aa r X i v : . [ m a t h . C T ] M a y Traces in monoidal categories
Stephan Stolz and Peter TeichnerOctober 29, 2018
Abstract
This paper contains the construction, examples and properties of a trace and a tracepairing for certain morphisms in a monoidal category with switching isomorphisms.Our construction of the categorical trace is a common generalization of the trace forendomorphisms of dualizable objects in a balanced monoidal category and the trace ofnuclear operators on a topological vector space with the approximation property. Ina forthcoming paper, applications to the partition function of super symmetric fieldtheories will be given.
Contents
Ban of Banach spaces . . . . . . . . . . . . . . . . . . . . 194.5 A Category of topological vector spaces . . . . . . . . . . . . . . . . . . 204.6 The Riemannian bordism category . . . . . . . . . . . . . . . . . . . . . 26
The results of this paper provide an essential step in our proof that the partitionfunction of a Euclidean field theory of dimension 2 | ST1]. While motivated by field theory, the two main results are the construction oftraces and trace pairings for certain morphisms in a monoidal category. Let C be amonoidal category with monoidal unit I ∈ C [McL]. Question.
What conditions on an endomorphism f ∈ C ( X, X ) allow us to constructa well-defined trace tr( f ) ∈ C ( I, I ) with the usual properties expected of a trace?Theorem 1.7 below provides an answer to this question. Our construction is acommon generalization of the following two well-known classical cases:1. If X ∈ C is a dualizable object (see Definition 4.17), then every endomorphism f has a well-defined trace [JSV, Proposition 3.1].2. If f : X → X is a nuclear operator (see Definitions 4.25 and 4.28) on a topologicalvector space X , then f has a well-defined trace provided X has the approximationproperty , i.e., the identity operator on X can be approximated by finite rankoperators in the compact open topology [Li].Let TV be the category of topological vector spaces (more precisely, these are assumedto be locally convex, complete, Hausdorff), equipped with the monoidal structure givenby the projective tensor product (see section 4.5). Then an object X ∈ TV is dualizableif and only if X is finite dimensional, whereas every Hilbert space has the approxima-tion property. Hence extending the trace from endomorphisms of dualizable objects of C to more general objects is analogous to extending the notion of trace from endomor-phisms of finite dimensional vector spaces to certain infinite dimensional topologicalvector spaces. In fact, our answer will involve analogues of the notions nuclear and approximation property for general monoidal categories which we now describe.The following notion is our analogue of a nuclear morphism. Definition 1.1.
A morphism f : X → Y in a monoidal category C is thick if it can befactored in the form X ∼ = I ⊗ X t ⊗ id X / / Y ⊗ Z ⊗ X id Y ⊗ b / / Y ⊗ I ∼ = Y (1.2)for morphisms t : I → Y ⊗ Z , b : Z ⊗ X → I .As explained in the next section, the terminology is motivated by considering thebordism category. In the category Vect of vector spaces, with monoidal structure givenby the tensor product, a morphism f : X → Y is thick if and only if it has finite rank(see Theorem 4.1). In the category TV a morphism is thick if and only if it is nuclear(see Theorem 4.27).If f : X → X is a thick endomorphism with a factorization as above, we attemptto define its categorical trace tr( f ) ∈ C ( I, I ) to be the composition I t / / X ⊗ Z s X,Z / / Z ⊗ X b / / I . (1.3)This categorical trace depends on the choice of a natural family of isomorphisms s = { s X,Y : X ⊗ Y → Y ⊗ X } for X, Y ∈ C . We don’t assume that s satisfies the relations(5.8) required for the braiding isomorphism of a braided monoidal category. Apparentlylacking an established name, we will refer to s as switching isomorphisms . We wouldlike to thank Mike Shulman for this suggestion.For the monoidal category Vect , equipped with the standard switching isomorphism s X,Y : X ⊗ Y → Y ⊗ X , x ⊗ y y ⊗ x , the categorical trace of a finite rank (i.e. thick) ndomorphism f : X → X agrees with its classical trace (see Theorem 4.1). Moregenerally, if X is a dualizable object of a monoidal category C , the above definitionagrees with the classical definition of the trace in that situation (Theorem 4.22). Ingeneral, the above trace is not well-defined , since it might depend on the factorizationof f given by the triple ( Z, b, t ) rather than just the morphism f . As we will see inSection 4.4, this happens for example in the category of Banach spaces.To understand the problem with defining tr( f ), let us write b tr( Z, t, b ) ∈ C ( I, I ) forthe composition (1.3), and Ψ(
Z, t, b ) ∈ C ( X, Y ) for the composition (1.2). There isan equivalence relation on these triples (see Definition 3.3) such that b tr( Z, t, b ) andΨ(
Z, t, b ) depend only on the equivalence class [
Z, t, b ]. In other words, there are well-defined maps b tr : b C ( X, X ) −→ C ( I, I ) Ψ : b C ( X, Y ) −→ C ( X, Y )where b C ( X, Y ) denotes the equivalence classes of triples (
Z, t, b ) for fixed
X, Y ∈ C .We note that by construction the image of Ψ consists of the thick morphisms from X to Y . We will call elements of b C ( X, Y ) thickened morphisms . If b f ∈ b C ( X, Y ) withΨ( b f ) = f ∈ C ( X, Y ), we say that b f is a thickener of f .Using the notation C tk ( X, Y ) for the set of thick morphisms from X to Y , it is clearthat there is a well-defined trace map tr : C tk ( X, X ) → C ( I, I ) making the diagram C tk ( X, X ) tr / / __________ C ( I, I ) b C ( X, X ) Ψ f f f f LLLLLLLLLL b tr : : tttttttttt (1.4)commutative if and only if X has the following property: Definition 1.5.
An object X in a monoidal category C with switching isomorphismshas the trace property if the map b tr is constant on the fibers of Ψ.For the category Ban of Banach spaces and continuous maps, we will show inSection 4.4 that the map Ψ can be identified with the homomorphismΦ : Y ⊗ X ′ −→ Ban ( X, Y ) w ⊗ f ( v wf ( v )) (1.6)where X ′ is the Banach space of continuous linear maps f : X → C equipped withthe operator norm and ⊗ is the projective tensor product. Operators in the imageof Φ are referred to as nuclear operators , and hence a morphism in Ban is thick ifand only if it is nuclear. It is a classical result that the trace property for a Banachspace X is equivalent to the injectivity of the map Φ which in turn is equivalent to the approximation property for X : the identity operator of X can be approximated by finiterank operators in the compact-open topology, see e.g. [Ko, § without the approximation property was found byEnflo only in 1973 [En]. Building on Enflo’s work, Szankowski showed in 1981 that theBanach space of bounded operators on an (infinite dimensional) Hilbert space does nothave the approximation property [Sz]. heorem 1.7. Let C be a monoidal category with switching isomorphisms, i.e. C comesequipped with a family of natural isomorphisms s X,Y : X ⊗ Y → Y ⊗ X . If X ∈ C is anobject with the trace property, then the above categorical trace tr( f ) ∈ C ( I, I ) is well-defined for any thick endomorphism f : X → X . This compares to the two classicalsituations mentioned above as follows:(i) If X is a dualizable object, then X has the trace property and any endomorphism f of X is thick. Moreover, the categorical trace of f agrees with its classical trace.(ii) In the category TV of topological vector spaces (locally convex, complete, Haus-dorff ), a morphism is thick if and only if it is nuclear, and the approximationproperty of an object X ∈ TV implies the trace property. Moreover, if f : X → X is a nuclear endomorphism of an object with the approximation property, then thecategorical trace of f agrees with its classical trace. The first part sums up our discussion above. Statements (i) and (ii) appear asTheorem 4.22 respectively Theorem 4.27 below. It would be interesting to find anobject in TV which has the trace property but not the approximation property.To motivate our second main result, Theorem 1.10, we note that a monoidal functor F : C → D preserves thick and thickened morphisms and gives commutative diagramsfor the map Ψ from (1.4). If F is compatible with the switching isomorphisms thenit also commutes with b tr. However, the trace property is not functorial in the sensethat if some object X ∈ C has the trace property then it is not necessarily inherited by F ( X ) (unless F is essentially surjective and full or has some other special property). Inparticular, when the functor F is a field theory, then, as explained in the next section,this non-functoriality causes a problem for calculating the partition function of F .We circumvent this problem by replacing the trace by a closely related trace pairing tr : C tk ( X, Y ) × C tk ( Y, X ) −→ C ( I, I ) (1.8)for objects X , Y of a monoidal category C with switching isomorphisms. Unlike thetrace map tr : C tk ( X, X ) −→ C ( I, I ) discussed above, which is only defined if X has thetrace property, no condition on X or Y is needed to define this trace pairing tr( f, g ) asfollows. Let b f ∈ b C ( X, Y ), b g ∈ b C ( Y, X ) be thickeners of f respectively g (i.e., Ψ( b f ) = f and Ψ( b g ) = g ). We will show that elements of b C ( X, Y ) can be pre-composed or post-composed with ordinary morphisms in C (see Lemma 3.9). This composition giveselements b f ◦ g and f ◦ b g in b C ( Y, Y ) which we will show to be equal in Lemma 3.11.Hence the trace pairing defined bytr( f, g ) := b tr( b f ◦ g ) = b tr( f ◦ b g ) ∈ C ( I, I )is independent of the choice of b f and b g . We note that Ψ( b f ◦ g ) = Ψ( f ◦ b g ) = f ◦ g ∈ C tk ( Y, Y ) and hence if Y has the trace property, thentr( f, g ) = tr( f ◦ g ) for f ∈ C tk ( X, Y ), g ∈ C tk ( Y, X ) . (1.9)In other words, the trace pairing tr( f, g ) is a generalization of the categorical trace of f ◦ g , defined in situations where this trace might not be well-defined.The trace pairing has the following properties that are analogous to properties oneexpects to hold for a trace. We note that the relationship (1.9) immediately impliesthese properties for our trace defined for thick endomorphism of objects satisfying thetrace property. heorem 1.10. Let C be a monoidal category with switching isomorphisms. Then thetrace pairing (1.8) is functorial and has the following properties:1. tr( f, g ) = tr( g, f ) for thick morphisms f ∈ C tk ( X, Y ) , g ∈ C tk ( Y, X ) . If Y hasthe trace property then tr( f, g ) = tr( f ◦ g ) and symmetrically for X .2. If C is an additive category with distributive monoidal structure (see Definition5.3), then the trace pairing is a bilinear map.3. tr( f ⊗ f , g ⊗ g ) = tr( f , g ) tr( f , g ) for f i ∈ C tk ( X i , Y i ) , g i ∈ C tk ( Y i , X i ) , pro-vided s gives C the structure of a symmetric monoidal category. More generally,this property holds if C is a balanced monoidal category . We recall that a balanced monoidal category is a braided monoidal category equippedwith a natural family of isomorphisms θ = { θ X : X → X } called twists satisfying a com-patibility condition (see Definition 5.12). Symmetric monoidal categories are balancedmonoidal categories with θ ≡ id. For a balanced monoidal category C with braidingisomorphism c X,Y : X ⊗ Y → Y ⊗ X and twist θ X : X → X , one defines the switchingisomorphism s X,Y : X ⊗ Y → Y ⊗ X by s X,Y := (id Y ⊗ θ X ) ◦ c X,Y .There are interesting examples of balanced monoidal categories that are not sym-metric monoidal, e.g., categories of bimodules over a fixed von Neumann algebra(monoidal structure given by Connes fusion) or categories of modules over quantumgroups. Traces in the latter are used to produce polynomial invariants for knots. Orig-inally, we only proved the multiplicative property of our trace pairing for symmetricmonoidal categories. We are grateful to Gregor Masbaum for pointing out to us theclassical definition of the trace of an endomorphism of a dualizable object in a balancedmonoidal category which involves using the twist (see [JSV]).The rest of this paper is organized as follows. In section 2 we explain the moti-vating example: we consider the d -dimensional Riemannian bordism category, explainwhat a thick morphism in that category is, and show that the partition function ofa 2-dimensional Riemannian field theory can be expressed as the relative trace of thethick operators that a field theory associates to annuli. Section 2 is motivational andcan be skipped by a reader who wants to see the precise definition of b C ( X, Y ), the con-struction of b tr and a statement of the properties of b tr which are presented in section3. In section 4 we discuss thick morphisms and their traces in various categories. Insection 5 we prove the properties of b tr and deduce the corresponding properties of thetrace pairing stated as Theorem 1.10 above.Both authors were partially supported by NSF grants. They would like to thankthe referee for many valuable suggestions. The first author visited the second author atthe Max-Planck-Institut in Bonn during the Fall of 2009 and in July 2010. He wouldlike to thank the institute for the support and for the stimulating atmosphere. A well-known aximatization of field theory is due to Graeme Segal [Se] who definesa field theory as a monoidal functor from a bordism category to the category TV oftopological vector spaces. The precise definition of the bordism category depends on thetype of field theory considered; for a d -dimensional topological field theory, the objectsare closed ( d − d -dimensional bordisms(more precisely, equivalence classes of bordisms where we identify bordisms if they are iffeomorphic relative boundary). Composition is given by gluing of bordisms, and themonoidal structure is given by disjoint union.For other types of field theories, the manifolds constituting the objects and mor-phisms in the bordism category come equipped with an appropriate geometric struc-ture; e.g., a conformal structure for conformal field theories, a Riemannian metric for
Riemannian field theories, or a
Euclidean structure (= Riemannian metric with van-ishing curvature tensor) for a
Euclidean field theory. In these cases more care is neededin the definition of the bordism category to ensure the existence of a well-defined com-position and the existence of identity morphisms.Let us consider the Riemannian bordism category d - RBord . The objects of d - RBord are closed Riemannian ( d − X to Y is a d -dimensionalRiemannian bordism Σ from X to Y , that is, a Riemannian d -manifold Σ with bound-ary and an isometry X ∐ Y → ∂ Σ. More precisely, a morphism is an equivalence classof Riemannian bordisms, where two bordisms Σ, Σ ′ are considered equivalent if thereis an isometry Σ → Σ ′ compatible with the boundary identifications. In order to havea well-defined compositon by gluing Riemannian bordisms we require that all metricsare product metrics near the boundary. To ensure the existence of identity morphisms,we enlarge the set of morphisms from X to Y by also including all isometries X → Y .Pre- or post-composition of a bordism with an isometry is the given bordism withboundary identification modified by the isometry. In particular, the identity isometry Y → Y provides the identity morphism for Y as object of the Riemannian bordismcategory d - RBord .A more sophisticated way to deal with the issues addressed above was developed inour paper [ST2]. There we don’t require the metrics on the bordisms to be a productmetric near the boundary; rather, we have more sophisticated objects consisting ofa closed ( d − d − E be d -dimensional Riemannian field theory, that is, a symmetric monoidalfunctor E : d - RBord −→ TV . For the bordism category d - RBord the symmetric monoidal structure is given by disjointunion; for the category TV it is given by the projective tensor product. Let X be aclosed Riemannian ( d − X to itself.Let Σ gl be the closed Riemannian manifold obtained by gluing the two boundary pieces(via the identity on X ). Both Σ and Σ gl are morphisms in d - RBord :Σ : X −→ X Σ gl : ∅ −→ ∅ We note that ∅ is the monoidal unit in d - RBord , and hence the vector space E ( ∅ ) can beidentified with C , the monoidal unit in TV . In particular, E (Σ gl ) ∈ Hom( E ( ∅ ) , E ( ∅ )) =Hom( C , C ) = C is a complex number. Question.
How can we calculate E (Σ gl ) ∈ C in terms of the operator E (Σ) : E ( X ) → E ( X )? e would like to say that E (Σ gl ) is the trace of the operator E (Σ), but to do sowe need to check that the conditions guaranteeing a well-defined trace are met. Fora topological field theory E this is easy: In the topological bordism category everyobject X is dualizable (see Definition 4.17), hence E ( X ) is dualizable in TV whichis equivalent to dim E ( X ) < ∞ . By contrast, for a Euclidean field theory the vectorspace E ( X ) is typically infinite dimensional, and hence to make sense of the trace ofthe operator E (Σ) associated to a bordism Σ from X to itself, we need to check thatthe operator E (Σ) is thick and that the vector space E ( X ) has the trace property.It is easy to prove (see Theorem 4.38) that every object X of the bordism category d - RBord has the trace property and that among the morphisms of d - RBord (consistingof Riemannian bordisms and isometries), exactly the bordisms are thick. The lattercharacterization motivated the adjective ‘thick’, since we think of isometries as ‘in-finitely thin’ Riemannian bordisms. It is straightforward to check that being thick isa functorial property in the sense that the thickness of Σ implies that E (Σ) is thick.Unfortunately, as already mentioned in the introduction, the trace property is notfunctorial and we cannot conclude that E ( X ) has the trace property.Replacing the problematical trace by the well-behaved trace pairing leads to thefollowing result. It is applied in [ST1] to prove the modularity and integrality of thepartition function of a supersymmetric Euclidean field theory of dimension 2. Theorem 2.1.
Suppose Σ is a Riemannian bordism of dimension d from X to Y ,and Σ is a Riemannian bordism from Y to X . Let Σ = Σ ◦ Σ be the bordism from Y to itself obtained by composing the bordisms Σ and Σ , and let Σ gl be the closedRiemannian d -manifold obtained from Σ by identifying the two copies of Y that makeup its boundary. If E is d -dimensional Riemannian field theory, then E (Σ gl ) = tr( E (Σ ) , E (Σ )) . Proof.
By Theorem 4.38 the bordisms Σ : X → Y and Σ : Y → X are thick mor-phisms in d - RBord , and hence the morphism tr(Σ , Σ ) : ∅ → ∅ is defined. Moreover,every object X ∈ d - RBord has the trace property (see Theorem 4.38) and hencetr(Σ , Σ ) = tr(Σ ◦ Σ ) = tr(Σ)In part (3) of Theorem 4.38 we will show that tr(Σ) = Σ gl . Then functoriality of theconstruction of the trace pairing implies E (Σ gl ) = E (tr(Σ , Σ )) = tr( E (Σ ) , E (Σ )) In this section we will define the thickened morphisms b C ( X, Y ) and the trace b tr( b f ) ∈ C ( I, I ) of thickened endomorphisms b f ∈ b C ( X, X ) for a monoidal category C equippedwith a natural family of isomorphisms s X,Y : X ⊗ Y → Y ⊗ X .We recall that a monoidal category is a category C equipped with a functor ⊗ : C × C → C alled the tensor product , a distinguished element I ∈ C and natural isomorphisms α X,Y,Z : ( X ⊗ Y ) ⊗ Z ∼ = −→ X ⊗ ( Y ⊗ Z ) (associator) ℓ X : I ⊗ X ∼ = −→ X (left unit constraint) r X : X ⊗ I ∼ = −→ X (right unit constraint)for objects X, Y, Z ∈ C . These natural isomorphisms are required to make two diagrams(known as the associativity pentagon and the triangle for unit ) commutative, see [McL].It is common to use diagrams to represent morphisms in C (see for example [JS1]).The pictures U V WX Yf X X ′ Y Y ′ g ⊗ g ′ = X X ′ Y Y ′ g g ′ represent a morphism f : U ⊗ V ⊗ W → X ⊗ Y and the tensor product of morphisms g : X → Y and g ′ : X ′ → Y ′ , respectively. The composition h ◦ g of morphisms g : X → Y and h : Y → Z is represented by the picture XZh ◦ g = XYZ gh
With tensor products being represented by juxtaposition of pictures, the isomor-phisms X ∼ = X ⊗ I ∼ = I ⊗ X suggest to delete edges labeled by the monoidal unit I from our picture. E.g., the pictures Y Z Z Xt b (3.1)represent morphisms t : I → Y ⊗ Z and b : Z ⊗ X → I , respectively. RephrasingDefinition 1.1 of the introduction in our pictorial notation, a morphism f : X → Y in C is thick if is can be factored in the form Y Z XXY = t bf (3.2)Here t stands for top and b for bottom . We will use the notation C tk ( X, Y ) ⊂ C ( X, Y )for the subset of thick morphisms. .1 Thickened morphisms It will be convenient for us to characterize the thick morphisms as the image of a mapΨ : b C ( X, Y ) −→ C ( X, Y ) , the domain of which we refer to as thickened morphisms . Definition 3.3.
Given objects
X, Y ∈ C , a thickened morphism from X to Y is anequivalence class of triples ( Z, t, b ) consisting of an object Z ∈ C , and morphisms t : I −→ Y ⊗ Z b : Z ⊗ X −→ I To describe the equivalence relation, it is useful to think of these triples as objectsof a category, and to define a morphism from (
Z, t, b ) to ( Z ′ , t ′ , b ′ ) to be a morphism g ∈ C ( Z, Z ′ ) such that t ′ = ZY Z ′ t g and b = ZZ ′ Xb ′ g (3.4)Two triples ( Z, t, b ) and ( Z ′ , t ′ , b ′ ) are equivalent if there is are triples ( Z i , t i , b i ) for i = 1 , . . . , n with ( Z, t, b ) = ( Z , t , b ) and ( Z ′ , t ′ , b ′ ) = ( Z n , t n , b n ) and morphisms g i between ( Z i , t i , b i ) and ( Z i +1 , t i +1 , b i +1 ) (this means that g i is either a morphism from( Z i , t i , b i ) to ( Z i +1 , t i +1 , b i +1 ) or from ( Z i +1 , t i +1 , b i +1 ) to ( Z i , t i , b i )). In other words,a thickened morphism is a path component of the category defined above. We write b C ( X, Y ) for the thickened morphisms from X to Y .As suggested by the referee, it will be useful to regard b C ( X, Y ) as a coend ; this willstreamline the proofs of some results. Let us consider the functor S : C op × C −→ Set given by ( Z ′ , Z ) C ( I, Y ⊗ Z ) × C ( Z ′ ⊗ X, I )Then the elements of ` Z ∈ C S ( Z, Z ) are triples (
Z, t, b ) with t ∈ C ( I, Y ⊗ Z ) and b ∈ C ( Z ⊗ X, I ). Any morphism g : Z → Z ′ induces maps C ( Z ′ , Z ) S ( g, id Z ) / / C ( Z, Z ) and C ( Z ′ , Z ) S (id Z ′ ,g ) / / C ( Z ′ , Z ′ )We note that for any ( t, b ′ ) ∈ S ( Z ′ , Z ) the two triples( Z, S ( g, id Z )( t, b ′ )) = ( Z, t, b ′ ◦ g ) and ( Z ′ , S (id Z ′ , g )( t, b ′ )) = ( Z ′ , g ◦ t, b ′ )represent the same element in b C ( X, Y ). In fact, by construction b C ( X, Y ) is the co-equalizer coequalizer a g ∈ C ( Z,Z ′ ) S ( Z ′ , Z ) ` S ( g, id Z ) / / ` S (id Z ′ ,g ) / / a Z ∈ C S ( Z, Z ) i.e., the quotient space of ` Z ∈ C S ( Z, Z ) obtained by identifying all image points ofthese two maps. his coequalizer can be formed for any functor S : C op × C → Set ; it is called the coend of S , and following [McL, Ch. IX, § Z Z ∈ C S ( Z, Z )for the coend. Summarizing our discussion, we have the following way of expressing b C ( X, Y ) as a coend: b C ( X, Y ) = Z Z ∈ C C ( I, Y ⊗ Z ) × C ( Z ⊗ X, I ) (3.5)
Lemma 3.6.
Given a triple ( Z, t, b ) as above, let Ψ( Z, t, b ) ∈ C ( X, Y ) be the composi-tion on the right hand side of equation (3.2) . Then Ψ only depends on the equivalenceclass [ Z, t, b ] of ( Z, t, b ) , i.e. the following map is well-defined: Ψ : b C ( X, Y ) −→ C ( X, Y ) [
Z, t, b ] Ψ( Z, t, b ) =
Y X ∨ Xt ev (3.7)We note that by construction the image of Ψ is equal to C tk ( X, Y ), the set of thickmorphisms from X to Y . As mentioned in the introduction, for f ∈ C ( X, Y ) we callany b f = [ Z, t, b ] ∈ b C ( X, Y ) with Ψ( b f ) = f a thickener of f . Remark 3.8.
The difference between an orientable versus an oriented manifold isthat the former is a property, whereas the latter is an additional structure on themanifold. In a similar vain, being thick is a property of a morphism f ∈ C ( X, Y ),whereas a thickener b f is an additional structure. To make the analogy between thesesituations perfect, we were tempted to introduce the words thickenable or thickable intomathematical English. However, we finally decided against it, particularly because thethick-thin distinction for morphisms in the bordism category is just perfectly suitedfor the purpose. Proof.
Suppose that g : Z → Z ′ is an equivalence from ( Z, t, b ) to ( Z ′ , t ′ , b ′ ) in the senseof Definition 3.3. Then the following diagram shows that Ψ( Z ′ , t ′ , b ′ ) = Ψ( Z, t, b ). Z ′ XY t ′ b ′ = ZZ ′ XY t b ′ g = Z XY t b hickened morphisms can be pre-composed or post-composed with ordinary mor-phisms to obtain again thickened morphisms as follows: b C ( Y, W ) × C ( X, Y ) ◦ / / b C ( X, W ) ([
Z, t, b ] , f ) [ Z, t, b ◦ (id Z ⊗ f )] C ( Y, W ) × b C ( X, Y ) ◦ / / b C ( X, W ) ( f, [ Z, t, b ]) [ Z, ( f ⊗ id Z ) ◦ t, b ]The proof of the following result is straightforward and we leave it to the reader. Lemma 3.9.
The composition of morphisms with thickened morphisms is well-definedand compatible with the usual composition via the map Ψ in the sense that the followingdiagrams commutes: b C ( Y, W ) × C ( X, Y ) Ψ × id (cid:15) (cid:15) ◦ / / b C ( X, W ) Ψ (cid:15) (cid:15) C ( Y, W ) × C ( X, Y ) ◦ / / C ( X, W ) C ( Y, W ) × b C ( X, Y ) id × Ψ (cid:15) (cid:15) ◦ / / b C ( X, W ) Ψ (cid:15) (cid:15) C ( Y, W ) × C ( X, Y ) ◦ / / C ( X, W ) Corollary 3.10. If f : X → Y is a thick morphism in a monoidal category C , thenthe compositions f ◦ g and h ◦ f are thick for any morphisms g : W → X , h : Y → Z . Lemma 3.11.
Let b f ∈ b C ( X, Y ) , b f ∈ b C ( U, X ) and f i = Ψ( b f i ) . Then b f ◦ f = f ◦ b f ∈ b C ( U, Y ) .Proof. Let b f i = [ Z i , t i , b i ]. Then b f ◦ f = [ Z , t , b ◦ (id Z ⊗ f )] and Z UXb f Z X Z Ub t b Z Z Ug b b ◦ (id Z ⊗ f ) = = =where Z X Z b t g =Similarly, f ◦ b f = [ Z , ( f ⊗ id Z ) ◦ t , b ], and Z XY t f Y Z X Z t b t Z Z Y gt ( f ⊗ id Z ) ◦ t = = =This shows that g is an equivalence from ( Z , t , b ◦ (id Z ⊗ f )) to ( Z , ( f ⊗ id Z ) ◦ t , b )and hence these triples represent the same element of b C ( U, Y ) as claimed. .2 The trace of a thickened morphism Our next goal is to show that for a monoidal category C with switching isomorphism s X,Z : X ⊗ Z → Z ⊗ X the map b tr : b C ( X, X ) −→ C ( I, I ) [
Z, t, b ] b tr( Z, t, b ) (3.12)is well-defined. We recall from equation 1.3 of the introduction that b tr( Z, t, b ) is definedby b tr( Z, t, b ) = b ◦ s X,Z ◦ t . In our pictorial notation, we write it as b tr( Z, t, b ) = tbX ZZ X where
X XZZ is shorthand for
XX ZZs
X,Z (3.13)We note that the naturality of s X,Y is expressed pictorially as X X Y Y g h = X X Y Y gh (3.14)for morphisms g : X → X , h : Y → Y . Lemma 3.15. b tr( Z, t, b ) depends only on the equivalence class of ( Z, t, b ) . In particular,the map (3.12) is well-defined.Proof. Suppose that g : Z → Z ′ is an equivalence from ( Z ′ , t ′ , b ′ ) to ( Z, t, b ) (see Defi-nition 3.3). Then b tr( Z ′ , t ′ , b ′ ) = = t ′ b ′ X Z ′ = tb ′ gX ZZ ′ = tb ′ gX ZZ ′ = b tr( Z, t, b ) tbX Z In terms of the coend description, this lemma is actually obvious because b tr is thecomposition b C ( X, X ) = Z Z ∈ C C ( I, X ⊗ Z ) × C ( Z ⊗ X, I ) → Z Z ∈ C C ( I, Z ⊗ X ) × C ( Z ⊗ X, I ) → C ( I, I )where the first map is given by switching and the second by composition. Traces in various categories
The goal in this section is to describe thick morphisms in various categories in classicalterms and relate our categorical trace to classical notions of trace. At a more technicallevel, we describe the map Ψ : b C ( X, Y ) → C ( X, Y ) and its image C tk ( X, Y ) in thesecategories.In the first subsection this is done for the category of vector spaces (not necessarilyfinite dimensional). In the second subsection we show that Ψ is a bijection if X isdualizable and hence any endomorphism of a dualizable object has a well-defined trace.Traces in categories for which all objects are dualizable are well-studied [JS2]. In thethird subsection we introduce semi-dualizable objects and describe the map Ψ in moreexplicit terms. In a closed monoidal category every object is semi-dualizable. In thefollowing subsection we apply these considerations to the category of Banach spaces.Then we discuss the category of topological vector spaces, and finally the Riemannianbordism category. Theorem 4.1.
Let C be the monoidal category Vect F of vector spaces (not necessarilyfinite dimensional) over a field F equipped with the usual tensor product and the usualswitching isomorphism s X,Y : X ⊗ Y → Y ⊗ X , x ⊗ y y ⊗ x .1. A morphism f : X → Y is thick if and only if it has finite rank.2. The map Ψ : b C ( X, Y ) → C ( X, Y ) is injective; in particular, every finite rankendomorphism f : X → X has a well-defined categorical trace tr( f ) ∈ C ( I, I ) ∼ = F .3. For a finite rank endomorphism f , its categorical trace tr( f ) agrees with its usual(classical) trace which we denote by cl-tr( f ) . For a vector space X , let X ∗ be the dual vector space, and define the morphismΦ : C ( F , Y ⊗ X ∗ ) −→ C ( X, Y ) (4.2)by sending t : F → Y ⊗ X ∗ to the composition X ∼ = F ⊗ X t ⊗ id X −→ Y ⊗ X ∗ ⊗ X id Y ⊗ ev −→ Y ⊗ F ∼ = Y. We note that the set C ( F , Y ⊗ X ∗ ) can be identified with Y ⊗ X ∗ via the evaluationmap t t (1). Using this identification, Φ maps an elementary tensor y ⊗ g ∈ Y ⊗ X ∗ to the linear map x yg ( x ) which is a rank ≤ y ⊗ g ) is defined to be g ( y ); by linearity this extends to a well-defined trace for all finite rank operators.The proof of Theorem 4.1 is based on the following lemma that shows that Ψ isequivalent to the map Φ. Lemma 4.3.
For any vector spaces X , Y the map α : C ( F , Y ⊗ X ∗ ) −→ b C ( X, Y ) t [ X ∗ , t, ev] (4.4) is a bijection and Ψ ◦ α = Φ . Here ev : X ∗ ⊗ X → F is the evaluation map g ⊗ x g ( x ) . n subsection 4.2, we will construct the map Φ and prove this lemma in the moregeneral context where X is a semi-dualizable object of a monoidal category. Proof of Theorem 4.1.
Statement (1) follows immediately from the Lemma. To provepart (2), it suffices to show that Φ is injective which is well known and elementary.For the proof of part (3) let f : X → X be a thick morphism. We recall that itscategorical trace tr( f ) is defined by tr( f ) = b tr( b f ) for a thickener b f ∈ b C ( X, X ). SoΨ( b f ) = f and this is well-defined thanks to the injectivity of Ψ. Using that α is abijection, there is a unique t ∈ C ( F , X ⊗ X ∗ ) with α ( t ) = [ X ∗ , t, ev] = b f . We note that b tr( α ( t )) = b tr([ X ∗ , t, ev]) ∈ C ( F , F ) = F is the composition F t −→ X ⊗ X ∗ s X,X ∗ −→ X ∗ ⊗ X ev −→ F In particular, if t (1) = y ⊗ g ∈ X ⊗ X ∗ , then1 y ⊗ g g ⊗ y g ( y )and hence tr( f ) = g ( y ) = cl-tr( f ) is the classical trace of the rank one operator Φ( t ) = f given by x yg ( x ). Since both the categorical trace of α ( t ) and the classical trace ofΦ( t ) depend linearly on t , this finishes the proof. Remark 4.5.
We can replace the monoidal category of vector spaces by the monoidalcategory
SVect of super vector spaces. The objects of this category are just Z / ⊗ . The grading involution on X ⊗ Y is thetensor product ǫ X ⊗ ǫ Y of the grading involutions on X respectively Y . The switchingisomorphism X ⊗ Y ∼ = Y ⊗ X is given by x ⊗ y ( − | x || y | y ⊗ x for homogeneouselements x ∈ X , y ∈ Y of degree | x | , | y | ∈ Z /
2. Then the statements above andtheir proof work for
SVect as well, except that the categorical trace of a finite rankendomorphism T : X → X is its super trace str( T ) := cl-tr( ǫ X T ). The goal of this subsection is to generalize Lemma 4.11 from the category of vec-tor spaces to general monoidal categories C , provided the object X ∈ C satisfies thefollowing condition. Definition 4.6.
An object X of a monoidal category C is semi-dualizable if the functor C op → Set , Z C ( Z ⊗ X, I ) is representable, i.e., if there is an object X ∨ ∈ C andnatural bijections C ( Z, X ∨ ) ∼ = C ( Z ⊗ X, I ) . (4.7)By Yoneda’s Lemma, the object X ∨ is unique up to isomorphism. It is usually referredto as the (left) internal hom and denoted by C ( X, I ).To put this definition in context, we recall that a monoidal category C is closed iffor any X ∈ C the functor C → C , Z Z ⊗ X has a right adjoint; i.e., if there is afunctor C ( X, ) : C −→ C and natural bijections C ( Z, C ( X, Y )) ∼ = C ( Z ⊗ X, Y ) Z, X, Y ∈ C (4.8) n particular, in a closed monoidal category every object X is semi-dualizable with X ∨ = C ( X, I ). The category C = Vect F is an example of a closed monoidal category; forvector spaces X , Y the internal hom C ( X, Y ) is the vector space of linear maps from X to Y . Hence every vector space is semi-dualizable with semi-dual X ∨ = C ( X, I ) = X ∗ .Other examples of closed monoidal categories is the category of Banach spaces (seesubsection 4.4) and the category of bornological vector spaces [Me]. As also discussed in[Me, p. 9], the symmetric monoidal category TV of topological vector spaces with theprojective tensor product is not an example of a closed monoidal category (no matterwhich topology on the space of continuous linear maps is used). Still, some topologicalvector spaces are semi-dualizable (e.g., Banach spaces), and so it seems preferable tostate our results for semi-dualizable objects rather than objects in closed monoidalcategories.In Lemma 4.11 we will generalize a statement about the category Vect F to a state-ment about a general monoidal category C . To do so, we need to construct the maps Φand α in the context of a general monoidal category. This is straightforward by usingthe same definitions as above, just making the following replacements:1. Replace F , the monoidal unit in Vect F , by the monoidal unit I ∈ C .2. Replace X ∗ , the vector space dual to X , by X ∨ , the semi-dual of X ∈ C . Here weneed to assume that X ∈ C is semi-dualizable which is automatic for any objectof a closed category like Vect F .3. For a semi-dualizable object X ∈ C , the evaluation map ev : X ∨ ⊗ X −→ I (4.9)is by definition the morphism that corresponds to the identity morphism X ∨ → X ∨ under the bijection (4.7) for Z = X ∨ . It is easy to check that this agrees withthe usual evaluation map X ∗ ⊗ X → F for C = Vect F .The map Φ : C ( I, Y ⊗ X ∨ ) −→ C ( X, Y ) is given by the picture
Y X ∨ t Y X ∨ Xt ev (4.10)The following Lemma is a generalization of Lemma 4.3. Lemma 4.11.
Let X be a semi-dualizable object of a monoidal category C . Then forany object Y ∈ C the map α : C ( I, Y ⊗ X ∨ ) −→ b C ( X, Y ) t [ X ∨ , t, ev] (4.12) is a natural bijection which makes the diagram C ( I, Y ⊗ X ∨ ) Φ ' ' NNNNNNNNNNN α ∼ = / / b C ( X, Y ) Ψ y y ssssssssss C ( X, Y ) ommutative. In particular, a morphism f : X → Y is thick if and only if it is in theimage of the map Φ (see Equation (4.10) ).Proof. The commutativity of the diagram is clear by comparing the definitions of themaps Φ (see Equation (4.10)), α (Equation (4.12)), and Ψ (Equation (3.7)).To see that α is a bijection we factor it in the following form C ( I, Y ⊗ X ∨ ) ∼ = −→ Z Z ∈ C C ( I, Y ⊗ Z ) × C ( Z, X ∨ ) ∼ = −→ Z Z ∈ C C ( I, Y ⊗ Z ) × C ( Z ⊗ X, I ) (4.13)Here the first map sends t ∈ C ( I, Y ⊗ X ∨ ) to [ X ∨ , t, id X ∨ ]. This map is a bijection bythe coend form of the Yoneda Lemma according to which for any functor F : C → Set the map F ( W ) −→ Z Z ∈ C F ( Z ) × C ( Z, W ) t [ W, t, id W ] (4.14)is a bijection (see [Ke, Equation 3.72]. The second map of Equation (4.13) is inducedby the bijection C ( Z, X ∨ ) ∼ = C ( Z ⊗ X, I ). By construction the identity map id X ∨ corresponds to the evaluation map via this bijection, and hence the composition ofthese two maps is α . Remark 4.15.
The referee observed that there is a neat interpretation of b C ( X, Y ) interms of the Yoneda embedding C −→ F := Fun ( C op , Set ) X C ( − , X )The monoidal structure ⊗ on C induces a monoidal structure ∗ on F , the convolutiontensor product [Day], defined by( M ∗ N )( S ) := Z V,W C ( S, V ⊗ W ) × M ( V ) × N ( W )for an object S ∈ C . Equipped with the convolution product, the functor category F is a closed monoidal category (see Equation (4.8)) with internal left hom F ( N, L ) ∈ F given by F ( N, L )( S ) = F ( N ( − ) , L ( S ⊗ − ))We can regard C as a full monoidal subcategory of F via the Yoneda embedding. Inparticular, every object X ∈ C has a left semi-dual X ∨ = F ( X, − ) ∈ F and hence by theprevious lemma we have a bijection b F ( X, Y ) ∼ = F ( I, Y ∗ X ∨ ). Explicitly, the semi-dual X ∨ is given by X ∨ ( S ) = F ( C ( − , X ) , C ( S ⊗ − , I )) = C ( S ⊗ X, I )The referee observed that the Yoneda embedding induces a bijection b C ( X, Y ) −→ b F ( X, Y ) ∼ = F ( I, Y ∗ X ∨ ) . (4.16)In particular, morphism f ∈ C ( X, Y ) is thick if and only if its image under the Yonedaembedding is thick. To see that the above map is a bijection, we evaluate the right and side F ( I, Y ∗ X ∨ ) = ( Y ∗ X ∨ )( I ) = Z V,W C ( I, V ⊗ W ) × C ( V, Y ) × X ∨ ( W )= Z W C ( I, Y ⊗ W ) × X ∨ ( W ) = Z W C ( I, Y ⊗ W ) × C ( W ⊗ X, I )which we recognize as the coend description of b C ( X, Y ) (Equation (3.5)). The secondequality is a consequence of (the coend form of) the Yoneda lemma (equation (4.14)).
As mentioned in the introduction, there is a well-known trace for endomophisms ofdualizable objects in a monoidal category (see e.g. [JSV, § Definition 4.17. [JS2, Def. 7.1] An object X of a monoidal category C is (left) du-alizable if there is an object X ∨ ∈ C (called the (left) dual of X ) and morphismsev : X ∨ ⊗ X → I (called evaluation map ) and coev : I → X ⊗ X ∨ ( coevaluation map )such that the following equations hold. X X ∨ X coev ev = XX id X X ∨ XX ∨ coevev = X ∨ X ∨ id X ∨ (4.18)If X is dualizable with dual X ∨ , then there is a family of bijections C ( Z, Y ⊗ X ∨ ) ∼ = −→ C ( Z ⊗ X, Y ) , (4.19)natural in Y, Z ∈ C , given by Y X ∨ Zf Y X ∨ XZf ev with inverse
Y X ∨ XZ g coev ←− [ Y XZ g
In particular, a dualizable object is semi-dualizable in the sense of Definition 4.6.
Example 4.20.
A finite dimensional vector space X is a dualizable object in thecategory Vect F : we take X ∨ to be the vector space dual to X , ev to be the usualevaluation map, and definecoev : F −→ X ⊗ X ∨ by 1 X i e i ⊗ e i here { e i } is a basis of X and { e i } is the dual basis of X ∨ . It is not hard to show thata vector space X is dualizable if and only if it is finite dimensional. Definition 4.21.
Let C be a monoidal category with switching isomorphisms. Let f : X → X be an endomorphism of a dualizable object X ∈ C . Then the classical trace cl-tr( f ) ∈ C ( I, I ) is defined bycl-tr( f ) := coevevfX X ∨ X ∨ X This definition can be found for example in section 3 of [JSV] for balanced monoidalcategories (see Definition 5.12 for the definition of a balanced monoidal category andhow the switching isomorphism is determined by the braiding and the twist of thebalanced monoidal category). In fact, the construction in [JSV] is more general: theyassociate to a morphism f : A ⊗ X → B ⊗ X a trace in C ( A, B ) if X is dualizable;specializing to A = B = I gives the classical trace described above. Theorem 4.22.
Let C be a monoidal category with switching isomorphisms, and let X be a dualizable object of C . Then1. The map Ψ : b C ( X, Y ) → C ( X, Y ) is a bijection. In particular, any morphismwith domain X is thick, and any endomorphism f : X → X has a well-definedcategorical trace tr( f ) ∈ C ( I, I ) .2. The categorical trace of f is equal to its classical trace cl-tr( f ) defined above. Remark 4.23.
Part (1) of the Theorem implies in particular that if X is (left) dual-izable, then the identity id X is thick. The referee observed that the converse holds aswell. To see this, assume that id X is thick. Then id X is in the image of Ψ : b C ( X, X ) → C ( X, X ) and hence by Lemma 4.11 in the image of Φ : C ( I, X ⊗ X ∨ ) → C ( X, X ). Ifcoev ∈ C ( I, X ⊗ X ∨ ) belongs to the preimage Φ − (id X ), then it is straightforward tocheck that Equations (4.10) hold and hence X is dualizable. The first equation holdsby construction of Φ; to check the second equation, we apply the bijection (4.7) (for Z = X ∨ ) to both sides and obtain ev for both. Proof.
To prove part (1), it suffices by Lemma 4.11 to show that the map Φ : C ( I, Y ⊗ X ∨ ) → C ( X, Y ) (see Equation (4.10)) is a bijection. Comparing Φ with the naturalbijection (4.19) for dualizable objects X ∈ C , we see that this bijection is equal to Φin the special case Z = I .To prove part (2) we recall that by definition tr( f ) = b tr( b f ) ∈ C ( I, I ) for any b f ∈ Ψ − ( f ) ⊂ b C ( X, X ) (see Equation (1.4)). In the situation at hand, Ψ is invertibleby part (1), and using the factorization Ψ = Φ ◦ α − provided by Lemma 4.11, we have b f = Ψ − ( f ) = α Φ − ( f ) = α (( f ⊗ id X ∨ ) ◦ coev) = [ X ∨ , ( f ⊗ id X ∨ ) ◦ coev , ev]Here the second equality follows from the explicit form of the inverse of Φ (whichagrees with the map (4.19) for Z = I ). Comparing b tr( b f ) (Equation (3.13)) and cl-tr( f ) Definition 4.21), we seetr( f ) = b tr( X ∨ , ( f ⊗ id X ∨ ) ◦ coev , ev) = cl-tr( f ) Ban of Banach spaces
Let X be a Banach space and f : X → X a continuous linear map. There are classicalconditions ( f is nuclear and X has the approximation property , see Definition 4.25)which guarantee that f has a well-defined (classical) trace which we again denote bycl-tr( f ) ∈ C . For example any Hilbert space H has the approximation property and acontinuous linear map f : H → H is nuclear if and only if it is trace class. The mainresult of this subsection is Theorem 4.26 which shows that these classical conditionsimply that f has a well-defined categorical trace and that the categorical trace of f agrees with its classical trace. Before stating this result, we review the (projective)tensor product of Banach spaces and define the notions nuclear and approximationproperty .The category Ban of Banach spaces is a monoidal category whose monoidal structureis given by the projective tensor product , defined as follows. For Banach spaces X , Y ,the projective norm on the algebraic tensor product X ⊗ alg Y is given by || z || := inf { X || x i || · || y i || | z = X x i ⊗ y i ∈ X ⊗ Y } where the infimum is taken over all ways of expressing z ∈ X ⊗ alg Y as a finite sumof elementary tensors. Then the projective tensor product X ⊗ Y is defined to be thecompletion of X ⊗ alg Y with respect to the projective norm.It is well-known that Ban is a closed monoidal category (see Equation (4.8)).For Banach spaces X , Y the internal hom space Ban ( X, Y ) is the Banach spaceof continuous linear maps T : X → Y equipped with the operator norm || T || :=sup x ∈ X, || x || = 1 || T ( x ) || . In particular, every Banach space has a left semi-dual X ∨ = Ban ( X, I ) in the sense of Definition 4.17 which is just the Banach space of con-tinuous linear maps X → C . The categorically defined evaluation map ev : X ∨ ⊗ X → C (see Equation(4.9)) agrees with the usual evaluation map defined by f ⊗ x b f ( x ).The map Y ⊗ X ∨ ∼ = Ban ( C , Y ⊗ X ∨ ) Φ −→ Ban ( X, Y )(see Equation (4.10)) is determined by sending y ∈ f ∈ Y ⊗ X ∨ to the map x yf ( x )(see the discussion after Theorem 4.1).We note that the morphism set Ban ( X ⊗ Y, Z ) is in bijective correspondence to thecontinuous bilinear maps X × Y → Z . This bijection is given by sending g : X ⊗ Y → Z to the composition g ◦ χ , where χ : X × Y → X ⊗ Y is given by ( x, y ) x ⊗ y . Inparticular, if X ∨ is the Banach space dual to X , we have a morphismev : X ∨ ⊗ X −→ C determined by g ⊗ x g ( x ) (4.24)called the evaluation map . Definition 4.25. [Sch, Ch. III, §
7] A continuous linear map between Banach spacesis nuclear if f is in the image of the mapΦ : Y ⊗ X ∨ −→ Ban ( X, Y ) determined by y ⊗ g ( x yg ( x )) Banach space X has the approximation property if the identity of X can be approx-imated by finite rank operators with respect to the compact-open topology. Theorem 4.26.
Let
X, Y ∈ Ban .1. A morphism f : X → Y is thick if and only if it is nuclear.2. If X has the approximation property, it has the trace property.3. If X has the approximation property and f : X → X is nuclear, then the categor-ical trace of f agrees with its classical trace.Proof. This result holds more generally in the category TV which we will prove inthe following subsection (see Theorem 4.27). For the proofs of statements (2) and (3)we refer to that section. For the proof of statement (1) we recall that a morphism f : X → Y in the category C = Ban is thick if and only if if is in the image of the mapΨ : b C ( X, Y ) → C ( X, Y ) and that it is nuclear if it is in the image of Φ : Y ⊗ X ∨ = C ( I, Y ⊗ X ∨ ) → C ( X, Y ). Hence part (1) is a corollary of Lemma 4.11 according towhich these two maps are equivalent for any semi-dualizable object X of a monoidalcategory C . In particular, since Ban is a closed monoidal category, every Banach spaceis semi-dualizable.
In this subsection we extend Theorem 4.26 from Banach spaces to the category TV whose objects are locally convex topological vector spaces which are Hausdorff andcomplete. We recall that the topology on a vector space X is required to be invariantunder translations and dilations. In particular, it determines a uniform structure on X which in turn allows us to speak of Cauchy nets and hence completeness; see [Sch,section I.1] for details. The morphisms of TV are continuous linear maps, and the pro-jective tensor product described below gives TV the structure of a symmetric monoidalcategory. It contains the category Ban of Banach spaces as a full subcategory.
Theorem 4.27.
Let
X, Y be objects in the category TV .1. A morphism f : X → Y is thick if and only if it is nuclear.2. If X has the approximation property, then it has the trace property.3. If X has the approximation property, and f : X → X is nuclear, then the cate-gorical trace of f agrees with its classical trace. Before proving this theorem, we define nuclear morphisms in TV and the approx-imation property. Then we’ll recall the classical trace of a nuclear endomorphism ofa topological vector space with the approximation property as well as the projectivetensor product. Definition 4.28.
A continuous linear map f ∈ TV ( X, Y ) is nuclear if it factors in theform X p −→ X f −→ Y j −→ Y, (4.29)where f is a nuclear map between Banach spaces (see Definition 4.25) and p , j arecontinuous linear maps. he definition of nuclearity in Schaefer’s book ([Sch, p. 98]) is phrased differently.We give his more technical definition at the end of this section and show that a con-tinuous linear map is nuclear in his sense if and only if it is nuclear in the sense of theabove definition. Approximation property.
An object X ∈ TV has the approximation property if theidentity of X is in the closure of the subspace of finite rank operators with respect tothe compact-open topology [Sch, Chapter III, section 9] (our completeness assumptionfor topological vector spaces implies that uniform convergence on compact subsets isthe same as uniform convergence on pre-compact subsets). The classical trace for nuclear endomorphisms.
Let f be a nuclear endomor-phism of X ∈ TV , and let I ν : X → X be a net of finite rank morphisms which convergesto the identity on X in the compact-open topology. Then f ◦ I ν is a finite rank operatorwhich has a classical trace cl-tr( f ◦ I ν ). It can be proved that the limit lim ν cl-tr( f ◦ I ν )exists [Li, Proof of Theorem 1] and is independent of the choice of the net I ν (this alsofollows from our proof of Theorem 4.27). The classical trace of f is defined bycl-tr( f ) := lim ν cl-tr( f ◦ I ν ) . Projective tensor product.
The projective tensor product of Banach spaces definedin the previous section extends to topological vector spaces as follows. For
X, Y ∈ TV the projective topology on the algebraic tensor product X ⊗ alg Y is the finest locallyconvex topology such that the canonical bilinear map χ : X × Y −→ X ⊗ alg Y ( x, y ) x ⊗ y is continuous [Sch, p. 93]. The projective tensor product X ⊗ Y is the topological vectorspace obtained as the completion of X ⊗ alg Y with respect to the projective topology. Itcan be shown that it is locally convex and Hausdorff and that the morphisms TV ( X ⊗ Y, Z ) are in bijective correspondence to continuous bilinear maps X × Y → Z ; thisbijection is given by sending f : X ⊗ Y → Z to f ◦ χ [Sch, Chapter III, section 6.2]. Semi-norms.
For checking the convergence of a sequence or continuity of a mapbetween locally convex topological vector spaces, it is convenient to work with semi-norms. For X ∈ TV and U ⊂ X a convex circled 0-neighborhood ( U is circled if λU ⊂ U for every λ ∈ C with | λ | ≤
1) one gets a semi-norm || x || U := inf { λ ∈ R > | x ∈ λU } (4.30)on X . Conversely, a collection of semi-norms determines a topology, namely the coars-est locally convex topology such that the given semi-norms are continuous maps. Forexample, if X is a Banach space with norm || || , we obtain the usual topology on X .As another example, the projective topology on the algebraic tensor product X ⊗ alg Y isthe topology determined by the family of semi-norms || || U,V parametrized by convexcircled 0-neighborhoods U ⊂ X , V ⊂ Y defined by || z || U,V := inf { n X i =1 || x i || U || y i || V | z = n X i =1 x i ⊗ y i } , where the infimum is taken over all ways of writing z ∈ X ⊗ alg Y as a finite sum ofelementary tensors. It follows from this description that the projective tensor product efined above is compatible with the projective tensor product of Banach spaces definedearlier (see [Sch, Chapter III, section 6.3]).Our next goal is the proof of Theorem 4.27, for which we will use the followinglemma. Lemma 4.31.
1. Any morphism t : C → Y ⊗ Z in TV factors in the form Y Zt = Y Zt jY where Y is a Banach space.2. Any morphism b : Z ⊗ X → C factors in the form Z Xb = Z Xb pX where X is a Banach space. For the proof of this lemma, we will need the following two ways to constructBanach spaces from a topological vector space X :1. Let U be a convex, circled neighborhood of 0 ∈ X . Let X U be the Banach spaceobtained from X by quotiening out the null space of the semi-norm || || U and bycompleting the resulting normed vector space. Let p U : X → X U be the evidentmap.2. Let B be a convex, circled bounded subset of X . We recall that B is bounded iffor each neighborhood U of 0 ∈ X there is some λ ∈ C such that B ⊂ λU . Let X B be the vector space X B := S ∞ n =1 nB equipped with the norm || x || B := inf { λ ∈ R > | x ∈ λB } . If B is closed in X , then X B is complete (by our assumptionthat X is complete), and hence X B is a Banach space ([Sch, Ch. III, §
7; p. 97]).The inclusion map j B : X B → X is continuous thanks to the assumption that B is bounded. Proof of Lemma 4.31.
To prove part (1) we use the fact (see e.g. Theorem 6.4 in Chap-ter III of [Sch]) that any element of the completed projective tensor product Y ⊗ Z , inparticular the element t (1), can be written in the form t (1) = ∞ X i =1 λ i y i ⊗ z i with y i → ∈ Y, z i → ∈ Z, X | λ i | < ∞ (4.32)Let B ′ := { y i | i = 1 , . . . } ∪ { } , and let B be the closure of the convex, circled hull of B ′ (the convex circled hull of B ′ is the intersection of all convex circled subsets of W ontaining B ′ ). We note that B ′ is bounded, hence its convex, circled hull is bounded,and hence B is bounded.We define j : Y → Y to be the map j B : Y B → Y . To finish the proof of part (1), itsuffices to show that t (1) is in the image of the inclusion map j B ⊗ id Z : Y B ⊗ Z ֒ → Y ⊗ Z .It is clear that each partial sum P ni =1 λ i y i ⊗ z i belongs to the algebraic tensor product Y B ⊗ alg Z , and hence we need to show that the sequence of partial sums is a Cauchysequence with respect to the semi-norms || || B,V on Y B ⊗ alg Z that define the projectivetopology (here V runs through all convex circled 0-neighborhoods V ⊂ Z ). Since y i ∈ B , it follows || y i || B ≤ || n X i =1 λ i y i ⊗ z i || B,V ≤ n X i =1 | λ i | || y i || B || z i || V ≤ n X i =1 | λ i | || z i || V Since z i →
0, we have || z i || V ≤ i , and this implies that thepartial sums form a Cauchy sequence.To prove part (2) we recall that the morphism b : Z ⊗ X → C corresponds to acontinuous bilinear map b ′ : Z × X → C . The continuity of b ′ implies that there areconvex, circled 0-neighborhoods V ⊂ Z , U ⊂ X such that | b ′ ( z, x ) | < z ∈ V , x ∈ U . It follows that | b ′ ( z, x ) | ≤ || z || V || x || U for all z ∈ Z , x ∈ X . Hence b extends toa continuous bilinear map e b ′ : Z × X U −→ C for the completion X U of X . This corresponds to the desired morphism b : Z ⊗ X U → C ; the property b = b ◦ (id Z ⊗ p U ) is clear by construction. Defining the map p : X → X to be p U : X → X U , we obtained the desired factorization of b . Remark 4.33.
The proofs above and below imply that for fixed objects
X, Y ∈ TV ,the thickened morphisms c TV ( X, Y ) actually form a set . Any triple (
Z, t, b ) can befactored into Banach spaces X , Y , Z as explained in the 3 pictures below. By theargument above, we may actually choose X = X U where U runs over certain subsetsof X . Since X is fixed, it follows that the arising Banach spaces X U range over acertain set. Finally, by Lemma 4.11, the Banach space Z may be replaced by X ∨ U without changing the equivalence class of the triple. Therefore, the given triple ( Z, t, b )is equivalent to a triple of the form ( X ∨ U , t ′ , b ′ ). Since the collection of objects X ∨ U ∈ TV forms a set, we see that c TV ( X, Y ) is a set as well.In this paper, we have not addressed the issue whether b C ( X, Y ) is a set because thisproblem does not arise in the examples we discuss: The argument above for C = TV is the hardest one, in all other examples we actually identify b C ( X, Y ) with some veryconcrete set.This problem is similar to the fact that presheaves on a given category do notalways form a category (because natural transformations do not always form a set).So we are following the tradition of treating this problem only if forced to.
Proof of Theorem 4.27.
The factorization (4.29) shows that a nuclear morphism f : X → Y factors through a nuclear map f : X → Y of Banach spaces. Then f is thick byTheorem 4.26 and hence f is thick, since pre- or post-composition of a thick morphismwith an any morphism is thick. To prove the converse, assume that f is thick, i.e.,that it can be factored in the form f = (id Y ⊗ b )( t ⊗ id X ) t : I → Y ⊗ Z b : Z ⊗ X → I. hen using Lemma 4.31 to factorize t and b , we see that f can be further factored inthe form f = Y Xt b j pY X Z This implies that f has the desired factorization X p −→ X f −→ Y j −→ Y , where f = Y X t b Z It remains to show that f is a nuclear map between Banach spaces. In the categoryof Banach spaces a morphisms is nuclear if and only if it is thick by Theorem 4.26. Atfirst glance, it seems that the factorization of f above shows that f is thick. However,on second thought one realizes that we need to replace Z by a Banach space to makethat argument. This can be done by using again our Lemma 4.31 to factorize t andhence f further in the form f = Y X t ′ b j ′ Z Z This shows that f is a thick morphism in the category Ban and hence nuclear byTheorem 4.26. The key for the proof of parts (2) and (3) of Theorem 4.27 will be thefollowing lemma.
Lemma 4.34.
For any b f ∈ c TV ( Y, X ) the map TV ( X, Y ) −→ TV ( I, I ) = C g b tr( b f ◦ g ) (4.35) is continuous with respect to the compact open topology on TV ( X, Y ) . o prove part (2) of Theorem 4.27, assume that X has the approximation propertyand let I ν : X → X be a net of finite rank operators converging to the identity of X inthe compact open topology. Then by the lemma, for any b f ∈ c TV ( X, X ), the net b tr( f ◦ b I ν ) = b tr( b f ◦ I ν ) converges to b tr( b f ◦ id X ) = b tr( b f ) . Here b I ν ∈ c TV ( X, X ) are thickeners of I ν . They exist since every finite rank morphismsis nuclear and hence thick by part (1). This implies the trace condition for X , since b tr( b f ) = lim ν b tr( f ◦ b I ν ) depends only on f .To prove part (3) let f ∈ C ( X, X ) be a nuclear endomorphism of X ∈ TV and let I ν be a net of finite rank operators converging to the identity of X in the compactopen topology. By part (1) f is thick, i.e., there is a thickener b f ∈ C ( X, X ). Then asdiscussed above, we havecl-tr( f ) = lim ν cl-tr( f ◦ I ν ) and tr( f ) = b tr( b f ) = lim ν b tr( f ◦ b I ν ) = lim ν tr( f ◦ I ν )For the finite rank operator f ◦ I ν its classical trace cl-tr( f ◦ I ν ) and its categoricaltrace tr( f ◦ I ν ) agree by part (3) of Theorem 4.1. Proof of Lemma 4.34.
Let b f = [ Z, t, b ] ∈ c TV ( Y, X ). Then b f ◦ g = [ Z, t, b ◦ (id Z ⊗ g )]and hence b tr( b f ◦ g ) is given by the composition C t −→ X ⊗ Z s X,Z −→ Z ⊗ X id Z ⊗ g −→ Z ⊗ Y b −→ C (4.36)As in Equation (4.32) we write t (1) ∈ X ⊗ Z in the form t (1) = ∞ X i =1 λ i x i ⊗ z i with x i → ∈ X, z i → ∈ Z, ∞ X i =1 | λ i | < ∞ This implies b tr( b f ◦ g ) = ∞ X i =1 λ i b ′ ( z i , g ( x i )) , where b ′ : Z × Y → C is the continuous bilinear map corresponding to b : Z ⊗ Y → C .To show that the map g b tr( b f ◦ g ) is continuous, we will construct for given ǫ >
0a compact subset K ⊂ X and an open subset U ⊂ Y such that for g ∈ U K,U := { g ∈ TV ( X, Y ) | g ( K ) ⊂ U } ⊂ TV ( X, Y )we have | b tr( b f ◦ g ) | < ǫ . To construct U , we note that the continuity of b ′ implies thatthere are 0-neighborhoods U ⊂ Y , V ⊂ Z such that y ∈ U , z ∈ V implies | b ′ ( z, y ) | < ǫ .Without loss of generality we can assume z i ∈ V for all i (by replacing z i by cz i and x i by x i /c for a sufficiently small number c ) and P | λ i | = 1 (by replacing λ i by λ i /s and x i by sx i for s = P | λ i | ). We define K := { x i | i ∈ N } ∪ { } ⊂ X . Then for g ∈ U K,U we have | b tr( b f ◦ g ) | = | X λ i b ′ ( z i , g ( x i ) | ≤ X | λ i || b ′ ( z i , g ( x i ) | < ( X | λ i | ) ǫ = ǫ. inally we compare our definition of a nuclear map between topological vectorspaces (see Definition 4.25) with the more classical definition which can be found e.g.in [Sch, p. 98]. A continuous linear map f ∈ TV ( X, Y ) is nuclear in the classical sense if there is a convex circled 0-neighborhood U ⊂ X , a closed, convex, circled, boundedsubset B ⊂ Y such that f ( U ) ⊂ B and the induced map of Banach spaces X U → Y B is nuclear. Lemma 4.37.
A morphism in TV is nuclear if and only if it is nuclear in the classicalsense.Proof. If f : X → Y is nuclear in the classical sense, it factors in the form X p U −→ X U f −→ Y B j B −→ Y, where f is nuclear. Hence f is nuclear.Conversely, let us assume that f is nuclear, i.e., that it factors in the form X p −→ X f −→ Y j −→ Y, where f is a nuclear map between Banach spaces. To show that f is nuclear in theclassical sense, we will construct a convex circled 0-neighborhood U ⊂ X , and a closed,convex, circled, bounded subset B ⊂ Y such that f ( U ) ⊂ B and we have a commutativediagram X p / / p U BBBBBBBB X f / / Y (cid:15) (cid:15) j / / YX U O O f ′ / / Y B j B > > }}}}}}}} Then the induced map f ′ factors through the nuclear map f , hence f ′ is nuclear and f is nuclear in the classical sense.We define U := p − ( ˚ B δ ), where ˚ B δ ⊂ X is the open ball of radius δ aroundthe origin in the Banach space X , and δ > B δ ⊂ f − ( ˚ B ). Thecontinuity of p implies that U is open. Moreover, p ( U ) ⊂ ˚ B δ def = δ ˚ B implies || p ( x ) || < δ for x ∈ U which in turn implies the estimate || p ( x ) || < δ || x || U for all x ∈ X . It followsthat the map p factors through p U .We define B := j ( B ) ⊂ Y , where B is the closed unit ball in Y . This is abounded subset of Y , since for any open subset U ⊂ Y the preimage j − ( U ) is an open0-neighborhood of Y and hence there is some ǫ > B ǫ ⊂ j − ( U ). Then ǫj ( B ) = j ( B ǫ ) is a subset of U . This implies j ( B ) ⊂ ǫ U and hence B is a boundedsubset of Y (it is clear that B is closed, convex and circled). By construction of B wehave the inequality || j ( y ) || B ≤ || y || which implies that j factors through j B . Also byconstruction, f ( U ) is contained in B and hence f induces a morphism f ′ : X U → Y B .It follows that the outer edges of the diagram form a commutative square. The factthat j B is a monomorphism and that the image of p U is dense in X U then imply thatthe middle square of the diagram above is commutative. We recall from section 2 that the objects of the d -dimensional Riemannian bordismcategory d - RBord are closed ( d − X, Y ∈ - RBord , the set d - RBord ( X, Y ) of morphisms from X to Y is the disjoint union ofthe set of isometries from X to Y and the set of Riemannian bordisms from X to Y (modulo isometry relative boundary), except that for X = Y = ∅ , the identity isometryequals the empty bordism. Theorem 4.38.
1. A morphism in d - RBord is thick if and only if it is a bordism.2. For any X, Y ∈ d - RBord the map
Ψ : \ d - RBord ( X, Y ) −→ d - RBord ( X, Y ) is injective. In particular, every object X ∈ d - RBord has the trace property andevery bordism Σ from X to X has a well-defined trace tr(Σ) ∈ d - RBord ( ∅ , ∅ ) .3. If Σ is a Riemannian bordism from X to X , then tr(Σ) = Σ gl , the closed Rie-mannian manifold obtained by gluing the two copies of X in the boundary of Σ .Proof. To prove part (1) suppose that f : X → Y is a thick morphism, i.e., it can befactored in the form X ∼ = ∅ ∐ X t ∐ id X / / Y ∐ Z ∐ X id Y ∐ b / / Y ∐ ∅ ∼ = Y (4.39)We note that the morphisms t : ∅ → Y ∐ Z and b : Z ∐ X → ∅ must both be bordisms (the only case where say t could possibly be an isometry is Y = Z = ∅ ; however thatisometry is the same morphism as the empty bordism). Hence the composition f is abordism.Conversely, assume that Σ is a bordism from X to Y . Then Σ can be decomposedas in the following picture: Z XY bt
Here t = Y × [0 , ǫ ] ⊂ Σ, Z = Y × { ǫ } and b = Σ \ ( Y × [0 , ǫ )) are bordisms, where ǫ > Y ⊂ Σ has a neighborhood isometric to Y × [0 , ǫ ) equippedwith the product metric. Regarding t as a Riemannian bordism from ∅ to Y ∐ Z , andsimilarly b as a Riemannian bordism from Z ∐ X to ∅ , it is clear from the constructionthat the composition (4.39) is Σ.To show that Ψ is injective, let [ Z ′ , t ′ , b ′ ] ∈ \ d - RBord ( X, Y ), and let Σ = Ψ([ Z ′ , t ′ , b ′ ]) ∈ d - RBord be the composition (1.2). In other words, we have a decomposition of the bor-dism Σ into two pieces t ′ , b ′ which intersect along Z ′ . Now let ( Z, t, b ) be the tripleconstructed in the proof of part (1) above. By choosing ǫ small enough, we can assumethat Z = Y × ǫ ⊂ Σ is in the interior of the bordism t ′ , and we obtain a decomposition f the bordism Σ as shown in the picture below. Z XY Z ′ b ′ t g | {z } t ′ b z }| { Regarding g as a bordism from Z to Z ′ we see that t ′ = (id Y ∐ g ) ◦ t and b = b ′ ◦ ( g ∐ id X ) , which implies that the triple ( Z, t, b ) and ( Z ′ , t ′ , b ′ ) are equivalent in the sense of Def-inition 3.3. If [ Z ′′ , t ′′ , b ′′ ] ∈ \ d - RBord is another element with Ψ([ Z ′′ , t ′′ , b ′′ ]) = Σ, thenby choosing ǫ small enough we conclude [ Z ′′ , t ′′ , b ′′ ] = [ Z, t, b ] = [ Z ′ , t ′ , b ′ ].For the proof of part (3), we decompose as in the proof of (1) the bordism Σ intotwo pieces t and b by cutting it along the one-codimensional submanifold Z = X × { ǫ } (here Y = X since Σ is an endomorphism ). We regard t as a bordism from ∅ to X ∐ Z and b as a bordism from Z ∐ X to ∅ . Then tr(Σ) is given by the composition ∅ t / / X ∐ Z s X,Z / / Z ∐ X b / / ∅ . which geometrically means to glue the two bordisms along X and Z . Gluing firstalong Z we obtain the Riemannian manifold Σ, then gluing along X we get the closedRiemannian manifold Σ gl . The goal of this section is the proof of our main theorem 1.10 according to which ourtrace pairing is symmetric, additive and multiplicative. There are three subsectionsdevoted to the proof of these three properties, plus a subsection on braided monoidaland balanced monoidal categories needed for the multiplicative property. Each proofwill be based on first proving the following analogous properties for the trace b tr( b f ) ofthickened endomorphisms b f ∈ b C ( X, X ): Theorem 5.1.
Let C be a monoidal category, equipped with a natural family of iso-morphisms s = s X,Y : X ⊗ Y → Y ⊗ X . Then the trace map b tr : b C ( X, X ) −→ C ( I, I ) has the following properties:1. (symmetry) b tr( b f ◦ g ) = b tr( g ◦ b f ) for b f ∈ b C ( X, Y ) , g ∈ C ( Y, X ) ;2. (additivity) If C is an additive category with distributive monoidal structure (seeDefinition 5.3), then b tr is a linear map; . (multiplicativity) b tr( b f ⊗ b f ) = b tr( b f ) ⊗ b tr( b f ) for b f ∈ b C ( X , X ) , b f ∈ b C ( X , X ) ,provided C is a symmetric monoidal category with braiding s . More generally,this property holds if C is a balanced monoidal category (see Definition 5.12). For the tensor product b f ⊗ b f ∈ b C ( X ⊗ X , X ⊗ X ) of the thickened morphisms b f i see Definition 5.15. Proof of part (1) of Theorem 5.1.
Let b f = [ Z, t, b ] ∈ b C ( X, Y ). Then b f ◦ g = [ Z, t, b ◦ (id Z ⊗ g )] and g ◦ b f = [ Z, ( g ⊗ id Z ) ◦ t, b ]and hence b tr( b f ◦ g ) = tb gXZY = tbg XZY = b tr( g ◦ b f )Here the second equality follows from the naturality of the switching isomorphism (seePicture (3.14)). Proof of part (1) of Theorem 1.10.
Let b g ∈ b C ( Y, X ) with Ψ( b g ) = g . Thentr( f, g ) = b tr( b f ◦ g ) = b tr( g ◦ b f ) = b tr( b g ◦ f ) = tr( f, g ) . Here the second equation is part (1) of Theorem 5.1, while the third is a consequenceof Lemma 3.11.
Throughout this subsection we will assume that the category C is an additive categorywith distributive monoidal structure (see Definitions 5.2 and 5.3 below). Often an addi-tive category is defined as a category enriched over abelian groups with finite products(or equally coproducts). However, the abelian group structure on the morphism setsis actually determined by the underlying category C , and hence a better point of viewis to think of ‘additive’ as a property of a category C , rather than specifying additionaldata. Definition 5.2.
A category C is additive if1. There is a zero object 0 ∈ C (an object which is terminal and initial);2. finite products and coproducts exist; . Given a finite collection of objects, the canonical morphism from their coproductto their product (given by identity maps on coordinates) is an isomorphism.4. Any morphism from X → Y has an additive inverse under the canonical addition(defined below) on C ( X, Y ).We remark that the requirement (1) is redundant since an initial object is a coproductof the empty family of objects, and a terminal object is the product of the same family;the map from the initial to the terminal object is an isomorphism by (3), thus makingthese objects zero objects.If C is an additive category, we will use the notation X ⊕· · ·⊕ X n for the coproduct ofobjects X i ∈ C (the term direct sum is used for finite coproducts in additive categories).Any map f : X = X ⊕ · · · ⊕ X n −→ Y = Y ⊕ · · · ⊕ Y m , amounts to an m × n matrix of morphisms f ij ∈ C ( X j , Y i ), 1 ≤ i ≤ m , 1 ≤ j ≤ n ,by identifying Y via property (3) with the product of the Y i ’s. In particular, there aremorphisms ∆ := (cid:0) idid (cid:1) : X −→ X ⊕ X ∇ := ( id id ) : Y ⊕ Y −→ Y referred to as diagonal map and fold map , respectively. For f, g ∈ C ( X, Y ) their sum f + g is defined to be the composition X ∆ −→ X ⊕ X f ⊕ g −→ Y ⊕ Y ∇ −→ Y This addition gives C ( X, Y ) the structure of an abelian group. Abusing notation wewrite 0 : X → Y for the additive unit which is given by the unique morphism thatfactors through the zero object. Identifying morphisms between direct sums with ma-trices, their composition corresponds to multiplication of the corresponding matrices. Definition 5.3.
A monoidal structure ⊗ on an additive category C is distributive iffor any objects X, Y ∈ C the functors C −→ C C −→ C Z Z ⊗ X Z Y ⊗ Z preserve coproducts. In particular, for objects Z , Z ∈ C we have canonical distribu-tivity isomorphisms( Z ⊗ X ) ⊕ ( Z ⊗ X ) ∼ = ( Z ⊕ Z ) ⊗ X and ( Y ⊗ Z ) ⊕ ( Y ⊗ Z ) ∼ = Y ⊗ ( Z ⊕ Z ) . In addition, these functors send the initial object 0 ∈ C (thought of as the coproductof the empty family of objects of C ) to an initial object of C ; in others words, there arecanonical isomorphisms 0 ⊗ X ∼ = 0 and Y ⊗ ∼ = 0. Definition 5.4.
Given triples ( Z , t , b ) and ( Z , t , b ) representing elements of b C ( X, Y )we define their sum by( Z , t , b ) + ( Z , t , b ) := (cid:0) Z ⊕ Z , (cid:0) t t (cid:1) , ( b b ) (cid:1) Here (cid:0) t t (cid:1) is a morphism from I to ( Y ⊗ Z ) ⊕ ( Y ⊗ Z ) ∼ = Y ⊗ ( Z ⊕ Z ) and ( b b ) isa morphism from ( Z ⊗ X ) ⊕ ( Z ⊗ X ) ∼ = ( Z ⊕ Z ) ⊗ X to I . The addition of triplesinduces a well-defined addition b C ( X, Y ) × b C ( X, Y ) + −→ b C ( X, Y ) ince, if g i : Z i → Z ′ i is an equivalence from ( Z i , t i , b i ) to ( Z ′ i , t ′ i , b ′ i ), then g ⊕ g : Z ⊕ Z → Z ′ ⊕ Z ′ is an equivalence from ( Z , t , b ) + ( Z , t , b ) to ( Z ′ , t ′ , b ′ ) + ( Z ′ , t ′ , b ′ ). Lemma 5.5.
The above addition gives b C ( X, Y ) the structure of an abelian group.Proof. We claim:1. The additive unit is represented by any triple (
Z, t, b ) with t = 0 or b = 0;2. the additive inverse of [ Z, t, b ] is represented by ( Z, − t, b ) or ( Z, t, − b ) (here − t , − b are the additive inverses to the morphisms t respectively b whose existence isguaranteed by axiom (4) in the definition of an additive category).It follows from the description of addition that [0 , ,
0] is an additive unit in b C ( X, Y ).It is easy to check that 0 : Z → Z, t,
0) to (0 , ,
0) and that0 : 0 → Z is an equivalence from (0 , ,
0) to ( Z, , b ), which proves the first claim.The diagonal map ∆ : Z → Z ⊕ Z is an equivalence from ( Z, t,
0) to (
Z, t, b ) +(
Z, t, − b ) = ( Z ⊕ Z, ( tt ) , ( b − b )). Similarly, the fold map ∇ : Z ⊕ Z → Z is an equivalencefrom ( Z, t, b ) + ( Z, − t, b ) = ( Z ⊕ Z, (cid:0) t − t (cid:1) , ( b b )) to ( Z, , b ). This proves the secondclaim. Lemma 5.6.
The map
Ψ : b C ( X, Y ) → C ( X, Y ) is a homomorphism.Proof. Using the notation from the definition above, we want to show that Ψ(
Z, t, b ) =Ψ( Z , t , b ) + Ψ( Z , t , b ). We recall (see Equation (1.2) of the introduction and theparagraph following it) that Ψ( Z, t, b ) ∈ C ( X, Y ) is given by the composition Y ∼ = Y ⊗ I id Y ⊗ b ←− Y ⊗ Z ⊗ X t ⊗ id X ←− I ⊗ X ∼ = X Here we write the arrows from right to left in order that composition corresponds tomatrix multiplication. Identifying Y ⊗ Z ⊗ X with ( Y ⊗ Z ⊗ X ) ⊕ ( Y ⊗ Z ⊗ X ), thesemaps are given by the following matrices:id Y ⊗ b = (cid:0) id Y ⊗ b id Y ⊗ b (cid:1) t ⊗ id X = (cid:18) t ⊗ id X t ⊗ id X (cid:19) Hence Ψ(
Z, t, b ) is given by the matrix product (cid:0) id Y ⊗ b id Y ⊗ b (cid:1) (cid:18) t ⊗ id X t ⊗ id X (cid:19) =(id Y ⊗ b ) ◦ ( t ⊗ id X ) + (id Y ⊗ b ) ◦ ( t ⊗ id X )=Ψ( Z , t , b ) + Ψ( Z , t , b ) Proof of part (2) of Theorem 5.1.
We recall from the definition of b tr (see Equation(3.12)) that b tr( Z, t, b ) ∈ C ( I, I ) is given by the composition I b ←− Z ⊗ X s X,Z ←− X ⊗ Z t ←− I Identifying Z ⊗ X with ( Z ⊗ X ) ⊕ ( Z ⊗ X ), and X ⊗ Z with ( X ⊗ Z ) ⊕ ( X ⊗ Z ),this composition is given by the following matrix product: (cid:0) b b (cid:1) (cid:18) s X,Z s X,Z (cid:19) (cid:18) t t (cid:19) = b ◦ s X,Z ◦ t + b ◦ s X,Z ◦ t = b tr( Z , t , b ) + b tr( Z , t , b ) o show that the additivity of b tr implies that the pairing tr( f, g ) is linear in f and g , we will need the following fact. Lemma 5.7.
The composition map b C ( X, Y ) × b C ( Y, X ) −→ b C ( Y, Y ) is bilinear. The proof of this lemma is analogous to the previous two proofs, and so we leaveit to the reader.
Proof of part (2) of Theorem 1.10.
Let f , f ∈ C ( X, Y ), b f , b f ∈ b C ( X, Y ) with Ψ( b f i ) = f i , and g ∈ C tk ( Y, X ). Then by Lemma 5.6 Ψ( b f + b f ) = f + f , and hencetr( f + f , g ) = b tr(( b f + b f ) ◦ g ) = tr( b f ◦ g + b f ◦ g )= b tr( b f ◦ g ) + b tr( b f ◦ g ) = tr( f , g ) + tr( f , g ) We recall from [JS2, Definition 2.1] that a braided monoidal category is a monoidalcategory equipped with a natural family of isomorphisms c = c X,Y : X ⊗ Y → Y ⊗ X such that XY YZ ZXc
X,Y ⊗ Z = XY YZ ZXc
X,Y c X,Z and
XZ YX ZYc X ⊗ Y,Z = XZ YX ZYc
X,Z c Y,Z (5.8)There is a close relationship between braided monoidal categories and the braid groups B n . We recall that the elements of B n are braids with n strands , consisting of isotopyclasses of n non-intersecting piecewise smooth curves γ i : [0 , → R , i = 1 , . . . , n withendpoints at { , . . . , n } × { } × { , } . One requires that the z -coordinate of each curveis strictly decreasing (so that strands are “going down”). Composition of braids isdefined by concatenation of their strands; e.g., in B we have the composition ◦ = =The last equality follows from the obvious isotopy in R , also known as the second Reidemeister move . The three Reidemeister moves are used to understand isotopies ofarcs in R via their projections to the plane R . Generically, the isotopy projects to anisotopy in the plane but there are three codimension one singularities where this doesnot happen: a cusp, a tangency and a triple point (of immersed arcs in the plane).The corresponding ‘moves’ on the planar projection are the Reidemeister moves; forexample, in the above figure one can see a tangency in the middle of moving the braid n the center to the (trivial) braid on the right. A cusp singularity cannot arise forbraids but a triple point can: the resulting (third Reidemeister move) isotopies are alsoknown as braid relations . A typical example would be the following isotopy (with atriple point in the middle): = (5.9)Thinking of the groups B n as categories with one object, we can form their disjointunion to obtain the braid category B := ` ∞ n =0 B n . The category B can be equippedwith the structure of a braided monoidal category [JS2, Example 2.1]; in fact, B isequivalent to the free braided monoidal category generated by one object (this is aspecial case of Theorem 2.5 of [JS2]).More generally, if C k is the free braided monoidal category generated by k objects,there is a braided monoidal functor F : C k → B which sends n -fold tensor productsof the generating objects to the object n ∈ B (whose automorphism group is B n ); onmorphisms, F sends the structure maps α X,Y,Z , ℓ X and r X (see beginning of Section 3)to the identity, and the braiding isomorphisms c X i ,X j for generating objects X i , X j tothe braid ∈ B This suggests to represent c X,Y by an overcrossing and c − Y,X by an undercrossing in thepictorial representations of morphisms in braided monoidal categories, a conventionbroadly used in the literature [JS1] that we will adopt. In other words, one defines:
YYX X := YXXYc
X,Y and
X XYY := YXXYc − Y,X
The key result, [JS2, Cor. 2.6], is that for any objects
X, Y ∈ C k the map F : C k ( X, Y ) → B ( F ( X ) , F ( Y ))is a bijection. One step in the argument is to realize that the Yang-Baxter relations 5.9follow from the relations 5.8 together with the naturality of the braiding isomorphism c . An immediate consequence is the following statement which we will refer to below. Proposition 5.10. (Joyal-Street)
Let f, g : X ⊗ · · · ⊗ X k → X σ (1) ⊗ · · · ⊗ X σ ( k ) bemorphisms of a braided monoidal category C which are in the image of the tautologicalfunctor T : C k → C which sends the i -th generating object of the free braided monoidalcategory C k to X i . If f = T ( e f ) , g = T ( e g ) and the associated braids F ( e f ) , F ( e g ) ∈ B k agree, then f = g . e note that a morphism f is in the image of T if and only if it can be written as acomposition of the isomorphisms α , ℓ , r , c and their inverses. For such a composition itis easy to read off the braid F ( e f ): in the pictorial representation of f we simply ignoreall associators and units and replace each occurrence of the braiding isomorphism c (respectively its inverse) by an overcrossing (respectively and undercrossing); then F ( b f )is the resulting braid. For example, for f = XZ YY ZXc − X,Y c X,Z c Y,Z F ( e f ) =Another result that we will need for the proof of the multiplicativity property arethe following isotopy relations. Roughly speaking, they say that if the unit I ∈ C isinvolved, then the isotopy does not need to be ‘relative boundary’ as in the previouspictures. Lemma 5.11.
Let V , W be objects of a braided monoidal category C , and let f : V → I and g : I → V be morphisms in C . Then there are the following relations: WWV f = WWVf = V fWWWV Vg = VVWg = V VW g
Proof.
WWV f = WV W f Ir W = WV Wf Ir W = WV WfI ℓ W = WWVf ere the first and last equality come from interpreting compositions involving the box f with no output (= morphism with range I ). The second equality is the naturalityof the braiding isomorphism, and the third equality is a compatibility between theunit constraints r W , ℓ W and the braiding isomorphism which is a consequence of therelations (5.8) [JS2, Prop. 2.1].This proves the first equality; the proofs of the other equalities are analogous. Definition 5.12. [JS2, Def. 6.1] A balanced monoidal category is a braided monoidalcategory C together with a natural family of isomorphisms θ X : X ∼ = → X , called twists ,parametrized by the objects X ∈ C . We note that the naturality means that for anymorphism g : X → Y we have g ◦ θ X = θ Y ◦ g . In addition it is required that θ X = id X for X = I , and for any objects X, Y ∈ C the twist θ X ⊗ Y is determined by the followingequation: XX YYθ X ⊗ Y = YXX Yθ Y θ X (5.13)If θ X = id X for all objects X ∈ C , this equality reduces to the requirement c Y,X ◦ c X,Y = id X ⊗ Y for the braiding isomorphism. This is the additional requirement in a symmetric monoidal category , so symmetric monoidal categories can be thought of asbalanced monoidal categories with θ X = id X for all objects X .For a balanced monoidal category C with braiding isomorphism c and twist θ , wedefine the switching isomorphism s = s X,Y : X ⊗ Y → Y ⊗ X by s X,Y := (id Y ⊗ θ X ) ◦ c X,Y ; pictorially:
X XYY := YYX Xθ X (5.14) In this subsection we will prove the multiplicativity of b tr and derive from it the multi-plicativity of the trace pairing (part (3) of Theorem 1.10). Our first task is to define atensor product of thickened morphisms. Definition 5.15.
Let b f i = [ Z i , t i , b i ] ∈ b C ( X i , Y i ) for i = 1 ,
2. If C is a braided monoidalcategory, we define b f ⊗ b f := [ Z, t, b ] ∈ b C ( X ⊗ X , Y ⊗ Y ) here Z = Z ⊗ Z , and t = t t Y Z Y Z b = Z X Z X b b We will leave it to the reader to check that this tensor product is well-defined. Acrucial property of this tensor product for thickened morphisms is its compatibilitywith the tensor product for morphisms:
Lemma 5.16.
The map
Ψ : b C ( X, Y ) → C ( X, Y ) is multiplicative, i.e. , Ψ( b f ⊗ b f ) =Ψ( b f ) ⊗ Ψ( b f ) .Proof. Ψ( b f ⊗ b f ) = Ψ([ Z, t, b ]) = t bZY X = t t b b Y Y X X = Y Y X X t t b b = Y Y X X t t b b == Y Y X X t t b b = Ψ([ Z , t , b ]) ⊗ Ψ([ Z , t , b ]) = Ψ( b f ) ⊗ Ψ( b f )Here the fourth equality is a consequence of Lemma 5.11 (applied to g = t : I → W = Y ⊗ Z , V = X ). The fifth equality follows from Proposition 5.10 and theobvious isotopy. The sixth equality is again a consequence of Lemma 5.11 (applied to f = b : W = Z ⊗ X → I , V = Y ). roof of part (3) of Theorem 5.1. Let b f i = [ Z i , t i , b i ] ∈ b C ( X i , X i ), i = 1 ,
2. Then b tr( b f ⊗ b f ) = b tr( Z, t, b ) = tc X,Z bθ X X ZZ XX (5.17)where Z := Z ⊗ Z , X := X ⊗ X , and t , b are as in Definition 5.15. We note that θ X = θ X ⊗ X can be expressed in terms of θ := θ X and θ := θ X as in Equation 5.13and that the braiding isomorphism c X,Z can be expressed graphically as c X,Z = c X ⊗ X ,Z ⊗ Z = Z Z X X It follows that b tr( b f ⊗ b f ) = θ θ t t b b X Z X Z Z X Z X = θ θ t t b b = θ θ t t b b = θ t t b b = θ θ t t b b = θ θ t t b b = b tr( b f ) ⊗ b tr( b f )Here the first equality is obtained from equation (5.17) by combining the pictures forthe morphisms t , c X,Z , θ X and b that b tr( b f ⊗ b f ) is a composition of. The second equalityfollows from the naturality of the braiding isomorphism (see picture (3.14)). The thirdequality is a consequence of Proposition 5.10 and an isotopy moving the strand withlabel Z to the right. The fourth equality is a consequence of an isotopy moving thestrand with label X left and down. The fifth equality follows from Lemma 5.11 with g = t and the relations (5.8). The sixth equality is a consequence of Proposition 5.10and an isotopy which moves the strands labeled X and Z to the left, and the strandlabeled X to the right. Proof of part (3) of Theorem 1.10.
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