David Penneys. Temperley Lieb Diagrams and 2-Categories 4/14/08

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1 Temperley Lieb iagrams and 2-ategories avid Penneys 4/14/08 (1) Simplicial Resolutions. Let : be a left adjoint to U :, and denote by σ (respectively δ) the unit (respectively counit) of the adjunction. So σ(d) (d, U (d)) corresponds to id (d) and δ(c) ((c), c) to id U(c). (a) or any object c, show that the following formulas define a simplicial object (c) in : Let n (c) = () n (c) and let d i = () i δ(() n i (c)): () n+1 (c) () n (c) s i = () i σ(u() n i (c)): () n+2 (c) () n+1 (c). (b) Show that δ(c) induces a simplicial map ɛ(c): (c) c, where the right hand side denotes the constant simplicial -object. (c) ssume that there is a functor K : Set such that KU( (c)) is Kan. Show that the augmentation ɛ(c) gives a weak equivalence KU(ɛ(c)), i.e. it induces an isomorphism on all homotopy groups (that vanish above dimension zero for the constant functor). (d) pply this to your favorite pair of adjoint functors and see what you get. or example, you could use = R Mod or Ring and = b or Set. If happens to be an abelian category, one can apply the (normalized) chain complex to ɛ(c) and get all resolutions we have studied in class so far! Solution. (a) We will prove a more general result which will imply the desired result. Our proof will rely on the following facts: acts: (i) If X,, and Z are categories, we have an equivalence un(x, un(, Z)) = un(x, Z) = un(, un(x, Z)). (ii) There is a 2-category at whose objects are small categories, 1-morphisms are functors, and 2-morphisms are natural transformations. Recall that there are two compositions and of 2-morphisms in at, given respectively by: B X θ η Z and X ζ ξ for natural transformations θ : B, η :, etc. urthermore, is strict in the sense that we have equality X X θ ζ η Z θ = X ζ Z ξ η Z. ξ We will construct a simplicial functor un(s, un(, )) = un(, un(s, )), i.e., (c) will be a simplicial object in. To define this simplicial functor, we will look at a 2-subcategory of 1

2 at whose objects are the categories and ; 1-morphisms are composites of, U,, and id ; and 2-morphisms are composites of identity natural transformations, σ : id U, and δ :. We will use the following suggestive notation: to say a 1-morphism is in, we will write, and to say there is a natural transformation η : B for, B, we will write η (, B). Using the definitions of composition of 2-morphisms in, one can easily prove the following lemma: Lemma 1: (i) onsider () n. Then () n () n δ id () n = δ() n, () n () n+1 i.e. δ id () n = δ() n. Similarly, id () n δ = () n δ, σ id (U ) n = σ(u ) n, and id (U ) n σ = (U ) n σ. (ii) onsider. Then id = δ, i.e. δ id = δ. Similarly, σ id U = σ. (iii) id id B = id B and id id = id for composable, B. lso, since we have an adjunction, we have the following lemma: Lemma 2: We have the following relations among, U, σ, δ, id, id U : (i) δ σ = id : σ = id, (ii) Uδ σu = id U. The diagram is similar. We now introduce a powerful tool: a graphical calculus for working in. Usually, is written as an arrow from its source to its target. One could instead write as a point and its source and target as arrows going in and out of. This diagram is the dual diagram: X X = X = where one often supresses the directions on the arrows when the convention is understood (all arrows will point left in these dual diagrams). One usually writes natural transformations as 2-cells. ually, we can write them as pictures from dual diagrams to dual diagrams. This gives the added benefit that we can denote an identity 2

3 2-morphisms as a string going from the bottom to the top: X id = X X id and we can denote the unit or counit by a cap, once more reading bottom to top: U σ = U id id δ = U Now -composition of 2-morphisms corresonds in the dual diagram language to splicing pictures together sideways, and -composition corresponds to stacking. Often we do not label the categories and 1-morphisms as the 2-morphisms encode this data. B X id id idb Z B = id id B When a cap appears, we sometimes do not label the 2-morphisms as they are completely determined. urthermore, we omit the bullet representing the identity functor: id = Using the convention of shading the regions of the diagram which have along the outer boundary, we get string diagrams in which the lemmas above correspond exactly to isotopy of the strings. In the following diagrams, shaded regions have a in them: 3

4 = igure 1: Lemma 1.ii, δ id = δ. = igure 2: Lemma 1.ii, σ id U = σ. = igure 3: Lemma 2.i, δ σ = id. = igure 4: Lemma 2.ii, Uδ σu = id U. Now that we have this graphical calculus at our disposal, we can build our simplicial functor. Set n = () n+1 and d i = () i δ() n i : () n+1 () n s i = () i σu() n i : () n+2 () n+1. We can represent these natural transformations diagramatically: igure 5: d 0, d 1 (() 2, ). 4

5 igure 6:s 0 (, () 2 ). To prove that this defines a simplicial functor, we must show the following relations: d i d j = d j 1 d i for i < j s i s j = s j+1 s i for i j s j d i 1 if i < j d i s j = id if i = j, j + 1 s j 1 d i if i > j + 1. However, these are straightforward from the earlier lemmas by drawing the appropriate diagrams. We will prove d 0 d 1 = d 0 d 0 : () 3 () 1 first using the usual diagrams in, and secondly using the graphical calculus, i.e. the dual diagrams. d 0 d 1 = = id id id id id = δ δ id id id = id id id = = d 0 d 0 () 2 = 5

6 igure 7: d 0 d 1 = d 0 d 0 (() 3, ). The only non-obvious relation is d i s i = id = d i+1 s i, but these follow directly from Lemma 2. We will prove d 1 s 1 = id = d 2 s 1 (() 2, () 2 ) using the dual diagrams. = = igure 8: d 0 s 0 = id = d 1 s 0 (, ). Thus, is a simplicial functor in un(, ), and evaluation at c gives a simplicial object (c) in. (b) efine 1 = and ɛ: () n by capping everything off, i.e. δ(d 0 ) n 1. Note that this is equivalent to δd i1 d in 1 by the dual diagrams. Next, define id as the constant simplicial functor, i.e. id n = and d i, s j are all id id. It is obvious that ɛ induces a simplicial map from id, i.e. d i ɛ = ɛd i, and s j ɛ = ɛs j, since ɛ is capping everything off. Once more, evaluation at c gives the simplicial map c. (c) omposition with U gives a simplicial functor U in un(, ). The dual diagrams (for the d i s and s j s) are altered by adding one more string to the left. One immediately observes the existence of an extra degeneracy s 1 : U() n U() n+1 by ual diagramatically, we have: s 1 = σu() n+1 (U() n+1, U() n+1 ). igure 9: s 1 (U, U() 2 ). Now when we apply the functor K, we have a simplicial functor KU in un(, Set) with an extra degeneracy. t this point, we must evaluate KU at an object c to get a simplicial set for which we can describe the homotopy groups. We show ɛ(c): KU (c) KU id (c) = KUc is a weak equivalence, i.e. all homotopy groups π n (KU (c), ) = 0 for n > 0 and π 0 (KU (c), ) = KU(c) for a basepoint KU(c). Suppose x, x Z n (KU (c), ) for n > 0. Recall that a horn h is given by (n + 1) n-simplices y 1,..., y i 1, ŷ i, y i+1,..., y n+2 such that d i y j = d j 1 y i if i < j. Note that we have a horn if n > 0 since for all i > 0, h = (ŷ 0, y 1 = s 1 (x), y 2 = s 1 (x ),,,..., ) KU n+1 (c) d i s 1 (x) = s 1 d i 1 (x) = = s 1 d i 1 (x ) = d i s 1 (x ). 6

7 Since KU (c) is Kan, we can fill this horn, so there is a y 0 KU n+1 (c) such that d i y 0 = for i > 1, d 0 y 0 = d 0 y 1 = d 0 s 1 (x) = x and d 1 y 0 = d 0 y 2 = d 0 s 1 (x ) = x. Hence y 0 is a homotopy x x, and π n (KU (c), ) = 0 for all n > 0. We show π = π 0 (KU (c), ) = KU(c). We have an extra degeneracy s 1 : KU(c) KU 0 (c) which satisfies ɛs 1 = id KU(c) : = igure 10: ɛs 1 = id U (U, U). Hence s 1 is injective and ɛ is surjective. Now Z = Z 0 (KU (c), ) = KU (c) and ɛ(z) = KU(c). Note that if x x for x, x Z, then there is a y KU 1 (c) such that d 0 (y) = x and d 1 (y) = x. Then ɛ(x) = ɛ(x ) as ɛd 1 = ɛd 0 (this is the capping off trick discussed in (b)). Thus, ɛ induces a map ɛ: π KU(c) which must be surjective as ɛ = ɛq where q : Z π is the canonical epimorphism. Z ɛ s 1 q eɛ KU() π Moreover ɛqs 1 = ɛs 1 = id, so qs 1 is injective. Let x Z and y = s 1 (x). Then d 0 (y) = d 0 s 1 (x) = x and d 1 (y) = d 1 s 1 (x) = s 1 ɛ(x). = igure 11: d 1 s 1 = s 1 ɛ (U, U). Hence x s 1 ɛ(x). Thus, qs 1 is bijective with inverse ɛ, so ɛ induces a bijection π = KU(c), and we are finished. 7

8 (d) We will illustrate two examples. irst, consider the example = ree: Set b and U = orget: b Set. Then δ(c): Z is evaluation of a formal finite linear combination of elements of b. We have a simplicial group Z s 0 d 0,d 1 Z Z The maps d i are evaluation of a formal linear combination done at the i th step. or example, if x Z Z, then we have N 1 N 2 x = n i m j a j, i=1 where n i, m j Z and a j, and we may not distribute the n i s into the sum over j. Then we have N 1 N 2 N 1 N 2 d 0 (x) = n i m j a j and d 1 (x) = n i b i where b j = m j a j. i=1 j=1 urthermore, σ(s): d U(Z S ) is inclusion by s 1 s, so if y Z, then we have y = j=1 i=1 N 3 N 3 ki a i and s 0 (y) = ki (1 a i ) where 1 a Z. It is clear that d 0 s 0 = d 1 s 0 = id in this case. onsider = Z : Group Ring, i.e. taking the group ring, and U = : Ring Group, i.e. taking the group of units. Then δ(r): Z(R ) R is once again evaluation and σ(g): G (ZG) is inclusion. We have a simplicial ring Z(R ) s 0 d 0,d 1 and maps d i and s j are defined similarly as before. Z(Z(R )) j=1 8