Revision as of 22:28, 28 April 2010

The geometry of interaction, GoI in short, was defined in the early nineties by Girard as an interpretation of linear logic into operators algebra: formulae were interpreted by Hilbert spaces and proofs by partial isometries.

This was a striking novelty as it was the first time that a mathematical model of logic (lambda-calculus) didn't interpret a proof of $A\limp B$ as a morphism from $A$ to $B$ ^[1], and proof composition (cut rule) as the composition of morphisms. Rather the proof was interpreted as an operator acting on $A\limp B$ , that is a morphism from $A\limp B$ to $A\limp B$ . For proof composition the problem was then, given an operator on $A\limp B$ and another one on $B\limp C$ to construct a new operator on $A\limp C$ . This problem was solved by the execution formula that bares some formal analogies with Kleene's formula for recursive functions. For this reason GoI was claimed to be an operational semantics, as opposed to traditionnal denotational semantics.

The first instance of the GoI was restricted to the $M E L L$ fragment of linear logic (Multiplicative and Exponential fragment) which is enough to encode lambda-calculus. Since then Girard proposed several improvements: firstly the extension to the additive connectives known as Geometry of Interaction 3 and more recently a complete reformulation using Von Neumann algebras that allows to deal with some aspects of implicit complexity

The GoI has been a source of inspiration for various authors. Danos and Regnier have reformulated the original model exhibiting its combinatorial nature using a theory of reduction of paths in proof-nets and showing the link with abstract machines; in particular the execution formula appears as the composition of two automata that interact one with the other through their common interface. Also the execution formula has rapidly been understood as expressing the composition of strategies in game semantics. It has been used in the theory of sharing reduction for lambda-calculus in the Abadi-Gonthier-Lévy reformulation and simplification of Lamping's representation of sharing. Finally the original GoI for the $M E L L$ fragment has been reformulated in the framework of traced monoidal categories following an idea originally proposed by Joyal.

The Geometry of Interaction as operators

The original construction of GoI by Girard follows a general pattern already mentionned in coherent semantics under the name symmetric reducibility and that was first put to use in phase semantics. First set a general space $P$ called the proof space because this is where the interpretations of proofs will live. Make sure that $P$ is a (not necessarily commutative) monoid. In the case of GoI, the proof space is a subset of the space of bounded operators on $\ell^2$ .

Second define a particular subset of $P$ that will be denoted by $\bot$ ; then derive a duality on $P$ : for $u,v\in P$ , $u$ and $v$ are dual^[2], iff $uv\in\bot$ .

For the GoI, two dualities have proved to work; we will consider the first one: nilpotency, ie, $\bot$ is the set of nilpotent operators in $P$ . Let us explicit this: two operators $u$ and $v$ are dual if there is a nonegative integer $n$ such that $(u v) n = 0$ . Note in particular that $uv\in\bot$ iff $vu\in\bot$ .

When $X$ is a subset of $P$ define $X\orth$ as the set of elements of $P$ that are dual to all elements of $X$ :

$X\orth = \{u\in P, \forall v\in X, uv\in\bot\}$ .

This construction has a few properties that we will use without mention in the sequel. Given two subsets $X$ and $Y$ of $P$ we have:

if $X\subset Y$ then $Y\orth\subset X$ ;
$X\subset X\biorth$ ;
$X\triorth = X\orth$ .

Last define a type as a subset $T$ of the proof space that is equal to its bidual: $T = T\biorth$ . This means that $u\in T$ iff for all operator $v\in T\orth$ , that is such that $u'v\in\bot$ for all $u'\in T$ , we have $uv\in\bot$ .

The real work^[3], is now to interpret logical operations, that is to associate a type to each formula, an object to each proof and show the adequacy lemma: if $u$ is the interpretation of a proof of the formula $A$ then $u$ belongs to the type associated to $A$ .

Preliminaries

We begin by a brief tour of the operations in Hilbert spaces that we use. In this article $H$ will stand for the Hilbert space $\ell^2(\mathbb{N})$ of sequences $(x_n)_{n\in\mathbb{N}}$ of complex numbers such that the series $\sum_{n\in\mathbb{N}}|x_n|^2$ converges. If $x = (x_n)_{n\in\mathbb{N}}$ and $y = (y_n)_{n\in\mathbb{N}}$ are two vectors of $H$ we denote by $\langle x,y\rangle$ their scalar product:

$\langle x, y\rangle = \sum_{n\in\mathbb{N}} x_n\bar y_n$ .

Two vectors of $H$ are othogonal if their scalar product is nul. We will say that two subspaces are disjoint when their vectors are pairwise orthogonal; this terminology is slightly misleading as disjoint subspaces always have $0$ in common.

The norm of a vector is the square root of the scalar product with itself:

$\|x\| = \sqrt{\langle x, x\rangle}$ .

Let us denote by $(e_k)_{k\in\mathbb{N}}$ the canonical hilbertian basis of $H$ : $e_k = (\delta_{kn})_{n\in\mathbb{N}}$ where $δ k n$ is the Kroenecker symbol: $δ k n = 1$ if $k = n$ , $0$ otherwise. Thus if $x=(x_n)_{n\in\mathbb{N}}$ is a sequence in $H$ we have:

$x = \sum_{n\in\mathbb{N}} x_ne_n$ .

An operator on $H$ is a continuous linear map from $H$ to $H$ . Continuity is equivalent to the fact that operators are bounded, which means that one may define the norm of an operator $u$ as the sup on the unit ball of the norms of its values:

$\|u\| = \sup_{\{x\in H,\, \|x\| = 1\}}\|u(x)\|$ .

The set of (bounded) operators is denoted by $\mathcal{B}(H)$ .

The range or codomain of the operator $u$ is the set of images of vectors; the kernel of $u$ is the set of vectors that are anihilated by $u$ ; the domain of $u$ is the set of vectors orthogonal to the kernel, ie, the maximal subspace disjoint with the kernel:

$\mathrm{Codom}(u) = \{u(x),\, x\in H\}$ ;
$\mathrm{Ker}(u) = \{x\in H,\, u(x) = 0\}$ ;
$\mathrm{Dom}(u) = \{x\in H,\, \forall y\in\mathrm{Ker}(u), \langle x, y\rangle = 0\}$ .

These three sets are closed subspaces of $H$ .

The adjoint of an operator $u$ is the operator $u *$ defined by $\langle u(x), y\rangle = \langle x, u^*(y)\rangle$ for any $x,y\in H$ . Adjointness if well behaved w.r.t. composition of operators:

(u v) * = v * u *

.

A projector is an idempotent operator of norm $0$ (the projector on the null subspace) or $1$ , that is an operator $p$ such that $p 2 = p$ and $\|p\| = 0$ or $1$ . A projector is auto-adjoint and its domain is equal to its codomain.

A partial isometry is an operator $u$ satisfying $u u * u = u$ ; this condition entails that we also have $u * u u * = u *$ . As a consequence $u u *$ and $u u *$ are both projectors, called respectively the initial and the final projector of $u$ because their codomain are respectively the domain and the codomain of $u$ . The restriction of $u$ to its domain is an isometry. Projectors are particular examples of partial isometries.

If $u$ is a partial isometry then $u *$ is also a partial isometry the domain of which is the codomain of $u$ and the codomain of which is the domain of $u$ .

If the domain of $u$ is $H$ that is if $u * u = 1$ we say that $u$ has full domain, and similarly for codomain. If $u$ and $v$ are two partial isometries, the equation $u u * + v v * = 1$ means that the codomains of $u$ and $v$ are disjoint and that their direct sum is $H$ .

Partial permutations and partial isometries

We will now define our proof space which turns out to be the set of partial isometries acting as permutations on a fixed basis of $H$ .

More precisely a partial permutation $\varphi$ on $\mathbb{N}$ is a function defined on a subset $D_\varphi$ of $\mathbb{N}$ which is one-to-one onto a subset $C_\varphi$ of $\mathbb{N}$ . $D_\varphi$ is called the domain of $\varphi$ and $C_\varphi$ its codomain. Partial permutations may be composed: if $ψ$ is another partial permutation on $\mathbb{N}$ then $\varphi\circ\psi$ is defined by:

$n\in D_{\varphi\circ\psi}$ iff $n\in D_\psi$ and $\psi(n)\in D_\varphi$ ;
if $n\in D_{\varphi\circ\psi}$ then $\varphi\circ\psi(n) = \varphi(\psi(n))$ ;
the codomain of $\varphi\circ\psi$ is the image of the domain.

Partial permutations are well known to form a structure of inverse monoid that we detail now.

A partial identitie is a partial permutation $1 D$ whose domain and codomain are both equal to a subset $D$ on which $1 D$ is the identity function. Partial identities are idempotent for composition.

Among partial identities one finds the identity on the empty subset, that is the empty map, that we will denote by $0$ and the identity on $\mathbb{N}$ that we will denote by $1$ . This latter permutation is the neutral for composition.

If $\varphi$ is a partial permutation there is an inverse partial permutation $\varphi^{-1}$ whose domain is $D_{\varphi^{-1}} = C_{\varphi}$ and who satisfies:

$\varphi^{-1}\circ\varphi = 1_{D_\varphi}$

$\varphi\circ\varphi^{-1} = 1_{C_\varphi}$

Given a partial permutation $\varphi$ one defines a partial isometry $u_\varphi$ by:

$u_\varphi(e_n) = \begin{cases} e_{\varphi(n)} & \text{ if }n\in D_\varphi,\\ 0 & \text{ otherwise.} \end{cases}$

In other terms if $x=(x_n)_{n\in\mathbb{N}}$ is a sequence in $\ell^2$ then $u_\varphi(x)$ is the sequence $(y_n)_{n\in\mathbb{N}}$ defined by:

$y_n = x_{\varphi^{-1}(n)}$ if $n\in C_\varphi$ ,

0

otherwise.

We will (not so abusively) write $e_{\varphi(n)} = 0$ when $\varphi(n)$ is undefined so that may shorten the definition of $u_\varphi$ into:

$u_\varphi(e_n) = e_{\varphi(n)}$ .

The domain of $u_\varphi$ is the subspace spanned by the family $(e_n)_{n\in D_\varphi}$ and the codomain of $u_\varphi$ is the subspace spanned by $(e_n)_{n\in C_\varphi}$ . As a particular case if $\varphi$ is $1 D$ the partial identity on $D$ then $u_\varphi$ is the projector on the subspace spanned by $(e_n)_{n\in D}$ .

Proposition

Let $\varphi$ and $ψ$ be two partial permutations. We have:

$u_\varphi u_\psi = u_{\varphi\circ\psi}$ .

The adjoint of $u_\varphi$ is:

$u_\varphi^* = u_{\varphi^{-1}}$ .

In particular the initial projector of $u_{\varphi}$ is given by:

$u_\varphi u^*_\varphi = u_{1_{D_\varphi}}$ .

and the final projector of $u_\varphi$ is:

$u^*_\varphi u_\varphi = u_{1_{C_\varphi}}$ .

Projectors generated by partial identities commute; in particular we have:

$u_\varphi u_\varphi^*u_\psi u_\psi^* = u_\psi u_\psi^*u_\varphi u_\varphi^*$ .

Definition

We call $p$ -isometry a partial isometry of the form $u_\varphi$ where $\varphi$ is a partial permutation on $\mathbb{N}$ . The proof space $\mathcal{P}$ is the set of all $p$ -isometries.

In particular note that $0$ is a $p$ -isometry. The set $\mathcal{P}$ is a submonoid of $\mathcal{B}(H)$ but it is not a subalgebra^[4]. In general given $u,v\in\mathcal{P}$ we don't necessarily have $u+v\in\mathcal{P}$ . However we have:

Proposition

Let $u, v\in\mathcal{P}$ . Then $u+v\in\mathcal{P}$ iff $u$ and $v$ have disjoint domains and disjoint codomains, that is:

$u+v\in\mathcal{P}$ iff

u u * v v * = u * u v * v = 0

.

Suppose for contradiction that $e n$ is in the domain of $u$ and in the domain of $v$ . As $u$ and $v$ are $p$ -isometries there are integers $p$ and $q$ such that $u (e n) = e p$ and $v (e n) = e q$ thus $(u + v)(e n) = e p + e q$ which is not a basis vector; therefore $u + v$ is not a $p$ -permutation.

As a corollary note that if $u + v = 0$ then $u = v = 0$ .

From operators to matrices: internalization/externalization

It will be convenient to view operators on $H$ as acting on $H\oplus H$ , and conversely. For this purpose we define an isomorphism $H\oplus H \cong H$ by $x\oplus y\rightsquigarrow p(x)+q(y)$ where $p:H\mapsto H$ and $q:H\mapsto H$ are partial isometries given by:

p (e n) = e 2 n

,

q (e n) = e 2 n + 1

.

From the definition $p$ and $q$ have full domain, that is satisfy $p * p = q * q = 1$ . On the other hand their codomains are disjoint, thus we have $p * q = q * p = 0$ . As the sum of their codomains is the full space $H$ we also have $p p * + q q * = 1$ .

Note that we have choosen $p$ and $q$ in $\mathcal{P}$ . However the choice is arbitrary: any two $p$ -isometries with full domain and disjoint codomains would do the job.

Given an operator $u$ on $H$ we may externalize it obtaining an operator $U$ on $H\oplus H$ defined by the matrix:

$U = \begin{pmatrix} u_{11} & u_{12}\\ u_{21} & u_{22} \end{pmatrix}$

where the $u i j$ 's are given by:

u 11 = p * u p

;

u 12 = p * u q

;

u 21 = q * u p

;

u 22 = q * u q

.

The $u i j$ 's are called the components of $u$ . The externalization is functorial in the sense that if $v$ is another operator externalized as:

$V = \begin{pmatrix} v_{11} & v_{12}\\ v_{21} & v_{22} \end{pmatrix} = \begin{pmatrix} p^*vp & p^*vq\\ q^*vp & q^*vq \end{pmatrix}$

then the externalization of $u v$ is $U V$ .

We also have:

u = (p p * + q q *) u (p p * + q q *) = p u 11 p * + p u 12 q * + q u 21 p * + q u 22 q *

which entails that externalization is reversible, its converse being called internalization.

Furthermore if we suppose that $u$ is a $p$ -isometry then so are the components $u i j$ 's. Thus the formula above entails that the four terms of the sum have pairwise disjoint domains and pairwise disjoint codomains. Proposition

If $u$ is a $p$ -isometry and $u i j$ are its external components then:

$u 1 j$ and $u 2 j$ have disjoint domains, that is $u_{1j}^*u_{1j}u_{2j}^*u_{2j} = 0$ for $j = 1,2$ ;
$u i 1$ and $u i 2$ have disjoint codomains, that is $u_{i1}u_{i1}^*u_{i2}u_{i2}^* = 0$ for $i = 1,2$ .

As an example of computation in $\mathcal{P}$ let us compute the product of the final projectors of $p u 11 p *$ and $p u 12 q *$ :

$\begin{align} (pu_{11}p^*)(pu^*_{11}p^*)(pu_{12}q^*)(qu_{12}^*p^*) &= (pp^*upp^*)(pp^*u^*pp^*)(pp^*uqq^*)(qq^*u^*pp^*)\\ &= pp^*upp^*u^*pp^*uqq^*u^*pp^*\\ &= pp^*u(pp^*)(u^*pp^*u)qq^*u^*pp^*\\ &= pp^*u(u^*pp^*u)(pp^*)qq^*u^*pp^*\\ &= pp^*uu^*pp^*u(pp^*)(qq^*)u^*pp^*\\ &= 0 \end{align}$

where we used the fact that all projectors in $\mathcal{P}$ commute, which is in particular the case of $p p *$ and $u * p p * u$ .

Interpreting the multiplicative connectives

Recall that when $u$ and $v$ are partial isometries in $\mathcal{P}$ we say they are dual when $u v$ is nilpotent, and that $\bot$ denotes the set of nilpotent operators. A type is a subset of $\mathcal{P}$ that is equal to its bidual. In particular $X\orth$ is a type for any $X\subset\mathcal{P}$ . We say that $X$ generates the type $X\biorth$ .

The tensor and the linear application

If $u$ and $v$ are two $p$ -isometries summing them doesn't in general produces a $p$ -isometry. However as $p u p *$ and $q v q *$ have disjoint domains and disjoint codomains it is true that $p u p * + q v q *$ is a $p$ -isometry. Given two types $A$ and $B$ , we thus define their tensor by:

$A\tens B = \{pup^* + qvq^*, u\in A, v\in B\}\biorth$

Note the closure by bidual to make sure that we obtain a type.

From what precedes we see that $A\tens B$ is generated by the internalizations of operators on $H\oplus H$ of the form:

$\begin{pmatrix} u & 0\\ 0 & v \end{pmatrix}$

Note that this so-called tensor resembles a sum rather than a product. We will stick to this terminology though because it defines the interpretation of the tensor connective of linear logic.

The linear implication is derived from the tensor by duality: given two types $A$ and $B$ the type $A\limp B$ is defined by:

$A\limp B = (A\tens B\orth)\orth$ .

Unfolding this definition we get:

$A\limp B = \{u\in\mathcal{P}\text{ s.t. } \forall v\in A, \forall w\in B\orth,\, u.(pvp^* + qwq^*) \in\bot\}$ .

The identity

The interpretation of the identity is an example of internalization. Given a type $A$ we are to find an operator $ι$ in type $A\limp A$ , thus satisfying:

$\forall u\in A, v\in A\orth,\, \iota(pup^* + qvq^*)\in\bot$ .

An easy solution is to take $ι = p q * + q p *$ . In this way we get $ι(p u p * + q v q *) = q u p * + p v q *$ . Therefore $(ι(p u p * + q v q *)) 2 = q u v q * + p v u p *$ , from which one deduces that this operator is nilpotent iff $u v$ is nilpotent. It is the case since $u$ is in $A$ and $v$ in $A\orth$ .

It is interesting to note that the $ι$ thus defined is actually the internalization of the operator on $H\oplus H$ given by the matrix:

$\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}$ .

We will see once the composition is defined that the $ι$ operator is the interpretation of the identity proof, as expected.

The execution formula, version 1: application

Definition

Let $u$ and $v$ be two operators; as above denote by $u i j$ the external components of $u$ . Assume that $u 11 v$ is nilpotent. We define a new operator $App(u, v)$ by:

App(u,v) = u₂₂ + u₂₁v	∑	(u₁₁v)^ku₁₁
	k

.

Note that the hypothesis that $u 11 v$ is nilpotent entails that the sum

∑	(u₁₁v)^k
k

is actually finite. It would be enough to assume that this sum converges. For simplicity we stick to the nilpotency condition, but we should mention that weak nilpotency would do as well.

Theorem

If $u$ and $v$ are $p$ -isometries such that $u 11 v$ is nilpotent, then $App(u, v)$ is also a $p$ -isometry.

Proof. Let us note $E k = u 21 v (u 11 v) k u 12$ . Recall that $u 22$ and $u 12$ being external components of the $p$ -isometry $u$ , they have disjoint domains. Thus it is also the case of $u 22$ and $E k$ . Similarly $u 22$ and $E k$ have disjoint codomains because $u 22$ and $u 21$ have disjoint codomains.

Let now $k$ and $l$ be two integers such that $k > l$ and let us compute for example the intersection of the codomains of $E k$ and $E l$ :

$E_kE^*_kE_lE^*_l = (u_{21}v(u_{11}v)^ku_{12})(u^*_{12}(v^*u^*_{11})^kv^*u^*_{21})(u_{21}v(u_{11}v)^lu_{12})(u^*_{12}(v^*u^*_{11})^lv^*u_{21}^*)$

As $k > l$ we may write $(v^*u_{11}^*)^l = (v^*u^*_{11})^{k-l-1}v^*u^*_{11}(v^*u^*_{11})^l$ . Let us note $E = u^*_{11}(v^*u^*_{11})^lv^*u_{21}^*u_{21}v(u_{11}v)^lu_{12}$ so that $E_kE^*_kE_lE^*_l = u_{21}v(u_{11}v)^ku_{12}u^*_{12}(v^*u^*_{11})^{k-l-1}v^*Eu^*_{12}(v^*u^*_{11})^lv^*u_{21}^*$ . We have:

$\begin{align} E &= u^*_{11}(v^*u^*_{11})^lv^*u_{21}^*u_{21}v(u_{11}v)^lu_{12}\\ &= (u^*_{11}u_{11}u^*_{11})(v^*u^*_{11})^lv^*u_{21}^*u_{21}v(u_{11}v)^lu_{12}\\ &= u^*_{11}(u_{11}u^*_{11})\bigl((v^*u^*_{11})^lv^*u_{21}^*u_{21}v(u_{11}v)^l\bigr)u_{12}\\ &= u^*_{11}\bigl((v^*u^*_{11})^lv^*u_{21}^*u_{21}v(u_{11}v)^l\bigr)(u_{11}u^*_{11})u_{12}\\ &= u^*_{11}(v^*u^*_{11})^lv^*u_{21}^*u_{21}v(u_{11}v)^lu_{11}u^*_{11}u_{12}\\ &= 0 \end{align}$

because $u 11$ and $u 12$ have disjoint codomains, thus $u^*_{11}u_{12} = 0$ .

Similarly we can show that $E k$ and $E l$ have disjoint domains. Therefore we have proved that all terms of the sum $App(u, v)$ have disjoint domains and disjoint codomains. Consequently $App(u, v)$ is a $p$ -isometry.

Theorem

Let $A$ and $B$ be two types and $u$ a $p$ -isometry. Then the two following conditions are equivalent:

$u\in A\limp B$ ;
for any we have:
- $u 11 v$ is nilpotent and
- $\mathrm{App}(u, v)\in B$ .

Proof. Let $v$ and $w$ be two $p$ -isometries. If we compute

$(u.(pvp^* + qwq^*))^n = \bigl((pu_{11}p^* + pu_{12}q^* + qu_{21}p^* + qu_{22}q^*)(pvp^* + qwq^*)\bigr)^n$

we get a finite sum of monomial operators of the form:

$p(u_{11}v)^{i_0}u_{12}w(u_{22}w)^{i_1}\dots u_{21}v(u_{11}v)^{i_m}p^*$
$p(u_{11}v)^{i_0}u_{12}w(u_{22}w)^{i_1}\dots u_{12}w(u_{22}w)^{i_m}q^*$ ,
$q(u_{22}w)^{i_0}u_{21}v(u_{11}v)^{i_1}\dots u_{21}v(u_{11}v)^{i_m}p^*$ or
$q(u_{22}w)^{i_0}u_{21}v(u_{11}v)^{i_1}\dots u_{12}w(u_{22}w)^{i_m}q^*$ ,

for all tuples of integers $(i_1,\dots, i_m)$ such that $i_0+\cdots+i_m+m = n$ .

Each of these monomial is a $p$ -isometry. Furthermore they have disjoint domains and disjoint codomains because their sum is the $p$ -isometry $(u .(p v p * + q w q *)) n$ .

Suppose $u 11 v$ is nilpotent and consider:

$\bigl(\mathrm{App}(u,v)w\bigr)^n = \biggl(\bigl(u_{22} + u_{21}v\sum_k(u_{11}v)^k u_{12}\bigr)w\biggr)^n$ .

Developping we get a finite sum of monomials of the form:

$(u_{22}w)^{l_0}u_{21}v(u_{11}v)^{k_1}u_{12}w(u_{22}w)^{l_1}\dots u_{21}v(u_{11}v)^{k_m}u_{12}w(u_{22}w)^{l_m}$

for all tuples $(l_0,\dots, l_m)$ such that $l_0\cdots l_m + m = n$ .

Again as these monomials are $p$ -isometries and their sum is the $p$ -isometry $(App(u, v) w) n$ , they have pairwise disjoint domains and pairwise disjoint codomains. Note that each of these monomial is equal to $q * M q$ where $M$ is a monomial of type 4 above.

Suppose now that $u\in A\limp B$ and $v\in A$ . Then, since $0\in B\orth$ ( $0$ belongs to any type) $u .(p v p *) = p u 11 v p *$ is nilpotent, thus $u 11 v$ is nilpotent.

Suppose further that $w\in B\orth$ . Then $u .(p v p * + q w q *)$ is nilpotent, thus there is a $N$ such that $(u .(p v p * + q w q *)) n = 0$ for any $n\geq N$ . This entails that all monomials of type 1 to 4 are null because they have disjoint domains and disjoint codomains. Therefore all monomials appearing in the developpment of $(App(u, v) w) N$ are null which proves that $App(u, v) w$ is nilpotent. Thus $\mathrm{App}(u,v)\in B$ .

Conversely suppose for any $v\in A$ and $w\in B\orth$ , $u 11 v$ and $App(u, v) w$ are nilpotent. Let $P$ and $N$ be their respective degrees of nilpotency and put $n = N (P + 1) + N$ . Then we claim that all monomials appearing in the development of $(u .(p v p * + q w q *)) n$ are null.

Consider for example a monomial of type 1:

$p(u_{11}v)^{i_0}u_{12}w(u_{22}w)^{i_1}\dots u_{21}v(u_{11}v)^{i_m}p^*$ .

If $i_{2k}\geq P$ for some $0\leq k\leq m/2$ then $(u_{11}v)^{i_{2k}}=0$ . Otherwise if $i 2 k < P$ for all $k$ then as we have:

$i_0+\cdots+i_m + m = n$

we deduce:

$i_1+i_3+\cdots +i_{m-1} + m/2 = n - m/2 - (i_0+i_2+\cdots +i_m)$

thus:

$i_1+i_3+\cdots +i_{m-1}\geq n - (m/2)(1+P)$ .

Now if $m/2\geq N$ then $i_1+\cdots+i_{m-1}+m/2 \geq N$ . Otherwise $m / 2 < N$ and

$i_1+i_3+\cdots +i_{m-1}\geq n - N(1+P) = N$ .

Since $N$ is the degree of nilpotency of $App(u, v) w$ we have that the monomial:

$(u_{22}w)^{i_1}u_{21}v(u_{11}v)^{i_2}u_{12}w\dots(u_{11}v)^{i_{m-2}}u_{12}w(u_{22}w)^{i_{m-1}}$

is null. Thus so is the monomial of type 1 we started with.

Corollary

If $A$ and $B$ are types then we have:

$A\limp B = \{u\in\mathcal{P} \text{ such that }\forall v\in A: u_{11}v\in\bot\text{ and } \mathrm{App}(u, v)\in B\}$ .

As an example if we compute the application of the interpretation of the identity $ι$ in type $A\limp A$ to the operator $v\in A$ then we have:

$\mathrm{App}(\iota, v) = \iota_{22} + \iota_{21}v\sum(\iota_{11}v)^k\iota_{12}$ .

Now recall that $ι = p q * + q p *$ so that $ι 11 = ι 22 = 0$ and $ι 12 = ι 21 = 1$ and we thus get:

App(ι, v) = v

as expected.

The tensor rule

Let now $A, A', B$ and $B'$ be types and consider two operators $u$ and $u'$ respectively in $A\limp B$ and $A\limp B'$ . We define an operator denoted by $u\tens u'$ by:

$\begin{align} u\tens u' &= ppp^*upp^*p^* + qpq^*upp^*p^* + ppp^*uqp^*q^* + qpq^*uqp^*q^*\\ &+ pqp^*u'pq^*p^* + qqq^*u'pq^*p^* + pqp^*u'qq^*q^* + qqq^*u'qq^*q^* \end{align}$

Once again the notation is motivated by linear logic syntax and is contradictory with linear algebra practice since what we denote by $u\tens u'$ actually is the internalization of the direct sum $u\oplus u'$ .

Indeed if we think of $u$ and $u'$ as the internalizations of the matrices:

$\begin{pmatrix}u_{11} & u_{12}\\ u_{21} & u_{22} \end{pmatrix}$ and $\begin{pmatrix}u'_{11} & u'_{12}\\ u'_{21} & u'_{22} \end{pmatrix}$

then we may write:

$\begin{align} u\tens u' &= ppu_{11}p^*p^* + qpu_{21}p^*p^* + ppu_{12}p^*q^* + qpu_{22}p^*q^*\\ &+ pqu'_{11}q^*p^* + qqu'_{21}q^*p^* + pqu'_{12}q^*q^* + qqu'_{22}q^*q^* \end{align}$

Thus the components of $u\tens u'$ are given by:

$(u\tens u')_{ij} = pu_{ij}p^* + qu'_{ij}q^*$ .

and we see that $u\tens u'$ is actually the internalization of the matrix:

$\begin{pmatrix} u_{11} & 0 & u_{12} & 0 \\ 0 & u'_{11} & 0 & u'_{12} \\ u_{21} & 0 & u_{22} & 0 \\ 0 & u'_{21} & 0 & u'_{22} \\ \end{pmatrix}$

We are now to show that if we suppose $u$ and $u'$ are in types $A\limp B$ and $A'\limp B'$ , then $u\tens u'$ is in $A\tens A'\limp B\tens B'$ . For this we consider $v$ and $v'$ in respectively in $A$ and $A'$ , so that $p v p * + q v' q *$ is in $A\tens A'$ , and we show that $\mathrm{App}(u\tens u', pvp^* + qv'q^*)\in B\tens B'$ .

Since $u$ and $u'$ are in $A\limp B$ and $A'\limp B'$ we have that $App(u, v)$ and $App(u', v')$ are respectively in $B$ and $B'$ , thus:

$p\mathrm{App}(u, v)p^* + q\mathrm{App}(u', v')q^* \in B\tens B'$ .

We know that both $u 11 v$ and $u' 11 v'$ are nilpotent. But we have:

$\begin{align} \bigl((u\tens u')_{11}(pvp^* + qv'q^*)\bigr)^n &= \bigl((pu_{11} + qu'_{11}q^*)(pvp^* + qv'q^*)\bigr)^n\\ &= (pu_{11}vp^* + qu'_{11}v'q^*)^n\\ &= p(u_{11}v)^np^* + q(u'_{11}v')^nq^* \end{align}$

Therefore $(u\tens u')_{11}(pvp^* + qv'q^*)$ is nilpotent. So we can compute $\mathrm{App}(u\tens u', pvp^* + qv'q^*)$ :

$\begin{align} &\mathrm{App}(u\tens u', pvp^* + qv'q^*)\\ &= (u\tens u')_{22} + (u\tens u')_{21}(pvp^* + qv'q^*)\sum\bigl((u\tens u')_{11}(pvp^* + qv'q^*)\bigr)^k(u\tens u')_{12}\\ &= pu_{22}p^* + qu'_{22}q^* + (pu_{21}p^* + qu'_{21}q^*)(pvp^* + qv'q^*)\sum\bigl((pu_{11}p^* + qu'_{11}q^*)(pvp^* + qv'q^*)\bigr)^k(pu_{12}p^* + qu'_{12}q^*)\\ &= p\bigl(u_{22} + u_{21}v\sum(u_{11}v)^ku_{12}\bigr)p^* + q\bigl(u'_{22} + u'_{21}v'\sum(u'_{11}v')^ku'_{12}\bigr)q^*\\ &= p\mathrm{App}(u, v)p^* + q\mathrm{App}(u', v')q^* \end{align}$

thus lives in $B\tens B'$ .

Other monoidal constructions

Contraposition

Let $A$ and $B$ be some types; we have:

$A\limp B = A\orth\limpinv B\orth$

Indeed, $u\in A\limp B$ means that for any $v$ and $w$ in respectively $A$ and $B\orth$ we have $u.(pvp^* + qwq^*)\in\bot$ which is exactly the definition of $A\orth\limpinv B\orth$ .

We will denote $u\orth$ the operator:

$u\orth = pu_{22}p^* + pu_{12}q^* + qu_{12}p^* + qu_{11}q^*$

where $u i j$ is given by externalization. Therefore the externalization of $u\orth$ is:

$(u\orth)_{ij} = u_{\bar i\,\bar j}$ where $\bar .$ is defined by $\bar1 = 2, \bar2 = 1$ .

From this we deduce that $u\orth\in B\orth\limp A\orth$ and that $(u\orth)\orth = u$ .

Commutativity

Let $σ$ be the operator:

σ = p p q * q * + p q p * q * + q p q * p * + q q p * p *

.

One can check that $σ$ is the internalization of the operator $S$ on $H\oplus H\oplus H\oplus H$ defined by: $S(x_1\oplus x_2\oplus x_3\oplus x_4) = x_4\oplus x_3\oplus x_2\oplus x_1$ . In particular the components of $σ$ are:

σ 11 = σ 22 = 0

;

σ 12 = σ 21 = p q * + q p *

.

Let $A$ and $B$ be types and $u$ and $v$ be operators in $A$ and $B$ . Then $p u p * + q v q *$ is in $A\tens B$ and as $σ 11 .(p u p * + q v q *) = 0$ we may compute:

$\begin{align} \mathrm{App}(\sigma, pup^* + qvq^*) &= \sigma_{22} + \sigma_{21}(pup^* + qvq^*)\sum(\sigma_{11}(pup^* + qvq^*))^k\sigma_{12}\\ &= (pq^* + qp^*)(pup^* + qvq^*)(pq^* + qp^*)\\ &= pvp^* + quq^* \end{align}$

But $pvp^* + quq^*\in B\tens A$ , thus we have shown that:

$\sigma\in (A\tens B) \limp (B\tens A)$ .

Distributivity

We get distributivity by considering the operator:

δ = p p p * p * q * + p q p q * p * q * + p q q q * q * + q p p p * p * + q p q p * q * p * + q q q * q * p *

that is similarly shown to be in type $A\tens(B\tens C)\limp(A\tens B)\tens C$ for any types $A$ , $B$ and $C$ .

Weak distributivity

We can finally get weak distributivity thanks to the operators:

δ 1 = p p p p * q * + p p q p * q * q * + p q q * q * q * + q p p * p * p * + q q p q * p * p * + q q q q * p *

and

δ 2 = p p p * p * q * + p q p q * p * q * + p q q q * q * + q p p p * p * + q p q p * q * p * + q q q * q * p *

.

Given three types $A$ , $B$ and $C$ then one can show that:

δ 1

has type $((A\limp B)\tens C)\limp A\limp (B\tens C)$ and

δ 2

has type $(A\tens(B\limp C))\limp (A\limp B)\limp C$ .

Execution formula, version 2: composition

Let $A$ , $B$ and $C$ be types and $u$ and $v$ be operators respectively in types $A\limp B$ and $B\limp C$ .

As usual we will denote $u i j$ and $v i j$ the operators obtained by externalization of $u$ and $v$ , eg, $u 11 = p * u p$ , ...

As $u$ is in $A\limp B$ we have that $\mathrm{App}(u, 0)=u_{22}\in B$ ; similarly as $v\in B\limp C$ , thus $v\orth\in C\orth\limp B\orth$ , we have $\mathrm{App}(v\orth, 0) = v_{11}\in B\orth$ . Thus $u 22 v 11$ is nilpotent.

We define the operator $Comp(u, v)$ by:

$\begin{align} \mathrm{Comp}(u, v) &= p(u_{11} + u_{12}\sum(v_{11}u_{22})^k\,v_{11}u_{21})p^*\\ &+ p(u_{12}\sum(v_{11}u_{22})^k\,v_{12})q^*\\ &+ q(v_{21}\sum(u_{22}v_{11})^k\,u_{21})p^*\\ &+ q(v_{22} + v_{21}\sum(u_{22}v_{11})^k\,u_{22}v_{12})q^* \end{align}$

This is well defined since $u 11 v 22$ is nilpotent. As an example let us compute the composition of $u$ and $ι$ in type $B\limp B$ ; recall that $ι i j = δ i j$ , so we get:

Comp(u,ι) = p u 11 p * + p u 12 q * + q u 21 p * + q u 22 q * = u

Similar computation would show that $Comp(ι, v) = v$ (we use $p p * + q q * = 1$ here).

Coming back to the general case we claim that $Comp(u, v)$ is in $A\limp C$ : let $a$ be an operator in $A$ . By computation we can check that:

App(Comp(u, v), a) = App(v,App(u, a))

.

Now since $u$ is in $A\limp B$ , $App(u, a)$ is in $B$ and since $v$ is in $B\limp C$ , $App(v,App(u, a))$ is in $C$ .

If we now consider a type $D$ and an operator $w$ in $C\limp D$ then we have:

Comp(Comp(u, v), w) = Comp(u,Comp(v, w))

.

Putting together the results of this section we finally have: Theorem

Let GoI(H) be defined by:

objects are types, ie sets $A$ of operators satisfying: $A\biorth = A$ ;
morphisms from $A$ to $B$ are operators in type $A\limp B$ ;
composition is given by the formula above.

Then GoI(H) is a star-autonomous category.

The Geometry of Interaction as an abstract machine

Cite error: <ref> tags exist, but no <references/> tag was found

[1]

[2]

[3]

[4]

Geometry of interaction