Twisted category

SupposeG, M are multiplicative abelian groups. ThenM can be regarded as a trivial

G-module (via the actionxµ=µforx∈G,µ∈M ). Anabelian3−cocycleforGwith coefficients in M is a pair (σ, τ) , where σ:G3 →M is a normalised 3-cocycle

σ(a,1, b) = 1

and τ :G2 _{→M is a function satisfying :}

σ(b, c, a)τ(a, bc)σ(a, b, c) =τ(a, c)σ(b, a, c)τ(a, b) (3.28)

σ(c, a, b)−1τ(ab, c)σ(a, b, c)−1 =τ(a, c)σ(a, c, b)−1τ(b, c). (3.29)

wherea, b, c∈G.

A 2-cochain is a function θ:G2→M satisfying :

θ(a,1) =θ(1, b) = 1.

The coboundary ofθ is the abelian 3-cocycle∂(θ) = (σ, τ) defined by the equations

σ(a, b, c) =θ(b, c)θ(ab, c)−1θ(a, bc)θ(a, b)−1 (3.30)

τ(a, b) =θ(a, b)θ(b, a)−1. (3.31)

ThenH3

ab(G, M) is the abelian group of 3-cocycles modulo the coboundaries. It might

look difficult to compute these groups but Theorem 3.6.4 comes for our rescue. After proving the theorem, we will, in particular, compute these groups when G is a cyclic group.

Suppose K is a commutative ring with unit. Suppose (σ, τ) is an abelian 3-cocycle on G with coefficients in the multiplicative group K×. We construct a twisted category

Aσ,τ_G ( or simply A) ofG-gradedK-modules. The objects ofA are

L= M

α∈G

Lα

where {Lα}α∈G is a family of K-modules. The arrows are direct sums of K-module

homomorphisms. The tensor on this category is given by the following formula:

(L⊗M)α= X β.γ=α

Lβ⊗Mγ.

braiding cinA

a((x⊗y)⊗z) =σ(α, β, γ)(x⊗(y⊗z))

c(x⊗y) =τ(α, β)y⊗x

wherex∈Lα, y ∈Mβ, z∈Nγ.We then extend the definition by linearity to define it on the objects of the category. In this way, Abecomes a monoidal category ofG-gradedK- modules. In other words, A is a monoidal category ofKG-comodules with associativity given byσ.

Proposition 3.6.3 Let (σ, τ) and (σ0, τ0) be any two representatives of their equivalence classes. If [σ, τ] = [σ0, τ0] ∈ H_ab3(G, K×), then we get a braided tensor isomorphism between the categories Aσ,τ_G and Aσ

0

,τ0 G .

Proof: Consider the identity functor I from the category Aσ,τG to the category A σ0,τ0 G .

We claim that it is in fact a braided tensor isomorphism. Let k:G2 _{→M be a function}

satisfying k(a,1) =k(1, b) = 1, such that

(σ, τ)(σ0, τ0)−1 =∂(k).

Define tensor isomorphism via the natural transformationφL,M :I(L)⊗I(M)→I(L⊗M)

asφL,M(Plβ⊗mγ) =Pk(β, γ)(lβ⊗mγ), wherelβ ∈Lβ and mγ ∈Mγ forα, β ∈π. Let

us first show that φis a tensor functor. Given any three objectsM, N, L in the category

Aσ,τ_G , we want the the following condition is satisfied

a·(φM⊗N,L)·(φM,N ⊗1L) = φM,(N⊗L)·(1M ⊗φN,L)·a

0

.

For anyb, c, e∈G, when a typical element P

above, the left hand side becomes : = a·(φM⊗N ,L)·(φM,N ⊗1L)( X (mb⊗nc)⊗le) = a·(φM⊗N ,L)( X φM,N(mb⊗nc)⊗le)) = a·(φM⊗N ,L)(Xk(b, c)(mb⊗nc)⊗le) = a(Xk(bc, e)(Xk(b, c)(mb⊗nc)⊗le) = Xσ(b, c, e)k(bc, e)k(b, c)(mb⊗(nc⊗le))

and after applying P

(mb ⊗nc)⊗le to the equation above, the right hand side of the

equation becomes : = φM,N⊗L·(1M ⊗φN,L)·a 0 (X(mb⊗nc)⊗le) = φM,N⊗L·(1M ⊗φN,L)( X σ0(b, c, e)mb⊗(nc⊗le)) = φM,N⊗L( X σ0(b, c, e)mb⊗φN,L(nc⊗le)) = φM,N⊗L( X σ0(b, c, e)k(c, e)mb⊗(nc⊗le)) = Xσ0(b, c, e)k(b, ce)k(c, e)(mb⊗(nc⊗le))

The coefficients of the left side σ(b, c, e)k(bc, e)k(b, c) equals the coefficients on the right side σ0(b, c, e)k(b, ce)k(c, e) using (3.15) and (σ, τ)(σ0, τ0)−1 =∂(k). To show that the isomorphism is also braided we need to show the following

φN,M ·c

0

M,N = cM,N ·φM,N

For any b, c ∈G and using the equation τ(b, c) = k(b, c)k−1(c, b) one can show that

the isomorphism is also braided. Applying a general element P

mb⊗nc, the left hand

side of the equation above becomes :

= φM,N ·c 0 M,N( X mb⊗nc) = φM,N(Xτ0(b, c)nc⊗mb) = Xτ0(b, c)k(c, b)nc⊗mb

Applying P

mb⊗nc, the right hand side of the equation above becomes :

= cM,N ·φM,N( X mb⊗nc) = cM,N( X k(b, c)mb⊗nc = Xτ(b, c)k(b, c)nc⊗mb

The coefficient of left sideτ0(b, c)k(c, b) equals the coefficient on right sideτ(b, c)k(b, c) using (3.16) and (σ, τ)(σ0, τ0)−1=∂(k). This completes the proof.

2

We recall the definition of a quadratic function. A function q:G→M between two abelian groupsG, M is called quadratic whenq(xy)q(x)−1q(y)−1 is a bilinear function of

x, y. This amounts to the following two conditions: (i) q(xyz) q(xy)−1q(yz)−1q(zx)−1q(x) q(y) q(z) = 1,

(ii) q(x) =q(x−1_).

Now to any abelian 3-cocycle (σ, τ) we assign the function q(x) = τ(x, x) ∈ M as

its trace. By (3.21), the trace of a coboundary is identity, and one can show that the

trace satisfies the above two identities. Thus traces are quadratic functions. We have the following result which appears in several papers without a proof. We discuss its proof here:

Theorem 3.6.4 ([Mac52], [Eil52], [EM53]) The function assigning to each abelian 3- cocycle its trace induces an isomorphism of H_ab3(G, M)to the group of all quadratic func-

tions on G toM.

Proof: For a fixed M, trace is natural in abelian groups G. Note that every abelian

group is a filtered colimit of finitely generated abelian groups. Moreover, every finitely generated abelian group is a finite direct sum of cyclic groups. So it suffices to verify the isomorphism whenG is cyclic as quadratic functions intoM preserve filtered colimits.

Surjectivity: Suppose q : G → M is a quadratic function. Put ν = q(1). Define

τ(x, y) = νxy. If G is infinite, (1, τ) is an abelian 3-cocycle whose trace is q. If G has ordern, notice thatν2n=νn2 = 1 in order forqto be well defined. Letσ be the 3-cocycle

defined with trace q. The standard form for such a σ is (cf. Chapter-2, Section 2.3) σ(x, y, z) =      1 fory+z < n νxn fory+z≥n (3.32)

Then (σ, τ) is an abelian 3-cocycle.

Injectivity: It is done in the following two cases.

Case 1 : G = Z. Let (σ, τ) be an abelian 3-cocycle such that τ(x, x) = 1. Since

H3(G, M) = 1, we haveσ =dθ for some 2-cochain θ. Let

(σ0, τ0) = (σ, τ)∂θ−1= (1, τ0).

Thenτ0(x, y) =τ(x, y)θ(x, y)−1θ(y, x).Substitutingσ0 = 1 in (3.28) and (3.29), we get

τ0(x, yz) =τ0(x, z)τ0(x, y),

τ0(xy, z) =τ0(x, z)τ0(y, z).

Equating for τ0(x, z) in the above two equations, we get

τ0(xy, z)τ0(y, z) =τ0(xy, z)τ0(x, y).

Substitutex= 1,z=y−1 in the above equation, we get

τ0(1,1)τ0(y, y−1) =τ0(y, y−1)τ0(1, y).

Thus,τ0(1, y) = 1. Similarly we can showτ0(y,1) = 1. Then using (3.28) and (3.29) and thatσ0 = 1, τ0(x, y) = 1 follows by induction.

Case 2 : G = Zn. Let (σ, τ) be an abelian 3-cocycle such that τ(x, x) = 1. Choose

a 2-cochain θ such that (σ0, τ0) = (σ, τ)∂θ−1 and σ0 is standard for some ν, as in (3.32). Similar to the Case (1), we have τ0(x, x) = 1. Substitute b = 1 in (3.29). We get

Since Gis abelian (3.32) implies that σ0(1, c, a) =σ0(1, a, c). Thus, from the above equation,τ0(a,1) = 1. Similarly we can show τ0(1, a) = 1. Then using again (3.28) and (3.29) and the standard form ofσ0, one deduces by induction thatτ0(x, y) = 1. Then the standard form for σ0 reduces to 1. This proves injectivity.

2

We are now in a better position to calculate H_ab3 (G, M) using the above theorem. In the case when G is an infinite cyclic group Z, let q : G → M be a quadratic function.

Then it is simply a group homomorphism. Indeed the group of all quadratic functions on

Z to any abelian group M is M itself, thus Hab3 (Z, M) =M. In the case when G=Zn

is a cyclic group of order n, let q : Zn → M be a quadratic function on G. Then it is

simply a group homomorphism such that q(1)2n =q(1)n2 = 1 for q to be a well defined quadratic function. Thus the group of all quadratic functions onZ/ntoM is isomorphic

to Hom Z/(n2,2n), M. Hence,

H_ab3 (Z/n, M)∼= Hom Z/(n2,2n), M.

Consider for example M asC× and n= 6 so that,

H_ab3(Z/6,C×)∼= Hom (Z/(n2,2n),C×) = Hom (Z/12,C×)∼=Z/12.

An explicit formula for defining a 3-cocycle can be worked out using the proof of Theorem (3.6.4). Accordingly, we define a 3-cocycle (h,c) as follows. Let ν = e2πι/12 ∈ _C× _then ν62 =ν2×6= 1. Define, h(x, y, z) =      1 foryz < n νxn foryz≥n c(x, y) =νxy.

where x, y, z ∈ _Z/n and ν = c(1,1). One can easily check that (h, c) so defined is an abelian 3-cocycle.