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<td style="text-align: center; vertical-align: top; color: rgb(0, 0, 102);"><big><span style="font-weight: bold;">About HAP: Explicit cocycles<br>
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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 <span style="font-style: italic;">n-cocycle</span> to be a ZG-homomorphism
f:R<sub>n </sub>→ A for which the composite homomorphism fd<sub>n+1</sub>:R<sub>n+1</sub>→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 R<sub>n</sub> are
indexed by n-tuples (g<sub>1</sub> | g<sub>2</sub> | ... | g<sub>n</sub>)
of elements g<sub>i</sub> in G. In this case we say that the n-cocycle
is a <spanstyle="font-style: italic;">standard n-cocycle </span>and
we think of it as a set-theoretic function <br>
<divstyle="text-align: center;">f : G × G × ...
× G → A<br>
<divstyle="text-align: left;">satisfying a certain algebraic
cocycle
condition. <br>
<br>
Bearing in mind that a standard n-cocycle really just assigns an
element f(g<sub>1</sub>,
... ,g<sub>n</sub>) 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.<br>
<br>
<divstyle="text-align: center;"><img alt="" src="cocycle.jpg" style="width: 576px; height: 257px;"><br>
<divstyle="text-align: left;"><br>
<br>
<br>
The definition of a cocycle clearly depends on the choice of
ZG-resolution R. However, the cohomology group H<sup>n</sup>(G,A),
which is a group of equivalence classes of n-cocycles, is independent
of the choice of R. <br>
<br>
There are some occasions when one needs explicit examples of standard
cocycles. For instance: <br>
<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 đ<sub>g</sub>:G→K,
h→đ<sub>g,h</sub> are both Hopf algebras. The tensor product F(G)
(×) K(G) also admits a Hopf algebra structure known as the <span style="font-style: italic;">quantum double</span> D(G). A <span style="font-style: italic;">twisted quantum double</span> D<sup>f</sup>(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 (đ<sub>g</sub> (×) x)(đ<sub>h</sub>
(×) x) = đ<sub>gx,xh</sub>ß<sub>g</sub>(x,y)(đ<sub>g</sub>
(×) xy) where ß<sub>a</sub> is defined by ß<sub>a</sub>(h,g)
= f(a,h,g) f(h,h<sup>-1</sup>ah,g)<sup>-1</sup> f(h,g,(hg)<sup>-1</sup>ahg)
. Although the algebraic structure of D<sup>f</sup>(G) depends
very much on the particular 3-cocycle f, representation theoretic
properties of D<sup>f</sup>(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 H<sup>2</sup>(G,A).</li>
<li>In work on coding theory and Hadamard matrices a number of
papers have investigated square matrices (a<sub>ij</sub>) whose
entries a<sub>ij</sub>=f(g<sub>i</sub>,g<sub>j</sub>) are the values of
a 2-cocycle f:G×G → Z<sub>2</sub> where G is a finite group
acting trivially on Z<sub>2</sub>. See for instance [K.J.
Horadam, "An introduction to cocyclic generalised Hadamard matrices", <span style="font-style: italic;">Discrete Applied Math, </span>102
(2000) 115-130].</li>
</ul>
</div>
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<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:G<sup>n</sup>
→ A. With the twisted quantum double in mind, we illustrate the
computation for n=3, G=S<sub>3</sub>, and A=U(1)
the group of complex numbers of modulus 1 with trivial G-action. <br>
<br>
We first compute a ZG-resolution R. The Universal Coefficient Theorem
gives an isomorphism H<sup>3</sup>(G,U(1)) = Hom<sub>Z</sub>(H<sub>3</sub>(G,Z),
U(1)), The multiplicative group U(1) can thus be viewed as Z<sub>m</sub>
where m is a multiple of the exponent of H<sub>3</sub>(G,Z).</td>
</tr>
<tr>
<td style="background-color: rgb(255, 255, 204); vertical-align: top;">gap>
G:=SymmetricGroup(3);;<br>
gap> R:=ResolutionFiniteGroup(G,4);;<br>
gap> TR:=TensorWithIntegers(R);;<br>
gap> Homology(TR,3);<br>
[ 6 ]<br>
<br>
gap> R.dimension(3);<br>
4<br>
gap> R.dimension(4);<br>
5<br>
</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 Z<sub>6</sub>.
Since the resolution R has 4 generators in degree 3, a homomorphism f:R<sub>3</sub>→U(1)
can be represented by a list f=[f<sub>1</sub>, f<sub>2</sub>, f<sub>3</sub>,
f<sub>4</sub>]
with f<sub>i</sub> is the image in Z<sub>6</sub> of the ith
generator. The cocycle condition on f can
be expressed as a matrix equation<br>
<br>
<divstyle="text-align: center;">Mf = 0 modulo 6<br>
</div>
<br>
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);;<br>
</td>
</tr>
<tr>
<td style="text-align: left; background-color: rgb(255, 255, 255); vertical-align: top;">A
particular cocycle f=[f<sub>1</sub>, f<sub>2</sub>, f<sub>3</sub>, f<sub>4</sub>]
can be got by
choosing a solution to the equation Mf=0.<br>
</td>
</tr>
<tr>
<td style="vertical-align: top; background-color: rgb(255, 255, 204);">gap>
SolutionsMod2:=NullspaceModQ(TransposedMat(M),2);<br>
[ [ 0, 0, 0, 0 ], [ 0, 0, 1, 1 ], [ 1, 1, 0, 0 ], [ 1, 1, 1, 1 ] ]<br>
<br>
gap> SolutionsMod3:=NullspaceModQ(TransposedMat(M),3);<br>
[ [ 0, 0, 0, 0 ], [ 0, 0, 0, 1 ], [ 0, 0, 0, 2 ], [ 0, 0, 1, 0 ],<br>
[ 0, 0, 1, 1 ], [ 0, 0, 1, 2 ], [ 0, 0, 2, 0 ], [ 0, 0, 2, 1 ],<br>
[ 0, 0, 2, 2 ] ]<br>
</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 <spanstyle="font-family: helvetica,arial,sans-serif;">StandardCocycle(R,f,n,q)</span>
. This command inputs R, integers n and q, and an n-cocycle f for the
resolution R. It returns a standard cocycle G<sup>n</sup> → Z<sub>q</sub>.<br>
</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.<br>
[ 3, 3, -1, -1 ]<br>
<br>
gap> Standard_f:=StandardCocycle(R,f,3,6);;<br>
<br>
gap> g:=Random(G); h:=Random(G); k:=Random(G);<br>
(1,2)<br>
(1,3,2)<br>
(1,3)<br>
<br>
gap> Standard_f(g,h,k);<br>
3<br>
</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 <br>
<br>
<divstyle="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 <br>
<br>
<divstyle="text-align: left;">for all g,h,k,l in G. In the above
example the group G=S<sub>3</sub>
acts trivially on A=Z<sub>6</sub>. The following commands show that the
standard 3-cocycle produced in the example really does satisfy this
3-cocycle condition. <br>
</div>
</div>
</td>
</tr>
<tr>
<td style="vertical-align: top; background-color: rgb(255, 255, 204);">gap>
sf:=Standard_f;;<br>
<br>
gap> Test:=function(g,h,k,l);<br>
> return sf(h,k,l) - sf(g*h,k,l) + sf(g,h*k,l) - sf(g,h,k*l) +
sf(g,h,k);<br>
> end;<br>
function( g, h, k, l ) ... end<br>
<br>
gap> for g in G do for h in G do for k in G do for l in G do<br>
> Print(Test(g,h,k,l),",");<br>
> od;od;od;od;<br>
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,<br>
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,<br>
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,<br>
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,<br>
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,<br>
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,<br>
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,6,0,6,6,0,0,6,<br>
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,<br>
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,<br>
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,<br>
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,<br>
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,<br>
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,<br>
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,<br>
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,6,0,0,6,0,6,0,0,0,0,0,0,<br>
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,6,6,6,6,6,0,<br>
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,<br>
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,<br>
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,<br>
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,<br>
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,<br>
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,<br>
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,<br>
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,<br>
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,<br>
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,<br>
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,<br>
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,<br>
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,<br>
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,<br>
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,<br>
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,<br>
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,<br>
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,<br>
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,<br>
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