Dear Forum,
On Jun 7, 2012, at 6/7/12 8:16, Cesar Neyit Galindo-Martinez wrote: > > Let H be an abelian group, and f:HxH-->F_q* a 2-cocycle (where F_q* > are the unities of the finite filed of q elements), if T:H-->H is an > automorphism the map f^T(a,b):=f(T(a),T(b)) is also a 2-cocycle. How > can I use GAP in order to decided if f and f^T are cohomologous and to > construct an explicit the coboundady? This is possible, but will require a bit of work. I did not work out a full solution (writing all the code would take a few hours, in particular if (q-1) is not prime), but here are some hints. Let M=F_q^*. For simplicity I will use the language of vector spaces for M or maps into M. This is not really true in most cases, but you can then perform the same operations using Hermite Normal Form computations. GAP has a built-in functionality for 2-comomology if the group is a pc group and the module a F_p vector space. In this case you could use some of the predefined functionality, but as you likely require modifications for F_q^* you are probably better off starting from scratch. First you need to represent maps HxH -> M. You could do so by enumerating the cartesian product HxH and representing the map as list of images. In fact it is not too hard to see that the values of such a map on AxA, where A is a generating set for H, already completely determine a map. So lists of length k=|A|^2 are sufficient. If M is not cyclic, in fact you probably want to chose a basis for M and write each element with respect to this basis, i.e. if the dimension of M is m, you will end up with lists of length k*m where each element of M is represented by m subsequent entries. (This is in fact what the GAP library does for pc groups.) Obviously the arrangement you chose for AxA and the basis is important here. Now you can consider such lists of length k*m as elements of the k*m dimensional row space over F_p, define subspaces and test membership with standard linear algebra. You thus write down the vectors for f and f^T, to test whether they are cohomologous you simply need to test whether the difference of the corresponding vectors is a coboundary. You obtain a basis of the coboundaries from the definition: For each basis element of M you determine the image under the derivation map. If you decompose coboundaries into these images you can trace back which element of M they belong to. Best wishes, Alexander Hulpke -- Colorado State University, Department of Mathematics, Weber Building, 1874 Campus Delivery, Fort Collins, CO 80523-1874, USA email: hul...@math.colostate.edu, Phone: ++1-970-4914288 http://www.math.colostate.edu/~hulpke _______________________________________________ Forum mailing list Forum@mail.gap-system.org http://mail.gap-system.org/mailman/listinfo/forum
