# Platonic Bell inequalities for all dimensions

Károly F. Pál<EMAIL_ADDRESS>Institute for Nuclear Research, P. O. Box 51,
H-4001 Debrecen, Hungary  Tamás Vértesi<EMAIL_ADDRESS>MTA Atomki
Lendület Quantum Correlations Research Group, Institute for Nuclear Research,
P. O. Box 51, H-4001 Debrecen, Hungary

###### Abstract

In this paper we study the Platonic Bell inequalities for all possible
dimensions. There are five Platonic solids in three dimensions, but there are
also solids with Platonic properties (also known as regular polyhedra) in four
and higher dimensions. The concept of Platonic Bell inequalities in the three-
dimensional Euclidean space was introduced by Tavakoli and Gisin [Quantum 4,
293 (2020)]. For any three-dimensional Platonic solid, an arrangement of
projective measurements is associated where the measurement directions point
toward the vertices of the solids. For the higher dimensional regular
polyhedra, we use the correspondence of the vertices to the measurements in
the abstract Tsirelson space [B. Tsirelson, J. Soviet Math. 36, 557 (1987)].
We give a remarkably simple formula for the quantum violation of all the
Platonic Bell inequalities, which we prove to attain the maximum possible
quantum violation of the Bell inequalities, i.e. the Tsirelson bound. To
construct Bell inequalities with a large number of settings, it is crucial to
compute the local bound efficiently. In general, the computation time required
to compute the local bound grows exponentially with the number of measurement
settings. We find a method to compute the local bound exactly for any
bipartite two-outcome Bell inequality, where the dependence becomes polynomial
whose degree is the rank of the Bell matrix. To show that this algorithm can
be used in practice, we compute the local bound of a 300-setting Platonic Bell
inequality based on the halved dodecaplex. In addition, we use a diagonal
modification of the original Platonic Bell matrix to increase the ratio of
quantum to local bound. In this way, we obtain a four-dimensional 60-setting
Platonic Bell inequality based on the halved tetraplex for which the quantum
violation exceeds the maximum quantum violation of the famous Clauser-Horne-
Shimony-Holt Bell inequality. This is the first example of a Platonic Bell
inequality exceeding the $\sqrt{2}$ ratio.

##  1 Platonic solids in quantum physics

Platonic solids [1] – convex polyhedra with equivalent faces composed of
congruent convex regular polygons – have been recently studied in the context
of Bell nonlocality [2, 3]. Specifically, Tavakoli and Gisin [4] constructed
bipartite two-outcome Bell inequalities [5, 6] whose maximal violation is
obtained with projective measurements pointing toward the vertices of three-
dimensional Platonic solids. These inequalities are called Platonic Bell
inequalites [4].

In addition to Platonic solids, semi-regular polyhedra, namely Archimedean
solids [1], were studied in Ref. [4]. A recent development is the realization
that starting from the fact that Platonic and Archimedean solids may be
generated as orbits of representations of groups makes it possible to derive
very interesting general properties of the Bell inequalities constructed [7].
The group-theoretical approach also allows one to derive further inequalities
sharing these properties based on three-dimensional bodies even without
apparent symmetries.

All the constructed Bell expressions involve only joint correlation terms and
they are denoted correlation-type Bell expressions [6] (also known as XOR
games [8]). The maximum quantum value of these Bell expressions can be
achieved by maximally entangled states [6]. The ratio $Q/L$ between the
optimal quantum value $Q$ for such a Bell expression and its local bound $L$
quantifies the amount of isotropic noise $p=1-(L/Q)$ which must be added to
the maximally entangled state such that the quantum value becomes smaller than
$L$. Therefore the $Q/L$ ratio is relevant from an experimental point of view
as well. As a matter of fact, no Bell inequalities arising from these three-
dimensional solids were found [4, 7] which have a higher $Q/L$ ratio than
$\sqrt{2}$, i.e., exhibit higher noise resistance to the isotropic noise than
the famous Clauser-Horne-Shimony-Holt (CHSH) Bell inequality [9]. Note,
however, that there exist correlation-type Bell inequalities overcoming the
ratio $Q/L=\sqrt{2}$ using maximally entangled states which are not based on
Platonic Bell inequalities, see e.g. [31, 19]. The first example using a two-
qubit maximally entangled state appeared in 2008 in Ref. [32].

Our present study is built upon and motivated by the recent work of Tavakoli
and Gisin [4]. We generalize the notion of Platonic Bell inequalities to
$d\neq 3$ dimensional Euclidean spaces. Obviously, if $d>3$, this space can
not be identified with the physical space such that the measurements point
towards the vertices of the solids. Nevertheless, even in such cases the
generalization does correspond to genuine quantum Bell scenarios. We have
studied all pairs of Platonic solids in all dimensions. In contrast to
previous works, we have also considered different local orientations of these
solids, especially to determine which relative orientation gives the maximum
$Q/L$ ratio. We have proven that the quantum value is independent of the
relative orientation and can be given in all cases by a very simple formula.
We have found a method to efficiently calculate the local bound when the rank
of the Bell matrix (which in this case is just $d$) is much smaller than the
number of measurement settings. Unfortunaly, we still could not find a single
Bell inequality where $Q/L$ is larger than $\sqrt{2}$. Next, we have also
studied Platonic Bell inequalities where the main diagonal entries of the Bell
matrix have been modified by a constant term. We have shown that this
modification does not change the measurement configuration corresponding to
the maximum quantum violation. Therefore, the diagonally modified inequalities
are still considered as Platonic Bell inequalities. Finally, starting from the
120-vertex tetraplex, with this diagonal modification we have found Platonic
Bell inequalities with $Q/L>\sqrt{2}$. Hence, these Bell inequalities are more
tolerant to isotropic noise than the CHSH inequality. In fact, they require a
state space of dimension $4\times 4$ to reach the maximum quantum violation,
i.e., the Tsirelson bound [5].

Besides Bell nonlocality, there is a number of other uses of Platonic solids
in quantum physics. Let us give just a few other examples. In Refs. [10, 11],
the directions of quantum measurements are chosen along the vertices of
Platonic solids to demonstrate Einstein-Podolsky-Rosen steering. In Ref. [12],
quantum circuits have been devised where the measurement directions point to
vertices of the Platonic solids. In Ref. [13], different types of single qubit
private quantum channels have been constructed and linked to three-dimensional
Platonic solids. In a more recent work [14] four-dimensional Platonic solids
were connected to qutrit-based private quantum channels. Such regular
polyhedra were also used in reference frame alignment [15] and quantum hashing
protocols [16]. More recently Platonic solids were considered in quantum
entanglement theory [17] and quantum contextuality [18].

The structure of the paper is as follows. Section 2 gives an introduction to
correlation-type Bell inequalities, and more specifically to Platonic Bell
inequalities introduced in Ref. [4]. Section 3 summarizes the basic properties
of Platonic solids in all the possible dimensions. In Section 4 the maximum
quantum value, that is the quantum bound of a special class of Bell
inequalities is given. This class is proved to include all Platonic (and
Archimedean) inequalities. We derive a remarkably simple formula for the
quantum violation of all members of this class. We note that the same formula
(5) for the quantum bound in the framework of group theory has been obtained
independently in Ref. [7]. However, more conditions have been used than
necessary for the derivation, and no proof has been provided that the formula
also defines the globally optimal quantum value, i. e., that it reaches the
Tsirelson bound of the Bell inequality. In contrast, we show that the choice
of the measurement vectors is optimal, proving that the measurement
configuration allows us to reach the Tsirelson bound.

To construct Bell inequalities with a large number of settings, it is crucial
to compute the local bound efficiently. In general, the computation time
required to compute the local bound grows exponentially with the number of
measurement settings. In Section 5, we give a numerical method that allows the
computation of the local bound in polynomial time (instead of exponential
scaling) in the number of measurement settings for any bipartite two-outcome
Bell inequality with constant rank of the Bell matrix. The dependence is
polynomial in terms of the rank of the Bell matrix, in particular, the degree
of the polynomial equals the rank of the Bell matrix. The considered bipartite
Bell inequalities include all correlation-type inequalities and all Platonic
Bell inequalities as a subset of correlation-type inequalities. In the next
sections we apply the developed tools to Platonic Bell inequalities in all
possible dimensions: In Section 6, we consider Platonic Bell inequalities
based on regular polygons (that is, the case of $d=2$). The three and higher
dimensional cases are discussed in Sections 7 and 8. In particular, the local
strategies for all $d\geq 3$ are given in Section 7 and the computation of the
local bounds for $d\geq 3$ is carried out in Section 8. Then, in Section 9, we
discuss Platonic Bell inequalities where the main diagonal entries of the Bell
matrix are modified with a constant term. In this way, we find (diagonally
modified) Platonic Bell inequalities that are more robust to isotropic noise
than the celebrated CHSH-Bell inequality. The obtained results are summarized
in Section 10. In Appendix A we discuss the known relationship between the
Euclidean and Hilbert space descriptions whose existence shows that higher
dimensional generalizations do indeed correspond to quantum Bell scenarios.
The proof of special properties of our classes of Bell inequalities is given
in Appendix B.

##  2 Introduction to Platonic Bell inequalities

Tavakoli and Gisin [4] have constructed two-party correlation-type Bell
inequalities [6] whose maximal quantum value can be reached by measurements
performed on a pair of one-half spins along directions of the vertices of
Platonic solids. In a two-party Bell scenario parts of identically prepared
physical systems are distributed between two distant parties, say Alice and
Bob. Each of them independently chooses a measurement setting from a pre-
decided set and performs the corresponding measurement on her/his subsystem. A
Bell inequality expresses the fact that, given the measurement settings chosen
by the parties, a linear combination of the conditional probabilities of the
possible outcomes cannot be greater than a certain value, the so-called local
bound, in any locally realistic theory. Quantum theory allows for the
violation of many Bell inequalities [3].

Correlation-type Bell inequalities [6] involve binary, that is two-outcome
measurements. Let the two outcomes be $+1$ and $-1$. In the present case, when
the system is a pair of one-half spins, the spin projections are measured
along pre-determined spatial directions. The possible result of such a
measurement is either $+\hbar/2$ (outcome $+1$) or $-\hbar/2$ (outcome $-1$).
The general form of such an inequality may be written as:

$\sum_{i=1}^{m_{A}}\sum_{j=1}^{m_{B}}M_{ij}\langle{a_{i}b_{j}}\rangle\leq L,$
(1)

where $m_{A}$ and $m_{B}$ are the number of measurement settings, while
$a_{i}=\pm 1$ and $b_{j}=\pm 1$ are the outcome of measurement $i$ and $j$ of
Alice and Bob, respectively. The $M$ is the Bell matrix whose real elements
are the Bell coefficients. The expectation value $\langle{a_{i}b_{j}}\rangle$
is $p(11|ij)+p(-1-1|ij)-p(1-1|ij)-p(-11|ij)$, that is it can be expressed as a
linear combination of $p(a_{i}b_{i}|ij)$ conditional probabilities. The value
$L$ is the maximum value the expression may take considering all possibilities
with deterministic outcomes (when the outcomes are the same in all rounds with
the same choices of settings), which is the maximum value allowed by any
locally realistic theory.

In quantum mechanics
$\langle{a_{i}b_{j}}\rangle=\mathrm{Tr}(\rho\hat{a}_{i}\otimes\hat{b}_{j})$ in
case of projective measurements, where $\hat{a}_{i}$ and $\hat{b}_{j}$ are the
Hermitian measurement operators with eigenvalues $\pm 1$ corresponding to the
$i$th and $j$th measurement of Alice and Bob, respectively, while $\rho$ is
the density operator of the system. For a pure state $|\psi\rangle$ this
becomes $\langle\psi|\hat{a}_{i}\otimes\hat{b}_{j}|\psi\rangle$. The maximal
quantum value, that is, the quantum bound, can always be achieved with a pure
state, in the case of a correlation-type Bell inequality with a maximally
entangled one [6, 19]. The quantum bound can be determined by semidefinite
programming applying the Navascués-Pironio-Acín (NPA) hierarchy of Ref. [20]
at its first level.

For a qubit the most general non-degenerate measurement operator is
$\alpha_{x}\hat{\sigma}_{x}+\alpha_{y}\hat{\sigma}_{y}+\alpha_{z}\hat{\sigma}_{z}\equiv\vec{\alpha}\cdot\vec{\sigma}$,
where $\hat{\sigma}_{x}$, $\hat{\sigma}_{y}$ and $\hat{\sigma}_{z}$ are the
Pauli-operators. If the qubit corresponds to a spin, then the unit vector
$\vec{\alpha}$ points towards the direction in the physical space along which
the spin projection is measured. It can be shown (for example [4]) that if the
measurement by Alice and Bob is performed on a pair of Bell state
$|\Phi^{+}\rangle=(|00\rangle+|11\rangle)/\sqrt{2}$ along $\vec{\alpha}$ and
$\vec{\beta}$, respectively, then the expectation value of the product of the
measurement outcomes
$\langle\Phi^{+}|(\vec{\alpha}\vec{\sigma})\otimes(\vec{\beta}\vec{\sigma})|\Phi^{+}\rangle=\vec{\alpha}\cdot\vec{\beta}^{*}$,
where $\beta^{*}_{x}\equiv\beta_{x}$, $\beta^{*}_{y}\equiv-\beta_{y}$ and
$\beta^{*}_{z}\equiv\beta_{z}$. Therefore, the quantum value of the Bell
inequality with the $|\Phi^{+}\rangle$ state is:

$Q=\sum_{i=1}^{m_{A}}\sum_{j=1}^{m_{B}}M_{ij}\vec{a}_{i}\cdot\vec{b}_{j},$ (2)

where $\vec{a}_{i}$ is Alice’s $i$th measurement direction, while
$\vec{b}_{j}$ is Bob’s $j$th measurement direction reflected through the $x-z$
plane.

The systematic method to construct the desired Bell inequalities Tavakoli and
Gisin [4] proposed was to take $M_{ij}=\vec{A}_{i}\cdot\vec{B}_{j}$, where
$\vec{A}_{i}$ and $\vec{B}^{*}_{j}$ are the unit vectors pointing towards the
desired measurement directions in Alice’s, and Bob’s side, respectively,
namely towards vertices of Platonic bodies. Then the quantum value for the
resulting Bell inequality with these measurement directions is
$\sum_{ij}(\vec{A}_{i}\cdot\vec{B}_{j})^{2}$. Although all terms in the
expression are positive, it itself does not ensure that the desired
measurement directions are really optimal. For a general choice of vectors
$\vec{A}_{i}$ and $\vec{B}_{j}$ this is usually not true. However, the authors
checked all pairs of Platonic solids, although only with one relative
orientation for each pair, by calculating the quantum bound using the NPA
method [20], and found that the desired measurement directions were optimal
for all of cases they checked. We will show here that all Platonic solids have
certain property that ensures this, moreover, the form of the quantum bound is
very simple and independent of the relative orientation of the two bodies.
However, the local bound, and hence the maximum violation of the inequality
does depend on this. We will discuss this dependence in the present paper.

Tavakoli and Gisin [4] have considered only the classical, three-dimensional
Platonic solids. The objects corresponding to the Platonic solids in two
dimensions are the regular convex polygons. There is an infinite number of
them. If the system is a pair of one-half spins, the Bell inequality, which is
constructed according to Ref. [4] from two such polygons in the same plane,
will be maximally violated by measurements towards the vertices of the
polygons. We will explore this generalization in Section 6. There are
generalizations of the Platonic solids in higher dimensions as well. In four
dimensions there are six such objects, while in each higher dimensional space
there are only three of them. Due to a theorem by Tsirelson [6], for any
Euclidean space there exists a Hilbert space $H$ and a state $|\Phi\rangle$ in
$H\otimes H$ such that for all unit vectors $\vec{a}$ and $\vec{b}$ in the
Euclidean space there exist measurement operators $\hat{a}$ and $\hat{b}$ in
$H$ such that
$\vec{a}\cdot\vec{b}=\langle\Phi|\hat{a}\otimes\hat{b}|\Phi\rangle$. For a
three dimensional Euclidean space the Hilbert space is the qubit space and the
operators are $\vec{a}\cdot\vec{\sigma}$ and $\vec{b}^{*}\cdot\vec{\sigma}$.
In Appendix A we discuss the relationship between the Euclidean and Hilbert
space description. The considerations there are based on the explicit
construction for all dimensions given for example in Refs. [19, 21]. From the
existence of such a construction, it follows that for any set of $\vec{a}_{i}$
and $\vec{b}_{j}$ vectors in any Euclidean space, the expression
$\sum_{ij}M_{ij}\vec{a}_{i}\cdot\vec{b}_{j}$ will give the quantum value of
the Bell expression with certain measurements performed on certain bipartite
quantum system. This means that the generalization of the construction of the
Bell inequalities to higher dimensional spaces does make sense, the result
will correspond to a genuine Bell scenario. We will call the $\vec{a}_{i}$ and
$\vec{b}_{j}$ the measurement vectors.

Unfortunately, the beauty of those measurement settings corresponding to
highly symmetric arrangements in the physical space is necessarily lost in
more than three dimensions. At the same time, in experimental realizations of
Bell scenarios the quantum systems are typically not pairs of one-half spins
even if they are pairs of qubits. In that case symmetry would only show up in
an abstract space anyway, just like in the generalization to higher
dimensions.

In more than two dimensions some Platonic solids have too many vertices, such
that the calculation of the local bound of Bell inequalities constructed from
them in the straightforward way may not be feasible. We have found a method
which makes it possible to calculate the local bound exactly whenever the Bell
coefficients are given as scalar products of low dimensional Euclidean
vectors, even if the number of measurement settings is large. As the vectors
need not be unit vectors in this case, the method may be useful whenever the
rank of the Bell matrix of a correlation-type inequality is low. In addition,
we show that the computation of the local bound can be done efficiently for
any bipartite two-outcome Bell inequality with low Bell matrix rank.

##  3 Platonic solids in $d$ dimensions

In this section we briefly summarize some well-known facts about Platonic
bodies [1]. All this knowledge can even be found on Wikipedia.

For $d>4$ there are just three types of Platonic solids: the simplex, the
cross polytope and the hypercube. A simplex has $d+1$ vertices, and if the
vectors pointing towards those vertices are normalized, the scalar product of
each pair of them is $-1/d$. A cross polytope has $2d$ vertices. In its
standard orientation, which is the most convenient one, these vertices are
given by $\vec{e}_{i}$ and $-\vec{e}_{i}$, where $\vec{e}_{i}$ are the
coordinate vectors. The $d$-dimensional hypercube or simply $d$-cube has
$m=2^{d}$ vertices $\vec{V}_{i}$, whose each coordinate $V_{ij}$ in the
standard orientation is either $+1/\sqrt{d}$ or $-1/\sqrt{d}$, such that all
variations of the signs are covered. Cross polytopes and hypercubes are
centrally symmetric, while simplices are not.

In two dimensions there are an infinite number of Platonic bodies, they are
the regular convex polygons. An $n$-sided regular polygon is centrally
symmetric if and only if $n$ is even. From our point of view if $n$ is odd,
the $n$-sided and the $2n$-sided polygons behave very similarly. The two-
dimensional simplex is the regular triangle, while both the cross polytope and
the 2-cube is the square, their standard orientation differs by an angle of
$\pi/4$.

In three dimensions the simplex is the tetrahedron, the cross polytope is the
octahedron, while the 3-cube is simply the cube. If we reflect the vertices of
the tetrahedron to its center and add the reflected points as new vertices, we
get a cube. As we will see later, for this reason the two bodies will behave
equivalently from our point of view. There are two additional three-
dimensional Platonic bodies, the icosahedron and the dodecahedron having $12$
and $20$ vertices, respectively. Both are centrally symmetric.

In four dimensions the simplex is the pentachoron, but it is also referred to
as 5-cell, pentatope, pentahedroid, or tetrahedral pyramid. The cross polytope
is the hexdecahedroid, also called 16-cell or hexadecachoron, while the 4-cube
is the tesseract, whose alternative names are 8-cell, octachoron, octahedroid,
cubic prism, and tetracube. There are three additional Platonic bodies in
$d$=4: the octaplex (24-cell, icositetrachoron, icosatetrahedroid, octacube,
hyper-diamond, polyoctahedron), the tetraplex (600-cell, hexacosichoron,
hexacosihedroid, polytetrahedron) and the dodecaplex (120-cell
hyperdodecahedron, polydodecahedron, hecatonicosachoron, dodecacontachoron,
hecatonicosahedroid) with 24, 120 and 600 vertices, respectively. These three
bodies are centrally symmetric. We note that in the name $n$-cell for each
$4$-dimensional body $n$ refers to the number of facets, while for us the
number of vertices is relevant.

##  4 Quantum bound for the special cases

Observation: The set of unit vectors $\vec{V}_{i}$ pointing towards the
vertices of a Platonic solid has the property that the columns of the matrix
whose elements $V_{ij}\equiv(\vec{V}_{i})_{j}$ are the components of the
vectors are orthogonal to each other and have equal norm, that is
$\sum_{i=1}^{m}V_{ij}V_{ik}=\delta_{jk}m/d$, where $m$ is the number of
vertices, $d$ is the dimensionality of the space and $j,k=1,\dots d$. In other
words, matrix $V$ is proportional to a semiorthogonal matrix.

The value of the norm of the columns follows from the unit lengths of
$\vec{V}_{i}$. If $\vec{V}_{i}$ has the property, then

$\sum_{i=1}^{m}(\vec{V}_{i}\cdot\vec{x})\vec{V}_{i}=\sum_{i=1}^{m}\sum_{j,k=1}^{d}(V_{ij}x_{j})(V_{ik}\vec{e}_{k})=\frac{m}{d}\vec{x},$
(3)

where $\vec{x}$ is any vector in the $d$-dimensional space, and $\vec{e}_{k}$
are the basis vectors. From this it follows that if $\vec{y}$ is also a
$d$-dimensional vector, then

$\sum_{i=1}^{m}(\vec{V}_{i}\cdot\vec{x})(\vec{V}_{i}\cdot\vec{y})=\frac{m}{d}\vec{x}\cdot\vec{y}.$
(4)

From the equation above it is clear that the property is independent of the
choice of the basis vectors ($\vec{x}$ and $\vec{y}$ may be members of another
basis), therefore, the property is invariant to any orthogonal transformation.

Now let the Bell coefficients be $M_{ij}=\vec{A}_{i}\cdot\vec{B}_{j}$
($i=1,\dots,m_{A}$; $j=1,\dots,m_{B}$). We will refer to $\vec{A}_{i}$ and
$\vec{B}_{j}$ as the construction vectors. Furthermore, let either
$\vec{A}_{i}$ or $\vec{B}_{j}$ obey the constraint given in the Observation.
Then the quantum value one gets when the measurement vectors are the same as
the respective construction vectors, i.e. $\vec{a}_{i}=\vec{A}_{i}$ and
$\vec{b}_{i}=\vec{B}_{i}$ is

$\sum_{i=1}^{m_{A}}\sum_{j=1}^{m_{B}}(\vec{A}_{i}\cdot\vec{B}_{j})(\vec{A}_{i}\cdot\vec{B}_{j})=\frac{m_{A}m_{B}}{d},$
(5)

which readily follows from Eq. (4) and from the fact that $\vec{A}_{i}$ and
$\vec{B}_{j}$ are unit vectors. It is trivial that the quantum value
calculated this way does not depend on the relative orientation of the two
sets of vectors, it is invariant to orthogonal transformations applied on
either side. The same formula has been obtained for three dimensions
independently in Ref. [7]. To derive it, the authors assumed that the vectors
on both sides correspond to orbits of the same group generated by the same
three-dimensional irreducible representation. Eqs. (15) and (16) of the paper
show that if the vectors generated this way, they will obey the constraint
given in the Observation on both sides, which is unnecessary. As we have
shown, if the condition holds on either side, then Eq. 5 is true. However,
this quantum value will usually be below the quantum bound unless the
constraint is obeyed on both sides, in which case the quantum bound is
reached, as we will prove later. If the vectors are generated as orbits, as in
Ref. [7], it is not necessary to use the same representation of the same group
on both sides; the quantum bound will be given by the simple formula even if
different groups are chosen.

The quantum values given by Tavakoli and Gisin [4] for 3-dimensional Platonic
solids all agree with $m_{A}m_{B}/3$ given above. They have shown for each
pairs of solids that taking the measurement vectors equal to the construction
vectors is actually an optimum choice, which gives the maximum quantum value.
For that they used the NPA hierarchy [20]. Proof was necessary, because
although such a choice must clearly be a good one, as each term in the Bell-
expression gives a positive contribution, but actually this is usually not the
optimum one if the construction vectors are chosen differently.

Now we will show that if both $\vec{A}_{i}$ and $\vec{B}_{i}$ obey the
constraint, this choice of the measurement vectors is optimal by proving that
the quantum value agrees with the upper bound derived by Epping et al. in Ref.
[22] for two-party binary-outcome correlation-type Bell inequalities. The
bound is given as $\sqrt{m_{A}m_{B}}||g||_{2}$, where $||g||_{2}$ is the
maximal singular value of the Bell matrix. The Bell matrix can be rewritten
as:

$M_{ij}=\vec{A}_{i}\cdot\vec{B}_{j}=\sum_{k=1}^{d}A_{ik}B^{T}_{kj}=\sum_{k=1}^{d}\bar{A}_{ik}S_{kk}\bar{B}^{T}_{kj},$
(6)

where $A_{ik}\equiv(\vec{A}_{i})_{k}$, $B_{jk}\equiv(\vec{B}_{j})_{k}$,
$\bar{A}_{ik}\equiv\sqrt{d/m_{A}}A_{ik}$,
$\bar{B}_{jk}\equiv\sqrt{d/m_{B}}B_{jk}$ and $S$ is a diagonal matrix with
nonzero elements $S_{kk}=\sqrt{m_{A}m_{B}}/d$. The constraint for
$\vec{A}_{i}$ and $\vec{B}_{i}$ means that matrices $\bar{A}$ and $\bar{B}$
are semiorthogonal. We may extend $\bar{A}$ and $\bar{B}$ with further
columns, which are normalized, orthogonal to the original columns and to each
other to get orthogonal matrices, and also we may extend matrix $S$ by zero
elements to size $m_{A}\times m_{B}$. Let us denote the respective extended
matrices by $\tilde{A}$, $\tilde{B}$ and $\tilde{S}$. Then

$M_{ij}=\sum_{k=1}^{m_{A}}\sum_{l=1}^{m_{B}}\tilde{A}_{ik}\tilde{S}_{kl}\tilde{B}_{lj}^{T}.$
(7)

The equation above is nothing else than a singular value decomposition of the
Bell-matrix. The singular values are the diagonal elements of $\tilde{S}$,
that is $\sqrt{m_{A}m_{B}}/d$ and zero. Therefore,
$||g||_{2}=\sqrt{m_{A}m_{B}}/d$, and the upper bound for the quantum value is
$m_{A}m_{B}/d$, which is achievable with the choice $\vec{a}_{i}=\vec{A}_{i}$
and $\vec{b}_{i}=\vec{B}_{i}$ (see Eq. 5). Therefore, Bell inequalities
constructed this way from vectors obeying the constraint in the Observation
are special cases of the ones discussed in Ref. [23]. Actually, Eq. (6) is
just the truncated singular value decomposition referred to in Theorem 1 in
Ref. [23], and the matrix denoted by $\alpha$ there, whose existence is
required, is $\sqrt{m_{A}/d}\delta_{ij}$, which leads to the optimum
measurement vectors given above, indeed.

In Ref. [23] the authors have also discussed modifications of the inequalities
leaving the upper bound for the maximum quantum value achievable. One is
twisting (Corollary 1i), which is nothing else for the present construction
than changing the relative orientation of sets $\vec{A}$ and $\vec{B}$. The
maximum quantum value remains the same, which is evident from the formula
giving it, as we have already noted. As the local value depends on it, we will
look for the optimum relative orientation minimizing the local value,
consequently, maximizing the quantum violation. The other modification changes
the singular values. If it is done appropriately as prescribed in Corollary
1ii, the bound for the maximum quantum value will change in a well-defined way
while it remains achievable. We will use this modification only in a very
special way and for a case when $\vec{A}_{i}=\vec{B}_{i}$. We subtract
$\lambda$, the same number from all diagonal elements of $\tilde{S}$. It is
allowed in this case if $\lambda<m_{A}/2d$. Then the modified Bell matrix will
be:

$\displaystyle M^{\prime}_{ij}$
$\displaystyle=\sum_{k}^{m_{A}}\tilde{A}_{ik}(\tilde{S}_{kk}-\lambda)\tilde{A}^{T}_{kj}$
$\displaystyle=\sum_{k=1}^{d}\bar{A}_{ik}S_{kk}\bar{A}^{T}_{kj}-\lambda\sum_{k=1}^{m_{A}}\tilde{A}_{ik}\tilde{A}^{T}_{kj}=M_{ij}-\lambda\delta_{ij}.$
(8)

Here we have used that $\tilde{A}_{ik}=\bar{A}_{ik}$ and
$\tilde{S}_{kk}=S_{kk}$ if $k\leq d$, and Eq. (6). The modification of the
Bell matrix is very simple, the same number is subtracted from each diagonal
element. The maximum quantum value becomes $m_{A}^{2}/d-\lambda m_{A}$ and it
can be achieved with measurement vectors
$\vec{a}_{i}=\vec{b}_{i}=\vec{A}_{i}$, the same ones as without the
modification. As it is easy to show that the local value decreases by the same
$\lambda m_{A}$ amount than the quantum maximum until the optimal local
strategy does not change, if the original inequality is violated, the
modification increases the violation.

We have shown above that $\vec{a}_{i}=\vec{A}_{i}$ and
$\vec{b}_{j}=\vec{B}_{j}$ is an optimal choice for the measurement vectors if
both sets of construction vectors obey the constraint of the Observation.
However this is not the only optimal choice. For example, the quantum value is
obviously the same if $\vec{a}_{i}=\hat{O}\vec{A}_{i}$ and
$\vec{b}_{j}=\hat{O}\vec{B}_{j}$, where $\hat{O}$ is an orthogonal
transformation. There may be less trivial examples, it may even happen that
measurement vectors in a lower dimensional subspace also lead to the maximum
quantum value.

For the five three-dimensional and for the six four-dimensional Platonic
solids the Observation may easily be verified explicitly. For more than four
dimensional spaces there are only three kinds of Platonic solids, the cross
polytope, the simplex and the hypercube (octahedron, tetrahedron and cube in
three dimensions, respectively). In Appendix B we prove that the Observation
is true for these bodies in all dimensions. We also prove that it is also true
for all convex polygons, which are the Platonic solids of two dimensions. The
13 Archimedean solids in three dimensions also obey the property, as it can
easily be checked.

The validity of the Observation is certainly connected to some symmetry
properties, therefore, it must be valid for many other symmetric objects. The
Observation involves $d(d+1)/2-1$ constraints. Finding the same number of
independent constraints for a set of unit vectors which can be derived from
the original ones is enough to show that the set has the property, with all of
its consequences. Such constraints can be found in case of symmetric objects.
For any $\hat{O}$ symmetry of $\hat{V}_{i}$ (that is when the set
$\hat{O}\vec{V}_{i}$ is the same as $\hat{V}_{i}$), it is true that
$\sum_{i=1}^{m}(\vec{V}_{i}\cdot\vec{x})(\vec{V}_{i}\cdot\vec{y})=\sum_{i=1}^{m}(\vec{V}_{i}\cdot\hat{O}^{-1}\vec{x})(\vec{V}_{i}\cdot\hat{O}^{-1}\vec{y})$.
From Eq. (33) in Appendix B we also show that if the vectors have an $l$-fold
symmetry for rotations in some plane, then for all vectors $\vec{x}$ in that
plane the value of the sum
$\sum_{i=1}^{m}(\vec{V}_{i}\cdot\vec{x})(\vec{V}_{i}\cdot\vec{x})$ are the
same, and for all orthogonal pairs of vectors $\vec{x}$ and $\vec{y}$ in that
plane $\sum_{i=1}^{m}(\vec{V}_{i}\cdot\vec{y})(\vec{V}_{i}\cdot\vec{x})=0$. If
there are several such planes and possibly some other symmetries, it may be
enough to ensure that all constraints are satisfied.

It is easy to see that if the set $\vec{V}_{i}$ is centrally symmetric and
obeys the constraints of the Observation, then the subset of $\vec{V}_{i}$
containing only one of each pairs pointing towards opposite directions also
obeys the constraints (the scalar products of the columns are just one half of
the scalar products corresponding to the original set).

##  5 The local bound

In this paper we are dealing with Bell inequalities whose coefficients are
given as scalar products of $d$-dimensional unit vectors from two sets. The
considerations of this section do not require that the sets are normalized,
which means that they are valid for any correlation-type bipartite Bell
inequality, as the Bell coefficients can always be written as scalar products
of two non-normalized sets of vectors using the singular value decomposition
of the Bell matrix. The dimensionality of the vectors is equal to the rank of
the Bell matrix. We further extend the validity of our analysis to generic
non-correlation-type Bell inequalities.

Normally, the computation time required for the exact calculation of the local
bound grows exponentially with the number of measurement settings. With the
method proposed in this section the dependence becomes polynomial whose degree
is the rank of the Bell matrix. Therefore, if this rank is considerably lower
than the number of settings on each side, a lot of computation effort can be
saved. Unfortunately, if the rank is not low enough, the computation effort
required still grows fast with the number of settings. Nevertheless, for low
ranks and not very many settings the method makes exact calculations of local
bounds affordable that would be intractable otherwise.

The local bound for any correlation-type Bell inequality can be determined as:

$L=\max\limits_{a_{i},b_{j}=\pm
1}\sum_{i=1}^{m_{A}}\sum_{j=1}^{m_{B}}a_{i}b_{j}(\vec{A}_{i}\cdot\vec{B}_{j}),$
(9)

where $a_{i}$ and $b_{j}$ are the respective components of the local
strategies of Alice and Bob, and $\vec{A}_{i}$ and $\vec{B}_{j}$ are not
necessarily normalized Euclidean vectors. The evaluation of the expression
would involve to check $2^{m_{A}+m_{B}}$ strategies. This can be reduced by
noticing that for any strategy of Bob there is an optimal strategy for Alice,
namely:

$a_{i}={\text{sgn}}\left(\vec{A}_{i}\cdot\sum_{j=1}^{m_{B}}b_{j}\vec{B}_{j}\right).$
(10)

Using this strategy for Alice, the local bound becomes:

$L=\max\limits_{b_{j}=\pm
1}\sum_{i=1}^{m_{A}}\sum_{j=1}^{m_{B}}|\vec{A}_{i}\cdot b_{j}\vec{B}_{j}|.$
(11)

Taking into account that strategies with components $b_{j}$ and $-b_{j}$ give
the same value, the evaluation of $2^{m_{B}-1}$ expressions is sufficient (for
example, one can consider only strategies with $b_{1}=+1$). Obviously, if
$m_{A}<m_{B}$, one can use an analogous formula and end up with $2^{m_{A}-1}$
evaluations. However, the present construction of the Bell coefficients may
lead to further and very significant reduction of the number of evaluations.
Instead of an exponential dependence on the number of settings we get a
polynomial dependence whose degree is $d-1$.

Furthermore, our result for bipartite correlation-type inequalities can be
extended to generic bipartite two-outcome inequalities. Indeed, in this case
we can write the Bell inequality

$\sum_{i=1}^{m_{A}}M^{A}_{i}\langle{a_{i}}\rangle+\sum_{j=1}^{m_{B}}M^{B}_{j}\langle{b_{j}}\rangle+\sum_{i=1}^{m_{A}}\sum_{j=1}^{m_{B}}M_{ij}\langle{a_{i}b_{j}}\rangle\leq
L.$ (12)

We show that the calculation of the local bound above can be traced back to
the calculation of the local bound of a correlation-type Bell inequality (1).
In particular, let us define the following $(m_{A}+1)\times(m_{B}+1)$-setting
correlation-type inequality:

$\displaystyle M_{m_{A}+1,m_{B}+1}\langle
a_{m_{A}+1}b_{m_{B}+1}\rangle+\sum_{i=1}^{m_{A}}M^{A}_{i}\langle{a_{i}}b_{m_{B}+1}\rangle$
$\displaystyle+\sum_{j=1}^{m_{B}}M^{B}_{j}\langle{a_{m_{A}+1}b_{j}}\rangle+\sum_{i=1}^{m_{A}}\sum_{j=1}^{m_{B}}M_{ij}\langle{a_{i}b_{j}}\rangle\leq
L^{\prime}.$ (13)

Let us further choose the constant

$M_{m_{A}+1,m_{B}+1}=\sum_{i=1}^{m_{A}}|M^{A}_{i}|+\sum_{j=1}^{m_{B}}|M^{B}_{j}|+\sum_{i=1}^{m_{A}}\sum_{j=1}^{m_{B}}|M_{ij}|.$
(14)

In this case, one can show that the local bound $L^{\prime}$ above is obtained
with a deterministic strategy, where $a_{m_{A}+1}b_{m_{B}+1}=1$. Then the
local bound $L$ in (12) can be obtained by a simple subtraction from the local
bound $L^{\prime}$ in (13): $L=L^{\prime}-M_{m_{A}+1,m_{B}+1}$. On the other
hand, the Bell matrix in Eq. (13) can have only one more rank than the Bell
matrix in Eq. (12). This follows from the definition of the matrix rank given
by the linearly independent number of rows or columns of the matrix. Thus, we
proved that the local bound $L$ of a generic two-outcome bipartite Bell
inequality (12) can be reduced to the calculation of the local bound
$L^{\prime}$ of the correlation-type Bell inequality (13). Using the following
algorithm to compute the local bound of a correlation-type Bell inequality
with small matrix rank $M$, however, we obtain an efficient algorithm for
computating the local bound of any two-outcome bipartite Bell inequality when
the matrix rank $M$ is much lower than the number of settings $m_{A}$ and
$m_{B}$.

Let us call strategy $a_{i}$ a geometrical strategy for Alice if there exists
some vector $\vec{q}$ such that $a_{i}={\text{sgn}}(\vec{A}_{i}\cdot\vec{q})$.
Looking at Eq. (10) it is clear that all strategies of Alice that are
potentially optimal are geometrical ones. From the analogous expression giving
the optimal $b_{j}$ for $a_{i}$ follows that the same is true for the
potentially optimal strategies of Bob. Therefore, it is enough to check only
the geometrical strategies when calculating the local bound. From Eq. (11) it
follows that applying a geometrical strategy of Bob means multiplying
$\vec{B}_{i}$ by $+1$ or $-1$ depending on which side of a hyperplane (the
normal plane of $\vec{q}$) it is, such that $b_{i}\vec{B}_{i}$ are all in the
same side.

Now we give a recipe to determine all geometrical strategies for a set of
vectors $\vec{A}_{i}$. We may take $\vec{q}$ to be a unit vector, so that it
may be thought of as a point on the surface of the unit sphere. It is clear
that a small change in $\vec{q}$ usually will not change the geometrical
strategy it defines: there is a whole region on the surface of the unit sphere
giving the same geometrical strategy. If we move around on the surface and
cross the normal hyperplanes of one of the $\vec{A}_{i}$ vectors, than $a_{i}$
will change sign. Until such does not happen, we remain in the same region.
Therefore, the borders of the regions corresponding to geometrical strategies
are the intersections of the normal hyperplanes of $\vec{A}_{i}$ with the unit
sphere, which are great (hyper)circles of the (hyper)sphere.

The question is how to find all regions, or at least the geometrical
strategies they define. In two dimensions the problem is easy. The unit sphere
is a circle and its intersections with the normal planes (now just lines),
that is the borders of the regions are points which are easy to find. There
are $2m_{A}$ such points, consequently the same number of regions and
geometrical strategies (if no two vectors point to the same or opposite
directions).

Let us consider the three-dimensional case, which is not difficult to
visualize and which can be generalized to higher dimensions. First let us
discuss the number of geometrical strategies, that is the number of regions on
the surface of the sphere whose borders are the great circles defined by the
normal planes of $\vec{A}_{i}$. If there is only one vector, then the one
great circle divides the surface into two regions. For two vectors, if they do
not point to the same or to the opposite directions, the number of regions is
four. If we add a third vector such that only two great circles cross each
other in all intersections, that is there are no multiple crossings, the
number of regions will be doubled again to eight. There is no gain so far, all
strategies are geometrical ones. However, when a fourth vector is added, the
number of regions grows only by six instead of eight, and the $(i+1)$th vector
will increase this number only by $2i$ instead of $2^{i}$. The reason is
simple. The new great circle crosses each existing one twice. Between each
consequtive crossings one region is cut into two parts. Since there are $2i$
new crossings, the number of regions grows by this amount. It is easy to prove
by induction that if there are $m_{A}$ vectors, then the number of regions is
$m_{A}(m_{A}-1)+2$. If there are multiple crossings, the number of regions
will be smaller: if two crossings occur in the same place, there is no region
cut between them. For symmetric arrangements, like the Platonic bodies,
multiple crossings occur. Although a smaller number of geometric strategies is
advantageous in itself, unfortunately the determination of them is more
difficult, as we will see later.

Our aim is to determine all geometrical strategies. To do that let us
concentrate on the intersections, which are the corners of the regions. Each
intersection is in a direction orthogonal to two vectors. All intersections
can be found in the directions of vectors $\pm\vec{A}_{k}\times\vec{A}_{l}$
taken for all pairs ($k,l=1,\dots,m_{A}$). We have supposed that there are no
pairs of vectors pointing towards opposite directions. Let us also suppose
that there are no multiple crossings. Then each intersection is on the border
of four regions. For the four strategies corresponding to these regions each
$a_{i}$ for $i\neq k;i\neq l$ are the same for all four regions and they are
given as the sign of the scalar product of $\vec{A}_{i}$ with the vector
pointing towards the intersection (these are well defined in case of no
multiple intersections), while for $i=k$ or $i=l$, for which this scalar
product is zero, the four variations of $+1$ and $-1$ has to be taken. This
way we get all geometrical strategies, but in multiple copies, each strategy
will occur in our list as many times as the number of corners of its region.
Therefore, the last step is to get rid of the multiple copies.

If $a_{i}$ represents a geometrical strategy so does $-a_{i}$. One can save
some effort when applying the algorithm above by looking only for one from
each such pair. Therefore, it is better to take only one of the intersections
from the pair in opposite directions (say $+\vec{A}_{k}\times\vec{A}_{l}$),
and to reverse all signs whenever a strategy derived has $a_{1}=-1$.

The algorithm may be extended to dimensions $d>3$. Now the corners of the
regions, that is the points to find are at the intersections of $d-1$
hyperplanes with the hypersphere. They are in directions which are orthogonal
to $d-1$ vectors. This is towards the null space of the matrix given by those
$d-1$ vectors, provided the null space is one-dimensional (if not, those
vectors do not define a corner). One may find all corners by taking all
combinations of $d-1$ vectors out of the $m_{A}$ ones. Each such point now is
on the border of $2^{d-1}$ regions. The strategies corresponding to these
regions can be determined similarly to the three-dimensional case. For $d=4$
the number of geometrical strategies is $m_{A}(m_{A}-1)(m_{A}-2)/3+2m_{A}$. We
note that if $m_{A}\leq d$ all strategies are geometrical ones if
$\vec{A}_{i}$ are linearly independent.

So far, we supposed that there are no pairs of vectors pointing towards
opposite directions and no multiple crossings, meaning that no more than the
minimum number of great hypercircles to define a point intersect in each
point. For randomly chosen $\vec{A}_{i}$ this is true. Unfortunately in the
case of Platonic solids, and when $\vec{A}_{i}$ shows some symmetries, it is
not the case. If $\vec{A}_{i}=-\vec{A}_{j}$ for some $i,j$ then in all
geometric strategies it is obviously true that $a_{i}=-a_{j}$. Therefore, we
may eliminate one of them and reduce the number of settings $m_{A}$ by one by
dropping one of the vectors and doubling the length of the other one. We note
that in this part we have not used that $A_{i}$ are normalized, all
considerations are true even if their lengths are different. For $d>2$ all
Platonic bodies are centrally symmetric except for the simplices, therefore
one half of the vertices can be eliminated. We may even leave the vectors
normalized and just multiply the local bound by two at the end.

The most obvious way to treat the problem of multiple crossings is to take all
variations of $+1$ and $-1$ for the undetermined strategy components, that is
if $\nu$ great hypercycles intersect, create $2^{\nu}$ strategies. However,
this way we would create spurious strategies. Let us look at the three-
dimensional case. If there is a $\nu$-fold intersection, we would create
$2^{\nu}$ strategies, while the intersection is on the border of only $2\nu$
regions. This recipe may even be intractable. For the four dimensional
Platonic body that has the most vertices (the dodecaplex) even 30-fold
intersections occur. It would probably be possible to work out an algorithm to
identify the truely existing regions, but it does not seem obvious. We have
chosen a different approach. We can dissolve the degeneracies of the
intersections by applying small perturbations to vectors $\vec{A}_{i}$. This
method will also introduce spurious strategies, but not as many as the one
discussed above. If the perturbations are small enough, the desired strategies
will always be found. The number of all strategies created this way, including
the spurious ones, will be the same as we would get for random $\vec{A}_{i}$.
The spurious strategies we get depend on the perturbation applied. Therefore,
we may get rid of them by repeating the calculation with different
perturbations several times and keeping the strategies present in the results
of all runs.

For the dodecaplex, the four dimensional Platonic solid that has 600 vertices,
the total number of strategies to consider is $2^{599}$, which can be reduced
to $2^{299}$ by eliminating half of the vertices, as explained earlier. For
the same object the number of geometric strategies to be checked is just
1787760.

For further considerations it is useful to introduce the notion of strategy
vectors that we define as $\sum_{i=1}^{m_{A}}a_{i}\vec{A}_{i}$ and
$\sum_{j=1}^{m_{B}}b_{i}\vec{B}_{i}$ for Alice and Bob, respectively, where
$a_{i}$ and $b_{j}$ are geometric strategies. The local bound in Eq. (9) may
be rewritten as:

$L=\max\limits_{a_{i},b_{j}{\text{geom.
strat.}}}\left(\sum_{i=1}^{m_{A}}a_{i}\vec{A}_{i}\right)\cdot\left(\sum_{j=1}^{m_{B}}b_{j}\vec{B}_{j}\right).$
(15)

This means that the local value is the largest of the scalar products of
Alice’s and Bob’s strategy vectors. If we vary the relative orientation of
sets $\vec{A}_{i}$ and $\vec{B}_{j}$ by applying orthogonal transformations to
one of them, when the longest strategy vectors overlap, the local value is
just the product of their length. This is obviously the maximum value of $L$
in any relative orientation. For symmetric arrangements, like in the case of
Platonic bodies, it may happen that there are many strategy vectors having the
same maximal length. In this case, if the dimensionality of the space is not
large, there will be no relative orientation with a local bound significantly
smaller than this maximum value, since there will always be a pair of such
vectors with a small angle between them.

##  6 The two dimensional case

The Platonic solids of two dimensions are the regular convex polygons. There
is an infinite number of them. Let $\vec{A}_{i}$ point towards the vertices of
an $m_{A}$-sided regular convex polygon. Then the lengths of all strategy
vectors will be the same. If $m_{A}$ is even, this value is
$l_{m_{A}}=2\sum_{k=1}^{m_{A}/2}\sin[(2k-1)\pi/m_{A}]$, and the vectors point
towards the midpoints of the sides if $m_{A}$ is divisible by four, and
towards the vertices, if not. If $m_{A}$ is odd, there are $2m_{A}$ strategy
vectors the same as for the $2m_{A}$-sided polygon, except that their length
is halved. Half of them point towards the vertices and half of them towards
the midpoints of the sides. The strategy vectors have length
$l_{m_{A}}=\sum_{k=1}^{m_{A}}\sin[(2k-1)\pi/2m_{A}]$.

If there are identically oriented identical polygons on both sides
($\vec{A}_{i}=\vec{B}_{i}$), the local bound of the Bell inequality
constructed is the square of the length of the strategy vector, while the
quantum bound with any relative orientation is $m_{A}^{2}/2$. Therefore, for
two squares the local bound is the same as the quantum bound, namely 8, so
there is no violation. However, if we rotate one of the squares by $\pi/4$ the
local bound is multiplied by $1/\sqrt{2}$ (the cosine of the largest angle
between pairs of strategy vecors from the two sides), therefore $Q/L$ becomes
$\sqrt{2}$ (the Bell inequality is like four CHSH ones [9] multiplied by
$1/\sqrt{2}$ or $-1/\sqrt{2}$). In case of two regular hexagons the maximum
local bound is $16$ (identical orientation) and the minimum one is
$8\sqrt{3}=13.856406$ ($\pi/6$ relative angle). As the quantum bound is 18,
the minimum $Q/L$ ratio is $9/8=1.125$, while the maximum one is
$9/(4\sqrt{3})=1.2990381$. The values are the same for regular triangles. If
there are two identical regular even-sided polygons, with one of them rotated
by angle $\varphi$, then for $-\pi/m\leq\varphi\leq\pi/m$ (where
$m=m_{A}=m_{B}$) the local bound is $l_{m}^{2}\cos\varphi$. At the ends of the
interval the function is minimal, and the pattern repeated $m$-times around
the circle. The ratio of the minimal and the maximal local bound is
$\cos(\pi/m)$. In Fig. 1 we show the $\varphi$-dependence of the local bound
for two squares and for two hexagons.

Figure 1: Dependence of the local bound on the relative orientation of the two
regular polygons used to construct the Bell inequality. The solid, the dash-
dotted and the dashed line correspond to two squares, two hexagons and a
square and a hexagon, respectively. At zero angle (at least) two vertices
align.

The behaviour is qualitatively the same for two non-identical even-sided
polygons, too, but the periodicity of the function is
$2\pi/\text{lcm}(m_{A},m_{B})$, where lcm denotes the least common multiple.
In Fig. 1 we also show the result for a square on one side and a hexagon on
the other side. The length of the period can be determined by finding out that
starting from a position when two strategy vectors overlap, at least how much
we have to rotate one of the polygons such that two other vectors overlap
again. Let the directions of the strategy vectors in the starting position be
given by angles $2\pi\mu/m_{A}$ ($\mu=0,\cdots,m_{A}-1$) and $2\pi\nu/m_{B}$
($\nu=0,\cdots,m_{B}-1$). Then the angles between the pairs are

$2\pi\left|\frac{\mu}{m_{A}}-\frac{\nu}{m_{B}}\right|=2\pi\frac{|\mu
m^{\prime}_{B}-\nu m^{\prime}_{A}|}{{\text{lcm}}(m_{A},m_{B})},$ (16)

where $m^{\prime}_{A}=m_{A}/{\text{gcd}}(m_{A},m_{B})$ and
$m^{\prime}_{B}=m_{B}/{\text{gcd}}(m_{A},m_{B})$, and gcd refers to the
greatest common divisor. As $m^{\prime}_{A}$ és $m^{\prime}_{B}$ are relative
primes, the minimal nonzero value of the numerator on the right hand side is
one, therefore, the smallest nonzero value of the expression is
$2\pi/\text{lcm}(m_{A},m_{B})$, indeed, which is the period of the function we
are looking for. The ratio between the minimal and the maximal local bound is
$\cos(\pi/\text{lcm}(m_{A},m_{B}))$. For odd-sided polygons the behaviour is
the same as for the ones with twice as many sides, only the local bound and
the quantum value has to be divided by two.

For large $m_{A}$ the length of the strategy vectors $l(m_{A})$ converges to
$2m_{A}/\pi$, because $l_{m_{A}}/m_{A}$ approaches the Riemann-sum of
$\int_{x=1}^{\pi}\sin(x)dx/\pi$. Therefore, $Q/L$ converges to
$\pi^{2}/8=1.2337006$ if both $m_{A}$ and $m_{B}$ tend to infinity. This is
obviously true for any relative orientations of the polygons. This value
recovers the $Q/L$ ratio of the inequality with a continuum infinite number of
settings corresponding to the special case $n=2$ in Refs. [24, 25].

##  7 Local strategies for $d\geq 3$

| $m$ | cs | $n_{\rm{sv}}$ | $n_{\rm{svg}}$ | $n_{\rm{svx}}$ | $m/\sqrt{3}/l_{x}$
---|---|---|---|---|---|---
tetrahedron | 4 | n | 14 | 2 | 6 | 1.0000000
octahedron | 6 | y | 8 | 1 | 8 | 1.0000000
cube | 8 | y | 14 | 2 | 6 | 1.0000000
icosahedron | 12 | y | 32 | 2 | 12 | 1.0704663
dodecahedron | 20 | y | 92 | 3 | 20 | 1.1026409
truncated tetrahedron | 12 | n | 134 | 5 | 8 | 1.1055416
cuboctahedron | 12 | y | 24 | 1 | 24 | 1.0954451
truncated cube | 24 | y | 134 | 6 | 6 | 1.0572365
truncated octahedron | 24 | y | 116 | 4 | 12 | 1.0954451
rhombicuboctahedron | 24 | y | 134 | 4 | 6 | 1.0978539
truncated cuboctahedron | 48 | y | 554 | 16 | 6 | 1.1084962
snub cube | 24 | n | 482 | 15 | 6 | 1.1074356
icosidodecahedron | 30 | y | 120 | 1 | 120 | 1.1387895
truncated dodecahedron | 60 | y | 872 | 12 | 20 | 1.1240927
truncated icosahedron | 60 | y | 872 | 12 | 12 | 1.1397231
rhombicosidodecahedron | 60 | y | 872 | 11 | 20 | 1.1414557
truncated icosidodecahedron | 120 | y | 3542 | 35 | 30 | 1.1387895
snub dodecahedron | 60 | n | 3422 | 37 | 30 | 1.1401438
Table 1: Properties of the strategy vectors for the three dimensional Platonic
and Archimedean solids. The number of vertices is $m$, cs shows if the solid
is centrally symmetric, $n_{\rm{sv}}$ is the number of strategy vectors,
$n_{\rm{svg}}$ is the number of groups of strategy vectors containing vectors
of the same lengths, $n_{\rm{svx}}$ is the size of the group containing the
longest strategy vectors of $l_{x}$ length. The $Q/L$ value for a Bell
inequality constructed from two solids in a relative orientation such that two
of their longest strategy vectors align is the product of the corresponding
$m/\sqrt{3}/l_{x}$ values shown in the last column. For any different relative
orientation $Q/L$ is larger.

First we discuss the three Platonic solids existing in all dimensions, the
cross polytope, the simplex and the hypercube. We note that in the present
context in $d=3$ the simplex (tetrahedron) and the cube are equivalent: one
can get the cube from the tetrahedron by reflecting its vertices to the
midpoint and adding the resulting points as new vertices. There is a one-to-
one correspondence between the geometrical strategies for the two bodies, in
the cube the component corresponding to the reflected vertex is minus one
times the value for its mirror image, therefore, all strategy vectors will be
the same only with double lengths. If used as construction vectors on one side
for a Bell inequality, both the classical and the quantum bound should simply
be multiplied by two. For $d>3$ there are no such equivalences.

From the present pont of view the simplest body is the cross polytope. As it
has been told before, it has $2d$ vertices, in the most convenient orientation
they are the basis vectors with both signs, that is $\vec{A}_{i}=\vec{e}_{i}$
and $\vec{A}_{i+d}=-\vec{e}_{i}$, ($i=1,\dots,d$). The geometrical strategy
defined by vector $\vec{q}$ is $a_{i}={\text{sgn}}(\vec{A}_{i}\cdot\vec{q})$,
which is $a_{i}=-a_{i+d}={\text{sgn}}(q_{i})$. The strategy vector
$\sum_{i=1}^{2d}a_{i}\vec{A}_{i}=2\sum_{i=1}^{d}\text{sgn}(q_{i})\vec{e}_{i}$.
All $\vec{q}$ whose every component has the same sign give the same strategy
vector. Therefore, each strategy vector can be written as
$2\sum_{i=1}^{d}\alpha_{i}\vec{e}_{i}$, where $\alpha_{i}$ takes the value of
$+1$ or $-1$, and all such combinations are to be taken to get all strategy
vectors. There are $2^{d}$ vectors, they point towards the vertices of a
hypercube and have the same length of $2\sqrt{d}=m_{A}/\sqrt{d}$. As the
quantum bound for a Bell inequality constructed from two identical bodies is
$m_{A}^{2}/d$, and the local bound is the square of the length of the longest
strategy vector if the orientation is the same in both sides, for two such
cross polytopes there is no violation.

A simplex has $d+1$ vertices. Each pair of vertices has the same scalar
product $\vec{A}_{i}\cdot\vec{A}_{j}=-1/d$, $i\neq j$. When constructing all
strategies, we now consider not the $\vec{q}_{i}$ vectors but their normal
hyperplanes. If such a plane cuts the $d$-dimensional space, any $1\leq k\leq
d$ of the $d+1$ vertices may get into one side and the rest into the other
side. Such a cut gives a strategy with $k$ plus one and $d+1-k$ minus one
components (from all strategies only the two strategies corresponding to all
signs equal are not geometrical ones). For a given $k$ there are
$\binom{d+1}{k}$ possibilities, corresponding to the same number of strategies
and strategy vectors of lengths $2\sqrt{k(d+1-k)/d}$. The reason is the
following. Due to the symmetry $\sum_{i=1}^{d+1}\vec{A}_{i}=0$, therefore, the
sum of the $k$ of them multiplied by one is the same as the sum of the $d+1-k$
of them multiplied by minus one. As
$(\sum_{i=1}^{k}\vec{A}_{i})\cdot(\sum_{j=1}^{k}\vec{A}_{j})=k-k(k-1)/d$ (any
$k$ vectors could have been used as the ones on the plus side), and the length
of the strategy vector is twice the square root of this value, what we get is
just what we have given above, indeed. The longest strategy vectors correspond
to the most even distribution of vectors, that is $k=(d+1)/2$ if $d$ is odd,
and $k=d/2$ and $k=d/2+1$ if $d$ is even, that is the maximum length is
$(d+1)/\sqrt{d}$ and $\sqrt{d+2}$ for odd and even $d$, respectively. When $d$
is odd (similarly to the cross polytope in all dimensions), the Bell
inequality constructed using two simplices of the same orientation can not be
violated (both the local and the quantum bound is $(d+1)^{2}/d$).

The $d$-dimensional hypercube, the $d$-cube has $m_{A}=2^{d}$ vertices
$\vec{A}_{i}$, whose each coordinate $A_{ij}$ in the most convenient
orientation is either $+1/\sqrt{d}$ or $-1/\sqrt{d}$, such that all variations
of the signs are covered. If $\hat{T}$ is an orthogonal transformation that
permutes the coordinates and change the sign of some of them, the set
$\hat{T}\vec{A}_{i}$ contains the same vectors as $\vec{A}_{i}$, only in a
different order. From this it follows that if $\vec{S}$ is a strategy vector,
so is $\hat{T}\vec{S}$. If $\vec{S}$ is a strategy vector there exists
$\vec{q}$ such that
$\vec{S}=\sum_{i=1}^{2^{d}}\text{sgn}(\vec{q}\cdot\vec{A}_{i})\vec{A}_{i}$.
Then the strategy vector belonging to $\hat{T}\vec{q}$ is
$\sum_{i=1}^{2^{d}}\text{sgn}(\hat{T}\vec{q}\cdot\vec{A}_{i})\vec{A}_{i}=\sum_{i=1}^{2^{d}}\text{sgn}(\hat{T}\vec{q}\cdot\hat{T}\vec{A}_{i})\hat{T}\vec{A}_{i}=\sum_{i=1}^{2^{d}}\text{sgn}(\vec{q}\cdot\vec{A}_{i})\hat{T}\vec{A}_{i}=\hat{T}\vec{S}$.
The first equality follows from the fact that sets $\vec{A}_{i}$ and
$\hat{T}\vec{A}_{i}$ are the same, while the second one from the orthogonality
of $\hat{T}$. From this it follows that the strategy vectors of the $d$-cube
belong to families, where the members of a family can be derived by taking all
permutations and all variations of the signs of the coordinates of one of its
members. For $d=3$ there are two families that may be represented by
$(0,0,8)/\sqrt{3}$ and $(4,4,4)/\sqrt{3}$ the members of the families point
towards the midpoints of the faces and the vertices, respectively. There are
altogether $6+8=14$ strategy vectors, each corresponding to a geometrical
strategy. For $d=4$ there are three families represented by $(0,0,0,8)$,
$(0,4,4,4)$ and $(2,2,2,6)$. The total number of strategy vectors is
$8+32+64=104$. For $d=5$ the number of families is $7$, while the number of
vectors is $10+80+320+160+960+320+32=1882$. The number of families for $d=6,7$
and $8$ are $21,135$ and $2470$, respectively, while for $d=9$ it is at least
$175428$. Unfortunately, we have not found a systematic way to get formulae
for all dimensions. However, it is true for all dimensions that the longest
strategy vectors point towards the midpoints of the facets, they are
$2^{d}\vec{e}_{i}/\sqrt{d}=\sum_{j=1}^{2^{d}}(\vec{e}_{i}\cdot\vec{A}_{j})\vec{A}_{j}$
(all components but the $i$th one add up to zero), which shows that they are
strategy vectors. As their length is $2^{d}/\sqrt{d}=m_{A}/\sqrt{d}$, the Bell
inequality constructed from two $d$-cubes of the same orientation can not be
violated. These strategy vectors are indeed the longest, since the local bound
cannot be greater than the quantum bound.

It is true for all cases we have considered so far except for simplices in
even dimensions that the length of the longest strategy vectors is
$m_{A}/\sqrt{d}$. Therefore, if a Bell inequality is constructed from any two
such objects, there exist relative orientations (when longest strategy vectors
are aligned), when the inequality cannot be violated. If one of the bodies is
a simplex in an even dimension, $Q/L$ is $(d+1)/\sqrt{d(d+2)}$ if the
orientation is such that the local bound is maximal, while it is the square of
this number if there are simplices in both sides.

For $d=3$ there are two further Platonic solids, the icosahedron and the
dodecahedron with 12 and 20 vertices, respectively, while for $d=4$ we have
the octaplex, the tetraplex and the dodecaplex with 24, 120 and 600 vertices,
respectively. We have calculated the geometrical strategies and the strategy
vectors using the algorithm presented in Section 5. All these bodies are
centrally symmetric, therefore when we applied the algorithm we eliminated one
half of their vertices.

In three dimensions for the icosahedron there are 32 geometrical strategies
and strategy vectors. The longest ones of length
$2\sqrt{6+2\sqrt{5}}=6.4721360$ point towards the 12 vertices, while the
shorter ones of lengths $2\sqrt{6+6/\sqrt{5}}=5.8934817$ point towards the
middle of the 20 faces (as the coordinates of the vertices are known
analytically, if we know a geometrical strategy, we may derive the
corresponding strategy vector analytically). If the Bell inequality is
constructed from two identical sets corresponding to the icosahedron, then the
local bound is $41.888544$, and as the maximum quantum value is $12^{2}/3=48$,
$Q/L=1.1458980$. For the dodecahedron there are $92$ geometric strategies. The
length of the $20$ longest strategy vectors is $10.472136$, and there are also
$60$ and $12$ further vectors of lengths $9.8863510$ and $9.8224695$. The
analytical forms are quite involved. For two identically oriented
dodecahedrons $L=109.66563$ and $Q/L=1.2158169$. Results for the strategy
vectors for all three dimensional Platonic solids and also for all Archimedean
solids are summarized in Table 1.

In four dimensions there are $192$ geometrical strategies for the octaplex,
and all strategy vectors have the same length of $4\sqrt{7}=10.583005$.
Therefore, for two identical bodies $L=112$ and $Q/L=144/112=1.2857143$. For
the tetraplex the $14400$ strategy vectors also have equal lengths of
$51.146605$, therefore for two such objects $L=2615.9752$ and $Q/L=1.3761599$.
For the dodecaplex there are $3575520$ strategy vectors. Surprisingly, they
can be arranged into $201$ groups of different lengths. One of the smallest
group of $1440$ members is the one containing the longest vectors of length
$255.71725$. Consequently, for two identical bodies $L=6539.1314$ and
$Q/L=1.3763296$. The shortest strategy vectors belong to one of the largest
groups of $129600$ members, and their length is $254.17151$. Although there
are many groups, the difference between the longest and shortest vectors is
just $0.6\%$.

As we have said, when two objects are oriented such that their longest
strategy vectors point into the same direction, the corresponding Bell
inequality has the maximum local bound. This can be reduced by changing the
relative orientation. If both objects obey the constraint given in the
Observation, reorientation will not change the quantum bound, therefore $Q/L$
will increase, like we have already discussed in the two-dimensional cases. We
explore in the next section how much improvement we can achieve this way for
$d>2$.

##  8 Minimal local bounds for $d\geq 3$

To find relative orientations where the local bound is minimal, consequently
$Q/L$ is maximal, we have used the Nelder-Mead method starting from many
random relative orientations. The transformation has been parameterized by the
$d(d-1)/2$ generalized Tait–Bryan angles. Unfortunately, the minimal values
found this way are not necessarily the global minima, it may happen that is
some cases there exists relative orientations with lower local bounds, but we
have missed them. Whenever the numerical result has shown some recognizable
analytical pattern, we have tried to find the analytical form (either for the
transformation or for the Bell inequality itself), and give $L$ analytically,
or as accurately as possible.

The simplest body from our point of view is the cross polytope, a centrally
symmetric object. If we eliminate half of its vertices in the most obvious
way, the remaining ones are just the basis vectors $\vec{e}_{i}$,
($i=1,\dots,d$). If these are the vectors on one side to construct the Bell
inequality, then the Bell coefficients will be nothing else than the vector
coordinates on the other side.

Cross polytope–cross polytope. Let another such halved cross polytope be on
the other side, but reoriented by an orthogonal transformation. An optimal
transformation we found consists of rotations in the $1-2$ the $3-4$, and so
on planes by $\pi/4$ angle. If $d$ is odd, the $d$th coordinate remains
unchanged. If $d$ is even, we get the sum of $d/2$ independent CHSH
inequalities multiplied by $1/\sqrt{2}$. Therefore, the local bound is
$d/\sqrt{2}$, while the quantum maximum is $d$, as it should be, giving
$Q/L=\sqrt{2}$. For odd $d$ there are $(d-1)/2$ independent inequalities
proportional to CHSH, and there is one additional nonzero Bell coefficient,
namely $M_{dd}=1$. Then $L=(d-1)/\sqrt{2}+1$, $Q=d$ and
$Q/L=\sqrt{2}/[1+(\sqrt{2}-1)/d]$. For two complete cross polytopes $L$ and
$Q$ should be multiplied by four.

Cross polytope–hypercube. Let there be a hypercube on Bob’s side. When $d=3$
an optimal transformation we found was a rotation around the third axis (that
is in the $1-2$ plane) by $\pi/3$, which leads to $L=8+8/\sqrt{3}=12.618802$
and $Q/L=1.2679492$. For $d=5$ the same transformation turned out to be
optimal, namely a rotation in the $1-2$ plane by $\pi/3$, which gives
$L=50.789699$ and $Q/L=1.2600980$. It is possible that this transformation is
optimal in all odd dimensions if there is a cross polytope on one side and a
hypercube on the other one. The local bound can be analytically calculated in
this case, the result is $L=4\binom{d-1}{(d-1)/2}(\sqrt{3}+d-2)/\sqrt{d}$, and
$Q/L=2^{d+1}/L$. The calculation can be done for even $d$ as well, but in that
case such transformation is never optimal. For $d=4$ the optimal
transformation is the same as in the case of two cross polytopes: $\pi/4$
rotations in the $1-2$ and in the $3-4$ planes, and then we obtain
$L=32/\sqrt{2}$ and $Q/L=\sqrt{2}$. If we eliminate appropriately one half of
vertices of both bodies, the Bell inequality we get will be like the sum of
four CHSH ones divided by $\sqrt{2}$. Although they are not independent, there
exist strategies and measurement settings that optimize all of them,
therefore, $Q/L=\sqrt{2}$, indeed. Surprisingly, the analogous transformation
is inferior for $d=6$. There the best we could find was to apply no rotation
at all. Then $L=40\sqrt{6}=97.979590$, and $Q/L=1.3063945$. The local bound
with no reorientation can be calculated for any $d$ analytically, the result
is $L=2\sqrt{d}\binom{d}{d/2}$ for even $d$. It is possible that this is the
best we can do for $d\geq 6$ even values, but we do not know. Actually, for
large $d$ the formula approaches $\sqrt{2/\pi}2^{d+1}$, consequently $Q/L$
converges to $\sqrt{\pi/2}$. For odd $d$ what we get without reorientations is
worse than what we obtained with the $\pi/3$ rotation in a single plane, as
discussed above.

3D Platonic solids | tetrahedron | octahedron | cube | icosahedron | dodecahedron
---|---|---|---|---|---
tetrahedron | 1.0000000 | 1.0000000 | 1.0000000 | 1.0704663 | 1.1026409
$m=4$ | 1.2360680 | 1.2679492 | 1.2360680 | 1.2584081 | 1.1803399
octahedron |  | 1.0000000 | 1.0000000 | 1.0704663 | 1.1026409
$m=6$ |  | 1.2426407 | 1.2679492 | 1.1969160 | 1.1755012
cube |  |  | 1.0000000 | 1.0704663 | 1.1026409
$m=8$ |  |  | 1.2360680 | 1.2584081 | 1.1803399
icosahedron |  |  |  | 1.1458980 | 1.1803399
$m=12$ |  |  |  | 1.2811529 | 1.2734640
dodecahedron |  |  |  |  | 1.2158169
$m=20$ |  |  |  |  | 1.2901781

Table 2: Maximum quantum values divided by local bounds $Q/L$ for Bell
inequalities constructed from unit vectors pointing towards the vertices of
three-dimensional (3D) Platonic solids. The upper and the lower numbers
correspond to relative orientations giving the maximal and the minimal local
bounds, respectively. In the first column the number of vertices for the
solids is also shown.

4D Platonic solids | pentacope | hexdecahedroid | tesseract | octaplex | tetraplex | dodecaplex
---|---|---|---|---|---|---
Orthographics |  |  |  |  |  |
pentacope | 1.0416667 | 1.0206207 | 1.0206207 | 1.1572751 | 1.1972885 | 1.1973624
$m=5$ | 1.3975425 | 1.1907107 | 1.2741137 | 1.2180011 | 1.2042164 | 1.2057998
hexdecahedroid |  | 1.0000000 | 1.0000000 | 1.1338934 | 1.1730984 | 1.1731707
$m=8$ |  | 1.4142136 | 1.4142136 | 1.4142136 | 1.2286820 | 1.1869546
tesseract |  |  | 1.0000000 | 1.1338934 | 1.1730984 | 1.1731707
$m=16$ |  |  | 1.4142136 | 1.4142136 | 1.2286820 | 1.1869546
octaplex |  |  |  | 1.2857143 | 1.3301686 | 1.33025058
$m=24$ |  |  |  | 1.4142136 | 1.3350119 | 1.34038194
tetraplex |  |  |  |  | 1.3761599 | 1.3762447
$m=120$ |  |  |  |  | 1.3776804 | 1.3842396
dodecaplex |  |  |  |  |  | 1.3763296
$m=600$ |  |  |  |  |  | 1.3834497

Table 3: Maximum quantum values divided by local bounds $Q/L$ for Bell
inequalities constructed from unit vectors pointing towards the vertices of
four-dimensional (4D) Platonic solids. The upper and the lower numbers
correspond to relative orientations giving the maximal and the minimal local
bounds, respectively. In the first column the number of vertices for the
solids is also shown. Orthographic projections of the vertices of the six 4D
Platonic solids are also shown below the names.

Cross polytope–simplex. Let there be a cross polytope on Alice’s side and a
simplex on Bob’ side. For $d=3$ the simplex is the tetrahedron that we can get
by appropriately eliminating half of the vertices of a cube. Therefore, $L$ is
one half that of we have got for the cube, namely $L=4+4/\sqrt{3}=6.3094011$
while $Q/L=1.2679492$ is the same. For $d=4$ and for $d=5$ the optimally
oriented simplex can also be determined from the numerical data. From those
forms it follows that $L=3+\sqrt{5}(1+\sqrt{2})=8.3983456$, $Q/L=1.1907107$,
and $4(\sqrt{0.1}+\sqrt{0.3}+2\sqrt{0.6})=9.6525746$, $Q/L=1.2431916$ for
$d=4$ and $d=5$, respectively. From the $d=4$ form we could give an ansatz for
$d=6$, but it turned out that it can further be improved by applying a further
rotation of a quite small angle in just a single plane defined by two basis
vectors. This way we get $L=11.639817$ and $Q/L=1.2027681$. For higher
dimensions we have no reliable results. For the cross polytope-simplex pair we
could not find any consistent behaviour that could be generalized.

Hypercube–hypercube. Let us consider $d$-cubes on both sides. In three
dimensions the matrix of an optimal transformation is:

$\frac{1}{4}\begin{pmatrix}1/2&\sqrt{5}+1&\sqrt{5}-1\\\
\sqrt{5}-1&-1/2&\sqrt{5}+1\\\ -\sqrt{5}-1&\sqrt{5}-1&1/2\end{pmatrix}.$ (17)

Then $L=16(\sqrt{5}+1)/3=17.259029$ and $Q/L=4/(\sqrt{5}+1)=1.2360680$. In
four dimensions the optimal transformation is the same as in the case of two
cross polytopes or the cross polytope $4$-cube pair. If we eliminate
appropriately one half of the simplices from both cubes, the Bell matrix we
get will correspond to the sum of eight CHSH Bell matrices divided by
$\sqrt{2}$. Like before, although they are not independent, there are
classical strategies and measurement settings optimal for all of them.
Therefore, $Q/L=\sqrt{2}$, and for the full cubes $L=64/\sqrt{2}$. For $d=5$
the smallest value we have got numerically was $L=165.01046$
($Q/L=1.2411334$), and we could not recognize pattern leading to some
analytical form. For $d=6$ the transformation we have found optimal was the
same as the one for the cross polytopes. However, the Bell inequality this
time did not come out as a sum of CHSH ones which could all be optimized by
the same strategy, therefore the $L=4256/6/\sqrt{2}=501.57441$ value is
slightly bigger than $4096/6/\sqrt{2}$, as it would be if the behaviour were
the same as for $d=4$. Now $Q/L=128\sqrt{2}/133=1.3610476$.

Hypercube–simplex. For a hypercube on one side and a simplex on the other in
$3$ dimensions the $L$ and $Q$ values are just one half of the corresponding
ones for the cube-cube pair, while $Q/L$ is the same. For $d>3$ we have only
got numerical results. Actually, we think that with more calculations they
could slightly be improved. The best values we got for $d=4$, $d=5$ and $d=6$
are $L=15.6971864$ ($Q/L=1.2741137$), $L=31.068002$ ($Q/L=1.2359984$) and
$L=60.254415$ ($Q/L=1.2391900$), respectively.

Simplex–simplex. For two simplices in $d=3$ (two tetrahedrons) the values of
$L$ and $Q$ are one quarter of the ones for two cubes. For $d=4$ we have been
able to figure out some analytical pattern from the numerical data, and we
have got $L=2\sqrt{5}=4.4721360$, and $Q/L=5\sqrt{5}/8=1.3975425$. For $d=5$
and $d=6$ we have only numerical results, which probably could be improved
with further calculations. What we have got is $L=5.8332724$ ($Q/L=1.2342986$)
and $L=6.6836480$ ($Q/L=1.2218876$) for $d=5$ and $d=6$, respectively.

Other pairs in three and four dimensions. In Tables 2 and 3 we show the
results for the ratios of the quantum and the local bounds, both the maximum
and the minimum ones for all pairs of Platonic bodies in $d=3$ and $d=4$,
respectively. The maximum values are calculated as the products of the longest
strategy vectors for the objects, as shown in the previous chapter, while the
minimum ones are numerical results we got using the Nelder-Mead method
starting from many random relative orientations. In Table 2 the results for
the cube and for the tetrahedron agree, as they must. One may notice in Table
3 that $Q/L=\sqrt{2}$ when one of the solids is the octaplex, while the other
one is either the hexdecahedroid (cross polytope), the octaplex or the
tesseract (4-cube). In all cases a number of CHSH inequalities are involved,
like in other similar cases. Let us take the 24 vectors one can get by
considering all permutations and all sign variations of the coordinates of
vector $(1,1,0,0)/\sqrt{2}$. These vectors correspond to the vertices of an
octaplex. If the other body is the cross polytope in standard orientation, and
we eliminate appropriately one half of the vertices of both bodies, the Bell
matrix is

$\left(\begin{array}[]{rrrrrrrrrrrr}1&1&1&1&1&1&0&0&0&0&0&0\\\
1&-1&0&0&0&0&1&1&1&1&0&0\\\ 0&0&1&-1&0&0&1&-1&0&0&1&1\\\
0&0&0&0&1&-1&0&0&1&-1&1&-1\\\ \end{array}\right)$

(18)

divided by $\sqrt{2}$. This involves six non-independent CHSH matrices such
that there exist local strategies and measurement settings maximizing all of
them. The inequality is the same as $I_{4,12}$ in Ref. [26]. For the octaplex
4-cube pair the behaviour is similar. For two octaplexes one of the bodies has
to be rotated by $\pi/4$ both in the $1-2$ and the $3-4$ planes. For two
pentacopes the number of settings is five for both parties, and the vectors
pointing towards the vertices of the pentacopes in the four dimensional space
are optimal measurement vectors. However, it turns out that one can find
measurement vectors within three-dimensional subspaces giving the same quantum
value. This means that the question raised in Ref. [24] about the existence of
correlation-type Bell inequality with five settings for each of the two
parties whose maximum violation requires four-dimensional measurement vectors
still remains open.

##  9 Diagonally modified Bell inequalities

In this section, we consider Platonic Bell inequalities in which the main
diagonal entries of the Bell matrix, formed by scalar product of Euclidean
vectors, are modified by subtracting the same constant from each diagonal
entry. This modification of the original Platonic Bell inequalities has been
performed in section 4, which, according to (8), looks as follows:

$M_{ij}(\lambda)=M_{ij}-\lambda\delta_{ij}=\vec{A}_{i}\cdot\vec{A}_{j}-\lambda\delta_{ij}.$
(19)

In section 4 we have also shown that the maximum quantum value, i.e., the
Tsirelson bound [5] becomes $(m_{A}^{2}/d)-\lambda m_{A}$, and this can be
obtained with the same measurement vectors as without the modification. Note
that in this case $M$ is a square matrix of size $m_{A}\times m_{A}$.
Therefore, the Bell inequality defined by the expression in Eq. (19) can still
be regarded as a Platonic Bell inequality. In this way, we find Platonic Bell
inequalities that are more robust to noise than the celebrated Clauser-Horne-
Shimony-Holt Bell inequality [9], i.e., the quantum ($Q$) per local bound
($L$) is greater than $\sqrt{2}$. However, this construction requires higher
dimensional Hilbert spaces to reach the Tsirelson bound.

This diagonal modification technique has been applied previously by Davie and
Reeds [27] to obtain the best lower bound for the Grothendieck constant,
$K_{G}\geq 1.67696$ [28, 29]. The Grothendieck constant of order $d$ [30],
denoted by $K_{G}(d)$, corresponds to the largest ratio $Q(M,d)/L(M)$ among
arbitrary $M$ real matrices, where $Q$ is defined by
$Q(M,d)=\sum_{i,j}M_{i,j}\vec{a}_{i}\cdot\vec{b}_{j}$, where
$\vec{a}_{i},\vec{b}_{j}$ are unit vectors in the $d$-dimensional Euclidean
space. This relation was first noticed by Tsirelson [6]. Note that
$K_{G}=\lim_{d\rightarrow\infty}K_{G}(d)$. Therefore the lower bound
$K_{G}\geq 1.67696$ corresponds to a correlation-type Bell expression $M$ with
$Q(M,d)/L(M)\geq 1.67696$. However, the constructed Bell inequality has an
infinite number of settings on both Alice’s and Bob’s side. We note that the
exact value of $K_{G}(d)$ is known only for $d=2$, $K_{G}(2)=\sqrt{2}$ with
the Bell matrix corresponding to the CHSH inequality. This proof is due to
Krivine [30].

The first example of $M$ with modest size exceeding the ratio $Q/L=\sqrt{2}$
was discovered by Reeds and Sloane in 1990. They used an $M$ matrix of size
$120\times 120$ of the family (19) with $\lambda=0$, where $\vec{A}_{i}$ are
formed by the 240 eight-dimensional unit vectors of the lattice $E_{8}$
resulting in $K_{G}(8)\geq 45/31=1.4516129$. A finite $M(\lambda)$ matrix size
in (19), where $\lambda>0$, was first used by Fishburn and Reeds [31]. Their
$20\times 20$ matrix $M(\lambda)$ provides a ratio $Q/L=10/7$, which implies
that $K_{G}(5)\geq 10/7=1.4285714$. Note, however, that the construction by
Fishburn and Reeds [31] does not give a lower bound on $K_{G}(4)$ greater than
$\sqrt{2}$.

In the following, we prove that

$K_{G}(4)\geq\frac{70}{5+27\phi}-4\times 10^{-6}=1.4377539$ (20)

using Platonic Bell inequalities, where $\phi$ is the golden ratio and the
dimension of the matrix $M(\lambda)$ is $60\times 60$. Note that there exist
better lower bounds in the literature. Namely, $K_{G}(4)\geq 1.445207$ [32],
$K_{G}(4)\geq 1.44566$ [33] and $K_{G}(4)\geq 1.4821664$ [34]. However,
neither of them is based on Platonic Bell inequalities. Let us also mention
the upper bound $K_{G}(4)\leq\pi/2$ [30].

To get the ratio $Q/L=1.4377539$ using the Platonic Bell inequalities, we
construct $M(\lambda)$ according to Eq. (19). In particular, we choose
$\\{\vec{A}_{i}\\}$ as the vertices of the halved tetraplex ($m_{A}=60$,
$d=4$). We explicitly define the 60 vertices $\vec{A}_{i}$ as follows: Four
vertices come from

$(0,0,0,1)$ (21)

by permuting the coordinates, eight vertices from

$(1/2,\pm 1/2,\pm 1/2,\pm 1/2)$ (22)

and the remaining 48 vertices are obtained by taking all even permutations of

$(\phi/2,\pm 1/2,\pm 1/(2\phi),0),$ (23)

where $\phi$ is the golden ratio. Note that the 120-vertex regular tetraplex
comprises the above $\vec{A}_{i}$ vertices and its antipodal ones
$-\vec{A}_{i}$. The quantum value $Q(\lambda)$ for the Bell matrix
$M(\lambda)$ in (19) is

$Q(\lambda)=m_{A}^{2}/d-\lambda m_{A}=450-60\lambda$ (24)

if $\lambda\leq m_{A}/(2d)=15/2$. This is due to the analysis of Section 4, in
which case the Tsirelson bound of the Bell inequality is given by the 60
vertices of the halved tetraplex shown above. Indeed, the 120 vertices of the
tetraplex readily obeys the constraints of the Observation in Section 4.
Moreover, the set of 120 vertices is centrally symmetric (i.e. every
$\vec{A}_{i}$ has an antipodal vector $-\vec{A}_{i}$ in the set). Therefore,
by the argument at the end of Section 4 the halved 60-vertex tetraplex also
obeys the constraints of the Observation and the maximum quantum bound is
given by (24).

However, the local bound is difficult to calculate. Recall that, for a generic
$M$ matrix the calculation of the local bound (9) is an instance of
$K_{m,n}$-quadratic programming [35]. In our case the matrix $M$ is replaced
by the one-parameter family of matrices $M(\lambda)$ in (19). However, unlike
the case $\lambda=0$ discussed in Section 5, we have not found an efficient
method to compute the local bound for $\lambda>0$. On the other hand, a brute-
force computation based on the enumeration of local deterministic strategies
would require the evaluation of $2^{m_{A}-1}=2^{59}$ instances. This task is
impossible to perform in a reasonable time even with supercomputers. However,
the algorithm based on the branch-and-bound (BB) technique [36] developed in
Ref. [34] makes the problem tractable for our particular case. This code is
publicly available [37], and supports parallel computation of the local bound
(9). With the latest parallel implementation available in [37], running our
problem on a 56-core workstation, the local bound $L(\lambda)$ is obtained for
an arbitrary $\lambda$ in about 2 minutes. The obtained ratio
$Q/L=(450-60\lambda)/L(\lambda)$ as a function of $\lambda$ is plotted from
$\lambda=0$ to $\lambda=7.5$ in Fig. 2. For comparison, we also plot
$Q(\lambda)$ and $L(\lambda)$ in the same figure. The abrupt change in the
$L(\lambda)$ function is observed just as the ratio $Q(\lambda)/L(\lambda)$
reaches its peak value at $\lambda=23/9$.

Figure 2: The $Q/L$ ratio (red dots) in the function of $\lambda$ for the
Platonic Bell inequality defined by the coefficients (19), where $\vec{A}_{i}$
are the vertices of the halved tetraplex. The functions $Q$ (blue dashed line)
and $L$ (blue solid line) are also plotted against $\lambda$. The maximum
ratio $Q/L=1.4377579-\epsilon$ is obtained at $\lambda=23/9$, where
$\epsilon=4\times 10^{-6}$. The value $1.4377579$ is analytically given by
$70/(5+27\phi)$, where $\phi=(1+\sqrt{5})/2$ is the golden ratio.

The BB-based code [37] computes exactly the local bounds of Bell matrices
whose coefficients are all integers.

Some technical notes on the computation of the local bound $L(\lambda)$ are in
order. The BB-based code [37] computes with exact arithmetics the local bound
of Bell matrices whose entries are all integers. Hence, for a matrix $M$ with
integer entries the calculated local bound is exact. However, the entries of
our special $M(\lambda)$ matrices are in general not integers. We avoid this
problem by the following method. This way we obtain a very good upper bound on
$L(M(\lambda))$ as follows. We multiply $M(\lambda)$ by a large number
($10^{6}$, in our particular case) and round each entry to the nearest integer
below it. That is, we have $10^{6}M=M_{\rm int}+\Delta M$, where $M_{\rm
int}=\lfloor 10^{6}M\rfloor$ is an integer matrix, and the entries of $\Delta
M$ are $\Delta M_{ij}\in[0,1]$. The number $10^{6}$ has been chosen in order
not to occur overflow, as the code is limited to Bell matrices where
$\sum_{i,j}|M_{ij}|\leq 2^{31}$. Then we have the following upper bound

$L(10^{6}M)\leq L(M_{\rm int})+L(\Delta M),$ (25)

where we used that for any two Bell matrices we have $L(M_{1}+M_{2})\leq
L(M_{1})+L(M_{2})$. Furthermore, due to all the positive entries in $\Delta
M_{i,j}$, we have $L(\Delta M)=\sum_{i,j}|\Delta M_{ij}|$. Then, if we divide
both sides of (25) by $10^{6}$, we get the following upper bound on $L(M)$:

$L(M)\leq\frac{L(M_{\rm int})}{10^{6}}+\frac{\sum_{i,j}|\Delta
M_{ij}|}{10^{6}},$ (26)

where $L(M_{\rm int})$ is the output of the BB-based code [37]. The maximum
$Q/L$ ratio of $M(\lambda)$ occurs at $\lambda=23/9$ (see Fig. 2). In this
case, we have $L(M_{\rm int})=519326904$ and $\sum_{i,j}|\Delta
M_{ij}|=1622.88706865$, which results in an upper bound $L(M)\leq
519.32852689$ and a lower bound $Q(M)/L(M)\geq 1.4377539$, where the latter
value equals $70/(5+27\phi)-\epsilon$, where $\epsilon=4\times 10^{-6}$. In
summary, using measurement directions pointing to the vertices of the four-
dimensional 60-vertex halved tetraplex and the diagonal modification of the
Bell matrix $M$, we obtain the ratio $Q/L\geq 1.4377539$ and the corresponding
lower bound $K_{G}(4)\geq 1.4377539$.

##  10 Discussion

Tavakoli and Gisin [4] constructed Bell inequalities with coefficients
$M_{ij}=\vec{A}_{i}\cdot\vec{B}_{j}$, where $\vec{A}_{i}$ and
$\vec{B}^{*}_{j}$ are the unit vectors pointing to the desired measurement
directions on the sides of Alice and Bob, respectively, i.e., towards the
vertices of Platonic bodies. Then the obtained quantum value of the Bell
expression with these measurement directions is
$\sum_{ij}(\vec{A}_{i}\cdot\vec{B}_{j})^{2}$. Using the NPA method [20], the
authors find that the above measurement directions provide optimal quantum
violation (i.e. the Tsirelson bound [5]). Such constructions have been called
Platonic Bell inequalities. Motivated by this construnction, we generalized
the Platonic Bell inequalities to all possible dimensions. Due to a theorem of
Tsirelson [6], the generalization of the construction of the Bell inequalities
to higher dimensional spaces does make sense, the result will correspond to a
genuine Bell scenario, however, the symmetry of the solids will show up in the
abstract Tsirelson space. To prove that this generalization is possible, we
have shown that all Platonic solids (and Archimedean solids) have some
property that ensures this, and that the form of the quantum bound is very
simple and independent of the relative orientation of the two solids on Alice
and Bob side. However, the local bound, and hence the maximum violation of the
inequality depends on this, which is discussed in detail in this paper.

In more than two dimensions some of the Platonic solids have too many
vertices, hence computing the local bounds of the Bell inequalities
constructed from them in a brute force way may not be feasible. We have found
an efficient method which makes it possible to calculate the local bound
exactly in all cases where the Bell coefficients are given as scalar products
of low-dimensional Euclidean vectors, even when the number of measurement
settings is large. The practical significance of the method is demonstrated by
the exact computation of the local bound for the 300-setting four-dimensional
Platonic Bell inequality. The Euclidean vectors in this method do not have to
be unit vectors, which allows us to prove that the method can be used not only
for Platonic Bell inequalities but also for all bipartite two-outcome Bell
inequalities where the rank of the Bell matrix is low.

Finally, we study Platonic Bell inequalities where the main diagonal entries
of the Bell matrix formed by the scalar product of Euclidean vectors are
modified by a constant term. In this way, we find Platonic Bell inequalities
that are more robust to noise than the celebrated Clauser-Horne-Shimony-Holt
Bell inequality, that is, the quantum ($Q$) per local bound ($L$) is greater
than $\sqrt{2}$. However, this construction requires higher dimensional
Hilbert spaces to achieve maximal quantum violation. In this regard, we leave
open the question whether the Tsirelson bound with a $Q/L$ ratio larger than
$\sqrt{2}$ can be achieved by a suitable symmetric Bell inequality using a
two-qubit maximally entangled state where the measurements point to the
vertices of some three-dimensional symmetric polyhedron.

_Acknowledgements._ We thank Péter Diviánszky for his help in writing the
parallel version of the $K_{m,n}$ quadratic programming code and Marco Túlio
Quintino for valuable discussions. TV acknowledges the support of the EU
(QuantERA eDICT) and the National Research, Development and Innovation Office
NKFIH (No. 2019-2.1.7-ERA-NET-2020-00003).

## References

  * [1] H. S. M. Coxeter, Regular Polytopes (New York: Dover Publications 1973).
  * [2] J. S. Bell, _On the Einstein-Poldolsky-Rosen paradox_ , https://doi.org/10.1103/PhysicsPhysiqueFizika.1.195 Physics 1, 195–200 (1964).
  * [3] N. Brunner, D. Cavalcanti, S. Pironio, V. Scarani, and S. Wehner, _Bell nonlocality_ , https://doi.org/10.1103/RevModPhys.86.419 Rev. Mod. Phys. 86, 419 (2014).
  * [4] A. Tavakoli and N. Gisin, _The Platonic solids and fundamental tests of quantum mechanics_ , https://doi.org/10.22331/q-2020-07-09-293 Quantum 4, 293 (2020).
  * [5] B. S. Cirel’son, _Quantum generalizations of Bell’s inequality_ , https://doi.org/10.1007/BF00417500 Letters in Mathematical Physics 4, 93–100 (1980).
  * [6] B. S. Tsirelson, _Quantum analogues of the Bell inequalities. The case of two spatially separated domains_ , https://doi.org/10.1007/BF01663472 J. Soviet Math. 36, 557 (1987).
  * [7] K. Bolonek-Lasoń, P. Kosiński, _Groups, Platonic solids and Bell inequalities_ , https://doi.org/10.22331/q-2021-11-29-593 Quantum 5, 593 (2021).
  * [8] R. Cleve, P. Hoyer, B. Toner, and J. Watrous, _Consequences and limits of nonlocal strategies_ , https://doi.org/10.1109/CCC.2004.1313847 in 19th IEEE Conference on Computational Complexity p. 236. (2004).
  * [9] J. F. Clauser, M. A. Horne, A. Shimony, and R. A. Holt. _Proposed experiment to test local hidden-variable theories_ , https://doi.org/10.1103/PhysRevLett.23.880 Phys. Rev. Lett. 23, 880 (1969).
  * [10] A.J. Bennet, D.A. Evans, D.J. Saunders, C. Branciard, E.G. Cavalcanti, H.M. Wiseman, and G.J. Pryde, _Arbitrarily loss-tolerant Einstein-Podolsky-Rosen steering allowing a demonstration over 1 km of optical fiber with no detection loophole_ , https://doi.org/10.1103/PhysRevX.2.031003 Phys. Rev. X 2, 031003 (2012).
  * [11] D. J. Saunders, S. J. Jones, H. M. Wiseman, G. J. Pryde, _Experimental EPR-Steering using Bell-local States_ , https://doi.org/10.1038/nphys1766 Nat. Phys. 76, 845-849 (2010).
  * [12] T. Decker, D. Janzing, T. Beth, _Quantum circuits for single-qubit measurements corresponding to platonic solids_ , https://doi.org/10.1142/S0219749904000298 Int. J. Quan. Inf. 02, 353 (2004).
  * [13] K. Jeong, J. S. Lee, J. T. Choi, S. M. Hong, M. G. Jung, G. B. Kim, J. K. Kim, and S. Kim, _Single Qubit Private Quantum Channels and 3-Dimensional Regular Polyhedra_ , https://doi.org/10.3938/NPSM.68.232 New Phys.: Sae Mulli 68 232-240 (2018).
  * [14] Junseo Lee, Kabgyun Jeong, _High-dimensional private quantum channels and regular polytopes_ , https://doi.org/10.15625/0868-3166/15762 Communications in Physics 31, 189 (2021).
  * [15] P. Kolenderski, R. Demkowicz-Dobrzanski, _Optimal state for keeping reference frames aligned and the Platonic solids_ , https://doi.org/10.1103/PhysRevA.78.052333 Phys. Rev. A 78, 052333 (2008).
  * [16] M. Burrello, H. Xu, G. Mussardo, X. Wan, _Quantum hashing with the icosahedral group_ , https://doi.org/10.1103/PhysRevLett.104.160502 Phys. Rev. Lett. 104, 160502 (2010).
  * [17] J. I. Latorre, G. Sierra, _Platonic Entanglement_ , https://doi.org/10.48550/arXiv.2107.04329 e-print arXiv:2107.04329 (2021).
  * [18] Y. Xiao, Z.-P. Xu, Q. Li, H.-Y. Su, K. Sun, A. Cabello, J.-S. Xu, J.-L. Chen, C.-F. Li, G.-C. Guo, _Experimental test of quantum correlations from Platonic graphs_ , https://doi.org/10.1364/OPTICA.5.000718 Optica 5, 718 (2018).
  * [19] A. Acín, N. Gisin, and B. Toner, _Grothendieck’s constant and local models for noisy entangled quantum states_ , https://doi.org/10.1103/PhysRevA.73.062105 Phys. Rev. A 73, 062105 (2006).
  * [20] M. Navascués, S. Pironio, and A. Acín, _Bounding the Set of Quantum Correlations_ , https://doi.org/10.1103/PhysRevLett.98.010401 Phys. Rev. Lett. 98, 010401 (2007).
  * [21] T. Vértesi and K. F. Pál, _Generalized Clauser-Horne-Shimony-Holt inequalities maximally violated by higher dimensional systems_ , https://doi.org/10.1103/PhysRevA.77.042106 Phys. Rev. A 77, 042106 (2008).
  * [22] M. Epping, H. Kampermann, D. Bruß, _Designing Bell inequalities from a Tsirelson bound_ , https://doi.org/10.1103/PhysRevLett.111.240404 Phys. Rev. Lett. 111 240404 (2013).
  * [23] M. Epping, H. Kampermann, D. Bruß, _Optimisation of Bell inequalities with invariant Tsirelson bound_ , https://doi.org/10.1088/1751-8113/47/42/424015 J. Phys. A bf 47 424015 (2014).
  * [24] T. Vértesi and K. F. Pál, _Bounding the dimension of bipartite quantum systems_ , https://doi.org/10.1103/PhysRevA.79.042106 Phys. Rev. A 79, 042106 (2009).
  * [25] J. Briët, H. Buhrman, and B. Toner, _A generalized Grothendieck inequality and nonlocal correlations that require high entanglement_ , https://doi.org/10.1007/s00220-011-1280-3 Commun. Math. Phys. 305, 827 (2011).
  * [26] M. Navascués, G. de la Torre, and T. Vértesi, _Characterization of quantum correlations with local dimension constraints and its device-independent applications_ , https://doi.org/10.1103/PhysRevX.4.011011 Phys. Rev. X 4, 011011 (2014).
  * [27] A. M. Davie (unpublished note, 1984) and J. A. Reeds (unpublished note, 1991).
  * [28] A. Grothendieck, _Résumé de la théorie métrique des produits tensoriels topologiques_ , Bol. Soc. Mat. São Paulo 8, 1–79 (1953).
  * [29] S. R. Finch, Mathematical Constants, ser. Encyclopedia of Mathematics and its Applications. Cambridge, UK: Cambridge University Press, 2003.
  * [30] J. L. Krivine, _Constantes de Grothendieck et fonctions de type positif sur les spheres_ , https://doi.org/10.1016/0001-8708(79)90017-3 Adv. Math. 31, 16 (1979).
  * [31] P.C. Fishburn and J.A. Reeds, _Bell inequalities, Grothendieck’s constant, and root two_ , https://doi.org/10.1137/S0895480191219350 SIAM Journal on Discrete Mathematics, 7, 48–56 (1994).
  * [32] T. Vértesi, _More efficient Bell inequalities for Werner states_ , https://doi.org/10.1103/PhysRevA.78.032112 Phys. Rev. A 78, 032112 (2008).
  * [33] B. Hua, M. Li, T. Zhang, C. Zhou, X. Li-Jost, S.-M. Fei, _Towards Grothendieck constants and LHV models in quantum mechanics_ , https://doi.org/10.1088/1751-8113/48/6/065302 J. Phys. A: Math. Theor. 48, 065302 (2015).
  * [34] P. Diviánszky, E. Bene, and T. Vértesi, _Qutrit witness from the Grothendieck constant of order four_ , https://doi.org/10.1103/PhysRevA.96.012113 Phys. Rev. A, 96, 012113 (2017).
  * [35] P. Raghavendra and D. Steurer, _Towards computing the Grothendieck constant_ , In Proceedings of the Twentieth Annual ACM-SIAM Symposium on Discrete Algorithms, 525 (2009).
  * [36] A. H. Land and A. G. Doig, _An automatic method of solving discrete programming problems_ , https://doi.org/10.2307/1910129 Econometrica 28, 497–520 (1960).
  * [37] https://github.com/divipp/kmn-programminghttps://github.com/divipp/kmn-programming

##  Appendix A Relationship between the Euclidean and the Hilbert space
description

In the present paper the quantum measurements performed by the parties are
represented by unit vectors of an Euclidean space. However, quantum states
live in Hilbert space and measurements correspond to appropriate Hermitian
operators in that Hilbert space. As shown for example in Ref. [4] and
discussed briefly in the main text, if the quantum system is a pair of qubits
in the maximally entangled $|\Phi^{+}\rangle=(|00\rangle+|11\rangle)/\sqrt{2}$
state, the quantum measurements performed by the parties on their respective
qubits can be represented by vectors in a three-dimensional Euclidean space
such that the quantum expectation values occuring in the Bell-expression are
simply the scalar products of the corresponding Euclidean vectors. Moreover,
if the qubits are spins, those vectors are very simply related to orientations
of the devices measuring the spin projection in the physical space, namely for
one of the parties they point towards the measurement directions themselves,
while for the other party one can get the measurement directions by reflecting
the vectors through the $x-z$ plane. We note that a similar relationship
exists if the state is the singlet one
$|\Psi^{-}\rangle=(|10\rangle-|01\rangle)/\sqrt{2}$. Then the measurement
vectors and the measurement directions point to the opposite directions for
both parties. In Ref. [7] this latter relationship has been exploited. A
beautiful feature of the Platonic Bell inequalities in three dimensions is
that if the system is a pair of one-half spins, the symmetries of the Platonic
solids show up directly in the orientations of the measurement settings in the
physical space. Unfortunately, this beauty is lost if the system is not like
that (even if it is a pair of qubits, but not spins). Therefore for the higher
dimensional generalization discussed in the present paper the symmetries are
only present in an abstract space, not in the physical arrangements concerning
the measurement settings.

However, it is not even obvious that the considerations in more than three
dimensional spaces are relevant at all for Bell scenarios. A theorem by
Tsirelson [6] ensures that this is the case. Due to this theorem, for any
Euclidean space there exists a Hilbert space $H$ and a state $|\Phi\rangle\in
H\otimes H$ such that for any unit vectors $\vec{a}$ and $\vec{b}$ in the
Euclidean space there exist binary measurement operators $\hat{a}$ and
$\hat{b}$ in the Hilbert space such that

$\vec{a}\cdot\vec{b}=\langle\Phi|\hat{a}\otimes\hat{b}|\Phi\rangle.$ (27)

Hence, for each party, any unit vector does represent some quantum measurement
performed on the respective subsystem of the quantum system $|\Phi\rangle$.
The statement can be proven by construction [19, 21]. The considerations we
will give here are based on this construction.

Let us consider a $d$-dimensional Eulidean space. If there exist $d$ mutually
anticommuting operators $\hat{\gamma}_{i}$ ($i=1,\dots,d$) whose square is the
identity operator (that is
$\hat{\gamma}_{i}\hat{\gamma}_{j}+\hat{\gamma}_{j}\hat{\gamma}_{i}=2\delta_{ij}\hat{I}$)
in the $D$-dimensional Hilbert space, then with
$\hat{a}=\sum_{i=1}^{d}a_{i}\hat{\gamma}_{i}$,
$\hat{b}=\sum_{i=1}^{d}b_{i}\hat{\gamma}_{i}^{t}$ ($t$ denotes the
transposition) and with the maximally entangled state
$|\Phi\rangle=\sum_{\mu=1}^{D}|\mu\mu\rangle/\sqrt{D}$ Eq. (27) is satisfied.
Indeed, $\langle\Phi|\hat{a}\otimes\hat{b}|\Phi\rangle={\rm
Tr}(\hat{a}\hat{b}^{t})/D=\sum_{i,j=1}^{d}a_{i}b_{j}{\rm
Tr}(\hat{\gamma}_{i}\hat{\gamma}_{j})/D=\vec{a}\cdot\vec{b}$. The last
equality follows from the fact that $\hat{\gamma}_{i}$ and $\hat{\gamma}_{i}$
anticommute if $i\neq j$, therefore ${\rm
Tr}(\hat{\gamma}_{i}\hat{\gamma}_{j})=0$, and that $\hat{\gamma}_{i}^{2}$ is
the identity operator whose trace is $D$. It can similarly be shown that the
square of both $\hat{a}$ and $\hat{b}$ is the identity operator, therefore
they do represent binary measurements with outcomes $+1$ and $-1$, as
required.

One can construct a set of $2n+1$ $\hat{\gamma}_{i}$ operators as tensor
products of $n$ Pauli operators and two-dimensional identity operators
$\hat{I}_{2}$ in appropriate orders. An example of such a construction for
arbitrary $n$ is given explicitly in the Appendix of [21]. This way one can
get the $d$ $\hat{\gamma}_{i}$ operators needed to assign quantum measurements
to the unit vectors of the $d$-dimensional Euclidean space as tensor products
of $\lfloor d/2\rfloor$ operators acting on qubit spaces. Therefore, the
Hilbert space $H$ of each subsystem has $D=2^{\lfloor d/2\rfloor}$ dimensions.

For $d=3$ the assignment is the same as we have already discussed. The three
$\hat{\gamma}_{i}$ operators are the Pauli operators themselves, and $D=2$.
For $d=2$ one can choose any two of the Pauli operators. With the choice of
$\hat{\sigma}_{x}$ and $\hat{\sigma}_{z}$, one arrives at a real Hilbert
space. For $d=5$ the five operators corresponding to the recipe given in Ref.
[21] are $\gamma_{1}=\hat{\sigma}_{x}\otimes\hat{I}_{2}$,
$\gamma_{2}=\hat{\sigma}_{y}\otimes\hat{I}_{2}$,
$\gamma_{3}=\hat{\sigma}_{z}\otimes\hat{\sigma}_{x}$,
$\gamma_{4}=\hat{\sigma}_{z}\otimes\hat{\sigma}_{y}$ and
$\gamma_{5}=\hat{\sigma}_{z}\otimes\hat{\sigma}_{z}$, and $D=4$. For $d=4$ any
four of these operators may be chosen. However only three of them are real,
therefore we can not get a real Hilbert space this way. Even if we do not
follow the recipe, we can not find four pairwise anticommuting operators among
all real ones which are tensor products of either two Pauli operators or one
Pauli operator and the identity.

Following the construction discussed above, the dimensionality of the Hilbert
space grows exponentially with $d$. We do not know the conditions for the
existence of some more economical assignment. In Ref. [24] almost a hundred
Bell inequalities that required four dimensional Euclidean vectors for maximum
violation have been considered, and neither of them could maximally be
violated in smaller than $4\times 4$ dimensional complex Hilbert spaces. We
have also investigated numerically some inequalities requiring $6$ and $8$
dimensional Euclidean spaces for maximum violation, and we could not violate
them maximally with less than $8\times 8$ and $16\times 16$ dimensional
quantum systems, respectively.

It has been shown that for all $d$-dimensional vector pairs $\vec{a}$ and
$\vec{b}$ one can assign operator pairs $\hat{a}$ and $\hat{b}$ from the
$D=2^{\lfloor d/2\rfloor}$ dimensional Hilbert space $H$ such that Eq. (27) is
satisfied with $|\Phi\rangle=\sum_{\mu=1}^{D}|\mu\mu\rangle$. The opposite
statement is not true for $d>3$. As all measurement operators constructed have
zero trace, other operators belonging to legitimate two-outcome measurements
are excluded. Not even all zero trace operators are covered: the $d-1$
independent parameters characterizing the unit vectors are not sufficient to
characterize all such operators. Moreover only Hilbert spaces with
dimensionalities of a power of two have occured.

Therefore, to show that for any quantum scenario one can find appropriate
Euclidean vectors for the two parties requires a different construction. This
relation is less important as far as this paper is concerned, but for the sake
of completenes we still sketch the argument here. Let $H_{A}$ and $H_{B}$ be
the $D_{A}$ and $D_{B}$ dimensional Hilbert spaces of the two parties,
respectively. Let the measurements be performed on the state $|\Phi\rangle\in
H_{A}\otimes H_{B}$. Then for any $\hat{a}$ and $\hat{b}$ for the parties the
$2D_{A}D_{B}$ dimensional real vectors $\vec{a}$ and $\vec{b}$ whose
components are the real and the imaginary parts of the components of
$\hat{a}\otimes\hat{I}_{B}|\Phi\rangle$ and
$\hat{I}_{A}\otimes\hat{b}|\Phi\rangle$, respectively satisfy Eq. 27, where
$I_{A}$ and $I_{B}$ are the identity operators in $H_{A}$ and $H_{B}$,
respectively. Using $\hat{a}^{2}=\hat{I}_{A}$ and $\hat{b}^{2}=\hat{I}_{b}$,
the statement is easy to prove. If the number of settings is $m_{A}$ and
$m_{B}$ for the respective parties, then an $(m_{A}+m_{B})$-dimensional
subspace is sufficient, as no more vectors are involved. Moreover, if one is
only interested in the maximum violation, no more than an $m_{A}$ or $m_{B}$
dimensional subspace is needed, whichever is the smaller. The reason is that
the optimum vectors on one side are in the subspace spanned by the vectors on
the other side.

##  Appendix B Proof of special property

In Section 4 we have proven that the maximum quantum value for the Bell
inequality whose Bell coefficients have been constructed as the scalar
products of two sets of $d$-dimensional unit vectors $\vec{A}_{i}$
($i=1,\cdots,m_{A})$ and $\vec{B}_{j}$ ($j=1,\cdots,m_{B})$ is $m_{A}m_{B}/d$
if both sets satisfy the constraints give in the Observation. This means that
the columns of matrix $A_{ij}$ whose rows are the coordinates of vectors
$\vec{A}_{i}$ are orthogonal to each other and have equal norm, which is
$\sqrt{m_{A}/d}$, and $\vec{B}_{j}$ also has this property. Here we will prove
that cross polytopes, simplices and $d$-cubes satisfy the constraints for all
$d$, and it is also true for the 2-dimensional regular polygones. The
additional two three-dimensional and the three four-dimensional Platonic
bodies has simply been checked, we do not consider them in this Appendix.

The cross polytope is simple. We get its matrix $A_{ij}$ if we write the
$d$-dimensional unit matrix above minus one times the same matrix. Then the
columns are trivially orthogonal and their norm is $\sqrt{2}$ (There is one
$1$ and one $-1$ in each column), as it should be.

Let us consider simplices in $d$-dimensions. They have $d+1$ vertices. Let the
elements of the matrix of size $(d+1)\times d$ be $A_{ij}=0$ for $j>i$,
$A_{ij}=\tau_{j}$ for $j=i$ and $A_{ij}=-\sigma_{i}$ for $j<i$, that is:

$\hat{A}=\begin{pmatrix}\tau_{1}&0&0&0&0&\cdots\\\
-\sigma_{1}&\tau_{2}&0&0&0&\cdots\\\
-\sigma_{1}&-\sigma_{2}&\tau_{3}&0&0&\cdots\\\
-\sigma_{1}&-\sigma_{2}&-\sigma_{3}&\tau_{4}&0&\cdots\\\
-\sigma_{1}&-\sigma_{2}&-\sigma_{3}&-\sigma_{4}&\tau_{5}&\cdots\\\
-\sigma_{1}&-\sigma_{2}&-\sigma_{3}&-\sigma_{4}&-\sigma_{5}&\cdots\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\end{pmatrix}$ (28)

where

$\displaystyle\tau_{j}$ $\displaystyle=\sqrt{\frac{(d+1)(d+1-j)}{d(d+2-j)}}$
$\displaystyle\sigma_{j}$ $\displaystyle=\sqrt{\frac{d+1}{d(d+1-j)(d+2-j)}}.$
(29)

it is easy to verify that $\tau_{1}=1$, and that
$\tau_{j}^{2}=\tau_{j+1}^{2}+\sigma_{j}^{2}$. From these it follows that

$1=\tau_{1}^{2}=\sigma_{1}^{2}+\tau_{2}^{2}=\sigma_{1}^{2}+\sigma_{2}^{2}+\tau_{3}^{2}=\cdots=\sum_{j=1}^{i-1}\sigma_{j}^{2}+\tau_{i}^{2},$
(30)

which is nothing else than the squares of the norms of the rows. It follows
from Eq. (28) that the scalar product of the $i$th and the $k$th rows for any
$k>i$ can be written as

$\sum_{j=1}^{i-1}\sigma_{j}^{2}-\sigma_{i}\tau_{i}=1-(\tau_{i}^{2}+\sigma_{i}\tau_{i})=1-\frac{d+1}{d}=-\frac{1}{d},$
(31)

where we used Eq. (30) and that $\tau_{i}^{2}+\sigma_{i}\tau_{i}=(d+1)/d$,
which can be verified using Eq. (29). From these it follows that the $d+1$
rows of the matrix are the coordinates of normalized vectors pointing towards
the vertices of a $d$-dimensional simplex. Now, let us look at the columns to
show that the constraints are met. The square of the norm of the $j$th column
is $\tau_{j}^{2}+(d+1-j)\sigma_{j}^{2}=(d+1)/d$, as required, which can be
checked by using Eq. (29). It can also be checked that
$\tau_{j}+(d+1-j)\sigma_{j}=0$, that is the sum of the elements in each column
is zero. From Eq. (28) it is obvious that the scalar product of columns $j$
and $l$ where $(l>j)$ is the sum of the elements of column $l$ multiplied by
$\sigma_{j}$, therefore, the columns are orthogonal to each other, so all
constraint are obeyed.

A simple way to generate the components of the $2^{d}$ vectors pointing
towards the simplices of a $d$-cube is to take the binary representation of
the integers between zero and $2^{d}-1$ (writing the necessary number of zeros
in front of the binary numbers), and replacing every zero by $-1$. If one
wants the vectors normalized, they should be multiplied by $1/\sqrt{d}$. To
show that the columns of the matrix whose rows are these vector components
have equal norm and they are orthogonal to each other, we may forget about the
normalization. Then the last column will have $-1$ and $+1$ values
alternating, starting with a $-1$. The column before that will have pairs of
$-1$ and $+1$ values alternating. In each column periods of $-1$s and $+1$s of
equal length alternate, where the length is some power of two, it is $2^{d-1}$
for the first column (the upper half of the column is -1, the lower one is +1)
and $2^{0}=1$ for the last one. It is easy to see that the scalar product of
any two different columns is zero: in each interval corresponding to a section
of equal signs in the column that has the longer sections there is an even
number of sections in the other column, therefore there are the same number of
$+1$ and $-1$ values in every such interval. The equality of the norms of the
column is trivial, all elements have the same absolute value.

Let us look at the Platonic solids of two-dimensional spaces, the convex
regular polygons. Let $\vec{A}_{i}$ be unit vectors pointing towards the
vertices of the regular $m$-sided polygon, and $\vec{x}$ and $\vec{y}$ be two
orthogonal vectors. Then

$\displaystyle\sum_{i=1}^{m}(\vec{A}_{i}\cdot\vec{x})(\vec{A}_{i}\cdot\vec{x})$
$\displaystyle=\sum_{i=1}^{m}\cos^{2}(\varphi_{i})$
$\displaystyle=\sum_{i=1}^{m}\frac{\cos(2\varphi_{i})+1}{2}=\frac{m}{2}$
$\displaystyle\sum_{i=1}^{m}(\vec{A}_{i}\cdot\vec{x})(\vec{A}_{i}\cdot\vec{y})$
$\displaystyle=\sum_{i=1}^{m}\cos(\varphi_{i})\sin(\varphi_{i})$
$\displaystyle=\sum_{i=1}^{m}\frac{\sin(2\varphi_{i})}{2}=0,$ (32)

where $\varphi_{i}$ the angle between $\vec{A}_{i}$ and $\vec{x}$ (measured
towards $\vec{y}$). A sufficient condition for the equations above to be
satisfied is that vectors $\vec{A}^{\prime}_{i}$, whose angles with $\vec{x}$
are two times the ones between $\vec{A}_{i}$ and $\vec{x}$ have a $k$-fold
symmetry. This is true in the present case, $k=m$ and $k=m/2$ for odd and even
sided polygons, respectively. There are cases when $\vec{A}$ has no such
symmetry at all, while $A^{\prime}_{i}$ does (for example, when $m$=2 and
$\vec{A}_{1}$ and $\vec{A}_{2}$ are two orthogonal vectors). We get the
constraints of the Observation if we write the basis vectors into the
equations.

The considerations above have relevance if $d>2$ as well, this is the reason
we have written the equations in the more general form. Let $\vec{A}_{i}$ have
an $l$-fold ($l>2$) symmetry for rotations in some plane. Let $\vec{x}$ and
$\vec{y}$ be orthogonal vectors in that plane. Then Eq. (32) is modified as

$\displaystyle\sum_{i=1}^{m}(\vec{A}_{i}\cdot\vec{x})(\vec{A}_{i}\cdot\vec{x})$
$\displaystyle=\sum_{i=1}^{m}\pi_{i}\cos^{2}(\varphi_{i})$
$\displaystyle=\sum_{i=1}^{m}\frac{\pi_{i}(\cos(2\varphi_{i})+1)}{2}=\sum_{i=1}^{m}\frac{\pi_{i}}{2}$
$\displaystyle\sum_{i=1}^{m}(\vec{A}_{i}\cdot\vec{x})(\vec{A}_{i}\cdot\vec{y})$
$\displaystyle=\sum_{i=1}^{m}\pi_{i}\cos(\varphi_{i})\sin(\varphi_{i})$
$\displaystyle=\sum_{i=1}^{m}\frac{\pi_{i}\sin(2\varphi_{i})}{2}=0,$ (33)

where $\pi_{i}$ is the length of the projection of $\vec{A}_{i}$ onto the
plane and $\varphi_{i}$ is the angle between the projection and $\vec{x}$.
Therefore, such symmetry ensures that some constraints are satisfied.