On Tue, Jan 20, 2015 at 02:05:30PM +0000, Dima Pasechnik wrote:
> 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.
Sorry, this last claim is wrong: to get a unitary g, you will need
to take D^(1/2) g D^(-1/2). I suppose this is still easier
to compute than taking square roots during the diagonalisation of X.
>
> Indeed, for computing D cyclotomics might not suffice, but we do not need D
> explicitly.
here I meant "computing D^(1/2)", certainly, not just D.
I shall never again write to Form from a mobile phone. :-)
Dima
>
> 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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