[GAP Forum] natural representation and orbits

Dima Pasechnik dmitrii.pasechnik at cs.ox.ac.uk
Thu Oct 22 16:58:40 BST 2015


Dear all,
On Thu, Oct 22, 2015 at 02:57:26PM +0200, Bill Allombert wrote:
> On Sun, Oct 18, 2015 at 08:49:15AM -0600, Alexander Hulpke wrote:
> > > Let G be a primitive transitive subgroup of S_n.
> > > I am interested by the links between:
> > > 1) the lengths of the orbits of {1,...,n} under the action of the stabilisator
> > > of 1 by G.
> > > 2) the degrees of the irreducible representations occuring in the natural
> > > representation of G.
> > 
> > Frobenius reciprocity and the fact that the permutation character (=natural
> > character) is the induced trivial representation of the point stabilizer show
> > that the permutation character has inner product m with itself, where m is
> > the number of orbits of the point stabilizer.
> > 
> > Thus the observation is true for doubly transitive groups:  the permutation
> > character has the form 1+chi with chi irreducible, so deg chi must be |
> > \Omega |-1, thus proving the statement.
> > 
> > It also is true (for trivial reasons) for regular groups and (an ad-hoc
> > observation) dihedral groups of prime degree.
> > 
> > This covers all but 22 of the primitive groups of degree up to 17. Of these,
> > 7 are Frobenius groups for which I think again an ad-hoc argument works,
> > leaving 15 groups, of which 9 fail (and 6 pass) the conjecture. So chances
> > are about 50%.
> > 
> > But 50% is still somewhat surprising. I suspect the reason is that character
> > degrees must divide up the group order and length of stabilizer orbits divide
> > the stabilizer order. Trying to write a smallish (in this case <=17) number
> > as sum of a few divisors leaves open only a few possibilities, making it
> > likely that the same numbers are involved.

There is more useful theory here known: namely, for a large class of such groups
these two sets of these degrees/lengths are indeed the same; namely, for groups with 
a regular normal Abelian subgroup (in particular for Frobenius groups).
This goes back to notion of S-ring, see e.g. the book on permutation groups by Wielandt, 
or Section II.6 in Bannai, Ito "Algebraic combinatorics I".
(Specifically, I refer to Thm 6.1 in the latter).

There are also results about recognising  a primitive G from the lengths of the orbits of the stabilizer
alone. Often under an assumption that the representation is multiplicity-free, i.e.
each irreducible character occurs at most once in the decomposition.

More generally, one can compute the multiplicites from the multiplication table of the
algebra of the orbitals of G.  (the orbitals are 0-1 matrices associated in the natural way,
i.e. as indicator functions, to the orbits of G on the ordered pairs).
(this is of course more infomation than the lengths of the orbits of the stabilizer
on {1..n}.)


> 
> Thanks for your very useful answer.  Maybe I should give some motivation:
> 
> Let K be a number field, L its Galois closure over Q (the rational field), and
> G = Gal(L/Q). The Dedekind Zeta function of K is equal to the Artin L function
> associated to the natural representation of G (seen as acting on the complex
> embedding of K). (the Artin L function is an arithmetic object functiorialy
> attached to representations of Galois groups of numbers field).
> 
> Thus it factors as a product of Artin L functions associated to the irreducible
> representations that occurs in the natural representation.
> 
> Computing this factorization is very important for computing the Dedekind Zeta
> function.
> 
> However computing the Galois group G is difficult (This is the GAP function
> GaloisType). But some property of G are easy to compute (e.g. primitiveness,
> and the orbits under the stabilisator of a point).

Is it also easy to compute orbits of stabilisers of pairs of points?  If yes,
then it should be easy to compute the orbitals, and thus the multiplicites, without knowing
the whole group, as I mentioned above.

HTH,
Dima

> 
> So any trick which can allow to compute the factorization without computing
> the exact Galois group is useful.
> 
> Cheers,
> Bill
> 
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