# Sums of even powers of polynomials

,  Tomasz Kowalczyk  Institute of Mathematics, Faculty of Mathematics and
Computer Science, Jagiellonian University, ul. Łojasiewicza 6, 30-348 Kraków,
Poland<EMAIL_ADDRESS>and  Julian Vill  Fakultät für Mathematik,
Otto-von-Guericke Universität Magdeburg, Universitätsplatz 2, 39106 Magdeburg,
Germany<EMAIL_ADDRESS>
###### Abstract.

We show that the higher Pythagoras numbers for the polynomial ring are
infinite $p_{2s}(K[x_{1},x_{2},\dots,x_{n}])=\infty$ provided that $K$ is a
real field, $n\geq 2$ and $s\geq 1$. This almost fully solves an old question
[16, Problem 8]. Moreover, we study in detail the cone of binary octics that
are sums of fourth powers of quadratic forms. We determine its facial
structure as well as its algebraic boundary. This can also be seen as sums of
fourth powers of linear forms on the second Veronese of ${\mathbb{P}}^{1}$. As
a result, we disprove a conjecture of Reznick [38, Conjecture 7.1].

###### Key words and phrases:

Higher Pythagoras number, Waring problem, binary forms, sums of squares,
convex cones.

###### 2020 Mathematics Subject Classification:

Primary:11P05, 14Q30

##  1\. Introduction

This paper is devoted to the qualitative study of sums of higher powers,
mainly in polynomial rings. In particular, we are interested in the problem of
computing the higher Pythagoras numbers of a given ring.

It is known that the theory of quadratic forms is well developed. Literature
devoted to that topic is extremely vast. Every quadratic form over a field of
characteristic not two is diagonalizable. This is not true for forms of higher
degree (cf. [23]). In particular considering forms of degree at least $3$ one
has to distinguish the cases of diagonal and general forms.

Cassels Theorem [15] states that if a quadratic form with coefficients from a
field $K$ represents a polynomial $f(x)\in K[x]$ in a single variable over the
field $K(x)$ then it already represents $f(x)$ over $K[x]$. Taking the above
quadratic form to be a sum of squares (sos) we deduce the equality of
$2$-Pythagoras numbers $p_{2}(K[x])=p_{2}(K(x))$. Cassels Theorem does not
generalize to higher powers [36]. In particular, a polynomial in a single
variable can be a sum of fourth powers of rational functions, but not a sum of
fourth powers of polynomials. Currently, this appears to be the main
obstruction for computing $p_{2s}({\mathbb{R}}[x])$ (cf. Section 5), however
it is known that $p_{2s}({\mathbb{R}}(x))$ is finite [3].

Let us also mention one more difference. Consider the usual $n$-dimensional
unit sphere $\mathbb{S}^{n}$. Obviously, it is the zero set of the quadratic
polynomial $\sum_{i=1}^{n+1}x_{i}^{2}-1=0$. If one further considers the group
of isometries of $\mathbb{S}^{n}$, then it is known that this action is
transitive. Consider now an algebraic set given by the zero set of
$\sum_{i=1}^{n+1}x_{i}^{2s}-1=0$. There are very few isometries of such a set
if $s>1$, in particular, the action of isometries is not transitive.
Necessarily, any point with some number of zero coordinates cannot be moved to
an element with a different number of zeros. Consequently, when dealing with
forms of higher degree, one cannot usually simplify things by the use of
isometries. One has to consider the general case.

There is a vast literature concerning to the problem of computation of the
$2$-Pythagoras number of a ring, see [5, 6, 19, 24, 29, 30, 31] to mention
just a few. Similarly there are recent results concerning Waring problems [1,
11, 12, 18, 27, 33, 34, 35, 43] and rings of $k$-regulous functions [2] as
well as henselian local rings [28]. On the other hand, there are very few
papers dealing with sums of higher powers in real rings, see [3, 17, 22, 41]
for higher Pythagoras numbers of fields and [14, 39, 20] for algebroid curves.
Beside those, it is known that $p_{2s}({\mathbb{Z}}[x])=\infty$ [16].

The goal of this paper is to study the higher Pythagoras numbers of the
polynomial rings. In particular we prove that
$p_{2s}(K[x_{1},x_{2},\dots,x_{n}])=\infty$ provided that $K$ is a real field
and $n>1$. This almost solves an open problem from the paper [16, Problem 8].
From this, we are able to deduce that for a large class of rings, all of its
higher Pythagoras numbers are infinite. The second part of the paper focuses
on the study of the 4-Pythagoras number. It follows from the first part, that
similarly as in the quadratic case, the $2s$-Pythagoras number in polynomial
rings tends to infinity if we increase the degree as long as we consider forms
in at least 3 variables. This led us to study binary forms, especially fourth
powers, as this is the first non quadratic case. For any fixed degree there
exists a full-dimensional semi-algebraic set such that any form in it has
4-length 4. If the degree is at least 10 this is also the smallest length that
appears in a full-dimensional subset. Writing binary forms as sums of powers
of linear forms has already been done by Reznick. Since we want to study the
4-Pythagoras number, we give a different proof of the fact that
$p_{4}(2,4)=3$. This has already been proven by Reznick by studying the dual
cone of $\Sigma_{2,4}^{4}$. We use a proof more similar to Hilbert’s and to
Scheiderer’s generalized version [40, §4] . Moreover, we will see that every
form on the boundary has length at most 2 and every form of length at most 2
is on the boundary. In particular, the boundary is the union of two
$\operatorname{GL}_{2}({\mathbb{R}})$ orbits. Our main results concern binary
octics. We determine the algebraic boundary of the cone $\Sigma_{2,8}^{4}$ by
studying its dual cone which can be seen as a linear slice of the cone
$\Sigma_{3,4}^{2}$ of ternary quartics that are sums of squares, or
equivalently are positive semidefinite (psd). Surprisingly, it turns out that
the boundary is irreducible and a generic point of the boundary has a
’special’ length 3 representation as a sum of fourth powers of quadratics.
Moreover, this is the unique representation as a sum of fourth powers. This
allows us to deduce that the 4-Pythagoras number is either 3 or 4. We are then
able to disprove a conjecture of Reznick [38, Conjecture 7.1] expecting that
sums of fourth powers are “doubly positive” meaning they can be written as
sums of squares of psd forms.

The structure of the paper is as follows. Section 2 contains the study of the
higher Pythagoras number of a variety of rings. The main result of this
section is the computation of the higher even Pythagoras numbers for the
polynomial ring, i.e, $p_{2s}(K[x_{1},x_{2},\dots,x_{n}])=\infty$. In Section
3 we examine the structure of the cone $\Sigma_{2,8}^{4}$ of binary octics
that are sums of fourth powers of quadratics as well as their $4$-Pythagoras
number. In the next section, we determine the facial structure of the
aforementioned cone and disprove Reznick’s Conjecture. We finish the paper
with some open problems.

Let us finish the introductory section with some basic definitions and
notation. Let $A$ be a commutative ring with identity and $s\geq 1$ be a
positive integer.

######  Definition 1.1 (non-homogeneous Pythagoras number).

We define the $s$-th Pythagoras number of $A$, $p_{s}(A)$, to be the least
positive integer $\ell$ such that any sum of $s$-th powers of elements of $A$
can be expressed as a sum of $\ell$ $s$-th powers of elements of $A$. If such
a number does not exist, we put $p_{s}(A)=\infty$.

Let $a\in A$.

######  Definition 1.2.

We define the $s$-length of $a$ as the smallest positive integer $\ell$ such
that $a$ can be written as a sum of $\ell$ $s$-th powers of elements of $a$.

For any $n,s,d\in{\mathbb{N}}$ we consider the cone
$\Sigma_{n,sd}^{s}\subseteq{\mathbb{R}}[x_{1},\dots,x_{n}]_{sd}$ of sums of
$s$-th powers of forms of degree $d$. If $s$ is even this is a closed, pointed
and full-dimensional convex cone. By definition any form $f$ in
$\Sigma_{n,sd}^{s}$ has a representation $f=\sum_{i=1}^{r}f_{i}^{s}$ for some
$r\in{\mathbb{N}}$ and $f_{i}\in{\mathbb{R}}[x_{1},\dots,x_{n}]_{d}$. The
smallest such $r$ is called the $s$-length of $f$.

######  Definition 1.3 (homogeneous Pythagoras number).

We define the $s$-Pythagoras number $p_{s}(n,sd)$ of
${\mathbb{R}}[x_{1},\dots,x_{n}]_{sd}$ as the minimal number $\ell$ such that
any $f\in\Sigma_{n,sd}^{s}$ has such a representation of length at most
$\ell$.

Elements of the vector space ${\mathbb{R}}[x_{1},\dots,x_{n}]_{sd}$ will be
called $n$-ary $sd$-ics.

######  Remark 1.4.

In this paper we consider only even lengths of homogeneous polynomials. This
is because of the following observation. Let $f$ be a homogeneous polynomial
of degree $sd$ and consider the $s$-length of $f$. If $s$ is even, then the
$s$-length of $f$ in the ring ${\mathbb{R}}[x_{1},\dots,x_{n}]$ will be the
same as in the vector space ${\mathbb{R}}[x_{1},\dots,x_{n}]_{sd}$. However,
if $s$ is odd, then those lengths will almost surely be different, since if
$f$ is a sum of odd powers of polynomials $f_{i}$, then the $f_{i}$’s do not
have to be homogeneous.

##  2\. Higher Pythagoras numbers

In this section we discuss higher Pythagoras numbers of rings. We are mainly
concerned with the polynomial rings over a real field. In particular,
$p_{2}(n,2d)$ corresponds to shortest sums of squares representations of
forms. Even these are not well-understood in general (see [40]). One of the
main difficulties in studying higher Pythagoras numbers is that we are no
longer considering quadratic forms which generally behave much nicer than
higher tensor powers.

Firstly, let us show a lower bound for $p_{2s}(n,2sd)$ by counting dimensions.
The dimension of ${\mathbb{R}}[x_{1},\dots,x_{n}]_{2sd}$ is equal to
$n+2sd-1\choose 2sd$ and this is also the dimension of $\Sigma^{2s}_{n,2sd}$
as this cone contains a generating set $\\{l^{2sd}\colon
l\in{\mathbb{R}}[x_{1},\dots,x_{n}]_{1}\\}$ of the real vector space
${\mathbb{R}}[x_{1},\dots,x_{n}]_{2sd}$. The $2s$-th Pythagoras number
$p_{2s}(n,2sd)$ has to satisfy

$p_{2s}(n,2sd){n+d-1\choose d}\geq{n+2sd-1\choose 2sd}.$

In other words,

$p_{2s}(n,2sd)\geq\frac{{n+2sd-1\choose 2sd}}{{n+d-1\choose d}}.$

A simple computation shows that if $d$ tends to infinity, then the right hand
side tends to a constant value $(2s)^{n-1}$. Upon dehomogenization, this can
be translated into the inequality

$p_{2s}({\mathbb{R}}[x_{1},\dots,x_{n-1}])\geq(2s)^{n-1}.$

In particular $p_{4}({\mathbb{R}}[x])\geq 4$.

Our first result is a generalization of [16, Proposition 4.5’] to higher even
powers. Let $s>1$ be a positive integer and $K$ be a formally real field.

######  Proposition 2.1.

Let $g(x,y)\in K[x,y]$ be a polynomial of $2s$-length $k$ and $r$ be a
positive integer such that $\deg_{x}g<2sr$. Then the polynomial
$G(x,y):=g(x,y)(y-x^{r})^{2s}+1$ has $2s$-length $k+1$.

###### Proof.

Assume by contrary that this is not the case, i.e.

$G(x,y)=\sum_{i=1}^{k}g_{i}^{2s}.$ (1)

By setting $y=x^{r}$ we obtain that each for each $i$,
$g_{i}=a_{i}+(y-x^{r})h_{i}$, where $a_{i}\in K$ and $h_{i}\in K[x,y]$. We may
now rewrite Eq. 1 as

$g(x,y)(y-x^{r})^{2s}+1=\sum_{i=1}^{k}a_{i}^{2s}+(y-x^{r}){2s\choose
1}\sum_{i=1}^{k}a_{i}^{2s-1}h_{i}+(y-x^{r})^{2}{2s\choose
2}\sum_{i=1}^{k}a_{i}^{2s-2}h_{i}^{2}+\dots+$
$+(y-x^{r})^{2s}\sum_{i=1}^{k}h_{i}^{2s}.$

Since $\sum_{i=1}^{k}a_{i}^{2s}=1$, both $1$’s cancel out. As a next step,
after dividing by $(y-x^{r})$, we see that $\sum_{i=1}^{k}a_{i}^{2s-1}h_{i}$
is divisible by $(y-x^{r})$. By the degree considerations, we see that
$\sum_{i=1}^{k}a_{i}^{2s-1}h_{i}\equiv 0$. If not, then the degree of some
$h_{i}$ would be at least $r$, hence the degree of the right-hand side of Eq.
1 would be at least $4sr$, and this is larger than the degree of the left-hand
side. Consequently, the above expression can be reduced to

$g(x,y)(y-x^{r})^{2s-2}={2s\choose
2}\sum_{i=1}^{k}a_{i}^{2s-2}h_{i}^{2}+\dots+(y-x^{r})^{2s-2}\sum_{i=1}^{k}h_{i}^{2s}.$

We may look at the polynomial $\sum_{i=1}^{k}a_{i}^{2s-2}h_{i}^{2}$. If this
polynomial is nonzero, then the $h_{i}$ are divisible by $(y-x^{r})$ and we
get a contradiction as previously. As a result, we get that

$\sum_{i=1}^{k}a_{i}^{2s-2}h_{i}^{2}\equiv 0$

and for each $i=1,2,\dots,k$, necessarily $a_{i}h_{i}=0$, as we are working in
a formally real field. In particular, for each $i=1,2,\dots,k$, $a_{i}=0$ or
$h_{i}=0$. Denote by $l$ the number of nonzero $h_{i}$ polynomials. After
enumerating, we can rewrite Eq. 1 as

$g(x,y)(y-x^{r})^{2s}+1=(y-x^{r})^{2s}\sum_{i=1}^{l}h_{i}^{2s}+1$

where we merged all nonzero $a_{i}$’s into $1$. As the final step we get

$(y-x^{r})^{2s}(g(x,y)-\sum_{i=1}^{l}h_{i}^{2s})\equiv 0,$

but this implies that the $2s$-length of $g(x,y)$ is less than $k$, a
contradiction. ∎

######  Theorem 2.2.

Let $n>1$ and $s>1$ be positive integers. Then
$p_{2s}(k[x_{1},x_{2},\dots,x_{n}])=\infty$ for any formally real field.

###### Proof.

It is enough to treat the case $n=2$. By the above Proposition, starting from
$G(x,y)=1$ we can inductively construct a sequence of polynomials of
arbitrarily large length. This finishes the proof. ∎

######  Theorem 2.3.

Let $s$ be a positive integer. Then

$p_{2s}(A[x_{1},x_{2},\dots,x_{n}])=\infty$

for any commutative ring with infinite second level and $n>1$.

###### Proof.

One can repeat proof of [16, Corollary 4.18] verbatim.

∎

By a similar use of the notion of homogeneous Pythagoras number as in [16] we
may easily deduce the following

######  Corollary 2.4.

Let $R$ be a regular local ring of dimension at least 3 with real residue
field. Then $p_{2s}(R)=\infty$ for all $s\geq 1$.

The above corollary allows us to deduce what are the $2s$-Pythagoras numbers
for the rings of regular functions on an algebraic set of dimension at least
3.

Let $X\subseteq{\mathbb{R}}^{n}$ be an algebraic set of dimension at least 3.
Denote by ${\mathcal{O}}(X)$ the ring of regular functions on $X$ [9,
Definition 3.2.1].

######  Corollary 2.5.

If $X$ is a real algebraic set of dimension at least 3, then for any positive
integer $d\geq 1$ we have $p_{2s}({\mathcal{O}}(X))=\infty$.

###### Proof.

Let $y$ be a regular point of $X$ and $\mathfrak{m}$ be its maximal ideal.
Then the basic properties of the Pythagoras number implie that
$p_{2s}({\mathcal{O}}(X))\geq p_{2s}({\mathcal{O}}(X)_{\mathfrak{m}})$.
However, the ring on the right hand side is a regular local ring of dimension
at least 3, hence $p_{2s}({\mathcal{O}}(X)_{\mathfrak{m}})=\infty$ and
$p_{2s}({\mathcal{O}}(X))=\infty$. ∎

We might also get an insight into a behaviour of higher Pythagoras numbers of
$k$-regulous functions ([2, Definition 1.1]. Denote by
$\mathcal{R}^{k}({\mathbb{R}}^{n})$ the ring of $k$-regulous functions on
${\mathbb{R}}^{n}$ for $k>0$ and $n>2$. Theorem 2.2 and [2, Theorem 3.7]
readily implies

######  Corollary 2.6.

For fixed positive integers $s>0$ and $n>2$ we have

$\limsup_{k\rightarrow\infty}p_{2s}(\mathcal{R}^{k}({\mathbb{R}}^{n})))=\infty.$

It is worth to mention that it is still not known whether the above numbers
are finite or not.

######  Corollary 2.7.

Let $s\geq 3$ and $d\in{\mathbb{N}}$. Then $p_{2s}(n,2sd)\to\infty$ as
$d\to\infty$.

This proof does not work for binary forms. We do not know if the
$4$-Pythagoras number is bounded in this case.

In [13] the first author and Błachut showed that the $2$-Pythagoras number of
coordinate rings of large class of hypersurfaces
$\\{z^{2}-f(x,y)=0\\}\subseteq{\mathbb{R}}^{3}$ is infinite. Here we are able
to generalize this result to the $2s$-Pythagoras numbers.

######  Definition 2.8.

Let $f(x,y)\in{\mathbb{R}}[x,y]$ be a non-constant polynomial such that
$f(0,0)=0$. Let $d=\deg_{y}f(x,y)$ and $\alpha x^{b}y^{d}$ be a monomial with
the largest possible $b$ and nonzero coefficient $\alpha$ among all monomials
in $f(x,y)$. We say that $f(x,y)$ is strictly admissible if:

  * a)

$\alpha>0$ and $b,d$ are even, or

  * b)

$b$ or $d$ is odd.

We say that $f(x,y)$ is an admissible polynomial if there exists an invertible
matrix $M\in\operatorname{GL}_{2}({\mathbb{R}})$ such that $f\circ M$ is a
strictly admissible polynomial.

See [13] for examples and further properties of admissible polynomials. In
particular, an admissible polynomial has to be positive on an open unbounded
set. We may now prove the following.

######  Theorem 2.9.

Let $f(x,y)$ be an admissible polynomial which is not a square. Then

$p_{2s}({\mathbb{R}}[x,y,\sqrt{f(x,y)}])=\infty.$

###### Proof.

We will carry out a proof for $s=2$, as this is the first non-quadratic case.
The proof for higher powers works similarly.

Clearly, for any invertible matrix $M\in GL_{2}({\mathbb{R}})$, the rings
${\mathbb{R}}[x,y,\sqrt{(f\circ M)(x,y)}]$ and
${\mathbb{R}}[x,y,\sqrt{f(x,y)}]$ are isomorphic, hence without loss of
generality we may assume that $f(x,y)$ is a strictly admissible polynomial.

We begin the proof by constructing a sequence of polynomials (cf. [13,
Definition 1.10]).

Consider a sequence of positive integers $(r_{m})_{m\geq 1}$ such that

  * •

$r_{1}>\deg f$ and $r_{i}>4\sum_{k=1}^{i-1}r_{k}$

  * •

$f(x,x^{r_{i}})$ is a polynomial in $x$, which is either of odd degree, or it
has even degree, and the leading coefficient is strictly positive.

We define a sequence of polynomials in the following way:

  * •

$F_{1}=1$

  * •

$F_{n}=F_{n-1}(y-x^{r_{n-1}})^{4}+1$

Fix a sequence of polynomial $(F_{m})_{m\geq 1}$ satisfying the above
properties. By the Proposition 2.1 $4$-length of $F_{m}$ is equal to $m$.

Assume by contrary that the $4$-Pythagoras number of
${\mathbb{R}}[x,y,\sqrt{f(x,y)}]$ is finite and equal to $L$. Consider the
polynomial $F_{L+1}$. By assumption, we have the following

$F_{L+1}=\sum_{i=1}^{L}(f_{i,1}+\sqrt{f}g_{i,1})^{4}$ (2)

which translates into the equation

$F_{L+1}=\sum_{i=1}^{L}f_{i,1}^{4}+6f\sum_{i=1}^{L}f_{i,1}^{2}g_{i,1}^{2}+f^{2}\sum_{i=1}^{L}g_{i,1}^{4}$
(3)

as the part that is divisible by $\sqrt{f}$ has to vanish. Let us now
substitute $y=x^{r_{L}}$. We assumed that $f(x,x^{r_{L}})$ is either of odd
degree or of even degree with positive leading coefficient and does not vanish
identically. This readily implies that $f_{i,1}=a_{i}+(y-x^{r_{L}})f_{i,2}$
for some $a_{i}\in{\mathbb{R}}$ such that $\sum_{i=1}^{L}a_{i}^{4}=1$ and
additionally $g_{i,1}=(y-x^{r_{L}})g_{i,2}$. We may now rewrite Eq. 3 as

$F_{L}(y-x^{r_{L}})^{4}+1=\sum_{i=1}^{L}a_{i}^{4}+4(y-x^{r_{L}})\sum_{i=1}^{L}a_{i}^{3}f_{i,2}+6(y-x^{r_{L}})^{2}\sum_{i=1}^{L}a_{i}^{2}f_{i,2}^{2}+4(y-x^{r_{L}})^{3}\sum_{i=1}^{L}a_{i}f_{i,2}^{3}+$
$(y-x^{r_{L}})^{4}\sum_{i=1}^{L}f_{i,2}^{4}+6f(y-x^{r_{L}})^{2}\sum_{i=1}^{L}(a_{i}+(y-x^{r_{L}})f_{i,2})^{2}g_{i,2}^{2}+f^{2}(y-x^{r_{L}})^{4}\sum_{i=1}^{L}g_{i,2}^{4}.$

After cancelling 1’s and dividing by $(y-x^{r_{L}})$ we see that the degree
considerations forces $\sum_{i=1}^{L}a_{i}^{3}f_{i,2}\equiv 0$. Again, by
dividing by $(y-x^{r_{L}})$ we get that

$(y-x^{r_{L}})|\sum_{i=1}^{L}a_{i}^{2}f_{i,2}^{2}+6f\sum_{i=1}^{L}a_{i}^{2}g_{i,2}^{2}.$

If the polynomial on the right is nonzero, then necessarily
$\sum_{i=1}^{L}f_{i,2}^{4}$ or $f^{2}\sum_{i=1}^{L}g_{i,2}^{4}$ have to be of
degree at least $2r_{L}$. Furthermore, the RHS of Eq. 3 will be of degree at
least $6r_{L}$, while the left-hand side is of degree $4\sum_{i=1}^{L}r_{i}$.
The contradiction follows from the assumption on the $r_{i}^{\prime}s$. As a
result
$\sum_{i=1}^{L}a_{i}^{2}f_{i,2}^{2}+6f\sum_{i=1}^{L}a_{i}^{2}g_{i,2}^{2}$ is
the zero polynomial. Even more, we know that for each $i=1,2,\dots,L$
$a_{i}f_{i,2}\equiv 0$ and $a_{i}g_{i,2}\equiv 0$, because if not, then $f$
would attain only nonpositive values, but this is a contradiction with the
assumption of admissibility. Furthermore, there exists at least one $i$ such
that $a_{i}\neq 0$ and for such, we have $f_{i,2}=g_{i,2}=0.$ By merging all
of the nonzero $a_{i}^{\prime}s$ into one, renumbering and adding artificial
zero polynomials if necessary, we can rewrite Eq. 2 as

$F_{L}(y-x^{r_{L}})^{4}+1=(y-x^{r_{L}})^{4}\sum_{i=1}^{L-1}(f_{i,1}+\sqrt{f}g_{i,1})^{4}+1$

which is equivalent to

$F_{L}=\sum_{i=1}^{L-1}(f_{i,2}+\sqrt{f}g_{i,2})^{4}.$

In this process, we managed to reduce the number of summands in Eq. 2 as well
as reduce the polynomial $F_{L+1}$ to $F_{L}$.

After repeating the above procedure $L-2$ times we get the following

$F_{2}=(y-x^{r_{1}})^{4}+1=f_{1,L}^{4}+ff_{1,L}^{2}g_{1,L}^{2}+f^{2}g_{1,L}^{4}.$

After repeating the above reasoning one last time, we get that $f_{1,L}=1$ and
$g_{1,L}=0$. This implies an equality of polynomials of the type
$(y-x^{r_{1}})^{4}=0$ which is absurd, hence the proof is done. ∎

We finish this section with a brief discussion of quartics in one variable. In
particular, we discuss the following: how does one decide whether a strictly
positive quartic polynomial is a sum of fourth powers of linear forms? By
Becker’s Valuative Criterion [3, Theorem 1.9], any strictly positive quartic
is a sum of fourth powers of rational functions. Let $f=f_{1}f_{2}$ be a
product of two strictly positive quadratics. As $f_{1}$ is strictly positive,
one can always find a linear change of variables $L$ such that $f_{1}\circ
L=x^{2}+1$. After an orthogonal change of coordinates applied to both $f_{1}$
and $f_{2}$ we may assume altogether that $f=(x^{2}+1)(x^{2}+b)$ for some
strictly positive $b$.

######  Proposition 2.10.

The polynomial $(x^{2}+1)(x^{2}+b)$ is a sum of fourth powers of linear forms
$\Leftrightarrow b\in[17-12\sqrt{2},17+12\sqrt{2}]=I$.

###### Proof.

Let $f_{b}=(x^{2}+1)(x^{2}+b)=x^{4}+(b+1)x^{2}+b$. Assume that $f_{b}$ can be
written a sum of fourth powers, i.e. $f_{b}=\sum_{i=1}^{k}(a_{i}+b_{i}x)^{4}$
for some positive integer $k$. In this way we get the following equations
$\sum_{i=1}^{k}a_{i}^{4}=1$, $6\sum_{i=1}^{k}a_{i}^{2}b_{i}^{2}=b+1$,
$\sum_{i=1}^{k}b_{i}^{4}=b$ and two equations which are of no interest for us.
Define two vectors $v=(a_{1}^{2},a_{2}^{2},\dots,a_{k}^{2})$ and
$w=(b_{1}^{2},b_{2}^{2},\dots,b_{k}^{2})$, clearly $||v||^{2}=1$ and
$||w||^{2}=b$. By the Cauchy-Bunyakowsky-Schwarz inequality applied to $v$ and
$w$ we obtain $\left(\frac{1+b}{6}\right)^{2}\leq b$, which becomes
$b^{2}-34b+1\leq 0$. Solution of this inequality is precisely the interval
$I$. By this reasoning we see that $f_{b}$ may be a sum of fourth powers only
for $b\in I$.

Now, for any $b\in I$ we have a following formula

$2f_{b}=(x^{2}+\sqrt{1+b-\sqrt{b}}x+\sqrt{b})^{2}+(x^{2}-\sqrt{1+b-\sqrt{b}}x+\sqrt{b})^{2}$

with the additional assumption that both quadratic polynomials are
nonnegative. The result follows readily from the formula
$(p_{1}^{2}+q_{1}^{2})^{2}=\frac{1}{18}((\sqrt{3}p_{1}+q_{1})^{4}+(\sqrt{3}p_{1}-q_{1})^{4}+16q_{1}^{4})$.
∎

The above proof shows that any quartic polynomial which is a sum of fourth
powers can be written as a sum of at most 6 fourth powers, however we will
later show that 3 is enough and for the boundary points 2 fourth powers
suffice. We encourage the reader to find a representation of length 3 of
$f_{b}$ for any $b\in\mathrm{int}(I)$ and of length 2 for the boundary points.

##  3\. Binary forms

For the rest of the paper we will work with binary forms. We do not know if
the higher Pythagoras numbers increase if the degree tends to infinity. This
is of course not the case for the 2-Pythagoras number as any positive
semidefinite binary form can be written as a sum of at most two squares. We
first study the cone $\Sigma_{2,4}^{4}$ of binary quartics that are sums of
fourth powers of linear forms. We then prove 3.4 which states that the
algebraic boundary of the cone $\Sigma_{2,8}^{4}$ is still irreducible.

We will often use the sum of $4$-th powers map

$\phi_{d}^{k}\colon\mathbb{R}[\underline{x}]_{d}^{k}\to\mathbb{R}[\underline{x}]_{4d},\quad(p_{1},\dots,p_{k})\mapsto\sum_{i=1}^{k}p_{i}^{4}$

and its differential at the point $\underline{p}=(p_{1},\dots,p_{k})$

$\operatorname{d\phi}_{d}^{k}(\underline{p})\colon\mathbb{R}[\underline{x}]_{d}^{k}\to\mathbb{R}[\underline{x}]_{4d},\quad(q_{1},\dots,q_{k})\mapsto\sum_{i=1}^{k}p_{i}^{3}q_{i}.$

The image of the differential is given by $\langle
p_{1}^{3},\dots,p_{k}^{3}\rangle_{4d}$.

Moreover, for $f_{1},\dots,f_{k}\in{\mathbb{R}}[x_{1},\dots,x_{n}]_{d}$ we
denote by
$\mathcal{V}(f_{1},\dots,f_{k})\subseteq{\mathbb{P}}^{n-1}={\mathbb{P}}^{n-1}{\mathbb{C}}$
the zero set of the forms $f_{1},\dots,f_{r}$, i.e. the set of all points
$\xi\in{\mathbb{P}}^{n-1}$ such that $f_{i}(\xi)=0$ for all $i=1,\dots,k$.

The following simple proposition shows that one could hope for a constant
4-Pythagoras number of binary forms and possibly prove $p_{4}(2,4d)=4$.
However, this set being full-dimensional is a very small indication. We give a
direct, slightly combinatorial argument here. One may also use powers of
linear forms instead of monomials and use known cases of Fröberg’s conjecture.

######  Proposition 3.1.

Let $n=2$. For every $d\geq 1$ the semi-algebraic set of all forms in
$\Sigma^{4}_{2,4d}$ of length at most 4 is a full-dimensional semi-algebraic
set.

###### Proof.

It suffices to show that the differential
$\operatorname{d\phi}^{4}_{d}(\underline{p})$ is surjective for one specific
(and therefore a generic) choice of
$\underline{p}\in\mathbb{R}[\underline{x}]^{4}_{d}$. Let $r\in\\{0,1,2\\}$
such that $d\equiv r\mod 3$. Choose $p_{1}=x^{d},p_{2}=y^{d}$ and
$p_{3},p_{4}$ depending on $r$. In every case we show that every monomial in
$\mathbb{R}[\underline{x}]_{4d}$ is contained in $I:=\langle
p_{1}^{3},\dots,p_{4}^{3}\rangle_{4d}$.

In every case, since $x^{3d},y^{3d}\in I$ we see that for $k\leq d$ and $k\geq
3d$ the monomial $x^{k}y^{4d-k}$ is contained in $I$.

$r=0\,$: $p_{3}=(x^{2}y)^{\frac{d}{3}},\,p_{4}=(xy^{2})^{\frac{d}{3}}$

We have $p_{3}^{3}=x^{2d}y^{d},\,p_{4}^{3}=x^{d}y^{2d}$ which implies that
$x^{k}y^{4d-k}\in I$ for $2d\leq k\leq 3d$ and $d\leq k\leq 2d$. Hence
$I_{4d}=\mathbb{R}[\underline{x}]_{4d}$.

$r=2\,$: $p_{3}=(x^{2}y)^{\frac{d-2}{3}}xy,\,p_{4}=(xy^{2})^{\frac{d-2}{3}}xy$

We have $p_{3}^{3}=x^{2d-1}y^{d+1},\,p_{4}^{3}=x^{d+1}y^{2d-1}$ which shows
$x^{k}y^{4d-k}\in I$ for $2d-1\leq k\leq 3d-1$ and $d+1\leq k\leq 2d+1$. Hence
$I_{4d}=\mathbb{R}[\underline{x}]_{4d}$.

$r=1\,$: In this case monomials do not suffice. Consider
$p_{3}=x^{\frac{d+2}{3}}y^{\frac{2d-2}{3}},\,p_{4}=x^{d}+y^{d}$.

Since $p_{3}^{3}=x^{d+2}y^{2d-2}$, the monomial $x^{k}y^{4d-k}$ is in $I$ if
$d+2\leq k\leq 2d+2$. We are therefore missing $k=d+1$ and $2d+3\leq k\leq
3d-1$. We have

$p_{3}^{3}=x^{3d}+y^{3d}+3x^{2d}y^{d}+3x^{d}y^{2d}\equiv
x^{2d}y^{d}+x^{d}y^{2d}\mod I_{3d}$

and therefore
$(x^{2d}y^{d}+x^{d}y^{2d})xy^{d-1}=x^{2d+1}y^{2d-1}+x^{d+1}y^{3d-1}\equiv
x^{d+1}y^{3d-1}\mod I_{4d}$. Lastly for every $0\leq s\leq d$ we see

$(x^{2d}y^{d}+x^{d}y^{2d})x^{s}y^{d-s}=x^{2d+s}y^{2d-s}+x^{d+s}y^{3d-s}\equiv
x^{2d+s}y^{2d-s}\mod I_{4}$

since $d+s\in\\{d,\dots,2d\\}$. ∎

###  3.1. Binary quartics

We now give a different proof that $p_{4}(2,4)$ equals 3. This has already
been proven by Reznick by studying the cone $(\Sigma_{2,4}^{4})^{\star}$ which
is isomorphic to $\Sigma_{2,4}^{2}$. Our proof is more similar to Hilbert’s
proof that every positive semidefinite ternary quartic has a sum of squares
representation of length at most 3.

######  Theorem 3.2 ([37, §5]).

The 4-Pythagoras number of binary quartics is 3, i.e. $p_{4}(2,4)=3$.

###### Proof.

For a generic choice of $l_{1},l_{2}\in{\mathbb{R}}[x,y]_{1}$ the image
$\operatorname{im}\operatorname{d\phi}_{1}^{2}(l_{1},l_{2})=\langle
l_{1}^{3},l_{2}^{3}\rangle_{4}$ has codimension 1 in ${\mathbb{R}}[x,y]_{4}$
since $l_{1},l_{2}$ form a regular sequence. This implies that the image of
$\phi_{1}^{2}$ is contained in a hypersurface. Especially $p_{4}(2,4)\geq 3$.

Let $l_{1},l_{2},l_{3}\in{\mathbb{R}}[x,y]_{1}$ be three pairwise linearly
independent linear forms. Hence $\mathcal{V}(l_{1},l_{2})=\varnothing$ and
$l_{1},l_{2}$ form a regular sequence and so do $l_{1}^{3},l_{2}^{3}$.
Therefore the Hilbert function of $\langle l_{1}^{3},l_{2}^{3}\rangle$ is
$(1,2,3,2,1)$. As this is a complete intersection and therefore Gorenstein
([25, Thm 1.44]), the socle has dimension 1 and is generated in degree 4.
Hence, the ideal $I:=\langle l_{1}^{3},l_{2}^{3},l_{3}^{3}\rangle$ has Hilbert
function $(1,2,3,1)$, i.e. $I_{4}={\mathbb{R}}[x,y]_{4}$.

This shows that $\operatorname{d\phi}_{1}^{3}(\underline{l})$ is surjective
whenever $\underline{l}=(l_{1},l_{2},l_{3})$ are pairwise linearly
independent. Let $\underline{l}$ such that
$\operatorname{d\phi}_{1}^{3}(\underline{l})$ is not surjective. We may assume
that $l_{2}=\lambda l_{1}$ for some $\lambda\in{\mathbb{R}}$. Hence
$\phi_{1}^{3}(\underline{l})=(1+\lambda^{4})l_{1}^{4}+l_{3}^{4}$ has length at
most 2.

One way to complete the proof now is to use an argument similar to [40, Thm
4.1]. We use a different observation. Denote by $X$ the Zariski closure of the
image of $\phi_{1}^{2}$. Every point $f\in\operatorname{im}\phi_{1}^{2}$
either has the form $f=l_{1}^{4}+l_{2}^{4}$ or $f=l_{1}^{4}$ with
$l_{1},l_{2}\in{\mathbb{R}}[x,y]_{1}$ linearly independent. Therefore,
$\operatorname{im}\phi_{1}^{2}$ is the union of the two
$\operatorname{GL}_{2}({\mathbb{R}})$ orbits of $x^{4}+y^{4}$ and $x^{4}$.
This immediately shows that any form in $\operatorname{im}\phi_{1}^{2}$ is on
the boundary of $\Sigma_{2,4}^{4}$. In particular
$\operatorname{im}\phi_{1}^{3}\smallsetminus\operatorname{im}\phi_{1}^{2}$ is
connected. Thus every form in the interior of $\Sigma_{2,4}^{4}$ has length 3
by the implicit function theorem. ∎

######  Corollary 3.3.

The algebraic boundary $\partial_{a}\Sigma^{4}_{4}$ is the Zariski closure of
the set

$X^{\circ}=\\{f\in\Sigma_{2,4}^{4}\colon\exists\,l_{1},l_{2}\in{\mathbb{R}}[x,y]_{1}\text{
linearly independent}\colon f=l_{1}^{4}+l_{2}^{4}\\}$

of all forms of length 2. In particular, $X^{\circ}$ is irreducible.

###  3.2. Binary octics

We study the cone of binary octics that are sums of fourth powers of quadratic
forms. Via the Veronese map and duality this is equivalent to studying a
certain intersection of the psd/sos cone of ternary quartics. It can also be
seen as the cone of sums of fourth powers of linear forms in
${\mathbb{R}}[a,b,c]/(b^{2}-ac)$, the coordinate ring of the second Veronese
of ${\mathbb{P}}^{1}$.

We show that the algebraic boundary $\partial_{a}\Sigma_{2,8}^{4}$ is an
irreducible hypersurface and consists of binary octics of length at most 3.
Using this we deduce that every binary octic that is a sum of fourth powers of
quadratic forms can be written as a sum of at most four such forms.

More precisely, we consider the hypersurface $\mathcal{G}$ in
${\mathbb{R}}[x,y]_{2}^{3}$ given as the Zariski-closure of

$\left\\{(q_{1},q_{2},q_{3})\in{\mathbb{R}}[x,y]_{2}^{3}\colon
q_{1},q_{2},q_{3}\text{ linearly independent},\langle
q_{1}^{3},q_{2}^{3},q_{3}^{3}\rangle_{8}\neq{\mathbb{R}}[x,y]_{8}\right\\}.$

######  Theorem 3.4.

The algebraic boundary of $\Sigma_{2,8}^{4}$ is an irreducible hypersurface.
It is given as the Zariski-closure of the image of the hypersurface
$\mathcal{G}$ under the sum of fourth powers map, i.e.

$\partial_{a}\Sigma_{2,8}^{4}=\overline{\phi_{2}^{3}(\mathcal{G})}.$

From this we can then bound the $4$-Pythagoras number as follows.

######  Theorem 3.5.

The 4-Pythagoras number of binary octics satisfies $p_{4}(2,8)\in\\{3,4\\}$.

###### Proof.

From the dimension count at the beginning of Section 2 we get $p_{4}(2,8)\geq
3$.

A general point of $\partial\Sigma_{2,8}^{4}$ is contained in
$\phi_{2}^{3}(\mathcal{G})$ and therefore has length at most 3. Since the
length is lower-semicontinuous (over the reals) every point of the boundary
has length at most 3.

Let $f\in\operatorname{int}\Sigma_{2,8}^{4}$ and let $K$ be a compact cone
basis of $\Sigma_{2,8}^{4}$ containing $x^{8}$. For some $\lambda>0$ we also
have $\lambda f\in K$. Consider the line through $x^{8}$ and $\lambda f$. This
line intersects the boundary of $K$ in $x^{8}$ and another point $g$. Then $g$
has length at most 3 and thus $\lambda f$ has length at most 4 as a convex
combination of $g$ and $x^{8}$. ∎

######  Remark 3.6.

Consider the set

$Z=\\{\underline{p}\in{\mathbb{R}}[x,y]_{2}^{3}\colon\langle\underline{p}^{3}\rangle_{8}\neq{\mathbb{R}}[x,y]_{8}\\}.$

This is the set of singular points of the map $\phi_{2}^{3}$. We can check
with a computer that $Z=Z_{1}\cup\mathcal{G}$ has two (geometrically)
irreducible components defined over ${\mathbb{R}}$ both of which are
hypersurfaces in the 9-dimensional affine space ${\mathbb{R}}[x,y]_{2}^{3}$.
Any point in $Z_{1}$ is a linearly dependent triple $\underline{p}$. Elements
of $\mathcal{G}$ satisfy a more complicated equation. The defining form has
degree 24 in the coefficients of the three binary quadratics. In 3.19 we give
a different interpretation of this component.

######  Remark 3.7.

We do not know if the Pythagoras number $p_{4}(2,8)$ is $3$ or $4$. The
situation is somewhat similar to the case of sums of squares of ternary octics
where the algebraic boundary is known but the exact 2-Pythagoras number is
unknown. It is either 5 or 6.

We now turn to prove 3.4. For any $n\in{\mathbb{N}}$ we consider the apolarity
pairing on ${\mathbb{R}}[x_{1},\dots,x_{n}]$ and on
${\mathbb{C}}\otimes{\mathbb{R}}[x_{1},\dots,x_{n}]$. Given two monomials
$\underline{x}^{\alpha},\,\underline{x}^{\beta}$,
$\alpha,\beta\in{\mathbb{Z}}_{\geq 0}^{n}$ we define

$\langle{\underline{x}^{\alpha}},{\underline{x}^{\beta}}\rangle=\begin{cases}\frac{\beta!}{\alpha!}\underline{x}^{\beta-\alpha}\quad&,\text{if
}\alpha\leq\beta\\\ 0\quad&,\text{else}\end{cases}$

where $\alpha!=\prod_{i=1}^{n}(\alpha_{i}!)$. This is the same as having the
first element act on the second via differentiation. For any
$d\in{\mathbb{N}}$ this gives a scalar product on
${\mathbb{R}}[x_{1},\dots,x_{n}]_{d}$. Over the complex numbers this is not
true, but it is still a perfect pairing.

Consider the dual cone

$(\Sigma_{2,8}^{4})^{\star}=\\{L\in{\mathbb{R}}[x,y]_{8}^{\scriptscriptstyle\vee}\colon
L(q^{4})\geq 0\,\forall\,q\in{\mathbb{R}}[x,y]_{2}\\}.$

Let $q(a,b,c)=ax^{2}+bxy+cy^{2}$ with new variables $a,b,c$ and let
$L=\sum_{i=0}^{8}a_{i}x^{i}y^{8-i}\in{\mathbb{R}}[x,y]_{8}$. By identifying
${\mathbb{R}}[x,y]_{8}^{\scriptscriptstyle\vee}$ and ${\mathbb{R}}[x,y]_{8}$
via the apolarity pairing we get

$\langle{L},{q(a,b,c)^{4}}\rangle=\sum_{i=0}^{8}a_{i}\langle{x^{i}y^{8-i}},{x^{i}y^{8-i}}\rangle
g_{i}(a,b,c)$ (4)

for some forms $g_{i}\in{\mathbb{R}}[a,b,c]_{4}$. Hence,
$L\in(\Sigma_{2,8}^{4})^{\star}\Leftrightarrow\langle{L},{q(a,b,c)^{4}}\rangle\in\Sigma_{3,4}^{2}\subseteq{\mathbb{R}}[a,b,c]_{4}$.
Note that it was proven by Hilbert that the cone of positive semidefinite
ternary quartics is equal to the sum of squares cone. This shows that the dual
cone $(\Sigma_{2,8}^{4})^{\star}$ is a linear slice of the sos cone
$\Sigma_{3,4}^{2}$ of ternary quartics. From Eq. 4 we see that the
coefficients of the ternary quartic $\langle{L},{q(a,b,c)^{4}}\rangle$ are not
linearly independent but that the ternary quartics are contained in a
9-dimensional subspace.

For $F\in{\mathbb{R}}[a,b,c]_{4}$ write
$F=\sum_{\alpha\in{\mathbb{Z}}^{3}}c_{\alpha}\underline{a}^{\alpha}$. We
calculate that the subspace $U$ of ${\mathbb{R}}[a,b,c]_{4}$ spanned by
ternary quartics of the above form is given by the vanishing of the six linear
forms

$\displaystyle\frac{2}{3}c_{220}-c_{301},\,3c_{130}-c_{211},\,\frac{2}{3}c_{022}-c_{103},\,3c_{031}-c_{112},\,12c_{040}-c_{121},\,6c_{040}-c_{202}.$

Equivalently, it is the orthogonal complement of $(b^{2}-ac)_{4}$ in
${\mathbb{R}}[a,b,c]_{4}$ with respect to apolarity. Understanding
$(\Sigma_{2,8}^{4})^{\star}$ is the same as studying $\Sigma_{3,4}^{2}\cap U$.
The facial structure of $\Sigma_{3,4}^{2}$ has been studied by Kunert in [32]
and is rather explicitly known.

Let $q_{1},\dots,q_{s}\in{\mathbb{R}}[x,y]_{2}$ and assume
$f=\sum_{i=1}^{s}q_{i}^{4}$ is on the boundary of $\Sigma_{2,8}^{4}$. Let
$L\in(\Sigma_{2,8}^{4})^{\star}$ be a linear functional in
${\mathbb{R}}[x,y]_{8}^{\scriptscriptstyle\vee}$ vanishing at $f$, then
$L(f)=0$ and thus $L(q_{i}^{4})=0$ ($i=1,\dots,s$) which shows that the points
in ${\mathbb{P}}^{2}{\mathbb{R}}={\mathbb{P}}{\mathbb{R}}[x,y]_{2}$
corresponding to $q_{1},\dots,q_{s}$ are zeros of the psd ternary quartic
$F=\langle{L},{q(a,b,c)^{4}}\rangle$.

The $\operatorname{GL}_{2}({\mathbb{R}})$ action on ${\mathbb{R}}[x,y]$ gives
rise to an embedding of $\operatorname{GL}_{2}({\mathbb{R}})$ into
$\operatorname{GL}_{3}({\mathbb{R}}[a,b,c]_{1})$ which allows some
automorphisms on ${\mathbb{R}}[a,b,c]_{4}$ that fix the subspace $U$ as well
as $\Sigma_{3,4}^{2}$.

Since we are dealing with real zeros of psd elements understanding the
behaviour of derivatives is essential. Let $L\in{\mathbb{R}}[x,y]_{8}$ and
$u=(u_{1},u_{2},u_{3})\in{\mathbb{R}}^{3}$. Via a direct calculation we see

$\partial_{u}\langle{L},{q(a,b,c)^{4}}\rangle=4\langle{L},{q(a,b,c)^{3}(u_{1}x^{2}+u_{2}xy+u_{3}y^{2})}\rangle$

where the partial derivatives are taken with respect to the variables $a,b,c$.

A different way to arrive at the same point is as follows. Let
$v_{2}\colon{\mathbb{P}}^{1}\to{\mathbb{P}}^{2}$ be the second Veronese map.
With coordinates $a,b,c$ on ${\mathbb{P}}^{2}$ the image is given as the
vanishing set of $b^{2}-ac$. This gives rise to an isomorphism
${\mathbb{R}}[x,y]_{2d}\cong{\mathbb{R}}[a,b,c]/(b^{2}-ac)_{d}$ for any $d\geq
1$. Moreover, under this isomorphism the cones $\Sigma_{2,8}^{4}$ and
$\Sigma^{4}{\mathbb{R}}[a,b,c]/(b^{2}-ac)_{1}$ are isomorphic. Here the second
cone consists of elements that are sums of fourth powers of linear forms.
Dualizing the second cone we get

$(\Sigma^{4}{\mathbb{R}}[a,b,c]/(b^{2}-ac)_{4})^{\star}\cong\Sigma_{3,4}^{2}\cap(b^{2}-ac)_{4}^{\perp}.$

The fact that the cone of sums of squares $\Sigma_{3,4}^{2}$ and the cone
$P_{3,4}$ of psd ternary quartics coincide will be essential for certain
calculations as testing for non-negativity is very hard.

Let $L\in\partial(\Sigma_{2,8}^{4})^{\star}$, then
$F:=\langle{L},{q(a,b,c)^{4}}\rangle$ is a psd ternary quartic in
${\mathbb{R}}[a,b,c]_{4}$ which therefore is also a sum of squares. Since $L$
is on the boundary there exists $g\in{\mathbb{R}}[x,y]_{2}$ such that
$L(g^{4})=0$. Thus $F$ has a real zero and lies on the boundary of the sos
cone $\Sigma_{3,4}^{2}$.

We claim that there are two cases. Either

  1. (i)

$|\mathcal{V}(F)({\mathbb{R}})|\leq 3$ or

  2. (ii)

$|\mathcal{V}(F)({\mathbb{R}})|=\infty$, $(L^{\perp})_{3}\neq\\{0\\}$ and
$F=l^{4}$ for some $l\in{\mathbb{R}}[a,b,c]_{1}$.

We use the characterization of faces of $\Sigma_{3,4}^{2}$ from Kunert [32].
In this thesis all faces of the sums of squares cone $\Sigma_{3,4}^{2}$ are
characterized in terms of the types of singularities of its relative interior
points. In particular, a quartic $F\in\Sigma_{3,4}^{2}$ has either infinitely
many real points or at most 4. Our main technical result is to prove that
there is no quartic in $\Sigma_{3,4}^{2}\cap U$ that has exactly four real
zeros.

For the rest of the paper we will slightly abuse notation and will see an
element of $(\Sigma_{2,8}^{4})^{\star}$ as either an element of
${\mathbb{R}}[x,y]_{8}$ or a ternary quartic in ${\mathbb{R}}[a,b,c]_{4}$. In
general, we use $f$ for binary octics, $F$ for ternary quartics and $L$ for
linear functionals in $({\mathbb{R}}[x,y]_{8})^{\scriptscriptstyle\vee}$. As
we have seen the latter two are both dual elements to ${\mathbb{R}}[x,y]_{8}$,
however both points of view have their separate advantages.

######  Remark 3.8.

Our results imply the following. There are the following types of ternary
quartics $F$ on the boundary of the dual cone $(\Sigma_{2,8}^{4})^{\star}$.

  1. (i)

$F=l^{4}$ with $l\in{\mathbb{R}}[a,b,c]_{1}$.

  2. (ii)

$\mathcal{V}(F)({\mathbb{R}})=\\{(\xi_{1}\colon\xi_{2}\colon\xi_{3})\\}$ with
$\xi_{2}^{2}-4\xi_{1}\xi_{3}>0$, i.e. the associated binary quadratic form is
indefinite. And for every such point $\xi$ there exists $F$ whose only real
zero is $\xi$.

  3. (iii)

$|\mathcal{V}(F)({\mathbb{R}})|=2,3$ and all points correspond to indefinite
binary quadratics. In this case not every pair and triple of points is
possible.

For $L\in{\mathbb{R}}[x,y]_{8}$ denote by $L^{\perp}$ the apolar ideal of $L$,
i.e. the ideal generated by all forms $g\in{\mathbb{R}}[x,y]$ such that
$\langle{g},{L}\rangle=0$. Note that ${\mathbb{R}}[x,y]/L^{\perp}$ is an
artinian Gorenstein ring with socle in degree 8. Moreover, since we are
working with binary forms the ideal $L^{\perp}$ is also a complete
intersection with the degrees of the two generators summing up to 10.
Considering this Gorenstein ideal already proved to be a main ingredient in
studying sum of squares cones, for example in [8].

######  Proposition 3.9.

Let $L\in(\Sigma_{2,8}^{4})^{\star}$ and $F=\langle{L},{q(a,b,c)^{4}}\rangle$.
If $|\mathcal{V}(F)({\mathbb{R}})|=\infty$, then $l^{\prime
3}\in(L^{\perp})_{3}$ for some $l^{\prime}\in{\mathbb{R}}[x,y]_{1}$ and every
element in $(L^{\perp})_{6}$ is divisible by $l^{\prime 2}$. Moreover, there
exists $l\in{\mathbb{R}}[a,b,c]_{1}$ such that $F=l^{4}$.

###### Proof.

By [32, §2.3.] either $F=p^{2}$ for some indefinite, irreducible
$p\in{\mathbb{R}}[a,b,c]_{2}$ or $F=l^{2}u$ for some
$u\in{\mathbb{R}}[a,b,c]_{2},l\in{\mathbb{R}}[a,b,c]_{1}$. Since $F$ is
singular at any real zero
$u=(u_{0}:u_{1}:u_{2})\in{\mathbb{P}}^{2}{\mathbb{R}}$ we see

$\langle{L},{q(u_{0},u_{1},u_{2})^{3}h}\rangle=0\quad\forall
h\in{\mathbb{R}}[x,y]_{2}.$

If $F=l^{2}u$, there exists a 2-dimensional subspace
$V\subseteq{\mathbb{R}}[x,y]_{2}$ such that for every $q\in V$, $q^{3}\in
L^{\perp}$, $V$ is given as the orthogonal of
$\operatorname{span}(l)\subseteq{\mathbb{R}}[a,b,c]_{1}\cong{\mathbb{R}}[x,y]_{2}$.
Since $\operatorname{span}(q^{3}\colon q\in V)=V^{3}$, we see
$V^{3}{\mathbb{R}}[x,y]_{2}\subseteq L^{\perp}$.

If $V$ is spanned by a regular sequence, the Hilbert function of the ideal
generated by $V$ is $(1,2,1)$. Since

$V^{3}{\mathbb{R}}[x,y]_{2}=V^{2}(V{\mathbb{R}}[x,y]_{2})=V^{2}({\mathbb{R}}[x,y]_{4})=V(V{\mathbb{R}}[x,y]_{4})=V{\mathbb{R}}[x,y]_{6}={\mathbb{R}}[x,y]_{8}$

and $\langle{\cdot},{L}\rangle$ vanishes on it we have a contradiction.
Therefore $V$ is spanned by $l^{\prime}l_{1},l^{\prime}l_{2}$ for some linear
forms $l^{\prime},l_{1},l_{2}\in{\mathbb{R}}[x,y]_{1}$. Then
$V=l^{\prime}{\mathbb{R}}[x,y]_{1}$ and $V^{3}{\mathbb{R}}[x,y]_{2}=l^{\prime
3}{\mathbb{R}}[x,y]_{5}$, hence $l^{\prime 3}$ is in the apolar ideal of $L$.
Additionally, since the apolar ideal is a complete intersection and there is
one generator in degree at most 3, the other generator has degree at least 7
(see [25, Thm 1.44.]). In particular, every form in degree 6 is divisible by
$l^{\prime 2}$. Thus every real zero of $F$ is a zero of $l$. Moreover,
$l^{\prime 8}\in L^{\perp}$. Using the calculation in the proof of 4.2 we see
that $F=l^{4}$.

Assume $F=p^{2}$ with $p$ indefinite, irreducible. Since $F\in U$ the
quadratic form $\langle{F},{b^{2}-ac}\rangle\in{\mathbb{R}}[a,b,c]_{2}$ has to
vanish identically. This gives six equations. One then easily checks on a
computer that the determinant of the quadratic form $p$ is contained in the
radical of the ideal generated by the six forms above, i.e. $p$ is reducible,
a contradiction. ∎

######  Lemma 3.10.

Let $q_{1},q_{2}\in{\mathbb{P}}{\mathbb{R}}[x,y]_{2}$ linearly independent
with a common factor and $q_{1}^{3},q_{2}^{3}\in L^{\perp}$. Then
$l^{3}\in(L^{\perp})_{3}$ for some $l\in{\mathbb{R}}[x,y]_{1}$.

###### Proof.

Let $q_{1}=ll_{1},q_{2}=ll_{2}$ with $l,l_{1},l_{2}\in{\mathbb{R}}[x,y]_{1}$.
By assumption
$(ll_{1})^{3}{\mathbb{R}}[x,y]_{2},(ll_{2})^{3}{\mathbb{R}}[x,y]_{2}\subseteq
L^{\perp}$. Since $l_{1},l_{2}$ are a regular sequence, the Hilbert function
of the ideal $\langle l_{1}^{3},l_{2}^{3}\rangle$ is $(1,2,3,2,1)$. Thus

$L^{\perp}\supseteq(ll_{1})^{3}{\mathbb{R}}[x,y]_{2}+(ll_{2})^{3}{\mathbb{R}}[x,y]_{2}=l^{3}\langle
l_{1}^{3},l_{2}^{3}\rangle_{5}=l^{3}{\mathbb{R}}[x,y]_{5}.$

∎

######  Theorem 3.11.

Let $F\in\partial(\Sigma_{2,8}^{4})^{\star}=\partial\Sigma_{3,4}^{2}\cap U$
with $\mathcal{V}(F)({\mathbb{R}})$ finite. Then
$|\mathcal{V}(F)({\mathbb{R}})|\leq 3$.

###### Proof.

By [32, Lemma 1.38.] we have $|\mathcal{V}(F)({\mathbb{R}})|\leq 4$ since
$\mathcal{V}(F)({\mathbb{R}})$ is finite. Assume
$|\mathcal{V}(F)({\mathbb{R}})|=4$. Since $\mathcal{V}(F)({\mathbb{R}})$ is
finite, no three of its real zeros are collinear. Let
$q_{1},\dots,q_{4}\in{\mathbb{R}}[x,y]_{2}$ be the corresponding quadratic
forms. If any two of them were not a complete intersection then by 3.10 there
exists $l\in{\mathbb{R}}[x,y]_{1}$ such that $l^{3}\in(L^{\perp})_{3}$ and
thus $(ll^{\prime})^{4}\in L^{\perp}$ for all
$l^{\prime}\in{\mathbb{R}}[x,y]_{1}$ which shows that $F$ has infinitely many
zeros.

Pick three of these points
$\xi_{1},\xi_{2},\xi_{3}\in{\mathbb{P}}^{2}{\mathbb{R}}$ and let
$I\subseteq{\mathbb{R}}[a,b,c]$ be the ideal of forms vanishing at
$\xi_{1},\xi_{2},\xi_{3}$. The degree 2 component $I_{2}$ is generated by
$q_{1}=l_{12}l_{13},q_{2}=l_{13}l_{23},q_{3}=l_{13}l_{23}$ where $l_{ij}$ is
the line through $\xi_{i},\xi_{j}$. Concretely, $l_{ij}$ is given as

$l_{ij}=\det\begin{pmatrix}\xi_{i}^{1}&\xi_{i}^{2}\\\
\xi_{j}^{1}&\xi_{j}^{2}\end{pmatrix}a+\det\begin{pmatrix}\xi_{i}^{2}&\xi_{i}^{0}\\\
\xi_{j}^{2}&\xi_{j}^{0}\end{pmatrix}b+\det\begin{pmatrix}\xi_{i}^{0}&\xi_{i}^{1}\\\
\xi_{j}^{0}&\xi_{j}^{1}\end{pmatrix}c$

where $\xi_{i}=(\xi_{i}^{j})_{j}$. Consider the $15\times 15$-matrix where the
first 9 rows correspond to a basis of $U$ (with respect to some basis of
${\mathbb{R}}[a,b,c]_{4}$) and the last 6 correspond to generators of
$(I_{2}^{2})_{4}$. Since $F$ is psd it vanishes to order 2 at
$\xi_{1},\xi_{2},\xi_{3}$ and is therefore contained in $U\cap(I_{2}^{2})_{4}$
which shows that the determinant of this matrix is zero.

Since $F\in(I_{2}^{2})_{4}$ we can represent
$F=(q_{1},q_{2},q_{3})G(q_{1},q_{2},q_{3})^{t}$ with a $3\times 3$-Gram matrix
with respect to the basis $(q_{1},q_{2},q_{3})$. Since $F$ is psd (and thus
sos since it is a ternary quartic) and vanishes at the real points
$\xi_{1},\xi_{2},\xi_{3}$, every sos representation of $F$ corresponds to such
a matrix $G$, i.e. there is no sos representation $F=\sum_{i=1}^{r}p_{i}^{2}$
where
$\operatorname{span}(p_{1},\dots,p_{r})\not\subseteq\operatorname{span}(q_{1},q_{2},q_{3})$.
Since $F$ has a fourth real zero $F$ is a sum of two squares and there exists
a psd matrix $G$ of rank 2. Note that $G$ cannot have rank 1 since
$\mathcal{V}(F)({\mathbb{R}})$ is finite.

We want to use Mathematica [26] to prove that this is not possible. Let $J$ be
the ideal generated by the determinant of the $15\times 15$-matrix and by the
determinant of the $3\times 3$-matrix $G$. Moreover, let $\chi_{G}(t)$ be the
characteristic polynomial of $G$, $\chi_{G}(t)=t^{3}+u_{2}t^{2}+u_{1}t+u_{0}$.
We saturate the ideal $J$ at $u_{1}$ to get $J_{1}:=(J\colon u_{1}^{\infty})$.
If $u_{1}$ was 0 then the matrix $G$ would have rank 1 which is impossible. We
can now calculate the minimal primes of $J_{1}$ of which there are 4.

For every minimal prime we use FindInstance in Mathematica to prove that there
are no real points $\xi_{1},\xi_{2},\xi_{3}$ such that the equations in the
prime ideal and the two psd conditions are satisfied. ∎

######  Remark 3.12.

Note that we cannot take the two equations defining $J$ and input them into
Mathematica instead of the minimal primes as the calculation does not finish.
Similarly, we can only compute the minimal primes after saturating.

To actually perform the calculations we mod out the action of
$\operatorname{GL}_{2}({\mathbb{R}})\subseteq\operatorname{GL}_{3}({\mathbb{R}})$
which fixes the subspace $U$. The embedding is given by the
$\operatorname{GL}_{2}({\mathbb{R}})$ action on the second Veronese of
${\mathbb{P}}^{1}$ induced by the $\operatorname{GL}_{2}({\mathbb{R}})$ action
on ${\mathbb{P}}^{1}$.

By 3.10 we may assume that any pair of $q_{i}$’s forms a complete intersection
and that the triple is linearly independent.

We have the following representatives for the orbits. If one $q_{i}$ is
positive definite then after a change of coordinates

$(1:0:1),(1:0:a),(b:1:c).$

If all $q_{i}$ would have rank 1 they are

$(1:0:0),(0:0:1),(1:2:1),$

but their third powers generate ${\mathbb{R}}[x,y]_{8}$, hence this case
cannot occur. Lastly, if none of the quadratic forms is positive definite but
at least one has rank 2, then we can achieve one of the following two forms

$(0:1:0),(1:a:1),(1:b:c)$

or

$(0:1:0),(1:a:-1),(1:b:c).$

Moreover, note that if we make the first point $(1:0:-1)$ in the last two
cases the saturation does not terminate any longer.

We thus have to do the calculations above for the three cases

  1. (i)

$(1:0:1),(1:0:a),(b:1:c)$,

  2. (ii)

$(0:1:0),(1:a:1),(1:b:c)$,

  3. (iii)

$(0:1:0),(1:a:-1),(1:b:c)$.

We can do the first case with a different argument as the above calculations
would not terminate.

######  Lemma 3.13.

Let $L\in(\Sigma_{2,8}^{4})^{\star}$ and $F=\langle{L},{q(a,b,c)^{4}}\rangle$.
$F$ does not vanishes at any real point $\xi\in{\mathbb{P}}^{2}{\mathbb{R}}$
corresponding to a definite binary quadratic form $q$.

###### Proof.

Assume otherwise. After a change of coordinates we can assume that
$q=x^{2}+y^{2}$. By [37, Thm 9.5.] $q^{4}$ is a sum of 5 fourth powers of real
linear forms. This shows that $F$ vanishes at least at 5 real points, hence
vanishes at infinitely many real points. By 3.9 there exists
$l\in{\mathbb{R}}[x,y]_{1}$ such that $l|q$ which is impossible if $q$ is
positive definite. ∎

######  Remark 3.14.

The code for the other two calculations can be found in the appendix.

###### Proof of 3.4.

Let $f\in\partial\Sigma_{2,8}^{4}$ and assume $f$ is not divisible by a fourth
power. Let $f=\sum_{i=1}^{r}q_{i}^{4}$ with
$q_{1},\dots,q_{r}\in{\mathbb{R}}[x,y]_{2}$. There exists
$L\in(\Sigma_{2,8}^{4})^{\star}$ with $L(f)=0$, hence $L(q_{i}^{4})=0$ for all
$i=1,\dots,r$. Let
$F:=\langle{L},{q(a,b,c)^{4}}\rangle\in{\mathbb{R}}[a,b,c]_{4}$. If $r>3$ then
$F$ has infinitely many real zeros and there exists
$l\in{\mathbb{R}}[x,y]_{1}$ such that $l^{3}\in L^{\perp}$. Especially, the
Gorenstein ideal $L^{\perp}$ is either a complete intersection generated in
degrees $3,7$ or $2,8$. Especially, its degree 6 component is given by
$u{\mathbb{R}}[x,y]_{6-\deg u}$ with $u$ of either degree 2 or 3. But
$q_{1}^{3},\dots,q_{r}^{3}$ are contained in $I_{6}$, which shows that all
$q_{i}$ have a common factor. Hence, $f$ is divisible by the fourth power of
this linear factor.

Therefore, every such $f\in\partial\Sigma_{2,8}^{4}$ has a length 3
representation. The complement $\\{l^{4}\Sigma_{2,4}^{4}\colon
l\in{\mathbb{R}}[x,y]_{1}\\}$ has dimension 6. Since the length is lower-
semicontinuous, every $f$ on the boundary has length at most 3.

Hence, we can compute the boundary as (some of the) singular values of the map

$\phi_{2}^{3}\colon{\mathbb{R}}[x,y]_{2}^{3}\to{\mathbb{R}}[x,y]_{8},\quad(q_{1},q_{2},q_{3})\mapsto
q_{1}^{4}+q_{2}^{4}+q_{3}^{4}.$

The differential at a point $(q_{1},q_{2},q_{3})$ is a linear map between
9-dimensional vector spaces of which we can compute the determinant with
respect to the monomial bases. Considering the coefficients of
$q_{1},q_{2},q_{3}$ as indeterminates, this determinant factors into two
irreducible factors (cf. 3.6). One is the determinant of the $3\times 3$
coefficient matrix of the quadratic forms $q_{1},q_{2},q_{3}$. The other is
the hypersurface $\mathcal{G}$ which we may check is irreducible over
${\mathbb{C}}$ with Macaulay2 [21].

If the three quadratic forms are linearly dependent the psd ternary quartic
$F$ vanishes at three collinear points and therefore has infinitely many
zeros. This shows that forms $f=q_{1}^{4}+q_{2}^{4}+q_{3}^{4}$ with
$q_{1},q_{2},q_{3}$ linearly dependent do not give rise to a component of the
algebraic boundary. (The next Proposition shows a stronger purely algebraic
statement.)

Therefore, the algebraic boundary is given as the Zariski-closure of the image
of the hypersurface $\mathcal{G}$ under the map $\phi_{2}^{3}$. ∎

######  Remark 3.15.

Since we know that the algebraic boundary is irreducible, it is also given as
the following dual variety by [42, Theorem 3.7.]. For some faces
$\operatorname{\mathfrak{F}}$ of $\Sigma_{3,4}^{2}$ whose relative interior
points consist of psd ternary quartics with exactly 3 real zeros the
intersection $\operatorname{\mathfrak{F}}\cap U$ has dimension 1. A generator
of this 1-dimensional cone spans an extreme ray of $\Sigma_{3,4}^{2}\cap U$.
If $X$ denotes the variety given as the Zariski-closure of all such extreme
rays, its dual variety $X^{\star}$ is precisely the algebraic boundary of
$\Sigma_{2,8}^{4}$.

######  Proposition 3.16.

Let $D\subseteq\mathbb{R}[\underline{x}]_{2}^{3}$ be the subvariety containing
linearly dependent triples. Then $\operatorname{codim}\phi_{2}^{3}(D)=2$.

###### Proof.

We parametrize the set of linearly dependent triples by
${\mathbb{R}}[x,y]_{2}^{2}\times{\mathbb{R}}^{2}$ as
$(p_{1},p_{2},\lambda_{1}p_{1}+\lambda_{2}p_{2})$.

$\phi^{\prime}\colon\mathbb{R}[\underline{x}]_{2}^{2}\times{\mathbb{R}}^{2}\to\mathbb{R}[\underline{x}]_{8},\quad(p_{1},p_{2},\lambda_{1},\lambda_{2})\mapsto
p_{1}^{4}+p_{2}^{4}+(\lambda_{1}p_{1}+\lambda_{2}p_{2})^{4}$

The differential is given by ($h:=\lambda_{1}p_{1}+\lambda_{2}p_{2}$)

$\displaystyle\operatorname{d\phi}^{\prime}(p_{1},p_{2},\lambda_{1},\lambda_{2})\colon$
$\displaystyle\mathbb{R}[\underline{x}]_{2}^{2}\times{\mathbb{R}}^{2}\to\mathbb{R}[\underline{x}]_{8},$
$\displaystyle(f,g,\mu_{1},\mu_{2})\mapsto
4f(p_{1}^{3}+h^{3}\lambda_{1})+4g(p_{2}^{3}+h^{3}\lambda_{2})+4h^{3}p_{1}\mu_{1}+4h^{3}p_{2}\mu_{2}.$

Hence, the image is $W:=\langle
p_{1}^{3}+\lambda_{1}h^{3},p_{2}^{3}+\lambda_{2}h^{3},h^{3}p_{1},h^{3}p_{2}\rangle_{8}$
which we show has dimension at most 7. The dimension is maximal for a generic
choice of $p_{1},p_{2},\lambda_{1},\lambda_{2}$, hence we may assume that
$p_{1},p_{2}$ are linearly independent. Let $q\in{\mathbb{R}}[x,y]_{2}$ be
such that $p_{1},p_{2},q$ form a basis of ${\mathbb{R}}[x,y]_{2}$ and let
$V:=\operatorname{span}(p_{1}^{3}+\lambda_{1}h^{3},p_{2}^{3}+\lambda_{2}h^{3})$.
Then

$\displaystyle W$
$\displaystyle=V\operatorname{span}(p_{1},p_{2},q)+h_{3}\operatorname{span}(p_{1},p_{2})$
$\displaystyle=qV+\operatorname{span}(p_{1},p_{2})\operatorname{span}(V,h^{3})$
$\displaystyle=qV+\operatorname{span}(p_{1},p_{2})\operatorname{span}(p_{1}^{3},p_{2}^{3},h^{3})$
$\displaystyle\subseteq qV+\operatorname{span}(p_{1},p_{2})^{4}$

As $\dim\operatorname{span}(p_{1},p_{2})^{4}\leq 5$ we see $\dim W\leq 7$.
Moreover, for a generic choice of $p_{1},p_{2},\lambda_{1},\lambda_{2}$, the
dimension of $W$ is equal to 7. ∎

######  Remark 3.17.

This proves that also the interior of
$\Sigma_{2,8}^{4}\smallsetminus\phi_{2}^{3}(D)$ is still connected. Thus if
there exist forms of length 4, these are separated from forms of length 3 by
$\overline{\phi_{2}^{3}(\mathcal{G})}$ only. From numerical calculations in
Julia [7] using the HomotopyContinuation.jl package [10] we know that there
are binary forms with different numbers of real length 3 representations.
Again, regions with different numbers of decompositions have to be separated
by $\overline{\phi_{2}^{3}(\mathcal{G})}$ only. This also proves that
$\Sigma_{2,8}^{4}\smallsetminus\overline{\phi_{2}^{3}(\mathcal{G})}$ does
indeed have several connected components (assuming that all solutions have
been found using Homotopy Continuation).

Up to sign and permutation of the summands $f_{i}$ has exactly $i$ real length
3 representations.

$\displaystyle f_{2}$
$\displaystyle=(4x^{2}-5xy+2y^{2})^{4}+(4x^{2}-4xy+2y^{2})^{4}+(4x^{2}+2xy-y^{2})^{4}$
$\displaystyle f_{4}$
$\displaystyle=(6x^{2}+10xy-2y^{2})^{4}+(3x^{2}-2xy+5y^{2})^{4}+(10x^{2}+4xy-6y^{2})^{4}$
$\displaystyle f_{6}$
$\displaystyle=(x^{2}+4xy+8y^{2})^{4}+(10x^{2}+xy+3y^{2})^{4}+(10x^{2}+10xy-10y^{2})^{4}$

The number of complex length 3 representations was constant at 76 for all
(generic) examples computed. The fact that all representations of the
$f_{i}$’s that have been found numerically, give correct decompositions can be
proven using the cert command in the HomotopyContinuation package. This does
not prove that all representations have been found even though this seems to
be the case here as the same number of complex solutions $29184=384\cdot 76$
has been found in every example calculated. This number is the number of
representations without accounting for permutations of the summands and
scaling summands by roots of unity.

######  Example 3.18.

One example of a psd ternary quartic in the cone
$(\Sigma_{2,8}^{4})^{\star}=(\Sigma_{3,4}^{2}\cap U)$ is given by the
following Gram matrix

$\tiny\left(\begin{array}[]{r}-918256c^{7}+14398014c^{6}-284749896c^{5}+34054823830c^{4}-389117333868c^{3}+384885454002c^{2}-323875629964c-248202\\\
-659350c^{7}+20329939c^{6}-68106915c^{5}-4687738385c^{4}+48361507226c^{3}-47864048337c^{2}+39815341637c+30859\\\
-95223c^{7}-2098390c^{6}+35378289c^{5}+1998413271c^{4}-24797139347c^{3}+24583084470c^{2}-20789333367c-15937\end{array}\right.$
$\tiny\begin{array}[]{r}-659350c^{7}+20329939c^{6}-68106915c^{5}-4687738385c^{4}+48361507226c^{3}-47864048337c^{2}+39815341637c+30859\\\
-823044c^{7}+6018130c^{6}+23370756c^{5}+30387414c^{4}-98730364c^{3}+120560334c^{2}-64520364c-50\\\
4786c^{7}-790488c^{6}+6085491c^{5}-99186325c^{4}+1409481342c^{3}-1392614400c^{2}+1196060035c+927\end{array}$
$\tiny\left.\begin{array}[]{r}-95223c^{7}-2098390c^{6}+35378289c^{5}+1998413271c^{4}-24797139347c^{3}+24583084470c^{2}-20789333367c-15937\\\
4786c^{7}-790488c^{6}+6085491c^{5}-99186325c^{4}+1409481342c^{3}-1392614400c^{2}+1196060035c+927\\\
-5663c^{7}-26893c^{6}-858726c^{5}+128759213c^{4}-1359409919c^{3}+1343865411c^{2}-1122516338c-859\end{array}\right)$

where $c$ is the real zero of

$c^{8}-c^{7}+c^{6}-133c^{5}-135542c^{4}+1550171c^{3}-1534319c^{2}+1290239c+1$

that is approximately $\text{-}21$ and with respect to the basis
$(q_{1},q_{2},q_{3})$ with

$\displaystyle q_{1}$ $\displaystyle=-(x+z)(cx-z)$ $\displaystyle q_{2}$
$\displaystyle=(x+z)((c-16)x-(c+1)y-17z)$ $\displaystyle q_{3}$
$\displaystyle=-(cx-z)((c-16)x-(c+1)y-17z).$

It vanishes at the real points $(0:1:0),(1:1:-1),(1:-16:c)$ and therefore
exposes the 3-dimensional face

$\mathrm{cone}((xy)^{4},(x^{2}+xy-y^{2})^{4},(x^{2}-16xy+cy^{2})^{4})$

of the cone $\Sigma_{2,8}^{4}$. The linear functional in
${\mathbb{R}}[x,y]_{8}^{\scriptscriptstyle\vee}$ is a generator of the degree
8 component of the Gorenstein ideal
$\langle(xy)^{3},(x^{2}+xy-y^{2})^{3},(x^{2}-16xy+cy^{2})^{3}\rangle_{8}$, a
complete intersection minimally generated in degree $5$.

We also give examples of psd ternary quartics in $U$ with exactly 1 and 2 real
zeros. Wrt the basis $x(x+z),xy,z(x+z),yz$ the following Gram matrix defines a
form in $(\Sigma_{2,8}^{4})^{\star}$ with real zeros $(0:1:0)$ and $(1:0:-1)$:

$\left(\begin{array}[]{rrrr}2&0&\text{-}1&0\\\ 0&3&0&0\\\ \text{-}1&0&2&0\\\
0&0&0&3\end{array}\right).$

In particular, taking the sum of the two quartics with 3 and 2 real zeros
gives a quartic in $(\Sigma_{2,8}^{4})^{\star}$ with the real zero $(0:1:0)$.

######  Remark 3.19.

Consider the irreducible hypersurface

$\mathcal{G}=\overline{\\{(q_{1},q_{2},q_{3})\in{\mathbb{R}}[x,y]_{2}^{3}\colon
q_{1},q_{2},q_{3}\text{ linearly
independent},\langle\underline{q}^{3}\rangle_{8}\neq{\mathbb{R}}[x,y]_{8}\\}}.$

The closure of the image $Z^{\prime}$ of the rational map
$\phi\colon\mathcal{G}\dashrightarrow{\mathbb{P}}{\mathbb{R}}[x,y]_{8},\,(q_{1},q_{2},q_{3})\mapsto\langle\underline{q}\rangle_{8}^{\perp}$
is also irreducible.

Consider the two subsets of ${\mathbb{R}}[x,y]_{8}$, $\operatorname{Gor}(5,5)$
and $\operatorname{Gor}(4,6)$ parametrizing Gorenstein ideals/complete
intersection of socle degree 8 with minimal generators of degree $5,5$ and
$4,6$ respectively. By [25] we have
$\operatorname{Gor}(4,6)\subseteq\overline{\operatorname{Gor}(5,5)}$ and
$\overline{\operatorname{Gor}(4,6)}\cap\operatorname{Gor}(5,5)=\varnothing$
and both are irreducible. Thus, either
$Z^{\prime}\subseteq\overline{\operatorname{Gor}(4,6)}$ or
$Z^{\prime}\cap\operatorname{Gor}(5,5)$ is dense in $Z^{\prime}$. Since
$Z^{\prime}\cap\operatorname{Gor}(5,5)$ is non-empty by Example 3.18 the
second case holds.

More concretely this means that if $\underline{q}\in\mathcal{G}$ is generic
then there exist elements $p_{1},p_{2}\in{\mathbb{R}}[x,y]_{5}$ such that
$\langle q_{1}^{3},q_{2}^{3},q_{3}^{3}\rangle_{d}=\langle
p_{1},p_{2}\rangle_{d}$ for all degrees $d\geq 6$.

##  4\. Facial structure

We study the facial structure of the cone $\Sigma_{2,8}^{4}$. We prove that
there are exactly four types of exposed faces on $\Sigma_{2,8}^{4}$. Three of
them are polyhedral, they are of the form
$F_{i}:=\operatorname{cone}(q_{1}^{4},\dots,q_{i}^{4})$ with
$i\in\\{1,2,3\\}$. And one is not polyhedral, it is of the form
$l^{4}\Sigma_{2,4}^{4}$ for some $l\in{\mathbb{R}}[x,y]_{1}$. Moreover, there
are some non-exposed faces contained in $l^{4}\Sigma_{2,4}^{4}$.

######  Proposition 4.1.

Every exposed face of $\Sigma_{2,8}^{4}$ is of one of the four types
$F_{1},F_{2},F_{3}$ or $l^{4}\Sigma_{2,4}^{4}$ for some
$l\in{\mathbb{R}}[x,y]_{1}$. Moreover, there are exposed faces of every such
type.

###### Proof.

Let $F\in\partial(\Sigma_{2,8}^{4})^{\star}$ be a psd ternary quartic. First
assume that $\mathcal{V}(F)({\mathbb{R}})$ is finite and $F$ has real zeros
corresponding to the binary quadratic forms $q_{1},\dots,q_{i}$ with
$i\in\\{1,2,3\\}$. Then the face exposed by $F$ contains
$\operatorname{cone}(q_{1}^{4},\dots,q_{i}^{4})$ and cannot contain any other
form as we know all zeros of $F$. By Example 3.18 there exist ternary quartics
in $\partial(\Sigma_{2,8}^{4})^{\star}$ with exactly one, two and three real
zeros.

Now, assume $F$ has infinitely many real zeros. If
$F(\sum_{i=1}^{r}q_{i}^{4})=0$ and $F$ is singular at any of its real zeros we
see that $F(q_{i}^{3}h)=0$ for all $i=1,\dots,r$ and all
$h\in{\mathbb{R}}[x,y]_{2}$. By 3.9 $q_{i}$ is divisible by $l$ for some
$l\in{\mathbb{R}}[x,y]_{1}$ and thus the face exposed by $F$ is contained in
$l^{4}\Sigma_{2,4}^{4}$. Moreover, since $l^{3}\in L^{\perp}$ any form
$l^{3}(lp)$ with $p\in\Sigma_{2,4}^{4}$ is contained in the face exposed by
$F$. ∎

######  Theorem 4.2.

Let $0\neq\operatorname{\mathfrak{F}}\subsetneq\Sigma_{2,8}^{4}$ be a face.
Then $\operatorname{\mathfrak{F}}$ is one of the following:

  * •

an exposed face of the form
$F_{i}=\operatorname{cone}(q_{1}^{4},\dots,q_{i}^{4})$ for some $i=1,2,3$
where all $q_{j}$ are indefinite binary quadratic forms,

  * •

an exposed face of type $l^{4}\Sigma_{2,4}^{4}$,

  * •

a non-exposed extreme ray of the form
$\operatorname{cone}(l^{8})={\mathbb{R}}_{+}l^{8}$ with
$l\in{\mathbb{R}}[x,y]_{1}$, or

  * •

a non-exposed face of the form $l^{4}\operatorname{cone}(l_{1}^{4},l_{2}^{4})$
with $l,l_{1},l_{2}\in{\mathbb{R}}[x,y]_{1}$ and $l_{1},l_{2}$ linearly
independent.

###### Proof.

Let $f\in\partial\Sigma_{2,8}^{4}$ with a representation
$f=\sum_{i=1}^{j}q_{i}^{4}$ and $j\leq 3$. Let
$\operatorname{\mathfrak{F}}\subseteq\Sigma_{2,8}^{4}$ be the supporting face
of $f$, i.e. the smallest face containing $f$ in its relative interior.

$j=3$: If the three quadratic forms $q_{1},q_{2},q_{3}$ are linearly
independent, they do not have a common factor and thus any ternary quartic
$F\in(\Sigma_{2,8}^{4})^{\star}$ that vanishes on
$\operatorname{\mathfrak{F}}$ has exactly 3 real zeros by 3.9. But then $F$
already exposes this face. If the three quadratic forms are linearly dependent
and are all divisible by $l\in{\mathbb{R}}[x,y]_{1}$, then $F$ already has to
vanish on the line through all three points by Bezout’s theorem, hence has
infinitely many zeros. Since $f$ has a length 3 representation $f$ is
contained in the interior of $l^{4}\Sigma_{2,4}^{4}$ and hence this face is
equal to $\operatorname{\mathfrak{F}}$.

$j=1$: Any form $l^{8}$ and $(ll^{\prime})^{4}$ defines an extreme ray since
this representation as a sum of fourth powers is unique. By 3.13 $q^{4}$,
$q\in{\mathbb{R}}[x,y]_{2}$ is not on the boundary of $\Sigma_{2,8}^{4}$
whenever $q$ is positive definite.

We claim that $l^{8}$ is not exposed and $(ll^{\prime})^{4}$ is exposed. After
a change of coordinates $l^{8}=x^{8}$. We calculate

$U\cap
I((1:0:0))_{2}^{2}=\operatorname{span}(xy^{3}+3x^{2}yz,y^{2}z^{2}+\frac{2}{3}xz^{3},y^{3}z+3xyz^{2},y^{4}+12xy^{2}z+6x^{2}z^{2},z^{4},yz^{3}).$

The Gram matrices of an element

$xy^{3}a_{1}+3x^{2}yza_{1}+y^{2}z^{2}a_{2}+\frac{2}{3}xz^{3}a_{2}+y^{3}za_{3}+3xyz^{2}a_{3}+y^{4}a_{4}+12xy^{2}za_{4}+6x^{2}z^{2}a_{4}+z^{4}a_{5}+yz^{3}a_{6}$

in the intersection have the form

$\left(\begin{array}[]{rrrrr}0&9a_{1}&3a_{1}&36a_{4}-\lambda_{1}&9a_{3}-\lambda_{2}\\\
9a_{1}&36a_{4}&\lambda_{1}&\lambda_{2}&2a_{2}\\\
3a_{1}&\lambda_{1}&6a_{4}&3a_{3}&\lambda_{3}\\\
36a_{4}-\lambda_{1}&\lambda_{2}&3a_{3}&6a_{2}-2\lambda_{3}&3a_{6}\\\
9a_{3}-\lambda_{2}&2a_{2}&\lambda_{3}&3a_{6}&6a_{5}\end{array}\right)$

for some $\lambda_{1},\lambda_{2},\lambda_{3}\in{\mathbb{R}}$ with respect to
the basis $xy,xz,y^{2},yz,z^{2}$. By checking all principle $2\times 2$
minors, it is easy to see that the only form in the intersection having a psd
Gram matrix is $a_{5}z^{4}$ with $a_{5}>0$. This shows that $x^{8}$ and
therefore $l^{8}$ is not exposed. Since exposed extreme rays are dense in the
set of all extreme rays and $(ll^{\prime})^{4}$ is one
$\operatorname{GL}_{2}({\mathbb{R}})$ orbit they have to be exposed..

$j=2$: Assume the two summands have a common factor. Write
$q_{1}=ll_{1},\,q_{2}=ll_{2}$. We claim that the supporting face of
$f=l^{4}(l_{1}^{4}+l_{2}^{4})$ is given as
$l^{4}\operatorname{cone}(l_{1}^{4},l_{2}^{4})$. This is true if every
$g\in\operatorname{cone}(l_{1}^{4},l_{2}^{4})$ has a unique representation as
a sum of fourth powers. Since $g\in\partial\Sigma_{2,4}^{4}$ all
representations are of length at most 2. By [37, Thm 5.1.] such
representations are unique up to permutation and sign.

Assume the summands do not have a common factor. Since $f$ is on the boundary
there exists a psd ternary quartic $F\in(\Sigma_{2,8}^{4})^{\star}$ vanishing
at the two points. Since the two do not have a common factor, $F$ has at most
3 real zeros. If $F$ has exactly these two real zeros, then
$\operatorname{\mathfrak{F}}=\operatorname{cone}(q_{1}^{4},q_{2}^{4})$ and
$\operatorname{\mathfrak{F}}$ is exposed by $F$. If $F$ has 3 real zeros, the
face exposed by $F$ is $\operatorname{cone}(q_{1}^{4},q_{2}^{4},q_{3}^{4})$
for some $q_{3}\in{\mathbb{R}}[x,y]_{2}$. This is a polyhedral face with $f$
on its boundary. Thus $\operatorname{\mathfrak{F}}$ has dimension 2 and is
given as $\operatorname{cone}(q_{1}^{4},q_{2}^{4})$. ∎

######  Remark 4.3.

Note that the structure of the extreme rays also implies that the algebraic
boundary of the dual cone $(\Sigma_{2,8}^{4})^{\star}$ is irreducible and is
given as the dual variety of the Zariski-closure of the set
$(ll^{\prime})^{4}$ over all $l,l^{\prime}\in{\mathbb{R}}[x,y]_{1}$ by [42,
Theorem 3.7.]

######  Remark 4.4.

$\Sigma_{2,8}^{4}$ is not a spectrahedron due to the existence of non-exposed
faces. It is of course a spectrahedral shadow since its dual cone is.

######  Remark 4.5.

-

$\Sigma_{2,8}^{4}$$l^{4}\Sigma_{2,4}^{4}$$F_{3}$$l^{4}(l_{1}^{4}+l_{2}^{4})$$F_{2}$$l^{8}$$(ll^{\prime})^{4}$95321

The numbers on the right denote the dimension of faces in that line.

On the left-hand side, the two faces $l^{4}(l_{1}^{4}+l_{2}^{4})$ and $l^{8}$
are not exposed, whereas $l^{4}\Sigma_{2,4}^{4}$ is exposed. On the right-hand
side $F_{3}$ and $(l_{1}l_{2})^{4}$ are exposed. There are faces of type
$F_{2}$ that are exposed, however we do not know if this is always the case.
Except for $l^{4}\Sigma_{2,4}^{4}$ and $\Sigma_{2,8}^{4}$ itself all faces are
polyhedral.

We see from our description of the boundary that Reznick’s Conjecture [38,
Conjecture 7.1] does not hold. It states that every $f\in\Sigma_{2,8}^{4}$ is
“doubly positive” in the sense that it can be written as
$f=f_{1}^{2}+f_{2}^{2}$ with $f_{1},f_{2}\in{\mathbb{R}}[x,y]_{4}$ both
themselves psd.

######  Corollary 4.6.

Let $f\in\partial\Sigma_{2,8}^{4}$ of length 3 and such that $f\notin
l^{4}\Sigma_{2,4}^{4}$ for any $l\in{\mathbb{R}}[x,y]_{1}$. Then $f$ cannot be
written as $f=f_{1}^{2}+f_{2}^{2}$ with $f_{1},f_{2}\in{\mathbb{R}}[x,y]_{4}$
psd.

In particular, this holds for a generic $f\in\partial\Sigma_{2,8}^{4}$.

###### Proof.

Assume $f=f_{1}^{2}+f_{2}^{2}$ for some $f_{1},f_{2}\in{\mathbb{R}}[x,y]_{4}$
psd. Write $f_{i}=p_{i}^{2}+q_{i}^{2}$ with
$p_{i},q_{i}\in{\mathbb{R}}[x,y]_{2}$. If both $f_{1},f_{2}$ are squares, we
may assume $f=p_{1}^{4}+p_{2}^{4}$, and thus $f$ has length at most 2.

Assume $f_{1}$ is not a square and consider the identity

$(p_{1}^{2}+q_{1}^{2})^{2}=\frac{1}{18}((\sqrt{3}p_{1}+q_{1})^{4}+(\sqrt{3}p_{1}-q_{1})^{4}+16q_{1}^{4}).$
(5)

The same identity holds also with the roles of $p_{1}$ and $q_{1}$ swapped.
Let $L\in(\Sigma_{2,8}^{4})^{\star}$ with $L(f)=0$. Then by Eq. 5 we see

$L(p_{1}^{4})=L(q_{1}^{4})=L((\sqrt{3}p_{1}+q_{1})^{4})=0.$

However, $p_{1}.q_{1},\sqrt{3}p_{1}+q_{1}$ are three distinct collinear points
in ${\mathbb{P}}^{2}={\mathbb{P}}{\mathbb{R}}[x,y]_{2}$. Therefore,
$F:=\langle{L},{q(a,b,c)}\rangle$ vanishes at infinitely many real points and
thus $f\in l^{4}\Sigma_{2,4}^{4}$ for some $l\in{\mathbb{R}}[x,y]_{1}$ by 4.2,
a contradiction. ∎

######  Remark 4.7.

From 4.6 we also see that there is indeed a full-dimensional semi-algebraic
set of forms in the interior of $\Sigma_{2,8}^{4}$ that can not be written in
this way. Indeed, the set of all forms that can be written in this way are
given as the image of the map

${\mathbb{R}}[x,y]_{2}^{4}\to\Sigma_{2,8}^{4},\quad(p_{1},p_{2},q_{1},q_{2})\mapsto(p_{1}^{2}+q_{1}^{2})^{2}+(p_{2}^{2}+q_{2}^{2})^{2}$

which is a closed semi-algebraic set. Since it is not equal to
$\Sigma_{2,8}^{4}$ their difference contains a full-dimensional semi-algebraic
set.

######  Corollary 4.8.

Let $f\in\partial\Sigma_{2,8}^{4}$ such that $f\notin
l^{4}\operatorname{int}(\Sigma_{2,4})$ for any $l\in{\mathbb{R}}[x,y]_{1}$.
Then $f$ has a unique sum of fourth powers decomposition (up to sign and
permutation of the summands).

###### Proof.

If $f=q^{4}$ then $q$ is indefinite by 3.13. Hence, this is the only
representation and we may now assume that $f$ has length at least 2. If
$f=l^{4}(q_{1}^{4}+q_{2}^{4})\in l^{4}\partial(\Sigma_{2,4}^{4})$ the same
holds by the proof of 4.2.

Thus we may now assume that there exists $F\in(\Sigma_{2,8}^{4})^{\star}$ with
finitely many zeros and such that $F(f)=0$. If $f=\sum_{i=1}^{r}p_{i}^{4}$
then $F$ vanishes at all points $p_{i}$. Let $q_{1},\dots,q_{k}$ be binary
quadratic forms corresponding to the zeros of $F$. Then the $p_{i}$ are scalar
multiples of some of the $q_{i}$. Then any representation of $f$ can only the
$q_{i}$ as summands. Assume $k=2$. If $f$ has at least 2 representations, then

$(\lambda_{1}q_{1})^{4}+(\lambda_{2}q_{2})^{4}+(\lambda_{3}q_{3})^{4}=(\mu_{1}q_{1})^{4}+(\mu_{2}q_{2})^{4}+(\mu_{3}q_{3})^{4}$

for some $\lambda_{i},\mu_{i}\in{\mathbb{R}}$. Without loss of generality
$\lambda_{1}^{4}\neq\mu_{1}^{4}$. Hence,

$q_{1}^{4}(\lambda_{1}^{4}-\mu_{1}^{4})=(\mu_{2}q_{2})^{4}+(\mu_{3}q_{3})^{4}-(\lambda_{2}q_{2})^{4}-(\lambda_{3}q_{3})^{4}=c_{1}q_{2}^{4}+c_{2}q_{3}^{4}.$

If both $c_{1},c_{2}$ have the same sign, both real linear factors of $q_{1}$
divide both $q_{2}$ and $q_{3}$, hence they are all the same and $f$ has
length 1. If they have different signs, then the right-hand-side factors over
the reals as

$c_{1}q_{2}^{4}+c_{2}q_{3}^{4}=\pm(\sqrt{|c_{1}|}q_{2}^{2}+\sqrt{|c_{2}|}q_{3}^{2})(\sqrt[4]{|c_{1}|}q_{2}-\sqrt[4]{|c_{2}|}q_{3})(\sqrt[4]{|c_{1}|}q_{2}+\sqrt[4]{|c_{2}|}q_{3}).$

However, now $q_{1}$ is a scalar multiple of the last two factors, showing
that the three points $q_{1},q_{2},q_{3}$ are collinear which implies that $F$
has infinitely many zeros, a contradiction. If one of the $c_{i}$ is equal to
zero then two of the $q_{i}$ coincide, i.e. $F$ has only two zeros which is
not true.

Lastly, consider the case $k=2$. Then any representation of $f$ can only use
$q_{1}$ and $q_{2}$ and we get

$(\lambda_{1}q_{1})^{4}-(\mu_{1}q_{1})^{4}=(\mu_{2}q_{2})^{4}-(\lambda_{2}q_{2})^{4}$

which shows that $q_{1}$ is a scalar multiple of $q_{2}$. ∎

######  Remark 4.9.

The assumption $f\notin l^{4}\operatorname{int}(\Sigma_{2,4}^{4})$ is
necessary as a general element of $\Sigma_{2,4}^{4}$ has a 1-dimensional
family of length 3 representation which can be read off the map

$\phi_{1}^{3}\colon{\mathbb{R}}[x,y]_{1}^{3}\to{\mathbb{R}}[x,y]_{4},\,(l_{1},l_{2},l_{3})\mapsto
l_{1}^{4}+l_{2}^{4}+l_{3}^{4}$

where the left-hand-side has dimension 6 and the right-hand-side has dimension
5.

##  5\. Open problems

In this last section we propose some open problems which might stimulate
further research in this area.

######  Question 5.1.

What is the value of $p_{2s}({\mathbb{R}}[x])$? In particular, what is the
value of $p_{4}({\mathbb{R}}[x])$?

So far the only known lower bounds is the $2s\leq p_{2s}({\mathbb{R}}[x]))$ as
shown at the beginning of the Section 2.

The next problem is a generalization of a theorem of Becker.

######  Conjecture 5.2.

Let $R$ be a commutative ring with identity. Then the following conditions are
equivalent.

  1. (i)

$p_{2}(R)<\infty$

  2. (ii)

$p_{2s}(R)<\infty$ for some $s$

  3. (iii)

$p_{2s}(R)<\infty$ for all $s$.

If the above conjecture is true, then the problem of computing higher
Pythagoras number could be reduced to the computation of the 2-Pythagoras
number. This is important and usually much easier, mostly because the behavior
of quadratic forms is well understood compared to forms of higher degrees. It
is worth mentioning that the above conjecture does hold for valuation rings.
This follows from the Colliot-Thélène Kneser result and [4, Theorem 5.17] with
the simple observation that for valuation rings it is enough to consider units
which are sums of $2s$-th powers.

######  Question 5.3.

We saw in 3.17 that a general element of $\Sigma_{2,8}^{4}$ has (up to
permutation and roots of unity) 76 complex representations of length 3. Where
does this number come from? Can we construct all 76 of them, and possibly even
show that some of them are real?

This might be similar to length three sum of squares representations of
ternary quartics. These are known to be in one-to-one correspondence (up to
orthogonal equivalence) to Steiner complexes associated to the complex curve.

######  Question 5.4.

The proof of 3.11 is a computation in the end. It would be nice to understand
the reason why such a ternary quartic cannot have 4 real zeros.

Acknowledgements. We want to thank Greg Blekherman for suggesting the
connection to ternary quartics which turned out to be a key point of view.
Moreover, we thank Paul Breiding for help with early computations in Julia.
Julian Vill was partially supported by DFG grant 314838170, GRK 2297,
MathCoRe. Part of this paper came to existence during the stay of Tomasz
Kowalczyk at the Otto von Guericke University in Magdeburg. The author is very
grateful for the hospitality.

## References

  * [1] Macarena Ansola, Antonio Díaz-Cano, and María Ángeles Zurro.  Waring problem for binary real forms.  Gac. R. Soc. Mat. Esp., 23(1):147–162, 2020.
  * [2] Juliusz Banecki and Tomasz Kowalczyk.  Sums of even powers of $k$-regulous functions.  Indag. Math., New Ser., 34(3):477–487, 2023.
  * [3] Eberhard Becker.  The real holomorphy ring and sums of $2n$-th powers.  Géométrie algébrique réelle et formes quadratiques, Journées Soc. Math. Fr., Univ. Rennes 1981, Lect. Notes Math. 959, 139-181 (1982)., 1982.
  * [4] Eberhard Becker and Victoria Powers.  Sums of powers in rings and the real holomorphy ring.  J. Reine Angew. Math., 480:71–103, 1996.
  * [5] Olivier Benoist.  On Hilbert’s 17th problem in low degree.  Algebra Number Theory, 11(4):929–959, 2017.
  * [6] Olivier Benoist.  Sums of squares in function fields over Henselian local fields.  Math. Ann., 376(1-2):683–692, 2020.
  * [7] Jeff Bezanson, Alan Edelman, Stefan Karpinski, and Viral B. Shah.  Julia: a fresh approach to numerical computing.  SIAM Rev., 59(1):65–98, 2017.
  * [8] Grigoriy Blekherman.  Nonnegative polynomials and sums of squares.  In Semidefinite optimization and convex algebraic geometry, pages 159–202. Philadelphia, PA: Society for Industrial and Applied Mathematics (SIAM), 2013.
  * [9] Jacek Bochnak, Michel Coste, and Marie-Françoise Roy.  Real algebraic geometry. Transl. from the French., volume 36 of Ergeb. Math. Grenzgeb., 3. Folge.  Berlin: Springer, rev. and updated ed. edition, 1998.
  * [10] Paul Breiding and Sascha Timme.  HomotopyContinuation.jl: A Package for Homotopy Continuation in Julia.  In International Congress on Mathematical Software, pages 458–465. Springer, 2018.
  * [11] Matej Brešar and Peter Šemrl.  The Waring problem for matrix algebras.  Isr. J. Math., 253(1):381–405, 2023.
  * [12] Matej Brešar and Peter Šemrl.  The Waring problem for matrix algebras. II.  Bull. Lond. Math. Soc., 55(4):1880–1889, 2023.
  * [13] Kacper Błachut and Tomasz Kowalczyk.  Sums of squares on hypersurfaces, 2023.
  * [14] Antonio Campillo and Jesús M. Ruiz.  Some remarks on pythagorean real curve germs.  J. Algebra, 128(2):271–275, 1990.
  * [15] J. W. S. Cassels.  On the representation of rational functions as sums of squares.  Acta Arith., 9:79–82, 1964.
  * [16] Man-Duen Choi, Zong Duo Dai, Tsin Yuen Lam, and Bruce Reznick.  The Pythagoras number of some affine algebras and local algebras.  J. Reine Angew. Math., 336:45–82, 1982.
  * [17] Man-Duen Choi, Tsin Yuen Lam, Alexander Prestel, and Bruce Reznick.  Sums of $2m$-th powers of rational functions in one variable over real closed fields.  Math. Z., 221(1):93–112, 1996.
  * [18] Yeşim Demiroğlu Karabulut.  Waring’s problem in finite rings.  J. Pure Appl. Algebra, 223(8):3318–3329, 2019.
  * [19] José F. Fernando.  Positive semidefinite analytic functions on real analytic surfaces.  J. Geom. Anal., 31(12):12375–12410, 2021.
  * [20] José F. Fernando and Ronan Quarez.  Some remarks on the computation of Pythagoras numbers of real irreducible algebroid curves through Gram matrices.  J. Algebra, 274(1):64–67, 2004.
  * [21] Daniel R. Grayson and Michael E. Stillman.  Macaulay2, a software system for research in algebraic geometry.  Available at http://www2.macaulay2.com.
  * [22] David Grimm.  Lower bounds for Pythagoras numbers of function fields.  Comment. Math. Helv., 90(2):365–375, 2015.
  * [23] David Harrison.  A Grothendieck ring of higher degree forms.  J. Algebra, 35:123–138, 1975.
  * [24] Detlev W. Hoffmann.  Pythagoras numbers of fields.  J. Am. Math. Soc., 12(3):839–848, 1999.
  * [25] Anthony Iarrobino and Vassil Kanev.  Power Sums, Gorenstein Algebras, and Determinantal Loci.  Lecture Notes in Mathematics. Springer Berlin Heidelberg, 1 edition, 1999\.
  * [26] Wolfram Research, Inc.  Mathematica, Version 13.3.  Champaign, IL, 2023.
  * [27] Krishna Kishore.  Matrix Waring problem.  Linear Algebra Appl., 646:84–94, 2022.
  * [28] Tomasz Kowalczyk and Piotr Miska.  On waring numbers of henselian rings, 2022.
  * [29] Jakub Krásenský.  A cubic ring of integers with the smallest Pythagoras number.  Arch. Math., 118(1):39–48, 2022.
  * [30] Jakub Krásenský, Martin Raška, and Ester Sgallová.  Pythagoras numbers of orders in biquadratic fields.  Expo. Math., 40(4):1181–1228, 2022.
  * [31] Jakub Krásenský and Pavlo Yatsyna.  On quadratic Waring’s problem in totally real number fields.  Proc. Am. Math. Soc., 151(4):1471–1485, 2023.
  * [32] Aaron Kunert.  Facial Structure of Cones of non-negative Forms.  PhD thesis, Universität Konstanz, Konstanz, 2014.
  * [33] Jarosław Mirek, Wojciech Słomian, and Błażej Wróbel.  On the solution of Waring problem with a multiplicative error term: dimension-free estimates.  Proc. Am. Math. Soc., 151(8):3365–3379, 2023.
  * [34] Javier Pliego.  On Waring’s problem in sums of three cubes.  Mathematika, 67(1):235–256, 2021.
  * [35] Ricardo A. Podestá and Denis E. Videla.  The Waring’s problem over finite fields through generalized Paley graphs.  Discrete Math., 344(5):13, 2021.  Id/No 112324.
  * [36] Alexander Prestel.  Model theory of fields: An application to positive semidefinite polynomials.  Mém. Soc. Math. Fr., Nouv. Sér., 16:53–65, 1984.
  * [37] Bruce Reznick.  Sums of even powers of real linear forms, volume 463 of Mem. Am. Math. Soc. Providence, RI: American Mathematical Society (AMS), 1992.
  * [38] Bruce Reznick.  Blenders.  In Notions of positivity and the geometry of polynomials. Dedicated to the memory of Julius Borcea, pages 345–373. Basel: Birkhäuser, 2011.
  * [39] Jesús M. Ruiz.  Pythagorean real curve germs.  J. Algebra, 94:126–144, 1985.
  * [40] Claus Scheiderer.  Sum of squares length of real forms.  Math. Z., 286(1-2):559–570, 2017.
  * [41] Joachim Schmid.  Sums of fourth powers of real algebraic functions.  Manuscr. Math., 83(3-4):361–364, 1994.
  * [42] Rainer Sinn.  Algebraic boundaries of convex semi-algebraic sets.  Res. Math. Sci., 2:18, 2015.  Id/No 3.
  * [43] Trevor D. Wooley.  On Waring’s problem for intermediate powers.  Acta Arith., 176(3):241–247, 2016.

##  Macaulay2 Code

    R=QQ[t,a0,a1,a2,b0,b1,b2,c0,c1,c2,e1,e2,e3,a,b,c];
    U={a^2*b^2+a^3*c*2/3,a*b^3+a^2*b*c*3,b^2*c^2+2/3*a*c^3,b^3*c+a*b*c^2*3,
    b^4+12*a*b^2*c+6*a^2*c^2,a^4,a^3*b,c^4,b*c^3}; --basis of the subspace U
    X={a^4,a^3*b,a^3*c,a^2*b^2,a^2*b*c,a^2*c^2,a*b^3,a*b^2*c,a*b*c^2,a*c^3,
    b^4,b^3*c,b^2*c^2,b*c^3,c^4}; --basis of R[a,b,c]_4
    lab=(a1*b2-a2*b1)*a+(a2*b0-a0*b2)*b+(a0*b1-a1*b0)*c;
    lac=sub(lab,{b0=>c0,b1=>c1,b2=>c2});
    lbc=sub(lab,{a0=>c0,a1=>c1,a2=>c2});
    q1=lab*lac; q2=lab*lbc; q3=lbc*lac; --generators of I_2
    L={q1^2,q1*q2,q1*q3,q2^2,q2*q3,q3^2}; --basis of I_2^2
    V=matrix{join(U,L)};
    (M,C)=coefficients(V,Variables=>{a,b,c},Monomials=>X);

    Adjugate = (G) -> (
    n := degree target G;
    m := degree source G;
    matrix table(n, n, (i, j) -> (-1)^(i+j) * det(submatrix(G,
                                    {0..j-1, j+1..n-1},{0..i-1, i+1..m-1})))
    );

    getEquations = (p1,p2,p3,p4,p5,p6,p7,p8,p9) -> (
    G1:=sub(C,{a0=>p1,a1=>p2,a2=>p3,b0=>p4,b1=>p5,b2=>p6,c0=>p7,c1=>p8,c2=>p9});
    G1ad := Adjugate G1;
    de := det G1;
    G := 1/2*matrix {{2*G1ad_(9,2), G1ad_(10,2), G1ad_(11,2) },{G1ad_(10,2),
         2*G1ad_(12,2), G1ad_(13,2)},{G1ad_(11,2), G1ad_(13,2), 2*G1ad_(14,2)}};
    --this is the 3x3 Gram matrix
    cp := det (t*id_(R^3)-G);
    cc := coefficients(cp,Variables=>t,Monomials=>{t^2,t});
    {de,det G,(cc#1)_(0,0),(cc#1)_(1,0)} --output the two determinants that should
    --vanish, the coefficient of the characteristic polynomial of G that needs to be
    --negative in order for G to be psd and the coefficient that we saturate with
    );

    allEq = (p1,p2,p3,p4,p5,p6,p7,p8,p9) -> (
    E := getEquations(p1,p2,p3,p4,p5,p6,p7,p8,p9);
    I := ideal(radical(ideal(E#0)),radical(ideal(E#1))); --we take radicals to
    --remove unnecessary powers of some factors, this is not relevant for
    --the minimal primes
    J := saturate(I,radical(ideal(E#3)));
    Jmin := minprimes J;
    {toExternalString Jmin,toExternalString E#2,toExternalString E#3}
    );
    E1=allEq(0,1,0,1,e1,1,1,e2,e3); --q_i lin.ind. gives condition e_3 != 1
    E2=allEq(0,1,0,1,e1,-1,1,e2,e3); --q_i lin.ind. gives condition e_3 != -1