11institutetext: Department of Physics and Astronomy, National Taiwan
University, Taipei 10617, Taiwan22institutetext: Physics Division, National
Center for Theoretical Sciences, Taipei 10617, Taiwan

# Into the EFThedron and UV constraints from IR consistency

Li-Yuan Chiang 1,2 Yu-tin Huang 1 Wei Li 1 Laurentiu Rodina 1 He-Chen Weng
<EMAIL_ADDRESS>
###### Abstract

Recently it was proposed that the theory space of effective field theories
with consistent UV completions can be described as a positive geometry, termed
the EFThedron. In this paper we demonstrate that at the core, the geometry is
given by the convex hull of product moment curves. This makes contact with the
well studied bi-variate moment problem, which in various instances has known
solutions, generalizing the Hankel matrices into moment matrices. In the
infinite derivative limit, this identification leads to a complete description
of all boundaries. In the finite case, the boundary structure is more
complicated, and we compare the analytic description we obtain to that given
by semi-definite programing. Furthermore, we demonstrate that crossing
symmetry in the IR imposes non-trivial constraints on the UV spectrum. In
particular, permutation invariance for identical scalar scattering requires
that any UV completion beyond the scalar sector must contain arbitrarily high
spins, including at least all even spins $\ell\leq 28$, with the ratio of
spinning spectral functions bounded from above, exhibiting large spin
suppression. The mass of spinning spectrum must also allow for at least one
state satisfying a bound $m^{2}_{\ell}<M^{2}B(\ell)$, with $M^{2}$ the mass of
the heaviest spin 2 state.

##  1 Introduction

Effective field theories (EFT) are a convenient tool to parameterize our
ignorance of the UV physics and at the same time provide a useful description
of the IR. Precisely because of its role of connecting the UV with the IR, it
has long been a fertile playground for understanding how consistency
conditions on one side impose non-trivial constraints on the other. A
prominent example is the positivity of the leading four derivative couplings
in EFT Adams:2006sv  relating to the $a$-theorem Komargodski:2011vj  as well
as the Weak Gravity Conjecture ArkaniHamed:2006dz .

In the original work Adams:2006sv , positivity bounds were derived from the
forward limit of the low energy amplitude, which is related to the cross-
section of the UV scattering amplitude through dispersion relations and the
optical theorem. As one extends away from the forward limit, the fact that the
imaginary contribution in the dispersion relation has a positive partial wave
expansion further constrains the structure of the derivative expansion. This
was first proposed in Nicolis:2009qm , and explored in a more systematic
framework recently Bellazzini:2017fep ; deRham:2017avq ; Bellazzini:2020cot ;
Arkani-Hamed:2020gyp . For identical states, crossing symmetry drastically
constrains the allowed space of EFT couplings, in general reducing it to
finite compact regionsTolley:2020gtv ; Caron-Huot:2020cmc ; Caron-Huot:2021rmr
; Sinha:2020win . Monodromy relations push this reduction even further,
seemingly converging to a single point corresponding to the low energy
expansion of the open superstring Huang:2020nqy .

A geometric characterization of the space of consistent EFTs was recently
given in Arkani-Hamed:2020gyp , where positivity bounds are given as the
boundaries of the EFThedron, which is a product of convex spaces. To be
concrete, let us begin with the low energy scattering amplitude, which can be
parametrized as

$M(s,t)^{\text{IR}}=\sum_{k,q}g_{k,q}s^{k-q}t^{q}\,,$ (1)

where the Taylor expansion in $s$ and $t$ reflects the derivative expansion
for the low energy description, and the couplings $g_{k,q}$ are the Wilson
coefficients defined on an on-shell basis. Via dispersion relations, the
couplings admit a representation as a positive sum over data provided by the
UV spectrum. In particular

$g_{k,q}=\sum_{i}p_{i}\frac{1}{m_{i}^{2k+2}}X_{\ell_{i},k,q}\,,$ (2)

where $i$ labels the UV state, and $\\{p_{i},m_{i},\ell_{i}\\}$ are the
spectral function, mass and spin of state $i$. The coefficient
$X_{\ell_{i},k,q}$ is related to the Taylor expansion of the Legendre
polynomial $P_{\ell}(x)$, as well as the precise channel ($s$ or $u$) in which
the state was exchanged.

Importantly, as unitarity requires the spectral function $p_{i}$ to be
positive, the dispersive representation of the couplings is in fact giving a
convex hull. In Arkani-Hamed:2020gyp  the geometry of the hull was explored in
full for the $s$-EFThedron, corresponding to the scenario where the UV states
appear only in the $s$-channel. There are no $t$-channel poles since the
dispersion relation is implemented for fixed $t$. The geometry was found to be
the product of a bounded moment curve and a cyclic polytope, whose individual
boundaries are known, and given in terms in terms of inequalities that the
couplings must satisfy. For a single moment curve, it is a well known fact the
boundary is given by asking the Hankel matrix of couplings to be positive
semi-definite

$H(g)=\begin{pmatrix}g_{0}&g_{1}&g_{2}\\\ g_{1}&g_{2}&g_{3}&\ldots\\\
g_{2}&g_{3}&g_{4}\\\ &\vdots&&\ddots\end{pmatrix}\geq 0$ (3)

However, as we will see, for the full product geometry of the EFThedron not
all boundaries were found. For the $su$-EFThedron, where both $s$\- and
$u$-channel thresholds are present, the cyclic polytope is replaced by a more
complicated object, and only a partial exploration was done. One of the goals
of this paper is to fill the gaps for both geometries, by showing they both
originate from a common structure, which is well understood in the mathematics
literature.

##### EFThedron from the moment problem

We begin by demonstrating that even the $s$-EFThedron can be expressed in
simpler terms. By a GL transformation, the cyclic polytope defined by
$v_{l,q}$, which are complete polynomials in $J^{2}=\ell(\ell+1)$ of degree
$q$, can be rotated into a discrete moment curve in $J^{2}$. Therefore,
instead of considering the $g$-geometry defined by eq.(2), we can instead
consider what we call the $a$-geometry,

$\displaystyle a_{k,q}=\sum_{i}p_{i}\frac{1}{m_{i}^{2k+2}}J_{i}^{2q}\,.$ (4)

and return to the $g$-geometry by a simple rotation of the couplings. For the
full EFThedron, the geometry is given by the Minkowski sum of the hulls
arising from the $s$\- and $u$-channels. Because the $u$-channel contribution
is just itself a GL rotation of the $s$-channel one, it turns out we can
obtain the full EFThedron geometry by a similar rotation, followed by
projecting out couplings $g_{k,q}$ with $k+q=\textrm{odd}$, reflecting the
$s-u$ channel symmetry. Thus we see both EFThedrons are unified into the
$a$-geometry

Figure 1: Deriving $s$\- and $su$-EFThedrons from $a$-geometry

The significance of this result is that the $a$-geometry is explicitly the
product of two moment curves. The complete characterization of its boundaries
can then be formulated as the “moment problem”, which is a rich and well
studied problem in mathematics (see Moment  for an overview). The problem lies
in finding both necessary and sufficient conditions that determine whether a
set of “moments” (in our terms – couplings) can be expressed as eq.(4).
Solutions to the problem are well established in the infinite dimension limit,
which for us corresponds to the limit where we have access to Wilson
coefficients of all derivative orders, that is arbitrarily high orders in $k$
and $q$. This allows us to propose the complete boundary of the $a$-geometry
in the infinite dimension limit. The crucial condition that we find, compared
to Arkani-Hamed:2020gyp , involves the generalization of the Hankel matrix
into a moment matrix

$M(a)=\begin{pmatrix}a_{0,0}&a_{1,0}&a_{0,1}&a_{2,0}\\\
a_{1,0}&a_{2,0}&a_{1,1}&a_{3,0}\\\ a_{0,1}&a_{1,1}&a_{0,2}&a_{2,1}&\ldots\\\
a_{2,0}&a_{3,0}&a_{2,1}&a_{4,0}\\\ &&\vdots&&\ddots\end{pmatrix}\geq 0$ (5)

The complete solution is given by eqns.(61)-(64). Having both necessary and
sufficient conditions implies the EFThedron is in fact equivalent to the
dispersion relations encoding locality, unitarity and analyticity!

In practice, we will only be privy to finite number of Wilson coefficients,
meaning we must truncate at some order $q,k<N$. In such case the geometry
arising from eq.(4) is referred to as the “truncated moment problem”, where
only partial explicit solutions exist, and the conditions derived in this
paper will only be necessary but not always sufficient. This eventually leads
to a mismatch from the true boundary, which we term “finite size” effects,
that can be seen by comparing with a semi-definite linear programing analysis.
However, we expect that the true boundary can be reached as we increase the
number of constraints involved.

In the second row in Figure 1, we considered further imposing the cyclic or
permutation invariance of the low energy four point amplitude. These
constraints imply linear relations between the couplings $g_{k,q}$ or
$a_{k,q}$, which define what we will call symmetry or symmetry planes.
Importantly, once we require the EFThedron to intersect with the symmetry
plane, the resulting space becomes bounded in all directions, as was also
observed in Caron-Huot:2020cmc ; Sinha:2020win  by using the dispersion
relation form of null constraints. The intersection allows us to project the
typically high dimension geometry of the EFThedron to a low dimensional finite
space. To clarify, we propose that cyclic or permutation invariance can be
fully taken into account simply by slicing the EFThedron by symmetry planes,
without having to assume any explicit dispersion relation form of the null
constraints. All bounds are obtained by working exclusively at the level of
couplings, that is with inequalities of the form $f(a_{k,q})\geq 0$ like
eq.(5), intersected with symmetry planes $n(a_{k,q})=0$. The equivalence is
guaranteed by having the necessary and sufficient conditions for eq.(4), at
least in the infinite dimension limit.

In Figure 2, we show ($\frac{g_{3,1}}{g_{2,0}},\frac{g_{3,0}}{g_{2,0}}$)-space
for both the $s$-EFThedron and
($\frac{g_{3,1}}{g_{2,0}},\frac{g_{4,0}}{g_{2,0}}$) space for the full
EFThedron, after intersecting with the $k=4$ and $k=5$ symmetry planes
respectively, which are the lowest $k$ order that appear in each case. An
analytic expression for $su$-EFThedron result is given in Appendix C.

(a) $s$-channel with $k=5$ null constraint.

(b) $su$-channels with $k=4$ null constraint

Figure 2: EFThedron compared to SDPB

In some regions we observe a mismatch between the geometry and functional
method, implying we are missing some conditions, which we expect to recover in
the infinite dimension limit.

##### UV constraints from IR symmetry

The reason “slices” of the EFThedron are finite is simple: the symmetry plane
is always defined for fixed derivative order ($k$), and only the “$J^{2}$”
moment curve at different ($q$) orders is relevant for the intersection. The
intersection geometry is therefore always between a polytope and a subplane,
and one can easily see that a minority of vertices, defined by spin $\ell$,
lies on one side of the plane. Thus the fact that the physical spectrum must
have an empty image in the directions orthogonal to the symmetry plane,
conventionally termed “null constraints”, requires a balancing act between the
low and the higher spin part of the spectrum.

Indeed we find null constraints, when now expressed in their dispersive
representation form, lead to conditions on the UV spectrum. In both $s$\- and
full EFThedron, if the UV completion involves non-scalar states, cyclic and
permutation invariance require the presence of an infinite number of massive
spin states. Permutation invariance in fact requires every even spin $\ell\leq
28$ to be present. While any massive UV completion will inevitably produce
states of all spins via the discontinuity of the two-particle threshold, if we
assume that the UV theory has a dimensionless coupling, for example $g_{s}$ in
string theory, such that massive loop corrections can be suppressed, this
constraint becomes non-trivial. Furthermore, we also find upper bounds for the
ratio of averaged spectral functions, defined as

$\langle
p_{k,\ell}\rangle\equiv\sum_{\\{i,\ell_{i}=\ell\\}}\frac{p_{\ell_{i}}}{m^{2(k{+1})}_{i}}\,.$
(6)

For example, for identical scalars we find

$\displaystyle\frac{\langle p_{4,\ell}\rangle}{\langle
p_{4,2}\rangle}\leq\frac{12}{\ell(\ell+1)\left(\ell^{2}+\ell-8\right)},~{}(\ell\geq
4)$ $\displaystyle\frac{\langle p_{5,\ell}\rangle}{\langle
p_{5,2}\rangle}\leq\frac{216}{\ell(\ell+1)(\ell(\ell+1)(2\ell(\ell+1)-43)+150)},~{}(\ell\geq
4)\,.$ (7)

The result for distinct $k$ can be plotted as follows

Thus we see high spin suppression, at the level of finite spins, observed in
Arkani-Hamed:2020gyp  and recently explored in Bern:2021ppb , can be in part
attributed to the combination of permutation invariance and unitarity.

Similarly, the mass distribution of the higher spin spectrum is constrained.
Taking the heaviest spin-$2$ state to have mass $M$, then we find amongst the
higher spin states, there must exists one satisfying

$m^{2}_{\ell}\leq\frac{2J^{4}{-}43J^{2}{+}150}{18J^{2}{-}144}M^{2}\,.$ (8)

The above results are just the simplest consequences of a small subset of null
constraints. The full geometry of $\\{p,\ell,m\\}$ is highly non-trivial, as
it must in fact satisfy an infinite number of null constraints. We expect many
more conditions can be obtained by a systematic analysis of this geometry.

The paper is organized as follows. In Section 2 we review the positive
geometry of moment curves, and present our proposed complete solution to the
bi-variate moment problem. In Section 3 we show the $s$-EFThedron can be
obtained by a GL rotation from the bi-variate moment form. We then present
results for the geometry, which we compare to a functional approach. In
Section 4 we perform a similar analysis for the $su$-EFThedron. Besides a
similar rotation, one also further needs to project out some couplings in
order to satisfy $s-u$ channel symmetry. We similarly compare results to a
functional approach. In Section 5 we begin a investigation into conditions
imposed by null constraints on the UV spectrum itself. We derive constraints
on spins, spectral functions, and masses. We conclude in Section 6.

##  2 Positive geometry of moment curves

In this section we give an overview of the geometry associated with the convex
hull of moment curves. The convex hull is the space associated the positive
sum of a fixed set of vectors $\\{v_{i}\\}$

$\textrm{Conv}[\\{v_{i}\\}]\equiv\left\\{\sum_{i}p_{i}v_{i}\,\Big{|}\,p_{i}\geq
0\right\\}\,.$ (9)

If a point is inside the hull, then there must be a corresponding solution
$\\{p_{i}\\}$. A moment curve is simply a curve whose image takes the form
$(1,x,x^{2},\cdots)$. The convex hull of points on the moment curve is a
space, parameterized by the coordinates $\\{a_{0},a_{1},\ldots a_{n}\\}$,
satisfying

$a_{k}=\sum_{i}p_{i}x^{k}_{i}\,,$ (10)

where $k=\\{0,1,2,\ldots\\}$. Stated in a continuous fashion,

$a_{k}=\int x^{k}d\mu$ (11)

where $d\mu$ represents a positive measure.

We are therefore interested in finding the necessary and sufficient conditions
on $a_{k}$ such that they satisfy eq.(10). Said in another way, what are the
constraints that ensure the existence of a solution $\\{p_{i},x_{i}\\}$ that
solves eq.(10)? For a familiar application of this problem,
$a_{1}=\sum_{i}^{n}p_{i}x_{i}$ can be interpreted as the coordinate for the
center of mass, for a system of $n$ particles of masses $p_{i}$, placed at
positions $x_{i}$. The possible location for the center of mass is given by a
convex region defined by the “outermost” particles. The geometry we are
looking for is a slight generalization of this problem, and is known as the
“moment problem”. This comes in different versions, depending on the support
of the variable $x$. the usual cases considered are $x\in\mathbb{R}$,
$x\in\mathbb{R}_{+}$, $x\in[0,1]$ and finally $x$ taking only discrete values
in some region.

The geometry that is our main focus is a further generalization of the above,
the convex hull of a product of two moment curves

$a_{k,q}=\sum_{i}p_{i}\;x_{i}^{k}y_{i}^{q}\quad\rightarrow\left(\begin{array}[]{cccc}a_{0,0}&a_{0,1}&a_{0,2}&\cdots\\\
a_{1,0}&a_{1,1}&a_{1,2}&\cdots\\\ a_{2,0}&a_{2,1}&a_{2,2}&\cdots\\\
\vdots&\vdots&\vdots&\vdots\end{array}\right)=\sum_{i}p_{i}\left(\begin{array}[]{c}1\\\
x_{i}\\\ x_{i}^{2}\\\
\vdots\end{array}\right)\otimes\left(\begin{array}[]{cccc}1&y_{i}&y^{2}_{i}&\cdots\end{array}\right)\,,$
(12)

where we have $x_{i}\in[0,1]$ (corresponding to the masses of the UV states),
$y_{i}=\ell(\ell+1)$, with $\ell\in\mathbb{N}$ (corresponding to the spins of
the UV states). For physical reasons, subtraction terms in the dispersion
relations, we will restrict ourselves to only consider $k\geq 2$ and $k-q\geq
2$. Finally, the complete geometry will be given by intersecting the geometry
with symmetry planes corresponding to cyclic or permutation invariance.

In the following we will first motivate the a set of necessary conditions for
eq. (12) to hold, some of which have already been described in Arkani-
Hamed:2020blm . Then, we will give the sufficient condition in the infinite
dimension limit, i.e. where we consider all degree of moments, and demonstrate
how most of the previous necessary condition becomes a corollary. Finally when
we truncate to finite degrees that is the truncated moment problem, the
previous set of sufficient conditions as well as their corollaries no longer
suffice, and there will be subtleties at the boundaries which we will comment
on.

###  2.1 Single moment problem

To motivate the general results we will begin by assuming $n=2$ and $k=0,1,2$
in eq.(10), which will allow to derive both necessary and sufficient
conditions in a straightforward way. The solutions generalize in a trivial way
to arbitrary $n$ and infinite moments.

Our starting point is then

$\begin{pmatrix}a_{0}\\\ a_{1}\\\
a_{2}\end{pmatrix}=\sum_{i=1}^{2}p_{i}\begin{pmatrix}1\\\ x_{i}\\\
x_{i}^{2}\end{pmatrix},$ (13)

and we will separately consider the cases for $x_{i}\in\mathbb{R}$,
$\mathbb{R_{+}}$, $[0,1]$, or a set of discrete integers.

Full moment

We can solve eq.(13) for $p_{1},p_{2}$ and $x_{1}$

$x_{1}=\frac{a_{2}-a_{1}x_{2}}{a_{1}-a_{0}x_{2}},\quad
p_{1}=\frac{(a_{1}-x_{2})^{2}}{a_{0}x_{2}^{2}-2a_{1}x_{2}+a_{2}},\quad
p_{2}=\frac{a_{0}a_{2}-a_{1}^{2}}{a_{0}x_{2}^{2}-2a_{1}x_{2}+a_{2}}\,,$ (14)

and demand $p_{1},p_{2}>0$, $\forall x_{2}\in\mathbb{R}$. Considering the
second equation, this means the denominator must be positive for all values of
$x_{2}$, implying

$a_{0}\geq 0\textrm{ and }a_{0}a_{2}-a_{1}^{2}\geq 0\,,$ (15)

Positivity of $p_{2}$ then follows automatically. To systematically arrange
these conditions, it will be helpful to assemble the couplings into a Hankel
matrix

$H_{2}(a)=\left(\begin{array}[]{cc}a_{0}&a_{1}\\\ a_{1}&a_{2}\\\
\end{array}\right)\,.$ (16)

Then the conditions we found in eqns.(15) correspond to the leading principle
minors of $H_{2}(a)$ being non-negative. Since the Hankel matrix is symmetric,
this means that it is positive semi definite (PSD), a condition we denote from
now on as

$H_{2}(a)\geq 0\,.$ (17)

We have shown that if $H_{2}(a)\geq 0$, then $(a_{0},a_{1},a_{2})$ can be
written as $a_{k}=\sum p_{i}x_{i}^{k}$ with $p_{i}\geq 0$. For the converse,
we can see that if $p_{i}>0$, then $a_{0}=p_{1}+p_{2}$ and
$a_{2}a_{0}-a_{1}^{2}=p_{1}p_{2}(x_{1}-x_{2})^{2}$ are both positive. We
conclude, for $k=0,1,2$ and $x\in\mathbb{R}$

$a_{k}=\sum p_{i}x_{i}^{k},\textrm{ if and only if }H_{2}(a)\geq 0$ (18)

In general, the solution for infinite moments $(a_{0},a_{1},\ldots)$ will
involve an $n\times n$ Hankel matrix being PSD

$H_{n}(a)\equiv\begin{pmatrix}a_{0}&a_{1}&&a_{n}\\\
a_{1}&a_{2}&\cdots&a_{n+1}\\\ &\vdots&\ddots&\\\
a_{n}&a_{n+1}&&a_{2n}\end{pmatrix}\geq 0$ (19)

where the above must hold for all values of $n$.

To see those conditions are necessary, we can compute the determinant,

${\rm
Det}[H_{n}(a)]=\sum_{\\{b_{1},b_{2},\cdots,b_{n}\\}}\left(\prod_{k=1}^{n}\;p_{b_{k}}\right)\prod_{1\leq
k<l\leq n}\left(x_{b_{k}}-x_{b_{l}}\right)^{2}$ (20)

where the summation is over all $n$-size subsets of the elements in the hull.
The expression is clearly positive if $p_{i}\geq 0$. Proving the condition is
also sufficient is more difficult, and we discuss this issue in the next
section.

Half moment

Next we consider restricting to $x_{i}\geq 0$, which should impose additional
constraints on $a_{k}$. Indeed it is easy to derive from eqns.(14) that
$x_{1},x_{2}\geq 0$ if and only if $a_{1}\geq 0$, in addition to the previous
constraints. In general, the solution is given by

$H_{n}(a)\geq 0\textrm{ and }H_{n}^{\textrm{shift}}(a)\equiv
H(a)|_{a_{i}\rightarrow a_{i+1}}\geq 0$ (21)

where $H_{n}^{\textrm{shift}}(a)$ is known as the “shifted” Hankel, because
the first entry is $a_{1}$, instead of $a_{0}$.

We can now illustrate the space carved out by $H_{2}(a)$ being totally
positive, corresponding to $x\in\mathbb{R}_{+}$. It is generally useful to
consider the projective geometry defined by $\frac{a_{1}}{a_{0}}$ and
$\frac{a_{2}}{a_{0}}$, or simply normalize $a_{0}=1$. The inequalities lead to
Figure 3(a).

(a) $x\in\mathbb{R_{+}}$

(b) $x\in[0,1]$

(c) $x\in\mathbb{N}$

Figure 3: Positive region of ($a_{1}$,$a_{2}$)

and we observe the moment curve $(x,x^{2})$ is the boundary of this convex
space.

Bounded moment

Imposing $x_{i}\leq 1$ on eqns.(14), we quickly find two extra conditions:
$a_{0}\geq a_{1}$ and $a_{1}\geq a_{2}$. Adding these constraints to our
previous example, we find this drastically reduces the allowed space to Figure
3(b).

It is useful to rewrite these conditions by considering “twisted” couplings
$a^{\prime}_{k}\equiv a_{k}-a_{k+1}$. It is easy to see that

$a^{\prime}_{k}\equiv a_{k}-a_{k+1}=\sum p_{i}(1-x_{i})x_{i}^{k}$ (22)

are positive if $x_{i}\leq 1$. In general, in addition to the original
conditions in eqns.(21), to impose $x\in[0,1]$ we must also have

$H_{n}(a)-H_{n}^{\textrm{shift}}(a)\geq 0\,.$ (23)

Discrete moment

Finally, we have the case where $x_{i}$ can only take discrete values in some
set ordered $S$. In this case the geometry is given by what is known as a
cyclic polytope, whose boundaries are given by adjacent vertices
$x_{i},x_{i+1}$, if the set is cyclicly ordered, so $x_{i}<x_{i+1}$, with
$x_{|S|+1}=x_{1}$. We can motivate this result for $a_{0},a_{1},a_{2}$ space
by directly computing the required boundaries just as before by assuming three
states. Solving $a_{k}=\sum_{i=1}^{3}p_{i}x_{i}^{k}$ for the three $p_{i}$,
and requiring $p_{i}\geq 0$ it is easy to see we must have

$\langle\vec{a},\vec{x}_{i},\vec{x}_{i+1}\rangle\equiv\textrm{det}\left(\begin{array}[]{ccc}a_{0}&1&1\\\
a_{1}&x_{i}&x_{i+1}\\\ a_{2}&x_{i}^{2}&x_{i+1}^{2}\end{array}\right)\geq
0,\,\forall x_{i}\in S$ (24)

where $\langle\vec{a},\vec{x}_{i},\vec{x}_{i+1}\rangle$ is called a polytope
wall, and is defined by the vertices $x_{i}$ and $x_{i+1}$. We can also check
the above condition is necessary. Writing it out:

$\langle\vec{a},\vec{x}_{i},\vec{x}_{i+1}\rangle=(x_{i+1}-x_{i})\sum_{j=1}^{3}p_{j}(x_{j}-x_{i})(x_{j}-x_{i+1})$
(25)

we see the importance of having boundaries given by only by adjacent ordered
vertices: it implies we can only have $x_{j}\leq x_{i}<x_{i+1}$, or
$x_{i}<x_{i+1}\leq x_{j}$, so that the $(x_{j}-x_{i})(x_{j}-x_{i+1})\geq 0$
for all $i,j$, so this sum is positive term by term. We plot the region for
$(a_{1}/a_{0},a_{2}/a_{0})$ in Figure 3(c).

If $\vec{a}$ is higher dimensional, with $\textrm{dim}[\vec{a}]=2d+1$, the
general boundaries will be given by

$\langle\vec{a},\vec{s}_{i_{1}},\vec{s}_{i_{1}+1},\ldots,\vec{s}_{i_{d}},\vec{s}_{i_{d}+1}\rangle=\underbrace{\left|\begin{array}[]{cccccc}a_{0}&1&1&\cdots&1&1\\\
a_{1}&s_{i_{1}}&s_{i_{1}+1}&\cdots&s_{i_{d}}&s_{i_{d}+1}\\\
a_{2}&s_{i_{1}}^{2}&s_{i_{1}+1}^{2}&\cdots&s_{i_{d}}^{2}&s_{i_{d}+1}^{2}\\\
a_{3}&s_{i_{1}}^{3}&s_{i_{1}+1}^{3}&\cdots&s_{i_{d}}^{3}&s_{i_{d}+1}^{3}\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
a_{2d}&s_{i_{1}}^{2d}&s_{i_{1}+1}^{2d}&\cdots&s_{i_{d}}^{2d}&s_{i_{d}+1}^{2d}\end{array}\right|}_{2d+1}\geq
0.$ (26)

For $\textrm{dim}[\vec{a}]=2d$ we have

$\displaystyle\langle\vec{0},\vec{a},\vec{s}_{i_{1}},\vec{s}_{i_{1}+1},\ldots,\vec{s}_{i_{d}},\vec{s}_{i_{d}+1}\rangle$
$\displaystyle\geq$ $\displaystyle 0,$
$\displaystyle\langle\vec{a},\vec{s}_{i_{1}},\vec{s}_{i_{1}+1},\ldots,\vec{s}_{i_{d}},\vec{s}_{i_{d}+1},\vec{\infty}\rangle$
$\displaystyle\geq$ $\displaystyle 0,$ (27)

where $\vec{0}=(1,0,\ldots,0)$ and $\vec{\infty}=(0,0,\ldots,1)$.

#### The bi-variate moment problem

Next we briefly turn to the bi-variate moment problem, and consider moments
given by

$a_{k,q}=\sum_{i}p_{i}x_{i}^{k}y_{i}^{q}$ (28)

We can derive some conditions by recombining the two moment curves into a
single one, and using the previous results to obtain some necessary
conditions. The general treatment of this problem is left to the following
section. We can select a subset (or combinations) of $a_{k,q}$ that lie on
single moments curves. For instance, if we pick fixed $k=1$ couplings
$a_{1,0},a_{1,1},a_{1,2}$,

$\left(\begin{array}[]{c}a_{1,0}\\\ a_{1,1}\\\
a_{1,2}\end{array}\right)=\sum_{i}p_{i}x_{i}\left(\begin{array}[]{c}1\\\
y_{i}\\\
y_{i}^{2}\end{array}\right)\equiv\sum_{i}p_{i}^{\prime}\left(\begin{array}[]{c}1\\\
y_{i}\\\ y_{i}^{2}\end{array}\right)$ (29)

where we redefined $p_{i}^{\prime}\equiv p_{i}x_{i}$, we recover a single
moment curve, and hence if $y_{i}\in\mathbb{R}$ we get the familiar Hankel
conditions, which we now denote as

$H_{2}(a)(y){=}\left(\begin{array}[]{cc}a_{k,0}&a_{k,1}\\\
a_{k,1}&a_{k,2}\end{array}\ \right)\geq 0$ (30)

We can also produce new non-trivial moments by collecting different degrees of
$x$ or $y$. For example, we can obtain a combined moment curve defined by
$a_{k,k}=\sum p_{i}\left(x_{i}y_{i}\right)^{k}$. This leads to the following
Hankel condition

$H_{2}(a)(xy)\equiv\left(\begin{array}[]{cc}a_{0,0}&a_{1,1}\\\
a_{1,1}&a_{2,2}\end{array}\ \right)\geq 0$ (31)

Such conditions were not previously considered in Arkani-Hamed:2020gyp , and
are also related to an inequality used in Caron-Huot:2020cmc  to analytically
obtain a bound we will also aim to study from the geometric perspective. Of
course there is nothing special about the choice $(xy)$. One could pick any
polynomial in $x,y$, and any such choice will lead to a corresponding Hankel
constraint like in eq.(31), which can get increasingly complex. For example, a
particularly strong constraint we will use is

$-a_{0,0}a_{1,1}^{2}a_{2,0}+2a_{0,0}a_{1,0}a_{1,1}a_{2,1}-a_{0,0}^{2}a_{2,1}^{2}-a_{0,0}a_{1,0}^{2}a_{2,2}+a_{0,0}^{2}a_{2,0}a_{2,2}\geq
0.$ (32)

which we derive in Appendix A using similar arguments as above. Besides linear
combinations of couplings, one can also take quadratic combinations, which
were originally termed “product Hankel constraints”. We also describe in
Appendix A several such constructions. We next continue to present the
complete solution, which will include all such possible constraints in a much
more compact and systematic way.

###  2.2 The solution in the infinite dimension limit

We now present the general solution to the product moment problem. To
facilitate the analysis, let us rephrase the single moment problem in a
fashion that can be straight forwardly generalized to the multi-variate case.

For the single moment, we have seen that in the infinite dimension limit, the
sufficient condition for $\vec{a}$ to reside in the hull is that the infinite-
dimensional Hankel matrix is a positive-semidefinite matrix. Observing that
the Hankel matrix can be written as

$H(a)=\sum_{i}p_{i}\mathbf{x}_{i}\mathbf{x}_{i}^{T}=\sum_{i}p_{i}\left(\begin{array}[]{c}1\\\
x_{i}\\\ x_{i}^{2}\\\
\vdots\end{array}\right)\left(\begin{array}[]{cccc}1&x_{i}&x^{2}_{i}&\cdots\end{array}\right),$
(33)

we immediately see that the Hankel matrix is positive semi-definite, since for
any vector $\mathbf{v}$,

$\mathbf{v}^{T}H\mathbf{v}=\sum_{i}p_{i}(\mathbf{v}^{T}\mathbf{x}_{i})^{2}\geq
0\,.$ (34)

This implies all the principle minors are non-negative.

Note that at this point, the Hankel matrix being positive-semidefinite is a
necessary condition, but not yet sufficient. To show that it is sufficient we
must show that satisfying the Hankel constraints implies the existence of a
positive solution $\\{p_{i}\\}$ that gives us $\\{a_{k}\\}$. First, notice
that $f(x)=\mathbf{v}^{T}\mathbf{x}$ is simply a polynomial in $x$. Let us
interpret the sum weighted by $p_{i}$ as computing an expectation value, and
the expectation value for the squared function $f(x)^{2}$ is positive

${\mathbb{E}}[f(x)^{2}]\equiv\sum_{i}p_{i}f(x_{i})^{2}\geq 0.$ (35)

Now since the vector $\mathbf{v}$ is infinite dimensional, one can generate
arbitrary polynomials in $x$ with any degree. Now in a finite interval, one
can approximate the delta function by an infinite sum of polynomials. In
particular, on the interval ${-}1\leq x\leq 1$ one can choose $f^{2}(x)$ to be
the Landau Kernal $L_{n}(x)$ with $n$ being a large even integer, and

$L_{n}(x)=(1-x^{2})^{n}\frac{(2n)!(2n{+}1)}{(n!)^{2}2^{2n{+}1}}\,.$ (36)

Choosing $\mathbf{v}$ appropriately, one sees that from eq.(35) one is ensured
with a positive $p_{i}$. Since the convex hull is by nature projective, one
can rescale $x\rightarrow Rx$ by any positive number $R$ and the geometry is
unchanged. Therefore, this argument holds for any bounded interval of $x$. A
more formal treatment is required for $R\rightarrow\infty$, given by Moment1 .

#### Extra conditions for restricted $\mathbf{x}$

If we further consider that $\\{x_{i}\\}$ are restricted on a closed interval
$[a,b]$, this implies

${\mathbb{E}}\left[f(x)^{2}(x-a)(b-x)\right]\geq 0$ (37)

since it gives us a positive sum of polynomial squared. Assuming eq.(37), we
can once again let $f(x)^{2}$ approach delta function, enforcing
$p_{i}(x_{i}-a)(b-x_{i})\geq 0.$ Since $p_{i}$ is positive, this guarantees
$(x_{i}-a)(b-x_{i})\geq 0$, so $x_{i}\in[a,b]$. Recalling that
$H(a)=\sum_{i}p_{i}\mathbf{x}_{i}\mathbf{x}_{i}^{T}$, we can traslate

${\mathbb{E}}\left[f(x)^{2}(x-a)(b-x)\right]=\mathbf{v}^{T}\left(\sum_{i}p_{i}(x_{i}-a)(b-x_{i})\mathbf{x}_{i}\mathbf{x}_{i}^{T}\right)\mathbf{v}\geq
0$ (38)

into a matrix which is a combination of the Hankel matrix and its shifted ones

$\displaystyle\sum_{i}p_{i}(x_{i}-a)(b-x_{i})\left(\begin{array}[]{c}1\\\
x_{i}\\\ x_{i}^{2}\\\
\vdots\end{array}\right)\left(\begin{array}[]{cccc}1&x_{i}&x_{i}^{2}&\cdots\end{array}\right)$
(44) $\displaystyle=$
$\displaystyle\begin{pmatrix}(-a_{2}+(a+b)a_{1}-aba_{0})&(-a_{3}+(a+b)a_{2}-aba_{1})&\cdots\\\
(-a_{3}+(a+b)a_{2}-aba_{1})&(-a_{4}+(a+b)a_{3}-aba_{2})&\cdots\\\
\vdots&\vdots&\ddots\end{pmatrix}\geq 0.$ (45)

Therefore, the necessary and sufficient condition is given by the matrices
eq.(19) and eq.(44) being PSD. We can now apply this result to the following
problems.

  1. 1.

Half moment problem: $a=0,b\rightarrow\infty$.

For $b\rightarrow\infty$, or simply $x\geq 0$, we have

${\mathbb{E}}\left[f(x)^{2}x\right]\geq
0\Rightarrow\begin{pmatrix}a_{1}&a_{2}&a_{3}&\cdots\\\
a_{2}&a_{3}&a_{4}&\cdots\\\ a_{3}&a_{4}&a_{5}&\cdots\\\
\vdots&\vdots&\vdots&\ddots\end{pmatrix}\geq 0,$ (46)

which is the same as eq.(21), i.e., the shifted Hankel matrix.

  2. 2.

Hausdorff moment problem: $a=0,b=1$.

In addition to $x\geq 0$, i.e., the positivity of the shifted Hankel, one also
needs to impose $x\leq 1$, or

${\mathbb{E}}[(1-x)f(x)^{2}]\geq
0\Rightarrow\begin{pmatrix}(a_{0}-a_{1})&(a_{1}-a_{2})&(a_{2}-a_{3})&\cdots\\\
(a_{1}-a_{2})&(a_{2}-a_{3})&(a_{3}-a_{4})&\cdots\\\
(a_{2}-a_{3})&(a_{3}-a_{4})&(a_{4}-a_{5})&\cdots\\\
\vdots&\vdots&\vdots&\ddots\end{pmatrix}\geq 0$ (47)

This is the twisted Hankel matrix, eq.(23).

  3. 3.

Discrete moment problem: $x\in\\{s_{i}\\}$

Let $s_{1},s_{2},...,s_{n}$ be monotonically increasing. The previous
constraints correspond to excluding regions $(-\infty,a)$ and $(b,\infty)$. If
we instead exclude $x\in(s_{i},s_{i+1})$ by imposing
${\mathbb{E}}\left[f(x)^{2}(x-s_{i})(x-s_{i+1})\right]\geq 0$ and then apply
this for every adjacent pair, it forces $x\in\\{s_{i}\\}$ and gives us the
discrete geometry.

$\displaystyle\begin{split}(a)\quad&{\mathbb{E}}\left[f(x)^{2}(x-s_{1})\right]\geq
0\Rightarrow\begin{pmatrix}(a_{1}-s_{1}a_{0})&(a_{2}-s_{1}a_{1})&\cdots\\\
(a_{2}-s_{1}a_{1})&(a_{3}-s_{1}a_{2})&\cdots\\\
\vdots&\vdots&\ddots\end{pmatrix}\geq 0.\\\
(b)\quad&{\mathbb{E}}\left[f(x)^{2}(x-s_{i})(x-s_{i+1})\right]\cdots\geq 0\\\
\Rightarrow&\begin{pmatrix}(a_{2}-(s_{i}+s_{i+1})a_{1}+s_{i}s_{i+1}a_{0})&(a_{3}-(s_{i}+s_{i+1})a_{2}+s_{i}s_{i+1}a_{1})&\cdots\\\
(a_{3}-(s_{i}+s_{i+1})a_{2}+s_{i}s_{i+1}a_{1})&(a_{4}-(s_{i}+s_{i+1})a_{3}+s_{i}s_{i+1}a_{2})&\cdots\\\
\vdots&\vdots&\ddots\end{pmatrix}\cdots\geq 0.\\\
(c)\quad&{\mathbb{E}}\left[f(x)^{2}(s_{n}-x)\right]\geq
0\Rightarrow\begin{pmatrix}(-a_{1}+s_{n}a_{0})&(-a_{2}+s_{n}a_{1})&\cdots\\\
(-a_{2}+s_{n}a_{1})&(-a_{3}+s_{n}a_{2})&\cdots\\\
\vdots&\vdots&\ddots\end{pmatrix}\geq 0.\end{split}$ (48)

Let us consider the first element of the matrix of (b) in eq.(48), which turns
out to be the polytopal wall in eq.(24), up to a positive factor

$a_{2}-(s_{i}+s_{i+1})a_{1}+s_{i}s_{i+1}a_{0}=\frac{1}{(s_{i+1}-s_{i})}\,\textrm{det}\begin{pmatrix}a_{0}&1&1\\\
a_{1}&s_{i}&s_{i+1}\\\ a_{2}&s_{i}^{2}&s_{i+1}^{2}\end{pmatrix}\,.$

Since the matrix in $(b)$ of eq.(48) is PSD, the above is positive and is
exactly the $\mathbb{P}^{2}$ polytopal constraint. Moving on to the larger
leading principal minors, we find them having the form of the usual Hankel
matrix with elements being the polytopal wall dotted into $\vec{a}$ up to a
positive factor.

#### The multi-variate problem

We can now extend these to the product of multiple moment curves. A product of
two moment curves, for example, requires a complete basis of polynomials in
$x$ and $y$, now given by $\\{1,x,y,x^{2},xy,y^{2},...\\}$, to generate
arbitrary function $f(x,y)$. Therefore, we can assemble the couplings into the
following matrix generalizing the Hankel matrix

$M(a)=\sum_{i}p_{i}\left(\begin{array}[]{c}1\\\ x_{i}\\\ y_{i}\\\ x_{i}^{2}\\\
x_{i}y_{i}\\\ y_{i}^{2}\\\
\vdots\end{array}\right)\left(\begin{array}[]{ccccccc}1&x_{i}&y_{i}&x^{2}_{i}&x_{i}y_{i}&y^{2}_{i}&\cdots\end{array}\right)=\begin{pmatrix}a_{0,0}&a_{1,0}&a_{0,1}&\cdots\\\
a_{1,0}&a_{2,0}&a_{1,1}&\cdots\\\ a_{0,1}&a_{1,1}&a_{0,2}&\cdots\\\
\vdots&\vdots&\vdots&\ddots\end{pmatrix}$ (49)

which is known as the moment matrix.

The constraints for $a_{k,q}$ from the previous subsection can now be obtained
by simply requiring $M(a)\geq 0$. For example, we have, up to $k,q\leq 2$,

$\begin{vmatrix}a_{0,0}&a_{1,0}\\\ a_{1,0}&a_{2,0}\\\
\end{vmatrix},\begin{vmatrix}a_{0,0}&a_{0,1}\\\ a_{0,1}&a_{0,2}\\\
\end{vmatrix},\begin{vmatrix}a_{0,0}&a_{1,1}\\\ a_{1,1}&a_{2,2}\\\
\end{vmatrix},\begin{vmatrix}a_{0,0}&a_{1,0}&a_{0,1}\\\
a_{1,0}&a_{2,0}&a_{1,1}\\\ a_{0,1}&a_{1,1}&a_{0,2}\end{vmatrix}\geq 0.$

Note the second to last determinant above is exactly eq.(31). A simple
derivation of the last determinant is also given in Appendix A.

Similarly, the additional constraints for restricted $x,y$ can be straight
forwardly derived. One imposes

${\mathbb{E}}\left[f(x,y)^{2}(x-x_{a})(x_{b}-x)\right]\geq 0$

to force $x\in[x_{a},x_{b}]$ and imposes

${\mathbb{E}}\left[f(x,y)^{2}(y-y_{a})(y-y_{b})\right]\geq 0$

to exclude $y\in(y_{a},y_{b})$. So for each case we have the following extra
constraints:

  1. 1.

Half moment problem

For the half moment problem in $x$, i.e., $x\in[0,\infty)$, we require

$\displaystyle{\mathbb{E}}\left[f(x,y)^{2}x\right]\geq 0$
$\displaystyle\Rightarrow$
$\displaystyle\sum_{i,j}p_{i,j}x_{i}\left(\begin{array}[]{c}1\\\ x_{i}\\\
y_{i}\\\
\vdots\end{array}\right)\left(\begin{array}[]{cccc}1&x_{i}&y_{i}&\cdots\end{array}\right)=\begin{pmatrix}a_{1,0}&a_{2,0}&a_{1,1}&\cdots\\\
a_{2,0}&a_{3,0}&a_{2,1}&\cdots\\\ a_{1,1}&a_{2,1}&a_{1,2}&\cdots\\\
\vdots&\vdots&\vdots&\ddots\end{pmatrix}\geq 0$ (55)

This is the two-dimensional generalization of the shifted Hankel matrix. In
short,

$M^{\text{shift},x}(a)\equiv M(a)|_{a_{k,q}\rightarrow a_{k+1,q}}\geq 0.$ (56)

If $y$ is also required to be positive, one has to impose

$M^{\textrm{shift},y}(a)\equiv M(a)|_{a_{k,q}\rightarrow a_{k,q+1}}\geq 0.$
(57)

  2. 2.

Hausdorff moment problem

If $x$ is further restricted on the closed interval $[0,1]$, then one has to
impose, besides eq.(56), positivity on the twisted Hankel matrix

${\mathbb{E}}\left[f(x,y)^{2}(1-x)\right]\geq
0\Rightarrow\begin{pmatrix}(a_{0,0}-a_{1,0})&(a_{1,0}-a_{2,0})&(a_{0,1}-a_{1,1})&\cdots\\\
(a_{1,0}-a_{2,0})&(a_{2,0}-a_{3,0})&(a_{1,1}-a_{2,1})&\cdots\\\
(a_{0,1}-a_{1,1})&(a_{1,1}-a_{2,1})&(a_{0,2}-a_{1,2})&\cdots\\\
\vdots&\vdots&\vdots&\ddots\end{pmatrix}\geq 0.$ (58)

This is the generalized version of eq.(47).

  3. 3.

Discrete moment problem

Suppose $y$ is discrete and can only take values $\\{s_{i}\\}$, then

${\mathbb{E}}\left[f(x,y)^{2}(y-s_{1})\right],\
{\mathbb{E}}\left[f(x,y)^{2}(y-s_{i})(y-s_{i+1})\right],{\mathbb{E}}\left[f(x,y)^{2}(s_{n}-y)\right]\geq
0.$ (59)

In other words,

$\displaystyle\begin{split}(a)\quad&{\mathbb{E}}\left[f(x,y)^{2}(y-s_{1})\right]\geq
0\Rightarrow\begin{pmatrix}(a_{0,1}-s_{1}a_{0,0})&(a_{1,1}-s_{1}a_{1,0})&\cdots\\\
(a_{1,1}-s_{1}a_{1,0})&(a_{2,1}-s_{1}a_{2,0})&\cdots\\\
\vdots&\vdots&\ddots\end{pmatrix}\geq 0.\\\
(b)\quad&{\mathbb{E}}\left[f(x,y)^{2}(y-s_{i})(y-s_{i+1})\right]\cdots\geq
0\\\
\Rightarrow&\begin{pmatrix}(a_{0,2}-(s_{i}+s_{i+1})a_{0,1}+s_{i}s_{i+1}a_{0,0})&(a_{1,2}-(s_{i}+s_{i+1})a_{1,1}+s_{i}s_{i+1}a_{1,0})&\cdots\\\
(a_{1,2}-(s_{i}+s_{i+1})a_{1,1}+s_{i}s_{i+1}a_{1,0})&(a_{2,2}-(s_{i}+s_{i+1})a_{2,1}+s_{i}s_{i+1}a_{2,0})&\cdots\\\
\vdots&\vdots&\ddots\end{pmatrix}\cdots\geq 0.\\\
(c)\quad&{\mathbb{E}}\left[f(x,y)^{2}(s_{n}-y)\right]\geq
0\Rightarrow\begin{pmatrix}(-a_{0,1}+s_{n}a_{0,0})&(-a_{1,1}+s_{n}a_{1,0})&\cdots\\\
(-a_{1,1}+s_{n}a_{1,0})&(-a_{2,1}+s_{n}a_{2,0})&\cdots\\\
\vdots&\vdots&\ddots\end{pmatrix}\geq 0.\end{split}$ (60)

The first element of the matrix (b) above being positive

$a_{0,2}-(s_{i}+s_{i+1})a_{0,1}+s_{i}s_{i+1}a_{0,0}=\frac{1}{(s_{i+1}-s_{i})}\,\textrm{det}\begin{pmatrix}a_{0,0}&1&1\\\
a_{0,1}&s_{i}&s_{i+1}\\\ a_{0,2}&s_{i}^{2}&s_{i+1}^{2}\end{pmatrix}\geq 0\,,$

is exactly the $\mathbb{P}^{2}$ polytopal wall, generalized to the bi-variate
case. The constraints in eq.(60) should give us the discrete-y geometry in the
infinite dimension limit.

##### Summary

For the bi-variate moment problem in the infinite dimension limit, we have the
following solutions

  * •

Full moment: $x,y\in\mathbb{R}$

$\displaystyle M(a)=\begin{pmatrix}a_{0,0}&a_{1,0}&a_{0,1}&\cdots\\\
a_{1,0}&a_{2,0}&a_{1,1}&\cdots\\\ a_{0,1}&a_{1,1}&a_{0,2}&\cdots\\\
\vdots&\vdots&\vdots&\ddots\end{pmatrix}\geq 0$ (61)

  * •

Half moment: $x\in\mathbb{R}^{+}$

$\displaystyle M(a)\geq 0,\quad M^{\text{shift},x}(a)\equiv
M(a)|_{a_{k,q}\rightarrow a_{k+1,q}}\geq 0$ (62)

  * •

Bounded moment: $x\in[0,1]$

$\displaystyle M(a)\geq 0,\quad M^{\text{shift},x}(a)\geq 0,\quad
M^{\text{twist,x}}(a)\equiv M(a)-M^{\text{shift,x}}(a)\geq 0$ (63)

  * •

Discrete moment: $y\in\\{s_{i}\\}$

$\displaystyle M(a)\geq 0$ (64)
$\displaystyle\begin{pmatrix}(a_{0,1}-s_{1}a_{0,0})&(a_{1,1}-s_{1}a_{1,0})&\cdots\\\
(a_{1,1}-s_{1}a_{1,0})&(a_{2,1}-s_{1}a_{2,0})&\cdots\\\
\vdots&\vdots&\ddots\end{pmatrix},\begin{pmatrix}(-a_{0,1}+s_{n}a_{0,0})&(-a_{1,1}+s_{n}a_{1,0})&\cdots\\\
(-a_{1,1}+s_{n}a_{1,0})&(-a_{2,1}+s_{n}a_{2,0})&\cdots\\\
\vdots&\vdots&\ddots\end{pmatrix}\geq 0$
$\displaystyle\begin{pmatrix}(a_{0,2}-(s_{i}+s_{i+1})a_{0,1}+s_{i}s_{i+1}a_{0,0})&(a_{1,2}-(s_{i}+s_{i+1})a_{1,1}+s_{i}s_{i+1}a_{1,0})&\cdots\\\
(a_{1,2}-(s_{i}+s_{i+1})a_{1,1}+s_{i}s_{i+1}a_{1,0})&(a_{2,2}-(s_{i}+s_{i+1})a_{2,1}+s_{i}s_{i+1}a_{2,0})&\cdots\\\
\vdots&\vdots&\ddots\end{pmatrix}\geq 0$

###  2.3 The truncated moment “problem”

The fact that constraints given in the infinite dimension limit are sufficient
conditions relies on the ability to represent the delta function as a sum over
polynomials of arbitrary degree. This is no longer possible when the moment
sequence is truncated at some $k<N$. Then the conditions in the previous
subsection may no longer be sufficient, and new conditions may need to be
imposed. In some cases the conditions are corollaries in the infinite
dimension limit, but become independent in the finite limit. In other words,
we can still reach the true boundary by considering an increasing number of
constraints, converging to the solution of the infinite limit. In other cases
the conditions are known as the flat extension conditions, and it is less
clear how to implement them practically article .

A simple example of the first kind of conditions can be seen in the emergence
of the twisted Hankel constraints from an alternative solution to infinite
dimensional bounded problem. A classical result states that sufficient and
necessary conditions for $a_{k}=\sum_{i}p_{i}x_{i}^{k}$, with $x_{i}\in[0,1]$,
are that the infinite difference sequence satisfies

$(-1)^{n}(\Delta^{n}a)_{k}\geq 0$ (65)

where $(\Delta a)_{k}=a_{k+1}-a_{k}$. The first few terms in the sequence are

$a_{k}-a_{k+1},\quad a_{k}-2a_{k+1}+a_{k+2},\quad
a_{k}-3a_{k+1}+3a_{k+2}-a_{k+3},$ (66)

In the infinite limit, that is arbitrary $n$ and $k$, the sequence is
equivalent to the twisted Hankel we wrote previously. However, if we truncate
at some finite order, and consider just constraints involving
$(a_{0},a_{1},a_{2})$, the difference sequence will not provide a sufficient
condition, unlike the Hankel constraint. But by increasing the order, to say
include $a_{3}$ constraints, and then projecting out $a_{3}$, we obtain an
improvement in the boundary of the original $(a_{0},a_{1},a_{2})$ space. In
this sense, by increasing the order of constraints, we are able to better
approximate the real space.

A more complicated example is the emergence of the cyclic polytope boundaries
from the infinite limit constraints. Recall that for $\mathbb{P}^{2d}$ space,
where $d$ is a positive integer, the $(2d+1)$-dimensional polytopal wall
constraint

$\langle\vec{a},\vec{s_{i_{1}}},\vec{s_{i_{1}+1}},\vec{s_{i_{2}}},\vec{s_{i_{2}+1}},...\rangle=\underbrace{\left|\begin{array}[]{cccccc}a_{0}&1&1&\cdots&1&1\\\
a_{1}&s_{i_{1}}&s_{i_{1}+1}&\cdots&s_{i_{d}}&s_{i_{d}+1}\\\
a_{2}&s_{i_{1}}^{2}&s_{i_{1}+1}^{2}&\cdots&s_{i_{d}}^{2}&s_{i_{d}+1}^{2}\\\
a_{3}&s_{i_{1}}^{3}&s_{i_{1}+1}^{3}&\cdots&s_{i_{d}}^{3}&s_{i_{d}+1}^{3}\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
a_{2d}&s_{i_{1}}^{2d}&s_{i_{1}+1}^{2d}&\cdots&s_{i_{d}}^{2d}&s_{i_{d}+1}^{2d}\end{array}\right|}_{2d+1}\geq
0.$ (67)

On the other hand, once eq.(48) is satisfied, we have a solution where the
$x$’s are given at the discrete sites. Then it immediately follows that one
also has

${\mathbb{E}}\left[f(x)^{2}(x-s_{i_{1}})(x-s_{i_{1}+1})...(x-s_{i_{d}})(x-s_{i_{d}+1})...\right]\geq
0.$ (68)

The first matrix element of the corresponding Hankel matrix is then

$\displaystyle\sum_{i}p_{i}(x_{i}-s_{i_{1}})(x_{i}-s_{i_{1}+1})...(x_{i}-s_{i_{d}})(x_{i}-s_{i_{d}+1})$
$\displaystyle=$ $\displaystyle\left|\begin{array}[]{ccccc}1&1&\cdots&1&1\\\
s_{i_{1}}&s_{i_{1}+1}&\cdots&s_{i_{d}}&s_{i_{d}+1}\\\
s_{i_{1}}^{2}&s_{i_{1}+1}^{2}&\cdots&s_{i_{d}}^{2}&s_{i_{d}+1}^{2}\\\
\vdots&\vdots&\ddots&\vdots&\vdots\\\
s_{i_{1}}^{2d-1}&s_{i_{1}+1}^{2d-1}&\cdots&s_{i_{2}}^{2d-1}&s_{i_{2}+1}^{2d-1}\end{array}\right|^{-1}\left|\begin{array}[]{cccccc}a_{0}&1&1&\cdots&1&1\\\
a_{1}&s_{i_{1}}&s_{i_{1}+1}&\cdots&s_{i_{d}}&s_{i_{d}+1}\\\
a_{2}&s_{i_{1}}^{2}&s_{i_{1}+1}^{2}&\cdots&s_{i_{d}}^{2}&s_{i_{d}+1}^{2}\\\
a_{3}&s_{i_{1}}^{3}&s_{i_{1}+1}^{3}&\cdots&s_{i_{d}}^{3}&s_{i_{d}+1}^{3}\\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
a_{2d}&s_{i_{1}}^{2d}&s_{i_{1}+1}^{2d}&\cdots&s_{i_{d}}^{2d}&s_{i_{d}+1}^{2d}\end{array}\right|\geq
0,$ (80)

where the first term is the inverse of the Vandermonde determinant, which is
positive. At finite dimensions we are no longer guaranteed that from eq.(48)
we have $x$’s at the discrete sites, and these are now no longer a corollary,
i.e. they become a new constraint to be imposed by hand.

For a single moment curve, the fact that for the truncated moment problem, PSD
of the Hankel matrix is necessary but not sufficient condition is well
understood. For example, from the work of Curto and Fialkow Curto  one can see
that If the Hankel matrix is non-vanishing, then there is a solution to the
moment problem. But when the Hankel matrix vanishes, then one needs to have an
extra requirement that there exists a flat extension, i.e. there exists an
extension of $H_{n}$ to $H_{n{+}1}$ such that it retains the same rank. The
corresponding constraint for the product moment is less understood, at least
in terms of concrete solutions that can be directly applied on the couplings,
and we leave this for future work.

###  2.4 Symmetry planes

An important ingredient in the geometry will be the symmetry planes, which
reflect some symmetry associated to the ordering of external particles in the
4 point amplitude. The symmetry plane can generically be described by any
linear function in $a_{k,q}$, $f(a_{k,q})=0$. From the UV point of view, the
null constraints act by imposing non-trivial conditions on the spectrum. This
reduces the allowed space of couplings, since now in couplings given by
$a_{k,q}=\sum p_{i}x_{i}^{k}y_{i}^{q}$, $p_{i}$, $x_{i}$ and $y_{i}$ are
further restricted. From the geometry point of view, the symmetries reduce the
allowed space by simply picking slices of the EFThedron. For now we will only
use the purely geometric aspect of the null conditions, and return to the UV
perspective in Section 5.

In general, a symmetry plane can combine couplings at both different $k$ and
$q$. This is the case of the monodromy plane, which was considered in
Huang:2020nqy . In our present discussion we will only consider cyclic and
permutation invariance, so it will turn out each symmetry plane only acts at
equal $k$, relating couplings at different $q$. While a plane acts in one
particular $k$ level, it also induces non trivial effect to both higher and
lower $k$ couplings. For instance, let us illustrate a simple example , that
will also be relevant to the $su$-EFThedron. We have the condition,
originating from permutation invariance

$a_{4,2}=8a_{4,1}$ (81)

The polytope boundaries at $k=4$ are
$\langle\vec{a}_{4},J_{i}^{2},J_{i+1}^{2}\rangle\geq 0$, $J^{2}=\ell(\ell+1)$,
where for $su$ $\ell\in\\{0,2,4,6,\ldots\\}$. The first few are

$-6a_{4,1}+a_{4,2}\geq 0,\quad 120a_{4,0}-26a_{4,1}+a_{4,2}\geq 0,\quad
840a_{4,0}-62a_{4,1}+a_{4,2}\geq 0$ (82)

We obtain the intersection in Figure 4,

Figure 4: 1\. $(a_{4,2},a_{4,1})$ polytope and $a_{4,2}=8a_{4,1}$ symmetry
plane

It is clear the symmetry plane intersects only one of the polytope walls,
namely the one defined by vertices $J^{2}_{i}=6$, $J^{2}_{i+1}=20$. This can
be directly explained by looking at the dispersive form of the null constraint
in eq.(81)

$n_{4}=\sum p_{i}\frac{1}{m_{i}^{10}}J_{i}^{2}\left(J_{i}^{2}-8\right)$ (83)

We observe the summand has a root at $J^{2}=8$, so the relevant polytope wall
is the one with vertices on either side of this root, namely $J^{2}_{i}=6$,
$J^{2}_{i+1}=20$ in our $su$ case when there is no $J_{i}^{2}=12$.

More generally, we can consider a 3D geometry of $(a_{3,1},a_{4,1},a_{4,2})$,
carved out by the constraints derived in the previous section, including the
polytope wall $\langle\vec{a}_{4},6,20\rangle$. In Figure 5 we plot the
intersection with the symmetry plane, now a finite region, now in
$(a_{3,1},a_{4,1})$ space

Figure 5: $(a_{3,1},a_{4,1},a_{4,2})$ space intersected with
$a_{4,2}=8a_{4,1}$ symmetry plane, and the resulting $(a_{3,1},a_{4,1})$ slice

##  3 The $s$-EFThedron

As a warm-up for latter analysis, let’s consider a toy-model EFT where the
light states are adjoint or bi-fundamental states that admit an ordering. That
is, we can decompose the amplitude of the massless scalars in terms of color
traces and focus on the single trace amplitudes

$M(\phi^{a}\phi^{b}\phi^{c}\phi^{d})tr(T^{a}T^{b}T^{c}T^{d})$ (84)

Due to its trace structure, at fixed $t$, the UV completion will only involve
$s$-channel discontinuities. Thus the dispersive representation of the EFT
coefficients will be given as

$\sum_{k,q}g^{(s)}_{k,q}s^{k-q}t^{q}=-\sum_{i}p_{i}\;\frac{P_{\ell_{i}}(1{+}2t/m_{i}^{2})}{s{-}m^{2}_{i}}\quad
p_{i}>0$ (85)

where $P_{\ell}(\cos\theta)$ is the Legendre polynomial and the equality is
understood in terms of Taylor expansion in $s,t$ keeping terms with $k{-}q\geq
2$. The latter constraint is required for the absence of subtraction terms in
the dispersion relation (see Arkani-Hamed:2020blm  for a general discussion).
In the following, we will study the geometry emerging for these dispersion
relations and their implications for the couplings.

###  3.1 The $s$-EFThedron as a rotation

Expanding the Legendre polynomial

$P_{\ell}(1+2\delta)=\sum_{q=0}^{\ell}v_{\ell,q}\delta^{q}\,,$ (86)

the EFT couplings $g_{k,q}$ through the dispersive representation are
therefore given by

$\left(\begin{array}[]{cccc}g^{(s)}_{2,0}&&&\\\
g^{(s)}_{3,0}&g^{(s)}_{3,1}&&\\\ g^{(s)}_{4,0}&g^{(s)}_{4,1}&g^{(s)}_{4,2}&\\\
\vdots&\vdots&\vdots&\vdots\end{array}\right)=\sum_{i}\frac{p_{i}}{m^{6}_{i}}\left(\begin{array}[]{c}1\\\
\frac{1}{m^{2}_{i}}\\\ \frac{1}{m^{4}_{i}}\\\ \frac{1}{m^{6}_{i}}\\\
\vdots\end{array}\right)\otimes(v_{\ell_{i},0},v_{\ell_{i},1},v_{\ell_{i},2},\cdots)\,.$
(87)

where $v_{\ell,q}$, in terms of $J^{2}=\ell(\ell+1)$, is

$\displaystyle
v_{\ell,q}=\frac{\prod_{a=1}^{q}(J^{2}{-}a(a{-}1))}{(q!)^{2}}\,.$ (88)

Importantly, as $v_{\ell,q}$ is a complete polynomial of degree $q$ in
$J^{2}$, we can invert the relation via a GL transformation, and so we can
write

${\bf{G}}_{J,v}\cdot\vec{v}=\vec{J^{2}}\,.$ (89)

For example, the $5\times 5$ rotation is given by

${\bf{G}}_{J,v}=\begin{pmatrix}1&0&0&0&0\\\ 0&1&0&0&0\\\ 0&2&4&0&0\\\
0&4&32&36&0\\\ 0&8&208&720&576\end{pmatrix}$ (90)

Writing $\vec{g}^{\;(s)}_{k}=(g^{(s)}_{k,0},g^{(s)}_{k,1},\cdots)$, we define
the new couplings $a_{k,q}$ via the same GL transformation

$\vec{a}_{k}={\bf{G}}_{J,v}\cdot\vec{g}^{\,(s)}_{k}\,,$ (91)

where now

$\left(\begin{array}[]{cccc}a_{2,0}&&&\\\ a_{3,0}&a_{3,1}&&\\\
a_{4,0}&a_{4,1}&a_{4,2}&\\\
\vdots&\vdots&\vdots&\vdots\end{array}\right)=\sum_{i}\frac{p_{i}}{m^{6}_{i}}\left(\begin{array}[]{c}1\\\
\frac{1}{m^{2}_{i}}\\\ \frac{1}{m^{4}_{i}}\\\ \frac{1}{m^{6}_{i}}\\\
\vdots\end{array}\right)\otimes(1,J^{2}_{i},J_{i}^{4},J_{i}^{6},\cdots)\,.$
(92)

Thus we see that the dispersion representation in eq.(85) translates to the
constraint that the EFT couplings $g^{(s)}_{k,q}$ must live in the space of
product moment curves. Moreover, one of the moments will be discrete since
$J^{2}$ can only take discrete values, $\\{0,2,6,12,20,\cdots\\}$, while the
other will only take values in the interval $[0,1]$, as we see next.

For a EFT to be well defined, there must be a gap between the UV states and
the low energy (massless) states. In other words we have that all $m_{i}\geq
M_{\rm gap}$. These observations are easily incorporated by considering the
rescaled couplings

$\left(\begin{array}[]{c}a_{k,q}\\\ a_{k{+}1,q}\\\ a_{k{+}2,q}\\\
\vdots\end{array}\right)=\left(\begin{array}[]{c}a_{k,q}\\\
a_{k{+}1,q}M^{2}_{\rm gap}\\\ a_{k{+}2,q}M^{4}_{\rm gap}\\\
\vdots\end{array}\right)=\sum_{a}p_{a}\left(\begin{array}[]{c}1\\\
\frac{M^{2}_{\rm gap}}{m^{2}_{a}}\\\ \frac{M^{4}_{\rm gap}}{m^{4}_{a}}\\\
\vdots\end{array}\right)=\sum_{a}p_{a}\left(\begin{array}[]{c}1\\\ x_{a}\\\
x^{2}_{a}\\\ \vdots\end{array}\right)\,,$ (93)

where $x_{a}$ is bounded from above and below, $0<x_{a}\leq 1$, so the results
for bounded intervals derived in the previous section can be used. We will
usually normalize $M_{\rm gap}=1$ in the rest of the paper.

Finally, since the amplitude is defined with a color trace, we can choose the
states in such a way that it is cyclic invariant, then we have

$M(s,t)=M(t,s)$ (94)

simply implying $g_{k,q}=g_{k,k-q}$, for $k\geq 2$ and $k-q\geq 2$. Couplings
satisfying the previous identities are said to live on the cyclic symmetry
plane, and the theory space for the EFT couplings is simply the intersection
of the cyclic plane with the product geometry.

In summary, we have shown that the $s$-EFThedron is simply a rotation of the
bi-variate moment problem defining the a geometry, intersected with the cyclic
symmetry plane!

###  3.2 Explicit region analysis

Having obtained the constraints, the allowed region for the couplings can be
computed, and in this section, we will be interested in the
$(g_{3,0},g_{3,1})$-space. We first carve out the $a$ space then rotate back
to the $g$ space. Since $a_{k,0}=g_{k,0}$ and $a_{k,1}=g_{k,1}$, mapping those
couplings is trivial. We will consider constraints by order in $k$.

  * •

$k=3$ The only conditions are

$\displaystyle a_{2,0},a_{3,0},a_{3,1}\geq 0$ $\displaystyle
a_{2,0}-a_{3,0}\geq 0$ (95)

  * •

$k=4$ We now have the moment matrix, and the shifted matrix, which in this
case contains just $a_{4,1}$

$\begin{pmatrix}a_{2,0}&a_{3,0}&a_{3,1}\\\ a_{3,0}&a_{4,0}&a_{4,1}\\\
a_{3,1}&a_{4,1}&a_{4,2}\end{pmatrix}\geq 0,a_{4,1}\geq 0$ (96)

Next, the twisted Hankel constraints are

$(a_{3,0}-a_{4,0}),\,(a_{3,1}-a_{4,1})\geq 0.$ (97)

Since for $s$-EFThedron $J^{2}_{i}=\ell_{i}(\ell_{i}+1)$ can only take
discrete values, $J^{2}_{i}=\\{0,2,6,12,20,\cdots\\}$, we only have the
following polytope constraints

$\langle\vec{a}_{4},J^{2}_{i},J^{2}_{i+1}\rangle\geq 0,$ (98)

with $\vec{a}_{4}=(a_{4,0},a_{4,1},a_{4,2})$. Eliminating
$a_{4,0},a_{4,1},a_{4,2}$ in these inequalities, we obtain the same simple
region carved out by the $k=3$ constraints alone

$1\geq a_{3,0}\geq 0,\,a_{3,1}\geq 0.$ (99)

As expected we see no improvement without the presence of a null constraint.

  * •

$k=5$ The new constraints are

$\begin{pmatrix}a_{3,0}&a_{4,0}&a_{4,1}\\\ a_{4,0}&a_{5,0}&a_{5,1}\\\
a_{4,1}&a_{5,1}&a_{5,2}\\\
\end{pmatrix},\,\begin{pmatrix}a_{3,1}&a_{4,1}&a_{4,2}\\\
a_{4,1}&a_{5,1}&a_{5,2}\\\ a_{4,2}&a_{5,2}&a_{5,3}\\\ \end{pmatrix}\geq 0,$
$\begin{pmatrix}a_{2,0}-a_{3,0}&a_{3,0}-a_{4,0}&a_{3,1}-a_{4,1}\\\
a_{3,0}-a_{4,0}&a_{4,0}-a_{5,0}&a_{4,1}-a_{5,1}\\\
a_{3,1}-a_{4,1}&a_{4,1}-a_{5,1}&a_{4,2}-a_{5,2}\\\
\end{pmatrix},\,\begin{pmatrix}a_{4,0}-a_{5,0}\end{pmatrix},\,\begin{pmatrix}a_{4,1}-a_{5,1}\end{pmatrix}\geq
0.$ (100)

Next, the $k=5$ polytope boundaries will be given by

$\displaystyle\langle 0,\vec{a}_{5},J^{2}_{i},J^{2}_{i+1}\rangle\geq 0$
$\displaystyle\langle\vec{a}_{5},J^{2}_{i},J^{2}_{i+1},\infty\rangle\geq 0$
(101)

with $\vec{a}_{5}=(a_{5,0},a_{5,1},a_{5,2},a_{5,3})$.

Crucially, at $k=5$ the first null constraint $g_{5,2}=g_{5,3}$ appears,
implying

$a_{5,3}=17a_{5,2}-30a_{5,1}$ (102)

Written in terms of UV states, the associated null constraint

$n_{5}=\sum_{i}\frac{p_{i}}{m_{i}^{12}}\left(J_{i}^{6}-17J_{i}^{4}+30J_{i}^{2}\right)=0$
(103)

has roots at $J^{2}=2,15$. This means the only relevant walls will be those
with vertices on either side of $J^{2}=15$, namely $\langle
0,\vec{a}_{5},12,20\rangle$ and $\langle\vec{a}_{5},12,20,\infty\rangle$.

The matrices in eqns.(100) make solving the inequalities difficult, so we will
resort to a numerical mapping of the space. In Appendix B we derive a weaker
analytic expression for the boundaries.

To solve the complete set of conditions, we discretize
$(a_{3,0},a_{3,1},a_{4,0},a_{5,0})$ into lattice points and check using the
FindInstance function in Mathematica whether there exist
$(a_{4,1},a_{4,2},a_{5,1},a_{5,2},a_{5,3})$ that meet all the conditions
above.111A cut-off time of 30 minutes was given to FindInstance to find a
solution per lattice point. We find that increasing the cut-off did not change
the boundary. The result is also consistent with the $su$ analysis, for which
we will present an analytic solution. This choice for couplings we put on the
lattice is important, since we already know all couplings with $q=0$ are
bounded, $0\leq a_{k0}\leq 1$, so given a lattice size there is a finite
number of points to check. We find a finite allowed region for $g_{3}$ space,
which we compare with an SDPB analysis Figure 6. We observe a small deviation,
which is a consequence of the truncation at $k=5$. Extracting the upper
bounds, we obtain

$\frac{g_{3,1}}{g_{2,0}}\leq 13.47$ (104)

from geometry compared to

$\frac{g_{3,1}}{g_{2,0}}\leq 13.33$ (105)

from SDPB.

Figure 6: Geometry bound and SDPB bound for $g_{3,1},g_{3,0}$ space.

We expect that including higher order constraints, such as those involving
couplings at order $k=6$, will improve our results and match the functional
result in the infinite limit.

##  4 The EFThedron

We now turn to the more general set up, the case where both $s$ and
$u$-channel singularities are allowed at fixed $t$. This set up will be
applicable to general $a,b\rightarrow a,b$ scattering where $a,b$ can be
distinct external states. If we consider the explicit amplitude $M(a,b,b,a)$,
such that the $t\rightarrow 0$ limit corresponds to forward scattering, the
low energy couplings, defined as

$M^{\rm IR}(a,b,b,a)=\sum_{k,q}\;g_{k,q}z^{k{-}q}t^{q}\,,$ (106)

where $s={-}t/2{+}z$, $u={-}t/2{-}z$. Due to the $2\leftrightarrow 3$ symmetry
of the setup, the couplings will be non-zero only for even powers of $z$, i.e.
$k{-}q\in even$. The dispersive representation equates

$\sum_{k,q}\;g_{k,q}z^{k{-}q}t^{q}=-\sum_{i}p_{i}\,P_{\ell_{i}}(1{+}\frac{2t}{m^{2}_{i}})\left(\frac{1}{-\frac{t}{2}-z-m_{i}^{2}}{+}\frac{1}{-\frac{t}{2}+z-m_{i}^{2}}\right)\,,$
(107)

where once again, the equality is understood in terms of a Taylor expansion in
$z,t$ on both sides and for terms with $k{-}q\geq 2$. The result is the
identity

$g_{k,q}=\sum_{i}p_{i}\frac{u_{\ell_{i},k,q}}{m^{2(k{+}1)}_{i}}$ (108)

where $u_{\ell,k,q}$ are related to the Legendre coefficients $v_{\ell,q}$ via
an $k$-dependent degenerate GL rotation ${\bf{G}}^{k}_{u,v}$

$\vec{u}_{\ell,k}=\left(\begin{array}[]{c}u_{\ell,k,0}\\\ 0\\\ u_{\ell,k,2}\\\
\vdots\\\ 0\\\
u_{\ell,k,k}\end{array}\right)=\left(\begin{array}[]{cccccc}1&0&0&0&0&0\\\
0&0&0&0&0&0\\\
\frac{(k{-}1)_{2}}{2}\frac{1}{2^{2}}&-\frac{(k{-}1)_{1}}{2}&1&0&0&0\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\\ 0&0&0&0&0&0\\\
\frac{(1)_{k}}{k!}2^{-k}&-\frac{(1)_{k{-}1}}{k{-}1!}2^{1-k}&\frac{(1)_{k{-}2}}{k{-}2!}2^{2-k}&\cdots&-\frac{1}{2}&1\end{array}\right)\left(\begin{array}[]{c}v_{\ell,0}\\\
v_{\ell,1}\\\ v_{\ell,2}\\\ \vdots\\\ v_{\ell,k{-}1}\\\
v_{\ell,k}\end{array}\right)\,.$ (109)

In the following, we will demonstrate how via a series of GL rotations, the
above geometry can be converted to our familiar $a$-geometry.

###  4.1 The EFThedron as a projection

Due to the $k$-dependence of ${\bf{G}}^{k}_{u,v}$, the convex hull is now of
an entangled product geometry. In Arkani-Hamed:2020blm , the boundaries of
this space were obtained by approximating the geometry with a series of
deformed moment curves.

Before the GL transformation, the hull is our previous $a$-geometry. Thus a
priori, simply acting with an inverse transform on $g_{k,q}$ in eq.(108), we
can revert to the $a$-geometry and apply the associated Hankel positivity. The
obstruction is of course that ${\bf{G}}^{k}_{u,v}$ is degenerate, of half-
rank. To this end, lets instead consider an auxiliary
$\widetilde{\bf{G}}^{k}_{u,v}$

$\widetilde{\bf{G}}^{k}_{u,v}=\begin{pmatrix}1&0&0&0&&0&&0\\\
0&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\textbf{1}}&0&0&&0&&0\\\
\frac{(k{-}1)_{2}}{2}\frac{1}{2^{2}}&-\frac{(k{-}1)_{1}}{2}&1&0&&0&&0\\\
0&0&0&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\textbf{1}}&&0&&0\\\
\vdots&\vdots&\vdots&\vdots&&\vdots&&\vdots\\\
0&0&0&0&&{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\textbf{1}}&&0\\\
\frac{(1)_{k}}{k!}2^{-k}&-\frac{(1)_{k{-}1}}{k{-}1!}2^{1-k}&\frac{(1)_{k{-}2}}{k{-}2!}2^{2-k}&\cdots&&-\frac{1}{2}&&1\end{pmatrix}\,.$
(110)

where we have inserted auxiliary rows such that $\widetilde{\bf{G}}^{k}_{u,v}$
becomes full rank. Defining

$\vec{u}^{\prime}_{\ell,k}=\widetilde{\bf{G}}^{k}_{u,v}\vec{v}_{\ell}\,,$
(111)

we see that $\vec{u}^{\prime}_{\ell,k}=\vec{u}_{\ell,k}$ for the components
where $k{-}q=even$, while $\vec{u}^{\prime}$ will have non-zero entries for
$k{-}q=odd$. Now let’s consider

$\vec{g}^{\prime}_{k}=\sum_{i}p_{i}\frac{\vec{u}^{\prime}_{\ell_{i},k}}{m^{2(k{+}1)}_{i}}=\sum_{i}p_{i}\frac{1}{m^{2(k{+}1)}_{i}}\widetilde{\bf{G}}^{k}_{u,v}\vec{v}_{\ell_{i}}\,.,$
(112)

where $\vec{g}^{\prime}_{k}$ now contains $k{-}q=odd$ auxiliary couplings.
Multiplying both sides by $(\widetilde{\bf{G}}^{k}_{u,v})^{{-}1}$ and
${\bf{G}}_{J,v}$, we can now recover the $a$-geometry

${\bf{G}}_{J,v}(\widetilde{\bf{G}}^{k}_{u,v})^{{-}1}\vec{g}^{\prime}_{k}=\sum_{i}\frac{p_{i}}{m^{6}_{i}}\left(\begin{array}[]{c}1\\\
\frac{1}{m^{2}_{i}}\\\ \frac{1}{m^{4}_{i}}\\\ \frac{1}{m^{6}_{i}}\\\
\vdots\end{array}\right)\otimes(1,J^{2}_{i},J_{i}^{4},J_{i}^{6},\cdots)\,.$
(113)

In other words, the full EFThedron is simply a projection of the $a$-geometry!

In practice, we first take the Hankel positivity constraints on the couplings
in the $a$-geometry, $a_{k,q}$, and GL rotate back to $g^{\prime}_{k,q}$
through the action of $\widetilde{\bf{G}}^{k}_{u,v}({\bf{G}}_{J,v})^{{-}1}$.
Second, project out the auxiliary couplings from the constraint, ie. those
which have $k{-}q=odd$. Importantly, the result is independent of how we
complete the rank for ${\bf{G}}^{k}_{u,v}$, i.e. after projecting out the
auxiliary couplings we retain a “gauge” invariant result. It is straight
forward to verify that this is indeed the case.

For $a,b\rightarrow b,a$ scattering, the constraint imposed by the
$a$-geometry is the final result. However, for identical scalars ($a=b$), the
low energy amplitude must now respect permutation invariance, i.e. the EFT
amplitude is a function of $\sigma_{2}=s^{2}+t^{2}+u^{2}$ and
$\sigma_{3}=stu$, implying linear relations between $g_{k,q}$ of equal $k$,
i.e. the couplings live on the permutation plane

$M(z,t)=M(\frac{3t}{4}{+}\frac{z}{2},z{-}\frac{t}{2}),\quad$ (114)

Thus for identical scalar scattering, the geometry we are interested in is the
intersection of the projected $a$-geometry, intersected with the permutation
plane.

###  4.2 Explicit region analysis

The conditions we will use are mostly identical to those for the
$s$-EFThedron. The two differences are related to polytopes, as now
$J^{2}=\\{0,6,20,\ldots\\}$, and the null constraint, which now first appears
at $k=4$. Because the summand of the null constraint, given by

$n_{4}=\sum_{i}\frac{p_{i}}{m_{i}^{10}}J^{2}_{i}(J_{i}^{2}-8)$ (115)

has a root at $J^{2}=8$, only the polytope wall
$\langle\vec{a}_{4},6,20\rangle$ will intersect the symmetry plane. Therefore
the only polytope condition we need to consider is

$\textrm{det}\begin{pmatrix}a_{4,0}&1&1\\\ a_{4,1}&6&20\\\
a_{4,2}&36&400\end{pmatrix}\geq 0$ (116)

Like for the $s$-EFThedron, wish to bound $(a_{4,0},a_{3,1})$ space. Because
we only need to go to $k=4$, we can actually solve all the constraints
explicitly. We can obtain a rough idea by first only solving

$\displaystyle a_{2,0}a_{4,2}-a_{3,1}^{2}$ $\displaystyle\geq$ $\displaystyle
0$ $\displaystyle\langle\vec{a}_{4},6,20\rangle$ $\displaystyle\geq$
$\displaystyle 0$ $\displaystyle a_{3,1}$ $\displaystyle\geq$ $\displaystyle
0$ (117)

which leads to a simple square root behavior

$0\leq\left(\frac{a_{3,1}}{a_{2,0}}\right)^{2}\leq\frac{160}{3}\frac{a_{4,0}}{a_{2,0}},\quad
0\leq a_{4,0}\leq 1$ (118)

We can improve this result further by adding the $3\times 3$ moment matrix
determinant

$-a_{3,1}^{2}a_{4,0}+2a_{3,0}a_{3,1}a_{4,1}-a_{2,0}a_{4,1}^{2}-a_{3,0}^{2}a_{4,2}+a_{2,0}a_{4,0}a_{4,2}\geq
0$ (119)

Solving the constraints now involving this equation, we find

$\displaystyle 0\leq a_{3,1}\leq 4\sqrt{a_{3,0}}+4a_{3,0},\quad\textrm{for
}0\leq a_{3,0}\leq 4/9,\quad$ $\displaystyle 0\leq
a_{3,1}\leq\frac{20}{3}a_{3,0}+\frac{4}{3}\sqrt{5}\sqrt{a_{3,0}-a_{3,0}^{2}},\quad\textrm{for
}4/9\leq a_{3,0}\leq 1$ (120)

where we have normalized $a_{2,0}=1$ for simplicity. We plot the two regions
corresponding to eqs.(118) and (120) in Figure 7.

Figure 7: ($a_{4,0}$, $a_{3,1}$) space, with and without $3\times 3$ moment
matrix

##### Real geometry

As in the s channel EFThedron, the last step requires us to rotate from the
$a$ to the $g$ couplings. Since $g_{k0}=a_{k0}$, most are trivial
replacements, except $a_{3,1}=\frac{3}{2}g_{3,0}+g_{3,1}$. Next we must
project out all couplings $g_{k,q}$ for which $k+q=\textrm{odd}$, that is
$g_{3,0}$ and $g_{4,1}$. Doing these operations we can explicitly find the
$(g_{3,1},g_{4,0}$) region, given in Appendix C, and shown in Figure 8, which
we also compare to results from SDPB Caron-Huot:2020cmc .

Figure 8: ($g_{3,1}$, $g_{4,0}$) space

We can also find a bound $\frac{g_{3,1}}{g_{2,0}}$

$-\frac{3}{2}\leq\frac{g_{3,1}}{g_{2,0}}\leq 1/12(31+\sqrt{1281})\approx 5.56$
(121)

This is compared to $-\frac{3}{2}\leq\frac{g_{3,1}}{g_{2,0}}\leq 5.30$ from
SDPB. Like for $s$-EFThedron, we expect that including higher order
constraints will converge to the real boundary in the infinite limit.

##  5 Constraints on the UV spectrum

As we have seen in previous analysis, while the space for the hull is in
general unbounded its intersection with symmetry planes yields a finite
region. The reason that it is finite is simple, in general only the low spins
are one side of the symmetry plane while the remaining majority are on the
other side. Indeed take for example the $su$-channel $k=4$ polytope
intersected with the symmetry plane $a_{4,2}=8a_{4,1}$, which we plot again in
Figure 9.

Figure 9: $(a_{4,1},a_{4,2})$ polytope and $a_{4,2}=8a_{4,1}$ symmetry plane

We see that only the spin 2 ($J^{2}=6$) vertex lives on one side of the plane.
This immediately tells us that the $k=4$ null constraint requires spin 2 to be
present. In other words, to live on the symmetry plane, certain distributions
in the spin spectrum are necessary.

A convenient way of formulating such conditions is the statement that the hull
must have zero components perpendicular to the symmetry plane. This condition
is often termed “null constraints”, which we denote as $n_{k}$. The subscript
$k$ indicates that it associated with symmetry constraints on the $k$-th level
couplings $g_{k,q}$. Once we substitute the dispersive representation for the
couplings, the condition translates into the vanishing of a sum over the UV
states. For example, the symmetry plane in Figure 9 corresponds to a null
condition

$n_{4}=a_{4,2}-8a_{4,1}=\sum_{i}\frac{p_{i}}{m_{i}^{10}}\left(J_{i}^{4}-8J^{2}_{i}\right)$
(122)

In general, null conditions will have a form

$n_{k}=\sum_{i}p_{i}\frac{1}{(m_{i}^{2})^{k{+}1}}\omega_{k}(\ell_{i})=0$ (123)

where $\omega_{k}(\ell)$ is a polynomial in $\ell$, and there may be several
null constraints at the same $k$ level. The fact that the above expression
must sum to zero imposes non-trivial conditions on the UV spectrum. We will
analyze these constraints in steps, first utilizing just the sign of
$\omega_{k}(\ell)$, then its magnitude, and finally linear combinations of
$\omega_{k}(\ell)$ with distinct $k$. We summarize our finding in the
following

  * •

Conditions on spins Given that $\omega_{k}(\ell)$ is a finite degree
polynomial, at sufficiently large spins its sign is fixed, which we can always
take to be positive. Thus if the spectrum only contains high spin states for
which $\omega_{k}(\ell)>0$, it is impossible for the null constraints to be
satisfied, since negative terms must also be present. This tells us that the
real roots of $\omega_{k}(\ell)$ encode the necessary conditions the spin
distribution must satisfy for $n_{k}$ to hold. Collecting the conditions from
different $k$ one obtains global constraints on the spectrum. For example, we
will find that assuming the existence of non-scalar states, the UV completion
of a theory containing identical scalars must include an infinite number of
arbitrarily high even spins, including all even spins $\ell\leq 28$.

  * •

Conditions on probabilities Beyond the sign pattern, the functional form
$\omega_{k}(\ell)$ tells us the magnitude of the contribution for each spin.
Since the contribution must cancel out, this imposes upper bounds on the
averaged spinning spectral function $\langle p_{\ell,k}\rangle$ for any fixed
$\ell$. In particular, for identical scalars we can deduce upper bounds for
the ratio $\langle p_{\ell,k}\rangle/\langle p_{2,k}\rangle$, such as

$\frac{\langle p_{4,\ell}\rangle}{\langle
p_{4,2}\rangle}\leq\frac{12}{\ell(\ell+1)\left(\ell^{2}+\ell-8\right)},~{}(\ell\geq
4)$ (124)

  * •

Conditions on masses Combining different $k$-order constraints, we now obtain
functions that are no longer homogenous in $m_{i}$. This will then allow us to
deduce that there must exists a state in the spectrum whose mass is below a
“gap”

$m_{\ell}^{2}\leq M^{2}B(\ell)$ (125)

where $M$ is the heaviest spin 2 particle.

In the following we will give the details for the $s$\- and full EFThedron, as
well as the case when the external states carry spin.

###  5.1 Implications for spin spectrum

Substituting the dispersive representation for the couplings in the null
constraints, we will in general arrive at

$n_{k}=\sum_{i}\frac{p_{\ell_{i}}}{(m_{i}^{2})^{k{+}1}}\omega_{k}(\ell_{i})=0\,,$
(126)

where the function $\omega_{k}(\ell)$ is a polynomial in $\ell$ and thus have
finite roots. Since $p_{\ell}$ is positive, this immediately tells us there
must be states residing in both the positive and negative regions of the
polynomial. We begin by first analyzing this sign pattern for the
$s$-EFThedron.

$s$-EFThedron

Imposing cyclic symmetry

$\displaystyle M(s,t){=}M(t,s)\quad\rightarrow\quad g_{k,q}=g_{k,k{-}q}\,,$
(127)

leads to the following null constraints

$\displaystyle
g_{k,q}{-}g_{k,k-q}=\sum_{i}\frac{p_{i}}{(m^{2}_{i})^{k{+}1}}\left(v_{\ell_{i},q}-v_{\ell_{i},k-q}\right)=0\,.$
(128)

Since the couplings are well defined only for $k\geq 2$ and $k-q\geq 2$, the
first valid cyclic null constraint appears at $k=5$, and is given by

$\displaystyle
n_{5}=g_{5,2}-g_{5,3}=\sum_{i}\frac{p_{i}}{m_{i}^{12}}(\ell_{i}{-}1)\ell_{i}(\ell_{i}+1)(\ell_{i}+2)(\ell_{i}^{2}{+}\ell_{i}{-}15)=0\,.$
(129)

First, if all spins in the theory are 0 or 1, the null constraint is trivially
satisfied. However, if there is at least one particle with spin $\ell>1$, the
null constraint becomes non-trivial. Importantly we must have states residing
in both positive and negative regions of the polynomial
$w_{5}(\ell)=(\ell{-}1)\ell(\ell+1)(\ell+2)(\ell^{2}{+}\ell-15)$, in order for
the null constraint to hold. Noting the last factor has a positive root at
$\ell\sim 3.4$, lets consider the sign pattern of this function for each spin

$\ell$ | 0 | 1 | 2 | 3 | 4 | 5 | $\ldots$
---|---|---|---|---|---|---|---
$\omega_{5}(\ell)$ | 0 | 0 | $-$ | $-$ | $+$ | + | +

We observe that for spins 2 and 3 it is negative, while for spins 4 and
greater it is positive. This means for a theory to satisfy cyclic invariance
and contain spins beyond spin 1, it must contain either a spin 2 or a spin 3,
as well as at least one spin 4 or higher.

The $k=6$ null constraint is weaker and does not add new information, while at
$k=7$ the null constraint $g_{7,3}=g_{7,4}$ yields

$\omega_{7}(\ell)=(\ell-2)(\ell-1)\ell(\ell+1)(\ell+2)(\ell+3)(\ell^{2}+\ell-28)\,,$
(130)

with a final root at $\ell\sim 4.8$, and the following sign pattern

$\ell$ | 0 | 1 | 2 | 3 | 4 | 5 | $\ldots$
---|---|---|---|---|---|---|---
$\omega_{7}(\ell)$ | 0 | 0 | 0 | $-$ | $-$ | + | +

This tells us that if we have states with $\ell\geq 3$, then the spectrum must
have either a spin 3 or 4 and at least one spin $\geq 5$.

We can continue this argument order by order at higher $k$. It is useful to
specifically consider constraints $g_{2a{+}1,a}=g_{2a{+}1,a{+}1}$ which takes
the form

$\displaystyle\omega_{2a+1}(\ell)=(\ell^{2}+\ell-
f_{c}(a))\prod_{i=0}^{a-1}(\ell-i)\prod_{j=1}^{a+1}(\ell+j)$ (131)

where $f_{c}(a)=(a+1)(2a+1)$, leading to a positive root at

$\ell_{c}(a)\equiv\frac{{-}1{+}\sqrt{1{+}4f_{c}(a)}}{2}$ (132)

which for integer $a>0$ is well approximated by

$\ell_{c}(a)\approx\sqrt{2}a+\frac{1}{2}\left(\frac{3}{\sqrt{2}}-1\right)$
(133)

Now $\omega_{2a+1}(\ell)$ has zeroes for $\ell\leq a{-}1$, and has only one
negative region between $a-1<\ell<\ell_{c}(a)$ , and becomes positive for
$\ell>\ell_{c}(a)$. This leads to the following conditions

$\displaystyle k$ $\displaystyle=5:\ell\in\\{2,3\\}\textrm{ and
}\ell\in\\{4,5,\dots\\}$ $\displaystyle k$
$\displaystyle=7:\ell\in\\{3,4\\}\textrm{ and }\ell\in\\{5,6,\dots\\}$
$\displaystyle k$ $\displaystyle=9:\ell\in\\{4,5,6\\}\textrm{ and
}\ell\in\\{7,8,\dots\\}$ $\displaystyle k$
$\displaystyle=11:\ell\in\\{5,6,7\\}\textrm{ and }\ell\in\\{8,9,\dots\\}\,.$
(134)

These constraints are valid for UV completions of colored scalars states
involving massive states beyond spin 1. From the above we can immediately
conclude that the spectrum must be unbounded in spin. As $k$ increases, we
observe that the conditions on spins become weaker. This is because the
interval between the last two roots grows as $\sim(\sqrt{2}-1)a$. The same
behavior will also occur for the $su$-EFThedron, but the conditions will be
stronger.

The full EFThedron

We now perform the same analysis with permutation invariance of identical
scalars. Recall that the permutation plane for the full EFThedron is defined
as

$\displaystyle M(z,t)=M(\frac{z}{2}{+}\frac{3t}{4},{-}\frac{t}{2}{+}z)\,.$
(135)

When translated into linear relations for the couplings, for the first few
$k$s they are given by (recall that we must have $k\geq 2$, $k-q\geq 2$ )

$\displaystyle k=4:\quad g_{4,2}$ $\displaystyle=\frac{2}{3}g_{4,0},\quad$
$\displaystyle k=5:$ $\displaystyle\quad g_{5,3}=\frac{g_{5,1}}{2}$
$\displaystyle k=6:\quad g_{6,4}$
$\displaystyle=\frac{45}{16}g_{6,0}{-}\frac{1}{2}g_{6,2},\quad$ $\displaystyle
k=7:$ $\displaystyle\quad g_{7,3}=\frac{4}{5}g_{7,1},\quad
g_{7,5}=\frac{16}{3}g_{7,1}\quad$ $\displaystyle k=8:\quad g_{8,4}$
$\displaystyle=\frac{21}{8}g_{8,0}{-}\frac{1}{4}g_{8,2},$ $\displaystyle\quad
g_{8,6}=\frac{21}{8}g_{8,0}{-}\frac{5}{16}g_{8,2}\,.$ (136)

Substituting the dispersive representation for the $k=4$ the null constraint
one finds

$\displaystyle
n_{4}=\sum_{i}\frac{p_{i}}{m^{10}}\ell_{i}(\ell_{i}+1)(\ell_{i}^{2}+\ell_{i}-8)=0\,.$
(137)

The sign pattern of the summand $\omega_{4}(\ell)=(\ell(\ell+1)-8)$ is given
as

$\ell$ | 0 | 2 | 4 | 6 | …
---|---|---|---|---|---
$\omega_{4}(\ell)$ | 0 | $-$ | + | + | +

This implies that if we have states other than scalars, then we must have a
spin $\ell=2$, and at least one state with $\ell\geq 4$. Going to higher $k$,
our strategy will be to take linear combinations of null constraints that
force roots for lower spins, are negative for one particular spin, and
positive for higher spins. At $k=5,6$ no new constraints in terms of spectrum
arise. At $k=7$ we have two independent null constraints

$n_{7}\equiv g_{7,3}-\frac{4}{5}g_{7,1}=0,\quad n_{7}^{\prime}\equiv
g_{7,5}-\frac{16}{3}g_{7,1}=0$ (138)

Both have zeros at $l=0$, and we would like to have a root at $l=2$. The
combination

$n_{7}+n_{7}^{\prime}=\sum_{i}\frac{p_{i}}{m_{i}^{7}}(\ell_{i}-2)\ell_{i}(\ell_{i}+1)(\ell_{i}+3)\left(\ell_{i}^{2}+\ell_{i}-\frac{49}{2}\right)=0$
(139)

achieves just that, and as desired we obtain sign pattern

$\ell$ | 0 | 2 | 4 | 6 | …
---|---|---|---|---|---
$\omega_{7}(\ell)$ | 0 | 0 | - | + | +

.

We observe it is negative only for $\ell=4$, and positive for $\ell\geq 6$.
Thus now we have the new condition that the spectrum must contain at least one
$\ell=4$ particle, as well even higher spin particles.

In general, it is easy to arrange null constraints at order $k=3a+1$ in the
form

$\omega_{3a+1}(\ell)=(\ell^{2}+\ell-
f_{p}(a))\prod_{i=0}^{a-1}(\ell-2i)\prod_{j=0}^{a-1}(\ell+2j+1)$ (140)

where $f_{p}(a)=(3a+1)^{2}/2$, leading to a root at

$\ell_{p}(a)\approx\frac{3a}{\sqrt{2}}+\frac{1}{2}\left(\sqrt{2}-1\right)$
(141)

Like in $s$-EFThedron, we have roots for $l\leq 2a-2$, negative values for
$2a-2<l<\ell_{p}(a)$, and positive values for $l>\ell_{p}(a)$. For constraints
up to $k\leq 43$, this implies the following sign pattern

$\ell$ | 0 | 2 | … | $2a-2$ | $2a$ | $2a+2$ | …
---|---|---|---|---|---|---|---
$\omega_{3a+1}(\ell)$ | 0 | 0 | 0 | 0 | - | + | +

which is negative only for $\ell=2a$, thus requiring the spin $\ell=2a$, as
well as some state with $\ell\geq 2a+2$. Combining the different $k$s, we
immediately conclude that for identical scalars coupled to at least a higher
spin-state, the UV completion must contain an unbounded number of even spins,
including all states below spin 28!

Going to $k=46$ we find that both spins 30 and 32 have negative contributions,
so the presence of either is sufficient, a behavior that continues as we
increase $k$. We see that permutation invariance imposes more stringent
constraint on the EFThedron spectrum than cyclic invariance on the
$s$-EFThedron, but the overall pattern is similar.

Note that pure scalar theories are not constrained, since scalars trivially
solve the null constraints. This allows theories such as

$M(s,t)=\frac{1}{s-m^{2}}{+}\frac{1}{t-m^{2}}{+}\frac{1}{u-m^{2}}\,$ (142)

to be compatible with permutation invariance while not containing any higher
spin.

###  5.2 Bounds on $\langle p_{k,\ell}\rangle$

The function $\omega(\ell)$ not only puts constraints on the spin-spectrum, it
also puts upper limits on how large the $p_{\ell}$s can take. It will be
convenient to consider the average spectral function $\langle
p_{k,\ell}\rangle$, defined as

$\langle
p_{k,\ell}\rangle\equiv\sum_{\\{i,\ell_{i}=\ell\\}}\frac{p_{\ell_{i}}}{m^{2(k{+1})}_{i}}\,.$
(143)

That is, we sum over contributions of all states with spin-$\ell$, weighted by
their mass. In terms of $\langle p_{k,\ell}\rangle$, the dispersive
representation take on a simple form

$g_{k,q}=\sum_{\ell}\;\langle p_{k,\ell}\rangle u_{\ell,q}\,.$ (144)

To illustrate how constraints on $\langle p_{k,\ell}\rangle$ can be derived
from symmetry considerations, starting with eq.(126) we write the following
sum rule

$n_{k}=\sum_{i}\frac{p_{\ell_{i}}}{(m_{i}^{2})^{k{+}1}}\omega_{k}(\ell_{i})=\sum_{\ell}\langle
p_{k,\ell}\rangle\omega_{k}(\ell)=0\,,$ (145)

where we have used the fact that $\omega_{k}(\ell)$ only depends on the spin
and not the mass.

From the previous discussion, we have seen that as we tend to higher $k$, we
have more than one null constraints $\omega_{k}(\ell)$. Suppose that we can
find a linear combination $\tilde{\omega}_{k}(\ell)$ of null constraints, such
that for some fixed spin $\ell^{*}$

$\tilde{\omega}_{k}(\ell^{*})<0,\quad\tilde{\omega}_{k}(\ell)\geq
0\;\forall\;\ell\neq\ell^{*}$ (146)

Now if we can conclude that $\ell^{*}$ must be part of the spectrum, then from
the sum rule

$\sum_{\ell}\;\langle p_{k,\ell}\rangle\tilde{\omega}_{k}(\ell)=0\,,$ (147)

we immediately find that there is an upper bound for the ratio $\langle
p_{k,\ell}\rangle/\langle p_{k,\ell^{*}}\rangle$, in order to cancel the
negative contribution from $\tilde{\omega}_{k}(\ell^{*})$. This leads to

$\frac{\langle p_{k,\ell}\rangle}{\langle
p_{k,\ell^{*}}\rangle}\leq\frac{|\tilde{\omega}_{k}(\ell^{*})|}{\tilde{\omega}_{k}(\ell)},\quad\forall\ell\geq\ell_{*}\,.$
(148)

Note that this functional approach is exactly what has been implemented in the
conformal bootstrap analysis Rattazzi:2008pe ; Caracciolo:2009bx , and in
particular to the problem of maximization of OPE coefficients. In some cases,
such a functional cannot be found, and the best one can do is having a subset
of spins being negative, with the remaining being non-negative. Then one has
instead

$\frac{\langle p_{k,\ell}\rangle}{\sum_{\ell^{*}}\langle
p_{k,\ell^{*}}\rangle|\tilde{\omega}_{k}(\ell^{*})|}\leq\frac{1}{\tilde{\omega}_{k}(\ell)}\,,$
(149)

where the sum is over the subset of spins whose $\tilde{\omega}_{k}(\ell^{*})$
is negative.

The $s$-EFThedron

For $s$-EFThedron, the first non-trivial null constraint starts at $k=5$, with

$n_{5}=\sum_{\ell}\langle p_{5,\ell}\rangle\omega_{5}(\ell)=0\,.$ (150)

The function $\omega_{5}(\ell)$ is negative for $\ell=\\{2,3\\}$ and positive
for $\ell\geq 4$, as illustrated in the following graph

The null constraint then translate to the sum of the positive region spin must
equal to those in the negative region, weighted by their $\langle
p_{5,\ell}\rangle$ respectively. For any given spin $\ell^{*}\geq 4$, the
maximum value for $\langle p_{5,\ell^{*}}\rangle$ is attained when it is the
sole source of positive contribution, which must match with the sum of that
from spin-$2$ and $3$. Thus we have an upper bound of the form,

$\frac{\langle p_{5,\ell}\rangle}{\langle
p_{5,2}\rangle|\tilde{\omega}_{5}(2)|{+}\langle
p_{5,3}\rangle|\tilde{\omega}_{5}(3)|}{=}\frac{\langle
p_{5,\ell}\rangle}{216\langle p_{5,2}\rangle{+}360\langle
p_{5,3}\rangle}\leq\frac{1}{(\ell{-}1)\ell(\ell{+}1)(\ell{+}2)(\ell^{2}{+}\ell{-}15)},\;(\ell\geq
4)\,.$ (151)

Moving to $k=6$, there is also only one null constraint in this case,

$n_{6}=\sum_{\ell\in\mathbb{Z}^{+}}\langle
p_{6,\ell}\rangle(\ell-1)\ell(\ell+1)(\ell+2)\left(\ell^{4}+2\ell^{3}-17\ell^{2}-18\ell-72\right)=0$
(152)

The spin pattern is,

$\ell$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | …
---|---|---|---|---|---|---|---|---
$\tilde{\omega}_{6}(\ell)$ | 0 | 0 | - | - | - | + | + | …

Since the $k=5$ null constraint already requires either a spin 2 or spin 3
state, the new constraint does not impose anything new in terms of which spins
should appear. However, it does yield a bound on $\langle p_{6,\ell}\rangle$:

$\frac{\langle p_{6,\ell}\rangle}{3\langle p_{6,2}\rangle{+}15\langle
p_{6,3}\rangle{+}10\langle
p_{6,4}\rangle}\leq\frac{1152}{(\ell{-}1)\ell(\ell{+}1)(\ell{+}2)\left(\ell^{4}{+}2\ell^{3}{-}17\ell^{2}{-}18\ell{-}72\right)},\quad(\ell\geq
5)\,.$ (153)

Starting from $k\geq 7$, at each level of $k$ there will be more than one null
constraint. Thus the general functionals $\tilde{\omega}_{k}(l)$ will take the
linear combination of these null constraints. One can then obtain further
bounds of the above form, where the spins of the positive region is bounded by
the total contribution of the negative region.

We can test the bound against two candidate amplitudes. The first is given as

$M(s,t)=-\frac{1}{(s{-}M^{2})(t{-}M^{2})}\,.$ (154)

Note that the above will have an infinite number of higher spin particles with
the same mass $M$, and thus does not make sense physically. However it can be
shown that this amplitude satisfies our requirements, in particular the
residue in the $s$-channel has a positive expansion, seeCaron-Huot:2016icg :

$-\frac{1}{(s-M^{2})(t-M^{2})}=\frac{1}{s-M^{2}}\sum_{\ell\in\mathbb{Z}^{+}}p_{M,\ell}P_{\ell}\Big{(}1+\frac{2t}{M^{2}}\Big{)}$
(155)

where

$p_{M,\ell}=\frac{2\ell+1}{M^{2}}\int_{-1}^{0}\mathrm{d}x\frac{P_{\ell}(1+2x)}{1-x}>0$
(156)

Thus $\langle p_{k,\ell}\rangle$ of this model is,

$\langle
p_{k,\ell}\rangle=\frac{2\ell+1}{M^{2(k+1)}}\int_{-1}^{0}\mathrm{d}x\frac{P_{\ell}(1+2x)}{1-x}\,.$
(157)

The second is the type-1 open string amplitude

$M(s,t)=\frac{\Gamma[{-}s]\Gamma[{-}t]}{\Gamma[1{-}s{-}t]}\,,$ (158)

where the residue is given as

$\text{Res}_{n}(t)=\frac{\prod_{i=1}^{n-1}(t+i)}{(n!)}\,.$ (159)

Since the higher level contributions are suppressed, for each $\langle
p_{k,\ell}\rangle$ we only sum over contributions from the first 50 levels
after the corresponding spin appears in the leading trajectory. In Figure 10
we show that both $s$-matrix will satisfy the bounds in eq.(151).

Figure 10: Here we present the $k=5$ bounds in the $s$-EFThedron ((151) &
(153)) with the $S$-matrix $\frac{-1}{(s-M^{2})(t-M^{2})}$ and the Type-I open
string amplitude

The full EFThedron

Based on the analysis of previous section 5.1, we know that for identical
scalars coupled to massive spin states, all the even spin particles must
exist, for $\ell\leq 28$. Thus it will be interesting to consider the ratio of
$\langle p_{k,\ell}\rangle$ for any higher spin state with respect to spin 2.
We first look at $k=4,5,6$ case where one only has one null constraint. From
(5.1) we get,

$\displaystyle n_{4}=\sum_{\ell\in\text{even}}\langle
p_{4,\ell}\rangle\ell(\ell+1)(\ell^{2}+\ell-8)=0$ $\displaystyle
n_{5}=\sum_{\ell\in\text{even}}\langle
p_{5,\ell}\rangle\ell(\ell+1)(\ell(\ell+1)(2\ell(\ell+1)-43)+150)=0$
$\displaystyle n_{6}=\sum_{\ell\in\text{even}}\langle
p_{6,\ell}\rangle(\ell-3)\ell(\ell+1)(\ell+4)\left(\ell(\ell+1)\left(\ell^{2}+\ell-32\right)+204\right)=0$
(160)

The sign pattern of these three null constraints is,

$\ell$ | 0 | 2 | 4 | 6 | 8 | …
---|---|---|---|---|---|---
$\tilde{\omega}_{4}(\ell)$ | 0 | - | + | + | + | …
$\tilde{\omega}_{5}(\ell)$ | 0 | - | + | + | + | …
$\tilde{\omega}_{6}(\ell)$ | 0 | - | - | + | + | …

Thus for $k=4,5$, we can use derive a valid upper bounds on $\langle
p_{k,\ell}\rangle/\langle p_{k,2}\rangle$,

$\displaystyle\frac{\langle p_{4,\ell}\rangle}{\langle
p_{4,2}\rangle}\leq\frac{12}{\ell(\ell+1)\left(\ell^{2}+\ell-8\right)},~{}(\ell\geq
4)$ $\displaystyle\frac{\langle p_{5,\ell}\rangle}{\langle
p_{5,2}\rangle}\leq\frac{216}{\ell(\ell+1)(\ell(\ell+1)(2\ell(\ell+1)-43)+150)},~{}(\ell\geq
4)\,.$ (161)

For $k=6$, since there are two spins with negative value, we can only obtain
bounds of the form similar to the $s$-EFThedron.

Starting from $k\geq 7$, one will get more than one null constraint. In order
to get the strongest upper bound for $\langle p_{k,\ell_{a}}\rangle/\langle
p_{k,2}\rangle$, we require the special functional to satisfy the following
conditions,

$\displaystyle\text{For}~{}\ell\geq
4,~{}l\in\text{even},~{}\tilde{\omega}_{k}(l)\geq 0$
$\displaystyle\text{Minimize}~{}-\frac{\tilde{\omega}_{k}(2)}{\tilde{\omega}_{k}(l_{a})}$
(162)

We call this kind of the special functional as optimal functional. At $k=7,9$,
one will have two null constraints. The optimal functionals take the form,

$\tilde{\omega}_{7}(\ell)={\rm
det}[\vec{G}^{(7)}_{4},\vec{G}^{(7)}_{\ell}],\quad\tilde{\omega}_{9}(\ell)={\rm
det}[\vec{G}^{(9)}_{4},\vec{G}^{(9)}_{\ell}]$ (163)

where vector $\vec{G}^{(k)}_{\ell}$ made by the independent null constraints
at the level of $k$. The sign patterns are,

$\ell$ | 0 | 2 | 4 | 6 | 8 | …
---|---|---|---|---|---|---
$\tilde{\omega}_{7}(\ell)$ | 0 | - | 0 | + | + | …
$\tilde{\omega}_{9}(\ell)$ | 0 | - | 0 | + | + | …

The upper bounds are,

$\displaystyle\frac{\langle p_{7,\ell}\rangle}{\langle
p_{7,2}\rangle}\leq\frac{604800}{(\ell-4)\ell(\ell+1)(\ell+5)(\ell(\ell+1)(7\ell(\ell+1)(2\ell(\ell+1)-115)+11112)-33516)},~{}(\ell\geq
6)$ $\displaystyle\frac{\langle p_{9,\ell}\rangle}{\langle
p_{9,2}\rangle}\leq\frac{152409600}{(\ell-4)\ell(\ell+1)(\ell+5)(\ell(\ell+1)(\ell(\ell+1)(\ell(\ell+1)}$
$\displaystyle\times\frac{1}{(\ell(\ell+1)(2\ell(\ell+1)-331)+19014)-448776)+3862728)-8899632)},\quad(\ell\geq
6)$ (164)

Starting from $k>11$, one cannot find any functional that gives bounds on
$\langle p_{k,\ell}\rangle/\langle p_{k,2}\rangle$. But one can mange to get
functional that give upper bound on $\langle p_{k,\ell}\rangle/\langle
p_{k,\ell*\geq 4}\rangle$.

We can also test the bound using the following two $S$-matrices. The first has
the simple form

$M(s,t,u)=\frac{1}{(s-M^{2})(t-M^{2})(u-M^{2})}\,.$ (165)

The $\langle p_{k,\ell}\rangle$ in this case is

$\langle
p_{k,\ell}\rangle=-\frac{2(2\ell+1)}{M^{2k-2}}\int_{-1}^{1}\mathrm{d}y\frac{1}{y^{2}-9}P_{\ell}(y)$
(166)

In Figure 11, one can see that the average spectral function of this
$S$-matrix is decaying exponentially as $\ell\to\infty$, while the bounds
eq.(5.2) are just polynomially suppressed.

Figure 11: Here we compare the different $k$ level upper bounds of
$\frac{\langle p_{k,\ell}\rangle}{\langle p_{k,2}\rangle}$ in the
$su$-EFThedron with the $S$-matrix $\frac{1}{(s-M^{2})(t-M^{2})(u-M^{2})}$

The second one is the closed-string amplitude,

$M_{\text{string}}(s,t,u)=\frac{\Gamma({-}s)\Gamma({-}t)\Gamma({-}u)}{\Gamma(1{+}s)\Gamma(1{+}t)\Gamma(1{+}u)}\,,$
(167)

whose residues are just the square of the open string residues:

$\text{Res}_{n}(t)=\frac{\prod_{i=1}^{n-1}(t+i)^{2}}{(n!)^{2}}\,.$ (168)

The result is displayed in Figure 12, where we see that it lies between our
bound and the result for eq.(165).

Figure 12: Here we compare the $k=4,5$ upper bounds of $\frac{\langle
p_{k,\ell}\rangle}{\langle p_{k,2}\rangle}$ with the $S$-matrix
$\frac{1}{(s-M^{2})(t-M^{2})(u-M^{2})}$ and the closed string theory

###  5.3 Implications for mass spectrum

Previously, we have considered the null constraints each $k$ at a time. This
imposes bounds on the spin content of the spectrum. Here instead we consider
various linear combinations of null constraints from different $k$, which will
result in constraints on the mass content.

Let us begin with the $k=4,5$ null constraint and combine it into a vector
equation,

$\begin{pmatrix}n_{4}\\\
n_{5}\end{pmatrix}=\sum_{i}p_{i}\begin{pmatrix}\frac{J_{i}^{2}(J_{i}^{2}-8)}{m_{i}^{10}}\\\
\frac{J_{i}^{2}(150+J_{i}^{2}(2J_{i}^{2}-43))}{m_{i}^{12}}\end{pmatrix}=\sum_{i,x}\frac{p_{i}}{m_{i}^{12}}\begin{pmatrix}J_{i}^{2}(J_{i}^{2}-8)m_{i}^{2}\\\
J_{i}^{2}(150+J_{i}^{2}(2J_{i}^{2}-43))\end{pmatrix}=0$ (169)

Now we assume that at $\ell=2$, the mass are bounded from above, and the mass
of heaviest spin-$2$ particle is $M$. Consider the following functional,

$\omega_{1}(m,J^{2})={\rm
det}[\vec{G}_{M,6},\vec{G}_{m,J^{2}}]=-12J^{2}M^{2}\Big{[}150-43J^{2}+2J^{4}-18(J^{2}-8)\Big{(}\frac{m}{M}\Big{)}^{2}\Big{]}$
(170)

where

$\vec{G}_{m_{i},J^{2}_{i}}=\begin{pmatrix}J_{i}^{2}(J_{i}^{2}-8)m_{i}^{2}\\\
J_{i}^{2}(150+J_{i}^{2}(2J_{i}^{2}-43))\end{pmatrix}$ (171)

By construction the functional will have a zero exactly at the largest massive
spin 2 state. The plot of eq.(170) for each spin is at Figure 13. For $\ell>2$
the function has a negative slope and only for spin $2$ does the slope becomes
positive. Since the zero occurs at the largest mass amongst the spin 2 states,
all $\ell=2$ particles have positive contribution. Thus in order for the sum
rule to be true, _there must exist a higher spin particle with mass_

$m^{2}_{\ell}\leq\frac{2J^{4}-43J^{2}+150}{18J^{2}-144}M^{2}\,.$ (172)

Figure 13: Here we present the functional $\omega_{1}(m,J^{2})$ for different
spin. The position of the largest zero eq.(172) will increase as $J^{2}$
increase.

This bound becomes stronger by adding $k=6$ constraint. The functional in this
case is,

$\displaystyle\omega_{2}(m,J^{2})$ $\displaystyle=$ $\displaystyle{\rm
det}[\vec{G}_{M,6},\vec{G}_{\sqrt{5}\frac{M}{2},42},\vec{G}_{m,J^{2}}]$ (173)
$\displaystyle\sim$
$\displaystyle\Big{[}(12{-}J^{2})((J^{2}{-}32)J^{2}{+}204){+}8(J^{2}(2J^{2}{-}43){+}150)\Big{(}\frac{m}{M}\Big{)}^{2}\Big{]}$

where the $\vec{G}_{m,J^{2}}$ now is determined by combining three null
constraints,

$\vec{G}_{m,J^{2}}=\begin{pmatrix}m^{4}J^{2}(J^{2}-8)\\\
m^{2}J^{2}(150+J^{2}(2J^{2}-43))\\\
J^{2}(J^{2}-12)(J^{2}(J^{2}-32)+204)\end{pmatrix}$ (174)

Using the same method, the mass gap now can be improved as,

$m^{2}_{\ell}\leq\frac{(J^{2}-12)((J^{2}-32)J^{2}+204)}{8(J^{2}(2J^{2}-43)+150)}M^{2}\,.$
(175)

Note that the dominant contribution to $\langle p_{k,\ell}\rangle$ is the
lightest state of a given spin. Thus an upper bound on $\langle
p_{k,\ell}\rangle$ is a constraint on this lightest state, which cannot become
arbitrary light without having its coupling $p_{\ell_{i}}$ being heavily
suppressed. Here the bound is in the opposite direction, where it requires
some state in the spectrum must be sufficiently “light”. Of course, if the
masses of the spin 2 states are unbounded, the condition is less useful.
However, the simplicity of these results suggests the complete geometry in
$(p,\ell,m)$ space is highly non-trivial.

###  5.4 Constraints on EM and gravitational EFTs

In this section we briefly consider some of the above analyses when the
external scalars are replaced by massless particles with spin, leading to
constraints for UV completions of electromagnetic or gravitational EFTs. For
external helicity states, dispersion relations for the couplings of an
amplitude $M({+}h,{+}h,{-}h,{-}h)$ are given as Arkani-Hamed:2020blm ;
Bern:2021ppb :

$\displaystyle[12]^{2h}\langle
34\rangle^{2h}(\sum_{k,q}g_{k,q}s^{k{-}q}t^{q})={-}[12]^{2h}\langle
34\rangle^{2h}\left(\sum_{\ell_{a}\geq
0}p_{i}\frac{d_{0,0}^{\ell_{i}{=}even}(\theta)}{s{-}m^{2}_{i}}{+}\sum_{\ell_{j}\geq
2h}p_{j}\frac{\tilde{d}_{2h,2h}^{\ell_{j}}(\theta)}{{-}t{-}s{-}m^{2}_{j}}\right)$

where $d^{\ell}_{a,b}(\theta)$ is the Wigner-$d$ matrix, while
$\tilde{d}_{2h,2h}^{\ell}(\theta)\equiv
d_{2h,2h}^{\ell}(\theta)/(\cos\theta/2)^{4h}$, and the $s$-channel residue
only allows for even spin. Once again the equality is understood as a Taylor
expansion on both sides. Since the spinor bracket prefactor behaves as
$s^{2h}$, the dispersive representation is valid for any $k,q$. The Wigner
$d$-matrices can be expanded in small $t$ by $\cos(\theta)=1+2t$ and expand in
$t$. It turns out that $d_{0,0}^{\ell=even}(\theta)$ is just our usual
Legendre polynomial so its expansion written in $J^{2}=\ell(\ell+1)$ is as
before. For $\tilde{d}_{2,2}^{\ell}(\theta)$, we have

$\tilde{d}_{2h,2h}^{\ell}(\theta)\bigg{|}_{\theta=\cos^{-1}(1+2t)}=\sum_{q}v_{\ell,q}t^{q}\,,$
(177)

with

$v_{\ell,q}=\frac{\prod_{a=1}^{q}(J^{2}-(a+2h)(a+2h-1))}{(q!)^{2}}\,.$ (178)

Note that our helicity configuration is invariant under $1\leftrightarrow 2$
exchange. Since our dispersive representation for the couplings are valid for
any $k,q$, the constraint that this symmetry imposes on the spectrum can be
obtained by Taylor expanding,

$\displaystyle M(s,t,u){-}M(s,u,t)=\left(\sum_{\ell_{i}\geq
0}p_{i}\frac{d_{0,0}^{\ell_{i}{=}even}(\theta)}{s{-}m^{2}_{i}}{-}\sum_{\ell_{j}\geq
2h}p_{j}\frac{\tilde{d}_{2h,2h}^{\ell_{j}}(\theta)}{t{+}s{+}m^{2}_{j}}\right){-}(t\leftrightarrow
u)=0\,.$ (179)

The constraint is then collecting the coefficients of each $s^{a}t^{b}$, which
must be zero. The null constraint is at level $k$ is,

$\sum_{\ell\in\text{even}}\langle
p^{s}_{\ell,k}\rangle\omega_{k}^{s}(\ell)+\sum_{\ell\geq 2h}\langle
p^{u}_{\ell,k}\rangle\omega_{k}^{u}(\ell)=0$ (180)

where $\langle p^{s}_{\ell,k}\rangle$ and $\langle p^{u}_{\ell,k}\rangle$ are
the average spectral function in $s$ and $u$-channels. The remaining steps are
identical with that of the scalar EFThedron. In the following we simply do the
analysis for $k=1,2$ when there is only one null constraint.

$\displaystyle\sum_{l\in\text{even}}\langle
p_{l,1}^{s}\rangle\ell(\ell+1)+\sum_{l\geq 2h}\langle
p_{\ell,1}^{u}\rangle(-4h^{2}-2h+\ell^{2}+\ell-1)=0$
$\displaystyle\sum_{l\in\text{even}}\langle
p_{l,2}^{s}\rangle\ell(\ell+1)(\ell+\ell^{2}-6)+\sum_{l\geq 2h}\langle
p_{\ell,2}^{u}\rangle 4-(2h-\ell)(2h-\ell+1)(2h+\ell+1)(2h+\ell+2)=0$ (181)

Notice that $\tilde{\omega}^{s}(\ell)$ are non-negative for $\ell$ given by
even integers. Thus we just have to look at the sign pattern for
$\tilde{\omega}^{u}(\ell)$,

$\ell$ | 2 | 3 | 4 | 5 | 6 | 7 | …
---|---|---|---|---|---|---|---
$h=1,\tilde{\omega}^{u}_{1}(\ell)$ | - | + | + | + | + | + | …
$h=2,\tilde{\omega}^{u}_{1}(\ell)$ |  |  | - | + | + | + | …
$h=1,\tilde{\omega}^{u}_{2}(\ell)$ | - | - | + | + | + | + | …
$h=2,\tilde{\omega}^{u}_{2}(\ell)$ |  |  | - | - | + | + | …

Thus when $k=1$, we know that there must be spin 2 and spin 4 particles in the
$u$-channels in the EM and gravitational EFTs respectively. We can then derive
upper bounds for average spectral functions in both $s$ and $u$-channels,

$\displaystyle h=1:\quad\frac{\langle p_{\ell,1}^{s}\rangle}{\langle
p_{2,1}^{u}\rangle}\leq\frac{1}{\ell(\ell+1)}~{}(\ell\geq
2),\quad\frac{\langle p_{\ell,1}^{u}\rangle}{\langle
p_{2,1}^{u}\rangle}\leq\frac{1}{\ell^{2}+\ell-7}~{}(\ell\geq 3)$
$\displaystyle h=2:\quad\frac{\langle p_{\ell,1}^{s}\rangle}{\langle
p_{4,1}^{u}\rangle}\leq\frac{1}{\ell(\ell+1)}~{}(\ell\geq
2),\quad\frac{\langle p_{\ell,1}^{u}\rangle}{\langle
p_{4,1}^{u}\rangle}\leq\frac{1}{\ell^{2}+\ell-21}~{}(\ell\geq 5)$ (182)

When $k=2$, from sign pattern table we know that there must be spin 2 or 3
particles and EM EFTs and spin 4 or 5 particles gravitational EFTs. Thus in
this case we cannot get an upper bound for $\langle p_{2,\ell}^{s,u}\rangle$
with respect to only a single average spectral function. The results are,

$\displaystyle h=1:\quad\langle p_{l,2}^{s}\rangle\leq 4\frac{\langle
p_{2,2}^{u}\rangle+\langle
p_{3,2}^{u}\rangle}{\ell(\ell+1)(\ell^{2}+\ell-6)}~{}(\ell\geq 4),\quad\langle
p_{l,2}^{u}\rangle\leq 4\frac{\langle p_{2,2}^{u}\rangle+\langle
p_{3,2}^{u}\rangle}{68+\ell(\ell+1)(\ell^{2}+\ell-18)}~{}(\ell\geq 4)$
$\displaystyle h=2:\quad\langle p_{l,2}^{s}\rangle\leq 4\frac{\langle
p_{4,2}^{u}\rangle+\langle
p_{5,2}^{u}\rangle}{\ell(\ell+1)(\ell^{2}+\ell-6)}~{}(\ell\geq 4),\quad\langle
p_{l,2}^{u}\rangle\leq 4\frac{\langle p_{4,2}^{u}\rangle+\langle
p_{5,2}^{u}\rangle}{596+\ell(\ell+1)(\ell^{2}+\ell-50)}~{}(\ell\geq 6)$ (183)

##  6 Conclusion and outlook

In this paper, we have explored in detail the analytic description of the
boundaries for the space of consistent EFT. This space is defined through the
dispersive representation of the Wilson coefficients. We have demonstrated
that at the core of the EFThedron, first introduced in Arkani-Hamed:2020blm ,
is the geometry given by the convex hull of product moment curves. In
particular, we have the product of a continuous bounded curve, associated with
the mass spectrum of the UV completion, and a discrete unbounded moment
associated with the spin content of the spectrum. The identification of this
geometry allows us to state the boundary of the EFThedron in the limit of
infinite dimensions, which corresponds to coefficients of higher dimension
operators to all orders in derivatives.

In a generic setup, we only have access to a finite number of Wilson
coefficients. In such case the boundaries in the infinite dimension limit are
no longer complete: they become necessary but not sufficient conditions. For
the “truncated” geometry we present a collection of new boundaries to be
included. These are constraints that were not independent in the infinite
dimension limit, but become independent when reduced to the truncated problem.
The space of consistent EFTs are then carved out by considering the
intersection of the symmetry plane, defined either by cyclic or permutation
invariance of the amplitude, and the EFThedron. We have compared the space
carved out from the geometry with that from SDP, with the results closely
matching except for some boundary regions at the corner, indicating the finite
truncation effects.

In general the intersection yields a finite region. The reason can be
attributed to the fact that for any symmetry plane, there are only finite
number of spins on one side of the plane, while the majority is on the other
side. This implies non-trivial constraint on the UV spectrum. In particular we
demonstrate that for permutation invariant amplitudes, the UV completion must
contain all even spins $\ell\leq 28$. We also derived upper bounds on the
ratios of averaged spinning spectral function, which demonstrates the
phenomenon of large spin suppression. Finally, we have shown that at least one
high spin particle satisfy an upper bound $m^{2}_{\ell}\leq M^{2}B(\ell)$,
where $M^{2}$ is the mass of the heaviest spin 2. All of these results have
their generalization to the case when the external states carry spins, i.e.
photons and gravitons. Much like the EFThedron, the UV parameters
$\\{p,\ell,m\\}$ form a highly non-trivial geometry (though no longer convex),
induced by the null conditions. We leave a more systematic study of this
problem to the future.

In all of the previous discussion, we have only required that $p_{i}$ is
positive. However, the spectral function is further bounded by unitarity to
lie between $[0,2]$. Indeed by exploiting this upper limit, the authors of
Caron-Huot:2020cmc  were able to obtain upper bounds for $g_{2,0}$ in the case
of identical scalars. It is an open problem how to implement such constraint
in the geometry, as well as in the numeric SDP setup.

Generically, one has more than one massless degree of freedom in an EFT. It
will be interesting to explore the constraint of positivity for the scattering
of arbitrary mixed states, and in particular, their implications on the UV
spectrum. Some analysis along these lines have recently been pursued in
Arkani-Hamed:2020blm ; Zhang:2020jyn ; Bern:2021ppb ; Li:2021cjv .

Now that the geometry of the EFThedron is settled, and new boundaries have
been obtained by using the moment matrix, it would be interesting if the
analysis performed in Huang:2020nqy  can be improved by obtaining a stronger
convergence to the open string solution. As in this paper we have also
obtained the complete boundary structure for the $su$-EFThedron, similar
analyses may now be carried out for closed strings (see also Green:2019tpt ).

We have seen that making contact with the mathematical moment problem was a
crucial step in better understanding the space of physical EFTs. However, many
other possibly fascinating connections to related results remain unexplored.
For instance, the moment problem has yet another generalization, known as the
“$K$ moment problem” (see eg. article ). This problem now deals precisely with
the situation where moments are related to each other by polynomial relations.
The symmetry planes we have encountered, and treated as independent geometric
intersection, may in fact be understood as part of the moment problem itself.
Relatedly, in our paper, the operations of slicing and projecting the
EFThedron are central, but they spoil the simple boundary structure of Hankels
or cyclic polytopes. Describing the resulting shapes of slices and shadows of
polytopes has also been previously investigated 10.1145/237218.237228 . As we
have mentioned previously, several aspects of the truncated moment problem are
still open problems. For instance, a much more efficient method to determine
the boundary of the truncated discrete moment problem was recently proposed in
INFUSINO2017443 . While not directly applicable to our case, as the discrete
moments are valued in $\\{0,1,2,\ldots\\}$, it suggests even the truncated
problems may have simple solutions.

Finally, the same convex hull of product moment curve also underlies the torus
partition function of 2D CFTs ModularHedron , where the Taylor coefficients
are associated with the expansion of the partition function around the self-
dual fix point under modular invariance. The symmetry subplane is then derived
from modular invariance. This suggests that there is a formal “duality”
between the space of consistent EFT and that of 2D torus partition functions.
It will be interesting to explore this connection in more detail.

###### Acknowledgements.

We would like to thank Ting-Chun Lin for useful discussions on the multi-
variate moment problem, and Nima Arkani-Hamed for suggestions on the spin-
suppression analysis. L-y Chiang, Y-t Huang, and H-c Weng are supported by
Taiwan Ministry of Science and Technology Grant No. 109-2112-M-002 -020 -MY3.
L Rodina is supported by Taiwan Ministry of Science and Technology Grant No.
109-2811-M-002-523.

##  Appendix A Product moment conditions from single moments

In this section we describe how some conditions can be obtained by casting the
two moment problem as the single moment problem. For example, we can form a
moment curve in $\left(x+rxy\right)$, where $r$ can be any real number.
Collecting the appropriate couplings that lie on this moment curve, we get

$H_{2}(a)\left(x+rxy\right)\equiv\left(\begin{array}[]{cc}a_{0,0}&a_{1,0}+ra_{1,1}\\\
a_{1,0}+ra_{1,1}&a_{2,0}+2ra_{2,1}+r^{2}a_{2,2}\end{array}\right)\geq
0,\quad\forall r\in\mathbb{R}$ (184)

Requiring this condition to hold for all values of $r$ can be accomplished by
requiring the discriminant of the above matrix to be negative,

$a_{0,0}a_{1,1}^{2}a_{2,0}-2a_{0,0}a_{1,0}a_{1,1}a_{2,1}+a_{0,0}^{2}a_{2,1}^{2}+a_{0,0}a_{1,0}^{2}a_{2,2}-a_{0,0}^{2}a_{2,0}a_{2,2}\leq
0$ (185)

together with the coefficient of $r^{2}$ to be positive, which is just eq.(31)
again

$a_{0,0}a_{2,2}-a_{1,1}^{2}\geq 0\,.$ (186)

Of course there is nothing special about the choice $\left(x+rxy\right)$. One
could also pick $\left(y+rxy\right)$ for example, or any polynomial in $x,y$
in fact. Any such choice will lead to a corresponding Hankel constraint like
in eq.(184), which could be a function of several parameters. We see the space
of (positive) polynomials is closely related to the moment problem.

##### ”Matryoshka” Hankel constraints

Besides taking linear combinations of couplings to form single moment curves,
we also find that Hankel determinants themselves form new single moment
curves, leading to yet new constraints. Consider for instance the following
Hankel matrices and their determinants

$\displaystyle H_{2}^{(q)}(a)$ $\displaystyle{=}$
$\displaystyle\left(\begin{array}[]{cc}a_{0,q}&a_{1,q}\\\
a_{1,q}&a_{2,q}\end{array}\right)$ (189)
$\displaystyle\textrm{det}\left[H_{2}^{(q)}(a)\right]$ $\displaystyle=$
$\displaystyle\sum_{i,j}p_{i}p_{j}(x_{i}-x_{j})^{2}(y_{i}y_{j})^{q}$ (190)

for $q=0,1,2$. By defining $p_{ij}\equiv p_{i}p_{j}(x_{i}-x_{j})^{2}$,
$y_{ij}^{\prime}\equiv y_{i}y_{j}$, we observe the determinants form a new
moment curve $\sum_{ij}p_{ij}y_{ij}^{k}$, so they must also satisfy Hankel
constraints

$\begin{pmatrix}\textrm{det}\left[H_{2}^{(0)}\right]&\textrm{det}\left[H_{2}^{(1)}\right]\\\\[8.5359pt]
\textrm{det}\left[H_{2}^{(1)}\right]&\textrm{det}\left[H_{2}^{(2)}\right]\end{pmatrix}\geq
0$ (191)

which we call the the row-column “Matryoshka” Hankel constraint. Similarly one
can obtain the column-row constraint, by starting with a fixed $k$ Hankel
instead of fixed $q$ in eq.(189).

As a final example, when the $y$ moments are discrete, we simply have to
replace the Hankels with polytopal walls

$\begin{pmatrix}\langle\vec{a}_{0},i,i+1\rangle&\langle\vec{a}_{1},i,i+1\rangle\\\\[8.5359pt]
\langle\vec{a}_{1},i,i+1\rangle&\langle\vec{a}_{2},i,i+1\rangle\end{pmatrix}$
(192)

where

$\langle\vec{a}_{k},y_{i},y_{i+1}\rangle=\left|\begin{array}[]{ccc}a_{k,0}&1&1\\\
a_{k,1}&y_{i}&y_{i+1}\\\ a_{k,2}&y_{i}^{2}&y_{i+1}^{2}\\\ \end{array}\right|$
(193)

and also the other way around, that is plug in Hankel determinants into
polytopal walls

$\langle\vec{H}_{2},y_{i},y_{+1}\rangle=\left(\begin{array}[]{ccc}\textrm{det}H_{2}^{(0)}(a)&1&1\\\\[8.5359pt]
\textrm{det}H_{2}^{(1)}(a)&y_{i}&y_{i+1}\\\\[8.5359pt]
\textrm{det}H_{2}^{(2)}(a)&y_{i}^{2}&y_{i+1}^{2}\\\ \end{array}\right)\geq 0$
(194)

The generalized and “Matryoshka” Hankel constraints found above are just some
examples of constraints for the bi-variate moment problem. In general there is
an infinite number of such constructions, involving increasingly complex
combinations, but we will see they are all included in the general solution to
the bi-variate moment problem, to be discussed next.

##  Appendix B $s$-EFThedron partial ($g_{3,1}$,$g_{3,0}$)-space

We can obtain a slightly weaker bound by some simple manipulations. We will
only use mass gap, polytope wall $\langle\vec{a}_{5},0,3,4\rangle$ and
$\langle\vec{a}_{5},3,4,\infty\rangle$, and the simple inequalities
$a_{2,0}a_{4,2}-a_{3,1}^{2}\geq 0$ and $a_{3,1}a_{53}-a_{4,2}^{2}\geq 0$, with
null constraint $a_{53}=17a_{52}-30a_{5,1}$. The latter can be combined into

$a_{3,1}^{4}\leq
a_{2,0}^{2}a_{3,1}a_{53}=a_{2,0}^{2}a_{3,1}(17a_{52}-30a_{5,1})$ (195)

Next we use the polytope walls to replace $a_{52}$ and $a_{5,1}$ to obtain

$a_{3,1}^{3}\leq\frac{8320}{3}a_{2,0}^{2}a_{5,0}$ (196)

Finally using mass gap

$\left(\frac{a_{3,1}}{a_{2,0}}\right)^{3}\leq\frac{8320}{3}\frac{a_{3,0}}{a_{2,0}}$
(197)

##  Appendix C $su$-EFThedron complete ($g_{3,1}$, $g_{4,0}$)-space

$\displaystyle\left(-\frac{3}{2}<g_{3,1}\leq
0\land\frac{4g_{3,1}^{2}}{9}<g_{4,0}<1\right)\lor$
$\displaystyle\left(0<g_{3,1}\leq\frac{57}{10}\land\frac{4g_{3,1}^{2}}{153}<g_{4,0}<1\right)\lor$
$\displaystyle\left(\frac{57}{10}<g_{3,1}<\frac{3}{20}\left(5\sqrt{17}+19\right)\land\frac{2(95g_{3,1}+48)}{1275}-\frac{8\sqrt{2}\sqrt{-50g_{3,1}^{2}+285g_{3,1}+72}}{1275}<g_{4,0}<\right.$
$\displaystyle\quad\left.<\frac{8\sqrt{2}\sqrt{-50g_{3,1}^{2}+285g_{3,1}+72}}{1275}+\frac{2(95g_{3,1}+48)}{1275}\right)$
(198)

##  Appendix D Using Semi-definite programming to carve out the space of EFT
coefficients

In Caron-Huot:2020cmc , the authors shows that one can use semi-definite
programming to help carve out the space of EFT coefficients. Here we briefly
review their method for self-containess.

One parameter space (Bounds of $\frac{M_{\text{gap}}^{2}g_{k,q}}{g_{2,0}}$)

If a EFT coefficient $g_{k,q}$ has a dispersive representation, we know that
generally it has the form,

$g_{k,q}=\sum_{i}p_{m_{i},\ell_{i}}\frac{F_{k,q}(\ell)}{m_{i}^{2(k+1)}}$ (199)

where $f_{k,q}(\ell)$ is polynomial of $\ell$. Also because EFThedron are
constrained in a sub-plane, which will generate infinite number of null
constraints which has the form,

$n_{k}=\sum_{i}p_{m_{i},\ell_{i}}\frac{B_{k}(\ell)}{m_{i}^{2(k+1)}}=0$ (200)

where $k$ labels the level of null constraints. With these two ingredients in
hand we can transform it to semi-definite programming problem. Let’s define
vector $\vec{E}_{m,\ell}$ as,

$\vec{E}_{m,\ell}=\begin{pmatrix}\frac{1}{m_{i}^{6}}\\\
\frac{M_{\text{gap}}^{2}F_{k,q}(\ell)}{m_{i}^{2(k+1)}}\\\
\frac{B_{k_{1}}(\ell)}{m_{i}^{2(k_{1}+1)}}\\\
\frac{B_{k_{2}}(\ell)}{m_{i}^{2(k_{2}+1)}}\\\ \vdots\end{pmatrix}$ (201)

Thus we can write down a vector equation,

$\sum_{m_{i},\ell_{i}}p_{m_{i},\ell_{i}}\vec{E}_{m_{i},\ell_{i}}=\begin{pmatrix}g_{2,0}\\\
M_{\text{gap}}^{2}g_{k,q}\\\ 0\\\ 0\\\ \vdots\end{pmatrix}$ (202)

Then we redefine the $\tilde{p}_{m_{i},\ell_{i}}$ as,

$\tilde{p}_{m_{i},\ell_{i}}=\frac{p_{m_{i},\ell_{i}}}{m_{i}^{2(\tilde{k}+1)}},\quad\text{where}~{}\tilde{k}=\text{max}(k,k_{1},k_{2},\dots)$
(203)

And make substitution $m_{i}^{2}\to M^{2}(1+x_{i})$. The vector equation now
becomes,

$\sum_{x_{i}>0,\ell_{i}}\tilde{p}_{x_{i},\ell_{i}}\vec{E}_{x_{i},\ell_{i}}=\begin{pmatrix}g_{2,0}\\\
M_{\text{gap}}^{2}g_{k,q}\\\ 0\\\ 0\\\ \vdots\end{pmatrix}$ (204)

Notice that now each entry in the vector $\vec{E}_{x_{i},\ell_{i}}$ is just
polynomial in $x_{i}$ and valid range is $x_{i}>0$. The corresponding semi-
definite programming problems are,

  * •

Upper bound ($\frac{M_{\text{gap}^{2}}g_{k,q}}{g_{2,0}}\leq A$)

$\displaystyle\text{Minimize}\quad A=\vec{\alpha}\cdot(1,0,0,\cdots)$
$\displaystyle\text{Such that}\quad\text{For}~{}x\geq
0,~{}\vec{\alpha}\cdot\vec{E}_{x,\ell}\geq
0,\quad\vec{\alpha}\cdot(0,-1,0,0,\dots)=1$ (205)

  * •

Lower bound ($B\leq\frac{M_{\text{gap}^{2}}g_{k,q}}{g_{2,0}}$)

$\displaystyle\text{Maximize}\quad B=\vec{\alpha}\cdot(-1,0,0,\cdots)$
$\displaystyle\text{Such that}\quad\text{For}~{}x\geq
0,~{}\vec{\alpha}\cdot\vec{E}_{x,\ell}\geq
0,\quad\vec{\alpha}\cdot(0,1,0,0,\dots)=1$ (206)

Two parameter space
($\frac{M_{\text{gap}}^{2}g_{k_{1},q_{1}}}{g_{2,0}}$,$\frac{M_{\text{gap}}^{2}g_{k_{2},q_{2}}}{g_{2,0}}$)

The strategy is that one can first use the method previous section to get the
upper and lower bound of $\frac{M_{\text{gap}}^{2}g_{k_{1},q_{1}}}{g_{2,0}}$

$A_{1}\leq\frac{M_{\text{gap}}^{2}g_{k_{1},q_{1}}}{g_{2,0}}\leq A_{2}$ (207)

Then at each value $\alpha\in[A_{1},A_{2}]$ we use semi-definite programming
again to get the upper and lower bound of
$\frac{M_{\text{gap}}^{2}g_{k_{2},q_{2}}}{g_{2,0}}$,

$\text{For
each}~{}\beta,~{}B_{1}(\beta)\leq\frac{M_{\text{gap}}^{2}g_{k_{2},q_{2}}}{g_{2,0}}\leq
B_{2}(\beta)$ (208)

Then we can carve out the 2D space of
($\frac{M_{\text{gap}}^{2}g_{k_{1},q_{1}}}{g_{2,0}}$,$\frac{M_{\text{gap}}^{2}g_{k_{2},q_{2}}}{g_{2,0}}$).
Below we show the optimization problem at some $\beta$. The vector
$\vec{E}_{x_{i},\ell_{i}}$ in this problem is,

$\sum_{x_{i}\geq
0,\ell_{i}}\tilde{p}_{x_{i},\ell_{i}}\vec{E}_{x_{i},\ell}=\begin{pmatrix}g_{2,0}\\\
M_{\text{gap}}^{2}g_{k_{1},q_{1}}\\\ M_{\text{gap}}^{2}g_{k_{2},q_{2}}\\\ 0\\\
0\\\ \vdots\end{pmatrix}$ (209)

  * •

Upper bound

$\displaystyle\text{Minimize}\quad
B_{2}(\beta)=\vec{\alpha}\cdot(1,\beta,0,0,\cdots)$ $\displaystyle\text{Such
that}\quad\text{For}~{}x\geq 0,~{}\vec{\alpha}\cdot\vec{E}_{x,\ell}\geq
0,\quad\vec{\alpha}\cdot(0,0,-1,0,\dots)=1$ (210)

  * •

Lower bound

$\displaystyle\text{Maximize}\quad
B_{1}(\beta)=\vec{\alpha}\cdot(-1,\beta,0,0,\cdots)$ $\displaystyle\text{Such
that}\quad\text{For}~{}x\geq 0,~{}\vec{\alpha}\cdot\vec{E}_{x,\ell}\geq
0,\quad\vec{\alpha}\cdot(0,1,0,0,\dots)=1$ (211)

This algorithm can be easily generalized to carving out higher dimensional
parameter space. And for SDP solver we are using SDPB Simmons-Duffin:2015qma ;
Landry:2019qug  to perform our computation.

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