Quinn's formula and abelian 3-cocycles for quadratic forms

In pointed braided fusion categories knowing the self-symmetry braiding of simples is theoretically enough to reconstruct the associator and braiding on the entire category (up to twisting by a braided monoidal auto-equivalence). We address the problem to provide explicit associator formulas given only such input. This problem was solved by Quinn in the case of finitely many simples. We reprove and generalize this in various ways. In particular, we show that extra symmetries of Quinn's associator can still be arranged to hold in situations where one has infinitely many isoclasses of simples.


Introduction
This note applies to both (a) pointed braided fusion categories as well as (b) braided categorical groups. Both are special types of braided monoidal categories. Both settings are closely related, yet a little different. We hope that we have found a way to formulate the introduction so that it is clear, irrespective of which of these applications the reader might have in mind. Among our four results below, Theorem B and Theorem C mainly reprove known results differently, while Theorem A and Theorem D appear to be new.
The problem motivating this note is the following: Suppose (C, ⊗) is a braided monoidal category of type either as in (a) or (b). Then this category has an associator a X,Y,Z : and a braiding s X,Y : X ⊗ Y ∼ −→ Y ⊗ X as part of its braided monoidal structure. One may find a different associator/braiding for the same bifunctor ⊗ : C × C → C such that the identity functor id : C → C can be promoted to a braided monoidal functor 1 . The "orbit" of all associators/braidings which can obtained by twisting with such braided monoidal self-equivalences is pinned down by very little data. In concrete terms: In the setting of (a) suppose X is a simple object, resp. in the setting of (b) an arbitrary object. It has the self-symmetry braiding s X,X : X ⊗ X ∼ −→ X ⊗ X, and moreover X is invertible in C, so we have the functor of tensoring with its inverse, (−) → (−) ⊗ X −1 . Thus, by functoriality, we get a well-defined automorphism (1.1) s X,X ⊗ X −1 ⊗ X −1 : 1 C where in the setting of (a) C simp is the groupoid of simple objects in C, and for (b) let C simp := C be the entire category. Thus, we get a map q : π 0 (C simp , ⊗) −→ π 1 (C simp , ⊗).
One can show that this map is a quadratic form. On the one hand, this form does not change under the aforementioned braided monoidal self-equivalences of C (which induce the identity map on π 0 and π 1 ). On the other hand, it also distinguishes different such orbits, i.e. it is a complete invariant. This note is concerned with the problem to provide an explicit formula for the associator and braiding for a given quadratic form q, i.e. if we only know the self-symmetry braidings of Equation 1.1, but have possibly no clue and no candidates what the associator and braiding should do on general objects X, Y, Z. This is a kind of integration problem: Find a valid choice of associators and braidings for the entire category such that, restricted to self-symmetries, it agrees with the given quadratic form.
Even stronger: One may ask whether there is a "simplest choice" of associators, e.g., an associator with additional symmetries (we propose a possible solution in Theorem D).
The problem can be attacked in concrete form as follows: First, since we only work up to braided monoidal equivalences inducing the identity map on π 0 and π 1 , it suffices to work with a skeleton of the category. Here the datum of an associator and braiding is encoded in an abelian 3-cocycle showing that this cohomology group is isomorphic to the group of quadratic forms on G with values in M , [ML52], [EML53]. This isomorphism underlies the above reconstruction procedure. Given only the "self-symmetries", i.e. only the right side, finding an associator and braiding amounts to finding a preimage under this isomorphism. (In particular, this paper provides a new proof of the surjectivity of the map in Equation 1.2 as a side result, see §11) 1.1. Application to explicit associator formulas. Unfortunately, solving the integration problem is not entirely trivial. The classical proof for the isomorphism in Equation 1.2 goes as follows: Do the cases G = Z and G = Z/nZ individually, then use that both sides of Equation 1.2 are quadratic functors in G, and exploit that any abelian group is a colimit of finitely generated ones. This sounds deceivingly simple, but note that if one wants an explicit formula, one needs a cocycle formula, i.e. a lift and once you consider such lifts, the nice functoriality properties break (as an illustration: A compatible family of cohomology classes in some directed system underlying the colimit need not come from a compatible family of cocycles, because there is no a priori control of the coboundaries along the system). We solve this as follows: In §3-5 we shall introduce the concept of 'optimal admissible presentations' and 'admissible liftings', and this can be thought of as machines to produce explicit cocycle formulas. Since these are in general necessarily non-linear maps and highly non-unique, it should not be surprising that admissible presentations and liftings also allow for a lot of variation. This will be Theorem 6.2. It underlies all other results of the paper. Quinn [Qui99] has given an explicit formula in the case G finite abelian and M := R × for R a commutative ring. We give a new proof for his formula as an application of Theorem A. In our generalized version M is arbitrary and G is allowed to be of a more general form. In particular, all finitely generated abelian groups are covered by our version: Theorem B (Generalized Quinn formula). Let M be any abelian group. Suppose for J 1 , J 2 any index sets, and n j ≥ 1 integers. Fix a total order on the disjoint union J := J 1∪ J 2 , say with J 1 < J 2 . Write (e j ) j∈J for the generator 1 in the j-th cyclic group. Let q ∈ Quad(G, M ) be a quadratic form and b(x, y) := q(x+y)−q(x)−q(y) its polarization. Define defines an abelian 3-cocycle such that the trace map of Equation 1.2 sends it to the given quadratic form q. Here x j (resp. y j , z j ) refers to coordinates with values x j ∈ Z for j ∈ J 1 resp. x j ∈ {0, 1, 2, . . . , n j − 1} for j ∈ J 2 . The map q → (h, c) is linear, so it provides a group homomorphism Quad(G, M ) → Z 3 ab (G, M ), which makes Diagram 1.3 commute. This will be Theorem 7.1. If G is finite abelian (so J 1 = ∅ and #J 2 < ∞) and M := R × , then up to rewriting the formula for h and c in the multiplicative notation customary for elements in the units R × , we recover precisely Quinn's formula given in [Qui99,§2.5.2].
For some applications, especially when wanting to do explicit computations in the setting of fusion categories over C, the following formulation might be more useful. We provide it with full details so that it can easily be referenced whenever needed: for n k ≥ 1 and J some (possibly infinite) totally ordered index set. Write (e k ) k∈J for the generator 1 of the k-th summand. Then there is a bijection between the following three sets: (1) All possible choices of values (2) All quadratic forms q ∈ Quad(G, C × ), uniquely described by the following properties where b is the polarization of q (and furthermore we necessarily then have b(e k , e l ) = b(e l , e k ) for k > l and b(e k , e k ) = 2q(e k ) as well).
(3) All abelian 3-cocycles (h, c) ∈ H 3 ab (G, C × ), uniquely pinned down by the cocycle representative Really, Quad(G, C × ) and H 3 ab (G, C × ) are abelian groups and the above bijections are abelian group isomorphisms, given in terms of the parameters p (k) , q (k,l) by elementwise addition in the quotient groups (i.e. Z/(n 2 k , 2n k ) for p (k) etc. 1.2. Application to normal forms of associators. In the setting of (a), i.e. pointed braided fusion categories, Quinn's formula is sufficent to describe an associator and braiding in all situations. This is because in this setting G := π 0 (C simp , ⊗) is a finite abelian group and thus safely covered by both Theorem B and Theorem C.
However, Quinn's formula has some special shape (e.g., far more symmetry than one might a priori expect!). Let us broaden the question: Suppose that (C, ⊗) is a pointed braided fusion category, but drop the assumption that there are only finitely many isomorphism classes of simple objects. You could think of finite-dimensional G-graded vector spaces Vect G k with some associator and braiding, but where the grading comes from any abelian group G, and not just a finite one. We will properly define this later and call it a big fusion category.
Suppose we want to bring (C, ⊗) into some particularly nice "normal form" under braided monoidal equivalence. As before, replace (C, ⊗) by a skeleton. Then the associator and braiding are merely automorphisms. If X, Y, Z are simple objects, we may read a X,Y,Z and s X,Y as elements of k × canonically. Now, the simplest conceivable normal form would be, through a braided monoidal equivalence, to make all associators and the braiding trivial. This is, however, an unrealistic hope (it cannot be achieved). Perhaps the following is the best possible normal form one can expect in general. 2 Theorem D (Extra symmetries). Suppose k is an algebraically closed field of any characteristic. Let (C, ⊗) be a k-linear pointed braided big fusion category. Then (C, ⊗) is braided monoidal equivalent to a skeletal big fusion category such that a X,Y,Z = s X,Y · s X,Z s X,Y ⊗Z and a Z,X,Y = s X⊗Y,Z s X,Z · s Y,Z hold for all simple objects X, Y, Z.
See Theorem 9.13. These properties are "extra symmetries" which are not visibly forced by the hexagon and pentagon axioms. I do not have a philosophical interpretation why such extra symmetries always exist (e.g., note that it follows from them that a X,Y,Z = a X,Z,Y ).
To restate the result in other words: The associator only measures the lack of "⊗linearity" of the braiding, in either variable. A tool to memorize the formulas: the first argument of the associator is the one argument which appears in all three factors on the other side of the equality sign.
The above result follows from Quinn's formula if (C, ⊗) is an ordinary pointed braided fusion category with G finite. We believe the above observation is new in the case of arbitrary G. It does not follow by a "colimit argument" from the case of finite G by the same problem as discussed around Figure 1.3. And at any rate our argument takes a different path and circumvents Quinn's formula or its siblings.

Abelian cohomology
Let us recall Eilenberg and MacLane's theory of abelian cohomology. We refrain from giving a careful motivation how this formalism arose. Instead, we may refer to [Bra19, §3.1] for some more background.
Let G, M be abelian groups. While there are elegant and systematic definitions of group cohomology and abelian cohomology, we will just work with an explicit presentation here, namely normalized inhomogeneous cochains. Also, we shall only need H 3 .
Write G n := G × · · · × G for the n-fold product of abelian groups. A group 3-cocycle is a map of sets holds for all x, y, z ∈ G (corresponding in tensor category language to the "pentagon axiom for associators", see §9, e.g., the proof of Theorem 9.2). A group 3-cocycle is called normalized if h(x 1 , x 2 , x 3 ) = 0 as soon as x i = 0 for some i ∈ {1, 2, 3}. A normalized group 3-coboundary is a group 3-cocycle of the shape for some map of sets k : G 2 → M such that k(x, 0) = 0 and k(0, y) = 0. It is easy to check that this is a normalized group 3-cocycle. These explicit expressions can directly be unravelled from [NSW08, Chapter I, §2] for example. An abelian 3-cocycle is a pair (h, c) consisting of a group 3-cocycle h : G 3 → M such that for all x, y, z ∈ G (corresponding to the two "hexagon axioms" in the dictionary with tensor categories). Equation 2.3 implies that h is normalized ([Bra19, Remark 3.5]). An abelian 3-coboundary is a pair (h, c), where h is a normalized group 3-coboundary coming from k : G 2 → M , and for the same k. Write Z 3 grp (resp. Z 3 ab ) to denote the group of normalized group 3-cocycles (resp. abelian 3-cocycles), resp. B 3 grp and B 3 ab for coboundaries.  By a quadratic form q : G → M (also known as 'quadratic map' or 'quadratic function' in various texts, depending on the taste of the various authors) we mean a map of sets such that q(x) = q(−x) and is Z-bilinear for all x, y ∈ G. The map b is known as the polarization form. Write Quad(G, M ) for the set of all quadratic forms. This is an abelian group under pointwise addition of maps.
Example 2.2. This definition may encompass more types of maps than a casual reader might expect. For example if G and M happen to be F 2 -vector spaces, every linear map G → M is a quadratic form. Concretely, the map might not 'look quadratic' as an algebraic expression, but it is a quadratic form. Thanks to (a + b) 2 = a 2 + b 2 in characteristic two rings, the polarization vanishes.
Example 2.3. If q is any quadratic form, we have q(nx) = n 2 q(x) for any x ∈ G and n ∈ Z.
To see this, note that both follow from Equation 2.5. The case n = 0 is clear. By induction, assuming the case n to be done, and by the Z-bilinearity of b and Equation 2.6, b(nx, x) = −nb(x, −x) = 2nq(x), and then (n 2 + 2n + 1)q(x) = q((n + 1)x), proving the claim for all n ≥ 0. It follows for negative n by q(−x) = q(x).
The key connection between abelian 3-cocycles and quadratic forms is the following theorem.
Theorem 2.4 (Eilenberg-Mac Lane). Let G, M be abelian groups. The so-called trace is an isomorphism of abelian groups.
We give an outline how this is proven in §11, including a new proof of surjectivity.

Admissible presentations
Let G, M be abelian groups. Suppose q ∈ Quad(G, M ) is a quadratic form. We write b for the polarization form of q as given in Equation 2.5.
Definition 3.1. A pre-admissible presentation for q is a triple (F 0 , π, C), where (1) F 0 is an abelian group and π a surjective group homomorphism π : F 0 ։ G; and write F 1 := ker(π); (2) C is a Z-bilinear form C : We speak of an admissible presentation when instead of (3) we have the stronger property that C(x, y) = 0 holds for all x, y ∈ F 1 . Axiom (3) just demands that the restriction C | F1 is an alternating form. For an admissible presentation, F 1 is an isotropic subgroup.
Given a pre-admissible presentation, we can lift the quadratic form from G to F 0 . To this end, we define Then Q ∈ Quad(F 0 , M ) is indeed a quadratic form. Here and henceforth write for its polarization form. Note that We have chosen our notation so that the uppercase letters refer to the lifts of their lowercase letter counterpart. There is a slightly more refined property one can demand (and always arrange) to hold: Definition 3.2. We call a (pre-)admissible presentation optimal if we have Example 3.3. The simplest example of a non-optimal admissible presentation is for the quadratic form q ∈ Quad(F 2 , F 2 ) given by q(x) = x 2 . For this form (F 2 , id F2 , C) with C(x, y) := 0 is a non-optimal admissible presentation with F 1 = 0. An optimal presentation is given by C(x, y) := xy.
Starting with any pre-admissible presentation (F 0 , π, C), in order to achieve optimality, one only needs to change the bilinear form C, while F 0 and π can remain the same. We prove this now. (1) If (F 0 , π, C) is a pre-admissible presentation, one can find an optimal pre-admissible presentation (F 0 , π, C ′ ).
Below, we write n M := {m ∈ M | nm = 0} for the n-torsion subgroup.
Proof. (Step 1) We first prove the first claim. It is immediate to see that Q ′ (x) := C(x, x) is a quadratic form on F 0 . Its polarization form is and by Equation 3.1 this is b(πx, πy), so Q ′ has the same polarization as Q by Equation 3.3-3.4. Thus, is a quadratic form in Quad(F 0 , M ) whose polarization vanishes. This means that L satisfies showing that L is a morphism of abelian groups, As a quadratic form, it also satisfies L(x) = L(−x), i.e. 2L(x) = 0 holds for all x ∈ F 0 . Thus, L descends to a group homomorphism L : F 0 /2F 0 → 2 M . The Z-module structure of either side now induces an F 2 -vector space structure, rendering L an F 2 -linear map. We next choose a special basis of F 0 /2F 0 . To this end, pick a direct sum splitting where we refer to the image coming from the inclusion F 1 ⊆ F 0 . Pick (γ i ) i∈I as a basis of F 0 /2F 0 first by picking a basis on the subspace im(F 1 ), say with indices in a subset I im F1 ⊆ I of the index set, and then prolong it to all of F 0 /2F 0 . Define a symmetric bilinear form on F 0 /2F 0 by where x i ∈ F 2 denotes the F 2 -coordinates of the vector x ∈ F 0 /2F 0 (resp. y i for y) with respect to the basis (γ i ) i∈I . As each L(x) lies in the 2-torsion group 2 M , the scalar multiplication with elements from F 2 is well-defined and indeed linear. For an arbitrary x ∈ F 0 /2F 0 we compute We keep the same name J for this lift. Returning to our definition of L in Equation 3.6, we now find Since both C and J are Z-bilinear forms, so is C − J. Define (3.10) We thus have Q(x) = C ′ (x, x), so Equation 3.5 holds, showing that C ′ is a promising candidate to satisfy the optimality property. ( Step 2) We need to check that (F 0 , π, C ′ ) is a pre-admissible presentation: Axiom (1) is clear; nothing about F 1 or π has changed. For axiom (2) we find since C satisfies axiom (2) by assumption and J is a symmetric form. However, J by construction takes values in the 2-torsion elements 2 M , so 2J(x, y) = 0 for any x, y. Hence, axiom (2) is satisfied. Axiom (3): From Equation 3.6 and Equation 3.2 we get Thus, if x ∈ F 1 then C(x, x) = 0 since C satisfies axiom (3), and q(πx) = 0 since F 1 = ker(π 1 ). We deduce that L | F1 = 0. Next, recall from Equation 3.9 that J was defined by where x i , y i were the respective F 2 -coordinates. If x, y ∈ F 1 , then thanks to our special choice of basis from Equation 3.8, we have x i = 0 for all i ∈ I \ I im F1 , and the same for y i . Thus, for x, y ∈ F 1 we have and this vanishes since C(x, x) = 0 (as axiom (3) holds for C) and we had already observed J | F1 = 0. This shows that axiom (3) holds for C ′ . This shows that (F 0 , π, C ′ ) is an optimal pre-admissible presentation (we had shown optimality in Step 1). This finishes the proof of the first claim. Finally, we want to show that if (F 0 , π, C) was an admissible presentation to start with, so is (F 0 , π, C ′ ). We already know the latter is optimal and pre-admissible. Now, as a direct variation of Equation 3.11, for x, y ∈ F 1 we get and since C was admissible, C(x, y) = 0, and we had already observed J(x, y) = 0 above.

Existence theorems for admissible presentations
We begin with the principal construction mechanism for pre-admissible presentations. This is a good construction whenever the quadratic form comes from a bilinear form, even if this is perhaps not possible on G, but only on a bigger group.
Lemma/Construction 4.1. Suppose G and M are arbitrary abelian groups.
(1) Assume one finds an abelian group F 0 with a surjection π : F 0 ։ G such that on F 0 one can exhibit a bilinear form C such that q(πx) = C(x, x) ("the lift of q comes from a bilinear form"). Then (F 0 , π, C) is an optimal pre-admissible presentation.
(2) If one can take F 0 = G and π = id G , then (G, id G , C) is an optimal admissible presentation and F 1 = 0.
Proof. We begin with the first claim. Axiom (1) is immediate. The polarization form b of q, written in terms of images of elements x, y from F 0 , is proving axiom (2). Finally, if x ∈ F 1 , then since F 1 = ker(π) we have C(x, x) = q(πx) = 0, so axiom (3) holds. Optimality holds by construction. For the second claim, note that F 1 = 0, so the strong form of axiom (3) is clear.
We should also mention a trivial case, where nothing much needs to be done at all.
This is an optimal admissible presentation.
Proof. Immediate. Since b is symmetric, so is C, take F 0 := G and then F 1 = 0. In particular, there is nothing to check for axiom (3). For x ∈ G the polarization form yields The hypothesis of M being a Z[ 1 2 ]-module is virtually never satisfied in applications though.
Example 4.3. Suppose (C, ⊗) is any k-linear braided fusion category. Then the group which appears for M in applications (see §9) is M := π 1 (C, ⊗) = k × since the tensor unit 1 C is a simple object. For this group to be 2-divisible one needs that k is closed under all quadratic field extensions. This is for example satisfied if k is algebraically closed. However, to be free from 2-torsion one also needs that x 2 = 1 implies x = 1, i.e. +1 = −1. This forces k to be of characteristic two.
Instead, a much more realistic hypothesis is that M is a divisible module. If k is any algebraically closed field, k × is a divisible group.
Lemma 4.4. Suppose G is an arbitrary abelian group and M a divisible abelian group. If (F 0 , π, C) is a pre-admissible presentation with F 0 free abelian, then one can replace C by a bilinear formC such that (F 0 , π,C) is an admissible presentation. If the presentation was optimal to start with, the newC can be taken optimal, too.
Proof. Consider the restriction C | F1 . We have C(x, x) = 0 for all x ∈ F 1 by axiom (3). Thus, C | F1 is an alternating bilinear form 3 , 3 just to be sure about nomenclature: Alternating means that C(x, x) = 0 for all x. This implies C(x, y) = −C(y, x), but is strictly more restrictive than the latter property.
Since F 0 is free, so is F 1 (Z is a hereditary ring). Then the injectivity F 1 ֒→ F 0 implies that the top horizontal arrow in the diagram is also injective (without too much harm it can be checked directly that exterior powers of free modules preserve injectivity, but a literature reference would be [Fla67, Theorem 1]). Since M is divisible, it is injective as a Z-module, so the dashed arrow A above exists and makes the diagram commute. Definẽ We claim that (F 0 , π,C) is an admissible presentation. We compute for x, y ∈ F 0 that .1 is still valid. This settles axioms (1) and (2). Moreover, for x, y ∈ F 1 we haveC The next construction will be the concrete input needed for establishing a generalized form of Quinn's formula.
Lemma/Construction 4.5. Suppose M is arbitrary. If G is a (possibly infinite) direct sum of various (possibly infinite) cyclic groups, an optimal admissible presentation for q exists. (A concrete construction is given in the proof ) Proof and Construction. (Step 1) A cyclic group is either of the form Z or Z/n for some n ≥ 1. Thus, each direct summand in G has a presentation of the shape Take the direct sum of these, i.e. we have found a presentation for suitable index sets I ⊆ J, where the first arrow is given by a diagonal matrix. We take this as F 1 ֒→ F 0 ։ G. This sets up (1) in an admissible presentation. Use the same notation Q and B as in Equation 3.2 and 3.3 now. ( Step 2) Write (e j ) j∈J for the basis vectors of F 0 . Fix a total order on J. Define and thus We find that the polarization forms of Q and Q ′ agree. Thus, L := Q − Q ′ is a quadratic form whose polarization is zero. With the same computation as in Equation 3.7 it follows that L : F 0 /2F 0 → 2 M is a group homomorphism. We compute just by unravelling the definition of Q ′ and using Equation 4.3. Since the (e j ) j∈J form a basis of F 0 and L is linear, it follows that L(x) = 0 for all x ∈ F 0 . Thus, for all x ∈ F 0 . This shows that Q comes from a bilinear form, so we may invoke Lemma 4.1 and learn that (F 0 , π, C) is an optimal pre-admissible presentation. (Step 3) Finally, we need to show the strong form of axiom (3). Note that by our special construction of the presentation in Equation 4.2, the group F 1 = ker(π) is generated by elements of the shape (n i e i ) i∈I with n i ∈ Z ≥1 , i.e. π(n i e i ) = 0. We compute for arbitrary i, j ∈ I (again using the fact from Example 2.3), but of course by the very definition of Q and B, these terms all vanish since we have π(n i e i ) = 0. This proves that (F 0 , π, C) is admissible.
Finally, let us show that optimal (pre-)admissible presentations always exist under very general hypotheses. However, the constructions in the proof cannot be carried out constructively usually, so the following theorem will not help when wanting to develop explicit formulas.
(1) Then for any quadratic form q : G → M an optimal pre-admissible presentation exists.
(2) If M is divisible, then for any quadratic form q : G → M an optimal admissible presentation exists.
Proof. (Step 1) We imitate the method of Lemma 4.5 for as long as possible. First, pick a free resolution of G, for suitable index sets I, J. This always exists (and has length 2 either because Z is a ring of global dimension one, or, more down to earth, since the kernel of π has to be free abelian itself). As before, write (e j ) j∈J for the basis vectors of F 0 , fix a total order on J, and let Q and B denote the lifts of q and b to F 0 (as in Equations 3.2-3.3). This replaces Step 1 in the proof of Lemma 4.5. ( Step 2) Define Now repeat the same arguments as in Step 2 of the proof of Lemma 4.5. This all goes through and proves that (F 0 , π, C) is an optimal pre-admissible presentation. This proves the first claim.
Step 3 in the cited proof does not adapt to the present setting. We do something else: ( Step 3) If M is divisible, we can invoke Lemma 4.4 and transform the construction from Step 2 into an admissible optimal presentation (F 0 , π,C). This settles the second claim.
Problem 1. Does any quadratic form q : G → M admit an optimal admissible presentation without assuming M divisible?
Thanks to Proposition 3.4 one would only need to exhibit an admissible presentation; the optimality can be achieved afterwards.
Example 4.7. We illustrate that Step 1 in the above proof cannot be expected to give admissible presentations right away. Consider the needlessly complicated free resolution For the quadratic form x → x 2 on Z, the procedure in the proof of Theorem 4.6 yields that (Z n , π, C) with is an optimal pre-admissible presentation. All the vectors e i −e k lie in the kernel F 1 = ker(π). For i < j < k we compute C(e i − e k , e j − e k ) = 1.

The lifting function
Suppose q ∈ Quad(G, M ) is a quadratic form and assume we have chosen an admissible presentation (F 0 , π, C) as in §3.
Definition 5.1. For any non-zero element x ∈ G pick once and for all a lift x ∈ F 0 , i.e. some element such that π( x) = x. For the neutral element we pick the special lift Call any such choice an admissible lift.
As π is surjective, it is clear that admissible lifts always exist.
Example 5.2. We stress that we have π x = x, but in general there is not much we can say about how (−) interacts with algebraic operations. For example, 2x = 2 x, (−x) = − (x) or x + y = x + y are all possible in suitably chosen examples, and in general π will not admit a splitting in terms of abelian groups, so in general we cannot avoid for these lifts to depend somewhat non-linearly on the input.
Having fixed an admissible lift, define a map Note that there is no reason why L would have to be bilinear in any way. We can record a few useful facts about L nonetheless: for all u, x, y, z ∈ G.
For merely pre-admissible presentations property (4) can fail.
where we have just used cancellations of terms. (4) Note that for all x, y ∈ G we have π( (x + y) − x − y) = 0, so (x + y) − x − y ∈ F 1 . This proves (4) since this applies to both arguments, so we can use the strong form of axiom (3) of an admissible presentation.

Constructing abelian 3-cocycles
Let q ∈ Quad(G, M ) be a quadratic form. Suppose we have fixed an admissible presentation (F 0 , π, C) as in Definition 3.1, alongside a choice of admissible lifts as in Definition 5.1. We use the notation F 0 , F 1 , π, C, B, Q as explained in §3.  We stress that we only need the above assumptions, i.e. the admissible presentation (F 0 , π, C) does not need to be optimal.
Next, by Equation 3.1 we have C(y, x) = B(x, y) − C(x, y) for all x, y ∈ F 0 . Thus, rewrite the preceding equation as However, we also have Equation 3.4, namely B(x, y) = b(πx, πy), so the first line simplifies to b(π( x + y), π( z)) − b(π x, π z) − b(π y, π z) Recall that if one only is given a non-optimal admissible presentation, one can always change it into an optimal one by Proposition 3.4.
Proof. This is easy now. Firstly, by Lemma 6.1 (h, c) is an abelian 3-cocycle. The trace maps it to the quadratic form

Generalized Quinn formula
We can now use the tools of §6 to reprove Quinn's formula in a generalized format. In particular, this gives an alternative approach to the original proof for G finite abelian [Qui99, §2.5.1-2.5.2]. Below, we intentionally stay close to the notation of Quinn's article so that the resulting formula has the same shape.
Theorem 7.1 (Generalized Quinn formula). Let M be any abelian group. Suppose for J 1 , J 2 any index sets, and n j ≥ 1 suitable integers. Fix a total order on the disjoint union J := J 1∪ J 2 , say with J 1 < J 2 . Write (e j ) j∈J for the standard generators (i.e. the element 1 Z resp. 1 Z/nj Z in the corresponding summand). Let q ∈ Quad(G, M ) be a quadratic form and Then the pair (h, c) with h(x, y, z) := j∈J2 with yj +zj ≥nj x j n j σ j,j and c(x, y) := i,j∈J with i≤j x i y j σ i,j defines an abelian 3-cocycle such that the trace map of Equation 1.2 sends it to the given quadratic form q. Here x j (resp. y j , z j ) refers to coordinates with values x j ∈ Z for j ∈ J 1 resp. x j ∈ {0, 1, 2, . . . , n j − 1} for j ∈ J 2 . The map q → (h, c) is linear, so it provides a group homomorphism Quad(G, M ) → Z 3 ab (G, M ), which makes Diagram 1.3 commute. Proof of Theorem 7.1. (Step 1) Given the format of our input abelian group G, we can use Lemma/Construction 4.5 to set up an optimal admissible presentation. Let us quickly walk through the relevant steps of the construction, adapted to our setting: Define .1) and this sets up a resolution Write (e j ) j∈J for the standard basis of F 0 , i.e. the j-th summand Z is spanned by e j (so that e j = πe j in terms of the elements in the statement of the theorem). As in Lemma 4.5, define B(x, y) = b(πx, πy) and Q(x) = q(πx) and then describes a Z-bilinear form on F 0 (this is the same as in the construction given loc. cit.). As guaranteed by the quoted lemma, (F 0 , π, C) is an optimal admissible presentation. (Step 2) Each element of G has a unique presentation as g = j∈J g j π(e j ) with g j ∈ {0, 1, . . . , n j − 1} if j ∈ J 2 and g j ∈ Z if j ∈ J 1 . Sending this g to the vector g := j∈J g j e j ∈ F 0 pins down an admissible lift in the sense of Definition 5.1 (including 0 = 0). With respect to the basis (e j ) j∈J we can write C(−, −) as We can write the admissible lift coordinate-wise for any vector x ∈ G as In particular, Now assume we are given x, y, z ∈ F 0 such that the coordinates satisfy the bound x j ∈ {0, 1, . . . , n j − 1} for all j ∈ J 2 (this will simplify the formulas). Demand the same for y j resp. z j . There is no condition if j ∈ J 1 . Invoke Theorem 6.2 to obtain that the pair (h, c) with h(x, y, z) := −C( x, L(y, z)) and c(x, y) := C( x, y) is an abelian 3-cocycle mapping to q under the trace map. Next, let us unravel these expressions. Expand h using Equation 7.4 to Since x j , y j ∈ {0, 1, . . . , n j − 1} for every j ∈ J 2 , we may rewrite the expression for h as and this simplifies to = i≤j with j∈J2 and yj +zj≥nj Finally, if i = j, we have by Equation 7.3 and the bilinearity of B that (7.9) n j σ i,j = n j B(e i , e j ) = B(e i , n j e j ) = b(πe i , π(n j e j )) = 0 since n j e j ∈ ker(π). Hence, This finishes the proof.
Example 7.2. Note that along the way, we have found some other possibly useful presentations of the 3-cocycle. For example, Equation 7.7 expresses the associator for arbitrary representatives/lifts x i , y i , z i ∈ Z.
Example 7.3 ([KS11]). A lively description how one attaches an abelian 3-cocycle to a quadratic form is also given by Kapustin and Saulina in [KS11, §3.2] (again in the situation with G finite). For readers familiar with their paper, let us note that A ⊙ B (in their notation) corresponds to our x + y. They obtain the formula at the end of [KS11,§3]. This is Equation 7.8, again with the summands i = j removed by the same argument as in Equation 7.9.

Abelian 3-cocycle formulas in exponential format
Aside from Quinn's formula, a lot of literature prefers to explicitly spell out the abelian 3-cocycle in terms of exponential functions when M := C × . Let us also provide this.
(2) All quadratic forms q ∈ Quad(G, C × ), uniquely described by the following properties q(e k ) = exp 2πi gcd(n 2 k , 2n k ) where b is the polarization of q (and further we necessarily then have b(e k , e l ) = b(e l , e k ) for k > l and b(e k , e k ) = 2q(e k ) as well). 2πiq (k,l) gcd(n k , n l ) x k y l (8.2) where x k (resp. y k , z k ) denotes the coordinates of vectors x, y, z ∈ G according to Equation 8.1. Here [−] n k refers to the remainder of division by n k , expressed as an element in {0, 1, . . . , n k − 1}.
Really, Quad(G, C × ) and H 3 ab (G, C × ) are abelian groups and the above bijections are abelian group isomorphisms, given in terms of the parameters p (k) , q (k,l) by elementwise addition in the quotient groups (i.e. Z/(n 2 k , 2n k ) for p (k) etc.). The map q → (h, c) is linear, so it provides a group homomorphism Quad(G, M ) → Z 3 ab (G, M ), which makes Diagram 1.3 commute. Before we prove this, let us first establish an explicit parametrization of the quadratic forms.
We apologize that the following repeats part of the statement of the above theorem, but we prefer to be clear about what we prove here amidst a lot of notation. (2) q (k,l) ∈ {0, 1, . . . , gcd(n k , n l ) − 1} for all k < l with k, l ∈ J. Once these choices are made, the corresponding quadratic form and its polarization satisfy b(e k , e l ) = exp 2πi gcd(n k , n l ) q (k,l) (for k < l) (and then necessarily b(e k , e l ) = b(e l , e k ) for k > l and b(e k , e k ) = 2q(e k ) as well).
Proof. (Step 0) It suffices to prove this for J finite, since we can write G as the colimit over all finite subsets J 0 ⊂ J, and correspondingly the subset of parameters p (k) , q (k,l) with k, l ∈ J 0 . This is compatible under inclusion of finite subsets of J.
(Step 1) So assume J finite. Let q ∈ Quad(G, C × ) be arbitrary. We first claim that q(e k ) must be a gcd(n 2 k , 2n k )-torsion element in C × , i.e.
for some uniquely determined p (k) ∈ {0, 1, . . . , gcd(n 2 k , 2n k ) − 1}. The proof for this goes as follows: The generator e k spans a subgroup ι : Z/n k Z ֒→ G, so we can pull the quadratic form back to this subgroup.
(8.4) Quad(G, C × ) ι * −→ Quad(Z/n k Z, C × ) ∼ = Hom(Z/(2n k , n 2 k )Z, C × ) The last isomorphism, the determination of quadratic forms on Z/n k Z, goes back to Whitehead. We explain the argument: One can further pull back along Z ։ Z/n k Z and it is easy to see that any quadratic form on Z must have the shape for some m ∈ C × (e.g., use Example 2.3). Here and for the rest of this sub-argument, we stick to the additive notation. One then only needs to check for which m such a q ′ descends to a quadratic form on the quotient Z/n k Z, which will then give Equation 8.4. We claim that this holds whenever m is a gcd(2n k , n 2 k )-torsion element in C × . Necessity: Suppose it descends. Then 0 = q ′ (0) = q ′ (n k ) = n 2 k q ′ (1), again by Example 2.3. Hence, we must have n 2 k m = 0 in C × . Moreover, if q ′ descends to Z/n k Z, so does the This forces 2n k m = 0. Thus, m must be both 2n k -and n 2 k -torsion, i.e. gcd(2n k , n 2 k )-torsion in C × . Conversely, suppose it is. Then q ′ (x + N n k ) = (x + N n k ) 2 m = x 2 m + 2n k N xm + N 2 n 2 k m ≡ x 2 m since the second and third summand vanish because of the torsion assumption. This proves the subclaim. That is: the valid choices for m ∈ C × in Equation 8.5 are precisely the gcd(2n k , n 2 k )-torsion elements. However, these are precisely such as described in Equation 8.3. Next, by bilinearity nb(e k , e l ) = 1 once gcd(n k , n l ) | n (because e k is n k -torsion and e l is n l -torsion), so we may write b(e k , e l ) = exp 2πi gcd(n k , n l ) q (k,l) for some uniquely determined q (k,l) ∈ {0, 1, . . . , gcd(n k , n l ) − 1}. ( Step 2) We have now seen that q(e k ) and b(e k , e l ) can only be of the described forms. This defines a set-theoretic map (8.6) Quad(G, C × ) −→ parameter values p (k) ,q (k,l) (in the ranges described) .
It is also clear that this map is injective. It remains to check that conversely any choice of parameters defines a quadratic form on G. Since we already have an injective map, it suffices to count elements on the right side. For any abelian groups A, B we have To see this, note that for Whitehead's universal quadratic functor Γ and the latter is a quadratic functor, i.e. and this unravels to For the first type of summands we have again used the isomorphism of Equation 8.4, and for the second type of summand note that any linear map of a torsion group to C × must have its image in the torsion of C × and thus certainly in U (1), and Hom(−, U (1)) is just the Pontryagin dual, which (non-canonically) can be identified with the input abelian group. Without spelling out the actual count, it is clear that this set has the same cardinality as our set of parameter values on the right side in Equation 8.6. This finishes the proof. Now the proof of Theorem 8.1 can be done in a similar fashion to the one we used to obtain Quinn's formula.
Proof of Theorem 8.1. We follow the same proof as for Theorem 7.1, so let us just describe how certain details need to be changed. Firstly, we are now in the special case M := C × .
Using our parametrization of quadratic forms of Lemma 8.2, we may rewrite Equation 7.3 in the concrete shape Write [x] n for the remainder in {0, 1, . . . , n − 1} of x ∈ Z under division by n. Rewrite the admissible lifting in Equation 7.5 in the shape This is an admissible lift in the sense of Definition 5.1 (and only optically different from the choice in the proof of Quinn's formula). We compute As in Equation 7.9, for k < l the expression [y l ] n l + [z l ] n l − [y l + z l ] n l is a multiple of n l , so the entire input to the exponential function lies in 2πiZ. Thus, which is what we had claimed.
9. Constructing associators and braidings 9.1. Recollections. As we had explained in the introduction, this note applies to both (a) pointed braided fusion categories over a field k, as well as (b) braided categorical groups. The reason for this is that both types of braided monoidal categories can be classified in terms of (a slight generalization of) pre-metric groups as in [JS93], [DGNO10].
where f, g are group homomorphisms. Write Quad for the category of quadratic triples. A symmetric triple is a quadratic triple such that the polarization of the quadratic form q vanishes. Write Quad sym for the full subcategory of symmetric triples.
Write BCG for the 1-category whose objects are braided categorical groups and whose morphisms are the equivalence classes of braided monoidal functors. 4 Theorem 9.2 (Joyal-Street [JS93,§3]). There is an equivalence of 1-categories where q is defined as follows: For any object X ∈ C the self-braiding s X,X : X ⊗ X ∼ → X ⊗ X induces an automorphism of the tensor unit, namely The idea is as follows: Firstly, one picks a skeleton of the category, using that every braided monoidal category is braided monoidal equivalent to any of its skeleta (with a suitably defined braided monoidal structure on the skeleton). On this skeleton, the associator and braiding s X,Y : X ⊗ Y ∼ −→ Y ⊗ X are automorphisms of objects, because in a skeleton any two mutually isomorphic objects must be the same objec. Thus, a X,Y,Z ∈ M and s X,Y ∈ M for any objects X, Y, Z. The pentagon and hexagon axioms of a braided monoidal structure then become equivalent to the axioms of an abelian 3-cocycle (h, c) (namely Equation 2.1 for the pentagon, and Equations A, A' for the hexagons).
This means that up to braided monoidal equivalence it suffices to work with the following explicit skeletal model in the situation G := π 0 (C, ⊗), M := π 1 (C, ⊗) = Aut C (1 C ) and how h, c enter is explained below: (1) The objects are the elements in G.
(2) The automorphisms of an object X ∈ G are Aut(X) := M , and their composition is the addition of M .
(3) There are no morphisms except for automorphisms (so no other composition of morphisms needs to be defined). (4) The monoidal structure is where addition is just addition in G (on objects) resp. in M (for f, f ′ ). (5) The associator is the automorphism defined by h(X, Y, Z) ∈ M . The associativity of G settles that the objects on either side are the same.
One then checks that changing the abelian 3-cocycle by an abelian 3-coboundary amounts to the structure rendering the identity functor id : T → T a braided monoidal selfequivalence, i.e. only the cohomology class of [(h, c)] ∈ H 3 ab (G, M ) is well-defined when regarding T as an object in BCG. Finally, the latter identifies uniquely with a quadratic form q ∈ Quad(G, M ) via the Eilenberg-Mac Lane isomorphism, Equation 1.2 (or §11). For details regarding the proof of Theorem 9.2 we refer to [JS93,§3] (or the slightly different treatment in [JS86]).
Let BCG sym ⊂ BCG denote the full subcategory of those braided categorical groups which are symmetric monoidal (the braided monoidal equivalences then automatically become symmetric monoidal equivalences). These are also known as Picard groupoids.
Proof. This was originally proven in Sính's thesis [Sín75], and this essentially started the entire subject from scratch. In particular, the formulation as a special case of the Joyal-Street equivalence is anachronistic. Sính defined her own functor and the right side was different (but equivalent to Quad sym ). We explain how the above arises as a special case within the Joyal-Street classification: If the braiding is symmetric, we have and in terms of the abelian 3-cocycle (h, c) built from the braiding and associator this amounts to c(y, x) + c(x, y) = 0. Among all abelian 3-cocycles, this additional constraint isolates the symmetric 3-cocycles, H 3 sym (G, M ) ⊆ H 3 ab (G, M ), but there is a commutative diagram, going back to Whitehead, whose top row is exact (see [Bra19,Lemma 4.10]) where "q → b" refers to the map sending a quadratic form to its polarization. In particular, H 3 sym (G, M ) identifies precisely with those quadratic forms whose polarization vanishes, which was the defining property for Quad sym on the right side and proves the claim. The embedding Hom(G/2G, M ) ֒→ Quad(G, M ) is based on the observation that any linear map G/2G → M satisfies the axioms of a quadratic form (see [Bra19,Lemma 4.10] for details). Now return to the concept of a quadratic triple as in Definition 9.1. Let k be a field. In the special case where G is finite, M := k × and g in Diagram 9.1 is constrained to be the identity map, the definition transforms into the concept of a pre-metric group. We will mildly generalize this and drop the finiteness assumption.
Definition 9.5. For us, a big fusion category is a semisimple rigid k-linear monoidal category (C, ⊗) with finite-dimensional Hom-spaces such that the monoidal unit 1 C is simple. It is pointed if all simple objects are invertible.
An ordinary (non-big) fusion category is a big fusion category such that there are only finitely many isomorphism classes in C; this is the standard definition. The definition of a semisimple category includes that every object decomposes as a finite direct sum of simple objects. Nonetheless, this would have to hold even if we only required that every object is a direct sum of simples: If X ≃ i∈I S i with each S i simple (or merely non-zero), the assumption dim k End(X) < ∞ already forces the index set I to be finite.
If X is a simple object in a pointed big fusion category, it is invertible and thus X −1 is also simple; and if X, Y are both simple, then X ⊗ Y can only have a single simple direct summand. Thus, X ⊗ Y must be simple, too.
Definition 9.6. If (C, ⊗) is a big pointed braided fusion category, let C simp be the full subcategory of simple objects, and keep only the isomorphisms as morphisms. Thus, C simp is a groupoid. Moreover, C simp is braided monoidal by restricting the braided monoidal structure to the subcategory.  (G, q), where G is an abelian group and q ∈ Quad(G, k × ) a quadratic form. A morphism (G, q) → (G ′ , q ′ ) is a commutative diagram where f is a group homomorphism. Write PM k for the category of big pre-metric groups. A symmetric big pre-metric group is a pair (G, q) such that the polarization of the quadratic form vanishes. Write PM k,sym for the corresponding full subcategory. The original definition without the word "big" refers to the same, except that we demand G to be a finite abelian group.
Write PB k for the 1-category whose objects are pointed braided k-linear big fusion categories and whose morphisms are the equivalence classes of k-linear braided monoidal functors. Write PS k ⊂ PB k for the full subcategory of those big fusion categories which are symmetric monoidal. where q is defined as follows: For any simple object X ∈ C the self-braiding s X,X : X ⊗ X ∼ → X ⊗ X induces an automorphism of the tensor unit, namely and q([X]) := (s X,X ⊗ X −1 ⊗ X −1 ) ∈ Aut(1 C ) ∼ = k × extends to a well-defined quadratic form on π 0 (C, ⊗).
Proof. This result essentially reduces to the classification of Joyal-Street of braided categorical groups. The main differences are as follows: (a) Since 1 C is simple by assumption, End(1 C ) must be a division algebra over k by Schur's Lemma. As k is algebraically closed, we must have End(1 C ) = k and thus Aut(1 C ) = k × . The category C simp thus is a braided categorical group with π 1 (C simp ) = k × . (b) Since the braided monoidal functors between big fusion categories are assumed k-linear, they correspond to morphisms of quadratic triples as in Figure 9.1, but must be the identity on k × .
See also [EGNO15,Theorem 8.4.12] for a direct proof which differs in certain parts. In analogy to Theorem 9.4 one obtains the following. Write π i := π i (C simp , ⊗) in situation (a), and π i := π i (C, ⊗) in situation (b). Then the following statements are equivalent: (1) (C, ⊗) is braided monoidal equivalent to a simultaneously skeletal and strictly associative braided monoidal category.
(3) The abelian 3-cocycle of (C, ⊗) admits a cocycle representative (h, c) such that h : π 0 × π 0 × π 0 −→ π 1 and c : π 0 × π 0 −→ π 1 are trilinear resp. bilinear in π 0 . (4) There exists a bilinear form S on π 0 such that Proof. We discuss the braided categorical group case. Suppose (G, M, q) is the quadratic triple attached to (C, ⊗) by Theorem 9.2 (1 ⇒ 2) Suppose (C, ⊗) is braided monoidal equivalent to a skeletal and strictly associative braided monoidal category. Being skeletal, it is of the form of a skeletal model T (G, M, (h, c)), Definition 9.3, where (h, c) is an abelian 3-cocycle. Since the model is strictly associative, the maps is an abelian 3-cocycle representative for the quadratic form q for any admissible lifting (−). Since F 0 = G, we can just take the identity as a lifting. In particular, L(x, y) = (x + y) − x − y = 0, so h(x, y, z) = 0 for all x, y, z ∈ G. It follows that the skeletal model T (G, M, (0, c)) is braided monoidal equivalent to (C, ⊗), and this is skeletal, and thanks to h = 0 the associators are trivial. In the big fusion category case, instead work with the triple (G, k × , q) of Theorem 9.8, and replace the skeletal model T (G, M, (h, c)) by its natural k-linear analogue.
In the symmetric monoidal situation the equivalent properties of Theorem 9.10 are always satisfied.
Theorem 9.11 (Johnson-Osorno [JO12]). Let (C, ⊗) be • a k-linear pointed symmetric fusion category with k algebraically closed, or • a Picard groupoid. Then (C, ⊗) is symmetric monoidal equivalent to a simultaneously skeletal and strictly associative symmetric monoidal category. More precisely, all characterizations in Theorem 9.10 are always met in this setting.
This was originally observed in the context of Picard groupoids [JO12, Theorem 2.2] with a different method.
Proof. We apply the equivalence of categories in the symmetric setting, Theorem 9.4, so (C, ⊗) corresponds to a symmetric triple (G, M, q) in Quad sym , or (G, k × , q) in the fusion category setting. Being in Quad sym , we know that the polarization of q is trivial, i.e.
showing that q is a linear map (the complete deduction of this is as we had done it around Equation 3.7). As the above correspondence of Theorem 9.4 is just a special case of the Joyal-Street equivalence of Theorem 9.2, we can also use Theorem 9.10 in this setting. The conclusion (4)⇒(1) shows that our claim is proven if we can show that q comes from a bilinear form. Define (9.4) S(x, y) = i x i y i q(γ i ) with respect to the coordinates x i , y i ∈ F 2 in any chosen basis (γ i ) i∈I of G/2G as an F 2vector space. Since q(−γ i ) = q(γ i ) holds for any quadratic form, the linearity of q yields 2q(γ i ) = 0, so the scalar multiplication with elements from F 2 in Equation 9.4 is well-defined and S is indeed bilinear. Then S(x, x) = x 2 i q(γ i ) ≡ x i q(γ i ) = q(x), proving the claim. The remaining characterizations follow since by Theorem 9.10 they are all equivalent.
As a further refinement of the characterization in Theorem 9.10 relating the possibility to trivialize associators to the linearity of the braiding, we can show the following.
Theorem 9.12. Suppose (C, ⊗) is a braided categorical group such that π 1 (C, ⊗) is a divisible group. Then (C, ⊗) is braided monoidal equivalent to a skeletal model as in Definition 9.3 such that h(x, y, z) = c(x, y) + c(x, z) − c(x, y + z) holds for the abelian 3-cocycle. That is: For any objects X, Y, Z we have (9.5) a X,Y,Z = s X,Y + s X,Z − s X,Y ⊗Z and a Z,X,Y = s X⊗Y,Z − s X,Z − s Y,Z .
We call this a normal form skeletal model.
We do not know whether this is also true for general π 1 (C, ⊗).
Analogously, we obtain: Theorem 9.13. Suppose k is an algebraically closed field. Let (C, ⊗) be a k-linear pointed braided big fusion category. Then (C, ⊗) is braided monoidal equivalent to a skeletal big fusion category such that a X,Y,Z = s X,Y · s X,Z s X,Y ⊗Z and a Z,X,Y = s X⊗Y,Z s X,Z · s Y,Z hold for all simple objects X, Y, Z.
Proof. The only difference to the proof for Theorem 9.12 is that we need to show that π 1 (C, ⊗) = k × is divisible, but this just amounts to the existence of n-th roots n √ x for any n ≥ 1 and x ∈ k × , which holds since k is algebraically closed. explicit coboundary formula would then be useful for, afterwards. So, we think the approach of Joyal and Street is just right.

Afterword
The following question appears very natural to me, but I have not been able to determine the answer (perhaps embarrassingly).
Problem 2. Suppose G is a torsion-free abelian group and M an arbitrary abelian group. Is any quadratic form q : G → M of the shape x → S(x, x) for S some bilinear form?