Incompatibility in General Probabilistic Theories, Generalized Spectrahedra, and Tensor Norms

Abstract

In this work, we investigate measurement incompatibility in general probabilistic theories (GPTs). We show several equivalent characterizations of compatible measurements. The first is in terms of the positivity of associated maps. The second relates compatibility to the inclusion of certain generalized spectrahedra. For this, we extend the theory of free spectrahedra to ordered vector spaces. The third characterization connects the compatibility of dichotomic measurements to the ratio of tensor crossnorms of Banach spaces. We use these characterizations to study the amount of incompatibility present in different GPTs, i.e. their compatibility regions. For centrally symmetric GPTs, we show that the compatibility degree is given as the ratio of the injective and the projective norm of the tensor product of associated Banach spaces. This allows us to completely characterize the compatibility regions of several GPTs, and to obtain optimal universal bounds on the compatibility degree in terms of the 1-summing constants of the associated Banach spaces. Moreover, we find new bounds on the maximal incompatibility present in more than three qubit measurements.

Appendices

Appendix A: Necessary Background

1.1 A.1. Convex cones

Let L be a finite-dimensional real vector space. A subset $L^+ \subseteq L$ is a convex cone if $\lambda x + \mu y \in L^+$ for all x, $y \in L^+$ and all $\lambda$, $\mu \in {\mathbb {R}}_+$. Often, we drop “convex” and talk simply about “cones”. To avoid pathologies, we assume all cones to be non-empty. A cone will be called generating if $L = L^+ - L^+$. Moreover, it is pointed if $L^+ \cap (-L^+) = \{0\}$ (sometimes this is called salient instead). A proper cone is a convex cone which is closed, pointed and generating. A base of a cone C is a convex set $K \subset C$ such that for every $x \in C$, there is a unique $\lambda \ge 0$ such that $x \in \lambda K$. The dual cone of $L^+$ is the closed convex cone

$$\begin{aligned} (L^+)^*:= \{f \in L^*: f(x) \ge 0 ~\forall x \in L \}, \end{aligned}$$

(42)

where $L^*$ is the dual vector space of L. Two cones $L^+_1$ and $L^+_2$ living in vector spaces $L_1$ and $L_2$, respectively, are isomorphic if there is a linear bijection $\Theta : L_1 \rightarrow L_2$ such that $\Theta (L^+_1) = L^+_2$. A cone $L^+$ is called simplicial if it is isomorphic to ${\mathbb {R}}^{d}_+$, where $d = \dim (V^+)$.

A preordered vector space is a tuple $(L, L^+)$, where L is a vector space and $L^+$ is a convex cone. If $L^+$ is additionally pointed, it is an ordered vector space. If $V^+$ is a proper cone, we call $(V, V^+)$ a proper ordered vector space. We write $y \ge x$ for x, $y \in V$ to mean that $y - x \in V^+$. If $(V, V^+)$ is a preordered vector space, then $(V^*, (V^+)^*)$ is its dual preordered vector space. If $(V, V^+)$ is a proper ordered vector space, so is $(V^*, (V^+)^*)$.

An order unit $\mathbbm {1} \in V^+$ is an element such that for every $v \in V$, there is a $\lambda > 0$ such that $v \in \lambda [-\mathbbm {1}, \mathbbm {1}]$. By [AT07, Lemma 1.7], $\mathbbm {1} \in V^+$ is an order unit if and only if $\mathbbm {1} \in {\text {int}}\, V^+$.

Lemma A.1

Let $(L, L^+)$ be a preordered vector space and let $\mathbbm {1}$ be an order unit. Then $L^+ = L$ if and only if $- \mathbbm {1} \in L^+$.

Proof

One implication is clear. Thus, let $- \mathbbm {1} \in L^+$. Let $v \in L$. Since $\mathbbm {1}$ is an order unit, there is a $\lambda > 0$ such that $v + \lambda \mathbbm {1} \in L^+$. Since $- \lambda \mathbbm {1} \in L^+$, it follows that $v \in L^+$. This proves the assertion since $v \in L$ was arbitrary. $\square$

The following facts about cones and their duals will be useful:

Theorem A.2

(Bipolar theorem, [Roc70, Theorem 14.1]). Every non-empty closed convex cone $C \subseteq L$ satisfies $C^{**} \cong C$, where we have identified $L \cong L^{**}$.

Lemma A.3

([Mul97]). Let $C \subseteq L$ be a convex cone. The cone C is generating if and only if ${\text {int}}\,{C} \ne \varnothing$. If ${\text {cl}}\,C$ is pointed, then $C^*$ is generating. If C is generating, then $C^*$ is pointed. In particular, if C is proper, then $C^*$ is proper.

1.2 A.2. Tensor products of cones

There are in general infinitely many natural ways to define the tensor product of two cones $L^+_1 \subseteq L_1$ and $L^+_2 \subseteq L_2$. Among these, there is a minimal and a maximal choice: The minimal tensor product of $L^+_1$ and $L^+_2$ is the cone

$$\begin{aligned} L^+_1 \otimes _{\min } L^+_2 := \mathrm {conv}\{x \otimes y: x \in L^+_1, y \in L^+_2 \}, \end{aligned}$$

(43)

whereas the maximal tensor product of $L^+_1$ and $L^+_2$ is defined as

$$\begin{aligned} L^+_1 \otimes _{\max } L^+_2 := ((L^+_1)^*\otimes _{\min } (L^+_2)^*)^*. \end{aligned}$$

(44)

It can be seen that if $L^+_1$ and $L^+_2$ are proper, $L^+_1 \otimes _{\min } L^+_2$ and $L^+_1 \otimes _{\max } L^+_2$ are proper as well [ALP19, Fact S23]. Moreover, $L_1 \otimes _{\min } L_2$ is closed if $L_1$ and $L_2$ are [AS17, Exercise 4.14]. We call C a tensor cone for $L_1^+$ and $L_2^+$ if

$$\begin{aligned} L_1^+ \otimes _{\min } L_2^+ \subseteq C \subseteq L_1^+ \otimes _{\max } L_2^+. \end{aligned}$$

From the recent work [ALPP19], we know that the tensor product of two cones is unique if and only if one of the cones is simplicial. This solves a longstanding open problem from [NP69, Bar81]:

Theorem A.4

([ALPP19, Theorem A]). Let $L^+_1$ and $L^+_2$ be proper cones. Then, $L^+_1 \otimes _{\min } L^+_2 = L^+_1 \otimes _{\max } L^+_2$ if and only if $L^+_1$ or $L^+_2$ is simplicial.

1.3 A.3. Positive maps

Let $(L,L^+)$ be a preordered vector space and let $(L^*,(L^+)^*)$ be its dual. Consider the identity map $\mathrm {id}_L : L \rightarrow L$ and the associated canonical evaluation tensor $\chi _L \in L\otimes L^*$, defined by the following remarkable property:

$$\begin{aligned} \forall v \in L, \, \forall \alpha \in L^*, \qquad \langle \chi _L, \alpha \otimes v \rangle = \alpha (v). \end{aligned}$$

(45)

Using coordinates, we have

$$\begin{aligned} \chi _L = \sum _{i=1}^{\dim L} v_i \otimes \alpha _i \in L \otimes L^*\end{aligned}$$

(46)

for $\{v_i\}_{i = 1}^{\dim L}$ a basis of L and $\{\alpha _i\}_{i = 1}^{\dim L}$ the corresponding dual basis in $L^*$ (we have $\alpha _i(v_j) = \delta _{ij}$). Let $(M,M^+)$ be another preordered vector space and let $\Phi :M\rightarrow L^*$ be a linear map. We define the linear functional $s_\Phi : M\otimes L\rightarrow {\mathbb {R}}$ as

$$\begin{aligned} s_\Phi (z)=\langle \chi _L, (\Phi \otimes \mathrm {id})(z)\rangle , \qquad z\in M\otimes L. \end{aligned}$$

(47)

We then have

$$\begin{aligned} \Phi (w)=\sum _{i=1}^{\dim L} s_\Phi (w\otimes v_i)\alpha _i,\qquad \forall w\in M. \end{aligned}$$

(48)

Since $(M\otimes L)^*\cong M^*\otimes L^*$, $s_\Phi$ corresponds to an element $\varphi ^\Phi \in M^*\otimes L^*$, determined as

$$\begin{aligned} \langle \varphi ^\Phi , w\otimes v\rangle = s_\Phi (w\otimes v)=\langle \Phi (w), v \rangle , \qquad w\in M,\ v\in L. \end{aligned}$$

(49)

One can check that

$$\begin{aligned} \varphi ^\Phi = (\mathrm {id} \otimes \Phi )(\chi _{M^*}) = (\Phi ^*\otimes \mathrm {id} )(\chi _{L}). \end{aligned}$$

(50)

Here, $\Phi ^*:L \rightarrow M^*$ is the dual linear map. Note that we have $\chi _L=\varphi ^{\mathrm {id}_{L^*}}$. In this paper, we shall often switch between the three different equivalent points of view:

$$\begin{aligned} \begin{array}{ccccc} \hbox {linear map }&{} \qquad \longleftrightarrow \qquad &{} \hbox {linear form} &{} \qquad \longleftrightarrow \qquad &{} \hbox {tensor} \\ \Phi :M\rightarrow L^*&{}&{} s_\Phi : M\otimes L\rightarrow {\mathbb {R}} &{}&{} \varphi ^\Phi \in M^*\otimes L^*\end{array} \end{aligned}$$

We say that $\Phi$ is a positive map $(M,M^+)\rightarrow (L^*,(L^+)^*)$ if $\Phi (M^+)\subseteq (L^+)^*$. It is quite clear that this happens if and only if $s_\Phi :(M\otimes L, M^+\otimes _{\mathrm {min}}L^+)\rightarrow {\mathbb {R}}$ is positive, equivalently, $\varphi ^\Phi \in (M^+)^*\otimes _{\mathrm {max}}(L^+)^*$; this gives the well-known correspondence between positive maps and the maximal tensor product. In particular, we have

Lemma A.5

The identity map $\mathrm {id}_L : (L,L^+) \rightarrow (L, L^+)$ is positive. Moreover, $\chi _L \in L^+\otimes _{\mathrm {max}}(L^+)^*$.

Proof

The first assertion is obvious. The second one follows from Eq. (45). $\square$

For a positive linear map $\Psi :(M,M^+)\rightarrow (L,L^+)$, the dual map $\Psi ^{*}:(L^*,(L^+)^*)\rightarrow (M^*,(M^+)^*)$ is positive as well. Moreover, it holds that for preordered vector spaces $(L_i, L_i^+)$, $(M_i, M_i^+)$ and positive maps $\Phi _i: (L_i, L_i^+) \rightarrow (M_i, M_i^+)$, $i \in [2]$, both

$$\begin{aligned} \Phi _1 \otimes \Phi _2: (L_1 \otimes L_2, L_1^+ \otimes _{\min } L_2^+) \rightarrow (M_1 \otimes M_2, M_1^+ \otimes _{\min } M_2^+) \end{aligned}$$

and

$$\begin{aligned} \Phi _1 \otimes \Phi _2: (L_1 \otimes L_2, L_1^+ \otimes _{\max } L_2^+) \rightarrow (M_1 \otimes M_2, M_1^+ \otimes _{\max } M_2^+) \end{aligned}$$

are positive [Jen18, Section II.C]. Both statements also follow from Lemma A.6 below.

The minimal tensor product is associated with a special kind of maps. We first prove the following lemma, clarifying what happens under reordering of tensor products of cones.

Lemma A.6

Let $A^+,B^+,C^+,D^+$ be cones. Then

$$\begin{aligned} (A^+ \otimes _{\mathrm {min}}B^+) \otimes _{\mathrm {min}}(C^+ \otimes _{\mathrm {max}}D^+) \subseteq (A^+ \otimes _{\mathrm {min}}C^+) \otimes _{\mathrm {max}}(B^+ \otimes _{\mathrm {min}}D^+). \end{aligned}$$

Proof

Consider arbitrary $a \in A^+$, $b \in B^+$ and $e \in C^+ \otimes _{\mathrm {max}}D^+$. We have to check that for any $\varphi \in (A^*)^+ \otimes _{\mathrm {max}}(C^*)^+$ and $\psi \in (B^*)^+ \otimes _{\mathrm {max}}(D^*)^+$, we have

$$\begin{aligned} \mathinner {\langle {a \otimes b \otimes e, \varphi \otimes \psi }\rangle } \ge 0. \end{aligned}$$

We shall use (twice) the following fact: given cones $X^+,Y^+$ in vector spaces X, Y, $z \in X^+ \otimes _{\mathrm {max}}Y^+$ and $\sigma \in (X^+)^*$, the vector $y = \mathinner {\langle {\sigma , z}\rangle } \in Y$ defined by

$$\begin{aligned} \mathinner {\langle {\tau , y}\rangle } = \mathinner {\langle {\sigma \otimes \tau ,z}\rangle }, \qquad \forall \tau \in Y^* \end{aligned}$$

is positive (i.e. $y \in \overline{Y^+}\cong (Y^+)^{**}$). This corresponds to the fact that the evaluation of a positive map at a positive element is positive, and we leave its proof as an exercise for the reader.

Using the fact above, and writing $\gamma := \mathinner {\langle {a,\varphi }\rangle } \in (C^+)^*$ and $\delta :=\mathinner {\langle {b,\psi }\rangle } \in (D^+)^*$, we have

$$\begin{aligned} \mathinner {\langle {a \otimes b \otimes e, \varphi \otimes \psi }\rangle } = \mathinner {\langle {e, \gamma \otimes \delta }\rangle } \ge 0, \end{aligned}$$

proving the statement of the lemma. $\square$

Proposition A.7

Let $\Phi : (M,M^+) \rightarrow (L^*, (L^+)^*)$ be a positive map between preordered vector spaces. The following conditions are equivalent:

• (1) The map $\Phi \otimes \mathrm {id}_{L} : (M \otimes L, M^+ \otimes _{\mathrm {max}}\overline{L^+}) \rightarrow (L^*\otimes L ,(L^+)^*\otimes _{\mathrm {min}}\overline{L^+})$ is positive.

• (2) The map $s_\Phi : (M \otimes L,M^+ \otimes _{\mathrm {max}}L^+) \rightarrow {\mathbb {R}}$ is positive.

• (3) $\varphi ^\Phi \in (M^+)^* \otimes _{\mathrm {min}}(L^+)^*$

• (4) The map $\Phi \otimes \mathrm {id}_N : (M \otimes N,M^+ \otimes _{\mathrm {max}}N^+) \rightarrow (L^*\otimes N,(L^+)^*\otimes _{\mathrm {min}}N^+)$ is positive, for any preordered vector space $(N,N^+)$ with $N^+$ closed.

Proof

Since both $(2) \iff (3)$ and $(4) \implies (1)$ are trivial, we only prove $(1) \implies (2)$ and $(3) \implies (4)$. The first implication follows immediately from Lemma A.5. For the second implication, use Lemmas A.5 and A.6 to prove that

$$\begin{aligned} \varphi ^\Phi&\otimes&\chi _N \in ((M^*)^+ \otimes _{\mathrm {min}}(L^+)^*) \otimes _{\mathrm {min}}((N^*)^+ \otimes _{\mathrm {max}}N^+) \subseteq (M^+ \otimes _{\mathrm {max}}N^+)^* \\&\otimes _{\mathrm {max}}((L^+)^*\otimes _{\mathrm {min}}N^+), \end{aligned}$$

which is precisely the desired conclusion, since $\varphi ^\Phi \otimes \chi _N$ is the tensor corresponding to the map $\Phi \otimes \mathrm {id}_N$. $\square$

Definition A.8

Let $\Phi :(M,M^+)\rightarrow (L^*,(L^+)^*)$ be a positive linear map between preordered vector spaces. We say that $\Phi$ is entanglement breaking (EB) if any of the equivalent conditions in Proposition A.7 holds.

Remark A.9

Point (1) of Proposition A.7 was used as a definition in [Jen18]. Note that this definition of entanglement breaking maps agrees with the usual one used in quantum mechanics [HSR03] for $L^*$, M being the vector space of Hermitian matrices and their respective cones being the cones of positive semidefinite elements. This can be seen from the point (3) of Proposition A.7, which states that the corresponding Choi matrix is separable [Wat18, Exercise 6.1].

We conclude this section with a small lemma connecting the trace of the composition of two maps and the inner product of the corresponding tensors. Recall that the trace of a linear operator $\Phi : L \rightarrow L$ is defined as

$$\begin{aligned} {\mathrm{Tr}} \Phi = \sum _{j=1}^{\dim L} \alpha _j(\Phi (v_j)) = \langle \chi _L, \varphi ^\Phi \rangle , \end{aligned}$$

where $\{v_j\}_{j = 1}^{\dim L}$, $\{\alpha _j\}_{j = 1}^{\dim L}$, are dual bases of L, $L^*$.

Lemma A.10

Let $\Psi : M \rightarrow L$, $\Phi : M^*\rightarrow L^*$ be two linear maps between vector spaces. Then,

$$\begin{aligned} {\mathrm{Tr}}[\Psi \Phi ^*] = \langle \varphi ^{\Phi }, \varphi ^{\Psi } \rangle . \end{aligned}$$

Proof

Using the definition of the trace and Eq. (50), we have

$$\begin{aligned} {\mathrm{Tr}}[\Psi \Phi ^*]&= \langle \chi _L, \varphi ^{\Psi \Phi ^*} \rangle \\&= \langle \chi _L, (\Phi \Psi ^* \otimes \mathrm {id}_L)(\chi _{L^*})\rangle \\&= \langle (\Phi ^*\otimes \mathrm {id}_{L^*}) (\chi _{L}), (\Psi ^*\otimes \mathrm {id}_{L})(\chi _{L^*}) \rangle \\&= \langle \varphi ^{\Phi }, \varphi ^{\Psi } \rangle . \end{aligned}$$

$\square$

1.4 A.4. Tensor products of Banach spaces

Besides the tensor product of convex cones, we will also need the tensor product of Banach spaces. See [Rya02] for a good introduction. Let X, Y be two Banach spaces with norms $\Vert \cdot \Vert _X$ and $\Vert \cdot \Vert _Y$, respectively. Then, there are usually infinitely many natural norms that can be used to turn the vector space $X \otimes Y$ into a Banach space. There are two choices of norms which are minimal and maximal in a sense as we shall see next.

Definition A.11

(Projective tensor norm). The projective norm of an element $z \in X \otimes Y$ is defined as

$$\begin{aligned} \Vert z\Vert _{X \otimes _\pi Y}:= \inf \left\{ \sum _i \Vert x_i\Vert _X \Vert y_i\Vert _Y: z = \sum _{i} x_i \otimes y_i \right\} . \end{aligned}$$

Let $X^*$ and $Y^*$ be the dual spaces of X and Y, respectively, and let their norms be $\Vert \cdot \Vert _{X^*}$ and $\Vert \cdot \Vert _{Y^*}$.

Definition A.12

(Injective tensor norm). Let $z = \sum _i x_i \otimes y_i \in X \otimes Y$. Then, its injective norm is

$$\begin{aligned} \Vert z\Vert _{X \otimes _\varepsilon Y}:= \sup \left\{ \left| \sum _i \varphi (x_i) \psi (y_i)\right| : \Vert \varphi \Vert _{X^*} \le 1, \Vert \psi \Vert _{Y^*} \le 1 \right\} . \end{aligned}$$

If X and Y are clear from the context, we will sometimes only write $\Vert \cdot \Vert _\pi$ and $\Vert \cdot \Vert _\varepsilon$, respectively. Importantly, $\Vert \cdot \Vert _\varepsilon$ and $\Vert \cdot \Vert _\pi$ are dual norms, i.e.

$$\begin{aligned} \Vert z\Vert _{X \otimes _\varepsilon Y} = \sup _{\Vert \varphi \Vert _{X^*\otimes _\pi Y^*} \le 1} |\varphi (z)| \end{aligned}$$

and vice versa.

In some cases, the projective and injective norms have simpler expressions. The projective norm $\ell ^{g}_1 \otimes _\pi X$ of a vector

$$\begin{aligned} {\mathbb {R}}^g \otimes X \ni z = \sum _{i=1}^g e_i \otimes z_i \end{aligned}$$

with $\{e_i \}_{i \in [g]}$ the standard basis is given by (see e.g. [Rya02, Example 2.6])

$$\begin{aligned} \Vert z\Vert _{\ell _1^g \otimes _\pi X} = \sum _{i=1}^g \Vert z_i\Vert _{X}. \end{aligned}$$

(51)

The injective norm $\ell ^g_1 \otimes _\varepsilon X$ is (see e.g. [Rya02, Example 3.4])

$$\begin{aligned} \Vert z\Vert _{\ell _1^g \otimes _\varepsilon X} = \sup _{\Vert y\Vert _{X^*} \le 1} \sum _{i = 1}^g |\langle y, z_i \rangle | = \sup _{\varepsilon \in \{\pm 1\}^g} \Vert \sum _{i = 1}^g \varepsilon _i z_i\Vert _X. \end{aligned}$$

(52)

We will later be interested in the maximal ratio between the projective and the injective norms introduced in [ALP+20, Equation (15)]:

$$\begin{aligned} \rho (X, Y) = \max _{\Vert z\Vert _{\varepsilon }\le 1} \Vert z\Vert _{\pi } \end{aligned}$$

(53)

where the maximum runs over all $z \in X \otimes Y$.

The injective and the projective norms both belong to a class of tensor norms which are the reasonable crossnorms.

Definition A.13

([Rya02]). Let X and Y be two Banach spaces. We say that a norm $\Vert \cdot \Vert _\alpha$ on $X \otimes Y$ is a reasonable crossnorm if it has the following properties:

• (1) $\Vert x \otimes y\Vert _\alpha \le \Vert x\Vert _X \Vert y\Vert _Y$ for all $x \in X$, $y \in Y$,

• (2) For all $\varphi \in X^*$, for all $\psi \in Y^*$, $\varphi \otimes \psi$ is bounded on $X \otimes Y$ and $\Vert \varphi \otimes \psi \Vert _{\alpha ^*} \le \Vert \varphi \Vert _{X^*} \Vert \psi \Vert _{Y^*}$,

where $\Vert \cdot \Vert _{\alpha ^*}$ is the dual norm to $\Vert \cdot \Vert _\alpha$.

The injective and projective norms are the smallest and largest reasonable crossnorms we can put on $X \otimes Y$, respectively:

Proposition A.14

([Rya02, Proposition 6.1]). Let X and Y be Banach spaces.

• (a) A norm $\Vert \cdot \Vert _\alpha$ on $X \otimes Y$ is a reasonable crossnorm if and only if

  $$\begin{aligned} \Vert z\Vert _{X \otimes _\varepsilon Y} \le \Vert z\Vert _\alpha \le \Vert z\Vert _{X \otimes _\pi Y} \end{aligned}$$

  for all $z \in X \otimes Y$.

• (b) If $\Vert \cdot \Vert _\alpha$ is a reasonable crossnorm on $X \otimes Y$, then $\Vert x \otimes y\Vert _\alpha = \Vert x\Vert _X \Vert y\Vert _Y$ for every $x \in X$ and every $y \in Y$. Furthermore, for all $\varphi \in X^*$ and all $\psi \in Y^*$, the norm $\Vert \cdot \Vert _{\alpha ^*}$ satisfies $\Vert \varphi \otimes \psi \Vert _{\alpha ^*} = \Vert \varphi \Vert _{X^*} \Vert \psi \Vert _{Y^*}$.

Appendix B: An Extension Theorem for Ordered Vector Spaces

In this section, we prove an extension theorem which we use in Sect. 5 to relate the compatibility of measurements in a GPT to properties of an associated map (in particular Proposition B.8). Along the way, we will use the extension theorem to study tensor products of certain cones.

Let $E\subseteq {\mathbb {R}}^d$ be a subspace containing a point with positive coordinates:

$$\begin{aligned} E \cap {\text {ri}}({\mathbb {R}}_+^d) \ne \varnothing . \end{aligned}$$

(54)

We set $E^+:=E \cap {\mathbb {R}}_+^d$. Note that $E^+$ is proper in E (see [Roc70, Corollary 6.5.1] to conclude that $E^+$ is generating).

We shall use the following key extension theorem (see e.g. [Cas05, Theorem 1]):

Theorem B.1

(M. Riesz extension theorem). Let $(X,X^+)$ be a preordered vector space, $Y \subseteq X$ a linear subspace, and $\varphi : Y \rightarrow {\mathbb {R}}$ a positive linear form on $(Y, Y^+)$, where $Y^+:=Y \cap X^+$. Assume that for every $x \in X$, there exists $y \in Y$ such that $x \le y$. Then, there exists a positive linear form ${\tilde{\varphi }}: X \rightarrow {\mathbb {R}}$ such that ${\tilde{\varphi }}|_Y = \varphi$.

Remark B.2

Note that [Cas05, Theorem 1] states the theorem only for ordered vector spaces. However, the same proof works if $X^+$ is an arbitrary convex cone.

We prove now the main result of this section; see also Remark B.7 for an equivalent formulation. Note that the tensor product of ${\mathbb {R}}_+^d$ with a proper cone $L^+$ is unique since ${\mathbb {R}}_+^d$ is simplicial.

Proposition B.3

Let $(L,L^+)$ be a proper ordered vector space, and $E \subseteq {\mathbb {R}}^d$ as in Eq. (54). Any positive linear form

$$\begin{aligned} \varphi : (E \otimes L,(E \otimes L) \cap ({\mathbb {R}}_+^d \otimes L^+)) \rightarrow {\mathbb {R}} \end{aligned}$$

(55)

can be extended to a positive linear form ${\tilde{\varphi }}: {\mathbb {R}}^d \otimes L \rightarrow {\mathbb {R}}$.

Proof

We shall use Theorem B.1 with $X = {\mathbb {R}}^d \otimes L$, $X^+ = {\mathbb {R}}^d_+ \otimes L^+$ and $Y = E \otimes L$. We have to show that for any $x \in {\mathbb {R}}^d \otimes L$, there is a $y \in E \otimes L$ such that $y-x \in {\mathbb {R}}^d_+ \otimes L^+$. It is enough to consider simple tensors of the form $x= r \otimes v$, where $r \in {\mathbb {R}}^d$ and $v \in L$; the general case will follow by linearity. Since $L^+$ is generating, there are $v_+$, $v_- \in L^+$ such that $v = v_+ -v_-$. Furthermore, from the assumption (54), E contains a vector with strictly positive coordinates e, hence there exist $\lambda _\pm > 0$ such that $\lambda _+ e - r \ge 0$ and $\lambda _- e + r \ge 0$. Then,

$$\begin{aligned} \lambda _+ e \otimes v_+ + \lambda _- e \otimes v_- - r \otimes v = \lambda _+ e \otimes v_+ + \lambda _- e\otimes v_- - r \otimes v_+ + r \otimes v_- \in {\mathbb {R}}^d_+ \otimes L^+ \end{aligned}$$

and $\lambda _\pm e \otimes v_\pm \in E \otimes L$. Thus, we can choose $y = \lambda _+ e \otimes v_+ + \lambda _- e \otimes v_-$. $\square$

We can now identify the dual proper ordered vector space $(E^*,(E^+)^*)$.

Proposition B.4

Let us identify the dual vector space $E^*\cong E$, with duality given by the standard inner product in ${\mathbb {R}}^d$. We then have $(E^+)^*=J({\mathbb {R}}^d_+)$, where $J:{\mathbb {R}}^d\rightarrow E$ is the orthogonal projection onto E. For a proper ordered vector space $(L,L^+)$, we have

$$\begin{aligned} (E^+)^*\otimes _{\mathrm {min}}L^+=(J\otimes {\text {id}}_L)({\mathbb {R}}^d_+\otimes L^+). \end{aligned}$$

Proof

Let $r\in {\mathbb {R}}^d_+$, then for $e\in E^+ \subseteq {\mathbb {R}}^d_+$

$$\begin{aligned} \langle J(r),e\rangle =\langle r, e\rangle \ge 0, \end{aligned}$$

since ${\mathbb {R}}^d_+$ is self-dual, so that $J({\mathbb {R}}^d_+)\subseteq (E^+)^*$. By Proposition B.3, any $\varphi \in (E^+)^*$ extends to a positive form ${\tilde{\varphi }}: {\mathbb {R}}^d\rightarrow {\mathbb {R}}$, so that $\varphi =J({\tilde{\varphi }})$, which implies the reverse inclusion. If $(L,L^+)$ is a proper ordered vector space, then $(E^+)^*\otimes _{\mathrm {min}}L^+$ is a cone of elements of the form

$$\begin{aligned} \sum _j \varphi _j\otimes v_j= \sum _j J({\tilde{\varphi }}_j)\otimes v_j= (J\otimes {\text {id}}_L)(\sum _j {\tilde{\varphi }}_j\otimes v_j) \end{aligned}$$

for $v_j\in L^+$ and ${\tilde{\varphi }}_j\in {\mathbb {R}}^d_+$, proving the last statement. $\square$

We now provide a useful characterization of the maximal tensor product of $E^+$ with $L^+$, identifying at the same time the cone appearing in Eq. (55).

Proposition B.5

For E as above and $(L,L^+)$ a proper ordered vector space, it holds that

$$\begin{aligned} E^+ \otimes _{\mathrm {max}} L^+ = (E \otimes L) \cap ({\mathbb {R}}_+^d \otimes L^+). \end{aligned}$$

Proof

The inclusion “$\subseteq$” follows from the monotonicity of the $\max$ tensor product with respect to each factor. To show the reverse inclusion “$\supseteq$”, we have to prove that for any $z \in (E \otimes L) \cap ({\mathbb {R}}_+^d \otimes L^+)$, and for any $\beta \in (E^+)^*$, $\alpha \in (L^+)^*$, we have that $\langle \beta \otimes \alpha , z \rangle \ge 0$. By Proposition B.4, $\beta \in (E^+)^*$ implies that $\beta =J({\tilde{\beta }})$ for a positive form ${\tilde{\beta }}: {\mathbb {R}}^d \rightarrow {\mathbb {R}}$. Since $z \in {\mathbb {R}}^d_+ \otimes L^+$, we have a decomposition $z = \sum _{i = 1}^d r_i \otimes v_i$, where $r_i \in {\mathbb {R}}^d_+$ and $v_i \in L^+$. This yields

$$\begin{aligned} \langle \beta \otimes \alpha , z \rangle = \langle {\tilde{\beta }} \otimes \alpha , z \rangle = \sum _{i = 1}^d {\tilde{\beta }}(r_i)\alpha ( v_i) \ge 0, \end{aligned}$$

finishing the proof. $\square$

Corollary B.6

For $E_1$, $E_2$ satisfying Eq. (54) (but not necessarily of the same dimension), it holds that

$$\begin{aligned} E_1^+ \otimes _{\mathrm {max}} E_2^+ = (E_1 \otimes E_2) \cap ({\mathbb {R}}_+^{d_1} \otimes {\mathbb {R}}_+^{d_2}). \end{aligned}$$

Proof

The proof is almost the same as for Proposition B.5, only that the functionals on both subspaces need to be extended. $\square$

Remark B.7

Using Proposition B.5, one can restate Proposition B.3 as follows: Any positive linear form $\varphi : (E \otimes L,E^+ \otimes _{\max } L^+) \rightarrow {\mathbb {R}}$ can be extended to a positive linear form ${\tilde{\varphi }}: {\mathbb {R}}^d \otimes L \rightarrow {\mathbb {R}}$ for proper $L^+$.

We now study extendability of general positive maps on $(E,E^+)$. The proof technique is inspired by the finite dimensional version of Arveson’s extension theorem [Pau03, Theorem 6.2].

Proposition B.8

Let $E \subseteq {\mathbb {R}}^d$ be a subspace such that $E \cap {\text {ri}}({\mathbb {R}}_+^d) \ne \varnothing$ and $E^+ = E \cap {\mathbb {R}}^d_+$. Let J be the orthogonal projection onto E and let $(L,L^+)$ be an proper ordered vector space. Finally, let $\Phi : E \rightarrow L^*$ be a linear map. The following are equivalent:

• (1) There exists a positive extension ${\tilde{\Phi }}: ({\mathbb {R}}^d, {\mathbb {R}}_+^d) \rightarrow (L^*, (L^+)^*)$ of $\Phi$.

• (2) The linear map $\Phi \otimes \mathrm {id}_L: (E \otimes L, E^+ \otimes _{\max } L^+) \rightarrow (L^*\otimes L, (L^+)^*\otimes _{\min } L^+)$ is positive.

• (3) The form $s_\Phi : (E \otimes L, E^+ \otimes _{\max } L^+) \rightarrow {\mathbb {R}}$ is positive.

• (4) $\varphi ^\Phi \in (J\otimes {\text {id}}_{L^*})({\mathbb {R}}^d_+\otimes (L^+)^*)$.

• (5) $\Phi$ is entanglement breaking.

Proof

By Propositions A.7 and B.4, the statements (2) - (5) are equivalent. It is therefore enough to show that (1) $\implies$ (2) and (3) $\implies$ (1). We start by showing that the existence of the positive extension (1) implies the positivity of the map $\Phi \otimes \mathrm {id}_L$ in (2). Both ${\tilde{\Phi }}$ and $\mathrm {id}_L$ are positive maps. Therefore, ${\tilde{\Phi }} \otimes \mathrm {id}_L: ({\mathbb {R}}^d \otimes L, {\mathbb {R}}_+^d \otimes _{\min } L^+ = {\mathbb {R}}_+^d \otimes _{\max } L^+) \rightarrow (L^*\otimes L, (L^+)^*\otimes _{\min } L^+)$ is positive. The claim follows by Proposition B.5, since $\Phi \otimes \mathrm {id}_L$ is a restriction of this map to $E \otimes L$.

It remains to show that $(3) \implies (1)$. Using Proposition B.3 and Remark B.7, we extend the form $s_\Phi$ to $\tilde{s}_\Phi : {\mathbb {R}}^d \otimes L \rightarrow {\mathbb {R}}$. Let ${\tilde{\Phi }}: ({\mathbb {R}}^d, {\mathbb {R}}_+^d) \rightarrow (L^*,(L^+)^*)$ be the positive map related to ${\tilde{s}}_\Phi$ as in Eq. (48). It remains to check that ${\tilde{\Phi }}$ is indeed an extension of $\Phi$. Let $v_j$ and $\alpha _j$ be elements of a basis of L and its dual basis for all $j \in [\dim L$]. For any $e \in E$, we compute

$$\begin{aligned} {\tilde{\Phi }}(e) = \sum _{j=1}^{\dim L} {\tilde{s}}_\Phi (e \otimes v_j) \alpha _j= \sum _{j=1}^{\dim L} s_\Phi (e \otimes v_j) \alpha _j=\Phi (e), \end{aligned}$$

finishing the proof. $\square$

Remark B.9

Note that the existence of a positive extension of $\Phi : E \rightarrow L^*$ can be checked using conic programming, see Sect. E.2 in the Appendix.

Appendix C: Generalized Spectrahedra

In this section, we will generalize some of the theory of (free) spectrahedra to the setting of ordered vector spaces by allowing for more general cones than the cone of positive semidefinite matrices. We reformulate the compatibility of measurements as an inclusion problem for generalized spectrahedra in Sect. 6. We define a generalized spectrahedron as a convex subset of some vector space which can be represented by positivity conditions with respect to some abstract cone.

Definition C.1

Let L, M be two finite-dimensional vector spaces and consider a cone $C \subseteq M \otimes L$. For a g-tuple of elements $a = (a_1, \ldots , a_g) \in M^g$, we define the generalized spectrahedron

$$\begin{aligned} {\mathcal {D}}_{a}(L,C) := \{(v_1, \ldots , v_g) \in L^g \, : \, \sum _{i=1}^g a_i \otimes v_i \in C\}. \end{aligned}$$

Remark C.2

It is easy to see that any generalized spectrahedron is a convex cone and that the generalized spectrahedron is closed if C is. In fact, note that the g-tuple $a\in M^g$ defines a linear map $a: {\mathbb {R}}^g\rightarrow M$, by $x\mapsto \sum _i x_ia_i$. The generalized spectrahedron ${\mathcal {D}}_{a}(L,C)$ is the largest cone in ${\mathbb {R}}^g\otimes L\cong L^g$ that makes the map

$$\begin{aligned} a\otimes \mathrm {id}_L : ({\mathbb {R}}^g \otimes L, {\mathcal {D}}_{a}(L,C)) \rightarrow (M \otimes L, C) \end{aligned}$$

positive. If a is a basis of M, then the corresponding map $a\otimes \mathrm {id}_L$ is an isomorphism through which the cones C and ${\mathcal {D}}_a(L,C)$ are affinely isomorphic.

Note that usual spectrahedra correspond to the choice $L = {\mathbb {R}}$, $M = \mathcal M^{\mathrm {sa}}_n(\mathbb C)$, and C being the positive semidefinite cone. Free spectrahedra are the union over $d \ge 1$ of generalized spectrahedra for $L = \mathcal M^{\mathrm {sa}}_d(\mathbb C$) and C being the PSD cone of $dn \times dn$ matrices; note that there exists no natural notion of generalized free spectrahedra, since there are no canonical sequences of cones $(C_d)_{d \ge 1}$, with $C_1=C$.

We will now consider generalized spectrahedra which are in some sense minimal and maximal. The definitions are inspired by the corresponding notions for matrix convex sets, see [DDOSS17, PSS18].

Definition C.3

Let $\mathcal C \subset \mathbb R^{g}$ be a closed convex cone. Let $(L,L^+)$ be a finite-dimensional preordered vector space. Then

$$\begin{aligned} {\mathcal {D}}_{\min }(\mathcal C; L, L^+){:=} \left\{ \sum _i x^{(i)}{\otimes } h_i\in {\mathbb {R}}^g{\otimes } L\cong L^g:\ x^{(i)} \in \mathcal C,~h_i \in L^+~\forall i\right\} {\cong } \mathcal C \otimes _{\min } L^+ \end{aligned}$$

is the minimal generalized spectrahedron corresponding to $\mathcal C$.

Definition C.4

Let $\mathcal C \subset {\mathbb {R}}^{g}$ be a closed convex cone. Let $(L,L^+)$ be a finite-dimensional preordered vector space with $L^+$ closed. Then,

$$\begin{aligned} {\mathcal {D}}_{\max }(\mathcal C; L, L^+):= & {} \left\{ (v_1, \ldots , v_g) \in L^g: \sum _{i = 1}^g h_i v_i \in L^+~\forall h \in {\mathbb {R}}^{g} \mathrm {~s.t.~} \right. \\&\quad \left. \sum _{i = 1}^g h_i c_i\ge 0~\forall c \in \mathcal {C}\right\} \cong \mathcal C \otimes _{\max } L^+ \end{aligned}$$

is the maximal generalized spectrahedron corresponding to $\mathcal C$.

In order to show that these sets are indeed generalized spectrahedra and to justify the identifications with minimal and maximal cones, we prove a lemma.

Lemma C.5

Let $(L, L^+)$ be a preordered vector space. Let $e = (e_1, \ldots , e_g)$ be the canonical basis of ${\mathbb {R}}^g$. Then,

$$\begin{aligned} {\mathcal {D}}_{\min }(\mathcal C; L, L^+) \cong {\mathcal {D}}_{e}(L, \mathcal C \otimes _{\min } L^+). \end{aligned}$$

If $L^+$ is closed, then

$$\begin{aligned} {\mathcal {D}}_{\max }(\mathcal C; L, L^+) \cong {\mathcal {D}}_{e}(L, \mathcal C \otimes _{\max } L^+). \end{aligned}$$

Proof

By definition, $(v_1, \ldots , v_g) \in {\mathcal {D}}_{\min }(\mathcal C; L, L^+)$ if and only if $v_j = \sum _{i} x_j^{(i)} h_i$ for some $x^{(i)} \in \mathcal C$ and $h_i \in L^+$. Moreover, $\sum _{j = 1}^g e_j \otimes v_j \in \mathcal C \otimes _{\min } L^+$ if and only if

$$\begin{aligned} \sum _{j = 1}^g e_j \otimes v_j = \sum _{i} x^{(i)} \otimes h_i \qquad x^{(i)} \in \mathcal C, h_i \in L^+. \end{aligned}$$

A comparison of coordinates proves the first assertion. Let now $\sum _{i = 1}^g e_i \otimes v_i \in \mathcal C \otimes _{\max } L^+$. By definition, this is true if and only if

$$\begin{aligned} \sum _{i = 1}^g \varphi (e_i) v_i \in L^+ \qquad \forall \varphi \in \mathcal C^*\end{aligned}$$

since $L^+$ is closed. Realizing that for $c = \sum _{i = 1}^g c_i e_i$, we can write $\varphi (c) = \sum _{i = 1}^g c_i h_i \ge 0$ with $h_i = \varphi (e_i)$ for all $i \in [g]$, proves one inclusion. The second assertion follows by identifying $(h_1, \ldots , h_g)$ as in the assertion with some $\varphi \in \mathcal C$. $\square$

The following proposition justifies that we speak of minimal and maximal cones.

Proposition C.6

Let $\mathcal C \subset {\mathbb {R}}^{g}$ be a closed convex cone. Let $(L,L^+)$, $(M,M^+)$ be two finite-dimensional preordered vector spaces with a tensor cone $C \subset M \otimes L$. Let $M^+$, $L^+$ be closed convex cones. If $a \in M^g$ such that ${\mathcal {D}}_a({\mathbb {R}}, M^+) = \mathcal C$, then

$$\begin{aligned} {\mathcal {D}}_{\min }(\mathcal C; L, L^+) \subseteq {\mathcal {D}}_a(L,C) \subseteq {\mathcal {D}}_{\max }(\mathcal C; L, L^+). \end{aligned}$$

Moreover, for a closed convex $\mathcal C^\prime$ such that $\mathcal C^\prime \subseteq \mathcal C$, it holds that

$$\begin{aligned} {\mathcal {D}}_{\min }(\mathcal C^\prime ; L, L^+) \subseteq {\mathcal {D}}_{\min }(\mathcal C; L, L^+) \quad \mathrm {and} \quad {\mathcal {D}}_{\max }(\mathcal C^\prime ; L, L^+) \subseteq {\mathcal {D}}_{\max }(\mathcal C; L, L^+). \end{aligned}$$

Proof

Let $v \in {\mathcal {D}}_{\min }(\mathcal C; L, L^+)$. Then, $v = (v_1, \ldots , v_g)$ where

$$\begin{aligned} v_j = \sum _i x^{(i)}_j h_i \qquad \forall j \in [g] \end{aligned}$$

(56)

for some $x^{(i)} \in \mathcal C$ and $h_i \in L^+$. Thus, we can rewrite

$$\begin{aligned} \sum _{j = 1}^g a_j \otimes v_j = \sum _{i} \left( \sum _{j = 1}^g x_j^{(i)} a_j \right) \otimes h_i. \end{aligned}$$

Since ${\mathcal {D}}_a({\mathbb {R}}, M^+) = \mathcal C$, it holds that

$$\begin{aligned} \sum _{j = 1}^g x_j^{(i)} a_j \in M^+ \qquad \forall i. \end{aligned}$$

Thus, since C is a tensor cone,

$$\begin{aligned} \sum _{j = 1}^g a_j \otimes v_j \in M^+ \otimes _{\min } L^+ \subseteq C. \end{aligned}$$

Therefore, we can conclude that $v \in {\mathcal {D}}_a(L,C)$, which proves the first assertion.

For the second assertion, let $v =(v_1, \ldots , v_g) \in D_a(L,C)$. Thus

$$\begin{aligned} \sum _{j = 1}^g a_j \otimes v_j \in C \subseteq M^+ \otimes _{\max } L^+, \end{aligned}$$

since C is a tensor cone. Thus,

$$\begin{aligned} \sum _{j = 1}^g \varphi (a_j) \psi (v_j) \ge 0 \qquad \forall \varphi \in (M^+)^*, \psi \in (L^+)^*. \end{aligned}$$

Using that $M^+ \cong (M^+)^{**}$ and that ${\mathcal {D}}_a({\mathbb {R}}, M^+) = \mathcal C$, it follows that $(\psi (v_1), \ldots , \psi (v_g)) \in \mathcal C$ for all $\psi \in (L^+)^*$. Hence, for all $\psi \in (L^+)^*$, it holds that

$$\begin{aligned} \sum _{i = 1}^g h_i \psi (v_i)\ge 0~\forall h \in {\mathbb {R}}^{g} \mathrm {~s.t.~} \sum _{i = 1}^g h_i c_i\ge 0~\forall c \in \mathcal C. \end{aligned}$$

Using that $L^+ \cong (L^+)^{**}$, it follows that $v \in {\mathcal {D}}_{\max }(\mathcal C; L, L^+)$. The inclusions in the last assertion follow directly from the Definitions C.3 and C.4. $\square$

Remark C.7

In view of Remark C.2, the first inclusion of the above proposition follows directly from the fact that the map $a: ({\mathbb {R}}^g,\mathcal C)\rightarrow (M,M^+)$ is positive and for any positive map, its tensor product with the identity map is positive with respect to the minimal tensor cone. Since we have $\mathcal C^*=a^*((M^+)^*)$, the second inclusion is a consequence of the same property of $a^*$.

Remark C.8

Assume that $M^+$ and $L^+$ are proper cones. By the results of [ALPP19], Proposition C.6 implies that the generalized spectrahedron ${\mathcal {D}}_a(L,C)$ does not depend on the choice of the tensor cone C if and only if at least one of the cones $\mathcal C={\mathcal {D}}_a({\mathbb {R}},M^+)$ or $L^+$ is simplicial.

As for free spectrahedra, we can connect the inclusion of generalized spectrahedra to the positivity of an associated map [HKM13, DDOSS17, HKMS19].

Proposition C.9

Let L, M and N be finite-dimensional vector spaces and let $C_M \subset M \otimes L$, $C_N \subset N \otimes L$ be two cones. Moreover, let $a \in M^g$, $b \in N^g$ be two tuples, where a consists of linearly independent elements which span the subspace $M^\prime \subseteq M$. We define a map $\Phi : M^\prime \rightarrow N$, $\Phi (a_i) = b_i$ for all $i \in [g]$. Then,

$$\begin{aligned} {\mathcal {D}}_a(L, C_M) \subseteq {\mathcal {D}}_b(L, C_N) \end{aligned}$$

if and only if $\Phi \otimes \mathrm {id}_L: (M^\prime \otimes L, M^\prime \otimes L \cap C_M) \rightarrow (N \otimes L, C_N)$ is positive.

Proof

Let us assume the inclusion. Consider $z \in M^\prime \otimes L$. Thus, we can write

$$\begin{aligned} z = \sum _{i = 1}^g a_i \otimes z_i, \end{aligned}$$

where $z_i \in L$ for all $i \in [g]$. If $z \in C_M$, then $(z_1, \ldots , z_g) \in {\mathcal {D}}_a(L, C_M)$ and hence

$$\begin{aligned} (\Phi \otimes \mathrm {id}_L)(z) = \sum _{i = 1}^g b_i \otimes z_i . \end{aligned}$$

The right hand side is in $C_N$ since ${\mathcal {D}}_a(L, C_M) \subseteq {\mathcal {D}}_b(L, C_N)$. Conversely, let $\Phi \otimes \mathrm {id}_L$ be positive. Let $(v_1, \ldots , v_g) \in {\mathcal {D}}_a(L, C_M)$. Then,

$$\begin{aligned} \sum _{i = 1}^g a_i \otimes v_i \in (M^\prime \otimes L) \cap C_M \end{aligned}$$

and the assertion follows from an application of $\Phi \otimes \mathrm {id}_L$ to this element. $\square$

As for free spectrahedra, it is possible to look at what inclusion with respect to some preordered vector space $(L,L^+)$ implies for the inclusion with respect to $({\mathbb {R}}, {\mathbb {R}}_+)$.

Proposition C.10

Let $(M,M^+)$, $(N,N^+)$ and $(L,L^+)$ be preordered vector spaces, where $N^+$ is closed and $L^+$ contains at least one element which is not in $\overline{-L^+}$. Moreover, let $C_M \subset M \otimes L$ and $C_N \subset N \otimes L$ be tensor cones and $a \in M^g$, $b \in N^g$ for some $g \in \mathbb N$. Then,

$$\begin{aligned} {\mathcal {D}}_a(L, C_M) \subseteq {\mathcal {D}}_b(L, C_N) \implies {\mathcal {D}}_a({\mathbb {R}}, M^+) \subseteq {\mathcal {D}}_b({\mathbb {R}}, N^+) \end{aligned}$$

Proof

Let $v \in L^+$, $v \not \in -\overline{L^+}$ and $x \in {\mathcal {D}}_a({\mathbb {R}}, M^+)$. Then, $(x_1 v, \ldots , x_g v) \in {\mathcal {D}}_a(L, C_M)$, since $C_M$ contains in particular $M^+ \otimes _{\min }L^+$. Thus,

$$\begin{aligned} \left( \sum _{i = 1}^g x_i b_i\right) \otimes v \in C_N. \end{aligned}$$

This implies that $\sum _{i = 1}^g x_i b_i \in (N^+)^{**} \cong N^+$, since $C_N \subseteq N^+ \otimes _{\max } L^+$ and we can find $\varphi \in (L^+)^*$ such that $\varphi (v) > 0$ since $\psi (v) = 0$ for all $\psi \in L^*$ implies $-v \in \overline{L^+}$. Hence $x \in {\mathcal {D}}_b({\mathbb {R}}, N^+)$. $\square$

In general, ${\mathcal {D}}_a({\mathbb {R}}, M^+) \subseteq {\mathcal {D}}_b({\mathbb {R}}, N^+)$ does not imply ${\mathcal {D}}_a(L, C_M) \subseteq {\mathcal {D}}_b(L, C_N)$. However, if M and N contain order units, the implication can be made true by shrinking the left hand side.

Definition C.11

Let $(M,M^+)$, $(N,N^+)$ and $(L,L^+)$ be preordered vector spaces where $M^+$ and $N^+$ contain order units $\mathbbm {1}_M$ and $\mathbbm {1}_N$, respectively. Moreover, let $C_M \subset M \otimes L$ and $C_N \subset N \otimes L$ be tensor cones and $a \in M^g$, $b \in N^g$ for some $g \in \mathbb N$. The set of inclusion constants for ${\mathcal {D}}_a(L, C_M)$ and $C_N$ is defined as

$$\begin{aligned} \Delta _a(L, C_M, C_N) :=&\{ s\in [0,1]^g: \forall b \in N^g,~ {\mathcal {D}}_{(\mathbbm {1}_M, a)}({\mathbb {R}}, M^+) \subseteq {\mathcal {D}}_{(\mathbbm {1}_N, b)}({\mathbb {R}}, N^+) \\&\implies (1,s) \cdot {\mathcal {D}}_{(\mathbbm {1}_M, a)}(L, C_M) \subseteq {\mathcal {D}}_{(\mathbbm {1}_N, b)}(L, C_N) \}. \end{aligned}$$

Here, $(1,s) \cdot {\mathcal {D}}_{(\mathbbm {1}_M, a)}(L, C_M) := \{(v_0, s_1 v_1, \ldots , s_g v_g): v \in {\mathcal {D}}_{(\mathbbm {1}_M, a)}(L, C_M)\}$.

Question C.12

Are there natural conditions which would entail $\{s: \sum _i s_i \le 1 \} \subseteq \Delta _a(L, C_M, C_N)$? This can be done for free spectrahedra, see [HKMS19, Theorem 1.4] and [DDOSS17, Section 8].

The following definition is motivated by the matrix range introduced in [Arv72] and generalized in [DDOSS17].

Definition C.13

Let $(L,L^+)$ be a preordered vector space with order unit $\mathbbm {1}_L$ and consider a tuple $a \in L^g$. Then, the functional range of a is the set

$$\begin{aligned} \mathcal W(a) = \{(\varphi (a_1), \ldots , \varphi (a_g)): \varphi \in (L^+)^*, \varphi (\mathbbm {1}_L) = 1 \} \subseteq {\mathbb {R}}^g. \end{aligned}$$

Proposition C.14

Let $(L,L^+)$ be a preordered vector space with order unit $\mathbbm {1}_L$ and let $a \in L^g$. If $V^+$ is proper, then $\mathcal W(a)$ is compact and convex. The set $\mathcal W(a)$ is non-empty if and only if $L^+ \ne L$.

Proof

For any $a_i$, $i \in [g]$, there is a $t_i \ge 0$ such that $a_i \in t_i [-\mathbbm {1}_L, \mathbbm {1}_L]$. Thus, $|\varphi (a_i)| \le t_i$ for all $\varphi \in (L^+)^*, \varphi (\mathbbm {1}_L) = 1$ and boundedness of $\mathcal W(a)$ follows. Let $x^{(n)}$ be a sequence in $\mathcal W(a)$ converging to x. With any $x^{(n)}$, we can associate a $\varphi _n \in (L^+)^*, \varphi _n(\mathbbm {1}_L) = 1$. Since $(L^+)^*$ is closed by definition, there is a map $\varphi \in (L^+)^*, \varphi (\mathbbm {1}_L) = 1$ such that $x = (\varphi (a_1), \ldots , \varphi (a_n))$ and $x \in \mathcal W(a)$. This follows from the Bolzano-Weierstrass theorem and the fact that the unital positive functionals form a compact set (consider the order unit norm on L). Convexity follows from the fact that the set of $\varphi$ as in the statement is convex.

The set $\mathcal W(a)$ is empty if and only if there are no unital functionals in $L^*$. Let $L = L^+$. Then, it is easy to see that $(L^+)^{*} = \{0\}$. Conversely, if $L^+ \ne L$, then $- \mathbbm {1}_L \not \in L^+$ by Lemma A.1. Thus, the functional $\varphi$ on ${\mathbb {R}}\mathbbm {1}_L$ such that $\varphi (\mathbbm {1}_L) = 1$ can be extended to an element in $(L^+)^*$ by Theorem B.1. $\square$

Proposition C.15

Let $(L,L^+)$ be a proper ordered vector space with order unit $\mathbbm {1}_L$. Furthermore, let $a \in L^{g}$. Let $\mathcal C_a := \{x \in {\mathbb {R}}^g: (1,-x) \in {\mathcal {D}}_{(\mathbbm {1}_L, a)}({\mathbb {R}}, L^+)\}$. Then, $\mathcal W(a)^\circ = \mathcal C_a$. If $0 \in \mathcal W(a)$, then $\mathcal C_a^\circ = \mathcal W(a)$.

Proof

Since $L^+$ is closed, $L^+ \cong (L^+)^{**}$ and

$$\begin{aligned} \mathbbm {1} + \sum _i x_i a_i \in L^+ \iff 1 + \sum _i x_i \varphi (a_i) \ge 0 \qquad \forall \varphi \in (L^+)^*, \varphi (\mathbbm {1}_L) = 1. \end{aligned}$$

Note that the only map in $(L^+)^*$ with $\varphi (\mathbbm {1}_L) = 0$ is the constant map, since $\mathbbm {1}_L$ is an order unit. Thus, it is enough to check unital maps for the $\Longleftarrow$ implication. This proves the first assertion. The second assertion follows from the bipolar theorem for convex sets ([AS17, Equation (1.10)]) since $\mathcal W(a)$ is closed. $\square$

Proposition C.16

Let $(L,L^+)$ be a proper ordered vector space with order unit $\mathbbm {1}_L$. Furthermore, let $a \in L^{g}$. Then, $\mathcal C_a$ is bounded if and only if $0 \in {\mathrm {int}}~\mathcal W(a)$.

Proof

This follows from the fact that for convex sets $K \subseteq {\mathbb {R}}^n$, $K^\circ$ is bounded if and only if $0 \in {\mathrm {int}}~K$ [AS17, Exercise 1.14], combined with Proposition C.15. $\square$

Proposition C.17

Let $(M,M^+)$ and $(N,N^+)$ be two proper ordered vector spaces containing order units $\mathbbm {1}_M$ and $\mathbbm {1}_N$. Let $a \in M^{k}$ and $b \in N^{l}$ be such that $\mathcal C_a$, $\mathcal C_b$ are polytopes for k, $l \in \mathbb N$. Then, for any closed tensor cone $C_{MN}$,

$$\begin{aligned} {\mathcal {D}}_{(\mathbbm {1}_M \otimes \mathbbm {1}_N, a \otimes \mathbbm {1}_N, \mathbbm {1}_M \otimes b)}({\mathbb {R}}, C_{MN}) = {\mathbb {R}}_+\{(1,-z): z\in \mathcal C_a \oplus \mathcal C_b\}. \end{aligned}$$

Proof

Let $-x \in C_{a}$, $-y \in \mathcal C_b$. Then, $(1,x,0) \in {\mathcal {D}}_{(\mathbbm {1}_M \otimes \mathbbm {1}_N, a \otimes \mathbbm {1}_N, \mathbbm {1}_M \otimes b)}({\mathbb {R}}, C_{MN})$, because $(\mathbbm {1}_M + \sum _i x_i a_i) \otimes \mathbbm {1}_N \in M^+ \otimes _{\min } N^+$, and likewise $(1,0,y) \in {\mathcal {D}}_{(\mathbbm {1}_M \otimes \mathbbm {1}_N, a \otimes \mathbbm {1}_N, \mathbbm {1}_M \otimes b)}$ $({\mathbb {R}},C_{MN})$, such that “$\supset$” holds in the assertion. Conversely, let $(c,x,y) \in {\mathcal {D}}_{(\mathbbm {1}_M \otimes \mathbbm {1}_N, a \otimes \mathbbm {1}_N, \mathbbm {1}_M \otimes b)}({\mathbb {R}}, C_{MN})$, where $x \in {\mathbb {R}}^k$ and $y \in {\mathbb {R}}^l$. Then,

$$\begin{aligned} c \mathbbm {1}_M \otimes \mathbbm {1}_N + (\sum _{i=1}^k x_i a_i)\otimes \mathbbm {1}_N + \mathbbm {1}_M \otimes (\sum _{j = 1}^l y_j b_j) \in C_{MN}. \end{aligned}$$

(57)

Boundedness of $\mathcal C_a$ and $\mathcal C_b$ implies by Proposition C.16 that $0 \in {\mathrm {int}}~\mathcal W(a)$, $0 \in {\mathrm {int}}~\mathcal W(b)$. Let $\varphi$, $\psi$ be positive unital functionals which send a and b to zero, respectively. Then, $\varphi \otimes \psi \in (C_{MN})^*$. An application of this map to Eq. (57) implies $c \ge 0$.

Let $c = 0$. Then, by applying $\varphi \otimes \beta$, $\alpha \otimes \psi$ for $\alpha \in (M^+)^*$ and $\beta \in (N^+)^*$, it holds that $\sum _i x_i a_i \in M^+$, $\sum _j y_j b_j \in N^+$. Since $\mathcal C_a$ is bounded, $\{a_i\}_i$ is a set of linearly independent elements. The same is true for $\{b_j\}_j$. Without loss of generality, let $x \ne 0$. Then $\sum _i x_i a_i \ne 0$ and $(1,- \lambda x) \in \mathcal C_a$ for all $\lambda \ge 0$, which is a contradiction to $\mathcal C_a$ being bounded. Thus $c = 0$ implies $x = 0 = y$.

Thus, we can set $c = 1$ without loss of generality. Let now for $\varphi ^\prime \in (M^+)^*$, $\psi ^\prime \in (N^+)^*$, $\varphi ^\prime (\mathbbm {1}_M) = 1 = \psi ^\prime (\mathbbm {1}_N)$. An application of $\varphi ^\prime \otimes \psi ^\prime$ to Eq. (57) implies $\mathcal C^\circ \supset W(a) \times \mathcal W(b)$, where $\mathcal C:= \{z:(1,-z) \in {\mathcal {D}}_{(\mathbbm {1}_M \otimes \mathbbm {1}_N, a \otimes \mathbbm {1}_N, \mathbbm {1}_M \otimes b)}({\mathbb {R}}, C_{MN})\}$. Since $\mathcal C$ is closed and contains 0, we obtain $\mathcal C \subset (\mathcal W(a) \times \mathcal W(b))^\circ = (\mathcal C_a^\circ \times \mathcal C_b^\circ )^\circ = \mathcal C_a \oplus \mathcal C_b$ with Proposition C.15. $\square$

Corollary C.18

Let $(M,M^+)$ and $(N,N^+)$ be two proper ordered vector spaces with closed cones containing order units $\mathbbm {1}_M$ and $\mathbbm {1}_N$. Let $a \in M^{k}$ and $b \in N^{l}$ be such that $\mathcal C_a$, $\mathcal C_b$ are polytopes for k, $l \in \mathbb N$. Let moreover $(L, L^+)$ be another preordered vector space with order unit $\mathbbm {1}_L$ and $h_1 \in L^{k}$, $h_2 \in L^{l}$. Then, for any closed tensor cone $C_{MN}$,

$$\begin{aligned} {\mathcal {D}}_{(\mathbbm {1}_M \otimes \mathbbm {1}_N, a \otimes \mathbbm {1}_N, \mathbbm {1}_M \otimes b)}({\mathbb {R}}, C_{MN}) \subseteq {\mathcal {D}}_{(\mathbbm {1}_L, h_1, h_2)}({\mathbb {R}}, L^+) \end{aligned}$$

if and only if

$$\begin{aligned} {\mathcal {D}}_{(\mathbbm {1}_M, a)}({\mathbb {R}}, M^+) \subseteq {\mathcal {D}}_{(\mathbbm {1}_L, h_1)}({\mathbb {R}}, L^+) \quad \wedge \quad {\mathcal {D}}_{(\mathbbm {1}_N, b)}({\mathbb {R}}, N^+) \subseteq {\mathcal {D}}_{(\mathbbm {1}_L, h_2)}({\mathbb {R}}, L^+). \end{aligned}$$

Appendix D: Basic Results on Compatibility Regions and Degree

In this section, we will prove some basic results on the relation between different compatibility regions, which complements Sect. 4. We will start by showing that the compatibillity degree decreases with the number of measurements.

Proposition D.1

Let $(V,V^+,\mathbbm {1})$ be a GPT and let g, $g^\prime \in \mathbb N$ be such that $g \le g^\prime$. Let ${\mathbf {k}}^\prime \in \mathbb N^{g^\prime }$ and ${\mathbf {k}} = (k_1^\prime , \ldots , k_g^\prime )$. Then,

$$\begin{aligned} \gamma ({\mathbf {k}}; V, V^+) \ge \gamma ({\mathbf {k}}^\prime ; V, V^+). \end{aligned}$$

Proof

Let $f^{(i)} \in A^{k_i^\prime }$, $i \in [g^\prime ]$, be a collection of measurements. Let $s = \gamma ({\mathbf {k}}^\prime ; V, V^+)$ and let ${\tilde{f}}^{(i)}$ be the corresponding noisy measurements defined by

$$\begin{aligned} {\tilde{f}}^{(i)}_j = s f^{(i)}_j + (1-s) \frac{\mathbbm {1}}{k_i^\prime } \qquad \forall j \in [k_i^\prime ],~i \in [g^\prime ]. \end{aligned}$$

Then, the ${\tilde{f}}^{(i)}$, $i \in [g^\prime ]$, are compatible with joint measurement $h^\prime$. Setting

$$\begin{aligned} h_{i_1, \ldots , i_g} = \sum _{i_j \in [k_j^\prime ], j \in [g^\prime ] \setminus [g]} h^\prime _{i_1, \ldots , i_{g^\prime }} \end{aligned}$$

yields a joint measurement for ${\bar{f}}^{(i)}$, $i \in [g]$. Since the measurements were arbitrary, it follows that $s \le \gamma ({\mathbf {k}}; V, V^+)$. $\square$

Intuitively, the more outcomes we have, the more difficult it becomes for all measurements to be compatible. This is the content of the next proposition, which is the GPT version of [BN20, Proposition 3.35]. The proof is very similar.

Proposition D.2

Let $g \in \mathbb N$, $\mathbf{k}^\prime$, $\mathbf{k}\in \mathbb N^g$ and let $(V, V^+, \mathbbm {1})$ be a GPT. Let $\mathbf{k}^\prime \ge \mathbf{k}$, where the inequality is meant to hold entrywise. Then

$$\begin{aligned} \Gamma ({\mathbf {k}}^\prime ; V, V^+) \subseteq \Gamma ({\mathbf {k}}; V, V^+). \end{aligned}$$

In particular, $\gamma ({\mathbf {k}}^\prime ; V, V^+) \le \gamma ({\mathbf {k}}; V, V^+)$.

Proof

Let $(f_1^{(i)}, \ldots , f_{k_i}^{(i)}) \in A^{k_i}$ be measurements for all $i \in [g]$. Let $s \in \Gamma ({\mathbf {k}}^\prime ; V, V^+)$. Then, the measurements

$$\begin{aligned} F^{(i)} {:=} \Bigg (s_if_1^{(i)}{+}(1-s_i)\frac{\mathbbm {1}}{k_i^\prime }, \ldots , s_i f_{k_i}^{(i)}{+}(1-s_i)\frac{\mathbbm {1}}{k_i^\prime }, \underbrace{(1-s_i)\frac{\mathbbm {1}}{k_i^\prime }, \ldots , (1-s_i)\frac{\mathbbm {1}}{k_i^\prime }}_{k_i^\prime {-} k_i}\Bigg )\in A^{k'_i} \end{aligned}$$

are compatible. We will show that this is still true if we replace, for some fixed $l \in [g]$, the l-th measurement $F^{(l)}$ by

$$\begin{aligned} G^{(l)} := \left( s_l f_1^{(l)}+(1-s_l)\frac{\mathbbm {1}}{k_l}, \ldots , s_l f_{k_l}^{(l)}+(1-s_l)\frac{\mathbbm {1}}{k_l}\right) \in A^{k_l}. \end{aligned}$$

An iterative application of this procedure then shows that $s \in \Gamma ({\mathbf {k}}; V, V^+)$ and the assertion follows.

Let $H_{i_1, \ldots , i_g}$ be the effects of the joint measurement for the $F^{(i)}$, where $i_j \in [k_j^\prime ]$, $j \in [g]$. Let us define

$$\begin{aligned} R_{i_1, \ldots , i_{l-1}, k_l + 1, i_{l+1}, \ldots , i_g} := \sum _{i_l = k_l + 1}^{k_l^\prime }H_{i_1, \ldots , i_g}. \end{aligned}$$

Then, let us define

$$\begin{aligned} h_{i_1, \ldots , i_g} = H_{i_1, \ldots , i_g} + \frac{1}{k_l} R_{i_1, \ldots , i_{l-1}, k_l + 1, i_{l+1}, \ldots , i_g} \qquad i_l \in [k_l],~i_j \in [k_j^\prime ]~\forall j \in [g] \setminus \{l\}. \end{aligned}$$

It is easy to verify that h is a measurement. Moreover, for $m \in [g]$, $m \ne l$, $q \in [k_m^\prime ]$,

$$\begin{aligned} \sum _{\begin{array}{c} i_j \in [k_j^\prime ], j \in [g]\setminus \{m,l\},\\ i_l \in [k_l], i_m = q \end{array}} h_{i_1, \ldots , i_g}= F^{(m)}_q \end{aligned}$$

and for $p \in [k_l]$,

$$\begin{aligned} \sum _{i_j \in [k_j^\prime ], j \in [g]\setminus \{l\}, i_l = p} h_{i_1, \ldots , i_g}&= \sum _{i_j \in [k_j^\prime ], j \in [g]\setminus \{l\}, i_l = p} H_{i_1, \ldots , i_g}\\ {}&+ \frac{1}{k_l} \sum _{i_j \in [k_j^\prime ], j \in [g]\setminus \{l\}} R_{i_1, \ldots , i_{l-1}, k_l + 1, i_{l+1}, \ldots , i_g} \\&= F_p^{(l)} + \frac{1}{k_l}\sum _{j = k_l +1}^{k_l^\prime }F_j^{(l)} \\&= s_l f_p^{(l)} + \frac{(1-s_l)}{k_l^\prime } \mathbbm {1} + \frac{(1-s_l)(k_l^\prime - k_l)}{k_l k_l^\prime } \mathbbm {1} \\&= s_l f_p^{(l)} +\frac{(1-s_l)}{k_l} \mathbbm {1} =G^{(l)}_p. \end{aligned}$$

Thus, h is the desired joint measurement for the $F^{(i)}$, $i \in [g] \setminus \{l\}$ and $G^{(l)}$. $\square$

We have seen in the last proposition how the compatibility regions are related for measurements with a different number of outcomes within the same theory. We will now show that sometimes the compatibility regions of different GPTs can be related.

Proposition D.3

Let $g \in \mathbb N$, ${\mathbf {k}} \in \mathbb N^g$. Let $(V_1, V_1^+, \mathbbm {1}_{A_1})$ and $(V_2, V_2^+, \mathbbm {1}_{A_2})$ be two GPTs. Let $\Psi :(V_1, V_1^+) \rightarrow (V_2,V_2^+)$ be a positive map such that $\Psi ^*(\mathbbm {1}_{A_2}) = \mathbbm {1}_{A_1}$. If $\Psi$ is a retraction, i.e. if there is a positive map $\Theta : (V_2, V_2^+) \rightarrow (V_1,V_1^+)$ such that $\Psi \circ \Theta = \mathrm {id}_{V_{2}}$, then the GPT $(V_1, V_1^+, \mathbbm {1}_{A_1})$ is “less compatible” than $(V_2, V_2^+, \mathbbm {1}_{A_2})$ in the measurement setting ${\mathbf {k}}$: $\Gamma ({\mathbf {k}}; V_1, V_1^+) \subseteq \Gamma ({\mathbf {k}}; V_2, V_2^+)$.

Proof

Let $s \in \Gamma ({\mathbf {k}}; V_1, V_1^+)$. Let $f^{(i)} \in A_2^{k_i}$, $i \in [g]$, be a collection of measurements. Then, the $\Psi ^*(f_j^{(i)}) \in A_1^+$, $j \in [k_i]$, form a collection of measurements as well. Moreover, the measurements given by

$$\begin{aligned} h_j^{(i)} = s_i \Psi ^*(f_j^{(i)}) + (1-s_i) \frac{\mathbbm {1}_{A_1}}{k_i} \qquad \forall i \in [g], ~j \in [k_i] \end{aligned}$$

are compatible by the assumption on s. Since, $\Psi$ is a retraction, it holds that $\Theta ^*\circ \Psi ^*= \mathrm {id}_{A_2}$. Since $\Psi ^*(\mathbbm {1}_{A_2}) = \mathbbm {1}_{A_1}$, also $\Theta ^*(\mathbbm {1}_{A_1}) = \mathbbm {1}_{A_2}$. Thus, the image of the joint measurement for the $h^{(i)}$ under $\Theta ^*$ is again measurement in $(V_2, V_2^+, \mathbbm {1}_{A_2})$. Furthermore, it is a joint measurement for the noisy measurements

$$\begin{aligned} s_i f_j^{(i)} + (1-s_i) \frac{\mathbbm {1}_{A_2}}{k_i} \qquad \forall i \in [g], ~j \in [k_i], \end{aligned}$$

since $\Theta ^*(\Psi ^*(f_j^{(i)})) = f_j^{(i)}$ for all $i \in [g]$, $j \in [k_i]$. $\square$

Remark D.4

An example for this situation is $(V_1, V_1^+, \mathbbm {1}_{A_1}) = \mathrm {QM_d}$ for some $d \in \mathbb N$ and $(V_2, V_2^+, \mathbbm {1}_{A_2}) =\mathrm {CM}_d$, where $\Psi ^*$ embeds the probability distributions in $\mathrm {CM}_d$ as diagonal matrices. Then, $\Theta ^*$ projects onto the diagonal entries of the matrix.

Appendix E: Conic Programming for Compatibility in GPTs

1.1 E.1. Background

To begin, let us briefly recapitulate the theory of conic programming. We follow [GM12, Section 4]. All vector spaces we consider will be finite dimensional. We will equip them with an inner product choosing a pair of dual bases for the vector space and its dual.

Definition E.1

(Conic program [GM12, Definition 4.6.1]). Let $L^+ \subseteq L$, $M^+ \subseteq M$ be closed convex cones, let $b \in M$, $c \in L^*$ and let $A: L \rightarrow M$ be a linear operator. A conic program is an optimization problem of the form

$$\begin{aligned} \mathrm {maximize} \qquad&\langle c, x \rangle \\ \mathrm {subject~to} \qquad&b -A(x) \in M^+ \\&x \in L^+ \end{aligned}$$

The dual problem is then given by [GM12, Section 4.7]

$$\begin{aligned} \mathrm {minimize} \qquad&\langle b, y \rangle \\ {\mathrm {subject~to}} \qquad&A^*(y) - c \in (L^+)^*\\&y \in (M^+)^*\end{aligned}$$

Weak duality always hold, i.e. the value of the primal problem is upper bounded by the value of the dual program if the dual program is feasible. A sufficient condition for strong duality to hold is the following version of Slater’s condition:

Theorem E.2

([GM12, Theorem 4.7.1]). If the conic program in Definition E.1 is feasible, has finite value $\gamma$ and has an interior point ${\tilde{x}}$, then the dual program is also feasible and has the same value $\gamma$.

If $M^+ \ne \{0\}$, ${\tilde{x}}$ is an interior point if ${\tilde{x}} \in {\mathrm {int}}(L^+)$ and $b - A({\tilde{x}}) \in {\mathrm {int}}(M^+)$ [GM12, Definition 4.6.4].

1.2 E.2. Map extension

In this section, we will show that the existence of a positive extension of a positive map can be checked using a conic program. We can give the following generalization of the results in [HJRW12].

Theorem E.3

Let $E \subseteq {\mathbb {R}}^d$ be a g-dimensional subspace such that $E \cap {\text {ri}}({\mathbb {R}}_+^d) \ne \varnothing$ and $E^+ = E \cap {\mathbb {R}}^d_+$. Moreover, let $(V,V^+, \mathbbm {1})$ be a GPT. Finally, let $\Phi : E \rightarrow A$ be given by a basis $\{e_i\}_{i \in [g]} \subset E$ of E, $g \in \mathbb N$, and $\{f_i\}_{i \in [g]} \subset A$ such that $\Phi (e_i) = f_i$ for all $i \in [g]$. Then, there exists a positive extension ${\tilde{\Phi }}: ({\mathbb {R}}^{d}, {\mathbb {R}}_+^{d}) \rightarrow (A, A^+)$ of $\Phi$ if and only if the conic program

$$\begin{aligned} {\mathrm {maximize}} \qquad&-\langle s, (h_1^+-h_1^-, \ldots , h_g^+-h_g^-) \rangle \\ {\mathrm {subject~to}} \qquad&\sum _{i \in [g]} e_i \otimes (h_i^+-h_i^-) \in E^+ \otimes _{\mathrm {max}} V^+ \\&1 - \langle \mathbbm {1}, h_i^\pm \rangle \ge 0 \\&h^{\pm }_i \in V^+ \qquad \forall i \in [g] \end{aligned}$$

has value 0. Here, $s: V^g \rightarrow {\mathbb {R}}$ is given as

$$\begin{aligned} \langle s, h_1, \ldots , h_g \rangle = \langle \chi _V, \sum _{i \in [g]} f_i \otimes (h^+_i-h_i^-) \rangle = \sum _{i \in [g]} f_i(h_i^+ - h_i^-). \end{aligned}$$

Proof

Let $z \in E \otimes V$. Using that the $e_i$ form a basis, we can write

$$\begin{aligned} z = \sum _{i \in [g]} e_i \otimes z_i, \end{aligned}$$

where $z_i \in V$ for all $i \in [g]$. Since $V^+$ is proper, we can decompose each $z_i = z_i^+ - z_i^-$, where $z_i^\pm \in V^+$. Then, $\langle s, (z_1, \ldots , z_g)\rangle = s_\Phi (z)$, where $s_\Phi$ is defined as in Proposition B.8. By linearity, it suffices to restrict to z such that $\langle \mathbbm {1}, z_i^\pm \rangle \le 1$ for all $i \in [g]$ in order to check positivity of $s_\Phi$. Thus, the conic program has value 0 if and only if $s_\Phi$ is positive. The assertion then follows from Proposition B.8. $\square$

Remark E.4

Of course, we could write down a trivial conic program for map extension just checking whether the corresponding tensor $\varphi ^\Phi$ is in $(E^+)^*\otimes _{\min } A^+$ using Proposition B.8. However, checking membership in this cone might be hard in practice. If $E^+$ is a polyhedral cone,

$$\begin{aligned} \sum _{i \in [g]} e_i \otimes (h_i^+-h_i^-) \in E^+ \otimes _{\mathrm {max}} V^+ \end{aligned}$$

can be checked as

$$\begin{aligned} \sum _{i \in [g]} \langle \alpha _j, e_i \rangle (h_i^+-h_i^-) \in V^+ \qquad \forall j, \end{aligned}$$

where the $\alpha _j$ are the extremal rays of $(E^+)^*$ (compare to Lemma 6.2). This is arguably easier since it does not involve checking the membership in a tensor cone directly. Moreover, it recovers the result that in quantum mechanics, where $V^+ = {\mathrm {PSD}}_d$: compatibility via map extension can be checked using a semidefinite program [HJRW12].

Theorem E.5

The conic program in Theorem E.3 is feasible and satisfies strong duality.

Proof

In the following, we identify $z = \sum _{i \in [2g]}z_i \otimes \mu _i \in L \otimes {\mathbb {R}}^{2g}$ with the vector $(z_1, \ldots z_{2g})$, where $\{\mu _i\}_{i \in [2g]}$ is an orthonormal basis of ${\mathbb {R}}^{2g}$. Comparing the conic program to Definition E.1, we identify

$$\begin{aligned} M&= (E \otimes V) \times {\mathbb {R}}^{2g} \\ M^+&= (E^+ \otimes _{\mathrm {max}} V^+) \times {\mathbb {R}}_+^{2g} \\ L&= V \otimes {\mathbb {R}}^{2g} \\ L^+&= (V^+)\otimes {\mathbb {R}}_+^{2g} \\ c&= -s \\ b&= (0, \underbrace{1, \ldots , 1}_{2g}) \in (E \otimes V) \times {\mathbb {R}}^{2g} \\ A(h_1^\pm , \ldots , h_g^\pm )&= \left( - \sum _{i \in [g]} e_i \otimes (h^+_i-h^-_i), \langle \mathbbm {1}, h^\pm _1 \rangle , \ldots , \langle \mathbbm {1}, h^\pm _g \rangle \right) . \end{aligned}$$

It can be verified that dual conic program is thus given by

$$\begin{aligned} \mathrm {minimize} \qquad&\sum _{i \in [2g]} y_i \\ {\mathrm {subject~to}} \qquad&s + \sum _{i \in [2g]} y_i \mathbbm {1} \otimes \mu _i^*+ B(z) \in A^+ \otimes {\mathbb {R}}_+^{2g} \\&z \in (E^+)^*\otimes _{\mathrm {min}} A^+ \\&y_i \in {\mathbb {R}}_+ \qquad \forall i \in [2g]. \end{aligned}$$

Here, $B(z) \in A \otimes {\mathbb {R}}^{2g}$ is given as $B(z)(h_1, \ldots , h_{g}) = \langle z, \sum _{i \in [g]} e_i \otimes h_i \rangle$ and $\mu ^*_i$ is the dual basis of $\varepsilon _i$. Letting $y_1 = \ldots = y_{2g}$ and realizing that $\mathbbm {1} \otimes (1, \ldots 1)$ is an order unit in $A^+ \otimes {\mathbb {R}}_+^{2g}$, for any $z \in (E^+)^*\otimes _{\mathrm {min}} A^+$ we can find a $y_1 > 0$ such that

$$\begin{aligned} s + y_1 \mathbbm {1} \otimes (1, \ldots , 1) + B(z) \in {\mathrm {int}}\left( A^+ \otimes {\mathbb {R}}_+^{2g}\right) . \end{aligned}$$

This is true, since the order unit is an interior point of $A^+ \otimes {\mathbb {R}}_+^{2g}$, hence there is a $y_1$ such that

$$\begin{aligned} \frac{1}{y_1} (s + B(z)) + \mathbbm {1} \otimes (1, \ldots , 1) \in {\mathrm {int}}\left( A^+ \otimes {\mathbb {R}}_+^g\right) . \end{aligned}$$

Since the interior points of $A^+ \otimes {\mathbb {R}}_+^g$ are those points w such that $\langle w, x \rangle > 0$ for all $x \in V^+ \otimes {\mathbb {R}}_+^{2g} \setminus \{0\}$, multiplication by $y_1$ preserves the fact that the point is in ${\mathrm {int}}\left( A^+ \otimes {\mathbb {R}}_+^{2g}\right)$. Therefore, the dual problem has an interior point. This also implies that the value of the dual program is finite, since it is lower bounded by 0. The remarks at the beginning of [GM12, Section 4.7] imply that Theorem E.2 still applies if we interchange the primal and the dual problem. Thus, the assertion follows. $\square$

1.3 E.3. Computing $\Vert \cdot \Vert _c$

Finally, we note that the norm $\Vert \cdot \Vert _c$, introduced in Theorem 7.2 and Proposition 7.4, can also be computed by a conic program, namely: for any ${\bar{\varphi }} \in E_g^*\otimes A$, $-\Vert \bar{\varphi }\Vert _c$ is the value of the following conic program

$$\begin{aligned} \mathrm {maximize}\qquad&- \lambda \\ {\mathrm {subject~to}} \qquad&\lambda {\check{1}}_g \otimes \mathbbm {1} - {\bar{\varphi }} \in (E^+_g)^*\otimes _{\min } A^+ \\&\lambda \in {\mathbb {R}}_+ \end{aligned}$$

Since $\Vert {\bar{\varphi }}\Vert _c$ is finite and since $\lambda \check{1}_g \otimes \mathbbm {1} - {\bar{\varphi }} \in {\text {int}}{(E^+_g)^*\otimes _{\min } A^+}$ for $\lambda$ large enough since ${\check{1}}_g \otimes \mathbbm {1} \in (E^+_g)^*\otimes _{\min } A^+$ is an order unit, strong duality holds by Theorem E.2. The dual conic program is

$$\begin{aligned} \mathrm {minimize}\qquad&- \langle {\bar{\varphi }}, y \rangle \\ {\mathrm {subject~to}} \qquad&1- \langle {\check{1}}_g \otimes \mathbbm {1}, y \rangle \ge 0 \\&y \in E_g^+ \otimes _{\max } V^+ \end{aligned}$$

The membership in $E_g^+ \otimes _{\max } V^+$ can be decided only evaluating the equations in Eq. (13) by Lemma 6.2. So we can give the alternative formulation of the dual conic program which only checks membership in $V^+$:

$$\begin{aligned} \mathrm {minimize}\qquad&- \langle {\bar{\varphi }}, 1_g \otimes y_0 + \sum _{i = 1}^g c_i \otimes (y_i^+ - y_i^-) \rangle \\ {\mathrm {subject~to}} \qquad&1- \langle \mathbbm {1}, y_0 \rangle \ge 0 \\&y_0 + \sum _{i = 1}^g \varepsilon _i(y_i^+ - y_i^-) \in V^+ \qquad \forall \varepsilon \in \{\pm 1\}^g\\&y_0, y_i^\pm \in V^+ \qquad \forall i \in [g]. \end{aligned}$$

Due to Theorem 7.2, this recovers the result that in quantum mechanics, where $V^+ = {\mathrm {PSD}}_d$, compatibility of effects can be checked using a semidefinite program (see e.g. [WPGF09]). It also recovers the result that compatibility in GPTs can be checked with a conic program [Plá16], although the one we give here is different.