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 > > _______________________________________________ Forum mailing list Forum@mail.gap-system.org http://mail.gap-system.org/mailman/listinfo/forum