Dear Sebastien,
Once again, your X can be written as X=L*DL, with D diagonal and real. The
group L^-1 G L preserves the Hermitian form x*Dx. In particular any g in this
group satisfies g*Dg=D. As it acts irreducibly, D is a scalar matrix, thus g is
unitary.
Indeed, for computing D cyclotomics might not suffice, but we do not need D
explicitly.
Dima
On 20 Jan 2015 10:07, Palcoux Sebastien <sebastienpalc...@yahoo.fr> wrote:
>
> Dear Dima and Forum.
>
> I don't understand how your answer solves my problem, perhaps there is a
> misunderstanding:
>
> What I want are the unitary matrices representing the elements of the group G
> for an irreducible representation V.
> For so, we should conjugate the non-unitary matrices (given by GAP) by the
> matrix R=S.P with S^{-2} the diagonalization D of the matrix X of the
> Hermitian positive definite form
> obtained by the averaging (or in some other way) and P the matrix of the
> change of basis (into the eigenvectors basis of X).
> In this process, we need the find the square root of D, i.e. the square root
> of positive cyclotomic numbers.
>
> Is there an other process for doing that without having to compute square
> root of positive cyclotomic numbers?
>
> Best regards,
> Sébastien
>
>
>
> Le Mardi 20 janvier 2015 14h29, Dima Pasechnik
> <dmitrii.pasech...@cs.ox.ac.uk> a écrit :
>
>
> On Tue, Jan 20, 2015 at 07:31:56AM +0000, Palcoux Sebastien wrote:
> > Dear Alexander and Forum,
> > If the cyclotomic number is the square of a cyclotomic number, is there an
> > easy way to find it?
> > The number I need are the eigenvalues of the matrix of the unitarized inner
> > product of an irreducible representation of a finite group (see the comment
> > of Paul Garett here: http://math.stackexchange.com/q/1107941/84284). This
> > matrix is positive, I guess its eigenvalues are always cyclotomic (true for
> > the examples I've looked, but I don't know in general), and I hope they are
> > square of cyclotomic. Thanks to these square roots I can compute the
> > unitary matrices for the irreducible representation.
>
> You don't need to take square roots. If H is the Hermitian positive definite
> form
> you obtained by the averaging (or in some other way) then H=LDL*, for
> L a lower-triangular matrix with 1s on the main diagonal, and D is a diagonal
> matrix.
> L and D can be computed without taking square roots (and so they will stay
> cyclotomic).
> Then conjugating by L gives you the unitary form.
>
> HTH,
> Dmitrii
>
>
>
> > Remark: a function on GAP computing the unitary irreducible representations
> > seems very natural, so if there is not such a function, this should means
> > that there are problems for computing them in general with GAP, isn't it?
> > Best regards,Sebastien Palcoux
> >
> > Le Mardi 20 janvier 2015 3h13, Alexander Hulpke <hul...@fastmail.fm> a
> >écrit :
> >
> >
> > Dear Forum,
> >
> > > On Jan 19, 2015, at 1/19/15 2:18, Palcoux Sebastien
> > > <sebastienpalc...@yahoo.fr> wrote:
> > >
> > > Hi,
> > > Is it possible to extend the function Sqrt on the cyclotomic numbers?
> >
> > How would you represent this root? In general the square root of a
> > cylotomic is not cyclotomic again. (You could form a formal
> > AlgebraicExtension, but then you lose the irrational cyclotomics for
> > operations.)
> >
> > Regards,
> >
> > Alexander Hulpke
>
>
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