begin{array}[]{l}{\mathcal{H}}_{\alpha}={\mathrm{Mat}}(n),\mbox{if}~{}\alpha\notin{\mathrm{Im}}\,\psi,\\\
{\mathcal{S}}^{-}(n),\mbox{if}~{}\alpha=\psi({t(\alpha)})\neq\psi({s(\alpha)}),\\\
{\mathcal{S}}^{+}(n),\mbox{if}~{}\alpha=\psi({s(\alpha)})\neq\psi({t(\alpha)}),\\\
\Lambda(n),\mbox{if}~{}\alpha=\psi({t(\alpha)})=\psi({s(\alpha)}).\end{array}\right.$
(1)

If $\alpha$ is a loop, we denote $q={s(\alpha)}={t(\alpha)}$. We have
$\textstyle{q\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha}$.
We define

${\mathcal{S}}_{\alpha}=\left\\{\begin{array}[]{l}{\mathcal{H}}_{\alpha}={\mathrm{Mat}}(n),~{}\mbox{if}~{}\alpha\neq\psi(q),\\\
{\mathcal{S}}^{-}(n),~{}\mbox{if}~{}\alpha=\psi(q).\\\ \end{array}\right.$ (2)

Proposition  2.1. The space ${\mathcal{H}}$ is a closure of the set

$\bigcup_{g\in U_{Q}}g.{\mathcal{S}}.$

_Proof_. Consider the open subset $\Omega\subset{\mathcal{H}}$ that consists
of all $h=(X_{\alpha})_{\alpha\in A}$ such that $D_{k}(X_{\alpha})\neq 0$ for
all $1\leqslant k\leqslant n$. Let us show that for any
$h=(X_{\alpha})_{\alpha\in A}\in\Omega$ there exist $g=(g_{v})_{v\in V}\in U$
such that $g.h\in{\mathcal{S}}$, i.e.

$(g.h)_{\alpha}=g_{t(\alpha)}X_{\alpha}g_{s(\alpha)}^{-1}\in{\mathcal{S}}_{\alpha}\quad\mbox{for~{}~{}each}\quad\alpha\in
A.$ (3)

If $\alpha\notin{\mathrm{Im}}\,\psi$, then
${\mathcal{S}}_{\alpha}={\mathrm{Mat}}(n)$, and (3) is true for all
$g=g=(g_{v})_{v\in V}$.

One knows the following property: for any $(n\times n)$-matrix $X$ obeying the
condition $D_{k}(X)\neq 0$, $1\leqslant k\leqslant n$, there exist matrices
$u_{1},u_{2}\in{\mathrm{UT}}(n)$ such that $u_{1}Xu^{-1}_{2}\in\Lambda(n)$ (it
implies $u_{1}X\in{\mathcal{S}}^{-}(n)$ and
$Xu^{-1}_{2}\in{\mathcal{S}}^{+}(n)$).

Consider the case when $\alpha$ connects different vertices. If
$\alpha=\psi({t(\alpha)})\neq\psi({s(\alpha)})$, then we have
${\mathcal{S}}_{\alpha}={\mathcal{S}}^{-}(n)$. Taking $X=X_{\alpha}$ and
$g_{t(\alpha)}=u_{1}$, we obtain
$g_{t(\alpha)}X_{\alpha}\in{\mathcal{S}}^{-}(n)$. Then
$g_{t(\alpha)}X_{\alpha}g_{s(\alpha)}^{-1}$ belongs to ${\mathcal{S}}^{-}(n)$
for any $g_{s(\alpha)}$. We get $(g.h)_{\alpha}\in{\mathcal{S}}_{\alpha}$.

If $\alpha=\psi({s(\alpha)})\neq\psi({t(\alpha)})$, then we have
${\mathcal{S}}_{\alpha}={\mathcal{S}}^{+}(n)$. Taking $g_{s(\alpha)}=u_{2}$,
we obtain $X_{\alpha}g_{s(\alpha)}^{-1}\in{\mathcal{S}}^{+}(n)$. Then
$g_{t(\alpha)}X_{\alpha}g_{s(\alpha)}^{-1}$ belongs to ${\mathcal{S}}^{+}(n)$
for any $g_{t(\alpha)}$. We get $(g.h)_{\alpha}\in{\mathcal{S}}_{\alpha}$.

If $\alpha=\psi({t(\alpha)})=\psi({s(\alpha)})$, then we have
${\mathcal{S}}_{\alpha}=\Lambda(n)$. Taking $g_{t(\alpha)}=u_{1}$ and
$g_{s(\alpha)}=u_{2}$, we obtain that
$g_{t(\alpha)}X_{\alpha}g_{s(\alpha)}^{-1}$ belongs to $\Lambda(n)$. We get
$(g.h)_{\alpha}\in{\mathcal{S}}_{\alpha}$.

Consider the case when $\alpha$ is a loop with vertex $q$. If
$\alpha=\psi(q)$, then we have ${\mathcal{S}}_{\alpha}={\mathcal{S}}^{-}(n)$.
Taking $g_{q}=u_{1}$, we obtain $g_{q}X_{\alpha}\in{\mathcal{S}}^{-}(n)$. Then
$g_{q}X_{\alpha}g_{q}^{-1}$ belongs to ${\mathcal{S}}^{-}(n)$. We get
$(g.h)_{\alpha}\in{\mathcal{S}}_{\alpha}$. $\Box$

Proposition 2 impies the corollary.
Corollary  2.2. The restriction map $\pi:K[{\mathcal{H}}]^{U}\to
K[{\mathcal{S}}]$ is an embedding. The embedding $\pi$ extends to the
embedding of fields

$\pi:K({\mathcal{H}})^{U}\hookrightarrow K({\mathcal{S}}).~{}~{}\Box$

In particular, the field $\pi({\mathcal{F}}_{\alpha})$ is a subfield of
$K({\mathcal{S}}_{\alpha})$. If ${\mathcal{S}}_{\alpha}$ equals to
${\mathcal{S}}^{\pm}(n)$ or $\Lambda(n)$, then $\pi({\mathcal{F}}_{\alpha})$
is generated by the anti-diagonal elements of the matrix
$X_{\alpha}\in{\mathcal{S}}_{\alpha}$.

##  3 $U$-invariants associated with arrows

In this section, we assign to each arrow $\alpha\in A$ a system of polynomials
${\mathbb{P}}_{\alpha}$. These polynomials are $U$-invariants. We show that
the system $\pi({\mathbb{P}}_{\alpha}\sqcup{\mathbb{D}}_{\alpha})$ is obtained
by a triangular transformation over the field $\pi({\mathcal{F}})$ from the
system of matrix entries $\\{x_{ab}\\}$ of the matrix $X_{\alpha}$, which form
the standard basis in the dual space of ${\mathcal{S}}_{\alpha}$. Moreover,
$\pi({\mathbb{D}}_{\alpha})$ is obtained by a triangular transformation over
the field $K$ from the matrix entries $\\{x_{a,a^{\prime}}\\}$ that belong to
the anti-diagonal of $X_{\alpha}$, and $\pi({\mathbb{P}}_{\alpha})$ is
obtained by a triangular transformation over the field $\pi({\mathcal{F}})$
from the other matrix entries of $X_{\alpha}$.

If $\alpha$ connects different vertices
$\textstyle{t(\alpha)}$$\textstyle{{s(\alpha)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha}$,
then we define
${\mathbb{P}}_{\alpha}={\mathbb{P}}_{s(\alpha)}\sqcup{\mathbb{P}}_{t(\alpha)}$,
where ${\mathbb{P}}_{s(\alpha)}$ (respectively, ${\mathbb{P}}_{t(\alpha)})$ is
a system of $U$-invariants associated with the source ${s(\alpha)}$
(respectively, with the target ${t(\alpha)}$) of the arrow $\alpha$ (see
formula (7)).

If ${\mathcal{S}}_{\alpha}={\mathcal{S}}^{+}(n)$ or
${\mathcal{S}}_{\alpha}=\Lambda(n)$, then we take
${\mathbb{P}}_{s(\alpha)}=\varnothing$. If
${\mathcal{S}}_{\alpha}={\mathrm{Mat}}(n)$ or
${\mathcal{S}}_{\alpha}={\mathcal{S}}^{-}(n)$, the system of standard
coordinate functions $\\{x_{ab}:a^{\prime}<b\\}$ (they lie below the anti-
diagonal of $X_{\alpha}$) is algebraically independent on
${\mathcal{S}}_{\alpha}$. The system ${\mathbb{P}}_{s(\alpha)}$ has the
property that its restriction $\pi({\mathbb{P}}_{s(\alpha)})$ on the section
${\mathcal{S}}$ is obtained by a triangular transformation over
$\pi({\mathcal{F}})$ from $\\{x_{ab}:a^{\prime}<b\\}$ (see Lemmas 3 and 3).

Similarly, we define ${\mathbb{P}}_{t(\alpha)}$. If
${\mathcal{S}}_{\alpha}={\mathcal{S}}^{-}(n)$ or
${\mathcal{S}}_{\alpha}=\Lambda(n)$, then we take
${\mathbb{P}}_{t(\alpha)}=\varnothing$. If
${\mathcal{S}}_{\alpha}={\mathrm{Mat}}(n)$ or
${\mathcal{S}}_{\alpha}={\mathcal{S}}^{+}n)$, then the system of standard
coordinate functions $\\{x_{ab}:a^{\prime}>b\\}$ (they lie above the anti-
diagonal of $X_{\alpha}$) is algebraically independent on
${\mathcal{S}}_{\alpha}$. The system ${\mathbb{P}}_{t(\alpha)}$ has the
property that its restriction $\pi({\mathbb{P}}_{t(\alpha)})$ on the section
${\mathcal{S}}$ is obtained by a triangular transformation over
$\pi({\mathcal{F}})$ from $\\{x_{ab}:a^{\prime}>b\\}$ (see Lemmas 3 and 3).

If $\alpha$ is a loop, then we take ${\mathbb{P}}_{t(\alpha)}=\varnothing$ and
${\mathbb{P}}_{\alpha}={\mathbb{P}}_{s(\alpha)}$.

In what follows, in Cases 1 and 2, for each arrow $\beta$ incident to its
source ${s(\alpha)}$, we construct a system of $U$-invariants
${\mathbb{P}}_{{s(\alpha)},\beta}$. We define
${\mathbb{P}}_{s(\alpha)}={\mathbb{P}}_{{s(\alpha)},\beta}$, where
$\beta=\psi({s(\alpha)})$ (see (5)).

Respectively, in Cases 3 and 4, for each $\gamma$ incident to its target
${t(\alpha)}$, we construct a system of $U$-invariants
${\mathbb{P}}_{\gamma,{t(\alpha)}}$. We define
${\mathbb{P}}_{t(\alpha)}={\mathbb{P}}_{\gamma,{t(\alpha)}}$, where
$\gamma=\psi({t(\alpha)})$ (see (6)).

Case 1. Let
$\textstyle{t(\alpha)}$$\textstyle{{s(\alpha)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha}$$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\beta}$,
i.e. $\beta$ terminates at ${s(\alpha)}$. We admit that $\alpha$ or $\beta$ is
a loop (i.e. ${t(\alpha)}={s(\alpha)}$ or ${s(\alpha)}=v$). It is possible
$\alpha$ and $\beta$ are loops and $\alpha=\beta$.

The action of $g\in{\mathrm{GL}}_{Q}$ on the pair of matrices $(X,Y)$, where
$X=X_{\alpha}$ and $Y=X_{\beta}$, is defined by the formula
$g.(X,Y)=(g_{t(\alpha)}Xg_{s(\alpha)}^{-1},g_{s(\alpha)}Yg_{v}^{-1})$.

For the pair $i^{\prime}\leqslant j$ (i.e. $(i,j)$ lies on or below the anti-
diagonal), we denote by $M_{ij}(X)$ the minor of order $i^{\prime}$ of the
matrix $X$ with the system of rows $[i,n]$ and columns
$[1,i^{\prime}-1]\sqcup\\{j\\}$.

For the pair $j\leqslant k$, we denote by $N_{jk}(Y)$ the minor of order
$k^{\prime}$ of the matrix $Y$ with the system of rows $\\{j\\}\sqcup[k+1,n]$
and columns $[1,k^{\prime}]$. Observe that

$D_{k}(X)=M_{k,k^{\prime}}(X),\qquad D_{k}(Y)=N_{k,k}(Y).$

Let $i^{\prime}<k$ (i.e. $(i,k)$ lies below the anty-diagonal). We define the
polynomial

$P_{ik}(X,Y)=\sum_{i^{\prime}\leqslant j\leqslant k}M_{ij}(X)N_{jk}(Y).$ (4)

The polynomials

${\mathbb{P}}(X,Y)=\\{P_{ik}(X,Y):~{}i^{\prime}<k\\}$

are $U$-invariants [9].

Lemma  3.1. 1) The polynomial $P_{ik}(X,Y)$, where $(i,k)$ lies below the
anti-diagonal, is obtained by a triangular transformation from $x_{ik}$ over
the field $K(Y)$.
2) Let $\beta$ obey the condition of Case 1, and $\beta=\psi({s(\alpha)})$.
Then the system of standard coordinate functions $\\{x_{ab}:a^{\prime}<b\\}$
(they lie below the anti-diagonal of $X_{\alpha}$) is algebraically
independent on ${\mathcal{S}}_{\alpha}$. The system
$\pi({\mathbb{P}}(X_{\alpha},X_{\beta}))$ is obtained by a triangular
transformation over $\pi({\mathcal{F}})$ from $\\{x_{ab}:a^{\prime}<b\\}$.

_Proof_. Statement 1) follows from the fact that each minor $M_{ij}(X)$,
$i^{\prime}\leqslant j\leqslant k$, from formula (4) is obtained by a
triangular transformation from $x_{ij}$.

Let $\beta$ obey the conditions of item 2). Then either
${\mathcal{S}}_{\alpha}$ equals to ${\mathrm{Mat}}(n)$ (if
$\alpha\neq\psi({t(\alpha)})$, or ${\mathcal{S}}^{-}(n)$ (if
$\alpha=\psi({t(\alpha)})$). The system of standard coordinate functions
$\\{x_{ab}:a^{\prime}<b\\}$ is algebraically independent on
${\mathcal{S}}_{\alpha}$. .

From $\beta=\psi({s(\alpha)})$, it implies that either
${\mathcal{S}}_{\beta}={\mathcal{S}}^{-}(n)$, or
${\mathcal{S}}_{\beta}=\Lambda(n)$. Let $Y=S^{-}$ be the matrix with entries
which are the standard coordinate functions on ${\mathcal{S}}_{\beta}$. Then
$N_{jk}(S^{-})=0$ for each $i^{\prime}\leqslant j<k$. Therefore,

$P_{ik}(X,S^{-})=M_{ik}(X)N_{k,k}(S^{-})=M_{ik}(X)D_{k}(S^{-}).$

The polynomial $P_{ik}(X,S^{-})$ is obtained by a triangular transformation
over $\pi({\mathcal{F}}_{\beta})$ from $x_{ik}$ , and the system of
polynomials ${\mathbb{P}}(X,S^{-})$ is obtained by a triangular transformation
over the field $\pi({\mathcal{F}}_{\beta})$ from $\\{x_{ab}\\}$ lying below
the anti-diagonal of $X$. This proves 2). $\Box$

Example 1

$X=\left(\begin{array}[]{ccc}x_{11}&x_{12}&x_{13}\\\ x_{21}&x_{22}&x_{23}\\\
x_{31}&x_{32}&x_{33}\end{array}\right),\qquad
Y=\left(\begin{array}[]{ccc}y_{11}&y_{12}&y_{13}\\\ y_{21}&y_{22}&y_{23}\\\
y_{31}&y_{32}&y_{33}\end{array}\right),\qquad
S^{-}=\left(\begin{array}[]{ccc}0&0&s_{13}\\\ 0&s_{22}&s_{23}\\\
s_{31}&s_{32}&s_{33}\end{array}\right),$ $D_{3}(X)=x_{31},\quad
D_{2}(X)=\left|\begin{array}[]{cc}x_{21}&x_{22}\\\
x_{31}&x_{32}\end{array}\right|,\quad D_{1}(X)=\det(X),$
$D_{3}(Y)=y_{31},\quad D_{2}(Y)=\left|\begin{array}[]{cc}y_{21}&y_{22}\\\
y_{31}&y_{32}\end{array}\right|,\quad D_{1}(Y)=\det(Y),$
$P_{32}(X,Y)=x_{31}\left|\begin{array}[]{cc}y_{11}&y_{12}\\\
y_{31}&y_{32}\end{array}\right|+x_{32}\left|\begin{array}[]{cc}y_{21}&y_{22}\\\
y_{31}&y_{32}\end{array}\right|,\qquad
P_{32}(X,S^{-})=x_{32}\left|\begin{array}[]{cc}0&s_{22}\\\
s_{31}&s_{32}\end{array}\right|=D_{2}(S^{-})x_{32},$

$P_{33}(X,Y)=x_{31}y_{11}+x_{32}y_{21}+x_{33}y_{31},\qquad
P_{33}(X,S^{-})=x_{33}s_{31}=D_{3}(S^{-})x_{33},$

$P_{23}(X,Y)=\left|\begin{array}[]{cc}x_{21}&x_{22}\\\
x_{31}&x_{32}\end{array}\right|y_{21}+\left|\begin{array}[]{cc}x_{21}&x_{23}\\\
x_{31}&x_{33}\end{array}\right|y_{31},\qquad
P_{23}(X,S^{-})=\left|\begin{array}[]{cc}x_{21}&x_{23}\\\
x_{31}&x_{33}\end{array}\right|s_{31}=D_{3}(S^{-})\left|\begin{array}[]{cc}x_{21}&x_{23}\\\
x_{31}&x_{33}\end{array}\right|.$

Case 2. Let
$\textstyle{t(\alpha)}$$\textstyle{{s(\alpha)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha}$$\scriptstyle{\beta}$$\textstyle{v}$,
i.e. $\beta$ starts at ${s(\alpha)}$. Here we assume that $\alpha\neq\beta$,
the arrow $\alpha$ may be a loop, and $\beta$ is not a loop (the case $\beta$
is a loop, is considered in Case 1).

The action of $g\in{\mathrm{GL}}_{Q}$ on the pair of matrices $(X,Y)$, where
$X=X_{\alpha}$ and $Y=X_{\beta}$, is defined by the formula
$g.(X,Y)=(g_{t(\alpha)}Xg_{s(\alpha)}^{-1},g_{v}Yg_{s(\alpha)}^{-1})$.

Consider the $(2n\times n)$-matrix $\left(\frac{X}{Y}\right)$. Let the pair
$(i,k)$ lie below the anti-diagonal (i.e. $i^{\prime}<k$). Decompose $k$ into
a sum of positive integers $k=i^{\prime}+(k-i^{\prime})$. Consider the
$(k\times k)$-matrix

$\left(\frac{X_{ik}}{Y_{ik}}\right),$

where $X_{ik}$ is the left lower block of the matrix $X$ of size
$i^{\prime}\times k$, and $Y_{ik}$ is the left lower block of the matrix $Y$
of size $(k-i^{\prime})\times k$. Let us show that the determinant of this
matrix

$R_{ik}^{-}(X,Y)=\det\left(\frac{X_{ik}}{Y_{ik}}\right)$

is a $U$-invariant. For an arbitrary matrix $A$, let $A_{t}$ stand for its
right lower $(t\times t)$-block, and $A^{(t)}$ is its left upper $(t\times
t)$-block.

In these notations

$g.\left(\frac{X_{ik}}{Y_{ik}}\right)=\left(\begin{array}[]{cc}\left(g_{t(\alpha)}\right)_{i^{\prime}}&0\\\
0&\left(g_{v}\right)_{k-i^{\prime}}\end{array}\right)\left(\frac{X_{ik}}{Y_{ik}}\right)\left(g^{-1}_{s(\alpha)}\right)^{(k)}.$

Since the matrices $g_{t(\alpha)}$, $g_{s(\alpha)}$, $g_{v}$ belong to
${\mathrm{UT}}(n)$, we have

$R_{ik}^{-}(g_{t(\alpha)}Xg_{s(\alpha)}^{-1},g_{v}Yg_{s(\alpha)}^{-1})=R_{ik}^{-}(X,Y).$

The polynomials $\\{R_{ik}^{-}(X,Y)\\}$ are $U$-invariants.

Introduce the system of $U$-invariants

${\mathbb{R}}^{-}(X,Y)=\\{R_{ik}^{-}(X,Y):~{}i^{\prime}<k\\}.$

Lemma  3.2. 1) The polynomial $R^{-}_{ik}(X,Y)$, where $(i,k)$ lies below the
anti-diagonal, is obtained by a triangular transformation over the field
$K(Y)$ from $x_{ik}$.
2) Let $\beta$ obey the condition of Case 2, and $\beta=\psi({s(\alpha)})$.
Then the system of standard coordinate functions $\\{x_{ab}:a^{\prime}<b\\}$
(they lie below the anti-diagonal of $X_{\alpha}$) is algebraically
independent on ${\mathcal{S}}_{\alpha}$. The system
$\pi({\mathbb{R}}^{-}(X_{\alpha},X_{\beta}))$ is obtained by a triangular
transformation over $\pi({\mathcal{F}})$ from $\\{x_{ab}:a^{\prime}<b\\}$.

_Proof_. The statement 1) follows from the fact that the element $x_{ik}$ lies
in the right upper corner of the determinant $R^{-}_{ik}(X,Y)$.

Let $\beta$ obey the conditions of item 2). Since $\beta\neq\alpha$ and
$\beta=\psi({s(\alpha)})$, we have $\alpha\neq\psi({s(\alpha)})$. Therefore,
${\mathcal{S}}_{\alpha}$ is either ${\mathrm{Mat}}(n)$, or
${\mathcal{S}}^{-}(n)$. The system of standard coordinate functions
$\\{x_{ab}:a^{\prime}<b\\}$ is algebraically independent on
${\mathcal{S}}_{\alpha}$.

It follows from $\beta=\psi({s(\alpha)})$ that either
${\mathcal{S}}_{\beta}={\mathcal{S}}^{+}(n)$, or
${\mathcal{S}}_{\beta}=\Lambda(n)$. Let $Y=S^{+}$ be the matrix with entries,
which are the standard coordinate functions on ${\mathcal{S}}_{\beta}$. Then
$R^{-}_{ik}(X,S^{+})$ decomposes into a product

$R^{-}_{ik}(X,S^{+})=\large{\left|\begin{array}[]{cccccc}x_{i1}&\ldots&x_{i,k-i^{\prime}}&x_{i,k-i^{\prime}+1}&\ldots&x_{ik}\\\
\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\\
x_{n1}&\ldots&x_{n,k-i^{\prime}}&x_{n,k-i^{\prime}+1}&\ldots&x_{nk}\\\
y_{i^{\prime}+k^{\prime},1}&\ldots&y_{i^{\prime}+k^{\prime},k-i^{\prime}}&0&\ldots&0\\\
\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\\ y_{n1}&\ldots&0&0&\ldots&0\\\
\end{array}\right|}=\pm M_{I}^{J}(X)D_{k^{\prime}+i^{\prime}}(S^{+}),$

where $M_{I}^{J}(X)$ is the minor of the matrix $X$ with the system of rows
$I=[i,n]$ and columns $J=[k-i^{\prime}+1,k]$. The element $x_{ik}$ lies in the
right upper corner of the minor. Therefore, $R^{-}_{ik}(X,S^{+})$ is obtained
by a triangular transformation over the field $\pi({\mathcal{F}}_{\beta})$
from $x_{ik}$ , and the system of polynomials ${\mathbb{R}}^{-}(X,S^{+})$ is
obtained by a triangular transformation over the field
$\pi({\mathcal{F}}_{\beta})$ from the system of coordinate functions
$\\{x_{ab}\\}$ lying below the anti-diagonal of $X$. This proves 2). $\Box$
Example 2. For $n=3$ we have

$\left(\frac{X}{Y}\right)=\left(\begin{array}[]{ccc}x_{11}&x_{12}&x_{13}\\\
x_{21}&x_{22}&x_{23}\\\ x_{31}&x_{32}&x_{33}\\\ \hline\cr
y_{11}&y_{12}&y_{13}\\\ y_{21}&y_{22}&y_{23}\\\ y_{31}&y_{32}&y_{33}\\\
\end{array}\right),\qquad\left(\frac{X}{S^{+}}\right)=\left(\begin{array}[]{ccc}x_{11}&x_{12}&x_{13}\\\
x_{21}&x_{22}&x_{23}\\\ x_{31}&x_{32}&x_{33}\\\ \hline\cr
s_{11}&s_{12}&s_{13}\\\ s_{21}&s_{22}&0\\\ s_{31}&0&0\end{array}\right),$

$R^{-}_{32}(X,Y)=\left|\begin{array}[]{cc}x_{11}&x_{32}\\\ y_{31}&y_{32}\\\
\end{array}\right|,\qquad
R^{-}_{32}(X,S^{+})=\left|\begin{array}[]{cc}x_{31}&x_{32}\\\ s_{31}&0\\\
\end{array}\right|=-D_{3}(S^{+})x_{32},$
$R^{-}_{33}(X,Y)=\left|\begin{array}[]{ccc}x_{31}&x_{32}&x_{33}\\\
y_{21}&y_{22}&y_{23}\\\ y_{31}&y_{32}&y_{33}\end{array}\right|,\qquad
R^{-}_{33}(X,S^{+})=\left|\begin{array}[]{ccc}x_{31}&x_{32}&x_{33}\\\
s_{21}&s_{22}&0\\\ s_{31}&0&0\end{array}\right|=D_{2}(S^{+})x_{33}$

,

$R^{-}_{23}(X,Y)=\left|\begin{array}[]{ccc}x_{21}&x_{22}&x_{23}\\\
x_{31}&x_{32}&x_{33}\\\ y_{31}&y_{32}&y_{33}\end{array}\right|,\qquad
R^{-}_{23}(X,S^{+})=\left|\begin{array}[]{ccc}x_{21}&x_{22}&x_{23}\\\
x_{31}&x_{32}&x_{33}\\\
s_{31}&0&0\end{array}\right|=D_{3}(S^{+})\left|\begin{array}[]{cc}x_{22}&x_{23}\\\
x_{32}&x_{33}\\\ \end{array}\right|.$

We define

${\mathbb{P}}_{{s(\alpha)},\beta}(X_{\alpha},X_{\beta})=\left\\{\begin{array}[]{l}{\mathbb{P}}(X_{\alpha},X_{\beta}),\mbox{if}~{}\beta~{}\mbox{obeys~{}
the~{} conditions~{} of~{} Case~{}1},\\\
{\mathbb{R}}^{-}(X_{\alpha},X_{\beta}),\mbox{if}~{}\beta~{}\mbox{obeys~{}
the~{} conditions~{} of~{} Case~{}2}.\end{array}\right.$

For $h=(X_{\alpha})\in{\mathcal{H}}$, we take

${\mathbb{P}}_{{s(\alpha)}}(h)=\left\\{\begin{array}[]{l}\varnothing,\quad\mbox{if}\quad\alpha=\psi({s(\alpha)}),\\\
{\mathbb{P}}_{{s(\alpha)},\beta}(X_{\alpha},X_{\beta}),~{}\mbox{where}~{}\beta=\psi({s(\alpha)}),\quad\mbox{if}\quad\alpha\neq\psi({s(\alpha)}).\end{array}\right.$
(5)

Let us formulate a statement that follows from Lemmas 3 and 3
Proposition  3.3. Let $\alpha$ obeys the conditions of Case 1 or 2, and
$\beta=\psi({s(\alpha)})$. Then the system of standard coordinate functions
$\\{x_{ab}:a^{\prime}<b\\}$ (they lie below the anti-diagonal of $X_{\alpha}$)
is algebraically independent on ${\mathcal{S}}_{\alpha}$. The system of
polynomials $\pi({\mathbb{P}}_{s(\alpha)}(X_{\alpha},X_{\beta}))$ is obtained
by a triangular transformation over $\pi({\mathcal{F}})$ from
$\\{x_{ab}:a^{\prime}<b\\}$.

Let us turn to a construction of the system of polynomials
${\mathbb{P}}_{t(\alpha)}$ associated with the target ${t(\alpha)}$ of the
arrow $\alpha$. Observe that if $\alpha$ is a loop (i.e.
${s(\alpha)}={t(\alpha)}=q$), then the pair $\alpha$ and $\beta=\psi(q)$
satisfies conditions of Case 1 or Case 2. In Cases 3 and 4, $\alpha$ is not a
loop.

Case 3. Let
$\textstyle{v}$$\textstyle{{t(\alpha)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\gamma}$$\textstyle{{s(\alpha)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha}$,
i.e. $\gamma$ starts at ${t(\alpha)}$. Here we assume that $\alpha$ and
$\gamma$ are not loops (i.e. ${t(\alpha)}\neq{s(\alpha)}$ and
$v\neq{t(\alpha)}$).

The action of $g\in{\mathrm{GL}}_{Q}$ on the pair of matrices $(Z,X)$, where
$Z=X_{\gamma}$ and $X=X_{\alpha}$, is defined by the formula
$g.(Z,X)=(g_{v}Zg_{t(\alpha)}^{-1},g_{t(\alpha)}Xg_{s(\alpha)}^{-1})$. The
polynomials

${\mathbb{P}}(Z,X)=\\{P_{ik}(Z,X):~{}i^{\prime}<k\\}$

are $U$-invariants (see Case 1 and [9]).

Lemma  3.4. 1) The polynomial $P_{ik}(Z,X)$, where $(i,k)$ lies below the
anti-diagonal, is obtained by a triangular transformation from
$x_{i^{\prime},k^{\prime}}$ over $K(Z)$ (observe that if $(i,k)$ lies below
the anti-diagonal, then $(i^{\prime},k^{\prime})$ lies above the anti-
diagonal).
2) Let $\gamma$ obey the conditions of Case 3, and $\gamma=\psi({t(\alpha)})$.
Then the system of standard coordinate functions $\\{x_{ab}:a^{\prime}>b\\}$
(they lie above the anti-diagonal of $X_{\alpha}$) is algebraically
independent on ${\mathcal{S}}_{\alpha}$. The system
$\pi({\mathbb{P}}(X_{\gamma},X_{\alpha}))$ is obtained by a triangular
transformation over $\pi({\mathcal{F}}_{\gamma})$ from
$\\{x_{ab}:a^{\prime}>b\\}$.

_Proof_. Statement 1) follows from the fact that each minor $N_{jk}(X)$,
$i^{\prime}\leqslant j\leqslant k$ from formula (4), where $i^{\prime}<k$, is
obtained by a triangular transformation from $x_{j,k^{\prime}}$, $j\leqslant
k$.

Let $\gamma$ obey the conditions of item 2). The condition
$\gamma=\psi({t(\alpha)})$ implies $\alpha\neq\psi({t(\alpha)})$. Therefore,
${\mathcal{S}}_{\alpha}$ equals to ${\mathrm{Mat}}(n)$ (if
$\alpha\neq\psi({s(\alpha)})$, or ${\mathcal{S}}^{+}(n)$ (if
$\alpha=\psi({s(\alpha)})$). The system of standard coordinate functions
$\\{x_{ab}:a^{\prime}>b\\}$ is algebraically independent on
${\mathcal{S}}_{\alpha}$.

It follows from $\gamma=\psi({t(\alpha)})$ that either
${\mathcal{S}}_{\gamma}={\mathcal{S}}^{+}(n)$ or
${\mathcal{S}}_{\gamma}=\Lambda(n)$. Let $Z=S^{+}$ be the matrix with entries,
which are the standard coordinate functions on ${\mathcal{S}}_{\gamma}$. Then
$M_{jk}(S^{+})=0$ for each $i^{\prime}<j\leqslant k$. Therefore,

$P_{ik}(S^{+},X)=N_{i,i^{\prime}}(S^{+})N_{i^{\prime},k}(X)=D_{i}(S^{+})N_{i^{\prime},k}(X).$

The polynomial $P_{ik}(S^{+},X)$ is obtained by a triangular transformation
over $\pi({\mathcal{F}}_{\gamma})$ from $x_{i^{\prime},k^{\prime}}$, and the
system of polynomials ${\mathbb{P}}(S^{+},X)$ is obtained by a triangular
transformation over $\pi({\mathcal{F}}_{\gamma})$ from $\\{x_{ab}\\}$ lying
above the anti-diagonal of $X$. This proves 2). $\Box$
Example 3. For $n=3$ we have

$Z=\left(\begin{array}[]{ccc}z_{11}&z_{12}&z_{13}\\\ z_{21}&z_{22}&z_{23}\\\
z_{31}&z_{32}&z_{33}\end{array}\right)\qquad
X=\left(\begin{array}[]{ccc}x_{11}&x_{12}&x_{13}\\\ x_{21}&x_{22}&x_{23}\\\
x_{31}&x_{32}&x_{33}\end{array}\right),\qquad,\qquad
S^{+}=\left(\begin{array}[]{ccc}s_{11}&s_{12}&s_{13}\\\ s_{21}&s_{22}&0\\\
s_{31}&0&0\end{array}\right),$
$P_{32}(Z,X)=z_{31}\left|\begin{array}[]{cc}x_{11}&x_{12}\\\
x_{31}&x_{32}\end{array}\right|+z_{32}\left|\begin{array}[]{cc}x_{21}&x_{22}\\\
x_{31}&x_{32}\end{array}\right|,\qquad
P_{32}(S^{+},X)=s_{31}\left|\begin{array}[]{cc}x_{11}&x_{12}\\\
x_{31}&x_{32}\end{array}\right|=D_{3}(S^{+})\left|\begin{array}[]{cc}x_{11}&x_{12}\\\
x_{31}&x_{32}\end{array}\right|,$

$P_{33}(Z,X)=z_{31}x_{11}+z_{32}x_{21}+z_{33}x_{31},\qquad
P_{33}(S^{+},X)=s_{31}x_{11}=D_{3}(S^{+})x_{11},$

$P_{23}(Z,X)=\left|\begin{array}[]{cc}z_{21}&z_{22}\\\
z_{31}&z_{32}\end{array}\right|x_{21}+\left|\begin{array}[]{cc}z_{21}&z_{23}\\\
z_{31}&z_{33}\end{array}\right|x_{31},\qquad
P_{23}(S^{+},X)=\left|\begin{array}[]{cc}s_{21}&s_{22}\\\
s_{31}&0\end{array}\right|x_{21}=D_{2}(S^{+})x_{21}.$

Case 4. Let
$\textstyle{v\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\gamma}$$\textstyle{t(\alpha)}$$\textstyle{{s(\alpha)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha}$,
i.e. $\gamma$ terminates at ${t(\alpha)}$. Here we assume that
$\alpha\neq\gamma$, $\alpha$ is not a loop, and $\gamma$ maybe a loop.

The action of $g\in{\mathrm{GL}}_{Q}$ on the pair of matrices $(Z,X)$, where
$Z=X_{\gamma}$ and $X=X_{\alpha}$, is defined by the formula
$g.(Z,X)=(g_{t(\alpha)}Zg_{v}^{-1},g_{t(\alpha)}Xg_{s(\alpha)}^{-1})$.

Consider the $(n\times 2n)$-matrix $(Z|X)$. Let the pair $(i,k)$ lie above the
anti-diagonal (i.e. $i^{\prime}>k$). Decompose $i^{\prime}$ into a sum of
positive integers $i^{\prime}=k+(i^{\prime}-k)$. Consider the
$(i^{\prime}\times i^{\prime})$-matrix $(Z_{ik}|X_{ik})$, where $Z_{ik}$ is
the left lower block of the matrix $Z$ of size
$i^{\prime}\times(i^{\prime}-k)$, and $X_{ik}$ is the left lower block of the
matrix $X$ of size $i^{\prime}\times k$.

Let us show that the determinant of this matrix

$R_{ik}^{+}(Z,X)=\det(Z_{ik}|X_{ik})$

is $U$-invariant. In notations we used in Case 2, we have

$g.(Z_{ik}|X_{ik})=\left(g_{t(\alpha)}\right)_{i^{\prime}}(Z_{ik}|X_{ik})\left(\begin{array}[]{cc}\left(g_{v}^{-1}\right)^{(i^{\prime}-k)}&0\\\
0&\left(g^{-1}_{s(\alpha)}\right)^{(k)}\end{array}\right).$

Since the matrices $g_{t(\alpha)}$, $g_{s(\alpha)}$, $g_{v}$ belong to
${\mathrm{UT}}(n)$, we have

$R_{ik}^{+}(g_{t(\alpha)}Zg_{v}^{-1},g_{t(\alpha)}Xg_{s(\alpha)}^{-1})=R_{ik}^{+}(Z,X).$

The polynomials $\\{R_{ik}^{+}(X,Y)\\}$ are $U$-invariants.

We introduce the system of $U$-invariants

${\mathbb{R}}^{+}(Z,X)=\\{R_{ik}^{+}(Z,X):~{}i^{\prime}>k\\}.$

Lemma  3.5. 1) The polynomial $R^{+}_{ik}(Z,X)$, where $(i,k)$ lies above the
anti-diagonal, is obtained by a triangular transformation from $x_{ik}$ over
the field $K(Z)$.
2) Let $\gamma$ obey the conditions of Case 4, and $\gamma=\psi({t(\alpha)})$.
Then the system of standard coordinate functions $\\{x_{ab}:a^{\prime}>b\\}$
(they lie above the anti-diagonal of $X_{\alpha}$) is algebraically
independent on ${\mathcal{S}}_{\alpha}$. The system
$\pi({\mathbb{R}}^{+}(X_{\gamma},X_{\alpha}))$ is obtained by a triangular
transformation over $\pi({\mathcal{F}})$ from $\\{x_{ab}:a^{\prime}>b\\}$.

_Proof_. The statement 1) follows from the fact that the element $x_{ik}$ lies
in the right upper corner of the matrix $R^{+}_{ik}(Z,X)$.

Let $\gamma$ obey the conditions of item 2). It follow from $\gamma\neq\alpha$
and $\gamma=\psi({t(\alpha)})$ that $\alpha\neq\psi({t(\alpha)})$. Therefore
${\mathcal{S}}_{\alpha}$ equals to either ${\mathrm{Mat}}(n)$ (if
$\alpha\neq\psi({s(\alpha)})$, or ${\mathcal{S}}^{+}(n)$ (if
$\alpha=\psi({s(\alpha)})$). The system of matrix standard coordinate
functions $\\{x_{ab}\\}$ lying above the anti-diagonal is algebraically on
${\mathcal{S}}_{\alpha}$.

From $\gamma=\psi({t(\alpha)})$, it implies that either
${\mathcal{S}}_{\gamma}={\mathcal{S}}^{-}(n)$, or
${\mathcal{S}}_{\gamma}=\Lambda(n)$. Let $Z=S^{-}$ be the matrix with entries,
which are the standard coordinate functions on ${\mathcal{S}}_{\gamma}$. Then
$R^{-}_{ik}(S^{-},X)$ decomposes into a product

$R^{+}_{ik}(S^{-},X)=\large{\left|\begin{array}[]{cccccc}0&\ldots&0&x_{i1}&\ldots&x_{ik}\\\
\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\\
0&\ldots&0&x_{i+k-1,1}&\ldots&x_{i+k-1,k}\\\
0&\ldots&z_{i+k,i^{\prime}-k}&x_{i+k,1}&\ldots&x_{i+k,k}\\\
\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\\
z_{n1}&\ldots&z_{n,i^{\prime}-k}&x_{n1}&\ldots&x_{nk}\\\
\end{array}\right|}=\pm D_{i+k}(S^{-})N_{I}^{J}(X),$

where $N_{I}^{J}(X)$is the minor of the matrix $X$ with the system of rows
$I=[i,i+k-1]$ and columns $J=[1,k]$. The element $x_{ik}$ lies in the right
upper cornor of this minor. Therefore, $R^{+}_{ik}(S^{-},X)$ is obtained by a
triangular transformation over $\pi({\mathcal{F}}_{\gamma})$ from $x_{ik}$,
and the system of polynomials ${\mathbb{R}}^{+}(S^{-},X)$ is obtained by a
triangular transformation over $\pi({\mathcal{F}}_{\gamma})$ from the system
of the standard coordinate functions $\\{x_{ab}\\}$ lying above the anti-
diagonal of $X$. This proves 2). $\Box$

Example 4. For $n=3$ we have

$(Z|X)=\left(\begin{array}[]{ccc|ccc}z_{11}&z_{12}&z_{13}&x_{11}&x_{12}&x_{13}\\\
z_{21}&z_{22}&z_{23}&x_{21}&x_{22}&x_{23}\\\
z_{31}&z_{32}&z_{33}&x_{31}&x_{32}&x_{33}\end{array}\right),\qquad(S^{-}|X)=\left(\begin{array}[]{ccc|ccc}0&0&s_{13}&x_{11}&x_{12}&x_{13}\\\
0&s_{22}&s_{23}&x_{21}&x_{22}&x_{23}\\\
s_{31}&s_{32}&s_{33}&x_{31}&x_{32}&x_{33}\end{array}\right),$

$R^{+}_{11}(Z,X)=\left|\begin{array}[]{ccc}z_{11}&z_{12}&x_{11}\\\
z_{21}&z_{22}&x_{21}\\\ z_{31}&z_{32}&x_{31}\end{array}\right|,\qquad
R^{+}_{11}(S^{-},X)=\left|\begin{array}[]{ccc}0&0&x_{11}\\\ 0&s_{22}&x_{21}\\\
s_{31}&s_{32}&x_{31}\end{array}\right|=D_{2}(S^{-})x_{11},$
$R^{+}_{12}(Z,X)=\left|\begin{array}[]{ccc}z_{11}&x_{11}&x_{12}\\\
z_{21}&x_{21}&x_{22}\\\ z_{31}&x_{31}&x_{32}\end{array}\right|,\qquad
R^{+}_{12}(S^{-},X)=\left|\begin{array}[]{ccc}0&x_{11}&x_{12}\\\
0&x_{21}&x_{22}\\\
s_{31}&x_{31}&x_{32}\end{array}\right|=D_{3}(S^{-})\left|\begin{array}[]{cc}x_{11}&x_{12}\\\
x_{21}&x_{22}\\\ \end{array}\right|,$
$R^{+}_{21}(Z,X)=\left|\begin{array}[]{cc}z_{21}&x_{21}\\\
z_{31}&x_{31}\end{array}\right|,\qquad
R^{+}_{21}(S^{-},X)=\left|\begin{array}[]{cc}0&x_{21}\\\
s_{31}&x_{31}\end{array}\right|=-D_{3}(S^{-})x_{21}.$

We define

${\mathbb{P}}_{\gamma,{t(\alpha)}}(X_{\gamma},X_{\alpha})=\left\\{\begin{array}[]{l}{\mathbb{P}}(X_{\gamma},X_{\alpha}),\mbox{if}~{}\gamma~{}\mbox{obeys~{}
the~{} conditions~{} of~{} Case~{}~{}3},\\\
{\mathbb{R}}^{+}(X_{\gamma},X_{\alpha}),\mbox{if}~{}\gamma~{}\mbox{obeys~{}
the~{} conditions~{} of~{} Case~{}~{}4}.\end{array}\right.$

For $h=(X_{\alpha})\in{\mathcal{H}}$, we take

${\mathbb{P}}_{{t(\alpha)}}(h)=\left\\{\begin{array}[]{l}\varnothing,\quad\mbox{if}\quad\alpha=\psi({t(\alpha)}),\\\
{\mathbb{P}}_{\gamma,{t(\alpha)}}(X_{\gamma},X_{\alpha}),~{}\mbox{where}~{}\gamma=\psi({t(\alpha)}),\quad\mbox{if}\quad\alpha\neq\psi({t(\alpha)}).\end{array}\right.$
(6)

Let us formulate a statement that follows from Lemmas 3 and 3
Proposition  3.6. Let $\alpha$ obeys the conditions of Case 3 or 4, and
$\gamma=\psi({t(\alpha)})$. Then the system of standard coordinate functions
$\\{x_{ab}:a^{\prime}<b\\}$ (they lie above the anti-diagonal of $X_{\alpha}$)
is algebraically independent on ${\mathcal{S}}_{\alpha}$. The system of
polynomials $\pi({\mathbb{P}}_{t(\alpha)}(X_{\gamma},X_{\alpha}))$ is obtained
by a triangular transformation over $\pi({\mathcal{F}})$ from
$\\{x_{ab}:a^{\prime}>b\\}$.

Introduce the notation

${\mathbb{P}}_{\alpha}={\mathbb{P}}_{s(\alpha)}\sqcup{\mathbb{P}}_{t(\alpha)}.$
(7)

Theorem  3.7. The field of $U$-invariants $K({\mathcal{H}})^{U}$ is freely
generated over $K$ by the system of polynomials

$\bigcup_{\alpha\in
A}\left({\mathbb{P}}_{\alpha}\sqcup{\mathbb{D}}_{\alpha}\right).$ (8)

_Proof_. For each $\alpha\in A$, the system $\pi({\mathbb{D}}_{\alpha})$ is
obtained by a triangular transformation over the field $K$ from the matrix
entries $\\{x_{a,a^{\prime}}\\}$ that belong to the anti-diagonal of
$X_{\alpha}$, and $\pi({\mathbb{P}}_{\alpha})$ is obtained by a triangular
transformation over the field $\pi({\mathcal{F}})$ from the other matrix
entries of $X_{\alpha}$ (see Propositions 3 and 3). Therefore, the system of
polynomials $\pi({\mathbb{P}}_{\alpha}\sqcup{\mathbb{D}}_{\alpha})$ freely
generates the field $K({\mathcal{S}}_{\alpha})$, and

$\bigcup_{\alpha\in A}\pi({\mathbb{P}}_{\alpha}\sqcup{\mathbb{D}}_{\alpha})$

freely generates the field $K({\mathcal{S}})$. It implies that the embedding
$\pi:K({\mathcal{H}})^{U}\hookrightarrow K({\mathcal{S}})$ (see Corollary 2)
is an isomorphism, and the system (8) freely generates $K({\mathcal{H}})^{U}$.
$\Box$
Example 3.8. Let us construct a system of free generators for the quiver
$Q=(V,A)$, defined by the diagram

---
$\textstyle{3\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha_{3}}$$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha_{1}}$$\textstyle{2\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha_{2}}$$\scriptstyle{\alpha_{4}}$$\textstyle{4}$

In this case, $V=\\{1,2,3,4\\}$ and
$A=\\{\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}\\}$. The linear space
${\mathcal{H}}$ is a direct sum of four subspaces
${\mathcal{H}}_{i}={\mathrm{Mat}}(n)$. The group
$GL_{Q}=\\{(g_{1},g_{2},g_{3},g_{4})\\}$ is a direct product of four copies of
${\mathrm{GL}}(n)$. The action of $g\in{\mathrm{GL}}_{G}$ on
$h=(X_{1},X_{2},X_{3},X_{3})$ is defined as follows

$g.h=(g_{1}X_{1}g_{1}^{-1},g_{1}X_{2}g_{2}^{-1},g_{2}X_{3}g_{3}^{-1},g_{4}X_{4}g_{2}^{-1}).$

Let us fix the map $\psi:V\to A$ taking

$\psi(1)=\alpha_{1},~{}~{}\psi(2)=\psi(4)=\alpha_{4},~{}~{}\psi(3)=\alpha_{3}.$

The choice of $\psi$ defines the section
${\mathcal{S}}=\oplus_{i=1}^{4}{\mathcal{S}}_{i}$ and the systems of
polynomials ${\mathbb{P}}=\sqcup_{i=1}^{4}{\mathbb{P}}_{i}$ by the formulas

$\left\\{\begin{array}[]{l}{\mathcal{S}}_{1}={\mathcal{S}}^{-}(n),\\\
{\mathcal{S}}_{2}={\mathrm{Mat}}(n),\\\
{\mathcal{S}}_{3}={\mathcal{S}}^{+}(n),\\\ {\mathcal{S}}_{4}=\Lambda(n)\\\
\end{array}\right.\quad\mbox{and}\quad\left\\{\begin{array}[]{l}{\mathbb{P}}_{1}={\mathbb{P}}(X_{1},X_{1}),\\\
{\mathbb{P}}_{2}={\mathbb{R}}^{+}(X_{1},X_{2})\sqcup{\mathbb{R}}^{-}(X_{2},X_{4}),\\\
{\mathbb{P}}_{3}={\mathbb{P}}(X_{4},X_{3}),\\\
{\mathbb{P}}_{4}=\varnothing.\\\ \end{array}\right.$

Denote by ${\mathbb{D}}_{i}={\mathbb{D}}_{\alpha_{i}}$ the system of left
lower corner minors of $X_{i}$.

The system of polynomials

$\bigsqcup_{i=1}^{4}~{}\\{{\mathbb{P}}_{i}\sqcup{\mathbb{D}}_{i}\\}$

freely generates the field $K({\mathcal{H}})^{U}$.

## References

  * [1] V. L. Popov, E. B. Vinberg,  Linear Algebraic Groups and Invariant Theory, Encyclopaedia of Mathematical Sciences, Springer-Verlag, Berlin Heidelberg New-York, 1995.
  * [2] C. Procesi, The Invariant theory of $n\times n$ Matrices, Adv. Math. 19 (1976) 306-381.
  * [3] L. Le Bruyn, C. Procesi, Semisimple representations of quivers, Trans. Amer. Math. Society, vol. 317, 1990, 585-598.
  * [4] S. Donkin, Invariants of several matrices, Invent. Math. 110(1992), 389-401
  * [5] K. Miyata, Invariants of certain groups. 1., Nagoya Math. J., vol.41, 1971, 69-73.
  * [6] S. Donkin, Polynomial invariants of representations of quivers, Comment. Math. Helvetici, vol. 69, 1994, 137-141.
  * [7] K.A. Vyatkina, A.N. Panov, Field of $U$-invariants of adjoint representation of the group ${\mathrm{GL}}(n,K)$, Mat. Notes. 93(1) (2013) 187-190.
  * [8] A.N. Panov, Fields of invariants for unipotent radicals of parabolic subgroups, Linear and Multilinear algebra, vol.71, no.15, 2023, 2499-2512.
  * [9] A.N. Panov, Fields of $U$-invariants of matrix tuples, Electronic Journal of Linear Algebra, 2023, 39, 117-123.


# Affine subspaces of antisymmetric matrices with constant rank

Elena Rubei

###### Abstract

For every $n\in\mathbb{N}$ and every field $K$, let $A(n,K)$ be the vector
space of the antisymmetric $(n\times n)$-matrices over $K$. We say that an
affine subspace $S$ of $A(n,K)$ has constant rank $r$ if every matrix of $S$
has rank $r$. Define

${\mathcal{A}}_{antisym}^{K}(n;r)=\\{S\;|\;S\;\mbox{\rm affine subsapce of
$A(n,K)$ of constant rank }r\\}$ $a_{antisym}^{K}(n;r)=\max\\{\dim S\mid
S\in{\mathcal{A}}_{antisym}^{K}(n;r)\\}.$

In this paper we prove the following formulas:

for $2r+2\leq n$

$a_{antisym}^{\mathbb{R}}(n;2r)=(n-r-1)r;$

for $n=2r$

$a_{antisym}^{\mathbb{R}}(n;2r)=r(r-1);$

for $n=2r+1$

$a_{antisym}^{\mathbb{R}}(n;2r)=r(r+1).$

Dipartimento di Matematica e Informatica “U. Dini”, viale Morgagni 67/A, 50134
Firenze, Italia

E-mail address<EMAIL_ADDRESS>
††footnotetext:  2010 Mathematical Subject Classification:  15A30 Key words:
affine subspaces, matrices, constant rank

##  1 Introduction

For every $m,n\in\mathbb{N}$ and every field $K$, let $M(m\times n,K)$ be the
vector space of the $(m\times n)$-matrices over $K$, let $S(n,K)$ be the
vector space of the symmetric $(n\times n)$-matrices over $K$ and let $A(n,K)$
be the vector space of the antisymmetric $(n\times n)$-matrices over $K$.
Moreover, denote the $\mathbb{R}$-vector space of the hermitian $(n\times
n)$-matrices by $H(n)$.

We say that an affine subspace $S$ of $M(m\times n,K)$ has constant rank $r$
if every matrix of $S$ has rank $r$ and we say that a linear subspace $S$ of
$M(m\times n,K)$ has constant rank $r$ if every nonzero matrix of $S$ has rank
$r$.

Define

${\mathcal{A}}^{K}(m\times n;r)=\\{S\;|\;S\;\mbox{\rm affine subsapce of
$M(m\times n,K)$ of constant rank }r\\}$
${\mathcal{A}}_{sym}^{K}(n;r)=\\{S\;|\;S\;\mbox{\rm affine subsapce of
$S(n,K)$ of constant rank }r\\}$
${\mathcal{A}}_{antisym}^{K}(n;r)=\\{S\;|\;S\;\mbox{\rm affine subsapce of
$A(n,K)$ of constant rank }r\\}$
${\mathcal{A}}_{herm}(n;r)=\\{S\;|\;S\;\mbox{\rm affine subsapce of $H(n)$ of
constant rank }r\\}$
${\mathcal{A}}_{sym}^{\mathbb{R}}(n;p,\nu)=\\{S\;|\;S\;\mbox{\rm affine
subsapce of $S(n,\mathbb{R})$ s.t. each $A\in S$ has signature }(p,\nu)\\}$
${\mathcal{A}}_{herm}(n;p,\nu)=\\{S\;|\;S\;\mbox{\rm affine subsapce of $H(n)$
s.t. each $A\in S$ has signature }(p,\nu)\\}$ ${\mathcal{L}}^{K}(m\times
n;r)=\\{S\;|\;S\;\mbox{\rm linear subsapce of $M(m\times n,K)$ of constant
rank }r\\}$ ${\mathcal{L}}_{sym}^{K}(n;r)=\\{S\;|\;S\;\mbox{\rm linear
subsapce of $S(n,K)$ of constant rank }r\\}$

Let

$a^{K}(m\times n;r)=\max\\{\dim S\mid S\in{\mathcal{A}}^{K}(m\times n;r)\\}$
$a_{sym}^{K}(n;r)=\max\\{\dim S\mid S\in{\mathcal{A}}_{sym}^{K}(n;r)\\}$
$a_{antisym}^{K}(n;r)=\max\\{\dim S\mid
S\in{\mathcal{A}}_{antisym}^{K}(n,r)\\}.$ $a_{herm}(n;r)=\max\\{\dim S\mid
S\in{\mathcal{A}}_{herm}(n;r)\\}$ $a_{sym}^{\mathbb{R}}(n;p,\nu)=\max\\{\dim
S\mid S\in{\mathcal{A}}_{sym}^{\mathbb{R}}(n;p,\nu)\\}$
$a_{herm}(n;p,\nu)=\max\\{\dim S\mid S\in{\mathcal{A}}_{herm}(n;p,\nu)\\}$
$l^{K}(m\times n;r)=\max\\{\dim S\mid S\in{\mathcal{L}}^{K}(m\times n;r)\\}$
$l_{sym}^{K}(n;r)=\max\\{\dim S\mid S\in{\mathcal{L}}_{sym}^{K}(n,r)\\}.$

There is a wide literature on linear subspaces of constant rank. In particular
we quote the following theorems:

######  Theorem 1.

(Westwick, [7]) For $2\leq r\leq m\leq n$, we have:

$n-r+1\leq l^{\mathbb{C}}(m\times n;r)\leq m+n-2r+1$

######  Theorem 2.

(Ilic-Landsberg, [5]) If $r$ is even and greater than or equal to $2$, then

$l_{sym}^{\mathbb{C}}(n;r)=n-r+1$

In case $r$ odd, the following result holds, see [5], [2], [3]:

######  Theorem 3.

If $r$ is odd, then

$l_{sym}^{\mathbb{C}}(n;r)=1$

We mention also that, in [1], Flanders proved that, if $r\leq m\leq n$, a
linear subspace of $M(m\times n,\mathbb{C})$ such that every of its elements
has rank less than or equal to $r$ has dimension less than or equal to $rn$.

In [6] we proved the following theorems:

######  Theorem 4.

Let $n,r\in\mathbb{N}$ with $r\leq n$. Then

$a_{sym}^{\mathbb{R}}(n;r)\leq\left\lfloor\frac{r}{2}\right\rfloor\left(n-\left\lfloor\frac{r}{2}\right\rfloor\right).$

######  Theorem 5.

Let $m,n,r\in\mathbb{N}$ with $r\leq m\leq n$. Then

$a^{\mathbb{R}}(m\times n;r)=rn-\frac{r(r+1)}{2}.$

We proved also a statement on the maximal dimension of affine subspaces with
constant signature in the space of symmetric real matrices and one on the
maximal dimension of affine subspaces of constant rank in the space of the
hermitian matrices.

In this paper we investigate on the maximal dimension of affine subspaces in
the space of the antisymmetric matrices of constant rank. Precisely we prove
the following theorem:

######  Theorem 6.

For $n\geq 2r+2$:

$a_{antisym}^{\mathbb{R}}(n;2r)=(n-r-1)r.$

For $n=2r$

$a_{antisym}^{\mathbb{R}}(n;2r)=r(r-1).$

For $n=2r+1$

$a_{antisym}^{\mathbb{R}}(n;2r)=r(r+1).$

##  2 Proof of the theorem

######  Notation 7.

Let $m,n\in\mathbb{N}-\\{0\\}$.

We denote the $n\times n$ identity matrix over $\mathbb{R}$ by $I_{n}$.

We denote $E_{i,j}^{n}$ the $n\times n$ matrix over $\mathbb{R}$ such that

$(E_{i,j}^{n})_{x,y}=\left\\{\begin{array}[]{ll}1&\mbox{\rm
if}\;(x,y)=(i,j)\\\ 0&\mbox{\rm otherwise}\end{array}\right.$

We omit the superscript when it is clear from the context.

We denote by $0_{m\times n}$ the $m\times n$ null real matrix. We omit the
subscript when it is clear from the context.

We denote $diag(d_{1},\ldots,d_{n})$ the diagonal matrix whose diagonal
entries are $d_{1},\ldots,d_{n}$.

For any $A\in M(m\times n,\mathbb{R})$ we denote the submatrix of $A$ given by
the rows $i_{1},,\ldots,i_{k}$ and the columns $j_{1},\ldots,j_{s}$ by
$A^{(j_{1},\ldots,j_{s})}_{(i_{1},\ldots,i_{k})}$.

We denote by $J$ the $(2\times 2)$-matrix

$\begin{pmatrix}0&1\\\ -1&0\end{pmatrix}$

and we denote by $\overline{J}_{2n}$ the $(2n\times 2n)$ block diagonal matrix
whose diagonal blocks are equal to $J$. We omit the subscript when it is clear
from the context.

We say that a $(2\times 2)$-matrix is pinco if and only if it is equal to

$\begin{pmatrix}a&b\\\ b&-a\end{pmatrix}$

for some $a,b\in\mathbb{R}$. For any pinco matrix $P=\begin{pmatrix}a&b\\\
b&-a\end{pmatrix}$, we denote by $\tilde{P}$ the matrix $JP$, that is the
matrix

$\begin{pmatrix}b&-a\\\ -a&-b\end{pmatrix}.$

We say that a $(2\times 2)$-matrix is antipinco if and only if it is equal to

$\begin{pmatrix}a&b\\\ -b&a\end{pmatrix}$

for some $a,b\in\mathbb{R}$.

Let $A_{i,j}$ for $i,j\in\\{1,\ldots,n\\}$ with $i<j$ be $2\times 2$ matrices.
We denote by $R(A_{i,j})$ the antisymmetric $(2n\times 2n)$-matrix

$\begin{pmatrix}0&A_{1,2}&A_{1,3}&.&.&.\\\
-{}^{t}\\!A_{1,2}&0&A_{2,3}&.&.&.\\\
-{}^{t}\\!A_{1,3}&-{}^{t}\\!A_{2,3}&0&.&.&.\\\ .&.&.&.&.&.\\\ .&.&.&.&.&.\\\
.&.&.&.&.&.\end{pmatrix}.$

We denote by $X(A_{i,j})$ the symmetric $(2n\times 2n)$-matrix

$\begin{pmatrix}0&A_{1,2}&A_{1,3}&.&.&.\\\ {}^{t}\\!A_{1,2}&0&A_{2,3}&.&.&.\\\
{}^{t}\\!A_{1,3}&{}^{t}\\!A_{2,3}&0&.&.&.\\\ .&.&.&.&.&.\\\ .&.&.&.&.&.\\\
.&.&.&.&.&.\end{pmatrix}.$

For any $l_{1},\ldots,l_{r}\in\mathbb{R}$, let $T(l_{1},\ldots,l_{r})$ be the
$2r\times 2r$ diagonal block matrix whose diagnal blocks are $l_{1}J$,….,
$l_{r}J$.

######  Lemma 8.

Let $m,r\in\mathbb{N}-\\{0\\}$. Let $V$ be vector subspace of
$\mathbb{R}^{3m}$.

Let $\pi_{1}:\mathbb{R}^{3m}\rightarrow\mathbb{R}^{2m}$ be the projection onto
the first $2m$ coordinates.

Let $\pi_{2}:\mathbb{R}^{3m}\rightarrow\mathbb{R}^{2m}$ be the projection onto
the last $2m$ coordinates.

Let $\pi_{3}:\mathbb{R}^{3m}\rightarrow\mathbb{R}^{2m}$ be the projection onto
the coordinates $1,\ldots,m,2m+1,\ldots 3m$.

If $\dim(\pi_{i}(V))\leq 2r$ for $i=1,2,3$, then $\dim(V)\leq 3r$.

###### Proof.

Suppose to the contrary that $\dim(V)\geq 3r+1$. Then we would have:

$\dim(Ker(\pi_{i}|_{V}))=\dim(V)-\dim(Im(\pi_{i}|_{V}))\geq 3r+1-2r=r+1$

for any $i=1,2,3$.

From $\dim(Ker(\pi_{1}|_{V}))\geq r+1$ we can deduce there would exist in $V$
at least $r+1$ linear independent vectors with the first $2m$ coordinates
equal to $0$.

From $\dim(Ker(\pi_{2}|_{V}))\geq r+1$ we can deduce there would exist in $V$
at least $r+1$ linear independent vectors with the last $2m$ coordinates equal
to $0$.

But then $\dim(\pi_{3}(V))$ would be greater than or equal to $2(r+1)$, which
is absurd.

∎

######  Lemma 9.

Let $n_{1},\ldots,n_{k},q_{1},\ldots,q_{k},m,r\in\mathbb{N}$. Let
$h=3m+n_{1}+\ldots+n_{k}$.

Let $\pi_{1}:\mathbb{R}^{h}\rightarrow\mathbb{R}^{2m}$ be the projection onto
the first $2m$ coordinates.

Let $\pi_{2}:\mathbb{R}^{h}\rightarrow\mathbb{R}^{2m}$ be the projection onto
the coordinates $m+1,\ldots,3m$.

Let $\pi_{3}:\mathbb{R}^{h}\rightarrow\mathbb{R}^{2m}$ be the projection onto
the coordinates $1,\ldots,m,2m+1,\ldots 3m$.

Let $\pi:\mathbb{R}^{h}\rightarrow\mathbb{R}^{3m}$ be the projection onto the
first $3m$ coordinates.

Finally, let $p_{1}:\mathbb{R}^{h}\rightarrow\mathbb{R}^{n_{1}}$ be the
projection onto the coordinates $3m+1,\ldots,3m+n_{1}$, let
$p_{2}:\mathbb{R}^{h}\rightarrow\mathbb{R}^{n_{2}}$ be the projection onto the
coordinates $3m+n_{1}+1,\ldots,3m+n_{1}+n_{2}$ and so on.

Let $V$ be a vector subspace of $\mathbb{R}^{h}$ such that
$\dim(\pi_{i}(V))\leq 2r$ for $i=1,2,3$ and $\dim(p_{j}(V))\leq q_{j}$ for
$j=1,\ldots,k$; then

$\dim(V)\leq\sum_{j=1,\ldots,k}q_{j}+3r.$

###### Proof.

Let $\tilde{\pi}_{1}:\mathbb{R}^{3m}\rightarrow\mathbb{R}^{2m}$ be the
projection onto the first $2m$ coordinates.

Let $\tilde{\pi}_{2}:\mathbb{R}^{3m}\rightarrow\mathbb{R}^{2m}$ be the
projection onto the coordinates $m+1,\ldots,3m$.

Let $\tilde{\pi}_{3}:\mathbb{R}^{3m}\rightarrow\mathbb{R}^{2m}$ be the
projection onto the coordinates $1,\ldots,m,2m+1,\ldots 3m$.

Obviously $\tilde{\pi}_{i}\circ\pi=\pi_{i}$; so, for $i=1,2,3$,

$\dim(\tilde{\pi}_{i}(\pi(V)))=\dim(\pi_{i}(V))\leq 2r,$

where the inequality holds by assumption. Hence, by Lemma 8, we have that

$\dim(\pi(V))\leq 3r.$

Moreover

$V\subseteq\pi(V)+p_{1}(V)+\ldots+p_{k}(V),$

hence

$\dim(V)\leq\dim(\pi(V))+\dim(p_{1}(V))+\ldots+\dim(p_{k}(V))\leq
3r+q_{1}+\ldots+q_{k}.$

∎

######  Lemma 10.

Let $r\in\mathbb{N}-\\{0\\}$ and let $b,c\in\mathbb{R}^{2r}$. Then

$\det\begin{pmatrix}\overline{J}_{2r}&c\\\
{}^{t}b&0\end{pmatrix}=c_{2}b_{1}-c_{1}b_{2}+c_{4}b_{3}-c_{3}b_{4}+\ldots$

###### Proof.

By Schur’s Complement Lemma

$\det\begin{pmatrix}\overline{J}_{2r}&c\\\
{}^{t}b&0\end{pmatrix}=\det(\overline{J}_{2r})\det\left(-{}^{t}b\,\overline{J}_{2r}^{-1}\,c\right)=\det(\overline{J}_{2r})\det\left({}^{t}b\,\overline{J}_{2r}\,c\right)=$
$=\det\left({}^{t}b\begin{pmatrix}c_{2}\\\ -c_{1}\\\ c_{4}\\\ -c_{3}\\\ .\\\
.\\\ .\end{pmatrix}\right)=c_{2}b_{1}-c_{1}b_{2}+c_{4}b_{3}-c_{3}b_{4}+\ldots$

∎

######  Corollary 11.

Let $r\in\mathbb{N}-\\{0\\}$, $b,c\in\mathbb{R}^{2r}-\\{0\\}$, $A\in
M(2r\times 2r,\mathbb{R})$. Then

$\det\begin{pmatrix}\overline{J}_{2r}+sA&sc\\\ s{}^{t}b&0\end{pmatrix}$

is a polynomial in $s$ whose lowest degree term is

$s^{2}(c_{2}b_{1}-c_{1}b_{2}+c_{4}b_{3}-c_{3}b_{4}+\ldots).$

###### Proof of Theorem 6.

First suppose $n\geq 2r+2$. Consider the following affine subspace:

$S=\left\\{\begin{pmatrix}\overline{J}_{2r}+R(A_{i,j})&C\\\
-{}^{t}C&0\end{pmatrix}|\;\begin{array}[]{l}A_{i,j}\in M(2\times
2,\mathbb{R})\;\mbox{\rm with the second row equal to zero }\\\
\;\;\;\;\;\mbox{\rm for }\;i,j\in\\{1,\ldots,r\\},\;\mbox{\rm with }i<j,\\\
C\in M(2r\times(n-2r),\mathbb{R})\;\mbox{\rm s.t.}\;C_{2i,j}=0\\\
\;\;\;\;\;\mbox{\rm for }i=1,\ldots,r,\;j=1,\ldots,n-2r\end{array}\right\\}.$

Observe that every element $A$ of $S$ has rank $2r$, in fact the submatrix

$A^{(1,\ldots,2r)}_{(1,\ldots,2r)}=\overline{J}_{2r}+R(A_{i,j})$

is invertible and every $(2r+1)\times(2r+1)$-submatrix
$A^{(1,\ldots,2r,i)}_{(1,\ldots,2r,j)}$ for $i,j\in\\{2r+1,\ldots,n\\}$ has
determinant equal to $0$. So $S\in{\mathcal{A}}_{antisym}^{\mathbb{R}}(n;2r)$.

Obviously the dimension of $S$ is $r(n-2r)+r(r-1)$, i.e. $r(n-r-1)$. So we get

$a_{antisym}^{\mathbb{R}}(n;2r)\geq(n-r-1)r.$

If $n=2r+1$, consider the following affine subspace:

$S=\left\\{\begin{pmatrix}\overline{J}_{2r}+R(A_{i,j})&C\\\
-{}^{t}C&0\end{pmatrix}|\;\begin{array}[]{l}A_{i,j}\in M(2\times
2,\mathbb{R})\;\mbox{\rm with the second row equal to zero }\\\
\;\;\;\;\;\mbox{\rm for }\;i,j\in\\{1,\ldots,r\\},\;\mbox{\rm with }i<j,\\\
C\in M(2r\times 1,\mathbb{R})\end{array}\right\\}.$

Observe that every element $A$ of $S$ has rank $2r$, in fact the submatrix

$A^{(1,\ldots,2r)}_{(1,\ldots,2r)}=\overline{J}_{2r}+R(A_{i,j})$

is invertible and obviously $A$ has determinant equal to $0$.

The dimension of $S$ is $r(r-1)+2r$, i.e. $r(r+1)$. So, in this case, we get

$a_{antisym}^{\mathbb{R}}(n;2r)\geq r(r+1).$

If $n=2r$, consider the following affine subspace:

$S=\left\\{\begin{pmatrix}\overline{J}_{2r}+R(A_{i,j})\\\
\end{pmatrix}|\;\begin{array}[]{l}A_{i,j}\in M(2\times
2,\mathbb{R})\;\mbox{\rm with the second row equal to zero }\\\
\;\;\;\;\;\mbox{\rm for }\;i,j\in\\{1,\ldots,r\\},\;\mbox{\rm with }i<j,\\\
\end{array}\right\\}.$

Every element $A$ of $S$ has rank $2r$ and the dimension of $S$ is $r(r-1)$.
So, in this case, we get

$a_{antisym}^{\mathbb{R}}(n;2r)\geq r(r-1).$

Now let us prove the other inequalities.

Suppose $n\geq 2r+2$.

Let $R\in{\mathcal{A}}_{antisym}^{\mathbb{R}}(n;2r)$. We want to prove that
$\dim(R)\leq r(n-r-1)$. We can write $R$ as $M+L$ where $M\in A(n,\mathbb{R})$
and $L$ is a linear subspace of $A(n,\mathbb{R})$. Let
$\tilde{J}=\begin{pmatrix}\overline{J}_{2r}&0\\\ 0&0\end{pmatrix}$. Let $Q$ be
an invertible matrix such that ${}^{t}QMQ$ is a matrix of the kind

$diag(d_{1},d_{1},d_{2},d_{2},\ldots,d_{r},d_{r},1,\ldots,1)\tilde{J}$

with $d_{i}\in\\{-1,1\\}$.

Let $V={}^{t}QLQ$ and $S={}^{t}QRQ=\tilde{J}+V$. Obviously
$S\in{\mathcal{A}}_{antisym}^{\mathbb{R}}(n;2r)$; moreover, $\dim(S)=\dim(R)$,
so, to prove that $\dim(R)\leq r(n-r-1)$, it is sufficient to prove that
$\dim(S)\leq r(n-r-1)$.

Suppose for simplicity $d_{1},\ldots,d_{r}=1$, i.e. ${}^{t}QMQ=\tilde{J}$. In
the other cases we can argue analogously.

Let

$P=\\{W\in A(n,\mathbb{R})|\;W_{i,j}=0\;\mbox{\rm
for}\;i,j\in\\{n-2r+1,\ldots,n\\}\\},$ $K=\left\\{\begin{pmatrix}0_{2r\times
2r}&C\\\ -{}^{t}C&0_{(n-2r)\times(n-2r)}\end{pmatrix}|\;C\in
M(2r\times(n-2r),\mathbb{R})\right\\}$

and, for any $j_{1},j_{2}\in\\{1,\ldots,n-2r\\}$,

$K_{j_{1},j_{2}}=\left\\{\begin{pmatrix}0_{2r\times 2r}&C\\\
-{}^{t}C&0_{(n-2r)\times(n-2r)}\end{pmatrix}|\;C\in
M(2r\times(n-2r),\mathbb{R})\;\mbox{\rm s.t.}\;C^{(j)}=0\;\mbox{\rm for }j\neq
j_{1},j_{2}\right\\}$

First let us prove that

$V\subseteq P.$ (1)

Suppose that there exists $v\in V$ such that $v_{i,j}\neq 0$ for some
$i,j\in\\{n-2r+1,\ldots,n\\}$. But in this case
$\det\left((\tilde{J}+sv)^{(1,\ldots,2r,j)}_{(1,\ldots,2r,i)}\right)$ would be
a polynomial in $s$ with coefficient of the term of degree $1$ equal to
$v_{i,j}$ (which is nonzero), so a nonconstant polynomial; thus there would
exist $s$ such that
$\det\left((\tilde{J}+sv)^{(1,\ldots,2r,j)}_{(1,\ldots,2r,i)}\right)\neq 0$.
So, for such an $s$, we would have ${\operatorname{rk}}(\tilde{J}+sv)>2r$, so
$S$ would not be of constant rank $2r$, which is contrary to our assumption.

Hence we have proved (1).

For any $j_{1},j_{2}\in\\{1,\ldots,n-2r\\}$ with $j_{1}\neq j_{2}$, let
$Z_{j_{1},j_{2}}$ be a $(2r)$-dimensional vector subspace of $K_{j_{1},j_{2}}$
such that the quadratic form (of signature $(2r,2r)$)

$\sum_{i=1,\ldots,r}c_{2i-1,j_{2}}c_{2i,j_{1}}-c_{2i-1,j_{1}}c_{2i,j_{2}}$ (2)

is positive definite on $Z_{j_{1},j_{2}}$.

Let

$\pi:V\longrightarrow K$

be the map

$\begin{pmatrix}A&C\\\
-{}^{t}C&0_{(n-2r)\times(n-2r)}\end{pmatrix}\longmapsto\begin{pmatrix}0_{2r\times
2r}&C\\\ -{}^{t}C&0_{(n-2r)\times(n-2r)}\end{pmatrix}$

For any $j_{1},j_{2}\in\\{1,\ldots,n-2r\\}$ with $j_{1}\neq j_{2}$, let

$\pi_{j_{1},j_{2}}:K\longrightarrow K_{j_{1},j_{2}}$

be the map sending every matrix $\begin{pmatrix}0&C\\\
-{}^{t}C&0_{(n-2r)\times(n-2r)}\end{pmatrix}$ in $K$ to the matrix

$\begin{pmatrix}0_{2r\times 2r}&\left(0\ldots 0,C^{(j_{1})},0\ldots
0,C^{(j_{2})},0\ldots 0\right)\\\ -{}^{t}\left(0\ldots 0,C^{(j_{1})},0\ldots
0,C^{(j_{2})},0\ldots 0\right)&0_{(n-2r)\times(n-2r)}\end{pmatrix}.$

Fix $j_{1},j_{2}\in\\{1,\ldots,n-2r\\}$ with $j_{1}\neq j_{2}$. We want to
prove that

$\pi_{j_{1},j_{2}}(\pi(V))\cap Z_{j_{1},j_{2}}=\\{0\\}$ (3)

Suppose on the contrary that there exists $x=\begin{pmatrix}A&C\\\
-{}^{t}C&0_{(n-2r)\times(n-2r)}\end{pmatrix}\in V$ such that
$\pi_{j_{1},j_{2}}(\pi(x))\in Z_{j_{1},j_{2}}\setminus\\{0\\}$. Then the
number in (2) would be positive. But, in this case, we would have, by
Corollary 11, that

$\det\left((\tilde{J}+sx)^{(1,\ldots,2r,j_{1}+2r)}_{(1,\ldots,2r,j_{2}+2r)}\right)$

is a polynomial in $s$ with coefficient of the term of degree $2$ equal to the
number in (2) (which is positive), hence a nonconstant polynomial. Thus there
would exist $s$ such that
$\det\left((\tilde{J}+sx)^{(1,\ldots,2r,j_{1}+2r)}_{(1,\ldots,2r,j_{2}+2r)}\right)\neq
0$

So, for such a $s$, we would have ${\operatorname{rk}}(\tilde{J}+sx)>2r$, so
$S$ would not be of constant rank $2r$, which is contrary to our assumption.

Hence we have proved (3); therefore

$\dim(\pi_{j_{1},j_{2}}(\pi(V)))\leq 2r.$ (4)

If $n-2r$ is even consider the projections
$\pi_{1,2},\ldots,\pi_{n-2r-1,n-2r}$. If $n-2r$ is odd consider the
projections
$\pi_{1,2},\pi_{1,3},\pi_{2,3},\pi_{4,5},\ldots,\pi_{n-2r-1,n-2r}$.

By Lemma 9 and from (4) we get that

$\dim(\pi(V))\leq r(n-2r).$

Hence there exists a $r(n-2r)$-dimensional vector subspace $Z$ in $K$ such
that

$\pi(V)\cap Z=\\{0\\}.$ (5)

Let

$U=\left\\{\begin{pmatrix}T(l_{1},\ldots,l_{r})+R(A_{i,j})&0_{2r\times(n-2r)}\\\
0_{(n-2r)\times
2r}&0_{(n-2r)\times(n-2r)}\end{pmatrix}|\;\begin{array}[]{l}A_{i,j}\;2\times
2\;\mbox{\rm antipinco matrix}\\\ \mbox{\rm
for}\;i,j\in\\{1,\ldots,r\\},\;i<j,\\\
l_{1},\ldots,l_{r}\in\mathbb{R}\end{array}\right\\}.$

Observe that

$\dim(U)=r(r-1)+r=r^{2}.$

We want to prove that

$V\cap(U+Z)=\\{0\\}$

Let $u\in U$ and $z\in Z$ be such that $u+z\in V$.

The element $z$ must be zero by formula (5).

Finally suppose $u\neq 0$. Let
$u=\begin{pmatrix}T(l_{1},\ldots,l_{2r-1})+R(A_{i,j})&0_{2r\times(n-2r)}\\\
0_{(n-2r)\times 2r}&0_{(n-2r)\times(n-2r)}\end{pmatrix}$ with $A_{i,j}$
$2\times 2$ antipinco matrices, $l_{1},\ldots,l_{r}\in\mathbb{R}$.

Observe that

$J\begin{pmatrix}a&b\\\ -b&a\end{pmatrix}=\begin{pmatrix}-b&a\\\
-a&-b\end{pmatrix}$,

$J\left(-{}^{t}\begin{pmatrix}a&b\\\
-b&a\end{pmatrix}\right)=\begin{pmatrix}-b&-a\\\ a&-b\end{pmatrix}\ $,

$J\begin{pmatrix}0&l\\\ -l&0\end{pmatrix}=-lI$.

Hence

$-\overline{J}_{2r}\left(\overline{J}_{2r}+s[T(l_{1},\ldots,l_{r})+R(A_{i,j})]\right)=I_{2r}+sA,$
(6)

where

$A=diag(l_{1},l_{1},l_{2},l_{2},\ldots,l_{r},l_{r})+X(T_{i,j})$

for some antipinco matrices $T_{i,j}$ for $i,j\in\\{1,\ldots,r\\}$ with $i<j$.

Since $A$ is symmetric (and nonzero since $u\neq 0$), it has a nonzero
eigenvalue $\lambda$; let $s=-\frac{1}{\lambda}$; we have:

$\det(I_{2r}+sA)=s^{2r}\left(\det\left(A+\frac{1}{s}I_{2r}\right)\right)=s^{2r}\left(\det\left(A-\lambda
I_{2r}\right)\right)=0.$

But
$\det(I_{2r}+sA)=\det\left(\overline{J}_{2r}+s\left[T(l_{1},\ldots,l_{r})+R(A_{i,j})\right]\right)$
by (6); hence we get

$\det\left(\overline{J}_{2r}+s\left[T(l_{1},\ldots,l_{r})+R(A_{i,j})\right]\right)=0$

and then
${\operatorname{rk}}(\tilde{J}+s(u+z))={\operatorname{rk}}(\tilde{J}+su)<2r$,
so $S$ would not be of constant rank $2r$, which is contrary to our
assumption. So also $u$ must be equal to $0$.

So we have proved that $V\cap(Z\cap U)=\\{0\\}$.

Hence $\dim(V)$ is less than or equal to

$\dim(P)-\dim(U+Z)=$ $=\dim(P)-\dim(U)-\dim(Z)=$
$=r(2r-1)+2r(n-2r)-r^{2}-r(n-2r)=r(n-r-1)$

In the cases $n=2r$ and $n=2r+1$, we can argue analogously.

∎

Acknowledgments. This work was supported by the National Group for Algebraic
and Geometric Structures, and their Applications (GNSAGA-INdAM).

## References

  * [1] H. Flanders, On spaces of linear transformations with bounded rank, J. London Math. Soc., 37 (1962), pp. 10-16.
  * [2] F.R. Gantmacher, The Theory of Matrices, vol. 2 Chelsea, Publishing Company New York 1959.
  * [3] W.V.D. Hodge, D. Pedoe, Methods of Algebraic Geometry, vol. 2 Cambridge University PressCambidge 1994.
  * [4] Z. Huang, X. Zhan, ACI-matrices all of whose completions have the same rank, Linear Algebra Appl., 434 (2011), pp. 2259-2271
  * [5] B. Ilic, J.M. Landsberg On symmetric degeneracy loci, spaces of symmetric matrices of constant rank and dual varieties, Mathematischen Annlen, 314 (1999), pp. 159–174.
  * [6] E. Rubei Affine subspaces of matrices with constant rank, Linear Algebra Appl., 644 (2022), pp. 259-269.
  * [7] R. Westwick Spaces of matrices of fixed rank, Linear and Multilinear Algebra, 20 (1987), pp. 171-174.