Revisions to Explicit 3-cocycles for the symmetric group $S_6$

improved the GAP example

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edited Apr 2, 2012 at 17:26

426
3
5

I hope it isit's ok to advertize some GAP code on this site.

$R_*: \cdots \rightarrow R_4 \rightarrow R_3 \rightarrow R_2 \rightarrow R_1 \rightarrow R_0$$$R_*: \cdots \rightarrow R_4 \rightarrow R_3 \rightarrow R_2 \rightarrow R_1 \rightarrow R_0$$

A 3-cocycle with coefficients in $U(1)$ is a $\mathbb ZG$-linear homomorphism $f\colon R_3 \rightarrow U(1)$$c\colon R_3 \rightarrow U(1)$ such that the composite $R_4 \rightarrow R_3 \rightarrow U(1)$ is trivial. Using the Universal Coefficient Theorem I thinkand the divisibility of $U(1)$ we get isomorphisms $H^3(G,U(1)) \cong Hom(H_3(G,\mathbb Z),U(1)) \cong Hom(H_3(G,\mathbb Z),A)$ where $A$ is the cyclic group generated by an $m$-th root of unity with $m$ the exponent of $H_3(G,\mathbb Z)$. We thus get a surjection $H^3(G,A) \rightarrow H^3(G,U(1))$ with kernel of order $|Ext(H_2(G,\mathbb Z),A)|$. In short, we can represent such athe $\mathbb ZG$-linear homomorphism homomorphism $c$ by a $\mathbb ZG$-linear homomorphism $ f\colon R_3 \rightarrow \mathbb Z/m \mathbb Z$ where$ f\colon R_3 \rightarrow A$. Here $m$$f$ is the exponenta 3-cocycle of $G$ with coefficients in the third homologycyclic group $H_3(G,\mathbb Z$)$A$.

Let's call the cocycle3-cocycle $f\colon R_3 \rightarrow \mathbb Z/m$$f\colon R_3 \rightarrow A$ a "standard cocycle"3-cocycle" in the case where the resolution $R_*$ is the standard bar resolution. A standard cocycle3-cocycle can then be thought of as a function $F\colon G\times G\times G\rightarrow \mathbb Z/m \mathbb Z$$F\colon G\times G\times G\rightarrow A$ .

EXAMPLE

As anThe following example constructs, let's construct a random cocycle $f\colon R_3 \rightarrow \mathbb Z/m\mathbb Z$ for each of the symmetric group96 cohomology classes in $G=S_5$$H^3(S_6,A)$, a representative standard 3-cocycle with coefficients in the cyclic group $U(1)$$A$ of order 12 (since we know that $H_3(S_6,\mathbb Z)$ has exponent 12). We use a small resolutionTo run the example the HAP package $R$ constructed by(v 1.10.1 http://hamilton.nuigalway.ie/Hap/www) needs to be loaded into GAP.

EXAMPLE

gap> G:=SymmetricGroup(56);;

gap> mA:=Lcm=Group(GroupHomology(G11,312,13,14,15,16,17,18,19,20,21,22));

12;;

gap> RA:=ResolutionFiniteGroup=TrivialGModuleAsGOuterGroup(G,4A);; #This is the cyclic group of order 12 encoded as a trivial G-module

gap> MR:=CocycleCondition=ResolutionFiniteGroup(RG,34);;

gap> CocycleBasisMod4C:=NullspaceModQ(TransposedMat=HomToGModule(M)R,4A);;

gap> CocycleBasisMod3CH:=NullspaceModQ(TransposedMat=CohomologyModule(M)C,3);;

gap> fclasses:=(Random=Elements(CocycleBasisMod4) + RandomActedGroup(CocycleBasisMod3CH)) mod 12;

[ 2, 2, 2, 2, 1, 1, 3, 2, 1, 2, 2, 3, 2, 3, 1, 3, 0, 2, 4, 0 ]

Now let's convert f to a standard cocycle $F\colon G\times G\times G \rightarrow \mathbb Z/12\mathbb Z$.;; #This is the list of cohomology classes

gap> F:=StandardCocycleLength(R,f,3,12classes);;; #This gives the number of distinct cohomology classes

And now let's evaluate $F(g,h,k)$ for, say, three random elements $g,h,k \in S_5$.96

gap> g:=Random(G); h:=Random(G); kc:=Random=CH!.representativeCocycle(Gclasses[2]);

(2,4)(3,5)

(1,3)(2,4,5) #This gives a 3-cocycle representing the second cohomology class

(1,5,2,4) Standard 3-cocycle

gap> Ff:=Mapping(g,h,kc);

7;; #A cocycle f:GxGxG-->A corresponding to the second cohomology class

I hope it is ok to advertize some GAP code on this site.

$R_*: \cdots \rightarrow R_4 \rightarrow R_3 \rightarrow R_2 \rightarrow R_1 \rightarrow R_0$

A 3-cocycle with coefficients in $U(1)$ is a $\mathbb ZG$-linear homomorphism $f\colon R_3 \rightarrow U(1)$ such that the composite $R_4 \rightarrow R_3 \rightarrow U(1)$ is trivial. Using the Universal Coefficient Theorem I think we can represent such a $\mathbb ZG$-linear homomorphism by a $\mathbb ZG$-linear homomorphism $ f\colon R_3 \rightarrow \mathbb Z/m \mathbb Z$ where $m$ is the exponent of the third homology $H_3(G,\mathbb Z$).

Let's call the cocycle $f\colon R_3 \rightarrow \mathbb Z/m$ a "standard cocycle" in the case where the resolution $R_*$ is the standard bar resolution. A standard cocycle can then be thought of as a function $F\colon G\times G\times G\rightarrow \mathbb Z/m \mathbb Z$ .

EXAMPLE

As an example, let's construct a random cocycle $f\colon R_3 \rightarrow \mathbb Z/m\mathbb Z$ for the symmetric group $G=S_5$ with coefficients in $U(1)$. We use a small resolution $R$ constructed by GAP.

gap> G:=SymmetricGroup(5);;

gap> m:=Lcm(GroupHomology(G,3));

12

gap> R:=ResolutionFiniteGroup(G,4);;

gap> M:=CocycleCondition(R,3);;

gap> CocycleBasisMod4:=NullspaceModQ(TransposedMat(M),4);;

gap> CocycleBasisMod3:=NullspaceModQ(TransposedMat(M),3);;

gap> f:=(Random(CocycleBasisMod4) + Random(CocycleBasisMod3)) mod 12;

[ 2, 2, 2, 2, 1, 1, 3, 2, 1, 2, 2, 3, 2, 3, 1, 3, 0, 2, 4, 0 ]

Now let's convert f to a standard cocycle $F\colon G\times G\times G \rightarrow \mathbb Z/12\mathbb Z$.

gap> F:=StandardCocycle(R,f,3,12);;

And now let's evaluate $F(g,h,k)$ for, say, three random elements $g,h,k \in S_5$.

gap> g:=Random(G); h:=Random(G); k:=Random(G);

(2,4)(3,5)

(1,3)(2,4,5)

(1,5,2,4)

gap> F(g,h,k);

7

I hope it's ok to advertize some GAP code on this site.

$$R_*: \cdots \rightarrow R_4 \rightarrow R_3 \rightarrow R_2 \rightarrow R_1 \rightarrow R_0$$

A 3-cocycle with coefficients in $U(1)$ is a $\mathbb ZG$-linear homomorphism $c\colon R_3 \rightarrow U(1)$ such that the composite $R_4 \rightarrow R_3 \rightarrow U(1)$ is trivial. Using the Universal Coefficient Theorem and the divisibility of $U(1)$ we get isomorphisms $H^3(G,U(1)) \cong Hom(H_3(G,\mathbb Z),U(1)) \cong Hom(H_3(G,\mathbb Z),A)$ where $A$ is the cyclic group generated by an $m$-th root of unity with $m$ the exponent of $H_3(G,\mathbb Z)$. We thus get a surjection $H^3(G,A) \rightarrow H^3(G,U(1))$ with kernel of order $|Ext(H_2(G,\mathbb Z),A)|$. In short, we can represent the $\mathbb ZG$-linear homomorphism $c$ by a $\mathbb ZG$-linear homomorphism $ f\colon R_3 \rightarrow A$. Here $f$ is a 3-cocycle of $G$ with coefficients in the cyclic group $A$.

Let's call the 3-cocycle $f\colon R_3 \rightarrow A$ a "standard 3-cocycle" in the case where the resolution $R_*$ is the standard bar resolution. A standard 3-cocycle can then be thought of as a function $F\colon G\times G\times G\rightarrow A$ .

The following example constructs, for each of the 96 cohomology classes in $H^3(S_6,A)$, a representative standard 3-cocycle with coefficients in the cyclic group $A$ of order 12 (since we know that $H_3(S_6,\mathbb Z)$ has exponent 12). To run the example the HAP package (v 1.10.1 http://hamilton.nuigalway.ie/Hap/www) needs to be loaded into GAP.

EXAMPLE

gap> G:=SymmetricGroup(6);;

gap> A:=Group((11,12,13,14,15,16,17,18,19,20,21,22));;

gap> A:=TrivialGModuleAsGOuterGroup(G,A);; #This is the cyclic group of order 12 encoded as a trivial G-module

gap> R:=ResolutionFiniteGroup(G,4);;

gap> C:=HomToGModule(R,A);;

gap> CH:=CohomologyModule(C,3);;

gap> classes:=Elements(ActedGroup(CH));; #This is the list of cohomology classes

gap> Length(classes); #This gives the number of distinct cohomology classes

96

gap> c:=CH!.representativeCocycle(classes[2]); #This gives a 3-cocycle representing the second cohomology class

Standard 3-cocycle

gap> f:=Mapping(c);; #A cocycle f:GxGxG-->A corresponding to the second cohomology class

added 2 characters in body

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edited Apr 1, 2012 at 14:29

Graham Ellis

426
3
5

I hope it is ok to advertize some GAP code on this site.

Let G be a finite group and let

$R_*: \cdots \rightarrow R_4 \rightarrow R_3 \rightarrow R_2 \rightarrow R_1 \rightarrow R_0$

be a free $\mathbb ZG$-resolution of $\mathbb Z$.

A 3-cocycle with coefficients in $U(1)$ is a $\mathbb ZG$-linear homomorphism $f\colon R_3 \rightarrow U(1)$ such that the composite $R_4 \rightarrow R_3 \rightarrow U(1)$ is trivial. Using the Universal Coefficient Theorem I think we can represent such a $\mathbb ZG$-linear homomorphism by a $\mathbb ZG$-linear homomorphism $ f\colon R_3 \rightarrow \mathbb Z/m \mathbb Z$ where $m$ is the exponent of the third homology $H_3(G,\mathbb Z$).

Let's call the cocycle $f\colon R_3 \rightarrow \mathbb Z/m$ a "standard cocycle" in the case where the resolution $R_*$ is the standard bar resolution. A standard cocycle can then be thought of as a function $F\colon G\times G\times G\rightarrow \mathbb Z/m \mathbb Z$ .

If we have any small resolution $R_*$, endowed with a contracting homotopy, then we can (in principle) construct a standard cocycle for each cohomology class. This can be done in GAP for many groups and provides examples of explicit cocycles which can be analyzed.

EXAMPLE

As an example, let's construct a random cocycle $f\colon R_3 \rightarrow \mathbb Z/m\mathbb Z$ for the symmetric group $G=S_5$ with coefficients in $U(1)$. We use a small resolution $R$ constructed by GAP.

gap> G:=SymmetricGroup(5);;

gap> m:=Lcm(GroupHomology(G,3));

12

gap> R:=ResolutionFiniteGroup(G,4);;

gap> M:=CocycleCondition(R,3);;

gap> CocycleBasisMod4:=NullspaceModQ(TransposedMat(M),4);;

gap> CocycleBasisMod3:=NullspaceModQ(TransposedMat(M),3);;

gap> f:=(Random(CocycleBasisMod4) + Random(CocycleBasisMod3)) mod 12;

[ 2, 2, 2, 2, 1, 1, 3, 2, 1, 2, 2, 3, 2, 3, 1, 3, 0, 2, 4, 0 ]

Now let's convert f to a standard cocycle $F\colon G\times G\times G \rightarrow \mathbb Z/12\mathbb Z$F\colon G\times G\times G \rightarrow \mathbb Z/12\mathbb Z$.

gap> F:=StandardCocycle(R,f,3,12);;;

And now let's evaluate $F(g,h,k)$ for, say, three random elements $g,h,k \in S_5$.

gap> g:=Random(G); h:=Random(G); k:=Random(G);

(2,4)(3,5)

(1,3)(2,4,5)

(1,5,2,4)

gap> F(g,h,k);

7

END OF EXAMPLE

I hope it is ok to advertize some GAP code on this site.

Let G be a finite group and let

$R_*: \cdots \rightarrow R_4 \rightarrow R_3 \rightarrow R_2 \rightarrow R_1 \rightarrow R_0$

be a free $\mathbb ZG$-resolution of $\mathbb Z$.

A 3-cocycle with coefficients in $U(1)$ is a $\mathbb ZG$-linear homomorphism $f\colon R_3 \rightarrow U(1)$ such that the composite $R_4 \rightarrow R_3 \rightarrow U(1)$ is trivial. Using the Universal Coefficient Theorem I think we can represent such a $\mathbb ZG$-linear homomorphism by a $\mathbb ZG$-linear homomorphism $ f\colon R_3 \rightarrow \mathbb Z/m \mathbb Z$ where $m$ is the exponent of the third homology $H_3(G,\mathbb Z$).

Let's call the cocycle $f\colon R_3 \rightarrow \mathbb Z/m$ a "standard cocycle" in the case where the resolution $R_*$ is the standard bar resolution. A standard cocycle can then be thought of as a function $F\colon G\times G\times G\rightarrow \mathbb Z/m \mathbb Z$ .

If we have any small resolution $R_*$, endowed with a contracting homotopy, then we can (in principle) construct a standard cocycle for each cohomology class. This can be done in GAP for many groups and provides examples of explicit cocycles which can be analyzed.

EXAMPLE

As an example, let's construct a random cocycle $f\colon R_3 \rightarrow \mathbb Z/m\mathbb Z$ for the symmetric group $G=S_5$ with coefficients in $U(1)$. We use a small resolution $R$ constructed by GAP.

gap> G:=SymmetricGroup(5);;

gap> m:=Lcm(GroupHomology(G,3));

12

gap> R:=ResolutionFiniteGroup(G,4);;

gap> M:=CocycleCondition(R,3);;

gap> CocycleBasisMod4:=NullspaceModQ(TransposedMat(M),4);;

gap> CocycleBasisMod3:=NullspaceModQ(TransposedMat(M),3);;

gap> f:=(Random(CocycleBasisMod4) + Random(CocycleBasisMod3)) mod 12;

[ 2, 2, 2, 2, 1, 1, 3, 2, 1, 2, 2, 3, 2, 3, 1, 3, 0, 2, 4, 0 ]

Now let's convert f to a standard cocycle $F\colon G\times G\times G \rightarrow \mathbb Z/12\mathbb Z.

gap> F:=StandardCocycle(R,f,3,12);

And now let's evaluate $F(g,h,k)$ for, say, three random elements $g,h,k \in S_5$.

gap> g:=Random(G); h:=Random(G); k:=Random(G);

(2,4)(3,5)

(1,3)(2,4,5)

(1,5,2,4)

gap> F(g,h,k);

7

END OF EXAMPLE

I hope it is ok to advertize some GAP code on this site.

Let G be a finite group and let

$R_*: \cdots \rightarrow R_4 \rightarrow R_3 \rightarrow R_2 \rightarrow R_1 \rightarrow R_0$

be a free $\mathbb ZG$-resolution of $\mathbb Z$.

A 3-cocycle with coefficients in $U(1)$ is a $\mathbb ZG$-linear homomorphism $f\colon R_3 \rightarrow U(1)$ such that the composite $R_4 \rightarrow R_3 \rightarrow U(1)$ is trivial. Using the Universal Coefficient Theorem I think we can represent such a $\mathbb ZG$-linear homomorphism by a $\mathbb ZG$-linear homomorphism $ f\colon R_3 \rightarrow \mathbb Z/m \mathbb Z$ where $m$ is the exponent of the third homology $H_3(G,\mathbb Z$).

Let's call the cocycle $f\colon R_3 \rightarrow \mathbb Z/m$ a "standard cocycle" in the case where the resolution $R_*$ is the standard bar resolution. A standard cocycle can then be thought of as a function $F\colon G\times G\times G\rightarrow \mathbb Z/m \mathbb Z$ .

If we have any small resolution $R_*$, endowed with a contracting homotopy, then we can (in principle) construct a standard cocycle for each cohomology class. This can be done in GAP for many groups and provides examples of explicit cocycles which can be analyzed.

EXAMPLE

As an example, let's construct a random cocycle $f\colon R_3 \rightarrow \mathbb Z/m\mathbb Z$ for the symmetric group $G=S_5$ with coefficients in $U(1)$. We use a small resolution $R$ constructed by GAP.

gap> G:=SymmetricGroup(5);;

gap> m:=Lcm(GroupHomology(G,3));

12

gap> R:=ResolutionFiniteGroup(G,4);;

gap> M:=CocycleCondition(R,3);;

gap> CocycleBasisMod4:=NullspaceModQ(TransposedMat(M),4);;

gap> CocycleBasisMod3:=NullspaceModQ(TransposedMat(M),3);;

gap> f:=(Random(CocycleBasisMod4) + Random(CocycleBasisMod3)) mod 12;

[ 2, 2, 2, 2, 1, 1, 3, 2, 1, 2, 2, 3, 2, 3, 1, 3, 0, 2, 4, 0 ]

Now let's convert f to a standard cocycle $F\colon G\times G\times G \rightarrow \mathbb Z/12\mathbb Z$.

gap> F:=StandardCocycle(R,f,3,12);;

And now let's evaluate $F(g,h,k)$ for, say, three random elements $g,h,k \in S_5$.

gap> g:=Random(G); h:=Random(G); k:=Random(G);

(2,4)(3,5)

(1,3)(2,4,5)

(1,5,2,4)

gap> F(g,h,k);

7

END OF EXAMPLE

Inserted latex commands; added 8 characters in body

Source Link

edited Apr 1, 2012 at 14:23

Graham Ellis

426
3
5

I hope it is ok to advertize some GAP code on this site.

Let G be a finite group and let

R_*: ... --> R_4 --> R_3 --> R_2 --> R_1 --> R_0$R_*: \cdots \rightarrow R_4 \rightarrow R_3 \rightarrow R_2 \rightarrow R_1 \rightarrow R_0$

be a free ZG$\mathbb ZG$-resolution of Z$\mathbb Z$.

A 3-cocycle with coefficients in U(1)$U(1)$ is a ZG$\mathbb ZG$-linear homomorpism f:R_3homomorphism --> U(1)$f\colon R_3 \rightarrow U(1)$ such that the composite R_4 --> R_3 --> U(1)$R_4 \rightarrow R_3 \rightarrow U(1)$ is trivial. Using the Universal Coefficient Theorem I think we can represent such a ZG$\mathbb ZG$-linear homomorphism by a ZG$\mathbb ZG$-linear homomorphism f:R_3 --> Z/mZ$ f\colon R_3 \rightarrow \mathbb Z/m \mathbb Z$ where m$m$ is the exponent of the third homology H_3(G,Z$H_3(G,\mathbb Z$).

Let's agree to call the cocycle f:R_3 --> Z/m$f\colon R_3 \rightarrow \mathbb Z/m$ a "standard cocycle" in the case where the resolution R_*$R_*$ is the standard bar resolution. A standard cocycle can then be thought of as a function F:GxGxG --> Z/mZ$F\colon G\times G\times G\rightarrow \mathbb Z/m \mathbb Z$ .

If we have aany small resolution R_*$R_*$, endowed with a contracting homotopy, then we can (in principle) construct a standard cocycle for each cohomology class. This can be done in GAP for many groups and provides examples of explicit cocycles which can be analyzed.

EXAMPLE

As an example, let's construct a fairly random cocycle f:R_3 --> Z/mZ$f\colon R_3 \rightarrow \mathbb Z/m\mathbb Z$ for the symmetric group G=S_5$G=S_5$ with coefficients in U(1)$U(1)$. We use a small resolution R$R$ constructed by GAP.

gap> G:=SymmetricGroup(5);; gap>

gap> m:=Lcm(GroupHomology(G,3)); 12 gap>

12

gap> R:=ResolutionFiniteGroup(G,4);; gap>

gap> M:=CocycleCondition(R,3);; gap>

gap> CocycleBasisMod4:=NullspaceModQ(TransposedMat(M),4);; gap>

gap> CocycleBasisMod3:=NullspaceModQ(TransposedMat(M),3);; gap>

gap> f:=(Random(CocycleBasisMod4) + Random(CocycleBasisMod3)) mod 12; [

[ 2, 2, 2, 2, 1, 1, 3, 2, 1, 2, 2, 3, 2, 3, 1, 3, 0, 2, 4, 0 ]

Now let's convert f to a standard cocycle F:GxGxG -->$F\colon G\times G\times G \rightarrow \mathbb Z/12Z12\mathbb Z.

gap> F:=StandardCocycle(R,f,3,12);

And now let's evaluate F(g$F(g,h,k)$ for,h say,k) for three random elements of S_5$g,h,k \in S_5$.

gap> g:=Random(G); h:=Random(G); k:=Random(G);

(2,4)(3,5)

(1,3)(2,4,5)

(1,5,2,4)

gap> F(g,h,k); 7

7

END OF EXAMPLE

I hope it is ok to advertize some GAP code on this site.

Let G be a finite group and let

R_*: ... --> R_4 --> R_3 --> R_2 --> R_1 --> R_0

be a free ZG-resolution of Z.

A 3-cocycle with coefficients in U(1) is a ZG-linear homomorpism f:R_3 --> U(1) such that the composite R_4 --> R_3 --> U(1) is trivial. Using the Universal Coefficient Theorem I think we can represent such a ZG-linear homomorphism by a ZG-linear homomorphism f:R_3 --> Z/mZ where m is the exponent of the third homology H_3(G,Z).

Let's agree to call the cocycle f:R_3 --> Z/m a "standard cocycle" in the case where the resolution R_* is the standard bar resolution. A standard cocycle can then be thought of as a function F:GxGxG --> Z/mZ .

If we have a resolution R_*, endowed with a contracting homotopy, then we can (in principle) construct a standard cocycle for each cohomology class.

EXAMPLE

As an example, let's construct a fairly random cocycle f:R_3 --> Z/mZ for the symmetric group G=S_5 with coefficients in U(1). We use a small resolution R.

gap> G:=SymmetricGroup(5);; gap> m:=Lcm(GroupHomology(G,3)); 12 gap> R:=ResolutionFiniteGroup(G,4);; gap> M:=CocycleCondition(R,3);; gap> CocycleBasisMod4:=NullspaceModQ(TransposedMat(M),4);; gap> CocycleBasisMod3:=NullspaceModQ(TransposedMat(M),3);; gap> f:=(Random(CocycleBasisMod4) + Random(CocycleBasisMod3)) mod 12; [ 2, 2, 2, 2, 1, 1, 3, 2, 1, 2, 2, 3, 2, 3, 1, 3, 0, 2, 4, 0 ]

Now let's convert f to a standard cocycle F:GxGxG --> Z/12Z.

gap> F:=StandardCocycle(R,f,3,12);

And now let's evaluate F(g,h,k) for three random elements of S_5.

gap> g:=Random(G); h:=Random(G); k:=Random(G); (2,4)(3,5) (1,3)(2,4,5) (1,5,2,4)

gap> F(g,h,k); 7

END OF EXAMPLE

I hope it is ok to advertize some GAP code on this site.

Let G be a finite group and let

$R_*: \cdots \rightarrow R_4 \rightarrow R_3 \rightarrow R_2 \rightarrow R_1 \rightarrow R_0$

be a free $\mathbb ZG$-resolution of $\mathbb Z$.

A 3-cocycle with coefficients in $U(1)$ is a $\mathbb ZG$-linear homomorphism $f\colon R_3 \rightarrow U(1)$ such that the composite $R_4 \rightarrow R_3 \rightarrow U(1)$ is trivial. Using the Universal Coefficient Theorem I think we can represent such a $\mathbb ZG$-linear homomorphism by a $\mathbb ZG$-linear homomorphism $ f\colon R_3 \rightarrow \mathbb Z/m \mathbb Z$ where $m$ is the exponent of the third homology $H_3(G,\mathbb Z$).

Let's call the cocycle $f\colon R_3 \rightarrow \mathbb Z/m$ a "standard cocycle" in the case where the resolution $R_*$ is the standard bar resolution. A standard cocycle can then be thought of as a function $F\colon G\times G\times G\rightarrow \mathbb Z/m \mathbb Z$ .

If we have any small resolution $R_*$, endowed with a contracting homotopy, then we can (in principle) construct a standard cocycle for each cohomology class. This can be done in GAP for many groups and provides examples of explicit cocycles which can be analyzed.

EXAMPLE

As an example, let's construct a random cocycle $f\colon R_3 \rightarrow \mathbb Z/m\mathbb Z$ for the symmetric group $G=S_5$ with coefficients in $U(1)$. We use a small resolution $R$ constructed by GAP.

gap> G:=SymmetricGroup(5);;

gap> m:=Lcm(GroupHomology(G,3));

12

gap> R:=ResolutionFiniteGroup(G,4);;

gap> M:=CocycleCondition(R,3);;

gap> CocycleBasisMod4:=NullspaceModQ(TransposedMat(M),4);;

gap> CocycleBasisMod3:=NullspaceModQ(TransposedMat(M),3);;

gap> f:=(Random(CocycleBasisMod4) + Random(CocycleBasisMod3)) mod 12;

[ 2, 2, 2, 2, 1, 1, 3, 2, 1, 2, 2, 3, 2, 3, 1, 3, 0, 2, 4, 0 ]

Now let's convert f to a standard cocycle $F\colon G\times G\times G \rightarrow \mathbb Z/12\mathbb Z.

gap> F:=StandardCocycle(R,f,3,12);

And now let's evaluate $F(g,h,k)$ for, say, three random elements $g,h,k \in S_5$.

gap> g:=Random(G); h:=Random(G); k:=Random(G);

(2,4)(3,5)

(1,3)(2,4,5)

(1,5,2,4)

gap> F(g,h,k);

7

END OF EXAMPLE

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created Apr 1, 2012 at 0:11

Graham Ellis

426
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