# Upper bounds for the rank of powers of quadrics

Cosimo Flavi  (Cosimo Flavi), Università degli Studi di Firenze, Dipartimento
di Matematica e Informatica ”Ulisse Dini”, Viale Giovanni Battista Morgagni
67/a, 50134 Florence, Italy<EMAIL_ADDRESS>
###### Abstract.

We establish an upper bound for the rank of every power of an arbitrary
quadratic form. Specifically, for any $s\in\mathbb{N}$, we prove that the
$s$-th power of a quadratic form of rank $n$ grows as $n^{s}$. Furthermore, we
demonstrate that its rank is subgeneric for all $n>(2s-1)^{2}$.

###### Key words and phrases:

Additive decompositions, quadratic forms, symmetric tensor, tensor rank

###### 1991 Mathematics Subject Classification:

Primary 14N07

##  Introduction

Given $n,d\in\mathbb{N}$ and a homogeneous polynomial
$f\in\mathbb{C}[x_{1},\dots,x_{n}]$ of degree $d$, a summation of the $d$-th
powers of $r$ distinct linear forms
$l_{1},\dots,l_{r}\in\mathbb{C}[x_{1},\dots,x_{n}]$, which can be viewed as
points in $\mathbb{C}^{n}$, such that

$f=\sum_{j=1}^{r}l_{j}^{d},$

is referred to as a Waring decomposition, or simply decomposition, of $f$ of
size $r$. The Waring rank (or simply rank) of $f$ is the minimum natural
number $r$ such that there exists a decomposition of $f$ of size $r$, that is,

$\operatorname{rk}f=\min{\Set{r\in\mathbb{N}}{\text{$f=\displaystyle\sum_{j=1}^{r}l_{j}^{d}$
: $l_{j}\in\mathbb{C}[x_{1},\dots,x_{n}]_{1}$}}}.$

The problem of determining the rank for a generic form was solved by J.
Alexander and A. Hirschowitz in [AH95].

######  Theorem (J. Alexander, A. Hirschowitz).

There exists a Zariski open set
$\Omega\subseteq\mathbb{C}[x_{1},\dots,x_{n}]_{d}$ such that, for every
$f\in\Omega$,

$\operatorname{rk}f=\biggl{\lceil}\frac{1}{n}\binom{d+n-1}{d}\biggr{\rceil},$

with the exceptions given by the following cases:

  1. •

if $d=2$, then $\operatorname{rk}f=n$;

  2. •

if $n=3$ and $d=4$, then $\operatorname{rk}f=6$;

  3. •

if $n=4$ and $d=4$, then $\operatorname{rk}f=10$;

  4. •

if $n=5$ and $d=3$, then $\operatorname{rk}f=8$;

  5. •

if $n=5$ and $d=4$, then $\operatorname{rk}f=15$.

More recently, this result has been carefully analyzed by M. C. Brambilla and
G. Ottaviani, who provided a shorter version of the proof in [BO08], which we
refer to for the details. However, although the Waring rank for generic forms
has been completely understood, obtaining the rank of a specific polynomial
still remains a challenging problem. Currently, there is no general efficient
method to solve it, or even to find suitable decompositions, regardless of the
form being considered. Nevertheless, many partial results and methods to
tackle the Waring problem for a polynomial have been developed over the years.
For a more detailed overview of Waring decompositions, there are numerous
texts and papers in the literature. We refer, for instance, to [BCC+18],
[BGI11], [CGO14], [Lan12], and [LO13].

For the special case of two variables, the determination of the rank is easier
to approach and has been completely analyzed (see [Syl51] or the more recent
work of G. Comas and M. Seiguer in [CS11]). In particular, there exists
several algorithms providing explicit decompositions. The first of these
algorithms is known as the Sylvester algorithm, which can be found in [Syl86]
or, in a more recent version, also in [CS11], [BGI11], and [BCMT10]. It has
been further analyzed over the years, with various other variants (see e.g.
[BGI11]*Algorithm 2).

However, of great interest are also the applications of Waring decompositions.
In [BGI11]*section 1, A. Bernardi, A. Gimigliano, and M. Idà briefly summarize
some of these applications, such as telecommunications in electrical
engineering (see e.g. [Che99] and [DC07]) or cumulant tensors in statistics
(see e.g. [McC87]). It is for this reason that the Waring rank of a
polynomial, or equivalently, the symmetric rank of a symmetric tensor, still
holds a special role even in more recent times, despite its classical origins.

As one could expect, many forms are relevant in both classical and modern
subjects and have appeared several times in the literature. The central
objects we will consider in this paper are the powers of the quadratic forms.
All the quadratic forms having rank equal to $n\in\mathbb{N}$ are equivalent.
Hence each of them can be written, after a linear change of variables, as

$q_{n}=x_{1}^{2}+\cdots+x_{n}^{2}.$

Therefore, we can focus our study on the polynomial $q_{n}^{s}$ for
$n,s\in\mathbb{N}$.

For the case where $n=2$, the problem of determining the rank of $q_{2}^{s}$
has been analyzed by B. Reznick, who proves in [Rez92]*Theorem 9.5 that
$\operatorname{rk}(q_{2}^{s})=s+1$. Actually, his analysis focuses on real
decompositions, but by the equivalence of the quadratic forms, it is possible
to extend this result to any quadratic form having rank $2$. However, we
cannot make the same claim for the general case with more variables, as there
is limited information available in the literature. To date, the most
comprehensive analysis on this subject has been conducted by B. Reznick in
[Rez92]*Chapters 8-9, providing a thorough survey of both classical and more
original results over $\mathbb{R}$. Given the greater relevance of real
numbers in applications compared to complex numbers, B. Reznick focuses on
real Waring decomposition in his notes, referring to them as representations.
However, to gain a new perspective on this problem, especially considering the
recent applications involving tensors, we also consider complex Waring
decompositions, which represents a departure from the classical point of view.

B. Reznick lists several applications of decompositions of powers of quadratic
forms in [Rez92]*Section 8, including their use in number theory for studying
the Waring problem and in functional analysis.

In section 1, we present some of the main decompositions analyzed by B.
Reznick, with a focus on closed formulas providing a family of decompositions
depending on the number of variables. In general, these formulas do not
correspond to minimal decompositions, but can offer insights into how the rank
grows as the number of variables tends to infinity. Moreover, we examine in
which cases we can establish that the rank of $q_{n}^{s}$ is subgeneric. A
basic example is given by the formula

$6q_{n}^{2}=\sum_{i_{1}<i_{2}}(x_{i_{1}}\pm
x_{i_{2}})^{4}+2(4-n)\sum_{i_{1}}x_{i_{1}}^{4},$

which is a decomposition of size $n^{2}$, for $n\neq 4$, presented by B.
Reznick in [Rez92]*formula (8.33). As we will see in section 1, this formula
does not provide minimal decompositions for every $n\in\mathbb{N}$. However,
it exhibits an elegant pattern and a structure that is invariant under the
action of the permutation group $\mathfrak{S}_{n}$. Furthermore, it yields
subgeneric decompositions for $n>17$. This formula has already been taken into
consideration by J. Buczyński, K. Han, M. Mella, and Z. Teitler in
[BHMT18]*section 4.5, where they provide a similar decomposition for the
exponent $3$, which follows the same pattern of points. It is given by the
formula

$60q_{n}^{3}=\sum_{i_{1}<i_{2}<i_{3}}(x_{i_{1}}\pm x_{i_{2}}\pm
x_{i_{3}})^{6}+2(5-n)\sum_{i_{1}<i_{2}}(x_{i_{1}}\pm
x_{i_{2}})^{6}+2(n^{2}-9n+38)\sum_{i_{1}}x_{i_{1}}^{6}.$

We generalize this pattern in section 2, providing other formulas for higher
powers. Unfortunately, describing the pattern is not so easy as one might
initially imagine, as linear forms involving only $1$ and $-1$ as coefficients
are not sufficient to cover the coefficients of all monomials of degree $2s$.
To obtain a general formula for every power $s$, it is necessary to consider
other types of linear forms. In order to do this, we need to recall the
classical notion of $k$-partition function for every $k\in\mathbb{N}$. This
function, denoted by $\operatorname{p}_{k}\colon\mathbb{N}\to\mathbb{N}$,
associates every natural number $s$ with the number of partitions of $s$ into
exactly $k$ parts. Based on this, we provide in section 3 an upper bound for
the rank of $q_{n}^{s}$ in section 3, which is summarized in the following
theorem.

######  Theorem.

For every $n,s\in\mathbb{N}$

$\operatorname{rk}(q_{n}^{s})\leq
2^{s-1}\binom{n}{s}+2^{s-2}\binom{n}{s-1}+\sum_{k=1}^{s-2}2^{k-1}k!\operatorname{p}_{k}(s)\binom{n}{k}.$

In particular, for any $s\in\mathbb{N}$, the rank of $q_{n}^{s}$ grows at most
as $n^{s}$.

Given any $s\in\mathbb{N}$, the generic rank of forms of degree $2s$ grows as
$n^{2s-1}$ when $n\to+\infty$. Hence, by the theorem of J. Alexander and A.
Hirschowitz ([AH95]), mentioned in the introduction, the last theorem
guarantees that, for any $s\in\mathbb{N}$, the rank can be generic or
supergeneric for only a finite number of possible values of $n$.

The catalecticant map

$\operatorname{Cat}_{f}\colon\mathbb{C}[y_{1},\dots,y_{n}]\to\mathbb{C}[x_{1},\dots,x_{n}]$

of a homogeneous polynomial $f$ is a graded linear map, defined on monomials
as

$\operatorname{Cat}_{f}\bigl{(}\mathbf{y}^{\bm{\alpha}}\bigr{)}=\frac{\mathop{}\\!\mathrm{\partial}^{\lvert\bm{\alpha}\rvert}f}{\mathop{}\\!\mathrm{\partial}\mathbf{x}^{\bm{\alpha}}}.$

We recall a well-known fact, classically attributed to J. J. Sylvester (see
[Syl51]), which states that, if $f$ is a homogeneous polynomial of degree $d$,
then

$\operatorname{rk}f\geq\operatorname{brk}f\geq\operatorname{rk}\bigl{(}\operatorname{Cat}_{f}^{k}\bigr{)},$

where

$\operatorname{Cat}_{f}^{k}\colon\mathbb{C}[y_{1},\dots,y_{n}]_{k}\to\mathbb{C}[x_{1},\dots,x_{n}]_{d-k}$

is the component of degree $k$ of the catalecticant map of $f$, for every
$k=1,\dots,d$. In particular, thanks to a result due to B. Reznick (see
[Rez92]), we know that all the catalecticant matrices are full-rank. This
implies that

$\operatorname{rk}(q_{n}^{s})\geq\operatorname{brk}(q_{n}^{s})\geq\binom{s+n-1}{s}.$

Putting together the lower and upper bounds obtained, we have the following
result, which is proved in section 3.

######  Corollary.

For every $s\in\mathbb{N}$,

$\lim_{n\to+\infty}\log_{n}\bigl{(}\operatorname{rk}(q_{n}^{s})\bigr{)}=\lim_{n\to+\infty}\log_{n}\bigl{(}\operatorname{brk}(q_{n}^{s})\bigr{)}=s.$

This fact perfectly agrees with the value of the border rank determined in
[Fla22]*Theorem 4.5 for ternary non-degenerate quadratic forms, which turns
out to be equal to the rank of the middle catalecticant matrix, that is,

$\operatorname{brk}(q_{3}^{s})=\binom{s+2}{2}.$

In section 4, we provide an estimate for the maximum value
$n_{s}\in\mathbb{N}$ for which the rank of $q_{n}^{s}$ is subgeneric for every
$n>n_{s}$. We state it in the following theorem.

######  Theorem.

For every $s\in\mathbb{N}$,

$\operatorname{rk}(q_{n}^{s})<\frac{1}{n}\binom{2s+n-1}{2s}.$

In particular, the rank of $q_{n}^{s}$ is subgeneric, whenever $n>(2s-1)^{2}$.

This result partially solves the problem posed by J. Buczyński, K. Han, M.
Mella, and Z. Teitler in [BHMT18]*section 4.5, who asked for the values of $n$
and $s$ for which the rank of $q_{n}^{s}$ is subgeneric. In particular, they
showed in [BHMT18]*Theorem 4.1 that, denoting by $W_{m}^{d}$ the maximum rank
locus for the space of $n$-ary forms of degree $d$, each of its irreducible
components has dimension at least $\binom{n+1}{2}-1$. Moreover, in the case
where equality holds for an irreducible component $W$, then $d$ is even and
$W$ must be the set of all the $(d/2)$-th powers of quadrics. With the result
of section 4, we improve this last statement and show in section 4 the
following result.

######  Corollary.

For every $s\in\mathbb{N}$, if $n>(2s-1)^{2}$ and $W$ is an irreducible
component of $W_{m}^{2s}$, then

$\dim(W)\geq\binom{n+1}{2}.$

##  Powers of quadrics and classical decompositions

Thorough the paper, we will use the notation

$\sum(a_{1}x_{1}\pm\cdots\pm a_{n}x_{n})^{m}$

to denote a summation of all the $2^{n-1}$ possible choices of the
coefficients $+1$ and $-1$, multiplying $a_{2},\dots,a_{n}$. In other words,

$\sum(a_{1}x_{1}\pm\cdots\pm
a_{n}x_{n})^{m}=\sum_{\begin{subarray}{c}j_{i}\in\\{0,1\\}\\\
i=2,\dots,n\end{subarray}}\bigl{(}a_{1}x_{1}+(-1)^{j_{2}}a_{2}x_{2}+\cdots+(-1)^{j_{n}}a_{n}x_{n}\bigr{)}^{m},$

where $a_{1},\dots,a_{n}\in\mathbb{C}$ and $m\in\mathbb{N}$. We will also use
the notation

$\sum_{i_{1}<\cdots<i_{k}}(a_{1}x_{i_{1}}\pm\cdots\pm a_{k}x_{i_{k}})^{m}$

to denote a summation over any choice of $k$ variables
$x_{i_{1}},\dots,x_{i_{k}}$ such that $1\leq i_{1}<\cdots<i_{k}\leq n$, where
$a_{1},\dots,a_{n}\in\mathbb{C}$ and $m\in\mathbb{N}$. In the case where
$k>n$, we assume the terms $x_{i_{n+1}},\dots,x_{i_{k}}$ to be $0$, so that in
this case, we have

$\sum_{i_{1}<\cdots<i_{k}}(a_{1}x_{i_{1}}\pm\cdots\pm
a_{k}x_{i_{k}})^{m}=\sum_{i_{1}<\cdots<i_{n}}(a_{1}x_{i_{1}}\pm\cdots\pm
a_{k}x_{i_{n}})^{m}.$

Determining decompositions of the polynomial $q_{n}^{s}$, for any
$n,s\in\mathbb{N}$, is a classical problem. Many decompositions appear in both
old and more recent literature. B. Reznick provides an overview of classical
results, along with new decompositions, in [Rez92]*chapters 8-9. Here are some
examples.

Using the notation $\mathbf{x}^{\bm{\alpha}}=x_{1}^{\alpha_{1}}\cdots
x_{n}^{\alpha_{n}}$, for every multi-index
$\bm{\alpha}=(\alpha_{1},\dots,\alpha_{n})\in\mathbb{N}^{n}$, the
catalecticant map of a homogeneous polynomial $f$ of degree $d$ is obtained by
extending the map defined on monomials as

${\operatorname{Cat}_{f}\colon\mathbb{C}[y_{1}\dots,y_{n}]}$${\mathbb{C}[x_{1}\dots,x_{n}]\hphantom{.}}$${{\hphantom{\operatorname{Cat}_{f}\colon{}}}\mathbf{y}^{\bm{\alpha}}}$${\displaystyle\frac{\mathop{}\\!\mathrm{\partial}^{\lvert\bm{\alpha}\rvert}f}{\mathop{}\\!\mathrm{\partial}\mathbf{x}^{\bm{\alpha}}}.}$$\leftarrow$$\rightarrow$$\mapsfrom$$\rightarrow$

Thanks to the catalecticant map, it is possible to establish a general
inequality that provides a well-known lower bound for the Waring rank, related
to catalecticant matrices and apolarity (see [IK99] for an overview on this
subject). This inequality is classically attributed to J. J. Sylvester (see
[Syl51]) and appears multiple times in the literature (see, e.g.,
[Lan12]*Proposition 3.5.1.1). We denote by $\operatorname{Cat}_{f}^{k}$ the
$k$-th component of catalecticant map of $f$, which is the restriction

$\operatorname{Cat}_{f}^{k}\colon\mathbb{C}[y_{1},\dots,y_{n}]_{k}\to\mathbb{C}[x_{1},\dots,x_{n}]_{d-k}$

of the catalecticant map of $f$ to the component of degree $k$ of
$\mathbb{C}[x_{1},\dots,x_{n}]$.

######  Proposition 1.0 (J. J. Sylvester).

Let $d,k\in\mathbb{N}$ be such that $d\geq k$ and let $f$ be a homogeneous
polynomial of degree $d$. Then

$\operatorname{rk}f\geq\operatorname{brk}f\geq\operatorname{rk}\bigl{(}\operatorname{Cat}_{f}^{k}\bigr{)}$

for every $f\in\mathbb{C}[x_{1},\dots,x_{n}]$ and for every $k=1,\dots,d$.

B. Reznick shows, in particular, that all the catalecticant matrices of
$q_{n}^{s}$ are full rank, using [Rez92]*Theorem 8.15 and referring to
[Rez92]*Theorems 3.7 and 3.16. Another proof of this fact is provided by F.
Gesmundo and J. M. Landsberg in [GL19]*Theorem 2.2. Moreover, it follows
directly from the structure of the kernel of the catalecticant map of
$q_{n}^{s}$, also known as the apolar ideal of $q_{n}^{s}$ and denoted by
$(q_{n}^{s})^{\perp}$, which has been determined in [Fla22]*Theorem 3.8. In
particular, we have that $(q_{n}^{s})^{\perp}=(\mathcal{H}_{n}^{s+1})$, where
$\mathcal{H}_{n}^{s+1}$ is the space of harmonic polynomials of degree $s+1$.
Therefore, as a direct consequence, we have that the $k$-th component of
$(q_{n}^{s})^{\perp}$ is equal to $0$ for every $k\leq s$, and hence, in
particular, $\operatorname{Cat}_{q_{n}^{s}}^{s}$ has maximum rank. Thus, we
immediately obtain the following proposition.

######  Proposition 1.0 (B. Reznick).

For every $n,s\in\mathbb{N}$,

$\operatorname{rk}(q_{n}^{s})\geq\operatorname{brk}(q_{n}^{s})\geq\operatorname{rk}\bigl{(}\operatorname{Cat}_{q_{n}^{s}}^{s}\bigr{)}=\binom{s+n-1}{s}.$
(1.1)

B. Reznick provides in [Rez92]*formulas (8.35)–(8.36) a decomposition of
$q_{n}^{2}$ for $3\leq n\leq 7$, based on a family of integration quadrature
formulas, which has been introduced by A. H. Stroud in [Str67a]. By verifying
the formulas, we can prove that the same formula is valid also for $n\geq 9$.
Hence, this gives an upper bound on the rank with the only exception of $n=8$,
for which we can only get an upper bound of $56$, that is $n(n-1)$, given by a
specific decomposition. This is given by

$\displaystyle 6q_{8}^{2}$ $\displaystyle=\sum_{j=1}^{4}(x_{j}\pm
x_{j+4})^{4}+\sum_{j=1}^{6}\Bigl{(}\psi^{-\frac{1}{4}}x_{j}\pm\psi^{\frac{1}{4}}x_{j+2}\Bigr{)}^{4}+\sum_{j=1}^{2}\Bigl{(}\psi^{\frac{1}{4}}x_{j}\pm\psi^{-\frac{1}{4}}x_{j+6}\Bigr{)}^{4}+\sum_{j=1}^{7}\Bigl{(}\psi^{\frac{1}{4}}x_{j}\pm\psi^{-\frac{1}{4}}x_{j+1}\Bigr{)}^{4}$
$\displaystyle\hphantom{{}={}}+\sum\Bigl{(}\psi^{-\frac{1}{4}}x_{1}\pm\psi^{\frac{1}{4}}x_{8}\Bigr{)}^{4}+\sum_{j=1}^{5}\Bigl{(}\psi^{\frac{1}{4}}x_{j}\pm\psi^{-\frac{1}{4}}x_{j+3}\Bigr{)}^{4}+\sum_{j=1}^{3}\Bigl{(}\psi^{-\frac{1}{4}}x_{j}\pm\psi^{\frac{1}{4}}x_{j+5}\Bigr{)}^{4},$
(1.2)

where $\psi=\frac{1}{2}\pm\mathrm{i}\frac{2\sqrt{2}}{3}$. In the cases where
$n\geq 9$, the decompositions are no longer real, but the size remains the
same and, in particular, they are not tight. We provide this family of
decompositions in the following theorem, whose proof is quite easy, as it is
obtained simply by solving a system of equations.

######  Theorem 1.2.

Let $n\in\mathbb{N}$ be such that $n\geq 3$ and $n\neq 8$ and let
$g\in\mathbb{C}$ such that $g^{4}=8-n$. Then, the form $q_{n}^{2}$ can be
decomposed as

$3e^{4}q_{n}^{2}=a\biggl{(}\sum_{j=1}^{n}x_{j}\biggr{)}^{4}+\sum_{k=1}^{n}\Biggl{(}b\biggl{(}\sum_{j=1}^{n}x_{j}\biggl{)}+cx_{k}\Biggr{)}^{4}+\sum_{j_{1}\neq
j_{2}}\Biggl{(}d\biggl{(}\sum_{j=1}^{n}x_{j}\biggr{)}+e(x_{j_{1}}+x_{j_{2}})\Biggr{)}^{4},$
(1.3)

which gives a decomposition of size $\binom{n+1}{2}+1$, where

$\displaystyle a=8(g^{4}-1)\bigl{(}g^{2}\pm 2\sqrt{2}\bigr{)}^{4},\quad
b=2g^{2}\pm 2\sqrt{2},\quad c=\mp 2\sqrt{2}g^{4}-8g^{2},\quad d=2g,\quad e=\mp
2\sqrt{2}g^{3}-8g.$

###### Proof.

To obtain the required formula, it is sufficient to equalize the coefficients
of each type of monomial in both the sides of the equation, obtaining a system
of $5$ equations. Indeed, the set of points appearing in formula (1.3) is
invariant under the action of the permutation group $\mathfrak{S}_{n}$ on the
set of variables $\\{x_{1},\dots,x_{n}\\}$. Therefore we can gather the
monomials that are in the same permutation class, obtaining the five
polynomials

$\displaystyle M_{(4)}=\sum_{i=1}^{n}x_{i}^{4},\qquad
M_{(3,1)}=\sum_{\begin{subarray}{c}1\leq i_{1},i_{2}\leq n\\\ i_{1}\neq
i_{2}\end{subarray}}x_{i_{1}}^{3}x_{i_{2}},\qquad M_{(2,2)}=\sum_{1\leq
i_{1}<i_{2}\leq n}x_{i_{1}}^{2}x_{i_{2}}^{2},$ $\displaystyle
M_{(2,1,1)}=\sum_{\begin{subarray}{c}1\leq i_{1},i_{2},i_{3}\leq n\\\
i_{1}\neq
i_{2},i_{3},\,i_{2}<i_{3}\end{subarray}}x_{i_{1}}^{2}x_{i_{2}}x_{i_{3}},\qquad
M_{(1,1,1,1)}=\sum_{1\leq 1_{1}<\cdots<i_{4}\leq
n}x_{i_{1}}x_{i_{2}}x_{i_{3}}x_{i_{4}}.$ (1.4)

Equalizing the coefficients of polynomials (1.4), we obtain the system

$\begin{cases}6b^{2}c^{2}+6(n-2)d^{2}e^{2}+4(n-1)de^{3}+(n-1)e^{4}+4bc^{3}+c^{4}=3e^{4},\vphantom{\dbinom{n}{2}}\\\
3b^{2}c^{2}+3(n-2)d^{2}e^{2}+(n+2)de^{3}+e^{4}+bc^{3}=0,\vphantom{\dbinom{n}{2}}\\\
2b^{2}c^{2}+2(n-2)d^{2}e^{2}+4de^{3}+e^{4}=e^{4},\vphantom{\dbinom{n}{2}}\\\
b^{2}c^{2}+(n-2)d^{2}e^{2}+2de^{3}=0,\vphantom{\dbinom{n}{2}}\\\
a+nb^{4}+4b^{3}c+\dbinom{n}{2}d^{4}+4(n-1)d^{3}e+6d^{2}e^{2}=0,\end{cases}$

by which we can compute the required values of the coefficients. In
particular, we have

$\begin{cases}c^{4}-g^{4}e^{4}=0,\vphantom{\dfrac{(g^{4}-8)(g^{4}-7)}{2}}\\\
2bc^{3}-2(g^{4}-4)de^{3}+2e^{4}=0,\vphantom{\dfrac{(g^{4}-8)(g^{4}-7)}{2}}\\\
b^{2}c^{2}-(g^{4}-6)d^{2}e^{2}+2de^{3}=0,\vphantom{\dfrac{(g^{4}-8)(g^{4}-7)}{2}}\\\
a-(g^{4}-8)b^{4}+4b^{3}c+\dfrac{(g^{4}-8)(g^{4}-7)}{2}d^{4}-4(g^{4}-7)d^{3}e+6d^{2}e^{2}=0,\\\
\end{cases}$

and several, but not difficult, calculations permit to verify that all the
equations are satisfied by the coefficients in the statement. ∎

The family of decompositions (1.3) is just one of the possible closed formulas
that provide us a family of decompositions for the power of a quadratic form.
We now generalize this pattern, defining a new set of generators.

######  Definition 1.5.

For every $s\in\mathbb{N}$, the set of $k$-partitions of $s$ is the set

$\mathcal{P}_{k}(s)=\Set{(m_{1},\dots,m_{k})\in\mathbb{N}^{k}}{\sum_{i=1}^{k}m_{i}=s,\,m_{1}\geq\cdots\geq
m_{k}>0}.$

The set of partitions of $s$ is the set

$\mathcal{P}(s)=\bigcup_{j=1}^{s}\mathcal{P}_{j}(s).$

Given a natural number $n\in\mathbb{N}$, a problem assuming great relevance in
both classical and recent literature is determining the number of partitions
of $n$.

######  Definition 1.6.

The partition function is defined as the map

${\operatorname{p}\colon\mathbb{N}}$${\mathbb{N}\hphantom{.}}$${{\hphantom{\operatorname{p}\colon{}}}s}$${\left\lvert\mathcal{P}(s)\right\rvert.}$$\leftarrow$$\rightarrow$$\mapsfrom$$\rightarrow$

For every $k\in\mathbb{N}$, the $k$-partition function is defined as the map

${\operatorname{p}_{k}\colon\mathbb{N}}$${\mathbb{N}\hphantom{.}}$${{\hphantom{\operatorname{p}_{k}\colon{}}}s}$${\left\lvert\mathcal{P}_{k}(s)\right\rvert.}$$\leftarrow$$\rightarrow$$\mapsfrom$$\rightarrow$

We immediately observe that the equality

$\operatorname{p}(s)=\sum_{k=1}^{s}\operatorname{p}_{k}(s)$

holds for every $s\in\mathbb{N}$. The partition function $\operatorname{p}(s)$
has always had great relevance in several branches of mathematics. To date,
although no closed formulas describing it are known, there are many asymptotic
approximations and also many recurrence formulas providing the exact value of
$\operatorname{p}(s)$ for every $s\in\mathbb{N}$. The first estimate of
$\operatorname{p}(s)$ was provided in 1918 by G. H. Hardy and S. Ramanujan in
[HR18]. In particular, they proved that, as $s\to+\infty$, it can be
approximated as

$\operatorname{p}(s)\sim\frac{1}{4s\sqrt{3}}\mathrm{e}^{\uppi\sqrt{\frac{2s}{3}}},$

which means that it grows as an exponential function of the square root of
$s$. To get further details about the partition function
$\operatorname{p}(s)$, we also refer to [Rad37], in which H. Rademacher
provides a convergent series for the partition function, improving the results
contained in [HR18]. For what concerns, instead, the $k$-partition function
$\operatorname{p}_{k}(s)$, there are some results regarding approximations and
lower or upper bounds. In [Mer14]*p. 299, M. Merca provides an upper bound for
the value $\operatorname{p}_{k}(s)$, from which we have the inequality

$\operatorname{p}_{k}(s)\leq\frac{1}{2}\binom{s-1}{k-1}+\frac{1}{2}\leq\binom{s-1}{k-1}.$
(1.7)

Another tighter upper bound is given by A. T. Oruç in [Oru16]*Corollary 1,
corresponding to the formula

$\operatorname{p}_{k}(s)\leq\frac{5.44}{s-k}\mathrm{e}^{\uppi\sqrt{\frac{2(s-k)}{3}}}.$

Denoting by $\mathfrak{S}_{n}$ the permutation group of $n$ elements, we
introduce a family of polynomials which are invariant under the action of
$\mathfrak{S}_{n}$, already met in (1.4). Fixed a partition
$\mathbf{m}\in\mathcal{P}_{k}(s)$, we define the polynomial

$M_{\mathbf{m}}=\frac{1}{\lvert(\mathfrak{S}_{k})_{\mathbf{m}}\rvert}\sum_{1\leq
t_{1}<\cdots<t_{k}\leq
n}\sum_{\sigma\in\mathfrak{S}_{k}}x_{t_{\sigma(1)}}^{m_{1}}\cdots
x_{t_{\sigma(k)}}^{m_{k}}=\frac{1}{\lvert(\mathfrak{S}_{n})_{\mathbf{m}}\rvert}\sum_{\sigma\in\mathfrak{S}_{n}}x_{\sigma(1)}^{m_{1}}\cdots
x_{\sigma(n)}^{m_{n}},$ (1.8)

where $(\mathfrak{S}_{k})_{\mathbf{m}}$ denotes the stabilizer of the $k$-uple
$\mathbf{m}$ under the action of $\mathfrak{S}_{k}$. These polynomials are
known as monomial symmetric polynomials, which are defined in [FH91]*formula
(A.2), and form a basis of the space of homogeneous symmetric polynomials. For
further details about symmetric functions, we refer to [FH91]*Appendix A.

In formula (1.8), we consider every $k$-partition to be viewed as an $n$-tuple
with the last entries equal to $0$. Clearly, the polynomial $M_{\mathbf{m}}$
is invariant under the action of $\mathfrak{S}_{n}$.

######  Example 1.8.

Let us consider the partition $(2,1,1)$ of $s=4$ in the case where $n=4$.
Then, the polynomial $M_{(2,1,1)}\in\mathbb{K}[x_{1},x_{2},x_{3},x_{4}]$,
already seen in (1.4), is given by

$M_{(2,1,1)}=\sum_{i=1}^{4}\sum_{\begin{subarray}{c}1\leq j_{1}<j_{2}\leq 4\\\
j_{1},j_{2}\neq i\end{subarray}}x_{i}^{2}x_{j_{1}}x_{j_{2}}.$

Instead, if we take the partition $(2,2,1,1)$ of $s=6$ in the case where
$n=5$, we get

$M_{(2,2,1,1)}=\sum_{\begin{subarray}{c}1\leq i_{1}<i_{2}\leq
5\end{subarray}}\sum_{\begin{subarray}{c}1\leq j_{1}<j_{2}\leq 5\\\
j_{1},j_{2}\neq
i_{1},i_{2}\end{subarray}}x_{i_{1}}^{2}x_{i_{2}}^{2}x_{j_{1}}x_{j_{2}}.$

The $s$-th power of the quadratic form $q_{n}$ can be developed as

$q_{n}^{s}=\sum_{i_{1}+\cdots+i_{n}=s}\binom{s}{i_{1},\dots,i_{n}}x_{1}^{2i_{1}}\cdots
x_{n}^{2i_{n}},$

for every $s\in\mathbb{N}$. Gathering all multi-indices of the same
permutation class, we can write

$\displaystyle q_{n}^{s}$
$\displaystyle=\sum_{\mathbf{m}\in\mathcal{P}(s)}\sum_{\sigma\in\mathfrak{S}_{n}}\binom{s}{m_{1},\dots,m_{n}}x_{\sigma(1)}^{2m_{1}}\cdots
x_{\sigma(n)}^{2m_{n}}$
$\displaystyle=\sum_{k=1}^{n}\sum_{\mathbf{m}\in\mathcal{P}_{k}(s)}\sum_{t_{1}\leq\cdots\leq
t_{k}}\sum_{\sigma\in\mathfrak{S}_{k}}\binom{s}{m_{1},\dots,m_{k}}x_{t_{\sigma(1)}}^{2m_{1}}\cdots
x_{t_{\sigma(k)}}^{2m_{k}}$
$\displaystyle=\sum_{k=1}^{n}\sum_{\mathbf{m}\in\mathcal{P}_{k}(s)}\binom{s}{m_{1},\dots,m_{k}}M_{2\mathbf{m}}.$
(1.9)

Now we need to remark a quite trivial fact.

######  Lemma 1.9.

For every $n$-uple $(a_{1},\dots,a_{n})\in\mathbb{C}^{n}$, the polynomial

$\sum(a_{1}x_{1}\pm\cdots\pm a_{n}x_{n})^{2k}$

contains only monomials with all exponents being even.

###### Proof.

The statement follows immediately from the fact that the polynomial

$\sum(a_{1}x_{1}\pm\cdots\pm a_{n}x_{n})^{2k}$

represents an even function in each coordinate. That is, it remains unchanged
by substituting the variable $x_{i}$ by the opposite $-x_{i}$ for every
$i=1,\dots,n$. ∎

Let us consider the decomposition

$6q_{n}^{2}=\sum_{i_{1}<i_{2}}(x_{i_{1}}\pm
x_{i_{2}})^{4}+2(4-n)\sum_{i_{1}}x_{i_{1}}^{4},$ (1.10)

provided by B. Reznick in [Rez92]*formula (10.35). By section 1, to obtain
decomposition (1.10), it is then sufficient to solve a linear system in two
variables, where the unknowns are the coefficients of each monomial. These are
the coefficients of the polynomials

$M_{(2,2)}=\frac{1}{2!(n-2)!}\sum_{\sigma\in\mathfrak{S}_{n}}x_{\sigma(1)}^{2}x_{\sigma(2)}^{2}=\sum_{i_{1}<i_{2}}x_{i_{1}}^{2}x_{i_{2}}^{2},\qquad
M_{(4)}=\frac{1}{(n-1)!}\sum_{\sigma\in\mathfrak{S}_{n}}x_{\sigma(1)}^{4}=\sum_{i_{1}}x_{i_{1}}^{4}.$

Thus, by comparing the coefficients in both sides of the equation

$q_{n}^{2}=c_{1}\sum_{i_{1}<i_{2}}(x_{i_{1}}\pm
x_{i_{2}})^{4}+c_{2}\sum_{i}^{n}x_{i}^{4},$

i.e., writing

$2M_{(2,2)}=2\binom{4}{2}c_{1}M_{(2,2)},\qquad
M_{(4)}=2(n-1)c_{1}M_{(4)}+c_{2}M_{(4)},$

we obtain that the complete matrix of the system is a $2\times 3$ matrix equal
to

$\pNiceMatrix[margin,vlines=3]\Block[fill=gray!10,borders={top,left,right,bottom}]{1-1}{}12&02\\\
2(n-1)\Block[fill=gray!10,borders={top,left,right,bottom}]{1-1}{}11,$ (1.11)

which provides coefficients $c_{1}=\frac{1}{6}$ and $c_{2}=\frac{4-n}{3}$.

A more recent decomposition has been provided by J. Buczyński, K. Han, M.
Mella, and Z. Teitler in [BHMT18]. This represents a natural generalization of
decomposition (1.10) and we give here a proof, as we generalize the procedure
to higher values of the exponent $s$.

######  Proposition 1.11 ([BHMT18]*Section 4.5).

For every $n\in\mathbb{N}$, the form $q_{n}^{3}$ can be decomposed as

$60q_{n}^{3}=\sum_{i_{1}<i_{2}<i_{3}}(x_{i_{1}}\pm x_{i_{2}}\pm
x_{i_{3}})^{6}+2(5-n)\sum_{i_{1}<i_{2}}(x_{i_{1}}\pm
x_{i_{2}})^{6}+2(n^{2}-9n+38)\sum_{i_{1}}x_{i_{1}}^{6},$ (1.12)

which is a decomposition of size

$4\binom{n}{3}+2\binom{n}{2}+n=\frac{2}{3}n^{3}-n^{2}+\frac{4}{3}n$

if $n\geq 3$ and $n\neq 5$ and of size $45$ if $n=5$.

###### Proof.

We can obtain the required equality by solving the linear system associated
with the equation

$q_{n}^{3}=c_{1}\sum_{i_{1}<i_{2}<i_{3}}(x_{i_{1}}\pm x_{i_{2}}\pm
x_{i_{3}})^{6}+c_{2}\sum_{i_{1}<i_{2}}(x_{i_{1}}\pm
x_{i_{2}})^{6}+c_{3}\sum_{i=1}^{n}x_{i}^{6}.$

By section 1 and the symmetry of each summand, we can suppose the development
of the decomposition to be a linear combination of the polynomials

$M_{(2,2,2)}=\sum_{i_{1}<i_{2}<i_{3}}x_{i_{1}}^{2}x_{i_{2}}^{2}x_{i_{3}}^{2},\quad
M_{(4,2)}=\sum_{i_{1}<i_{2}}\bigl{(}x_{i_{1}}^{4}x_{i_{2}}^{2}+x_{i_{1}}^{2}x_{i_{2}}^{4}\bigr{)},\quad
M_{(6)}=\sum_{i_{1}}x_{i_{1}}^{6}.$

By comparing their coefficients, we get a linear system whose associated
matrix is

$\pNiceMatrix[vlines=4,cell-space-
limits=2pt]2^{2}\binom{6}{2,2,2}&00\binom{3}{1,1,1}\\\
2^{2}(n-2)\binom{6}{4,2}2\binom{6}{4,2}0\binom{3}{2,1}\\\
2^{2}\binom{n-1}{2}2(n-1)11,$

which we can transform as

$\pNiceMatrix[margin,vlines=4,cell-space-
limits=2pt]\Block[fill=gray!10,borders={top,left,right,bottom}]{1-1}{}4&00\frac{1}{15}\\\
4(n-2)\Block[fill=gray!10,borders={top,left,right,bottom}]{1-1}{}20\frac{1}{5}\\\
4\binom{n-1}{2}2(n-1)\Block[fill=gray!10,borders={top,left,right,bottom}]{1-1}{}11$
(1.13)

By solving this linear system, we get the required decomposition. ∎

In this case, the size of decompositions given by formula (1.12) grows as
$n^{3}$ for $n\to+\infty$. Therefore, we can find a number
$n_{3}\in\mathbb{N}$ such that

$\operatorname{rk}(q^{3}_{n})<\frac{1}{n}\binom{n+5}{6}$

for every $n>n_{3}$. In particular, we have

$\frac{2}{3}n^{3}-n^{2}+\frac{4}{3}n<\frac{1}{n}\binom{n+5}{6}$

if and only $n>11$. Thus we get $n_{3}=11$, implying the following
proposition.

######  Proposition 1.13.

Let $n\in\mathbb{N}$ be such that $n>11$. Then

$\operatorname{rk}(q_{n}^{3})<\frac{1}{n}\binom{n+5}{6}.$

In particular, for $n>11$, $\operatorname{rk}(q_{n}^{3})$ is subgeneric.

##  Further examples of closed formulas

The most natural idea to generalize this pattern for higher exponents would be
to consider in the summation just the same kind of linear forms with
additional variables, that is, linear forms of the type

$\sum(x_{i_{1}}\pm\cdots\pm x_{i_{s}})^{2s}.$

Unfortunately, this condition is not sufficient to obtain powers of quadratic
forms in general. We can see this in the following example concerning the case
of $s=4$. In this case, we have five polynomials, namely, $M_{(2,2,2,2)}$,
$M_{(4,2,2)}$, $M_{(4,4)}$, $M_{(6,2)}$, and $M_{(8)}$. Assuming an
hypothetical decomposition as

$q_{n}^{4}=c_{1}\sum_{i_{1}<i_{2}<i_{3}<i_{4}}(x_{i_{1}}\pm x_{i_{2}}\pm
x_{i_{3}}\pm x_{i_{4}})^{8}+c_{2}\sum_{i_{1}<i_{2}<i_{3}}(x_{i_{1}}\pm
x_{i_{2}}\pm x_{i_{3}})^{8}+c_{3}\sum_{i_{1}<i_{2}}(x_{i_{1}}\pm
x_{i_{2}})^{8}+c_{4}\sum_{i_{1}}x_{i_{1}}^{8},$

we have four unknown coefficients $c_{1},c_{2},c_{3},c_{4}\in\mathbb{K}$.
However, the matrix associated with the system obtained by comparing the
coefficients of each monomial is

$\pNiceMatrix[vlines=5,cell-space-
limits=2pt]2^{3}\binom{8}{2,2,2,2}&000\binom{4}{1,1,1,1}\\\
2^{3}(n-3)\binom{8}{4,2,2}2^{2}\binom{8}{4,2,2}00\binom{4}{2,1,1}\\\
2^{3}\binom{n-2}{2}\binom{8}{4,4}2^{2}(n-2)\binom{8}{4,4}2\binom{8}{4,4}0\binom{4}{2,2}\\\
2^{3}\binom{n-2}{2}\binom{8}{6,2}2^{2}(n-2)\binom{8}{6,2}2\binom{8}{6,2}0\binom{4}{3,1}\\\
2^{3}\binom{n-1}{3}2^{2}\binom{n-1}{2}2(n-1)11,$

whose rank is the same as the matrix

$\pNiceMatrix[vlines=5,cell-space-limits=2pt]8&000\frac{1}{105}\\\
8(n-3)400\frac{1}{35}\\\ 4(n-2)(n-3)4(n-2)20\frac{3}{35}\\\
4(n-2)(n-3)4(n-2)20\frac{1}{7}\\\ 4(n-1)(n-2)(n-3)6(n-1)(n-2)6(n-1)33.$

Comparing the third and the fourth rows of the matrix, we can see that the
system does not have any solution. This demonstrates that the condition of
considering the same kind of linear forms with additional variables is not
sufficient to obtain powers of quadratic forms in general for the case of
$s=4$.

So, to obtain suitable decompositions for the case of exponent $4$, we need
another set of points while maintaining the symmetry among all of the
variables.

######  Proposition 2.0.

For every $n\in\mathbb{N}$, the form $q_{n}^{4}$ can be decomposed as

$\displaystyle 840q_{n}^{4}$
$\displaystyle=\sum_{i_{1}<i_{2}<i_{3}<i_{4}}(x_{i_{1}}\pm x_{i_{2}}\pm
x_{i_{3}}\pm x_{i_{4}})^{8}+2(6-n)\sum_{i_{1}<i_{2}<i_{3}}(x_{i_{1}}\pm
x_{i_{2}}\pm x_{i_{3}})^{8}$
$\displaystyle\hphantom{{}={}}+\frac{2}{3}(3n^{2}-33n+76)\sum_{i_{1}<i_{2}}(x_{i_{1}}\pm
x_{i_{2}})^{8}+\frac{2}{3}\sum_{i_{1}<i_{2}}\bigl{[}(2x_{i_{1}}\pm
x_{i_{2}})^{8}+(x_{i_{1}}\pm 2x_{i_{2}})^{8}\bigr{]}$
$\displaystyle\hphantom{{}={}}-\frac{4}{3}\bigl{(}n^{3}-15n^{2}+317n-918\bigr{)}\sum_{i_{1}}x_{i_{1}}^{8},$
(2.1)

which is a decomposition of size

$8\binom{n}{4}+4\binom{n}{3}+6\binom{n}{2}+n=\frac{1}{3}(n^{4}-4n^{3}+14n^{2}-8n),$

for $n\geq 4$ and $n\neq 6$, and of size $216$ for $n=6$.

###### Proof.

As before, we consider the generic decomposition

$\displaystyle q_{n}^{4}$
$\displaystyle=c_{1}\sum_{i_{1}<i_{2}<i_{3}<i_{4}}(x_{i_{1}}\pm x_{i_{2}}\pm
x_{i_{3}}\pm x_{i_{4}})^{8}+c_{2}\sum_{i_{1}<i_{2}<i_{3}}(x_{i_{1}}\pm
x_{i_{2}}\pm x_{i_{3}})^{8}+c_{3}\sum_{i_{1}<i_{2}}(x_{i_{1}}\pm
x_{i_{2}})^{8}$
$\displaystyle\hphantom{{}={}}+c_{4}\sum_{i_{1}<i_{2}}\bigl{[}(2x_{i_{1}}\pm
x_{i_{2}})^{8}+(x_{i_{1}}\pm
2x_{i_{2}})^{8}\bigr{]}+c_{5}\sum_{i_{1}}x_{i_{1}}^{8},$ (2.2)

and the five different polynomials

$\displaystyle
M_{(2,2,2,2)}=\sum_{i_{1}<i_{2}<i_{3}<i_{4}}x_{i_{1}}^{2}x_{i_{2}}^{2}x_{i_{3}}^{2}x_{i_{4}}^{2},\quad
M_{(4,2,2)}=\sum_{i_{1}<i_{2}<i_{3}}\bigl{(}x_{i_{1}}^{4}x_{i_{2}}^{2}x_{i_{3}}^{2}+x_{i_{1}}^{2}x_{i_{2}}^{4}x_{i_{3}}^{2}+x_{i_{1}}^{2}x_{i_{2}}^{2}x_{i_{3}}^{4}\bigr{)},$
$\displaystyle M_{(4,4)}=\sum_{i_{1}<i_{2}}x_{i_{1}}^{4}x_{i_{2}}^{4},\quad
M_{(6,2)}=\sum_{i_{1}<i_{2}}\bigl{(}x_{i_{1}}^{6}x_{i_{2}}^{2}+x_{i_{1}}^{2}x_{i_{2}}^{6}\bigr{)},\quad
M_{(8)}=\sum_{i_{1}}x_{i_{1}}^{8}.$

By comparing the coefficients of these last elements with those of formula
(1), we obtain the matrix

$\pNiceMatrix[vlines=6,cell-space-
limits=2pt]2^{3}\binom{8}{2,2,2,2}&0000\binom{4}{1,1,1,1}\\\
2^{3}(n-3)\binom{8}{4,2,2}2^{2}\binom{8}{4,2,2}000\binom{4}{2,1,1}\\\
2^{3}\binom{n-2}{2}\binom{8}{4,4}2^{2}(n-2)\binom{8}{4,4}2\binom{8}{4,4}2(2^{4}+2^{4})\binom{8}{4,4}0\binom{4}{2,2}\\\
2^{3}\binom{n-2}{2}\binom{8}{6,2}2^{2}(n-2)\binom{8}{6,2}2\binom{8}{6,2}2(2^{6}+2^{2})\binom{8}{6,2}0\binom{4}{3,1}\\\
2^{3}\binom{n-1}{3}2^{2}\binom{n-1}{2}2(n-1)2(n-1)(2^{8}+1)11,$

which, for the system, is equivalent to

$\pNiceMatrix[margin,vlines=6,cell-space-
limits=2pt]\Block[fill=gray!10,borders={top,left,right,bottom}]{1-1}{}8&0000\frac{1}{105}\\\
8(n-3)\Block[fill=gray!10,borders={top,left,right,bottom}]{1-1}{}4000\frac{1}{35}\\\
4(n-2)(n-3)4(n-2)\Block[fill=gray!10,borders={top,left,right,bottom}]{2-2}{}2640\frac{3}{35}\\\
4(n-2)(n-3)4(n-2)21360\frac{1}{7}\\\
4(n-1)(n-2)(n-3)6(n-1)(n-2)6(n-1)1542(n-1)\Block[fill=gray!10,borders={top,left,right,bottom}]{1-1}{}33.$
(2.3)

Since the highlighted blocks in matrix (2.3) are invertible matrices, the
linear system admits a unique solution, which corresponds to the coefficients
of decomposition (2.0). ∎

The crucial fact in determining decomposition (3) is that the entries of the
blocks in matrix (2.3) do not depend on $n$. This allows to obtain an explicit
decomposition for an arbitrary number of variables. However, to generalize
this pattern, it is necessary for all of the blocks to have non-zero
determinant.

As observed in section 1 for decomposition (1.12), one could verify that

$\frac{1}{3}(n^{4}-4n^{3}+14n^{2}-8n)<\frac{1}{n}\binom{n+7}{8}$

if and only if $n>10$. Consequently, we have the following proposition.

######  Proposition 2.3.

Let $n\in\mathbb{N}$ be such that $n>10$. Then

$\operatorname{rk}(q_{n}^{4})<\frac{1}{n}\binom{n+7}{8}.$

In particular, for $n>10$, $\operatorname{rk}(q_{n}^{4})$ is subgeneric.

Before examining another example for the case of exponent $s=5$ we observe
that decomposition (2.0) is not optimal in general. By considering a different
kind of linear forms for the fourth coefficient $c_{4}$, we can determine
another decomposition of lower size, involving roots of unity. Indeed, since
$8$-th powers of linear forms are invariant under the action of $4$-rooth of
unity, we have, in particular,

$(x_{i_{1}}\pm\mathrm{i}x_{i_{2}})^{8}=(\mathrm{i}x_{i_{1}}\pm
x_{i_{2}})^{8}.$

Thus, replacing the fourth summand of equation (2) with this last linear form,
we can repeat the same procedure and obtain the matrix

$\pNiceMatrix[margin,vlines=6,cell-space-
limits=2pt]\Block[fill=gray!10,borders={top,left,right,bottom}]{1-1}{}8&0000\frac{1}{105}\\\
8(n-3)\Block[fill=gray!10,borders={top,left,right,bottom}]{1-1}{}4000\frac{1}{35}\\\
4(n-2)(n-3)4(n-2)\Block[fill=gray!10,borders={top,left,right,bottom}]{2-2}{}220\frac{3}{35}\\\
4(n-2)(n-3)4(n-2)2-20\frac{1}{7}\\\
4(n-1)(n-2)(n-3)6(n-1)(n-2)6(n-1)6(n-1)\Block[fill=gray!10,borders={top,left,right,bottom}]{1-1}{}33.$
(2.4)

We therefore obtain decomposition of size

$8\binom{n}{4}+4\binom{n}{3}+4\binom{n}{2}+n=\frac{1}{3}n^{4}-\frac{4}{3}n^{3}+\frac{11}{3}n^{2}-\frac{5}{3}n,$

which represents a slight improvement over decomposition (2.0).

To generalize this behavior, we provide one more example for the case of
exponent $s=5$. The proof is structured exactly like section 2, but in this
case, we have two blocks of size $2$ along the diagonal of the matrix.

######  Proposition 2.4.

For every $n\in\mathbb{N}$, the form $q_{n}^{5}$ can be decomposed as

$\displaystyle 15120q_{n}^{5}$
$\displaystyle=\sum_{i_{1}<i_{2}<i_{3}<i_{4}<i_{5}}(x_{i_{1}}\pm x_{i_{2}}\pm
x_{i_{3}}\pm x_{i_{4}}\pm
x_{i_{5}})^{10}-2(n-7)\sum_{i_{1}<i_{2}<i_{3}<i_{4}}(x_{i_{1}}\pm x_{i_{2}}\pm
x_{i_{3}}\pm x_{i_{4}})^{10}$
$\displaystyle\hphantom{{}={}}+2(n^{2}-13n+36)\sum_{i_{1}<i_{2}<i_{3}}(x_{i_{1}}\pm
x_{i_{2}}\pm
x_{i_{3}})^{10}+\frac{2}{3}\sum_{i_{1}<i_{2}<i_{3}}\bigl{[}(2x_{i_{1}}\pm
x_{i_{2}}\pm x_{i_{3}})^{10}$ $\displaystyle\hphantom{{}={}}+(x_{i_{1}}\pm
2x_{i_{2}}\pm x_{i_{3}})^{10}+(x_{i_{1}}\pm x_{i_{2}}\pm
2x_{i_{3}})^{10}\bigr{]}+\frac{4}{3}(n^{3}-18n^{2}+90n-226)\sum_{i_{1}<i_{2}}(x_{i_{1}}\pm
x_{i_{2}})^{10}$
$\displaystyle\hphantom{{}={}}-\frac{4}{3}(n-4)\sum_{i_{1}<i_{2}}\bigl{[}(2x_{i_{1}}\pm
x_{i_{2}})^{10}+(x_{i_{1}}\pm 2x_{i_{2}})^{10}\bigr{]}$
$\displaystyle\hphantom{{}={}}+\frac{2}{3}(n^{4}-22n^{3}+2195n^{2}-15086n+35592)\sum_{i_{1}}x_{i_{1}}^{10},$
(2.5)

which is a decomposition of size

$16\binom{n}{5}+8\binom{n}{4}+16\binom{n}{3}+6\binom{n}{2}+n=\frac{2}{15}n^{5}-n^{4}+\frac{16}{3}n^{3}-8n^{2}+\frac{68}{15}n$

for $n\geq 5$ and $n\neq 7$, and of size $1029$ for $n=7$.

###### Proof.

In this case, the polynomials involved for the decomposition are seven. These
are

$\displaystyle
M_{(2,2,2,2,2)}=\sum_{i_{1}<i_{2}<i_{3}<i_{4}<i_{5}}x_{i_{1}}^{2}x_{i_{2}}^{2}x_{i_{3}}^{2}x_{i_{4}}^{2}x_{i_{5}}^{2},$
$\displaystyle
M_{(4,2,2,2)}=\sum_{i_{1}<i_{2}<i_{3}<i_{4}}(x_{i_{1}}^{4}x_{i_{2}}^{2}x_{i_{3}}^{2}x_{i_{4}}^{2}+x_{i_{1}}^{2}x_{i_{2}}^{4}x_{i_{3}}^{2}x_{i_{4}}^{2}+x_{i_{1}}^{2}x_{i_{2}}^{2}x_{i_{3}}^{4}x_{i_{4}}^{2}+x_{i_{1}}^{2}x_{i_{2}}^{2}x_{i_{3}}^{2}x_{i_{4}}^{4}),$
$\displaystyle
M_{(4,4,2)}=\sum_{i_{1}<i_{2}<i_{3}}(x_{i_{1}}^{4}x_{i_{2}}^{4}x_{i_{3}}^{2}+x_{i_{1}}^{4}x_{i_{2}}^{2}x_{i_{3}}^{4}+x_{i_{1}}^{2}x_{i_{2}}^{4}x_{i_{3}}^{4}),\quad
M_{(6,2,2)}=\sum_{i_{1}<i_{2}<i_{3}}(x_{i_{1}}^{6}x_{i_{2}}^{2}x_{i_{3}}^{2}+x_{i_{1}}^{2}x_{i_{2}}^{6}x_{i_{3}}^{2}+x_{i_{1}}^{2}x_{i_{2}}^{2}x_{i_{3}}^{6}),$
$\displaystyle
M_{(6,4)}=\sum_{i_{1}<i_{2}}(x_{i_{1}}^{6}x_{i_{2}}^{4}+x_{i_{1}}^{4}x_{i_{2}}^{6}),\quad
M_{(8,2)}=\sum_{i_{1}<i_{2}}(x_{i_{1}}^{8}x_{i_{2}}^{2}+x_{i_{1}}^{2}x_{i_{2}}^{8}),\quad
M_{(10)}=\sum_{i_{1}}x_{i_{1}}^{10}.$

We can determine seven values $c_{1},\dots,c_{7}$ such that

$\displaystyle q_{n}^{5}$
$\displaystyle=c_{1}\sum_{i_{1}<i_{2}<i_{3}<i_{4}<i_{5}}(x_{i_{1}}\pm
x_{i_{2}}\pm x_{i_{3}}\pm x_{i_{4}}\pm
x_{i_{5}})^{10}+c_{2}\sum_{i_{1}<i_{2}<i_{3}<i_{4}}(x_{i_{1}}\pm x_{i_{2}}\pm
x_{i_{3}}\pm x_{i_{4}})^{10}$
$\displaystyle\hphantom{{}={}}+c_{3}\sum_{i_{1}<i_{2}<i_{3}}(x_{i_{1}}\pm
x_{i_{2}}\pm
x_{i_{3}})^{10}+c_{4}\sum_{i_{1}<i_{2}<i_{3}}\bigl{(}(2x_{i_{1}}\pm
x_{i_{2}}\pm x_{i_{3}})^{10}+(x_{i_{1}}\pm 2x_{i_{2}}\pm x_{i_{3}})^{10}$
$\displaystyle\hphantom{{}={}}+(x_{i_{1}}\pm x_{i_{2}}\pm
2x_{i_{3}})^{10}\bigr{)}+c_{5}\sum_{i_{1}<i_{2}}(x_{i_{1}}\pm
x_{i_{2}})^{10}+c_{6}\sum_{i_{1}<i_{2}}\bigl{(}(2x_{i_{1}}\pm
x_{i_{2}})^{10}+(x_{i_{1}}\pm
2x_{i_{2}})^{10}\bigr{)}+c_{7}\sum_{i_{1}}x_{i_{1}}^{10}.$

By comparing the coefficients, we obtain a linear system associated to the
matrix

$\pNiceMatrix[margin,vlines=8]\Block[fill=gray!10,borders={top,left,right,bottom}]{1-1}{}16&000000\binom{10}{2,2,2,2,2}^{-1}\binom{5}{1,1,1,1,1}\\\
16(n-4)\Block[fill=gray!10,borders={top,left,right,bottom}]{1-1}{}800000\binom{10}{4,2,2,2}^{-1}\binom{5}{2,1,1,1}\\\
16\binom{n-3}{2}8(n-3)\Block[fill=gray!10,borders={top,left,right,bottom}]{2-2}{}4144000\binom{10}{4,4,2}^{-1}\binom{5}{2,2,1}\\\
16\binom{n-3}{2}8(n-3)4288000\binom{10}{6,2,2}^{-1}\binom{5}{3,1,1}\\\
16\binom{n-2}{3}8\binom{n-2}{2}4(n-2)324(n-2)\Block[fill=gray!10,borders={top,left,right,bottom}]{2-2}{}21600\binom{10}{6,4}^{-1}\binom{5}{3,2}\\\
16\binom{n-2}{3}8\binom{n-2}{2}4(n-2)1044(n-2)25200\binom{10}{8,2}^{-1}\binom{5}{4,1}\\\
16\binom{n-1}{4}8\binom{n-1}{3}4\binom{n-1}{2}2052\binom{n-1}{2}2(n-1)2050(n-1)\Block[fill=gray!10,borders={top,left,right,bottom}]{1-1}{}11$
(2.6)

and solving it, we get the required decomposition. ∎

Again, we can compute the minimum number of variables that guarantees the size
of the decomposition is subgeneric.

######  Proposition 2.6.

Let $n\in\mathbb{N}$ be such that $n>8$. Then

$\operatorname{rk}(q_{n}^{5})<\frac{1}{n}\binom{n+9}{10}.$

In particular, for $n>8$, $\operatorname{rk}(q_{n}^{5})$ is subgeneric.

##  Asymptotic growth and upper bounds for Waring rank

To obtain a pattern similar to what we have seen in matrices (2.3), (2.4), and
(2.6), we need to construct suitable decompositions that provide us with a
sufficiently high number of equations. This will ensure the existence of
square matrices along the diagonal with non-zero determinants, i.e., the
highlighted blocks we have seen before in matrices (1.11), (1.13), (2.3),
(2.4), and (2.6). To prove this, we start with a lemma for matrices of
polynomials.

######  Lemma 3.0.

For every $n,s\in\mathbb{N}$, let
$f_{1},\dots,f_{s}\in\mathbb{C}[x_{1},\dots,x_{n}]$ be linearly independent
polynomials and let

$g\in\mathbb{C}[x_{ij}]_{i=1,\dots,s,\,j=1,\dots,n}$

be the polynomial defined as

$g(x_{11},\dots,x_{1n},\dots,x_{s1},\dots,x_{sn})=\det\pNiceMatrix
f_{1}(x_{11},\dots,x_{1n})&\Cdots f_{1}(x_{s1},\dots,x_{sn})\\\
\Vdots\Ddots\Vdots\\\ f_{s}(x_{11},\dots,x_{1n})\Cdots
f_{s}(x_{s1},\dots,x_{sn}).$

Then there exist $s$ points
$\mathbf{a}_{1},\dots,\mathbf{a}_{s}\in\mathbb{C}^{n}$ such that

$g(\mathbf{a}_{1},\dots,\mathbf{a}_{s})\neq 0,$

namely, $g\not\equiv 0$.

###### Proof.

We prove the statement by induction on $s$. For $s=1$, the proof is trivial,
since by linear independence we must have $f_{1}\not\equiv 0$. So, let us
suppose the statement is true for $s-1$ and consider the polynomial matrix

$A(x_{11},\dots,x_{1n},\dots,x_{s1},\dots,x_{sn})=\pNiceMatrix
f_{1}(x_{11},\dots,x_{1n})&\Cdots f_{1}(x_{s1},\dots,x_{sn})\\\
\Vdots\Ddots\Vdots\\\ f_{s}(x_{11},\dots,x_{1n})\Cdots
f_{s}(x_{s1},\dots,x_{sn}).$

Let us further consider the polynomials

$g_{k}\in\mathbb{C}[x_{ij}]_{i=1,\dots,s-1,\,j=1,\dots,n},$

defined as

$g_{k}=(-1)^{k+1}\det
A_{kn}(x_{11},\dots,x_{1n},\dots,x_{(s-1)1},\dots,x_{(s-1)n})$

for every $k=1,\dots,s$, where $A_{kn}$ is the $(s-1)\times(s-1)$ matrix
obtained by removing the $k$-th row and the $s$-th column from the matrix $A$.
By the inductive hypothesis, we have that $g_{k}\not\equiv 0$, which means

$Z(g_{k})\neq\mathbb{C}^{n(s-1)},$

and hence,

$D(g_{k})=\mathbb{C}^{n(s-1)}\setminus Z(g_{k})$

is a non-empty open set for every $k=1,\dots,s$. Therefore, we have

$\bigcap_{k=1}^{s}D(g_{k})\neq\varnothing$

and we can select a point

$\mathbf{a}=(\mathbf{a}_{1},\cdots,\mathbf{a}_{s-1})\in\mathbb{C}^{n(s-1)}$

such that $g_{k}(\mathbf{a})\neq 0$ for every $k=1,\dots,s$. Now, considering
the polynomial

$h(x_{s1},\dots,x_{sn})=g(\mathbf{a}_{1},\dots,\mathbf{a}_{s-1},x_{s1},\dots,x_{sn})=\sum_{k=1}^{s}g_{k}(\mathbf{a}_{1},\dots,\mathbf{a}_{s-1})f_{k}(x_{s1},\dots,x_{sn}),$

we have by the linear independence of the polynomials $f_{1},\dots,f_{s}$,
that

$h(x_{s1},\dots,x_{sn})\not\equiv 0.$

Therefore, we can select a point $\mathbf{a}_{s}\in\mathbb{C}^{n}$ such that

$h(\mathbf{a}_{s})=g(\mathbf{a}_{1},\dots,\mathbf{a}_{s})\neq 0,$

proving the statement. ∎

The main blocks we want to consider have sizes equal to the different numbers
of $k$-partitions of $s$, as they correspond to the number of monomials with
exactly $k$ variables. In the following, we denote by $\mathbb{N}_{0}^{k}$ the
set of $k$-tuple having strictly positive coordinates, for every
$k\in\mathbb{N}$.

######  Theorem 3.0.

For every $n,s\in\mathbb{N}$

$\operatorname{rk}(q_{n}^{s})\leq\sum_{k=1}^{s}2^{k}k!\operatorname{p}_{k}(s)\binom{n}{k}.$

###### Proof.

As we have observed for the initial cases above, we want to consider a sum of
powers of linear form while maintaining strong symmetry among all the the
variables. For every $k=1,\dots,s$ and for every point
$\mathbf{a}=(a_{1},\dots,a_{k})\in\mathbb{C}^{k}$, we introduce the polynomial
$f_{k,\mathbf{a}}\in\mathbb{K}[x_{1},\dots,x_{n}]$, defined as

$f_{k,\mathbf{a}}=\frac{1}{\lvert(\mathfrak{S}_{k})_{\mathbf{a}}\rvert}\sum_{t_{1}<\cdots<t_{k}}\sum_{\sigma\in\mathfrak{S}_{k}}(a_{\sigma(1)}x_{t_{1}}\pm\cdots\pm
a_{\sigma(k)}x_{t_{k}})^{2s}.$

Here, we divide the sum by the quantity
$\lvert(\mathfrak{S}_{k})_{\mathbf{a}}\rvert$, as the action of the elements
in $\mathfrak{S}_{k}$ on $\mathbf{a}$ provides all the possible permutations
of $\mathbf{a}$ without repetitions. This structure was observed in formula
(2), where we used the points

$(1,1,1,1),\quad(1,1,1),\quad(1,1),\quad(2,1),\quad(1),$

and in formula (2.4), where we used the points

$(1,1,1,1,1),\quad(1,1,1,1),\quad(1,1,1),\quad(2,1,1),\quad(1,1),\quad(2,1),\quad(1).$

We develop the summations to separate monomials having the exponents in the
same permutation class. We get

$\displaystyle f_{k,\mathbf{a}}$
$\displaystyle=\frac{1}{\lvert(\mathfrak{S}_{k})_{\mathbf{a}})\rvert}\sum_{t_{1}<\cdots<t_{k}}\sum_{\sigma\in\mathfrak{S}_{k}}\sum_{\begin{subarray}{c}\mathbf{m}\in\mathbb{N}^{k}\\\\[1.0pt]
\lvert\mathbf{m}\rvert=s\end{subarray}}2^{k-1}\binom{2s}{2m_{1},\dots,2m_{k}}\prod_{i=1}^{k}a_{\sigma(i)}^{2m_{i}}x_{t_{i}}^{2m_{i}}$
$\displaystyle=\frac{1}{\lvert(\mathfrak{S}_{k})_{\mathbf{a}}\rvert}\sum_{t_{1}<\cdots<t_{k}}\sum_{\sigma\in\mathfrak{S}_{k}}\sum_{\lambda=1}^{k}\sum_{\begin{subarray}{c}\mathbf{m}\in\mathbb{N}_{0}^{\lambda}\\\
\lvert\mathbf{m}\rvert=s\end{subarray}}\sum_{1\leq
s_{1}<\cdots<s_{\lambda}\leq
k}2^{k-1}\binom{2s}{2m_{1},\dots,2m_{\lambda}}\prod_{i=1}^{\lambda}a_{\sigma(s_{i})}^{2m_{i}}x_{t_{s_{i}}}^{2m_{i}},$

where in the second equality, we have separated the monomials having a
different number $\lambda$ of non-zero values appearing in the $k$-tuple
$(m_{1},\dots,m_{k})$. We can also permute some summations, giving

$\displaystyle f_{k,\mathbf{a}}$
$\displaystyle=\frac{1}{\lvert(\mathfrak{S}_{k})_{\mathbf{a}}\rvert}\sum_{\lambda=1}^{k}\sum_{\begin{subarray}{c}\mathbf{m}\in\mathbb{N}_{0}^{\lambda}\\\
\lvert\mathbf{m}\rvert=s\end{subarray}}\sum_{t_{1}<\cdots<t_{k}}\sum_{1\leq
s_{1}<\cdots<s_{\lambda}\leq
k}\sum_{\sigma\in\mathfrak{S}_{k}}2^{k-1}\binom{2s}{2m_{1},\dots,2m_{\lambda}}\prod_{i=1}^{\lambda}a_{\sigma(s_{i})}^{2m_{i}}x_{t_{s_{i}}}^{2m_{i}}$
$\displaystyle=\sum_{\lambda=1}^{k}\sum_{\begin{subarray}{c}\mathbf{m}\in\mathbb{N}_{0}^{\lambda}\\\
\lvert\mathbf{m}\rvert=s\end{subarray}}\sum_{t_{1}<\cdots<t_{k}}\sum_{1\leq
s_{1}<\cdots<s_{\lambda}\leq
k}2^{k-1}\binom{2s}{2m_{1},\dots,2m_{\lambda}}\biggl{(}\frac{1}{\lvert(\mathfrak{S}_{k})_{\mathbf{a}}\rvert}\sum_{\sigma\in\mathfrak{S}_{k}}\prod_{i=1}^{\lambda}a_{\sigma(s_{i})}^{2m_{i}}\biggr{)}\prod_{j=1}^{\lambda}x_{t_{s_{j}}}^{2m_{j}}.$

The variables $x_{t_{s_{1}}},\cdots,x_{t_{s_{\lambda}}}$ appear among the
variables $x_{t_{1}},\dots,x_{t_{k}}$ a number of times equal to the binomial
coefficient

$\binom{n-\lambda}{k-\lambda}.$

Thus, we can remove the summation depending on $s_{1},\dots,s_{\lambda}$ and
write

$\displaystyle f_{k,\mathbf{a}}$
$\displaystyle=\sum_{\lambda=1}^{k}\sum_{\begin{subarray}{c}\mathbf{m}\in\mathbb{N}_{0}^{\lambda}\\\
\lvert\mathbf{m}\rvert=s\end{subarray}}\sum_{t_{1}<\cdots<t_{\lambda}}2^{k-1}\binom{n-\lambda}{k-\lambda}\binom{2s}{2m_{1},\dots,2m_{\lambda}}\biggl{(}\frac{1}{\lvert(\mathfrak{S}_{k})_{\mathbf{a}}\rvert}\sum_{\sigma\in\mathfrak{S}_{k}}\prod_{i=1}^{\lambda}a_{\sigma(i)}^{2m_{i}}\biggr{)}\prod_{i=1}^{\lambda}x_{t_{i}}^{2m_{i}}$
$\displaystyle=\sum_{\lambda=1}^{k}\sum_{\begin{subarray}{c}\mathbf{m}\in\mathbb{N}_{0}^{\lambda}\\\
\lvert\mathbf{m}\rvert=s\end{subarray}}2^{k-1}\binom{n-\lambda}{k-\lambda}\binom{2s}{2m_{1},\dots,2m_{\lambda}}\biggl{(}\frac{1}{\lvert(\mathfrak{S}_{k})_{\mathbf{a}}\rvert}\sum_{\sigma\in\mathfrak{S}_{k}}\prod_{i=1}^{\lambda}a_{\sigma(i)}^{2m_{i}}\biggr{)}\sum_{t_{1}<\cdots<t_{\lambda}}\prod_{i=1}^{\lambda}x_{t_{i}}^{2m_{i}}.$

Finally, we can gather the multi-indices representing the same partition of
order $\lambda$ of the number $k$. To do this, it is sufficient to consider an
element $\mathbf{m}=(m_{1},\dots,m_{\lambda})$, such that $m_{1}\geq\cdots\geq
m_{\lambda}>0$, and its orbit under the action of the permutation group
$\mathfrak{S}_{\lambda}$. We can decompose the polynomial $f_{k,\mathbf{a}}$
into more summands by distinguishing the order of each partition of $s$.
Hence, we get

$f_{k,\mathbf{a}}=\sum_{\lambda=1}^{k}\sum_{\mathbf{m}\in\mathcal{P}_{\lambda}(s)}\sum_{\upgamma\in\mathfrak{S}_{\lambda}}\frac{2^{k-1}}{\lvert(\mathfrak{S}_{\lambda})_{\mathbf{m}}\rvert}\binom{n-\lambda}{k-\lambda}\binom{2s}{2m_{1},\dots,2m_{\lambda}}\biggl{(}\frac{1}{\lvert(\mathfrak{S}_{k})_{\mathbf{a}}\rvert}\sum_{\sigma\in\mathfrak{S}_{k}}\prod_{i=1}^{\lambda}a_{\sigma(i)}^{2m_{\gamma(i)}}\biggr{)}\sum_{t_{1}<\cdots<t_{\lambda}}\prod_{i=1}^{\lambda}x_{t_{i}}^{2m_{\gamma(i)}}.$

Considering the summation

$\sum_{\sigma\in\mathfrak{S}_{k}}\prod_{i=1}^{\lambda}a_{\sigma(i)}^{2m_{\gamma(i)}},$

since the permutation $\gamma$ just permutes the elements $1,\dots,\lambda$,
it varies among all the possible choices of the elements $a_{1},\dots,a_{k}$
and, in particular, it is invariant under the action of
$\mathfrak{S}_{\lambda}$. Therefore, we have

$\sum_{\sigma\in\mathfrak{S}_{k}}\prod_{i=1}^{\lambda}a_{\sigma(i)}^{2m_{\gamma(i)}}=\sum_{\sigma\in\mathfrak{S}_{k}}\prod_{i=1}^{\lambda}a_{\sigma\gamma^{-1}(i)}^{2m_{i}}=\sum_{\sigma\in\mathfrak{S}_{k}}\prod_{i=1}^{\lambda}a_{\sigma(i)}^{2m_{i}}.$

and, by formula (1.8), we obtain

$\displaystyle f_{k,\mathbf{a}}$
$\displaystyle=\sum_{\lambda=1}^{k}\sum_{\mathbf{m}\in\mathcal{P}_{\lambda}(s)}\frac{2^{k-1}}{\lvert(\mathfrak{S}_{\lambda})_{\mathbf{m}}\rvert}\binom{n-\lambda}{k-\lambda}\binom{2s}{2m_{1},\dots,2m_{\lambda}}\biggl{(}\frac{1}{\lvert(\mathfrak{S}_{k})_{\mathbf{a}}\rvert}\sum_{\sigma\in\mathfrak{S}_{k}}\prod_{i=1}^{\lambda}a_{\sigma(i)}^{2m_{i}}\biggr{)}\sum_{\gamma\in\mathfrak{S}_{\lambda}}\sum_{t_{1}<\cdots<t_{\lambda}}\prod_{i=1}^{\lambda}x_{t_{i}}^{2m_{\gamma(i)}}$
$\displaystyle=\sum_{\lambda=1}^{k}\sum_{\mathbf{m}\in\mathcal{P}_{\lambda}(s)}2^{k-1}\binom{n-\lambda}{k-\lambda}\binom{2s}{2m_{1},\dots,2m_{\lambda}}\biggl{(}\frac{1}{\lvert(\mathfrak{S}_{k})_{\mathbf{a}}\rvert}\sum_{\sigma\in\mathfrak{S}_{k}}\prod_{i=1}^{\lambda}a_{\sigma(i)}^{2m_{i}}\biggr{)}M_{2\mathbf{m}}.$
(3.1)

Thus, we have written the polynomial $f_{k,\mathbf{a}}$ as a linear
combination of the ones given by the sum of monomials belonging to the same
permutation class. Since these are the same appearing in the development of
the form $q_{n}^{s}$ in formula (1), we need to satisfy exactly
$\operatorname{p}(s)$ conditions.

We establish these conditions by defining a polynomial $f_{k,\mathbf{a}_{j}}$
for every $j=1,\dots,\operatorname{p}_{k}(s)$ and repeating the same process
for every $k=1,\dots,s$. This gives us a number of equations equal to

$\operatorname{p}(s)=\sum_{k=1}^{s}\operatorname{p}_{k}(s).$

In other words, we need to guarantee the existence, for every $k=1,\dots,s$,
of a set of points

$\mathcal{A}_{k}=\\{\mathbf{a}_{k,1},\dots,\mathbf{a}_{k,\operatorname{p}_{k}(s)}\\},$

such that

$q_{n}^{s}=\sum_{k=1}^{s}\sum_{j=1}^{\operatorname{p}_{k}(s)}c_{k,j}f_{k,\mathbf{a}_{k,j}},$
(3.2)

for some coefficients $c_{k,j}\in\mathbb{C}$. To find these points, we denote
each coordinate of the points by considering
$\mathbf{a}_{k,j}=(a_{k,j,1},\dots,a_{k,j,k})$ for every
$j=1,\dots,\operatorname{p}_{k}(s)$. Formula (1) tells us that

$q_{n}^{s}=\sum_{k=1}^{s}\sum_{\mathbf{m}\in\mathcal{P}_{k}(s)}\binom{s}{m_{1},\dots,m_{k}}M_{2\mathbf{m}}.$

Hence, for every $k=1,\dots,s$, we combine formulas (1), (3) and (3.2), to
obtain a linear equation for each partition $\mathbf{m}\in\mathcal{P}_{k}(s)$.
This equation arises by equalizing the coefficients of the polynomial
$M_{2\mathbf{m}}$, leading to

$\sum_{\lambda=k}^{s}\sum_{j_{\lambda}=1}^{\operatorname{p}_{\lambda}(s)}c_{\lambda,j_{\lambda}}\biggl{(}\frac{1}{\lvert(\mathfrak{S}_{k})_{\mathbf{a}_{\lambda,j_{\lambda}}}\rvert}\sum_{\sigma\in\mathfrak{S}_{\lambda}}\prod_{i=1}^{\lambda}a_{\lambda,j_{\lambda},\sigma(i)}^{2m_{i}}\biggr{)}=2^{1-k}\binom{n-\lambda}{k-\lambda}^{-1}\binom{2s}{2m_{1},\dots,2m_{\lambda}}^{-1}\binom{s}{m_{1},\dots,m_{\lambda}}.$
(3.3)

Formula (3.2) provides a linear system of exactly $\operatorname{p}(s)$
equations of the type of formula (3.3). The system has $\operatorname{p}(s)$
unknowns, corresponding to the coefficients $c_{k,j_{k}}$ for $k=1,\dots,s$
and $j_{k}=1,\dots,\operatorname{p}_{k}(s)$. Denote each of the values
multiplying coefficients $c_{k,j_{k}}$ as

$h_{k,j_{k},\lambda}=h_{k,j_{k}}(\mathbf{m}_{\lambda})=\frac{1}{\lvert(\mathfrak{S}_{k})_{\mathbf{a}_{k,j}}\rvert}\sum_{\sigma\in\mathfrak{S}_{k}}\prod_{i=1}^{k}a_{k,j,\sigma(i)}^{2m_{\lambda,i}},$
(3.4)

where
$\mathbf{m}_{1},\dots\mathbf{m}_{\operatorname{p}_{k}(s)}\in\mathbb{N}_{0}^{k}$
are the $k$-partitions in $\mathcal{P}_{k}(s)$. Now, the aim is to establish
in which cases the linear system admits a solution. Thus we want to determine
whenever the associated matrix is invertible. This is given by

$\pNiceMatrix[margin,parallelize-
diags=false]\Block[fill=gray!10,borders={top,left,right,bottom}]{1-1}{}h_{s,1,1}&0\Cdots
0\\\
*\Block[fill=gray!10,borders={top,left,right,bottom}]{1-1}{}h_{s-1,1,1}0\Cdots
0\\\ *\ddots\\\
\Vdots\Vdots\Block[fill=gray!10,borders={top,left,right,bottom}]{3-3}{}h_{k,1,1}\Cdots
h_{k,\operatorname{p}_{k}(s),1}\Vdots\\\ \Vdots\Ddots\Vdots\\\
h_{k,1,\operatorname{p}_{k}(s)}\Cdots
h_{k,\operatorname{p}_{k}(s),\operatorname{p}_{k}(s)}\\\ \ddots 0\\\
**\Cdots*\Block[fill=gray!10,borders={top,left,right,bottom}]{1-1}{}h_{1,1,1}$

with several blocks on the diagonal of order $\operatorname{p}_{k}(s)$ for
each $k=1,\dots,s$. In particular, the matrix is invertible if and only if
each block has a non-zero determinant. For simplicity, we can assume that the
coordinates of the points involved are pairwise distinct. This is possible by
the fact that the set of points not annihilating each determinant is open.
Nonetheless, the number of points used to obtain a decomposition of this kind
can be much higher than the minimal ones. For each $k=1,\dots,s$, the elements
$h_{k,j_{k},\lambda}$ can be treated as polynomials with variables
$a_{k,j_{k},1},\dots,a_{k,j_{k},k}$. Now, let us consider the polynomials

$H_{k,\lambda}(x_{1},\dots,x_{k})=\frac{1}{\lvert(\mathfrak{S}_{k})_{\mathbf{a}_{k,j}}\rvert}\sum_{\sigma\in\mathfrak{S}_{k}}\prod_{i=1}^{k}x_{\sigma(i)}^{2m_{\lambda,i}},$

with $\lambda=1,\dots,\mathrm{p}_{k}(s)$, representing the $k$-partitions in
$\mathcal{P}_{k}(s)$. These polynomials obtained by substituting the variables
$x_{1},\dots,x_{k}$ to $a_{k,j_{k},1},\dots,a_{k,j_{k},k}$, that is,

$h_{k,j_{k},\lambda}=H_{k,\lambda}(a_{k,j_{k},1},\dots,a_{k,j_{k},k}),$

for every point $\mathbf{a}_{k,j_{k}}\in\mathbb{C}^{k}$, with
$j_{k}=1,\dots,k$. We want to prove that the polynomials $H_{k,\lambda}$ are
linearly independent. So, let $c_{1},\dots,c_{\mathrm{p}_{k}(s)}\in\mathbb{C}$
be coefficients such that

$c_{1}H_{k,1}+\cdots+c_{\mathrm{p}_{k}(s)}H_{k,\mathrm{p}_{k}(s)}=0.$

Every monomial appearing in this linear combination involves exactly $k$
variables. Furthermore, by definition of $h_{k,j_{k},\lambda}$, seen in
formula (3.4), each of these monomials appears in only one of the polynomials
$H_{k,\lambda}$. Indeed, every $H_{k,\lambda}$ only involves monomials of the
same permutations class induced by the $k$-partition
$\mathbf{m}_{\lambda}\in\mathcal{P}_{k}(s)$. Therefore, as the permutation
classes of monomials form a partition of the set of monomials with $k$
variables, the zero polynomial is obtained only if

$c_{1}=\cdots=c_{\mathrm{p}_{k}(s)}=0,$

which means that the polynomials $H_{k,\lambda}$ are linearly independent.
Hence, we obtain the statement simply by applying section 3. ∎

The upper bound provided by section 3 can be improved by considering just one
point for the partitions of order $s$ and $s-1$, as these are unique and hence
have just one condition to satisfy. By choosing the points

$\mathbf{a}_{s,1}=\mathbf{a}_{s-1,1}=(1,\dots,1),$

we obtain the following corollary.

######  Corollary 3.4.

For every $n,s\in\mathbb{N}$

$\operatorname{rk}(q_{n}^{s})\leq
2^{s-1}\binom{n}{s}+2^{s-2}\binom{n}{s-1}+\sum_{k=1}^{s-2}2^{k-1}k!\operatorname{p}_{k}(s)\binom{n}{k}.$
(3.5)

In particular, for any $s\in\mathbb{N}$, the rank of $q_{n}^{s}$ grows at most
as $n^{s}$ when $n\to+\infty$.

By combining section 1 and section 3, we obtain the following corollary.

######  Corollary 3.5.

For every $s\in\mathbb{N}$,

$\lim_{n\to+\infty}\log_{n}\bigl{(}\operatorname{rk}(q_{n}^{s})\bigr{)}=\lim_{n\to+\infty}\log_{n}\bigl{(}\operatorname{brk}(q_{n}^{s})\bigr{)}=s.$

###### Proof.

We have

$\binom{s+n-1}{s}=\frac{1}{s!}\prod_{j=1}^{s}(n-1+j)=\frac{1}{s!}\bigl{(}n^{s}+f(n)\bigr{)},$

where $f\in\mathbb{C}[n]$ is a polynomial of degree $s-1$. Thus, since

$\lim_{n\to+\infty}\frac{n^{s}}{n^{s}+f(n)}=1,$

we get

$\lim_{n\to+\infty}\log_{n}\Biggl{(}\binom{s+n-1}{s}\Biggr{)}=\lim_{n\to+\infty}\log_{n}\biggl{(}\frac{n^{s}}{s!}\biggr{)}=s.$

If we consider now the second member of inequality (3.5), we get

$2^{s-1}\binom{n}{s}+2^{s-2}\binom{n}{s-1}+\sum_{k=1}^{s-2}2^{k-1}k!\operatorname{p}_{k}(s)\binom{n}{k}\leq
2^{s-1}\binom{n}{s}+2^{s-2}s!\binom{n}{s-1}\leq\frac{2^{s-1}}{s!}n^{s}+2^{s-2}sn^{s-1}.$

As above, we have

$\lim_{n\to+\infty}\frac{n^{s}}{n^{s}+\dfrac{s}{2}s!n^{s-1}}=1$

and hence

$\lim_{n\to+\infty}\log_{n}\Biggl{(}2^{s-1}\binom{n}{s}+2^{s-2}\binom{n}{s-1}+\sum_{k=1}^{s-2}2^{k-1}k!\operatorname{p}_{k}(s)\binom{n}{k}\Biggr{)}\leq\lim_{n\to+\infty}\log_{n}\biggl{(}\frac{2^{s-1}}{s!}n^{s}\biggr{)}=s.$

Therefore, by section 1 and section 3, we get the statement. ∎

##  Subgeneric rank

We have observed from section 2 that the size of decomposition (2.4) is
subgeneric if $n>8$. Although it is not easy to determine the exact minimal
value for which the higher numbers only provide subgeneric rank, we can
estimate that this value is quite low. Indeed, according to section 3 and
formula (1.7), we obtain the inequality

$\displaystyle\operatorname{rk}(q_{n}^{s})$ $\displaystyle\leq
2^{s-1}\binom{n}{s}+2^{s-2}\binom{n}{s-1}+\sum_{k=1}^{s-2}2^{k-1}k!\operatorname{p}_{k}(s)\binom{n}{k}$
$\displaystyle\leq
2^{s-1}\binom{n}{s}+2^{s-2}\binom{n}{s-1}+\sum_{k=1}^{s-2}\frac{2^{k-1}}{(k-1)!}\frac{(s-1)!n!}{(s-k)!(n-k)!}.$
(4.1)

Meanwhile, the generic rank is equal to

$\frac{(2s+n-1)!}{(2s)!n!}.$

Hence, in the case of $n\to+\infty$, the value of the generic rank grows as
$n^{2s-1}$. However, the upper bound in formula (4) tells us that the rank of
$q_{n}^{s}$ grows more slowly than $n^{s}$ as $n\to+\infty$.

Following the proof of section 3, the objective is to determine some points
$\mathbf{a}_{k,1},\dots,\mathbf{a}_{k,\operatorname{p}_{k}(s)}$ for every
block matrix

$\pNiceMatrix g_{k,1}(\mathbf{a}_{k,1})&\Cdots
g_{k,1}\bigl{(}\mathbf{a}_{k,\operatorname{p}_{k}(s)}\bigr{)}\\\
\Vdots\Ddots\Vdots\\\ g_{k,\operatorname{p}_{k}(s)}(\mathbf{a}_{k,1})\Cdots
g_{k,\operatorname{p}_{k}(s)}\bigl{(}\mathbf{a}_{k,\operatorname{p}_{k}(s)}\bigr{)}$

such that its determinant is non-zero. Obviously, it would be convenient to
choose as few distinct points as possible to get the smallest possible size of
such decompositions. Although it can be easy for small values of $k$, it
becomes more difficult for higher sizes.

Let us consider the upper bound of section 3, given by

$\sum_{k=1}^{s}2^{k}k!\operatorname{p}_{k}(s)\binom{n}{k}.$ (4.2)

All the elements involved in the summation are quantities that have been
extensively studied in the literature and for which many estimates and
approximations have been provided. A more precise estimate of the values
involved in formula (4.2) can be made using the classical and well-known
Stirling’s approximation of the factorial of an arbitrary natural number $n$
(see [Sti30], or [Twe03] for a more recent version of the proof). This
approximation shows that

$\sqrt{2\uppi}n^{n+\frac{1}{2}}\mathrm{e}^{-n+\frac{1}{12n+1}}<n!<\sqrt{2\uppi}n^{n+\frac{1}{2}}\mathrm{e}^{-n+\frac{1}{12n}}$
(4.3)

and hence, considering just the inverse of $n!$, we have

$\frac{1}{\sqrt{2\uppi}}n^{-n-\frac{1}{2}}\mathrm{e}^{n-\frac{1}{12n}}<\frac{1}{n!}<\frac{1}{\sqrt{2\uppi}}n^{-n-\frac{1}{2}}\mathrm{e}^{n-\frac{1}{12n+1}}.$
(4.4)

Now, considering the binomial coefficient

$\binom{n}{k}=\frac{n!}{k!(n-k)!},$

we can use relations (4.3) and (4.4) to provide the bounds

$\frac{1}{\sqrt{2\uppi
k}}\biggl{(}\frac{n}{n-k}\biggr{)}^{n+\frac{1}{2}}\biggl{(}\frac{n-k}{k}\biggr{)}^{k}\mathrm{e}^{\frac{1}{12n+1}-\frac{n}{12k(n-k)}}<\left(\begin{array}[]{c}n\\\
k\end{array}\right)<\frac{1}{\sqrt{2\uppi
k}}\biggl{(}\frac{n}{n-k}\biggr{)}^{n+\frac{1}{2}}\biggl{(}\frac{n-k}{k}\biggr{)}^{k}\mathrm{e}^{\frac{1}{12n}-\frac{12n+1}{(12k+1)(12(n-k)+1)}}.$
(4.5)

Finally, we also provide an upper bound for the partition function
$\operatorname{p}(s)$. D. M. Kane shows in [Kan06]*Remark 1 that for every
$n\in\mathbb{N}$, we get

$\frac{C^{-}_{1,1}}{n}\mathrm{e}^{\uppi\sqrt{\frac{2n}{3}}}\leq\operatorname{p}(n)\leq\frac{C^{+}_{1,1}}{n}\mathrm{e}^{\uppi\sqrt{\frac{2}{3}}},$

where $C_{1,1}^{-}$ and $C_{1,1}^{+}$ are some specified constant values.
These inequalities are used also by A. Y. Oruç in [Oru16], where the author
provides some upper bounds for the function $\operatorname{p}_{k}(n)$, to
which we refer for further details about this. In particular, as explained for
[Oru16]*formula (4), we can suppose $C_{1,1}^{+}=6$, getting

$\operatorname{p}(n)\leq\frac{6}{n}\mathrm{e}^{\uppi\sqrt{\frac{2n}{3}}}$
(4.6)

for every $n\in\mathbb{N}$.

######  Theorem 4.6.

For every $s\in\mathbb{N}$,

$\operatorname{rk}(q_{n}^{s})\leq\frac{1}{n}\binom{2s+n-1}{2s}.$

In particular, the rank of $q_{n}^{s}$ is subgeneric, whenever
$n\geq(2s-1)^{2}$.

###### Proof.

According to section 3, we have the inequality

$\operatorname{rk}(q_{n}^{s})\leq\sum_{k=1}^{s}2^{k}k!\operatorname{p}_{k}(s)\binom{n}{k}\leq
2^{s}s!\binom{n}{s}\sum_{k=1}^{s}\operatorname{p}_{k}(s)=\frac{2^{s}n!}{(n-s)!}\operatorname{p}(s).$
(4.7)

By employing relations (4.3), (4.4), (4.5) and (4.6), we can write

$\frac{2^{s}n!}{(n-s)!}\operatorname{p}(s)\leq\frac{6\cdot
2^{s}}{s}\biggl{(}\frac{n}{n-s}\biggr{)}^{n+\frac{1}{2}}(n-s)^{s}\mathrm{e}^{\uppi\sqrt{\frac{2s}{3}}+\frac{1}{12n}-\frac{1}{12(n-s)+1}-s}.$
(4.8)

Furthermore, by relation (4.5), we have inequality

$\frac{1}{n}\binom{2s+n-1}{2s}>\frac{1}{\sqrt{4\uppi
s}}\biggl{(}\frac{2s+n-1}{n-1}\biggr{)}^{2s+n-\frac{1}{2}}\biggl{(}\frac{n-1}{2s}\biggr{)}^{2s}\mathrm{e}^{-\frac{12n^{2}+72ns-23n+48s^{2}-70s+11}{24s(n-1)(12n+24s-11)}}$
(4.9)

Now, we want to determine for which cases the second term of inequality (4.8)
is lower than the second term of inequality (4.9). So, we have

$\frac{6\cdot
2^{s}}{s}\biggl{(}\frac{n}{n-s}\biggr{)}^{n+\frac{1}{2}}(n-s)^{s}\mathrm{e}^{\uppi\sqrt{\frac{2s}{3}}+\frac{1}{12n}-\frac{1}{12(n-s)+1}-s}<\frac{1}{\sqrt{4\uppi
s}}\biggl{(}\frac{2s+n-1}{n-1}\biggr{)}^{2s+n-\frac{1}{2}}\biggl{(}\frac{n-1}{2s}\biggr{)}^{2s}\mathrm{e}^{\frac{1}{12(2s+n-1)+1}-\frac{2s+n-1}{24s(n-1)}}$

if and only if

$\frac{12\cdot
2^{s}\sqrt{\uppi}(n-1)}{\sqrt{s}(2s+n-1)}\biggl{(}\frac{n(n-1)}{(n-s)(2s+n-1)}\biggr{)}^{n+\frac{1}{2}}\biggl{(}\frac{2s\sqrt{n-s}}{2s+n-1}\biggr{)}^{2s}\mathrm{e}^{\uppi\sqrt{\frac{2s}{3}}+\frac{1}{12n}+\frac{2s+n-1}{24s(n-2s)}-\frac{1}{12(2s+n-1)+1}-\frac{1}{12(n-s)+1}-s}<1.$

By performing some computations, we observe that

$\frac{1}{2n}+\frac{2s+n-1}{24s(n-1)}-\frac{1}{12(2s+n-1)+1}<1$

for every $s>1$ and $n>(2s-1)^{2}$. Thus, we can also write

$\frac{12\cdot
2^{s}\sqrt{\uppi}(n-1)}{\sqrt{s}(2s+n-1)}\biggl{(}\frac{n(n-1)}{(n-s)(2s+n-1)}\biggr{)}^{n+\frac{1}{2}}\biggl{(}\frac{2s\sqrt{n-s}}{2s+n-1}\biggr{)}^{2s}\mathrm{e}^{\uppi\sqrt{\frac{2s}{3}}-s+1}<1,$

which leads to

$\frac{(n-1)}{\sqrt{s}(2s+n-1)}2^{s}12\sqrt{\uppi}\mathrm{e}^{\uppi\sqrt{\frac{2s}{3}}-s+1}\biggl{(}\frac{n(n-1)}{(n-s)(2s+n-1)}\biggr{)}^{n+\frac{1}{2}}\biggl{(}\frac{2s\sqrt{n-s}}{2s+n-1}\biggr{)}^{2s}<1.$

Now we need to analyze each factor and determine when these are lower than
$1$. First we observe that for $n>(2s-1)^{2}$ we have

$\frac{(n-1)}{\sqrt{s}(2s+n-1)}<1,\quad\frac{n(n-1)}{(n-s)(2s+n-1)}=\frac{n(n-1)}{n(n-1)+s(n-2s+1)}<1.$

Regarding the element,

$\frac{2s\sqrt{n-s}}{2s+n-1},$

we can compute that this is lower than $1$ if and only if

$4s^{2}(n-s)<4s^{2}+n^{2}+1+4sn-4s-2n,$

which simplifies to

$n^{2}-2(2s^{2}-2s+1)n+4s^{3}+4s^{2}-4s+1>0.$

This inequality holds, in particular, if

$n>(2s^{2}-2s+1)+2s\sqrt{s^{2}-3s+1},$

and hence also if

$n\geq(2s^{2}-2s+1)+2s\sqrt{s^{2}-2s+1}=(2s^{2}-2s+1)+2s(s-1)=(2s-1)^{2}.$

It remains to analyze the last factor
$2^{s}12\sqrt{\uppi}\mathrm{e}^{\uppi\sqrt{\frac{2s}{3}}-s+1}$, but it is
possible to determine that

$2^{s}12\sqrt{\uppi}\mathrm{e}^{\uppi\sqrt{\frac{2s}{3}}-s+1}<1$

whenever $s\geq 95$. Despite this estimation proves the statement by $s\geq
95$, it is possible to compute that, for $6\leq s\leq 94$, the upper bound
provided by section 3 is lower than generic rank, i.e.,

$\sum_{k=1}^{s}2^{k}k!\operatorname{p}_{k}(s)\binom{n}{k}<\frac{1}{n}\binom{2s+n-1}{2s}$
(4.10)

whenever $n\geq(2s-1)^{2}$. ∎

As already mentioned in section 1 through section 1, the problem of
determining the rank of $q_{n}^{s}$ has also been analyzed recently by J.
Buczyński, K. Han, M. Mella, and Z. Teitler in [BHMT18]. They ask, in
particular, whether the rank is greater than the generic rank (see
[BHMT18]*section 4.5). The importance of the powers of quadrics is related to
the maximal rank locus with respect to any Veronese variety. This set is
defined, given a finite dimensional vector space $V$ with $\dim V=n$, as

$W_{m}^{d}=\overline{{\Set{f\in\mathbb{P}(S^{d}V)}{\operatorname{rk}f=m}}}$

and, in particular, we have the following theorem.

######  Theorem 4.10 ([BHMT18]*Theorem 4.1).

Let $V$ be a finite dimensional vector space with $\dim V=n$ such that $n\geq
3$. If $W$ is an irreducible component of the rank locus $W_{m}$, then

$\dim(W)\geq\binom{n+1}{2}-1.$

Moreover, if

$\dim(W)=\binom{n+1}{2}-1,$

then $d$ is even, and $W$ is the set of all the $\bigl{(}d/2\bigr{)}$-th
powers of quadrics.

By section 4, we can state, in particular, that the second condition of
section 4 does not hold for $n>(2s-1)^{2}$. Thus, we get the following
corollary.

######  Corollary 4.10.

For every $s\in\mathbb{N}$, if $n\geq 3$, $n>(2s-1)^{2}$ and $W$ is an
irreducible component of $W_{m}^{2s}$, then

$\dim(W)\geq\binom{n+1}{2}.$

### Acknowledgements.

This paper is the natural continuation of the author’s Ph.D. thesis, which was
pursued at Alma Mater Studiorum – Università di Bologna, under the patient co-
supervision of Alessandro Gimigliano and Giorgio Ottaviani, to whom sincere
thanks are due. The author also extends special gratitude to Enrique Arrondo
and Jarosław Buczyński for their invaluable assistance and insightful
suggestions during the author’s Ph.D. program.

## References