[GAP Forum] Recombining irreducible representations

[Jerry Swan]

Dear Stefan, Marc and Max,

Thanks very much for your help.

Best wishes,

Jerry.

On Sun, Jun 18, 2017 at 4:16 PM, Max Horn <max at quendi.de> wrote:

> Dear Jerry,
>
> > On 18 Jun 2017, at 15:05, Jerry Swan <dr.jerry.swan at gmail.com> wrote:
> >
> > Dear all,
> >
> > For some element g of a group G for which irr :=
> > IrreducibleRepresentations(G) have been obtained, is it possible to
> recover
> > g from images := List(irr,r->Image(r,g)) ?
>
> The representations are homomorphisms, and as such, you can compute
> preimages -- which in general are of course not unique, but rather cosets
> of the kernel. But for a finite group, the intersection of the kernel of
> all irreducible representations is trivial, so you can recover g like this:
>
> pre:=Intersection(List([1..Length(irr)], i -> PreImagesElm(irr[i],
> images[i])));
>
> Applied to a concrete example:
>
> gap> G:=SymmetricGroup(5);;
> gap> irr:=IrreducibleRepresentations(G);;
> gap> g:=Random(G);
> (1,4,3)
> gap> images := List(irr,r->Image(r,g));;
> gap> pre:=Intersection(List([1..Length(irr)], i -> PreImagesElm(irr[i],
> images[i])));
> [ (1,4,3) ]
>
>
> In a later email, you clarified that your group G is always a symmetric
> group S_n. In that case, at least for n>=5, most irreducible
> representations are actually faithful, the exception being the trivial and
> the sign representation. In that case, you can simply take a preimage of
> one of the faithful representations, like so:
>
> gap> List(irr, IsInjective);
> [ false, true, true, true, true, true, false ]
> gap> irr[1];   # this is the sign representation
> [ (1,2,3,4,5), (1,2) ] -> [ [ [ 1 ] ], [ [ -1 ] ] ]
> gap> irr[7];   # this is the trivial representation
> [ (1,2,3,4,5), (1,2) ] -> [ [ [ 1 ] ], [ [ 1 ] ] ]
> gap> PreImage(irr[2], images[2]);
> (1,4,3)
> gap> PreImage(irr[3], images[3]);
> (1,4,3)
>
>
> Hope that helps,
> Max