Lecture 15: Duality. Next we spell out the answer to Exercise It is part of the definition of a TQFT.

Transcription

1 Lecture 15: Duality We ended the last lecture by introducing one of the main characters in the remainder of the course, a topological quantum field theory (TQFT). At this point we should, of course, elaborate on the definition and give examples, background, motivation, etc. I will not do so in these notes. Instead I refer you to the expository paper [F1] as well as to the beginning sections of [L1]. There are many other references with great expository material. In this lecture we explore the finiteness property satisfied by a TQFT, which is encoded via duality in symmetric monoidal categories. Some categorical preliminaries We begin with a standard notion which you ll find in any book which contains a chapter on categories, including books on category theory. Definition Let C,D be categories. A functor F: C D is an equivalence if there exist a functor G: D C, and natural isomorphisms G F id C and F G id D. Proposition A functor F : C D is an equivalence if and only if it satisfies: (i) For each d D there exist c C and an isomorphism ( f(c) d ) D; and (ii) For each c 1,c 2 C the map of hom-sets F: C(c 1,c 2 ) D ( F(c 1 ),F(c 2 ) ) is a bijection. If F satisfies (i) it is said to be essentially surjective and if it satisfies (ii) it is fully faithful. Exercise Prove Proposition Next we spell out the answer to Exercise It is part of the definition of a TQFT. Definition Let C,D be symmetric monoidal categories and F,G: C D symmetric monoidal functors. Then a symmetric monoidal natural transformation η: F G is a natural transformation such that the diagrams (15.5) F(1 C ) 1 D G(1C) η(1 C ) and (15.6) F(y 1 ) F(y 2 ) ψ η η F(y 1 y 2 ) η G(y 1 ) G(y 2 ) ψ G(y 1 y 2 ) commute for all y 1,y 2 C. Bordism: Old and New (M392C, Fall 12), Dan Freed, October 27,

2 2 D. S. Freed TQFT s as a symmetric monoidal category Fix a bordism category B = Bord X(n) and a symmetric monoidal category C. We now explain that topological quantum field theories F: B C are objects in a symmetric monoidal category. A morphism F G is as defined in Definition The monoidal product of theories F 1,F 2 is defined by (15.7) (F 1 F 2 )(Y) = F 1 (Y) F 2 (Y) (F 1 F 2 )(X) = F 1 (X) F 2 (X) for all objects Y B and morphisms (X: Y 0 Y 1 ) B. The tensor unit 1 is the trivial theory (15.8) 1(Y) = 1 C 1(X) = id 1C for all Y B and (X: Y 0 Y 1 ) B. We denote the symmetric monoidal category of TQFT s as (15.9) TQFT n = TQFTn X(n) [C] = Hom (Bord X(n),C). The short form of the notation is used if the tangential structure X(n) and codomain category C are clear. (15.10) Endomorphisms of the trivial theory. Suppose η: 1 1 in TQFT n. Then for all Y Bord X(n) we have η(y) C(1 C,1 C ) = End(1 C ). Note that if C = Ab is the category of abelian groups, then End(1 C ) = Z. So η is a numerical invariant of closed (n 1)-manifolds. Furthermore, if X: Y 0 Y 1 then by the naturality condition (13.18) we find that η(y 0 ) = η(y 1 ). This shows that η factors down to a homomorphism of monoids (15.11) η: Ω X(n) n 1 End(1 C). Now by Lemma 1.30, and its generalization to manifolds with tangential structure, we know that every element of Ω X(n) n 1 is invertible. It follows that the image of η consists of invertible elements. We say, simply, that η is invertible. In other words, an endomorphism of 1 is a bordism invariant of the type studied in the first half of the course. A topological quantum field theory, then, is a categorified bordism invariant. Exercise What is a topological quantum field theory whose codomain category has as objects the set of integers and only identity arrows? The invertibility observed in (15.10) is quite general. Theorem A morphism (η: F G) TQFT n is invertible. TQFT n is a groupoid. The two statements are equivalent. We prove Theorem at the end of this lecture.

3 Bordism: Old and New (Lecture 15) 3 (15.14) Central problem. Given a dimension n, a tangential structure X(n), and a codomain category C we can ask to compute the groupoid TQFTn X(n) [C]. This is a vague problem whose solution is an equivalent groupoid which is simpler than the groupoid of topological quantum field theories. It has a nice answer when n = 1. In the oriented case it is a generalization of the theorem that Ω SO 0 is the free abelian group with a single generator pt +. There is also a nice answer in the oriented case for n = 2. Finiteness in TQFT To motivate the abstract formulation of finiteness in symmetric monoidal categories, we prove the following simple proposition. For simplicity we omit any tangential structure. Proposition Let F: Bord Vect C be a TQFT. Then for all Y Bord the vector space F(Y) is finite dimensional.! Figure 26. Some elementary bordisms Figure 27. The S-diagram Proof. Fix Y Bord and let V = F(Y). Let c: n 1 Y Y and e: Y Y n 1 be the bordisms pictured in Figure 26. The manifold Y is depicted as a point, and each bordism has underlying manifold with boundary [0,1] Y. The composition depicted in Figure 27 is diffeomorphic to the identity bordism id Y : Y Y. Under F it maps to id V : V V (see (13.40)). On the other hand, the composition maps to (15.16) V id V F(c) V V V F(e) id V V Let the value of F(c): C V V on 1 C be v i v i for some finite set of vectors v i,v i V. i Then equating (15.16) with the identity map we find that for all ξ V we have (15.17) ξ = i e(ξ,v i)v i, and so the finite set of vectors {v i } spans V. This proves that V is finite dimensional. Exercise Prove that F(c) and F(e) are inverse bilinear forms.

4 4 D. S. Freed Duality data and dual morphisms We abstract the previous argument by singling out those objects in a symmetric monoidal category which obey a finiteness condition analogous to that of a finite dimensional vector space. Definition Let C be a symmetric monoidal category and y C. (i) Duality data for y is a triple of data (y,c,e) in which y is an object of C and c,e are morphisms c: 1 C y y, e: y y 1 C. We require that the compositions (15.20) y c idy y y idy e y y and (15.21) y id y c y y y e id y y be identity maps. If duality data exists for y, we say that y is dualizable. (ii) A morphism of duality data (y,c,e) (ỹ, c,ẽ) is a morphism y f ỹ such that the diagrams (15.22) y y c 1 C id y f c y ỹ and (15.23) y y e f id y 1 C ỹ y ẽ commute. c is called coevaluation and e is called evaluation. We now express the uniqueness of duality data. As duality data is an object in a category, as defined in Definition 15.19, we cannot say there is a unique object. Rather, here we have the strongest form of uniqueness possible in a category: duality data is unique up to unique isomorphism. Definition Let C be a category. (i) Ifforeach pairy 0,y 1 C thehom-setc(y 0,y 1 ) iseitherempty orcontains auniqueelement, we say that C is a discrete groupoid. (ii) If for each pair y 0,y 1 C the hom-set C(y 0,y 1 ) has a unique element, we say that C is contractible.

5 Bordism: Old and New (Lecture 15) 5 A discrete groupoid is equivalent to a set (Example 13.10). A contractible groupoid is equivalent to a category with one object and one morphism, the categorical analog of a point. Proposition Let C be a symmetric monoidal category and y C. Then the category of duality data for y is either empty or is contractible. The proof is a homework problem (Problem Set #2). A morphism between dualizable objects has a dual. Definition Let y 0,y 1 C be dualizable objects in a symmetric monoidal category and f: y 0 y 1 a morphism. The dual morphism f : y1 y 0 is the composition (15.27) y 1 id y 1 c 0 y 1 y 0 y 0 id y 1 f id y 0 y 1 y 1 y 0 e 1 id y 0 y 0 In the definition we choose duality data (y 0,c 0,e 0 ), (y 1,c 1,e 1 ) for y 0,y 1. Exercise Check that this definition agrees with that of a dual linear map for C = Vect. Also, spell out the consequence of Proposition for the dual morphism. Duality in bordism categories We already encountered dual manifolds and dual bordisms in Definition 1.22, Remark 1.24, and (2.20). In this subsection we prove the following. Theorem Every object in a bordism category Bord X(n) is dualizable. Proof. If X(n) is the trivial tangential structure BO(n) BO(n), so Bord X(n) = Bord is the bordism category of (unoriented) manifolds, then for any closed (n 1)-manifold Y we have Y = Y with coevaluation and evaluation as in Figure 26. In the general case, an object (Y,θ) Bord X(n) is a closed (n 1)-manifold Y equipped with a classifying map (15.30) R TY θ S(n) Y X(n) to the universal bundle (9.59); cf. (9.60). Its dual (Y,θ) = (Y,θ ) has the same underlying manifold and classifying map θ the composition (15.31) R TY 1 id TY R TY θ S(n) Y id Y Y X(n)

6 6 D. S. Freed The coevaluation and evaluation (X,p,θ 0,θ 1 ) are as depicted in Figure 26. In both cases X = [0,1] Y. For the coevaluation p: X {0,1} is the constant function 1, and for the evaluation it is the constant function 0. For the evaluation (15.32) θ 0 : [0,1) Y [0,1) Y [0,1] Y (t,y) (t/4, y) (t,y) (1 t/4,y) with natural lifts to the X(n)-structures. The formula for θ 1 for the coevaluation is similar. Exercise Write the map on X(n)-structures explicitly. Note the sign in the differential of the last formula in (15.32) matches the 1 in the first map of (15.31). Proof of Theorem We first prove the following. Proposition Let B,C be symmetric monoidal categories, F,G: B C symmetric monoidal functors, and y B dualizable. Then (i) F(y) C is dualizable. (ii) If η: F G is a symmetric monoidal natural transformation, then η(y): F(y) G(y) is invertible. Proof. If (y,c,e) is duality data for y, then ( F(y ),F(c),F(e) ) is duality data for F(y). This proves (i). For (ii) we claim that η(y ) is inverse to η(y). Note that by Definition 15.26, η(y ) is a map G(y ) F(y ), and since G(y ) = G(y) it may be interpreted as a map G(y) F(y). Let c: 1 B y y and e: y y 1 B be coevaluation and evaluation. Consider the diagram (15.35) G(y) id F(c) G(y) F(y ) F(y) id η(y ) id G(y) G(y ) F(y) G(e) id F(y) id G(c) id η(y ) η(y) id id η(y) G(y) G(y ) G(y) G(e) id η(y) G(y) We claim it commutes. The left triangle commutes due to the naturality of η applied to the coevaluation c: 1 B y y. The next triangle and the right square commute trivially. Now starting on the left, the composition along the top and then down the right is the composition η(y) η(y ). The composition diagonally down followed by the horizontal map is the identity, by G applied to the S-diagram relation (15.20) (and using (13.40)). A similar diagram proves that η(y ) η(y) = id. Theorem is an immediate consequence of Theorem and part (ii) of (11.71). Part (i) implies the following.

7 Bordism: Old and New (Lecture 15) 7 Theorem Let C be a symmetric monoidal category and F: Bord X(n) C be a topological quantum field theory. Then for all Y Bord X(n), the object F(Y) C is dualizable. References [F1] D. S. Freed, The cobordism hypothesis. [L1] Jacob Lurie, On the classification of topological field theories, Current developments in mathematics, 2008, Int. Press, Somerville, MA, 2009, pp arxiv: