Hi Dima,
On Feb 18, 6:26 am, Dima Pasechnik <dimp...@gmail.com> wrote:
> I am curious to know, how you are doing this. IMHO for this you need
> to know
> each irreducible representation explicitly --- but then you can just
> stack up the right
> number of copies of each irreducible.
>
> Or you rather mean a weaker decomposition, into direct sums of
> "homogeneous components", where the latter are isomorphic to direct
> sums of copies of the same irreducible?
The latest, indeed. The elements
e_i = n_i / n * \sum_{g\in G] X_i(g)g
provide a family of orthogonal idempotents (where X_i is the i-th
irreducible character and n_i = X_i(1) the size of the corresponding
irrep). These idempotents provide a decomposition
A = \oplus e_i A e_i
which is not the complete AW decomposition but might suffice for my
needs. Ultimately what I want to do is test some property in PSU(3,4)
(order 62400) that requires computing the determinant of an operator
in the group algebra. Since the matrix is too big to fit in memory, my
idea is to restrict the operator to each of the homogeneous components
and then multiply the resulting determinants, so replacing computing
one 62400x62400 determinant by around 20 determinants of 3000x3000
matrices.
>From the above decomposition I can now how many copies of each irrep
are there, just have to divide the dimension of the block by the
dimension of the irrep, thought that doesn't provide an explicit
description of the basis.
Cheers
J
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