Geometry of interaction
m (→The Geometry of Interaction as operators: typo) 
(Execution formula : false assertion corrected) 

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These three sets are closed subspaces of <math>H</math>. 
These three sets are closed subspaces of <math>H</math>. 

−  The ''adjoint'' of an operator <math>u</math> is the operator <math>u^*</math> defined by <math>\langle u(x), y\rangle = \langle x, u^*(y)\rangle</math> for any <math>x,y\in H</math>. 
+  The ''adjoint'' of an operator <math>u</math> is the operator <math>u^*</math> defined by <math>\langle u(x), y\rangle = \langle x, u^*(y)\rangle</math> for any <math>x,y\in H</math>. Adjointness if well behaved w.r.t. composition of operators: 
+  : <math>(uv)^* = v^*u^*</math>. 

A ''projector'' is an idempotent operator of norm <math>0</math> (the projector 
A ''projector'' is an idempotent operator of norm <math>0</math> (the projector 

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{{Definition 
{{Definition 

−  The ''proof space'' <math>\mathcal{P}</math> is the set of partial isometries of the form <math>u_\varphi</math> for partial permutations <math>\varphi</math> on <math>\mathbb{N}</math>. 
+  We call ''<math>p</math>isometry'' a partial isometry of the form <math>u_\varphi</math> where <math>\varphi</math> is a partial permutation on <math>\mathbb{N}</math>. The ''proof space'' <math>\mathcal{P}</math> is the set of all <math>p</math>isometries. 
}} 
}} 

−  In particular note that <math>0\in\mathcal{P}</math>. The set <math>\mathcal{P}</math> is a submonoid of <math>\mathcal{B}(H)</math> but it is not a subalgebra: in general given <math>u,v\in\mathcal{P}</math> we don't necessarily have <math>u+v\in\mathcal{P}</math>. However we have: 
+  In particular note that <math>0</math> is a <math>p</math>isometry. The set <math>\mathcal{P}</math> is a submonoid of <math>\mathcal{B}(H)</math> but it is not a subalgebra<ref><math>\mathcal{P}</math> is the normalizing groupoid of the maximal commutative subalgebra of <math>\mathcal{B}(H)</math> consisiting of all operators ''diagonalizable'' in the canonical basis.</ref>. In general given <math>u,v\in\mathcal{P}</math> we don't necessarily have <math>u+v\in\mathcal{P}</math>. However we have: 
{{Proposition 
{{Proposition 

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: <math>u+v\in\mathcal{P}</math> iff <math>uu^*vv^* = u^*uv^*v = 0</math>. 
: <math>u+v\in\mathcal{P}</math> iff <math>uu^*vv^* = u^*uv^*v = 0</math>. 

}} 
}} 

+  Suppose for contradiction that <math>e_n</math> is in the domain of <math>u</math> and in the domain of <math>v</math>. As <math>u</math> and <math>v</math> are <math>p</math>isometries there are integers <math>p</math> and <math>q</math> such that <math>u(e_n) = e_p</math> and <math>v(e_n) = e_q</math> thus <math>(u+v)(e_n) = e_p + e_q</math> which is not a basis vector; therefore <math>u+v</math> is not a <math>p</math>permutation. 

−  Also note that if <math>u+v=0</math> then <math>u=v=0</math>. 
+  As a corollary note that if <math>u+v=0</math> then <math>u=v=0</math>. 
=== From operators to matrices: internalization/externalization === 
=== From operators to matrices: internalization/externalization === 

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: <math>q(e_n) = e_{2n+1}</math>. 
: <math>q(e_n) = e_{2n+1}</math>. 

−  From the definition <math>p</math> and <math>q</math> have full domain, that is satisfy <math>p^* p = q^* q = 1</math>. On the other hand their codomains are orthogonal, thus we have <math>p^* q = q^* p = 0</math>. Note that we also have <math>pp^* + qq^* = 1</math>. 
+  From the definition <math>p</math> and <math>q</math> have full domain, that is 
+  satisfy <math>p^* p = q^* q = 1</math>. On the other hand their codomains are 

+  disjoint, thus we have <math>p^*q = q^*p = 0</math>. As the sum of their 

+  codomains is the full space <math>H</math> we also have <math>pp^* + qq^* = 1</math>. 

−  The choice of <math>p</math> and <math>q</math> is actually arbitrary, any two partial isometries with full domain and orthogonal codomains would do the job. 
+  Note that we have choosen <math>p</math> and <math>q</math> in <math>\mathcal{P}</math>. However the choice is arbitrary: any two <math>p</math>isometries with full domain and disjoint codomains would do the job. 
−  Let <math>U</math> be an operator on <math>H\oplus H</math>. We can write <math>U</math> as a matrix: 
+  Given an operator <math>u</math> on <math>H</math> we may ''externalize'' it obtaining an operator <math>U</math> on <math>H\oplus H</math> defined by the matrix: 
: <math>U = \begin{pmatrix} 
: <math>U = \begin{pmatrix} 

u_{11} & u_{12}\\ 
u_{11} & u_{12}\\ 

u_{21} & u_{22} 
u_{21} & u_{22} 

\end{pmatrix}</math> 
\end{pmatrix}</math> 

−  where each <math>u_{ij}</math> operates on <math>H</math>. 
+  where the <math>u_{ij}</math>'s are given by: 
−  
−  Now through the isomorphism <math>H\oplus H\cong H</math> we may transform <math>U</math> into the operator <math>u</math> on <math>H</math> defined by: 

−  
−  : <math>u = pu_{11}p^* + pu_{12}q^* + qu_{21}p^* + qu_{22}q^*</math>. 

−  
−  We call <math>u</math> the ''internalization'' of <math>U</math>. Internalization is compatible with composition (functorial so to speak): if <math>V</math> is another operator on <math>H\oplus</math> then the internalization of the matrix product <math>UV</math> is the product <math>uv</math>. 

−  
−  Conversely given an operator <math>u</math> on <math>H</math> we may externalize it obtaining an operator <math>U</math> on <math>H\oplus H</math>: 

: <math>u_{11} = p^*up</math>; 
: <math>u_{11} = p^*up</math>; 

: <math>u_{12} = p^*uq</math>; 
: <math>u_{12} = p^*uq</math>; 

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: <math>u_{22} = q^*uq</math>. 
: <math>u_{22} = q^*uq</math>. 

−  The <math>u_{ij}</math>'s are called the ''components'' of <math>u</math>. Note that if <math>u</math> is generated by a partial permutation, that is if <math>u\in\mathcal{P}</math> then so are the <math>u_{ij}</math>'s. Moreover we have: 
+  The <math>u_{ij}</math>'s are called the ''components'' of <math>u</math>. The externalization is functorial in the sense that if <math>v</math> is another operator externalized as: 
+  : <math>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} 

+  </math> 

+  then the externalization of <math>uv</math> is <math>UV</math>. 

+  
+  We also have: 

: <math>u = (pp^*+qq^*)u(pp^*+qq^*) = pu_{11}p^* + pu_{12}q^* + qu_{21}p^* + qu_{22}q^*</math> 
: <math>u = (pp^*+qq^*)u(pp^*+qq^*) = pu_{11}p^* + pu_{12}q^* + qu_{21}p^* + qu_{22}q^*</math> 

−  which entails that the four terms of the sum have pairwise disjoint domains and pairwise disjoint codomains. This can be verified for example by computing the product of the final projectors of <math>pu_{11}p^*</math> and <math>pu_{12}q^*</math>: 
+  which entails that externalization is reversible, its converse being called ''internalization''. 
+  
+  Furthermore if we suppose that <math>u</math> is a <math>p</math>isometry then so are the components <math>u_{ij}</math>'s. Thus the formula above entails that the four terms of the sum have pairwise disjoint domains and pairwise disjoint codomains. 

+  {{Proposition 

+  If <math>u</math> is a <math>p</math>isometry and <math>u_{ij}</math> are its external components then: 

+  * <math>u_{1j}</math> and <math>u_{2j}</math> have disjoint domains, that is <math>u_{1j}^*u_{1j}u_{2j}^*u_{2j} = 0</math> for <math>j=1,2</math>; 

+  * <math>u_{i1}</math> and <math>u_{i2}</math> have disjoint codomains, that is <math>u_{i1}u_{i1}^*u_{i2}u_{i2}^* = 0</math> for <math>i=1,2</math>. 

+  }} 

+  
+  As an example of computation in <math>\mathcal{P}</math> let us compute the product of the final projectors of <math>pu_{11}p^*</math> and <math>pu_{12}q^*</math>: 

: <math>\begin{align} 
: <math>\begin{align} 

(pu_{11}p^*)(pu^*_{11}p^*)(pu_{12}q^*)(qu_{12}^*p^*) 
(pu_{11}p^*)(pu^*_{11}p^*)(pu_{12}q^*)(qu_{12}^*p^*) 

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=== The tensor and the linear application === 
=== The tensor and the linear application === 

−  Given two types <math>A</math> and <math>B</math>, we define their tensor by: 
+  If <math>u</math> and <math>v</math> are two <math>p</math>isometries summing them doesn't in general produces a <math>p</math>isometry. However as <math>pup^*</math> and <math>qvq^*</math> have disjoint domains and disjoint codomains it is true that <math>pup^* + qvq^*</math> is a <math>p</math>isometry. Given two types <math>A</math> and <math>B</math>, we thus define their ''tensor'' by: 
: <math>A\tens B = \{pup^* + qvq^*, u\in A, v\in B\}\biorth</math> 
: <math>A\tens B = \{pup^* + qvq^*, u\in A, v\in B\}\biorth</math> 

−  Note the closure by bidual to make sure that we obtain a type. From what precedes we see that <math>A\tens B</math> is generated by the internalizations of operators on <math>H\oplus H</math> of the form: 
+  Note the closure by bidual to make sure that we obtain a type. 
+  
+  From what precedes we see that <math>A\tens B</math> is generated by the internalizations of operators on <math>H\oplus H</math> of the form: 

: <math>\begin{pmatrix} 
: <math>\begin{pmatrix} 

u & 0\\ 
u & 0\\ 

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\end{pmatrix}</math> 
\end{pmatrix}</math> 

−  This is an abuse of notations as this operation is more like a direct sum than a tensor. We will stick to this notation though because it defines the interpretation of the tensor connective of linear logic. 
+  Note that this socalled 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 <math>A</math> and <math>B</math> the type <math>A\limp B</math> is defined by: 
The linear implication is derived from the tensor by duality: given two types <math>A</math> and <math>B</math> the type <math>A\limp B</math> is defined by: 

: <math>A\limp B = (A\tens B\orth)\orth</math>. 
: <math>A\limp B = (A\tens B\orth)\orth</math>. 

−  Unfolding this definition we see that we have: 
+  Unfolding this definition we get: 
−  : <math>A\limp B = \{u\in\mathcal{P}\text{ such that } \forall v\in A, \forall w\in B\orth,\, u.(pvp^* + qwq^*) \in\bot\}</math>. 
+  : <math>A\limp B = \{u\in\mathcal{P}\text{ s.t. } \forall v\in A, \forall w\in B\orth,\, u.(pvp^* + qwq^*) \in\bot\}</math>. 
=== The identity === 
=== The identity === 

−  The interpretation of the identity is an example of the internalization/externalization procedure. Given a type <math>A</math> we are to find an operator <math>\iota</math> in type <math>A\limp A</math>, thus satisfying: 
+  The interpretation of the identity is an example of internalization. Given a type <math>A</math> we are to find an operator <math>\iota</math> in type <math>A\limp A</math>, thus satisfying: 
: <math>\forall u\in A, v\in A\orth,\, \iota(pup^* + qvq^*)\in\bot</math>. 
: <math>\forall u\in A, v\in A\orth,\, \iota(pup^* + qvq^*)\in\bot</math>. 

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=== The execution formula, version 1: application === 
=== The execution formula, version 1: application === 

−  Let <math>A</math> and <math>B</math> be two types and <math>u</math> an operator in <math>A\limp B</math>. By definition this means that given <math>v</math> in <math>A</math> and <math>w</math> in <math>B\orth</math> the operator <math>u.(pvp^* + qwq^*)</math> is nilpotent. 
+  {{Definition 
+  Let <math>u</math> and <math>v</math> be two operators; as above denote by <math>u_{ij}</math> the external components of <math>u</math>. Assume that <math>u_{11}v</math> is nilpotent. We define a new operator <math>\mathrm{App}(u,v)</math> by: 

+  : <math>\mathrm{App}(u,v) = u_{22} + u_{21}v\sum_k(u_{11}v)^ku_{11}</math>. 

+  }} 

−  Let us define <math>u_{11}</math> to <math>u_{22}</math> by externalization as above. If we compute <math>(u.(pvp^* + qwq^*))^n</math> we see that this is a finite sum of operators of the form: 
+  Note that the hypothesis that <math>u_{11}v</math> is nilpotent entails that the sum <math>\sum_k(u_{11}v)^k</math> 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. 
−  # <math>q(u_{22}w)^{k_0}u_{21}v(u_{11}v)^{k_1}u_{12}w\dots u_{12}w(u_{22}w)^{k_{p+1}}q^*</math>, 

−  # <math>p(u_{11}v)^{k_1}u_{12}w\dots u_{12}w(u_{22}w)^{k_{p+1}}q^*</math>, 

−  # <math>q(u_{22}w)^{k_0}u_{21}v(u_{11}v)^{k_1}u_{12}w\dots (u_{11}v)^{k_p}p^*</math> or 

−  # <math>p(u_{11}v)^{k_1}u_{12}w\dots (u_{11}v)^{k_p}p^*</math> 

−  where each of these monimials has exactly <math>n</math> factors of the form <math>u_{i1}v</math> or <math>u_{i2}w</math>. 

−  From the nilpotency of <math>u.(pvp^* + qwq^*)</math> we deduce that <math>u_{11}v</math> is nilpotent by considering the particular case where <math>w=0</math>. We also have that <math>q^*(u.(pvp^* + qwq^*))^nq</math> is null for <math>n</math> big enough, <strike>which means that monomials of type 1 above are null as soon as their length (the number of factors of the form <math>u_{i1}v</math> or <math>u_{i2}w</math>) is bigger than <math>n</math></strike>. 
+  {{Theorem 
+  If <math>u</math> and <math>v</math> are <math>p</math>isometries such that <math>u_{11}v</math> is nilpotent, then <math>\mathrm{App}(u,v)</math> is also a <math>p</math>isometry. 

+  }} 

−  This implies that the two following operators are nilpotent: 
+  {{Proof 
−  : <math>u_{11}v</math> and 
+  Let us note <math>E_k = u_{21}v(u_{11}v)^ku_{12}</math>. Recall that <math>u_{22}</math> and <math>u_{12}</math> being external components of the <math>p</math>isometry <math>u</math>, they have disjoint domains. Thus it is also the case of <math>u_{22}</math> and <math>E_k</math>. Similarly <math>u_{22}</math> and <math>E_k</math> have disjoint codomains because <math>u_{22}</math> and <math>u_{21}</math> have disjoint codomains. 
−  : <math>\bigl(u_{22} + u_{21}v\sum_k(u_{11}v)^k u_{12}\bigr)w</math>. 

−  Conversely if these two operators are nilpotent then one can show that so is <math>u.(pvp^* + qwq^*)</math>. Moreover we have: 
+  Let now <math>k</math> and <math>l</math> be two integers such that <math>k>l</math> and let us compute for example the intersection of the codomains of <math>E_k</math> and <math>E_l</math>: 
−  : <math>q^*\sum_n\bigl(u.(pvp^* + qwq^*)\bigr)^nq = \sum_n\bigl((u_{22} + u_{21}v\sum_k(u_{11}v)^k u_{12})w\bigr)^n</math>. 
+  : <math> 
+  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}^*) 

+  </math> 

+  As <math>k>l</math> we may write <math>(v^*u_{11}^*)^l = (v^*u^*_{11})^{kl1}v^*u^*_{11}(v^*u^*_{11})^l</math>. Let us note <math>E = u^*_{11}(v^*u^*_{11})^lv^*u_{21}^*u_{21}v(u_{11}v)^lu_{12}</math> so that <math>E_kE^*_kE_lE^*_l = u_{21}v(u_{11}v)^ku_{12}u^*_{12}(v^*u^*_{11})^{kl1}v^*Eu^*_{12}(v^*u^*_{11})^lv^*u_{21}^*</math>. We have: 

+  : <math>\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}</math> 

+  because <math>u_{11}</math> and <math>u_{12}</math> have disjoint codomains, thus <math>u^*_{11}u_{12} = 0</math>. 

−  We define the ''application of <math>u</math> to <math>v</math>'' as: 
+  Similarly we can show that <math>E_k</math> and <math>E_l</math> have disjoint domains. Therefore we have proved that all terms of the sum <math>\mathrm{App}(u,v)</math> have disjoint domains and disjoint codomains. Consequently <math>\mathrm{App}(u,v)</math> is a <math>p</math>isometry. 
−  : <math>\mathrm{App}(u, v) = u_{22} + u_{21}v\sum(u_{11}v)^ku_{12}</math>. 
+  }} 
−  Note that this is well defined as soon as <math>u_{11}v</math> is nilpotent. 

−  
−  We summarize what has just been shown in the following theorem: 

{{Theorem 
{{Theorem 

−  Let <math>u</math> be an operator, <math>A</math> and <math>B</math> be two types; the following conditions are equivalent: 
+  Let <math>A</math> and <math>B</math> be two types and <math>u</math> a <math>p</math>isometry. Then the two following conditions are equivalent: 
−  * <math>u\in A\limp B</math>; 
+  # <math>u\in A\limp B</math>; 
−  * for any <math>v\in A</math>, we both have: 
+  # for any <math>v\in A</math> we have: 
−  :: <math>u_{11}v</math> is nilpotent and 
+  #* <math>u_{11}v</math> is nilpotent and 
−  :: <math>\mathrm{App}(u, v)\in B</math>. 
+  #* <math>\mathrm{App}(u, v)\in B</math>. 
+  }} 

+  
+  {{Proof 

+  Let <math>v</math> and <math>w</math> be two <math>p</math>isometries. If we compute 

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

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

+  # <math>p(u_{11}v)^{i_0}u_{12}w(u_{22}w)^{i_1}\dots u_{21}v(u_{11}v)^{i_m}p^*</math> 

+  # <math>p(u_{11}v)^{i_0}u_{12}w(u_{22}w)^{i_1}\dots u_{12}w(u_{22}w)^{i_m}q^*</math>, 

+  # <math>q(u_{22}w)^{i_0}u_{21}v(u_{11}v)^{i_1}\dots u_{21}v(u_{11}v)^{i_m}p^*</math> or 

+  # <math>q(u_{22}w)^{i_0}u_{21}v(u_{11}v)^{i_1}\dots u_{12}w(u_{22}w)^{i_m}q^*</math>, 

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

+  
+  Each of these monomial is a <math>p</math>isometry. Furthermore they have disjoint domains and disjoint codomains because their sum is the <math>p</math>isometry <math>(u.(pvp^* + qwq^*))^n</math>. 

+  
+  Suppose <math>u_{11}v</math> is nilpotent and consider: 

+  : <math>\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</math>. 

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

+  : <math>(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}</math> 

+  for all tuples <math>(l_0,\dots, l_m)</math> such that <math>l_0\cdots l_m + m = n</math>. 

+  
+  Again as these monomials are <math>p</math>isometries and their sum is the <math>p</math>isometry <math>(\mathrm{App}(u,v)w)^n</math>, they have pairwise disjoint domains and pairwise disjoint codomains. Note that each of these monomial is equal to <math>q^*Mq</math> where <math>M</math> is a monomial of type 4 above. 

+  
+  Suppose now that <math>u\in A\limp B</math> and <math>v\in A</math>. Then, since <math>0\in B\orth</math> (<math>0</math> belongs to any type) <math>u.(pvp^*) = pu_{11}vp^*</math> is nilpotent, thus <math>u_{11}v</math> is nilpotent. 

+  
+  Suppose further that <math>w\in B\orth</math>. Then <math>u.(pvp^*+qwq^*)</math> is nilpotent, thus there is a <math>N</math> such that <math>(u.(pvp^* + qwq^*))^n=0</math> for any <math>n\geq N</math>. 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 <math>(\mathrm{App}(u,v)w)^N</math> are null which proves that <math>\mathrm{App}(u,v)w</math> is nilpotent. Thus <math>\mathrm{App}(u,v)\in B</math>. 

+  
+  Conversely suppose for any <math>v\in A</math> and <math>w\in B\orth</math>, <math>u_{11}v</math> and <math>\mathrm{App}(u,v)w</math> are nilpotent. Let <math>P</math> and <math>N</math> be their respective degrees of nilpotency and put <math>n=N(P+1)+N</math>. Then we claim that all monomials appearing in the development of <math>(u.(pvp^*+qwq^*))^n</math> are null. 

+  
+  Consider for example a monomial of type 1: 

+  : <math>p(u_{11}v)^{i_0}u_{12}w(u_{22}w)^{i_1}\dots u_{21}v(u_{11}v)^{i_m}p^*</math>. 

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

+  : <math>i_0+\cdots+i_m + m = n</math> 

+  we deduce: 

+  : <math>i_1+i_3+\cdots +i_{m1} + m/2 = n  m/2  (i_0+i_2+\cdots +i_m)</math> 

+  thus: 

+  : <math>i_1+i_3+\cdots +i_{m1}\geq n  (m/2)(1+P)</math>. 

+  Now if <math>m/2\geq N</math> then <math>i_1+\cdots+i_{m1}+m/2 \geq N</math>. Otherwise <math>m/2<N</math> and 

+  : <math>i_1+i_3+\cdots +i_{m1}\geq n  N(1+P) = N</math>. 

+  Since <math>N</math> is the degree of nilpotency of <math>\mathrm{App}(u,v)w</math> we have that the monomial: 

+  : <math>(u_{22}w)^{i_1}u_{21}v(u_{11}v)^{i_2}u_{12}w\dots(u_{11}v)^{i_{m2}}u_{12}w(u_{22}w)^{i_{m1}}</math> 

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

}} 
}} 

{{Corollary 
{{Corollary 

−  Under the hypothesis of the theorem we have: 
+  If <math>A</math> and <math>B</math> are types then we have: 
: <math>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\}</math>. 
: <math>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\}</math>. 

}} 
}} 
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 (lambdacalculus) didn't interpret a proof of 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 , that is a morphism from to . For proof composition the problem was then, given an operator on and another one on to construct a new operator on . 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 MELL fragment of linear logic (Multiplicative and Exponential fragment) which is enough to encode lambdacalculus. 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 proofnets 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 lambdacalculus in the AbadiGonthierLévy reformulation and simplification of Lamping's representation of sharing. Finally the original GoI for the MELL fragment has been reformulated in the framework of traced monoidal categories following an idea originally proposed by Joyal.
Contents

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 .
Second define a particular subset of P that will be denoted by ; then derive a duality on P: for , u and v are dual^{[2]}, iff .
For the GoI, two dualities have proved to work; we will consider the first one: nilpotency, ie, 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 (uv)^{n} = 0. Note in particular that iff .
When X is a subset of P define as the set of elements of P that are dual to all elements of X:
 .
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 then ;
 ;
 .
Last define a type as a subset T of the proof space that is equal to its bidual: . This means that iff for all operator , that is such that for all , we have .
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 of sequences of complex numbers such that the series converges. If and are two vectors of H we denote by their scalar product:
 .
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:
 .
Let us denote by the canonical hilbertian basis of H: where δ_{kn} is the Kroenecker symbol: δ_{kn} = 1 if k = n, 0 otherwise. Thus if is a sequence in H we have:
 .
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:
 .
The set of (bounded) operators is denoted by .
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:
 ;
 ;
 .
These three sets are closed subspaces of H.
The adjoint of an operator u is the operator u^{ * } defined by for any . Adjointness if well behaved w.r.t. composition of operators:
 (uv)^{ * } = 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 or 1. A projector is autoadjoint and its domain is equal to its codomain.
A partial isometry is an operator u satisfying uu^{ * }u = u; this condition entails that we also have u^{ * }uu^{ * } = u^{ * }. As a consequence uu^{ * } and uu^{ * } 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 uu^{ * } + vv^{ * } = 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 on is a function defined on a subset of which is onetoone onto a subset of . is called the domain of and its codomain. Partial permutations may be composed: if ψ is another partial permutation on then is defined by:
 iff and ;
 if then ;
 the codomain of 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 that we will denote by 1. This latter permutation is the neutral for composition.
If is a partial permutation there is an inverse partial permutation whose domain is and who satisfies:
Given a partial permutation one defines a partial isometry by:
In other terms if is a sequence in then is the sequence defined by:
 if , 0 otherwise.
We will (not so abusively) write when is undefined so that may shorten the definition of into:
 .
The domain of is the subspace spanned by the family and the codomain of is the subspace spanned by . As a particular case if is 1_{D} the partial identity on D then is the projector on the subspace spanned by .
Proposition
Let and ψ be two partial permutations. We have:
 .
The adjoint of is:
 .
In particular the initial projector of is given by:
 .
and the final projector of is:
 .
Projectors generated by partial identities commute; in particular we have:
 .
Definition
We call pisometry a partial isometry of the form where is a partial permutation on . The proof space is the set of all pisometries.
In particular note that 0 is a pisometry. The set is a submonoid of but it is not a subalgebra^{[4]}. In general given we don't necessarily have . However we have:
Proposition
Let . Then iff u and v have disjoint domains and disjoint codomains, that is:
 iff uu^{ * }vv^{ * } = u^{ * }uv^{ * }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 pisometries 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 ppermutation.
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 , and conversely. For this purpose we define an isomorphism by where and are partial isometries given by:
 p(e_{n}) = e_{2n},
 q(e_{n}) = e_{2n + 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 pp^{ * } + qq^{ * } = 1.
Note that we have choosen p and q in . However the choice is arbitrary: any two pisometries 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 defined by the matrix:
where the u_{ij}'s are given by:
 u_{11} = p^{ * }up;
 u_{12} = p^{ * }uq;
 u_{21} = q^{ * }up;
 u_{22} = q^{ * }uq.
The u_{ij}'s are called the components of u. The externalization is functorial in the sense that if v is another operator externalized as:
then the externalization of uv is UV.
We also have:
 u = (pp^{ * } + qq^{ * })u(pp^{ * } + qq^{ * }) = pu_{11}p^{ * } + pu_{12}q^{ * } + qu_{21}p^{ * } + qu_{22}q^{ * }
which entails that externalization is reversible, its converse being called internalization.
Furthermore if we suppose that u is a pisometry then so are the components u_{ij}'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 pisometry and u_{ij} are its external components then:
 u_{1j} and u_{2j} have disjoint domains, that is for j = 1,2;
 u_{i1} and u_{i2} have disjoint codomains, that is for i = 1,2.
As an example of computation in let us compute the product of the final projectors of pu_{11}p^{ * } and pu_{12}q^{ * }:
where we used the fact that all projectors in commute, which is in particular the case of pp^{ * } and u^{ * }pp^{ * }u.
Interpreting the multiplicative connectives
Recall that when u and v are partial isometries in we say they are dual when uv is nilpotent, and that denotes the set of nilpotent operators. A type is a subset of that is equal to its bidual. In particular is a type for any . We say that X generates the type .
The tensor and the linear application
If u and v are two pisometries summing them doesn't in general produces a pisometry. However as pup^{ * } and qvq^{ * } have disjoint domains and disjoint codomains it is true that pup^{ * } + qvq^{ * } is a pisometry. Given two types A and B, we thus define their tensor by:
Note the closure by bidual to make sure that we obtain a type.
From what precedes we see that is generated by the internalizations of operators on of the form:
Note that this socalled 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 is defined by:
 .
Unfolding this definition we get:
 .
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 , thus satisfying:
 .
An easy solution is to take ι = pq^{ * } + qp^{ * }. In this way we get ι(pup^{ * } + qvq^{ * }) = qup^{ * } + pvq^{ * }. Therefore (ι(pup^{ * } + qvq^{ * }))^{2} = quvq^{ * } + pvup^{ * }, from which one deduces that this operator is nilpotent iff uv is nilpotent. It is the case since u is in A and v in .
It is interesting to note that the ι thus defined is actually the internalization of the operator on given by the matrix:
 .
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_{ij} the external components of u. Assume that u_{11}v is nilpotent. We define a new operator App(u,v) by:
App(u,v) = u_{22} + u_{21}v  ∑  (u_{11}v)^{k}u_{11} 
k 
.
Note that the hypothesis that u_{11}v is nilpotent entails that the sum
∑  (u_{11}v)^{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 pisometries such that u_{11}v is nilpotent, then App(u,v) is also a pisometry.
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 pisometry 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}:
As k > l we may write . Let us note so that . We have:
because u_{11} and u_{12} have disjoint codomains, thus .
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 pisometry.
Theorem
Let A and B be two types and u a pisometry. Then the two following conditions are equivalent:
 ;
 for any we have:
 u_{11}v is nilpotent and
 .
Proof. Let v and w be two pisometries. If we compute
we get a finite sum of monomial operators of the form:
 ,
 or
 ,
for all tuples of integers such that .
Each of these monomial is a pisometry. Furthermore they have disjoint domains and disjoint codomains because their sum is the pisometry (u.(pvp^{ * } + qwq^{ * }))^{n}.
Suppose u_{11}v is nilpotent and consider:
 .
Developping we get a finite sum of monomials of the form:
for all tuples such that .
Again as these monomials are pisometries and their sum is the pisometry (App(u,v)w)^{n}, they have pairwise disjoint domains and pairwise disjoint codomains. Note that each of these monomial is equal to q^{ * }Mq where M is a monomial of type 4 above.
Suppose now that and . Then, since (0 belongs to any type) u.(pvp^{ * }) = pu_{11}vp^{ * } is nilpotent, thus u_{11}v is nilpotent.
Suppose further that . Then u.(pvp^{ * } + qwq^{ * }) is nilpotent, thus there is a N such that (u.(pvp^{ * } + qwq^{ * }))^{n} = 0 for any . 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 .
Conversely suppose for any and , 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.(pvp^{ * } + qwq^{ * }))^{n} are null.
Consider for example a monomial of type 1:
 .
If for some then . Otherwise if i_{2k} < P for all k then as we have:
we deduce:
thus:
 .
Now if then . Otherwise m / 2 < N and
 .
Since N is the degree of nilpotency of App(u,v)w we have that the monomial:
is null. Thus so is the monomial of type 1 we started with.
Corollary
If A and B are types then we have:
 .
As an example if we compute the application of the interpretation of the identity ι in type to the operator then we have:
 .
Now recall that ι = pq^{ * } + qp^{ * } 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 and . We define an operator denoted by by:
Once again the notation is motivated by linear logic syntax and is contradictory with linear algebra practice since what we denote by actually is the internalization of the direct sum .
Indeed if we think of u and u' as the internalizations of the matrices:
 and
then we may write:
Thus the components of are given by:
 .
and we see that is actually the internalization of the matrix:
We are now to show that if we suppose uand u' are in types and , then is in . For this we consider v and v' in respectively in A and A', so that pvp^{ * } + qv'q^{ * } is in , and we show that .
Since u and u' are in and we have that App(u,v) and App(u',v') are respectively in B and B', thus:
 .
We know that both u_{11}v and u'_{11}v' are nilpotent. But we have:
Therefore is nilpotent. So we can compute :
thus lives in .
Other monoidal constructions
Contraposition
Let A and B be some types; we have:
Indeed, means that for any v and w in respectively A and we have which is exactly the definition of .
We will denote the operator:
where u_{ij} is given by externalization. Therefore the externalization of is:
 where is defined by .
From this we deduce that and that .
Commutativity
Let σ be the operator:
 σ = ppq^{ * }q^{ * } + pqp^{ * }q^{ * } + qpq^{ * }p^{ * } + qqp^{ * }p^{ * }.
One can check that σ is the internalization of the operator S on defined by: . In particular the components of σ are:
 σ_{11} = σ_{22} = 0;
 σ_{12} = σ_{21} = pq^{ * } + qp^{ * }.
Let A and B be types and u and v be operators in A and B. Then pup^{ * } + qvq^{ * } is in and as σ_{11}.(pup^{ * } + qvq^{ * }) = 0 we may compute:
But , thus we have shown that:
 .
Distributivity
We get distributivity by considering the operator:
 δ = ppp^{ * }p^{ * }q^{ * } + pqpq^{ * }p^{ * }q^{ * } + pqqq^{ * }q^{ * } + qppp^{ * }p^{ * } + qpqp^{ * }q^{ * }p^{ * } + qqq^{ * }q^{ * }p^{ * }
that is similarly shown to be in type for any types A, B and C.
Weak distributivity
We can finally get weak distributivity thanks to the operators:
 δ_{1} = pppp^{ * }q^{ * } + ppqp^{ * }q^{ * }q^{ * } + pqq^{ * }q^{ * }q^{ * } + qpp^{ * }p^{ * }p^{ * } + qqpq^{ * }p^{ * }p^{ * } + qqqq^{ * }p^{ * } and
 δ_{2} = ppp^{ * }p^{ * }q^{ * } + pqpq^{ * }p^{ * }q^{ * } + pqqq^{ * }q^{ * } + qppp^{ * }p^{ * } + qpqp^{ * }q^{ * }p^{ * } + qqq^{ * }q^{ * }p^{ * }.
Given three types A, B and C then one can show that:
 δ_{1} has type and
 δ_{2} has type .
Execution formula, version 2: composition
Let A, B and C be types and u and v be operators respectively in types and .
As usual we will denote u_{ij} and v_{ij} the operators obtained by externalization of u and v, eg, u_{11} = p^{ * }up, ...
As u is in we have that ; similarly as , thus , we have . Thus u_{22}v_{11} is nilpotent.
We define the operator Comp(u,v) by:
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 ; recall that ι_{ij} = δ_{ij}, so we get:
 Comp(u,ι) = pu_{11}p^{ * } + pu_{12}q^{ * } + qu_{21}p^{ * } + qu_{22}q^{ * } = u
Similar computation would show that Comp(ι,v) = v (we use pp^{ * } + qq^{ * } = 1 here).
Coming back to the general case we claim that Comp(u,v) is in : 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 , App(u,a) is in B and since v is in , App(v,App(u,a)) is in C.
If we now consider a type D and an operator w in 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: ;
 morphisms from A to B are operators in type ;
 composition is given by the formula above.
Then GoI(H) is a starautonomous category.
The Geometry of Interaction as an abstract machine
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