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 <td
style="text-align: center; vertical-align: top; color: rgb(0, 0, 102);"><big>About HAP: Explicit cocycles 
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 <td
style="vertical-align: top; background-color: rgb(255, 255, 255); text-align: left;">Given
a ZG-resolution R and a ZG-module A, one defines an n-cocycle to be a ZG-homomorphism
f:Rn → A for which the composite homomorphism fdn+1:Rn+1→A
is zero. If R happens to be the standard bar resolution (i.e. the
cellular chain complex of the nerve of the group G considered as a one
object category) then the free ZG-generators of Rn are
indexed by n-tuples (g1 | g2 | ... | gn)
of elements gi in G. In this case we say that the n-cocycle
is a standard n-cocycle and
we think of it as a set-theoretic function 
 <div style="text-align: center;">f : G × G × ...
× G → A 
 <div style="text-align: left;">satisfying a certain algebraic
cocycle
condition. 
 
Bearing in mind that a standard n-cocycle really just assigns an
element f(g1,
... ,gn) of A to an n-simplex in the nerve of G, the cocycle
condition is a very natural one which states that "f must vanish on the
boundary of a certain
(n+1)-simplex". For n=2 the condition is that a 2-cocycle f(g₁,g₂)
must satisfy g.f(h,k) + f(g,hk) = f(gh,k) + f(g,h) for all g,h,k in G.
This equation is explained by the following picture. 
 
 <div style="text-align: center;"><img alt="" src="cocycle.jpg"
style="width: 576px; height: 257px;"> 
 <div style="text-align: left;"> 
 
 
The definition of a cocycle clearly depends on the choice of
ZG-resolution R. However, the cohomology group Hn(G,A),
which is a group of equivalence classes of n-cocycles, is independent
of the choice of R. 
 
There are some occasions when one needs explicit examples of standard
cocycles. For instance: 
 <ul>
 <li>Let G be a finite group and K a field of characteristic 0.
The group algebra K(G) and the algebra F(G) of functions đg:G→K,
h→đg,h are both Hopf algebras. The tensor product F(G)
(×) K(G) also admits a Hopf algebra structure known as the quantum double D(G). A twisted quantum double Df(G)
was introduced in [R. Dijkraaf, V. Pasquier & P. Roche, "Quasi-Hopf
algebras, group cohomology and orbifold models", Nuclear Phys. B
Proc. Suppl. 18B (1991) 60-72]. The twisted double is a quasi-Hopf
algebra depending on a 3-cocycle f:G×G×G→K. The
multiplication is given by (đg (×) x)(đh
(×) x) = đgx,xhßg(x,y)(đg
(×) xy) where ßa is defined by ßa(h,g)
= f(a,h,g) f(h,h-1ah,g)-1 f(h,g,(hg)-1ahg)
. Although the algebraic structure of Df(G) depends
very much on the particular 3-cocycle f, representation theoretic
properties of Df(G) depend only on the cohomology class of f.</li>
 <li>An explicit 2-cocycle f:G×G→A is needed to construct
the multiplication (a,g)(a',g') = (a + g·a' + f(g,g'),
gg') in the extension a group G by a ZG-module A determined by the
cohomology class of f in H2(G,A).</li>
 <li>In work on coding theory and Hadamard matrices a number of
papers have investigated square matrices (aij) whose
entries aij=f(gi,gj) are the values of
a 2-cocycle f:G×G → Z2 where G is a finite group
acting trivially on Z2. See for instance [K.J.
Horadam, "An introduction to cocyclic generalised Hadamard matrices", Discrete Applied Math, 102
(2000) 115-130].</li>
 </ul>
 </div>
 </div>
 </div>
 </div>
 </td>
 </tr>
 <tr>
 <td
style="vertical-align: top; background-color: rgb(255, 255, 255);">Given
a ZG-resolution R (with contracting homotopy) and a ZG-module A one can
use HAP commands to compute explicit standard n-cocycles f:Gn
→ A. With the twisted quantum double in mind, we illustrate the
computation for n=3, G=S3, and A=U(1)
the group of complex numbers of modulus 1 with trivial G-action. 
 
We first compute a ZG-resolution R. The Universal Coefficient Theorem
gives an isomorphism H3(G,U(1)) = HomZ(H3(G,Z),
U(1)), The multiplicative group U(1) can thus be viewed as Zm
where m is a multiple of the exponent of H3(G,Z).</td>
 </tr>
 <tr>
 <td
style="background-color: rgb(255, 255, 204); vertical-align: top;">gap>
G:=SymmetricGroup(3);; 
gap> R:=ResolutionFiniteGroup(G,4);; 
gap> TR:=TensorWithIntegers(R);; 
gap> Homology(TR,3); 
[ 6 ] 
 
gap> R.dimension(3); 
4 
gap> R.dimension(4); 
5 
 </td>
 </tr>
 <tr>
 <td
style="background-color: rgb(255, 255, 255); vertical-align: top;">We
thus replace the very infinite group U(1) by the finite cyclic group Z6.
Since the resolution R has 4 generators in degree 3, a homomorphism f:R3→U(1)
can be represented by a list f=[f1, f2, f3,
f4]
with fi is the image in Z6 of the ith
generator. The cocycle condition on f can
be expressed as a matrix equation 
 
 <div style="text-align: center;">Mf = 0 modulo 6 
 </div>
 
where the matrix M is got from the following command. </td>
 </tr>
 <tr>
 <td
style="background-color: rgb(255, 255, 204); vertical-align: top;">gap>
M:=CocycleCondition(R,3);; 
 </td>
 </tr>
 <tr>
 <td
style="text-align: left; background-color: rgb(255, 255, 255); vertical-align: top;">A
particular cocycle f=[f1, f2, f3, f4]
can be got by
choosing a solution to the equation Mf=0. 
 </td>
 </tr>
 <tr>
 <td
style="vertical-align: top; background-color: rgb(255, 255, 204);">gap>
SolutionsMod2:=NullspaceModQ(TransposedMat(M),2); 
[ [ 0, 0, 0, 0 ], [ 0, 0, 1, 1 ], [ 1, 1, 0, 0 ], [ 1, 1, 1, 1 ] ] 
 
gap> SolutionsMod3:=NullspaceModQ(TransposedMat(M),3); 
[ [ 0, 0, 0, 0 ], [ 0, 0, 0, 1 ], [ 0, 0, 0, 2 ], [ 0, 0, 1, 0 ], 
 [ 0, 0, 1, 1 ], [ 0, 0, 1, 2 ], [ 0, 0, 2, 0 ], [ 0, 0, 2, 1 ], 
 [ 0, 0, 2, 2 ] ] 
 </td>
 </tr>
 <tr>
 <td
style="vertical-align: top; background-color: rgb(255, 255, 255);">A
non-standard 3-cocycle f can be converted to a standard one using the
command StandardCocycle(R,f,n,q)
. This command inputs R, integers n and q, and an n-cocycle f for the
resolution R. It returns a standard cocycle Gn → Zq. 
 </td>
 </tr>
 <tr>
 <td
style="vertical-align: top; background-color: rgb(255, 255, 204);">gap>
f:=3*SolutionsMod2[3] -
SolutionsMod3[5];
#An example solution to Mf=0 mod 6. 
[ 3, 3, -1, -1 ] 
 
gap> Standard_f:=StandardCocycle(R,f,3,6);; 
 
gap> g:=Random(G); h:=Random(G); k:=Random(G); 
(1,2) 
(1,3,2) 
(1,3) 
 
gap> Standard_f(g,h,k); 
3 
 </td>
 </tr>
 <tr>
 <td
style="vertical-align: top; background-color: rgb(255, 255, 255);">A
function f: G×G×G → A is a standard 3-cocycle if and only
if 
 
 <div style="text-align: center;">g·f(h,k,l) -
f(gh,k,l) + f(g,hk,l) - f(g,h,kl) + f(g,h,k) = 0 
 
 <div style="text-align: left;">for all g,h,k,l in G. In the above
example the group G=S3
acts trivially on A=Z6. The following commands show that the
standard 3-cocycle produced in the example really does satisfy this
3-cocycle condition. 
 </div>
 </div>
 </td>
 </tr>
 <tr>
 <td
style="vertical-align: top; background-color: rgb(255, 255, 204);">gap>
sf:=Standard_f;; 
 
gap> Test:=function(g,h,k,l); 
> return sf(h,k,l) - sf(g*h,k,l) + sf(g,h*k,l) - sf(g,h,k*l) +
sf(g,h,k); 
> end; 
function( g, h, k, l ) ... end 
 
gap> for g in G do for h in G do for k in G do for l in G do 
> Print(Test(g,h,k,l),","); 
> od;od;od;od; 
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,6,0,6,6,0,0,6, 
0,0,0,0,0,6,6,6,0,6,0,12,12,6,12,6,0,12,6,0,6,6,0,0,0,0,0,0,0,12,12,6,6,6,0, 
6,6,0,6,6,0,0,-6,0,0,0,0,0,0,0,0,0,0,6,6,6,6,6,0,0,0,0,0,0,0,6,0,0,6,6,0,6,6, 
0,6,0,0,6,6,6,0,0,0,0,0,0,0,-6,0,0,-6,0,-6,0,0,0,0,0,0,0,0,6,6,0,6,0,0,6,0,0, 
0,0,0,6,6,6,0,0,0,6,6,6,0,0,0,0,-6,0,6,6,0,0,0,0,0,0,0,12,6,6,0,6,0,0,0,0,12, 
6,0,0,0,0,0,0,0,6,6,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,6,0,0,6,0,0,6,0,0,0,0,0,6,6, 
6,0,0,0,6,12,6,6,0,0,0,-6,0,0,6,0,0,0,0,0,0,0,12,12,6,6,6,0,0,0,0,6,6,0,0,0, 
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,6,0,0,6,0,6,0,0,0,0,0,0, 
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,6,6,6,6,6,0, 
6,6,0,6,6,0,12,12,6,12,12,0,0,0,0,0,0,0,6,6,0,0,0,0,6,6,6,12,12,0,-6,-6,0,0, 
0,0,6,6,0,0,6,0,0,6,0,6,6,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,6,0,6,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,6,6, 
0,6,0,0,6,0,0,0,0,0,0,0,0,0,0,0,6,6,0,6,0,0,6,0,0,0,0,0,0,0,0,0,0,0,0,0,6,0, 
0,0,0,0,0,0,0,0,0,0,0,0,0,0,6,0,0,6,6,0,6,6,0,6,0,0,6,6,6,0,0,0,0,0,0,0,0,0, 
0,0,0,0,0,0,0,0,0,0,0,0,0,6,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 
0,6,6,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,6,6,0,0,0,0,0,0,0,6,6,0,0,0,0,0,0,0,0, 
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 
0,0,0,0,0,0,0,0,0,0,-6,0,6,0,6,0,6,0,0,0,0,0,0,0,12,12,6,12,12,0,6,6,0,6,6,0, 
0,0,0,0,0,0,12,12,6,12,12,0,6,6,0,6,6,0,0,0,0,0,0,0,0,0,0,0,0,0,6,6,6,6,6,0, 
0,0,0,0,0,0,6,0,0,6,6,0,6,6,0,6,0,0,6,6,6,0,0,0,-6,0,0,0,-6,0,0,-6,0,-6,0,0, 
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,6,6,6,6,6,0,6,6,0,0,0,0,0,0,0,6,6,0,0,0, 
0,0,0,0,6,6,0,-6,0,0,-6,0,0,12,6,0,-6,-6,0,0,0,0,6,6,0,0,6,0,0,6,0,6,6,0,0,0, 
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 
0,0,0,-6,0,0,0,0,0,0,0,0,0,0,6,6,6,6,6,0,6,12,0,6,0,0,6,0,0,0,6,0,0,0,0,0,0, 
0,6,12,0,0,0,0,0,0,0,6,6,0,-6,-6,0,0,0,0,0,0,0,0,6,0,0,6,0,6,6,0,0,0,0,0,0,0, 
6,0,0,0,6,0,0,6,0,6,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,6, 
0,0,0,0,0,0,0,0,6,0,0,0,0,0,0,0,6,6,0,6,6,0,6,6,6,12,12,0,0,0,0,0,0,0,6,6,0, 
6,6,0,6,6,6,12,12,0,0,0,0,0,0,0,6,6,0,0,6,0,0,6,0,6,6, 
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