Javier,
In fact,
http://brauer.maths.qmul.ac.uk/Atlas/v3/clas/U34/
provides you all almost you need.
If you take the sum of all the representations given there, it's
exactly
1_G+"the irreducibles", so each irreducible comes with multiplicity 1.
So you can just take the (GAP) data given there, and
On Feb 18, 2:09 am, javier wrote:
> Hi Dima,
>
> On Feb 18, 6:26 am, Dima Pasechnik 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
Hi Dima,
On Feb 18, 6:26 am, Dima Pasechnik 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 d
On Feb 17, 7:57 am, javier wrote:
> Hi all,
>
> I am trying to use sage to compute the Artin-Wedderburn decomposition
> of a group algebra.
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
Hi all,
thanks for the tip-off in CombinatorialFreeModule, I have been trying
to use this, but cannot find any sensible way to make it work.
sage: G = SymmetricGroup(3)
sage: B = sorted(list(G))
sage: n = len(B)
sage: K = CyclotomicField(n)
sage: A = GroupAlgebra(G,K)
sage: V = CombinatorialFreeM
>> I don't think you can have automated conversion like C(a^2 + b^2) since it
>> makes sense to define:
>> sage: C = CombinatorialFreeModule(QQ, [ a^2, b^2, a*b, a^2+b^2 ])
>> sage: 2*C.basis()[a^2] + C.basis()[b^2]
>> B[b^2] + 2*B[a^2]
>> sage: 2*C.basis()[a^2] + C.basis()[b^2 + a^2]
>> 2*B[a^2] +
I don't think you can have automated conversion like C(a^2 + b^2)
since it
makes sense to define:
sage: C = CombinatorialFreeModule(QQ, [ a^2, b^2, a*b, a^2+b^2 ])
sage: 2*C.basis()[a^2] + C.basis()[b^2]
B[b^2] + 2*B[a^2]
sage: 2*C.basis()[a^2] + C.basis()[b^2 + a^2]
2*B[a^2] + B[a^2 + b^2]
The
Hi Nick,
> With an old version of sage, this is unfortunately not all that useful for
> my purpose. Not sure how to address this, since there is not a standard
> way to convert a multivariate polynomial into such an expression. Perhaps
> things have improved since this version was
On 17-Feb-10, at 10:03 AM, Nicolas M. Thiery wrote:
On Wed, Feb 17, 2010 at 08:53:23AM -0800, Nick Alexander wrote:
PS: FWIW, in this kind of problem having a nice
"VectorSpaceWithBasis"
so that I could define a vector space with basis given by the group
elements, would come really fancy.
On Wed, Feb 17, 2010 at 08:53:23AM -0800, Nick Alexander wrote:
> >PS: FWIW, in this kind of problem having a nice "VectorSpaceWithBasis"
> >so that I could define a vector space with basis given by the group
> >elements, would come really fancy.
>
> I also want this! Various people in the combina
On Feb 17, 8:53 am, Nick Alexander wrote:
> > PS: FWIW, in this kind of problem having a nice "VectorSpaceWithBasis"
> > so that I could define a vector space with basis given by the group
> > elements, would come really fancy.
>
> I also want this! Various people in the combinat group suggested
PS: FWIW, in this kind of problem having a nice "VectorSpaceWithBasis"
so that I could define a vector space with basis given by the group
elements, would come really fancy.
I also want this! Various people in the combinat group suggested they
had it/were working on it, but I don't know the s
Hi Nicolas,
> What do you mean by exact? I am using CyclotomicFields on a regular
> basis for similar things, and this works well. And I would expect it
> to be faster than QQbar.
You are right, I guess I could just use CyclotomicField(n) where n is
the order of the group and everything should wo
On 17-Feb-10, at 8:27 AM, Nicolas M. Thiery wrote:
On Wed, Feb 17, 2010 at 07:50:57AM -0800, javier wrote:
Apparently I was assuming too much. The result of the evaluation of
the character belongs to some cyclotomic field, so apparently the
problem is that there is not a coercion between cyclo
On Wed, Feb 17, 2010 at 07:50:57AM -0800, javier wrote:
> Apparently I was assuming too much. The result of the evaluation of
> the character belongs to some cyclotomic field, so apparently the
> problem is that there is not a coercion between cyclotomic fields and
> QQbar. I can circumvent the pro
On Feb 17, 2:57 pm, javier wrote:
> Observe that in this situation the numbers I am trying to coerce into
> K are rational:
Apparently I was assuming too much. The result of the evaluation of
the character belongs to some cyclotomic field, so apparently the
problem is that there is not a coercion
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