aainstitutetext: Department of Physics, ETH Zürich, 8093 Zürich,
Switzerlandbbinstitutetext: Department of Physics, University of California,
Santa Barbara, CA 93106, USA

# The Symmetrized Holographic Entropy Cone

Matteo Fadel<EMAIL_ADDRESS>b , Sergio Hernández-Cuenca
<EMAIL_ADDRESS>
###### Abstract

The holographic entropy cone (HEC) characterizes the entanglement structure of
quantum states which admit geometric bulk duals in holography. Due to its
intrinsic complexity, to date it has only been possible to completely
characterize the HEC for at most $n=5$ numbers of parties. For larger $n$, our
knowledge of the HEC falls short of incomplete: almost nothing is known about
its extremal elements. Here, we introduce a symmetrization procedure that
projects the HEC onto a natural lower dimensional subspace. Upon
symmetrization, we are able to deduce properties that its extremal structure
exhibits for general $n$. Further, by applying this symmetrization to the
quantum entropy cone, we are able to quantify the typicality of holographic
entropies, which we find to be exponentially rare quantum entropies in the
number of parties.

##  1 Introduction

From the microscopic degrees of freedom of gravity to the macroscopic
emergence of spacetime itself, the holographic principle has provided deep
insight into many aspects of quantum gravity tHooft:1993dmi ; Susskind:1994vu
. Realizations thereof like AdS/CFT Maldacena_1999 ; Witten1998  may be used
to define a bulk theory of quantum gravity non-perturbatively in terms of a
lower-dimensional boundary field theory – a hologram. Elucidating quantum
gravity then amounts to decoding the holographic dictionary between bulk and
boundary physics. A remarkably powerful entry in this dictionary concerns the
correspondence between boundary entanglement and bulk geometry, and is often
captured by the slogan that spacetime emerges from entanglement
VanRaamsdonk:2010pw .

At the heart of this idea lies the Ryu-Takayanagi (RT) formula Ryu:2006bv
(see Hubeny:2007xt ; Faulkner:2013ana ; Engelhardt:2014gca  for
generalizations thereof), which geometrizes the von Neumann entropy of
boundary regions into areas of certain surfaces in the bulk. By RT, the
emergence of a classical spacetime from entanglement relies on strong
constraints on the entanglement structure of boundary states: only so-called
holographic states with certain patterns of quantum entanglement admit such
geometrizations. When formulated as inequalities for the von Neumann entropy,
these constraints define a polytope in the space of entropies of all
subsystems one can form involving $n$ parties. In Bao:2015bfa , this object
was proven to be a rational polyhedral cone for any given $n$ and coined the
holographic entropy cone (HEC). The work of Bao:2015bfa  also pioneered a
graph-theoretic reformulation of RT entropies, later formalized by
Avis:2021xnz , that made the HEC amenable to systematic study. In particular,
Bao:2015bfa  proved that the HEC could be equivalently defined as the space of
entropies that can be realized by the weights of minimum cuts on weighted
graphs. Despite this breakthrough, deciphering the general structure of the
HECn for general $n$ remains a formidable challenge: its complete description
is only known for up to $n=5$ parties HernandezCuenca:2019wgh , and despite
the systematic computational algorithms of Avis:2021xnz  for arbitrary $n$,
results for $n\geq 6$ are only partial and highly incomplete Bao:2015bfa ;
n6wip .

The remarkably complicated structure the HECn exhibits as $n$ increases
suggests that an explicit description of all its rich details, if attainable
at all, would probably not be very illuminating. Hence, rather than trying to
explicitly describe the HEC in full glory, some attempts at studying it have
focused on trying to understand its basic structural properties Hubeny:2018trv
; Hubeny:2018ijt ; Hernandez-Cuenca:2019jpv ; He:2019ttu ; He:2020xuo . In
pursuing the latter strategy, the present work identifies an interesting
projection of the HEC which is able to wash out its uninteresting fine details
while preserving non-trivial features of the extremal structure of the HECn at
arbitrary $n$. Since this projection is inspired by the underlying symmetries
of the HEC, we will be referring to it as a symmetrization. Similar approaches
have been extremely successful in the context of Bell nonlocality, where a
projection of the local polytope onto the subspace invariant under permutation
of the particles allowed to find Bell inequalities valid for arbitrary $n$
TuraSci ; FadelPRL . Importantly, our symmetrization of the HEC seems to take
a simple enough form that we are able to propose a general solution to it, and
thereby conjecture non-trivial properties that all extreme rays and facets of
the HECn must satisfy for all $n$.

We begin in section 2 with a brief review of basic quantum and holographic
inequalities that will be appear throughout, and a self-contained introduction
to the graph models of holographic entanglement of Bao:2015bfa . Having laid
out our notation, in Section 3 we then carefully explain the symmetrization
operations that are central to our explorations. Section 4 presents our
explicit results and conjectures for the general form of the HEC upon
symmetrizations. By looking at analogous general-$n$ results for arbitrary
quantum states in Section 5, we are then able to quantify the typicality of
holographic entropies in Section 6: with respect to the uniform volume measure
of entropy space, the latter are exponentially rare in the number of parties.
We conclude in Section 7 with a summary of results and a discussion of open
questions.

##  2 Basics

###  2.1 Inequalities

Let us collect here some basic inequalities which will be useful for later
reference. The simplest universal quantum inequality involves $2$ parties and
is known as subadditivity (SA). Proven by lanford1968mean , for arbitrary
disjoint subsystems $I$ and $J$, SA reads

$S_{I}+S_{J}\geq S_{I\cup J}\;.$ (1)

This inequality holds trivially in holography. At $3$ parties, there appears
another well-known universal quantum inequality: strong subadditivity (SSA).
Proven by lieb1973proof , SSA for arbitrary overlapping subsystems $I$ and $J$
reads111To clarify, this is a $3$-party inequality because it involves $3$
disjoint subsystems: $I\smallsetminus J$, $J\smallsetminus I$, and $I\cap J$.

$S_{I}+S_{J}\geq S_{I\cap J}+S_{I\cup J}\;.$ (2)

The proof of this inequality in holography is non-trivial and was established
by Headrick:2007km ; Wall:2012uf . The first genuinely holographic inequality
(i.e., valid in holography, but not in quantum theory) appears for $3$ parties
and is known as the monogamy of mutual information (MMI). Proven by
Hayden:2011ag ; Wall:2012uf , MMI for arbitrary disjoint subsystems $I$, $J$,
and $K$ reads

$S_{I\cup J}+S_{I\cup K}+S_{J\cup K}\geq S_{I}+S_{J}+S_{K}+S_{I\cup J\cup
K}\;.$ (3)

###  2.2 Graphs

For any positive integer $k$, introduce the notation
$[k]\equiv\\{1,\dots,k\\}$. Let $G=(V,E)$ be an undirected graph with finite
vertex set $V$ and edge set $E\subset V\times V$ consisting of unordered pairs
$(i,j)\in E$ with $i\neq j$. We make $G$ into a weighted graph with
nonnegative capacities by endowing it with a weight vector
$w\in\mathbb{R}_{\geq 0}^{\absolutevalue{E}}$ encoding a map
$E\to\mathbb{R}_{\geq 0}$. Select some vertex subset $\partial V\subseteq V$
and call them boundary vertices. One then defines a coloring as a surjective
map $b:\partial V\to[n+1]$. The elements $i\in[n+1]$ are called parties, and
the non-empty subsets $I\subseteq[n+1]$ are called subsystems. Altogether,
this structure defines a graph model of holographic entanglement.

Any subset $W\subseteq V$ characterizes a cut of $G$, defined by a set of cut
edges $C(W)\subseteq E$ as

$C(W)=\left\\{(v,v^{\prime})\in E\,:\,v\in W,\,v^{\prime}\in
W^{\complement}\right\\}.$ (4)

The cut weight is defined as the total weight of its edges
$\norm{C(W)}=\sum_{e\in C(W)}w_{e}$. A set $W\subseteq V$ is a cut for a
subsystem $I$ if it contains precisely the boundary vertices colored by $I$,
i.e., if $W\cap\partial V=b^{-1}(I)$. Any cut $W$ for $I$ with minimum cut
weight $\norm{C(W)}$ defines a min-cut for $I$. The min-cut weight for a
subsystem $I$ defines its entropy $S_{I}$ via

$S_{I}=\min\,\left\\{\norm{C(W)}\,:\,W\cap\partial V=b^{-1}(I)\right\\}.$ (5)

In the context of holography, the above can indeed be understood as computing
the von Neumann entropy of a subsystem $I$ of some pure holographic state on
$[n+1]$ Bao:2015bfa . Such a pure state can be used to encode an arbitrary
$n$-party, mixed holographic state on $[n]$. Due to its quantum mechanical
role, the special party $n+1$ is thus referred to as the purifier. Notice in
particular that (5) indeed reproduces the desired purification property that
$S_{I}=S_{I^{\complement}}$ for complementary subsystems $I$ and
$I^{\complement}=[n+1]\smallsetminus I$. For this reason, the entropy vector
of the full graph model is defined excluding the purifier by

$S=\\{S_{I}\,:\,\varnothing\neq I\subseteq[n]\\},$ (6)

where the conventional choice of ordering is by cardinality first, and then
lexicographically. The set of all entropy vectors $S\in\mathbb{R}^{2^{n}-1}$
obtained this way defines the holographic entropy cone (HEC) for $n$ parties,
or HECn.

More generally, the definition of an entropy vector in (6) applies to any
$n$-partite quantum system, with $S_{I}$ denoting the von Neumann entropy of a
subsystem $I\subseteq[n]$. If instead of holographic states one considers
completely arbitrary quantum states, the resulting set of all $n$-party
entropy vectors defines the quantum entropy cone (QEC) for $n$ parties, or
QECn pippenger2003inequalities . Both the HEC and the QEC are convex
cones,222To be precise, it is actually only the topological closure of the
latter which is a convex cone pippenger2003inequalities . This technicality
will not be important for us. and the former is additionally known to be
polyhedral Bao:2015bfa . Consistent with the fact that holographic states
define a special class of quantum states, the HEC is a subcone of the QEC
Nezami:2016zni . In the following, we will be interested in studying the HEC
and how it relates to the QEC.

##  3 Symmetrizations

Both entropy cones of interest in this paper are clearly symmetric under
permutations of the colors $[n]$. To see a larger symmetry, recall that
subsystems $I\subseteq[n+1]$ obey $S_{I}=S_{I^{\complement}}$. Hence these
entropy cones are symmetric not only under permutations of $[n]$, but also
under the extended symmetric group $Sym_{n+1}$ of permutations of $[n+1]$
involving the purifier $n+1$.

We would like to simplify the structure of our entropy cones by defining a
symmetrization, i.e., an operation $P$ on their elements that is invariant
under the action of $Sym_{n+1}$. More explicitly, if $x$ and $y$ are two
elements of an entropy cone in the same symmetry orbit, then they should have
the same symmetrized form $P(x)=P(y)$. By asking that this symmetrization be a
linear map, we will be guaranteed that the symmetrization of a full convex
cone remain so. The symmetrized versions of the HEC and QEC will respectively
be referred to as SHEC and SQEC.

For an entropy vector $S\in\mathbb{R}^{2^{n}-1}$, our symmetrization will be
some linear map $P:S\mapsto\tilde{S}$. To build it, first note that by the
$Sym_{n}$ symmetry of permutations of $[n]$, $P$ can only depend on the
cardinality $\absolutevalue{I}$ of the coordinate $I\subseteq[n]$.
Additionally, by the full action of $Sym_{n+1}$ on $[n+1]$ with the
identification $S_{I^{\complement}}=S_{I}$, one has that coordinates of
cardinalities $\absolutevalue{I}$ and
$|I^{\complement}|=n+1-\absolutevalue{I}$ are also related. Altogether, this
means that $P$ should just depend on at most $\lceil n/2\rceil$ different
variables, associated to the distinct possible cardinalities of subsets of
$[n+1]$ with complements identified. The codomain of $P$ is thus
$\mathbb{R}^{\lceil n/2\rceil}$ and one can label the symmetric variables
$\tilde{S}_{k}$ of the symmetrized vector $\tilde{S}\in\mathbb{R}^{\lceil
n/2\rceil}$ by cardinalities $1\leq k\leq\lceil n/2\rceil$.

A natural way of defining $P$ thus employs subsets of $[n+1]$ of fixed
cardinality $k\in\left[\lceil n/2\rceil\right]$,

$Q_{n}(k)=\\{I\subseteq[n+1]\,:\,\absolutevalue{I}=k\\}.$ (7)

Then the symmetric variables can be obtained by just summing over coordinates
in $Q_{n}(k)$,

$\tilde{S}_{k}=\frac{1}{\absolutevalue{Q_{n}(k)}}\sum_{I\in
Q_{n}(k)}S_{I},\qquad\absolutevalue{Q_{n}(k)}=\binom{n+1}{k},$ (8)

where we have introduced a natural normalization factor accounting for the
number of terms in the sum. For instance, for $n=3$, the symmetric variables
in (8) are

$\displaystyle\tilde{S}_{1}$ $\displaystyle=\frac{1}{4}(S_{\text{\tiny
1}}+S_{\text{\tiny 2}}+S_{\text{\tiny 3}}+S_{\text{\tiny
4}})=\frac{1}{4}(S_{\text{\tiny 1}}+S_{\text{\tiny 2}}+S_{\text{\tiny
3}}+S_{\text{\tiny 123}}),$ (9) $\displaystyle\tilde{S}_{2}$
$\displaystyle=\frac{1}{6}(S_{\text{\tiny 12}}+S_{\text{\tiny
13}}+S_{\text{\tiny 14}}+S_{\text{\tiny 23}}+S_{\text{\tiny
24}}+S_{\text{\tiny 34}})=\frac{1}{3}(S_{\text{\tiny 12}}+S_{\text{\tiny
13}}+S_{\text{\tiny 23}}),$

where in the last equalities we used $S_{I^{\complement}}=S_{I}$. Crucially,
notice that these sets $Q_{n}(k)$ automatically take care of coordinates
$I\subseteq[n]$ of cardinality $\absolutevalue{I}=n+1-k$, since
$I^{\complement}\in Q_{n}(k)$. Hence the operation
$P:\mathbb{R}^{2^{n}-1}\to\mathbb{R}^{\lceil n/2\rceil}$ defined this way is a
valid linear symmetrization invariant under $Sym_{n+1}$. For any number of
parties $n$, the SHEC and SQEC will thus be convex cones defined by the sets
of all symmetrized entropy vectors $\tilde{S}\in\mathbb{R}^{\lceil n/2\rceil}$
obtained from symmetrizations of their parent entropy vectors
$S\in\mathbb{R}^{2^{n}-1}$ of the HEC and QEC, respectively.

In the case of the HEC, the symmetrization of entropy vectors that we just
described can also be easily arrived at from a symmetrization of graph models
themselves. Indeed, since weighted graphs are ultimately the object which
defines the HEC, one may consider performing a symmetrization already at this
level. Given a graph model $G$ for $n$ parties, consider all possible $(n+1)!$
graphs $G^{\sigma}$ (some of which may be equal) obtained by permutations
$\sigma\in Sym_{n+1}$ of the colors $[n+1]$. If $S$ and $S^{\sigma}$ are
respectively the entropy vectors from $G$ and $G^{\sigma}$, one obviously has
$S_{I}=S^{\sigma}_{\sigma(I)}$, where $\sigma(I)=\\{\sigma(i)\,:\,i\in I\\}$,
or, equivalently, $S^{\sigma}_{I}=S_{\sigma^{-1}(I)}$. Suppose one now
combines two permuted graphs in a disjoint manner or glued together by
identifying all vertices of the same color in both. Either way, it is
straightforward to see that the resulting graph $G^{\sigma}\oplus
G^{\sigma^{\prime}}$ yields $S^{\sigma}+S^{\sigma^{\prime}}$ as its entropy
vector. With a suggestive choice of notation, let’s then define a permutation-
averaged graph $\tilde{G}$ by

$\tilde{G}=\frac{1}{(n+1)!}\bigoplus_{\sigma\in Sym_{n+1}}G^{\sigma},$ (10)

where the multiplication shall be understood as rescaling the weights of the
graphs. The computation of the entropy of $I$ in this graph $\tilde{G}$ gives

$\frac{1}{(n+1)!}\sum_{\sigma\in
Sym_{n+1}}S^{\sigma}_{I}=\frac{k!(n+1-k)!}{(n+1)!}\sum_{J\in
Q_{n}(k)}S_{J}=\tilde{S}_{k},\qquad
k=\min\\{\absolutevalue{I},n+1-\absolutevalue{I}\\},$ (11)

which recovers precisely the symmetric variables in (8). In other words, the
entropies of $\tilde{G}$ only depend on the cardinality of the subsystem $I$
and are given by the entries of the symmetric entropy vector $\tilde{S}$ of
the original graph $G$. In this sense, we see that given a graph $G$, the
operations of symmetrization and computation of entropies commute.

Having dealt with entropy vectors, we would now like to have an analogous
operation that we can directly apply to the vectors $q\in\mathbb{R}^{2^{n}-1}$
which define entropy inequalities $qS\geq 0$. To do so, consider constructing
valid inequalities that we can write down using only the symmetric variables
in (8). This can be accomplished by, given an inequality $qS\geq 0$, adding up
all inequalities in its symmetry orbit so as to form a new inequality
$q^{\prime}S\geq 0$ which will be valid by convexity. The coefficients of the
new vector $q^{\prime}$ can be easily computed to be

$q^{\prime}_{I}=k!(n+1-k)!\sum_{J\in Q_{n}(k)}q_{J},\qquad
k=\min\\{\absolutevalue{I},n+1-\absolutevalue{I}\\},$ (12)

where the combinatorial factor accounts for the permutations in $Sym_{n+1}$
which fix each $J$. As expected, $q^{\prime}_{I}$ has the right dependence on
cardinality only and we can now write $q^{\prime}S\geq 0$ using just
$\tilde{S}_{k}$ variables. Declaring
$q^{\prime}_{I^{\complement}}=q^{\prime}_{I}$ for convenience, one finds

$q^{\prime}S=\sum_{\varnothing\neq
I\subseteq[n]}q^{\prime}_{I}S_{I}=\sum_{k=1}^{\lceil n/2\rceil}\sum_{I\in
Q_{k}(n)}q^{\prime}_{I}S_{I}=(n+1)!\sum_{k=1}^{\lceil
n/2\rceil}\tilde{q}_{k}\tilde{S}_{k},$ (13)

where in the last equality we have identified the desired symmetric
coefficients of our symmetrized inequality $\tilde{q}\tilde{S}\geq 0$ to be

$\tilde{q}_{k}=\sum_{J\in Q_{n}(k)}q_{J}.$ (14)

Formally, one can understand this operation as the right inverse of the one in
(8). More explicitly, let $M$ and $N$ respectively be the rectangular matrices
implementing (8) and (14) as linear maps, such that $\tilde{S}=MS$ and
$\tilde{q}=Nq$. Then one easily verifies that $N^{T}=M^{+}$, where
$M^{+}=M^{T}(MM^{T})^{-1}$ is the canonical Moore–Penrose right inverse of
$M$. For instance, for $n=3$ these matrices are

$M=\begin{bmatrix}\frac{1}{4}&\frac{1}{4}&\frac{1}{4}&0&0&0&\frac{1}{4}\\\
0&0&0&\frac{1}{3}&\frac{1}{3}&\frac{1}{3}&0\end{bmatrix},\qquad
N=\begin{bmatrix}1&1&1&0&0&0&1\\\ 0&0&0&1&1&1&0\end{bmatrix},$ (15)

and one clearly has $MN^{T}=\mathds{1}_{2}$. Note in passing that these
matrices can also be used to rephrase our symmetrization operations as
projections of vectors $S\in\mathbb{R}^{2^{n}-1}$ onto a symmetric subspace
within $\mathbb{R}^{2^{n}-1}$. In particular, such a projection is implemented
by the matrix $N^{T}M$, which one can easily verify is idempotent as a
projection should be. The result of applying $N^{T}M$ to any vector
$S\in\mathbb{R}^{2^{n}-1}$ is to project it onto a subspace where, for all
$I\subseteq[n]$, the coordinates attain values
$S_{I}=\tilde{S}_{\absolutevalue{I}}$ if $\absolutevalue{I}\leq\lceil
n/2\rceil$ and $S_{I}=\tilde{S}_{n+1-\absolutevalue{I}}$ otherwise. For the
$n=3$ example in (15), the projection matrix just mentioned reads

$N^{T}M=\begin{bmatrix}\frac{1}{4}&\frac{1}{4}&\frac{1}{4}&0&0&0&\frac{1}{4}\\\
\frac{1}{4}&\frac{1}{4}&\frac{1}{4}&0&0&0&\frac{1}{4}\\\
\frac{1}{4}&\frac{1}{4}&\frac{1}{4}&0&0&0&\frac{1}{4}\\\
0&0&0&\frac{1}{3}&\frac{1}{3}&\frac{1}{3}&0\\\
0&0&0&\frac{1}{3}&\frac{1}{3}&\frac{1}{3}&0\\\
0&0&0&\frac{1}{3}&\frac{1}{3}&\frac{1}{3}&0\\\
\frac{1}{4}&\frac{1}{4}&\frac{1}{4}&0&0&0&\frac{1}{4}\end{bmatrix}.$ (16)

##  4 The Symmetrized HEC

Figure 1: General star graphs realizing the extreme rays of the SHEC. Black
edges have unit weight, while the blue “purification” edge has positive
integer weight. This is $w=1,3,\dots,n$ for $n$ odd, and $w=2,4,\dots,n$ for
$n$ even. Dashed red lines represent the two possible min-cuts for subsystem
$AB$. An analogous binary choice of including or excluding the central vertex
exists for every subsystem.

Full descriptions of the HEC in terms of extreme rays and facets are known for
$n\leq 5$ Bao:2015bfa ; HernandezCuenca:2019wgh ; Avis:2021xnz . From this
data, it is straightforward to compute the SHEC for $n\leq 5$ by applying (8)
to extreme rays and (14) to facets (see Appendix A). One can then easily check
that the resulting descriptions of the SHEC are indeed dual to each other.

At this point, one already notices that the extreme rays of the SHEC for all
$n\leq 5$ can be generated by a very simple family of graphs.333Intuitively,
by “generated” here we mean that these graphs yield entropy vectors whose
symmetrizations through (8) match the extreme rays of the SHEC. In particular,
they can all be captured by star graphs with $n+1$ edges, one for each
boundary vertex, where the weight of all but one of them can be set to unity.
The remaining edge, which by symmetry and without loss of generality can be
picked to be the one associated to the purifier $n+1$, then carries some other
integer weight $w\geq 1$, as in Fig. 1. We observe that this family of star
graphs turns out to yield precisely all extreme rays at each $n$ when the
weight $w$ is chosen to be of the same parity as $n$ and obey $1\leq w\leq n$.
These observations lead to the following conjecture:

######  Conjecture 1 (SHEC extreme rays).

The extreme rays of the SHECn can be generated by star graphs with $n$
weight-$1$ edges and another edge of positive weight $w=n,n-2,\dots$.

As a sanity check, notice that this proposal is consistent with the
dimensionality of the SHEC, since the number of conjectured extreme rays
$\lceil n/2\rceil$ precisely matches it. In turn, this suggests that the SHEC
may be a simplicial cone for all $n$. To be more explicit, and building upon
Conjecture 1, we now work out a closed-form expression for all conjectured
extreme rays of the SHECn. Firstly, from the star graph in Fig. 1, every
entropy $S_{I}$ is by definition given by

$S_{I}=\min\\{\absolutevalue{I},n-\absolutevalue{I}+w\\}\;.$ (17)

Since $S_{I}$ only depends on the subsystem cardinality $k=\absolutevalue{I}$,
using momentarily the notation $S_{I}=S_{k}$, it is then straightforward to
apply (8) and obtain the coordinates of the symmetrized entropy vectors as

$\displaystyle\tilde{S}_{k}$
$\displaystyle=\binom{n+1}{k}^{-1}\left(\binom{n}{k}S_{k}+\binom{n}{n+1-k}S_{n+1-k}\right)$
(18)
$\displaystyle=\frac{1}{n+1}\left((n+1-k)\min\\{k,n-k+w\\}+k\min\\{n+1-k,w+k-1\\}\right)\;.$

The particular set of star graphs we are interested in for each $n$ can be
conveniently parameterized by setting $w=n-2(l-1)$ for $l=1,2,\dots,\lceil
n/2\rceil$. This way, $l$ labels each of the $\lceil l/2\rceil$ conjectured
extreme rays, whose entropies in (18) can be further simplified to

$\tilde{S}_{k}^{(l)}=\frac{2k}{n+1}(n+1-\max\\{k,l\\}).$ (19)

We can now use this result in order to obtain explicit results for the
complete set of extreme rays of the SHEC for any number of parties $n$
according to Conjecture 1.

The $\lceil n/2\rceil$ entropy vectors of dimension $\lceil n/2\rceil$ we
obtain for each $n$ this way can be arranged as columns of a square matrix.
For instance, for $n=10$ (and with entropy vectors conveniently renormalized
by a factor of $(n+1)/2$), this matrix can be written as

$M_{10}^{\text{SHEC}}=\begin{bmatrix}10&9&8&7&6\\\ 18&18&16&14&12\\\
24&24&24&21&18\\\ 28&28&28&28&24\\\ 30&30&30&30&30\end{bmatrix}\;.$ (20)

One can easily check that these matrices are non-singular for every $n$,
thereby demonstrating that the conjectured SHEC indeed is a simplicial cone of
full dimension in $\mathbb{R}^{\lceil n/2\rceil}$. As such, its dual facet
description can be immediately obtained from the rows of the inverse of the
matrices of extreme rays just described. For the $n=10$ example above, this
leads to

$(M_{10}^{\text{SHEC}})^{-1}=\begin{bmatrix}1&-\frac{1}{2}&0&0&0\\\
-1&~{}~{}1&-\frac{1}{3}&0&0\\\
0&-\frac{1}{2}&~{}~{}\frac{2}{3}&-\frac{1}{4}&0\\\
0&0&-\frac{1}{3}&~{}~{}\frac{1}{2}&-\frac{1}{5}\\\
0&0&0&-\frac{1}{4}&~{}~{}\frac{7}{30}\end{bmatrix},$ (21)

where from e.g. the first row one reads off the inequality

$2\tilde{S}_{1}-\tilde{S}_{2}\geq 0\;.$ (22)

This expression can be easily checked to come from a symmetrization of SA in
(1) for singletons, and turns out to define a facet of the conjectured SHEC
for all $n$. Another inequality which shows up for every $n$ and involves
symmetrized subsystems of just two different cardinalities is the following
(cf. the last row in (21)):

$-\left(1-\frac{1}{\left\lfloor\frac{n+1}{2}\right\rfloor}\right)^{-1}\tilde{S}_{\lceil
n/2\rceil-1}+\left(1+\frac{1}{\left\lceil\frac{n+1}{2}\right\rceil}\right)\tilde{S}_{\lceil
n/2\rceil}\geq 0\;.$ (23)

Finally, the remaining set of inequalities completing the list of $\lceil
n/2\rceil$ facets are captured by

$-l(l+1)\tilde{S}_{l-1}+2(l-1)(l+1)\tilde{S}_{l}-(l-1)l\tilde{S}_{l+1}\geq
0\qquad\text{for}\quad l=2,\dots,\lceil n/2\rceil-1\;.$ (24)

This last family can in fact be extended to $l=\lceil n/2\rceil$ in order to
also reproduce (23) by declaring that $\tilde{S}_{\lceil
n/2\rceil+1}=\tilde{S}_{\lfloor n/2\rfloor}\;.$ To summarize, we have the
following result:

######  Corollary 1 (SHEC facets).

If Conjecture 1 holds, the facets of the SHECn are defined by the set of
inequalities given by (22), (23), and (24).

Because all extreme rays in Conjecture 1 are realizable by star graphs, it
follows that the conjectured SHEC is contained in the true SHEC. Whether or
not Conjecture 1 is true thus depends on whether the conjectured facets of the
SHEC are actually obtainable from symmetrizations of valid HEC inequalities.
In other words, proving our conjecture requires finding and proving valid HEC
inequalities for arbitrary $n$, which is a hard problem. Our explicit
knowledge of the HECn for small $n$ allows us to perform some checks on the
conjectured SHEC. We will be able to provide a complete proof of our
conjecture for up to $n=6$, but only a partial one for larger $n$.

As already mentioned, the conjectured SHEC facet (22) arises from singleton
SA. Since this is a valid inequality of the HEC for all $n$, it follows that
(22) is a valid facet of the SHEC for all $n$. Consider now (24) for $l=2$,
which yields

$-3\tilde{S}_{1}+3\tilde{S}_{2}-\tilde{S}_{3}\geq 0\;.$ (25)

It is a simple exercise to show that this follows from the MMI inequality in
(3) for singletons when symmetrized for $n\geq 5$.444For $n=3$ and $n=4$, as
shown in Appendix A, (25) respectively collapses down to
$-4\tilde{S}_{1}+3\tilde{S}_{2}\geq 0$ and $-3\tilde{S}_{1}+2\tilde{S}_{2}\geq
0$, which take the form of (23) (see comment below (24)). Again, since MMI is
a well-known HEC inequality which holds for all $n$, it follows that (25) is a
valid facet of the SHEC for all $n$. Further, (24) for $l=3$ gives

$-6\tilde{S}_{2}+8\tilde{S}_{3}-3\tilde{S}_{4}\geq 0\;.$ (26)

This inequality can be seen to arise from any one of the last three $5$-party
inequalities in Appendix A when symmetrized for $n\geq 7$.555As was the case
for MMI in the footnote above, if symmetrized for $n=5$ or $n=6$, these
instead give rise to the special case of (23) corresponding to (24) applied to
$l=\lceil n/2\rceil$. For $n=5$, as shown in Appendix A, these symmetrize down
to $-9\tilde{S}_{2}+8\tilde{S}_{3}\geq 0$, while for $n=6$ they give
$-6\tilde{S}_{2}+5\tilde{S}_{3}\geq 0$.

These results suffice to prove validity of our proposal for the extremal
elements of the SHECn for up to $n=6$. Unfortunately, our incomplete knowledge
of the HECn for larger $n$ is insufficient to do so for $n\geq 7$, for which a
different approach that bypasses full knowledge of the HEC will be needed.

##  5 The Symmetrized QEC

The QEC is poorly understood and only known exactly for up to $n=3$. At this
number of parties, its facets belong to the two orbits associated to the
following inequalities:

$\displaystyle S_{1}+S_{2}$ $\displaystyle\geq S_{12}\;,$ (27) $\displaystyle
S_{12}+S_{23}$ $\displaystyle\geq S_{2}+S_{123}\;.$

The first one can be recognized as the SA inequality in (1), while the last
one corresponds to SSA as in (2). Going to higher $n$, different instances of
the same type of inequality associated to inequivalent choices of subsystems
can define algebraically independent orbits of inequalities. This way, one can
apply SA and SSA to all possible subsystems in order to construct a rich
polyhedral cone bounded by multiple orbits of these two types of inequalities.
Because both are universal quantum inequalities, the resulting cone defines an
outer approximation to the QEC. A neat study of precisely which of these
orbits of SA and SSA constitute facets of the resulting cone, and which are
redundant, was performed by Nicholas Pippenger in pippenger2003inequalities .

Additionally, for arbitrary number of parties $n$, Pippenger studied the cone
that results under $Sym_{n}$ symmetrizations. More specifically, rather than
symmetrizing over all $I\subseteq[n+1]$ for each
$\absolutevalue{I}=k\leq\lceil n/2\rceil$, Pippenger considered
symmetrizations only over $I\subseteq[n]$ for each $\absolutevalue{I}=k\leq
n$, thus not including the purification symmetry. Even for this milder
$Sym_{n}$ symmetrizations, he was able to prove that every extreme ray of the
resulting cone is realizable by some quantum state.666There is a minor typo
associated to Theorem $5.5$ of pippenger2003inequalities : in equation
$(5.20)$ therein, the condition $n-a+1\leq b\leq n$ should instead read
$n-a\leq b\leq n$. This way, he established that the cone of SA and SSA
instances collapses down to precisely match the QEC under respective $Sym_{n}$
symmetrizations. As we did for the HEC, here we will instead consider
$Sym_{n+1}$ symmetrizations, i.e., we further symmetrize over the purifier.
This symmetrization of the QEC defines the SQEC, which by
pippenger2003inequalities  will of course continue to match the cone of SA and
SSA instances accordingly symmetrized under $Sym_{n+1}$. In other words, using
the results of pippenger2003inequalities , we will be able to present complete
descriptions of the SQEC for all $n$.

Like the SHEC, the SQEC turns out to be simplicial. For each $n$, the facets
of the SQEC are given by the following set of $\lceil n/2\rceil$ inequalities:

$-\tilde{S}_{l-1}+2\tilde{S}_{l}-\tilde{S}_{l+1}\geq 0\qquad\text{for}\quad
1\leq l\leq\lceil n/2\rceil\;,$ (28)

where we have declared $\tilde{S}_{0}=0$ and $\tilde{S}_{\lceil
n/2\rceil+1}=\tilde{S}_{\lceil n/2\rceil}$. For $l=1$, (28) yields (22), which
recall comes from the SA facet that the QEC shares with the HEC. The rest come
from symmetrizations of SSA inequalities as in (2) for subsystems
$I,J\subseteq[n]$ obeying $\absolutevalue{I\cap J}=2,3,\dots,\lceil
n/2\rceil$.

One can easily see that the SQEC is a simplicial cone. Writing out its facet-
defining vectors as rows in a matrix, the extreme rays of the SQEC can thus be
obtained as the column vectors of the inverse matrix. For instance, for
$n=10$, the matrix of row inequalities reads

$M_{10}^{\text{SQEC}}=\begin{bmatrix}2&-1&0&0&0\\\ -1&2&-1&0&0\\\
0&-1&2&-1&0\\\ 0&0&-1&2&-1\\\ 0&0&0&-1&1\end{bmatrix}\;,$ (29)

and its inverse of column extreme rays yields

$(M_{10}^{\text{SQEC}})^{-1}=\begin{bmatrix}1&1&1&1&1\\\ 1&2&2&2&2\\\
1&2&3&3&3\\\ 1&2&3&4&4\\\ 1&2&3&4&5\\\ \end{bmatrix}\;.$ (30)

For general $n$, letting $l=1,2,\dots,\lceil n/2\rceil$ enumerate each of the
resulting vectors, the $k^{\text{th}}$ component of the $l^{\text{th}}$
extreme ray can be easily seen to be given by

$\tilde{S}^{(l)}_{k}=\min\\{k,l\\}\;.$ (31)

Figure 2: The SQEC and SHEC for $n=4,5$.

Figure 3: Cross-sections of the SQEC and SHEC through the hyperplane defined
by $\sum_{k=1}^{\lceil n/2\rceil}\tilde{S}_{k}=1$ for $n=5,6,7,8$. The cross-
section of the positive orthant is not shown because it is too large. Instead,
for orientation we show arrows pointing in the direction in which the
intersections with the canonical basis vectors occur – these would define the
extreme points of the positive orthant in the cross-section.

In Figure 2, we illustrate the SQEC and SHEC full-dimensionally for $n=4,5$.
In Figure 3, we illustrate the SQEC and SHEC via codimension-$1$ radial cross-
sections for $n=5,6,7,8$. A question naturally arising is now how these two
cones relate to each other. In particular, we will be interested in comparing
their “size”, and the scaling with $n$.

##  6 Volumes of Symmetrized Entropy Cones

The volume of a pointed polyhedral cone is infinite. However, if the cone lies
within a small solid angle inside the positive orthant, then a useful notion
of how large it is can be obtained as the volume of the polytope it forms up
to some choice of cross-section. When this cone is simplicial, as is the case
for both the SQEC and the SHEC, this polytope is a simplex whose vertices are
the origin and the intersection of its extreme rays with the cross-section.
The volume of such a simplex can be easily computed using the determinant of
the matrix of extreme rays $\hat{M}$ suitably normalized to match the vertices
on the cross-section. Since the determinant actually computes the volume of
the parallelotope formed by the vectors involved, one has to divide it by a
combinatorial factor of $d!$ to obtain the volume of the corresponding
$d$-dimensional simplex. Denoting a general matrix of normalized extreme rays
of a $d$-dimensional simplicial cone by $\hat{M}$, we will thus define its
volume by

$\operatorname{vol}\left(\operatorname{cone}\,\\{\hat{M}\\}\right)\equiv\frac{1}{d!}|\det\hat{M}|\;.$
(32)

When considering the SQEC and the SHEC, our starting point will be a matrix
$M$ of unnormalized, integral extreme rays. A natural choice of cross-section
is the hyperplane defined by $\sum_{k=1}^{\lceil n/2\rceil}\tilde{S}_{k}=1$,
as in Fig. 3. The vertices on the cross-section can then be easily obtained by
normalizing the extreme rays of the desired cone linearly to unity, i.e.,
using the taxicab norm. Explicit numerical results for small $n$ are given in
Appendix C. More generally, we are interested in the scaling of volumes as
$n\rightarrow\infty$. Our strategy for obtaining this will be as follows:

  1. 1.

Derive a simple general expression for the determinant of $M$.

  2. 2.

Derive an expression for the product $\Pi(M)$ of the norms of all extreme rays
in $M$.

  3. 3.

Obtain the desired (normalized) determinant as
$|\det\hat{M}|=\frac{\absolutevalue{\det M}}{\Pi(M)}$.

For the SQECn, the extreme rays are given component-wise by (31), which we
will take to define our unnormalized matrix of extreme rays
$M^{\text{SQEC}}_{n}$. The determinant of the matrix formed by these vectors
is always unity for all $n$, i.e., $\det M^{\text{SQEC}}_{n}=1$, which takes
care of Step 1. Letting $d_{n}=\lceil n/2\rceil$ for convenience, the
normalization factor $\Pi(M^{\text{SQEC}}_{n})$ resulting from Step 2 can also
be obtained exactly and reads

$\Pi(M^{\text{SQEC}}_{n})=\frac{\left(2d_{n}\right)!}{2^{d_{n}}}.$ (33)

We can thus conclude that the volume of the SQEC for $n$ parties is exactly
given by

$\operatorname{vol}(\text{SQEC}_{n})=\frac{1}{d_{n}!}\frac{2^{d_{n}}}{\left(2d_{n}\right)!},$
(34)

Since we will be ultimately interested only in the large-$n$ scaling of this
volume, we may consider keeping track of just the leading order behavior of
(34) as $n\to\infty$. Undergoing numerically coarser approximations at large
$n$, and taking $n$ to be even (such that $d_{n}=n/2$), one obtains

$\operatorname{vol}(\text{SQEC}_{n})\sim\frac{1}{(n/2)!}\frac{e^{\frac{13}{10}n}}{n^{\frac{1}{2}+n}}\sim\frac{e^{\frac{11}{5}n}}{n^{1+\frac{3}{2}n}}.$
(35)

Although slightly more complicated, the same approach can be applied to the
SHEC. We will use (19) with an $(n+1)/2$ prefactor for the unnormalized
extreme rays in $M^{\text{SHEC}}_{n}$. Remarkably, the determinants of the
resulting matrix are easily computable and read

$\det
M^{\text{SHEC}}_{n}=\left\lceil{\frac{n+1}{2}}\right\rceil!\times\begin{cases}1&\quad\text{if
$n$ is even,}\\\ \frac{n+1}{2}&\quad\text{if $n$ is odd.}\end{cases}$ (36)

The exact result for $\Pi(M^{\text{SHEC}}_{n})$ can be written

$\Pi(M^{\text{SHEC}}_{n})=\left(-\frac{1}{6}\right)^{d_{n}}\prod_{k=1}^{3}(1-x_{k})_{d_{n}}\;,$
(37)

where the subscripted brackets denote the Pochhammer symbol
$(a)_{k}=\frac{\Gamma(a+k)}{\Gamma(a)}$, and $x_{k}$ stands for the
$k^{\text{th}}$ root of the following $3^{\text{rd}}$-order polynomial:

$\prod_{k=1}^{3}(x-x_{k})=x^{3}-x-d_{n}(d_{n}+1)(3n-2(d_{n}-1)).$ (38)

Albeit exact, (37) still relies on an implicit expression for the $x_{k}$
roots. We could of course use the general formulae for the roots of a
$3^{\text{rd}}$-order polynomial, but this would not be very illuminating.
Instead, we will take advantage of large-$n$ approximations in order to get a
more explicit result for this. A key observation is to notice that the roots
of (38) scale linearly with $n$. Hence, to simplify this polynomial non-
trivially at large $n$, we will first have to factor out this uninteresting
$n$-dependence. We can do so by letting $x=d_{n}\hat{x}$ and introducing
renormalized roots $x_{k}=d_{n}\hat{x}_{k}$. Plugging these into (38), at
leading order in $n$ we obtain

$\prod_{k=1}^{3}(\hat{x}-\hat{x}_{k})\approx\hat{x}^{3}-4\;.$ (39)

The renormalized roots no longer depend on $n$ and are simply the cubic roots
of $4$. In terms of these, we can re-express (37) as

$\Pi(M^{\text{SHEC}}_{n})\approx\left(-\frac{1}{6}\right)^{d_{n}}\prod_{k=1}^{3}(1-\hat{x}_{k}d_{n})_{d_{n}}\;,$
(40)

and expand the Pochhammer symbols at large $n$. For each root, this expansion
yields

$(1-d_{n}\hat{x}_{k})_{d_{n}}\approx\left(1-\frac{1}{\hat{x}_{k}}\right)^{\frac{1}{2}+(1-\hat{x}_{k})d_{n}}\left(-\frac{\hat{x}_{k}d_{n}}{e}\right)^{d_{n}}.$
(41)

Taking the profuct over all roots in (40) and making some numerical
approximations, we arrive at

$\Pi(M^{\text{SHEC}}_{n})\approx\frac{\sqrt{3}}{2}\left(\frac{3}{5}\right)^{d_{n}}d_{n}^{3d_{n}}\sim
e^{-\frac{13}{10}n}n^{\frac{3}{2}n}.$ (42)

Having completed Step 2, we finally obtain our desired expression for the
volume of the SHEC at large $n$:

$\operatorname{vol}(\text{SHEC}_{n})\sim\frac{1}{(n/2)!}\frac{e^{\frac{9}{2}n}}{n^{-1+n}}\sim\frac{e^{\frac{13}{10}n}}{n^{-\frac{1}{2}+\frac{3}{2}n}}\;.$
(43)

We are finally ready to compare the asymptotic volumes of the SQEC and the
SHEC. Using the expressions from (35) and (43), we obtain

$\frac{\operatorname{vol}(\text{SHEC}_{n})}{\operatorname{vol}(\text{SQEC}_{n})}\sim
n^{\frac{3}{2}}e^{-\frac{9}{10}n}.$ (44)

We thus conclude that, in terms of volumes in entropy space, the SHECn
constitutes an exponentially small fraction of the SQECn. This is particularly
interesting considering that holographic states are in fact expected to be
typical when randomly sampling some Hilbert space of quantum states with
respect to an invariant measure on it.777We thank Bartek Czech for comments on
this points. We can also compare the volumes of these two cones to the volume
of the unit simplex that the positive orthant defines. Using $n/2$ as the
dimensionality at large $n$, the volume of this object is simply given by
$1/(n/2)!$. Comparing this to (34), we see that the SQECn itself also occupies
a very small fraction of the positive orthant which decreases faster than
exponential as $n$ increases.

##  7 Conclusion and Outlook

We motivated and presented a projection of the HEC via a natural
symmetrization which results in a much simpler (but still non-trivial) entropy
cone: the SHEC. By inspecting its extremal elements, we were able to
conjecture a full characterization of this cone for an arbitrary number of
parties $n$ in terms of both full sets of extreme rays and facet inequalities.
Our proof of this proposal for the SHEC for small $n\leq 6$ was based on the
known complete descriptions of the HEC for $n\leq 5$. In the future, however,
since the SHEC is by construction a coarser and much simpler version of the
HEC, one would hope to make progress in proving the general form of the SHEC
without necessarily knowing such a description for the HEC. Indeed, already at
the conjectural level, here we have provided a plausible general form of the
SHEC for arbitrary $n$, while such an achievement for the HEC seems
unattainable as of now.

For instance, the partial description of the HEC6 known to date n6wip  already
exhibits an unprecedented level of complexity compared to $n\leq 5$. While the
HEC5 consists of just $8$ distinct orbits of facets and $19$ orbits of extreme
rays, the current description of the HEC6 already involves at least $182$
orbits of facets and at least $4122$ orbits of extreme rays n6wip .888Data to
date suggests that, in fact, these numbers are still highly underestimating
the complexity of a complete description of the HEC6. According to Conjecture
1 and Corollary 1, all this extremely rich structure that arises for $n\geq 6$
turns out to be symmetrized away at the level of the SHEC, which nonetheless
still preserves some non-trivial properties of the HEC in the form of $\lceil
n/2\rceil$ facets and extreme rays at each $n$.

Symmetrizations provide an organizing principle for the complicated structure
of the HEC. The usefulness of this organization is already evidenced by $n=5$
e.g. at the level of facets, where we observe that $1)$ not all facets of the
HEC remain facets of the SHEC, $2)$ those which remain facets do so in non-
trivial families of multiple facets, and $3)$ it takes genuinely new facets of
the HEC at each $n$ to generate higher-$n$ facets of the SHEC. Observation
$1)$ is consistent with our expectations for the SHEC: for larger $n$ there
arise combinatorially large numbers of new orbits of facets for the HEC, but
the simplicial SHEC should only involve $\lceil n/2\rceil$ as per our
conjecture. Observation $2)$ implies that there are distinguished families of
facets of the HEC which remain “extremal” upon symmetrization. Observation
$3)$ textitasizes the fact that the SHEC retains non-trivial information about
the HEC for all $n$, thus stressing its usefulness as a prerequisite to
understand the HEC. To elaborate further on this, let us textitasize that none
of the $182$ facet orbits that we currently known of the HEC6 happens to prove
the form of the SHEC for $n=7$ – namely, none of them symmetrizes down to the
$l=4$ inequality coming from (24). This suggests that genuinely $n=7$
information about the HEC may be needed to characterize the SHEC7, and
similarly for any larger $n$. It is also interesting to note that even though
the $l=3$ inequality coming from (24) can be realized by HEC5 inequalities,
there also appear new HEC6 inequalities falling into the same family. For
instance,

$\displaystyle-S_{12}-S_{13}-S_{14}-$ $\displaystyle
S_{25}-S_{26}-S_{35}+S_{123}+S_{124}+S_{125}+S_{126}$ (45) $\displaystyle+$
$\displaystyle S_{134}+S_{135}+S_{235}+S_{256}-S_{1234}-S_{1235}-S_{1256}\geq
0\;,$

is a genuinely HEC6 inequality (i.e., not obtainable as a lift from lower-$n$
inequalities), whose symmetrization for $n\geq 7$ matches precisely (26) as
well.

More generally, the symmetrizations studied here have proven to be useful in
distilling non-trivial structure at arbitrary $n$ for both the quantum and
holographic entropy cones. Other interesting classes of quantum states and
constructs have been studied in the past, which suffer from the same rapidly
increasing complexity as $n$ increases. It would be interesting to explore if,
upon symmetrizations, one can gain some further knowledge about the
general-$n$ structure of the cone of stabilizer states Linden:2013kal , the
cone of linear rank inequalities Dougherty2009 , the cone of hypergraph
entropies Bao:2020zgx ; Walter:2020zvt ; Bao:2020mqq , the cone of topological
links Bao:2021gzu , or even the HEC under quantum corrections from bulk matter
fields Akers:2021lms .

###### Acknowledgements.

It is a pleasure to thank David Avis, Max Rota and Jordi Tura for useful
discussions. MF was supported by the Swiss National Science Foundation and by
The Branco Weiss Fellowship – Society in Science, administered by the ETH
Zürich. SHC was supported by NSF grant PHY-1801805 and UCSB during the early
stages of this project, and is currently supported by NSF grant PHY-2107939
and a Len DeBenedictis Graduate Fellowship. Note – During the completion of
this work we became aware of Czech:2021rxe , which applies the same projection
to the HEC and obtains results consistent with ours.

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##  Appendix A Extremal structure of the SHEC for $n\leq 5$

The HEC has been completely characterized for $n\leq 5$ Bao:2015bfa ;
HernandezCuenca:2019wgh ; Avis:2021xnz , while for larger $n$ only partial
results are known due to its rapidly increasing complexity n6wip . We list
here all the extreme rays and facets of the HEC for $n\leq 5$, and compute
their symetrization according to the prescription presented in Section 3.

### Extreme Rays

In the tables below, the column “extremal?” refers to whether the symmetrized
ray is extremal for the associated SHEC, while the column “$w$” indicates the
weight of the purification edge in the star graph realizing the corresponding
ray (see Fig. 1).

$n=2$
$\\#$ HEC SHEC extremal? $w$ 1 (11; 0) (1) yes any

$n=3$
$\\#$ HEC SHEC extremal? $w$ 1 (110; 011; 0) (3 4) yes 3 2 (111; 222; 1) (1 2)
yes 1

$n=4$
$\\#$ HEC SHEC extremal? $w$ 1 (1100; 011110; 0011; 0) (2 3) yes 4 2 (1110;
221211; 1222; 1) (1 2) yes 2 3 (1111; 222222; 3333; 2) (1 2) yes 2

$n=5$
$\\#$ HEC SHEC extremal? $w$ 1 (11000; 0111111000; 0001111110; 00011; 0) (5 8
9) yes 5 2 (11100; 2211211110; 1222212211; 11222; 1) (5 10 12) 3 (11110;
2221221211; 3323223222; 23333; 2) (5 10 12) 4 (11111; 2222222222; 2233333323;
22222; 1) (10 20 27) 5 (11111; 2222222222; 3332233323; 22222; 1) (10 20 27) 6
(11112; 2223223233; 3343443444; 43333; 2) (10 20 27) 7 (11111; 2222222222;
2222333333; 22222; 1) (5 10 13) 8 (11111; 2222222222; 2233223333; 22222; 1) (5
10 13) 9 (33333; 6666666666; 5777799999; 66666; 3) (5 10 13) 10 (33333;
6666666666; 5777979999; 66666; 3) (5 10 13) 11 (33333; 6666666666; 5777999799;
66666; 3) (5 10 13) 12 (11111; 2222222222; 1333333333; 22222; 1) (5 10 14) 13
(11111; 2222222222; 3333333333; 22222; 1) (1 2 3) yes 1 14 (11111; 2222222222;
3333333333; 44444; 3) (4 8 9) yes 3 15 (11112; 2223223233; 3323443444; 43333;
2) (20 40 51) 16 (11122; 2233233334; 3444454455; 55444; 3) (25 50 63) 17
(11222; 2333333444; 4445355534; 44433; 2) (25 50 63) 18 (22223; 4445445455;
4656576777; 65555; 3) (7 14 18) 19 (22223; 4445445455; 4656756777; 65555; 3)
(7 14 18)

### Facets

Below, the table entry “facet?” refers to whether the symmetrized inequality
is extremal for the associated SHEC.

$n=2$
$\\#$ HEC SHEC facet? 1 $S_{\text{\tiny A}}+S_{\text{\tiny B}}-S_{\text{\tiny
AB}}\geq 0$ $\tilde{S}_{1}\geq 0$ yes

$n=3$
$\\#$ HEC SHEC facet? 1 $S_{\text{\tiny A}}+S_{\text{\tiny B}}-S_{\text{\tiny
AB}}\geq 0$ $2\tilde{S}_{1}-\tilde{S}_{2}\geq 0$ yes 2 $S_{\text{\tiny
AB}}+S_{\text{\tiny AC}}+S_{\text{\tiny BC}}-S_{\text{\tiny A}}-S_{\text{\tiny
B}}-S_{\text{\tiny C}}-S_{\text{\tiny ABC}}\geq 0$
$-4\tilde{S}_{1}+3\tilde{S}_{2}\geq 0$ yes

$n=4$
$\\#$ HEC SHEC facet? 1 $S_{\text{\tiny A}}+S_{\text{\tiny B}}-S_{\text{\tiny
AB}}\geq 0$ $2\tilde{S}_{1}-\tilde{S}_{2}\geq 0$ yes 2 $S_{\text{\tiny
AB}}+S_{\text{\tiny AC}}+S_{\text{\tiny BC}}-S_{\text{\tiny A}}-S_{\text{\tiny
B}}-S_{\text{\tiny C}}-S_{\text{\tiny ABC}}\geq 0$
$-3\tilde{S}_{1}+2\tilde{S}_{2}\geq 0$ yes

$n=5$
$\\#$ HEC SHEC facet? 1 $S_{\text{\tiny A}}+S_{\text{\tiny B}}-S_{\text{\tiny
AB}}\geq 0$ $2\tilde{S}_{1}-\tilde{S}_{2}\geq 0$ yes 2 $S_{\text{\tiny
AB}}+S_{\text{\tiny AC}}+S_{\text{\tiny BC}}-S_{\text{\tiny A}}-S_{\text{\tiny
B}}-S_{\text{\tiny C}}-S_{\text{\tiny ABC}}\geq 0$
$-3\tilde{S}_{1}+3\tilde{S}_{2}-\tilde{S}_{3}\geq 0$ yes 3 $S_{\text{\tiny
AB}}+S_{\text{\tiny ACD}}+S_{\text{\tiny BCD}}-S_{\text{\tiny
A}}-S_{\text{\tiny B}}-S_{\text{\tiny CD}}-S_{\text{\tiny ABCD}}\geq 0$
$-2\tilde{S}_{1}-\tilde{S}_{2}+2\tilde{S}_{3}\geq 0$ 4 $S_{\text{\tiny
AD}}+S_{\text{\tiny BC}}+S_{\text{\tiny ABE}}+S_{\text{\tiny
ACE}}+S_{\text{\tiny ADE}}+S_{\text{\tiny BDE}}+S_{\text{\tiny
CDE}}-S_{\text{\tiny A}}-S_{\text{\tiny B}}-S_{\text{\tiny C}}-S_{\text{\tiny
D}}-S_{\text{\tiny AE}}-S_{\text{\tiny DE}}-S_{\text{\tiny
BCE}}-S_{\text{\tiny ABDE}}-S_{\text{\tiny ACDE}}\geq 0$
$-2\tilde{S}_{1}-\tilde{S}_{2}+2\tilde{S}_{3}\geq 0$ 5 $S_{\text{\tiny
ABC}}+S_{\text{\tiny BCD}}+S_{\text{\tiny CDE}}+S_{\text{\tiny
ADE}}+S_{\text{\tiny ABE}}-S_{\text{\tiny AB}}-S_{\text{\tiny
BC}}-S_{\text{\tiny CD}}-S_{\text{\tiny DE}}-S_{\text{\tiny
AE}}-S_{\text{\tiny ABCDE}}\geq 0$
$-\tilde{S}_{1}-5\tilde{S}_{2}+5\tilde{S}_{3}\geq 0$ 6 $2S_{\text{\tiny
ABC}}+S_{\text{\tiny ABD}}+S_{\text{\tiny ABE}}+S_{\text{\tiny
ACD}}+S_{\text{\tiny ADE}}+S_{\text{\tiny BCE}}+S_{\text{\tiny
BDE}}-S_{\text{\tiny AB}}-S_{\text{\tiny AC}}-S_{\text{\tiny
AD}}-S_{\text{\tiny BC}}-S_{\text{\tiny BE}}-S_{\text{\tiny
DE}}-S_{\text{\tiny ABCD}}-S_{\text{\tiny ABCE}}-S_{\text{\tiny ABDE}}\geq 0$
$-9\tilde{S}_{2}+8\tilde{S}_{3}\geq 0$ yes 7 $S_{\text{\tiny
ABC}}+S_{\text{\tiny ABD}}+S_{\text{\tiny ABE}}+S_{\text{\tiny
ACD}}+S_{\text{\tiny ACE}}+S_{\text{\tiny ADE}}+S_{\text{\tiny
BCE}}+S_{\text{\tiny BDE}}+S_{\text{\tiny CDE}}-S_{\text{\tiny
AB}}-S_{\text{\tiny AC}}-S_{\text{\tiny AD}}-S_{\text{\tiny
BE}}-S_{\text{\tiny CE}}-S_{\text{\tiny DE}}-S_{\text{\tiny
BCD}}-S_{\text{\tiny ABCE}}-S_{\text{\tiny ABDE}}-S_{\text{\tiny ACDE}}\geq 0$
$-9\tilde{S}_{2}+8\tilde{S}_{3}\geq 0$ yes 8 $3S_{\text{\tiny
ABC}}+3S_{\text{\tiny ABD}}+3S_{\text{\tiny ACE}}+S_{\text{\tiny
ABE}}+S_{\text{\tiny ACD}}+S_{\text{\tiny ADE}}+S_{\text{\tiny
BCD}}+S_{\text{\tiny BCE}}+S_{\text{\tiny BDE}}+S_{\text{\tiny
CDE}}-S_{\text{\tiny AD}}-S_{\text{\tiny AE}}-S_{\text{\tiny
BC}}-S_{\text{\tiny DE}}-S_{\text{\tiny ABDE}}-S_{\text{\tiny
ACDE}}-2S_{\text{\tiny AB}}-2S_{\text{\tiny AC}}-2S_{\text{\tiny
BD}}-2S_{\text{\tiny CE}}-2S_{\text{\tiny ABCD}}-2S_{\text{\tiny ABCE}}\geq 0$
$-9\tilde{S}_{2}+8\tilde{S}_{3}\geq 0$ yes

##  Appendix B Extremal structure of the SQEC for $n\leq 5$

The extreme rays of the SQEC are very simple and given by (31). In contrast,
the extreme rays of its parent cone defined by all possible instances of SA
and SSA turn out to be extremely complicated (e.g. there are millions of them
for $n=5$). Hence we have decided to not include this data. Instead, we list
here explicitly only the facets of the cone involving instances of SA and SSA
and their symmetrizations, indicating which of the latter yield facets of the
SQEC.

$n=2$
$\\#$ SA+SSA SQEC facet? 1 $S_{\text{\tiny A}}+S_{\text{\tiny
B}}-S_{\text{\tiny AB}}\geq 0$ $\tilde{S}_{1}\geq 0$ yes

$n=3$
$\\#$ SA+SSA SQEC facet? 1 $S_{\text{\tiny A}}+S_{\text{\tiny
B}}-S_{\text{\tiny AB}}\geq 0$ $2\tilde{S}_{1}-\tilde{S}_{2}\geq 0$ yes 2
$S_{\text{\tiny A}}-S_{\text{\tiny B}}+S_{\text{\tiny AB}}\geq 0$
$\tilde{S}_{2}\geq 0$ 3 $S_{\text{\tiny AC}}+S_{\text{\tiny
BC}}+S_{\text{\tiny C}}-S_{\text{\tiny ABC}}\geq 0$
$-\tilde{S}_{1}+\tilde{S}_{2}\geq 0$ yes

$n=4$
$\\#$ SA+SSA SQEC facet? 1 $S_{\text{\tiny A}}+S_{\text{\tiny
B}}-S_{\text{\tiny AB}}\geq 0$ $2\tilde{S}_{1}-\tilde{S}_{2}\geq 0$ yes 2
$S_{\text{\tiny A}}-S_{\text{\tiny B}}+S_{\text{\tiny AB}}\geq 0$
$\tilde{S}_{2}\geq 0$ 3 $S_{\text{\tiny A}}+S_{\text{\tiny BC}}-S_{\text{\tiny
ABC}}\geq 0$ $\tilde{S}_{1}\geq 0$ 4 $-S_{\text{\tiny A}}+S_{\text{\tiny
BC}}+S_{\text{\tiny ABC}}\geq 0$ $-\tilde{S}_{1}+2\tilde{S}_{2}\geq 0$ 5
$S_{\text{\tiny AC}}+S_{\text{\tiny BC}}-S_{\text{\tiny C}}-S_{\text{\tiny
ABC}}\geq 0$ $-\tilde{S}_{1}+\tilde{S}_{2}\geq 0$ yes

$n=5$
$\\#$ SA+SSA SQEC facet? 1 $S_{\text{\tiny A}}+S_{\text{\tiny
B}}-S_{\text{\tiny AB}}\geq 0$ $2\tilde{S}_{1}-\tilde{S}_{2}\geq 0$ yes 2
$S_{\text{\tiny A}}-S_{\text{\tiny B}}+S_{\text{\tiny AB}}\geq 0$
$\tilde{S}_{2}\geq 0$ 3 $S_{\text{\tiny A}}+S_{\text{\tiny BC}}-S_{\text{\tiny
ABC}}\geq 0$ $\tilde{S}_{1}+\tilde{S}_{2}-\tilde{S}_{3}\geq 0$ 4
$S_{\text{\tiny A}}-S_{\text{\tiny BC}}+S_{\text{\tiny ABC}}\geq 0$
$\tilde{S}_{1}-\tilde{S}_{2}+\tilde{S}_{3}\geq 0$ 5 $-S_{\text{\tiny
A}}+S_{\text{\tiny BC}}+S_{\text{\tiny ABC}}\geq 0$
$-\tilde{S}_{1}+\tilde{S}_{2}+\tilde{S}_{3}\geq 0$ 6 $-S_{\text{\tiny
C}}+S_{\text{\tiny BC}}+S_{\text{\tiny AC}}-S_{\text{\tiny ABC}}\geq 0$
$-\tilde{S}_{1}+2\tilde{S}_{2}-\tilde{S}_{3}\geq 0$ yes 7 $-S_{\text{\tiny
A}}-S_{\text{\tiny B}}+S_{\text{\tiny AC}}+S_{\text{\tiny BC}}\geq 0$
$-\tilde{S}_{1}+\tilde{S}_{2}\geq 0$ 8 $-S_{\text{\tiny D}}+S_{\text{\tiny
CD}}+S_{\text{\tiny ABD}}-S_{\text{\tiny ABCD}}\geq 0$
$-\tilde{S}_{1}+\tilde{S}_{3}\geq 0$ 9 $-S_{\text{\tiny CD}}+S_{\text{\tiny
BCD}}+S_{\text{\tiny ACD}}-S_{\text{\tiny ABCD}}\geq 0$
$-\tilde{S}_{2}+\tilde{S}_{3}\geq 0$ yes

##  Appendix C Exact Volumes for Small $n$

From the extreme rays of the SHEC and SQEC we can compute the volume of these
cones as explained in Section 6. For small number of parties $n$, this volume
can be easily computed exactly. In Fig. 4, and in the following table, we show
the exact relative volume between the two cones for $n\leq 10$. Note that the
SHEC becomes exponentially small compared to the SQEC as $n$ increases, see
(44).

Figure 4: Exact ratio between the volumes of the SHECn and SQECn for $2\leq n\leq 10$. $n$ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10
---|---|---|---|---|---|---|---|---|---
$(\operatorname{vol}\text{SHEC}_{n})^{-1}$ | 1 | 21 | 30 | 2772 | 4536 | 1127000 | 1982400 | 1036103250 | 1906410000
$(\operatorname{vol}\text{SQEC}_{n})^{-1}$ | 1 | 12 | 12 | 540 | 540 | 60480 | 60480 | 13608000 | 13608000
$\text{SHEC}_{n}/\text{SQEC}_{n}$ | 1 | 0.571 | 0.4 | 0.195 | 0.119 | 0.0537 | 0.0305 | 0.0131 | 0.00714