From: Jyrki Lahtonen <lahtonen@utu.fi>
Subject: Re: Representations of S_5
Date: Mon, 05 Feb 2001 10:58:10 +0200
Newsgroups: sci.math
Summary: Construction representations of symmetric groups

ahmedfares@my-deja.com wrote:
> 
> Hello, I wanted to see a specific construction of these
> representations.
> 
> Regards,
> Ahmed
> 
> In article <95kkdr$ji2$1@wisteria.csv.warwick.ac.uk>,
>   mareg@primrose.csv.warwick.ac.uk () wrote:
> > In article <95hhuc$dmn$1@nnrp1.deja.com>,
> >       ahmedfares@my-deja.com writes:
> > >What are the irreducible representations of the symmetric group
> > >S_5 ?
> > >
> >
> > You have been given a couple of references for the general theory of  the
> > representations of the symmetric group. For S_5, there are 6 - the degrees
> > are 1,1,4,4,5,5,6. Were you hoping for specific matrices?
> >
> > Derek Holt.
> >
> >
> 
> Sent via Deja.com
> http://www.deja.com/

The Young tableaux are really the way to go, if you want to study
this topic further. If your ambition is limited to S_5 only, then
you may find the following "hands on"-computation will give you 
the character tables (and some hints to the actual construction).

The 1D irreducibles are obviously the trivial representation chi0
and the sign representation chi1. To construct the others you might 
do the following:

Let S_5 act "naturally" on the basis vectors x_1,x_2,x_3,x_4,x_5.
This 5D representation (call it eeta) is not irreducible as it has 
chi0 as a component (spanned by the sum of the basis vectors). The
other component (call it chi2) is an irreducible 4D representation.
You get the other 4D-irreducible chi3 by tensoring chi2 with chi1.

We need other constructions to get the remaining irreducibles.
Consider the space of antisymmetric rank 2 tensor (let * denote
a tensor product here): x_i*x_j-x_j*x_i, i<j. There are 10 such
basic antisymmetric tensors and S_5 acts naturally on this 10-
dimensional space. Computations with characters show that this
10-dimensional representation has 2 irreducibles components and
one of them can be identified as chi2. The other (call it chi4) is
then the sole 6-dimensional irreducible (as an exercise check
that tensoring it with chi1 won't change its character at all).
I leave it to you the identify the 4-dimensional and 6-dimensional
components as vector spaces.

The group S_5 also acts on the space of symmetric tensors (=
homogeneous degree two polynomials in the 5 unknowns) spanned
by x_i^2, all i, and x_i x_j, i<j. This 15-dimensional representation
has as components 2 copies of chi0, 2 copies of chi2 and a new
irreducible (call it chi5). We see that chi5 is, indeed, 5-dimensional.
We get the other 5-dimensional irreducible representation (chi6) as the
tensor product chi5*chi1.

Have fun constructing the actual matrix representations!!

-- 
Jyrki Lahtonen, Ph.D.
Department of Mathematics,
University of Turku,
FIN-20014 Turku, Finland

http://users.utu.fi/lahtonen