Given a relation R : A ⊗ U → B ⊗ U ∈NomLinRel, let us defineTrUA,B(R) by:
TrUA,B( R )= {r ˆA × ˆB | r ∈ ˆR ∧ r ˆA × ˆB ∈ A × B ∧ r ˆU × ˆU ∈idcU}. (3.24)
The use of strict substitutions allows us to equate some names in the relation. However, the part in ˆB × ˆC must still have the property that the atomes are name-distinct on each projection, condition that is enforced through r ˆB × ˆC ∈ B × C. One can see thatNomLinList
is a subcategory of NomLinRel. We write I for the inclusion functor, each morphism of
defined inNomLinRelextends the one ofNomLinList.
Proposition 3.19. Letφ : A ⊗ U → B ⊗ U ∈NomLinList, then:
I(TrUA,B, NomLinList(φ))=TrUA,B, NomLinRel(I(φ)) .
Proof. We first show that the inclusionTrUA,B, NomLinRel(I(φ)) ⊆ I(TrUA,B, NomLinList(φ)). Let x ∈ I(φ), and consider any y such that ∃e ∈Ξ. y = e · x, y ˆU × ˆU ∈idcU, and y ˆA × ˆB ∈ A × B. Then, without loss of generality, we can look at a canonical form of e given x, use the fact that I(φ) is closed under permutations, and consider that the only action of e is to equate names present in x.
Now let us consider, that, as defined in Section 3.1.2.1 above, we attribute a number to each location, and we will now compute the function ψ attributed to y A × B. So let us consider a location l in A, and the associated name yl in y, corresponding to xl in x. As e · x ˆA =
y ˆA, e · x ˆA ∈ A, and A is separated, then this entails there is a π ∈ Perm(A) such that
(π · y) A = x A. Furthermore, applying a permutation to y does not change the function ψ it defines. Hence one can assume without loss of generality that for all locations l in A, xl = yl.
We decompose x, y into x1, x2, y1, y2, where x1 = x A ? U (respectively y1 = y ˆA × ˆU),
x2= x A ? U (resp y2 = y ˆB × ˆU).
Let l in A, and consider that φ(l) = l0 ∈ B. Then, as x1,l = x2,l0 and as y is obtained from x
by strict substitutions, then y2,l0 = y1,l. Therefore, if φ(l) ∈ B then ψ(l)= φ(l). Writing n for the
number of locations of A (and hence, B as well), this sums up to: Given i ≤ n, ψ(i)= φ(i) if φ(i) ≤ n.
Now, let us suppose that φ(l) ∈ U. Then y1,l = y2,φ(l). But, as y ˆU × ˆU ∈ cidU, we also have
that y2,φ(l) = y1,φ(l). As y2,φ(φ(l) = y1,φ(l), we obtain y1,l = y2,φ◦φ(l). Then, either y2,φ◦φ(l)is in B or
we repeat the routine. That is, if i > n, then ψ(i)=feedback(φ)(i), where:
feedback(φ)(i)= φ(i) if φ(i) ≤ n
feedback(φ)(φ(i)) otherwise.
Therefore, ψ(i)=Tr(φ)(i), andTrUA,B,NomLinPol(I(φ)) ⊆ I(TrUA,B,NomLinList(φ)).
We now tackle the reverse inclusion. Let y ∈ I(TrUA,B, NomLinList(φ)), let us show that this y belongs inTrUA,B, NomLinPol(I(φ)). So we start by picking an x ∈ I(φ), such that y A = x A. We build a strict substitution in n-steps, corresponding to the n locations of A.
Given i ≤ n, let x1,i the ithelement of x1. Suppose that φ(i)= j ≤ n, then ei = ε (the empty
substitution). On the other hand, suppose that φ(i) = j > n. Then we set ei = [x1, j/x1,i]. That
where l corresponds to the occurrence of the literal in the jthposition of A ⊗ U. Inductively, if
φ( j) = k > n, we set ei::= [x1,k/x1, j].ei. We repeat this process, until φ(l)= m ≤ n.
At this stage, one needs an important property of traces of permutations: each location l of Uis used at most once when tracing (see [1] proposition 1). That is, if ∃i ≤ n, such that ∃k ≥ 1 φk(i) = l and φ1(i), φ2(i), ..., φk−1(i) > n, then this i is unique. As a result, if a name x
lsuch that
lin U is in one of the support of an ei, then this i is unique.
Now, let us consider all names whose locations are in U and that are not in the support of any ei, 1 ≤ i ≤ n. Then we consider the substitution en+1whose only action is to send all those
names to a fresh name d. Finally, we consider e = e1.e2. ... .en.en+1. Then e · x A → B = y.
This can be checked as in the above part of the proof. Furthermore, e · x U → U ∈ cidU. We
conclude that y ∈TrUA,B,NomLinRel(I(φ)).
This property allows us to prove that tracing a linear relation indeed results in a linear relation.
Proposition 3.20. Let R : A ⊗ U → B ⊗ U. Let x ∈TrUA,B( R ), then ν(x A) = ν(x B).
Proof. The proof is based on the above proposition 3.19. Let e · x an element that appears as witness of the trace, and hence such that x ∈ R . Then let us consider φ the morphism of
NomLinListassociated to x. Then e · x corresponds, as explained above, to tracing φ. Therefore, e · x A → B belongs in a graph of a function ψ ∈NomLinList(A → B) and therefore has same
support in A and B.
We can now state with confidence that this definition of trace seems appropriate, as it is a simple extension of the one previously defined in the case without biproduct, and acts in a compatible way with linear nominal relations. Only remains to prove that this family of functions formally defines a trace.
Proposition 3.21. The family of functionsTrUA,Bdefined in 3.24 forms a trace of the symmetric monoidal categoryNomLinRel.
Proof. We start the proof with the naturality of the trace. We treat only one of the three cases, as the two others could be dealt with along the same lines. Let R : A ⊗ U → B ⊗ U and Q : A0 → A. Then one must check that :
TrUA,B( R ) ◦ Q =TrUA0,B( R ◦ ( Q ⊗idU)).
{u A0× B | u ∈ A0× ˆA × ˆU × ˆB × ˆU), u A0× ˆA ∈ Q, u ( ˆA × ˆU) × ( ˆB × ˆU) ∈ bR, u ˆ
U × ˆU ∈ cidU, u ˆA × ˆB ∈ A × B}
whereas the second can only be easily presented if we use the functoriality of b(.), that is, [
R ; Q = bR ; bQ :
[
R ; Q = {u ˆA × ˆC | u ∈ ˆA × ˆB × ˆC | u ˆA × ˆB ∈ bR, u ˆB × ˆC ∈ bQ }
as well asQ[?idU = Q × cb idU, which follows from the monoidality of the functor b(.) which is
proven in proposition 3.17. Using it, the second term unfolds to:
{w A0× B | w ∈ ( ˆA0× ˆU
1) × ( ˆA × ˆU2) × ( ˆB × ˆU3), w ˆA0× ˆA ∈ bQ, w ( ˆA × ˆU2) × ( ˆB × ˆU3) ∈
b
R, w ˆU1× ˆU3∈ cidU, w ˆU2× ˆU3∈ cidU, w ˆA0× ˆB ∈ A × B}
In the second term, as w ˆU1× ˆU3 ∈ cidU and w ˆU2 × ˆU3 ∈ cidU, one can devise that
u ˆU1× ˆU2 = u ˆU2× ˆU3 = u ˆU1× ˆU3. The two terms only differ by the presence, in the second one, of U1× U2, which is insignificant (as w ˆU1× ˆU2 ∈ cidU), and the fact that
w ˆA0× ˆA ∈ bQ whereas u A0× A ∈ Q in the first one. At this stage, it is useful to note the
following property. Let P : A → B a nominal linear relation. Suppose u ∈ bP and u bA ∈ A. Then u ∈ P. This is simply proven by noticing that as u bA ∈ A, then no names have been merged. We can apply this property in our case, as w bA0 ∈ A0. As a result, w
ˆA0× ˆA ∈ Q
and the two sets are equal.
Next, we check the first part of vanishing property, namely that
TrIB,C( R )= R which translates into:
{u B × C | u ∈ ˆR ×idI, u ˆB × ˆC ∈ B × C} = {u ∈ R}
The right to left inclusion is straightforward. For the left to right one, we use the property that r ∈ bR ∧ r ˆB ∈ B ⇒ r ∈ R , as explained above. This entails the left to right inclusion.
The second part of the vanishing property is consists in proving that given R : A ⊗ U ⊗ V → B ⊗ U ⊗ V, thenTrU⊗VA,B ( R )=TrUA,B(TrVA⊗U,B⊗U( R )). This first term can be described by:
{r A × B | r ∈ ˆR , r ˆA × ˆB ∈ A × B, r ∈ ˆU × ˆV × ˆU × ˆV ∈ [idU⊗V}
The second term unfolds in two steps. First, we considerTrVA⊗U,B⊗U( R ).
TrVA⊗U,B⊗U( R ))= {u (A ? U) × (B ? U) | u ∈ ˆR, u ( ˆA × ˆU) × ( ˆB × ˆU) ∈ (A ? U) × (B ? U), u ˆV × ˆV ∈ cidV}.
Then we can consider its closureTrVA⊗U,B⊗U( R )
V
,
TrVA⊗U,B⊗U( R )
V
= {u ( ˆA × ˆU) × ( ˆB × ˆU) | u ∈ ˆR, u ˆV × ˆV ∈idcV} in order to devise the whole second term.
TrUA,B(TrVA⊗U,B⊗U( R ))= {u A × B | u ∈TrVA⊗U,B⊗U( R )
V
, u ˆA × ˆB ∈ A × B, u ˆU × ˆU ∈idcU}
= {u A × B | u ∈ ˆR, u ˆV × ˆV ∈idcV, u ˆU × ˆU ∈idcU, u ˆA × ˆB ∈ A × B}
= {u A × B | u ∈ ˆR, u ( ˆU × ˆV) × ( ˆU × ˆV) ∈ [idU?V, u ˆA × ˆB ∈ A × B}
=TrU⊗VA,B ( R ) We now deal with superposing:
TrUC⊗A,D⊗B( Q ⊗ R )= Q ⊗TrUA,B( R ), for R : A ⊗ U → B ⊗ U, and Q : C → D. This translates into:
{u (C ? A) × (D ? B) | u ∈ Q ? R
V
, u ˆU × ˆU ∈idcU, u ( ˆC × ˆA) × ( ˆD × ˆB) ∈ (C ? A) × (D ? B)} ?
= {w (C ? A) × (D ? B) | w ∈ Q ? bR, w ˆA × ˆB ∈ A × B, w ˆU × ˆU ∈idcU}
The fact that the second set is included in the first one is the easiest inclu- sion to prove. It follows from Q ? bR ⊆ Q ? R
V
. Let us prove the reverse in- clusion. As u ∈ Q ? R
V
∧ u ˆC × ˆD ∈ C × D then u C × D ∈ Q ∧ u ∈idC× ˆR , how-
ever, nothing imposes that u ∈idC? ˆR , the names in the idC and R part of u couldˆ
be the same. However, as u ( ˆC × ˆA) × ( ˆD × ˆB) ∈ (C ? A) × (D × B), one gathers that ν(u ( ˆA × ˆU) × ( ˆB × ˆU)) ∩ ν(u C × D) = ν(u ˆU × ˆU) ∩ ν(u C × D). Let us suppose that this set is not empty, and let a be a name appearing in it. Then, by definition, this name does not appear in u ˆA × ˆB. Therefore, by applying a permutation (a, c) to u ( ˆA × ˆU) × ( ˆB × ˆU), where c is fresh, we can remove the name a from the intersection. That is, let u0= (u C × D) × ((a, c) · u ( ˆA × ˆU) × ( ˆB × ˆU)). Then u0 ( ˆC × ˆA) × ( ˆD × ˆB) = u ( ˆA × ˆC) × ( ˆB × ˆD), u0 ( ˆA × ˆU) × ( ˆB × ˆU) ∈ bR , and u0 ˆU × ˆU ∈ cidU. Furthermore,
ν(u0
( ˆA × ˆU) × ( ˆB × ˆU)) ∩ ν(u0 C × D) = ν(u ( ˆA × ˆU) × ( ˆB × ˆU)) ∩ ν(u C × D) \ {a}. Since the support of u is finite, we can repeat the procedure for every name in the intersection. Finally, we obtain that an element u00 such that u00 ( ˆC × ˆA × ˆD × ˆB)= u ( ˆC × ˆA) × ( ˆD × ˆB), u00 ∈ Q ? bR , and such that u00 satisfies all the conditions of the second set. Hence the two sets are equal.
Finally, we must prove the yanking property:
Let us remind that sA,Ais the isomorphism coming from the symmetry of the monoidal category,
sA,A= {(u.v, v.u) | u, v ∈ A, u#v}. Hence
TrAA,A= {(u, w) ∈ A × A | ∃(v, x) ∈A × bb A.(u.v, w.x) ∈sdA,A.(v, x) ∈idcA}
As (v, x) ∈ cidA, v= x. Furthermore, as (u.v, w.v) ∈sdA,A, u= v, and v = w. As a result:
trAA,A(sA,A)= {(u, u) | u ∈ A} =idA