[GAP Forum] Decomposing rational modules

[Asst. Prof. Dmitrii (Dima) Pasechnik]

Dear Willem, Peter, all,

2008/9/26  <degraaf at science.unitn.it>:
> Dear Dima, Peter, All,
>
> In order to decompose a rational module, one can
> compute its endomorphism algebra (i.e., all linear maps that
> comute with the action of G). In this algebra one can find the
> central idempotents. (That particular step is also implemented in GAP).
What I wrote coincides with this, up to terminology.
As the question was about Q-modules of permutation groups, I adhered
to the terminology
used in this area. The endomorphism algebra in this case is known e.g.
as "coherent configuration", name introduced by D.Higman (there are
other names, too).
It has the basis of 0-1 matrices (so-called 2-orbits) corresponding to
the orbits of the group G on ordered 2-tuples, allowing for a very
quick computation, not only of this basis, but of the
multiplication table, too.

> Those will give one the components that are direct sums of isotypical
> modules.
The centre of a coherent configuration of 2-orbits is generated by
sums C_i=g_i^G of conjugacy classes of G.
The isotypical component corresponding to an irreducible character x
is given by the projection sum_i x(g_i)* C_i.

Computing C_i is very quick, once 2-orbits are known, as one just has to expand
C_i as a linear combination of 2-orbits (one views each 2-orbit as the
adjacency matrix
of a digraph, and cycles of g_i give rise to cycles in these digraphs
- the rest is elementary counting)

Computing a decomposition of each isotypical component into
G-irreducibles can be easily done if the irreducible representation of
G corresponding to this component is known.
Formulae for this can be found in Serre's book "Linear representations
of finite groups." Springer GTM, Vol. 42.



> However, if the endomorphism algebra contains subalgebras
> that are isomorphic to full matrix algebras, then one would have to
> find complete sets of idempotents also in those algebras. That is an
> extremely hard problem in general.
Well, as I mentioned, in this case the knowledge of the complex
irreducible representations
of the group in question would suffice.
Group theory can be very helpful sometimes ;-)

Best,
Dima

>
> I hope the above makes sense.
>
> All the best,
>
> Willem
>
>
>
>