Counterexamples to Indexing System Conjectures

Transcription

1 to Indexing System Conjectures January 5, 2016 Contents 1 Introduction 1 2 N Operads Barratt-Eccles operad Computing Stabilizers The Model Conjecture A Counterexample Property (E) and the Model Conjecture A Sufficient Condition Counterexample to the Suboperad Conjecture 8 5 Future Work 8 1 Introduction We provide counterexamples to some conjectures about indexing systems presented in [BH15]. 2 N Operads Let us fix a finite group G. We recall some definitions: Definition 1. Given a group G, a collection F of subgroups is called a family if F is closed under passage to subgroups and under conjugacy. Given such a family F, the universal space for F is a G-space EF such that for all subgroups H G, { (EF ) H H F H / F These exist for all families of finite groups, and there exist models which are G-CW complexes [Lue05]. Definition 2. A G-operad O consists of a sequence of G Σ n spaces O n for all n 0, such that i. There is a G-fixed identity element 1 O(1) ii. We have G-equivariant composition maps O n O k1... O kn O k1+...+k n satisfying the the usual associativity, unitality, and symmetric group equivariance. 1

2 Definition 3. A G-operad O is an N operad if i. The space O 0 is G-contractible ii. The action of Σ n on O n is free iii. O n is a universal space for a family F n (O) of subgroups of G Σ n which contains all subgroups of the form H {1}. We can combine (ii) and (iii) by saying O n is a universal space for a family F n (O) of subgroups Λ G Σ n such that Λ Σ n = 1 and all H {1} are in F. Lastly, we say a map O O between N operads is a weak equivalence it induces G Σ n -equivalence level-wise. That is, O and O have the same associated families. We would like to transpose this geometric information into a different setting. Blumberg and Hill used categorical coefficient systems: Definition 4. A categorical coefficient system is a contravariant functor C from the orbit category O G of G to the category of small categories. A coefficient system is symmetric monoidal if it lands in the category of symmetric monoidal categories and strong monoidal functors. The exemplary one is Set with disjoint union, defined by Set(H) = Set H, the category of H-sets and H-maps, where it sends morphisms to restrictions and conjugations. A sub-symmetric coefficient system C of Set is called an indexing system if C is closed under direct products, subobjects, and self-induction: T C(K) and H/K C(H) implies H K T C(H). Given a sequence of families {F n } of subgroups of G Σ n, we define a subcoefficient system of Set, called the system C O of admissible sets, defined as follows: we declare an H-set T to be admissible iff Γ(ρ) F n O for any choice of permutative representation ρ : G Σ n of T. In particular, given an N -operad, we have an associated coefficient system C O. Blumberg and Hill prove: Proposition 5. The coefficient system associated to an N operad is an indexing system. Moreover: Proposition 6 ([BH15], Theorem 4.18). This association is functorial, and the descends to a full and faithful functor from the homotopy category of N operads to the poset category of indexing systems. Conversely, given a coefficient system C, we define an associated sequence of families F n of subgroups of G Σ n : we declare Γ(ρ) is in Fn C iff there exists T C(H) which realizes the representation ρ. This leads to a natural conjecture: Conjecture 7 ([BH15]). An indexing system uniquely (up to homotopy equivalence) determines an N operad. In particular, the universal spaces for the families of subgroups generated by an indexing system form an N operad. This would imply that the homotopy category of N operads is equivalent to the poset category of indexing systems. The question now becomes, how do we realizes these spaces as an operad? sub-operads of some well-known operad? Can we display them as 2

3 2.1 Barratt-Eccles operad We recall the categorical Barratt-Eccles operad O, defined by O n = i Set(G, Σ n ). Here, i : Set Cat is the right adjoint to the objects forgetful functor; so i (X) is the category with objects X and a unique arrow between any two objects (Guillou-May called i X the chaotic category generated by X ). Guillou- May-Merling show in [GMM12] that each O n is a categorical model for the universal (G, Σ n )-bundle; that is, they are a universal space for the complete graph family F om of G Σ n. Moreover, in [GM14], they show that these form a categorical model for a G-E -operad, i.e. a complete N -operad. Since we are trying to form universal spaces of smaller families, we would like to construct, for each family F associated to an indexing system C, a suboperad O F of this Barratt-Eccles operad for which O F n is a universal space for the family F n. The natural choice is to define Set F (G, Σ n ) := {f Set(G, Σ n ) Stab(f) F n }, and let O n F be the subcategory i Set F (G, Σ n ) i Set(G, Σ n ). Conjecture 8 ([BH15], The Model Conjecture). The realization of O F n is a model for EF n. Moreover: Conjecture 9 ([BH15], The Suboperad Conjecture). If the families F are generated by an indexing system, then the categories O F form a suboperad of the Barratt-Eccles operad O. A direct consequence of these two conjectures would be our original conjecture; sending C to the operad O F would be a clear inverse of the functor described in the above section. However, we have some problems. Proposition 10. Conjecture 8 is false for generic groups G. Moreover, even if this conjecture holds for a particular group G, the resulting symmetric G-sequences may not be a suboperad: Proposition 11. There exist groups G for which, given any sequence of families F, O F n is a model for EF n, but there also exist indexing systems C over G such that, for F = F (C), O F is not a suboperad of O. 2.2 Computing Stabilizers Due to our definition of O F, much of the discussion will revolve around calculations of stabilizers. We have: Lemma 12. For f Set(G, Σ n ), Stab(f) = { (h, f(h) 1 f(1)) f(hx) = f(x)f(1) 1 f(h) for all x G }. Proof. Suppose (g, σ) Stab(f), so f(x) = ((g, σ).f)(x) = f(g 1 x)σ 1 for all x G. In particular, taking x = g, we find σ = f(g) 1 f(1), and then taking x = gx, we produce f(gx) = f(x)f(1) 1 f(g). We write H f = { h G f(hx) = f(x)f(1) 1 f(h) for all x G } ; then Stab(f) = Γ(f Hf ). Remark 13. Two warnings: 1. Even if f(1) = 1, this H f is not necessarily the largest subgroup H of G such that f H is a group anti-homomorphism; clearly H f H, but they do not have to be equal. Consider the example where 3

4 G = C 4 = { 1, t, t 2, t 3}, k >> 0 so that {z n 0 n 7} = C 8 Σ k, and we have a map f : G Σ k defined by 1 1 t z t 2 z 4 t 3 z. Then the largest H f such that f(hg) = f(g)f(h) for all h H f, g G is the trivial subgroup. Indeed, f(t t) f(t)f(t), f(t 2 t) f(t)f(t 2 ), and f(t 3 t) f(t)f(t 3 ). However, f {1,t 2 } is clearly a group (anti-)homomorphism. 2. The stabilizer of f K for some subgroup K G can again be larger than H f K; the previous example also shows this, with K = { 1, t 2}. We will also need to know when set maps f are fixed by subgroups Λ: Lemma 14. For ρ Hom(H, Σ n ) and f Set(G, Σ n ), Γ(ρ) Stab(f) iff f(hx) = f(x)ρ(h) 1 for all x G and h H. Proof. Assuming f is stabilized by Γ(ρ), we have ρ(h) = f(h) 1 f(1) for all h H. Thus, for all h H, f(hx) = f(x)f(1) 1 f(h) = f(x)f(1) 1 f(1)ρ(h) 1 = f(x)ρ(h) 1. Conversely, we have f(h 1) = f(1)ρ(h) 1, so ρ(h) 1 = f(1) 1 f(h), and thus f(hx) = f(x)ρ(h) 1 = f(x)f(1) 1 f(h), as desired. We call a set map f with Γ(ρ) Stab(f) a stabilizer extension of ρ. Moreover: Lemma 15. If f Set(G, Σ n ) is a stabilizer extension of ρ Hom(H, Σ n ), then g i H f iff Hg i H f. Proof. We have f(g i x) = f(x)f(1) 1 f(g i ) for all x G, and moreover f(hx) = f(x)ρ(h) 1 for all x G and h H. Thus f(hg i x) = f(g i x)ρ(h) 1 = f(x)f(1) 1 f(g i )ρ(h) 1 = f(x)f(hg i ). Further: Lemma 16. If there exists f Set(G, Σ n ) with Stab(f) = Λ, then there exists f Set(G, Σ n ) with f(1) = 1 and Stab( f) = Stab(f) = Λ. Proof. We let f(x) = f(1) 1 f(x). The verification is straight-forward. 4

5 Remark 17. These last three lemmas imply that if we are trying to build a stabilizer extension of ρ, we only need to choose values for f on a transversal {g i } of H\G with g 1 = 1 and f(1) = 1; indeed, we then must define f by f(kg i ) = f(g i )ρ(k) 1. That is, f must repeat itself (shifted by the values of f(g i )) on cosets of H. 3 The Model Conjecture 3.1 A Counterexample For determining whether O F n is a universal space for the family F n, it suffices to check that Set F (G, Σ n ) Λ iff Λ F n. Indeed, since O F n is a chaotic category, it is a connected groupoid where every element has trivial automorphism group; thus its realization is contractible iff it is non-empty. Moreover, since fixed points commute with geometric realization and right adjoints, we have O F n Λ = i Set F (G, Σ n ) Λ = i (Set F (G, Σ n ) Λ ). We now give a counterexample to Conjecture 8, proving Proposition 10: Example 18. Let G = C 2 C 2, and consider the indexing system defined by letting C(1 1) = C(1 C 2 ) = C(C 2 C 2 ) be just the trivial H-sets, and C(C 2 1) = { n C 2 1} n N N. Then, in particular Λ = Γ(ρ) = {((1, 1), 1), ((τ, 1), τ)} is in F 2, where τ is the non-trivial element in C 2 = Σ 2 and ρ the obvious associated non-trivial homomorphism C 2 1 Σ 2. However, any map ρ Set(C 2 C 2, Σ 2 ) having Λ = Γ(ρ) Stab( ρ) has a strictly larger stabilizer. Indeed, no matter where we send (1, τ) and (τ, τ), they will both be in the stabilizer: (1, 1) > 1 (τ, 1) > τ (1, τ) > x (τ, τ) > τ x; a ρ((1, τ) a) ρ(a) ρ((1, τ)) (1, 1) x 1 x = x (τ, 1) τx τ x (1, τ) 1 x x = 1 (τ, τ) τ τx x = τ for x either 1 or τ. Thus any map in Set(G, Σ n ) Λ has stabilizer Γ(β) for some β Hom(G, Σ 2 ), and hence cannot be in F 2. Thus Set F (G, Σ n ) Λ = for some Λ F 2, and hence O F 2 is not a universal space for F 2. So, our original guess does not work for all groups G and all the sequences of families F we want to consider. 5

6 3.2 Property (E) and the Model Conjecture Let us try to salvage something from O F n. In particular, we would like to know when O F n is in fact a universal space for F n. Our counterexample shows what can go wrong: the map f Stab(f) might not have a large enough target. This is the defining feature of groups that will work. Definition 19. We say a group G satisfies Property (E) if for all ρ Hom(H, Σ n ) non-trivial, there exists f Set(G, Σ n ) such that Stab(f) = Γ(ρ) Equivalently, G satisfies Property (E) if the image of the map Set(G, Σ n ) {Λ G Σ n Λ Σ n = 1} given by f Stab(f) contains the subset {Λ G Σ n Λ Σ n = 1} \ {H {1} G Σ n } for all n N. Proposition 20. G satisfies Property (E) iff Conjecture 8 holds for G. Proof. We just need to show that Set F (G, Σ n ) Λ is non-empty. But since Λ F n there exists f Set(G, Σ n ) with Stab(f) = Λ, and hence f Set F (G, Σ n ) Λ. Suppose we have ρ Hom(H, Σ n ) non-trivial such that for all f Set(G, Σ n ) with Stab(f) Γ(ρ) (that is, f Set(G, Σ n ) Γ(ρ) ), Stab(f) is strictly larger than Γ(ρ). Then we construct the family F generated by Γ(ρ) by collecting all subgroups, all conjugates, and all conjugates of subgroups. In particular, H and it s conjugates are maximal elements in the lattice of F. But since all f Set(G, Σ n ) Γ(ρ) have stabilizers in a strictly higher stratum of the lattice, Set F (G, Σ n ) Γ(ρ) is empty. Remark 21. Example 18 above exactly says that C 2 C 2 does not have Property (E). Moreover, a similar argument shows that G C 2, G abelian and non-trivial, never has Property (E). So the question now becomes: when does a group satisfy Property (E)? We start by looking at some extension properties: for fixed ρ Hom(H, Σ n ), are there some properties of the pair (G, H) that can allow us to construct an f Set(G, Σ n ) with Stab(f) = Γ(ρ)? By Lemma 17, we only need to define our new set map f on a transversal of H\G: f(kg i ) := f(g i )ρ(k) 1, k H and g i in our transversal A Sufficient Condition Suppose we have a given fixed ρ Hom(H, Π). Lemma 22. Suppose there exists π 0 Π such that π 2 0 1, and there exists h 0 Z(G) such that ρ(h 0 ) 1. Then there exists a map f Set(G, Σ n ) such that Stab(f) = Γ(ρ). Proof. We will build our function f Set(G, Π) coset by coset, by choosing our representatives and their images carefully by induction, again setting f(kg i ) = f(g i )ρ(k) 1 for k H and {g i } our chosen transversal. For each g i 1, we will show that g i / H f, and by the above lemmas that will be enough. We start by letting g 1 = 1 and letting f(1) = 1. Now, by induction, suppose we have choose g 1,..., g n 1 such that g i H f iff i = 1, and let g n G \ ( n 1 i=1 Hg i) be arbitrary. 6

7 Case I Hg 1 n = Hg n. Let h n be defined by g 1 n = h n g n. Case IA ρ(h n ) 1. Then define f(g n ) = 1. We observe that g n / H f : 1 = f(g n gn 1 ) f(gn 1 )f(g n ) = ρ(h 2 ) 1 1, Case IB ρ(h n ) = 1. Then define f(g n ) = π 0. We observe that g n / H f : 1 = f(g n gn 1 ) f(gn 1 )f(g n ) = 1 π 0. Case II Hg2 1 Hg i for any i {1,..., n 1}. Then define g n+1 such that gn 1 = h 0 g n+1 (i.e. g n+1 = h 1 observe that neither g n nor g n+1 are in H f : 0 g 1 n 1 = f(g n gn 1 ) f(gn 1 )f(g n ) = f(h 0 g n+1 )f(g n ) = 1 ρ(h 0 ) 1 1; ), and let f(g n ) = f(g n+1 ) = 1. We 1 = f(g n+1 g 1 n+1 ) f(g 1 n+1 )f(g n+1) = f(g n h 0 )f(g n+1 ) = f(h 0 g n )f(g n+1 ) = 1 ρ(h 0 ) 1 1. Case III Hg 1 n = Hg i for some i {1,..., n 1}; say g 1 n = h n g i. Case IIIA ρ(h n ) 1. Define f(g n ) = f(g i ) 1. We observe that g n / H f : f(gn 1 )f(g n ) = f(h n g i )f(g i ) 1 = f(g i )ρ(h 2 )f(g i ) 1, and this equals 1 = f(g n gn 1 ) iff ρ(h n ) = 1, a contradiction. Case IIIB ρ(h n ) = 1. Define f(g n ) to be 1 if f(g i ) = π 0 or π 1 0, and π 0 if f(g i ) = 1. We observe that g n / H f : 1 = f(g n gn 1 ) f(gn 1 )f(g n ) = f(h n g i )f(g n ) = f(g i )f(g n ). This again may seem restrictive, but for abelian groups it simplifies matters greatly: Corollary 23. If G is abelian, we only need to check that Property (E) holds for n = 2. Proposition 24. Cyclic groups satisfy Property (E). Proof. Let G = C n =< t >, and H G, say H =< t m >, with m minimal, and let b be such that n = mb. Let ρ Hom(H, Σ 2 ). Case I: m = 1. Choose f = inv ρ. 7

8 Case II: m 1. Since ρ is non-trivial, ρ(t m ) = π 1 and ρ((t m ) k ) = π k. We have a transversal { 1, t, t 2,..., t m 1} for G/H, and we define f(t l ) = π if l 0, and f(t 0 ) = f(1) = 1. Thus, globally, we have f(t mk+l ) = π ɛ π k, for 0 l m, 0 a b, with ɛ = 0 if l = 0 and ɛ = 1 if l 0. In particular, this satisfies f(hx) = f(x)ρ(h) 1 for all h H, x G (since every element of C 2 is its own inverse). By Lemma 15, it suffices to check that these t l are not in H f unless l = 0. If t l were in H f, then in particular we would have f(t l t m l ) = f(t m 1 )f(t l ). Since l 0, both f(t l ) and f(t m l ) are equal to π, and hence the right hand side is equal to π 2 = 1. However, the left hand side is f(t m ) = π. Hence no non-trivial t l are in H f, and hence H f is precisely H, so Stab(f) = Γ(ρ), as desired. 4 Counterexample to the Suboperad Conjecture We can now describe a family of counterexamples to Conjecture 9 which proves Proposition 11: Example 25. Let G = C 2N be any even-ordered cyclic group. Consider the map φ Set(C 2N, Σ 3 ) which sends t 2m to 1, and t 2m+1 to σ := (1 3 2) for all 0 m < N. Then, a straightforward calculation shows that Stab(φ) =< t 2 > 1, where < x > is the subgroup generated by the element x. Now, let f 1, f 2, and f 3 be trivial maps from C 2 to Σ 5, Σ 3, and Σ 2, respectively, and γ := γ(φ; f 1, f 2, f 3 ) be their Barratt- Eccles operadic composition. We compute that γ(t 2m ) equals 1, and γ(t 2m+1 ) equals the block permutation τ := (1 3 2)(5, 3, 2). However, since τ 2 = 1, H γ is all of C 2N, and in particular is the graph of a non-trivial homomorphism out of G. Thus, if C is any indexing system for G = C 2N such that C(G) contains only the trivial G-sets, then this lands outside Set F (C 2N, Σ ), and hence Set F (C 2N, Σ ) is not a suboperad of the Barratt-Eccles operad Set(C 2N, Σ ). Remark 26. This example is fairly ad-hoc, and we can create many other similar families of counterexamples. Moreover, this counterexample came from composing set functions with trivial graphs as stabilizers. This just emphasizes the fact that the stabilizers exert very little control over the composition. Even more troubling, we can show: Proposition 27. Let p be an odd prime, q = 2, and let G = C p 2 q and H = C pq. Consider the indexing system where C(K) is generated only by K/K if K H, and is generated by K/K and K itself if K H. This indexing system cannot be realized as Set F (G, Σ n ), nor D(U) or L(U) for any representation G-universe U, nor P (U) or Q(U) for any G-set universe U, where D is the little disks operad, L is the linear isometries operad, P is the coproduct-embeddings operad, and Q is the product-embeddings operad (see [GM14] for definitions of the last two). 5 Future Work We would like to find a categorical operad which realizes any indexing system. We have possibilities for realizing universal spaces, but not yet for suboperads. The goal would be to use our solution to this problem 8

9 to build a standard N operad for any indexing system, which could possibly allow us to mimic some of Fiedorowicz-Vogt [FV15] to show that N -operads are closed under tensor product in a nice way, or to try to extend the work of Guillou-May to build N spaces/spectra out of incomplete G-permutative categories. References [BH15] A. J. Blumberg and M. A. Hill. Operadic multiplications in equivariant spectra, norms, and transfers. Advances in Mathematics, 285: , November [FV15] Z. Fiedorowicz and R. M. Vogt. An Additivity Theorem for the interchange of En structures. Advances in Mathematics, 273: , [GM14] B. J. Guillou and J. P. May. Permutative G-categories in equivariant infinite loop space theory. ArXiv preprint: math/ v2, [GMM12] B. J. Guillou, J. P. May, and M. Merling. Categorical models for equivariant classifying spaces. ArXiv preprint: math/ , [Lue05] W. Lueck. Survey on classifying spaces for families of subgroups. In Infinite groups: geometric, combinatorial and dynamical aspects, pages Springer,