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
On Feb 18, 2:09 am, javier vengor...@gmail.com wrote:
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
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 17, 2:57 pm, javier vengor...@gmail.com 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
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
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
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 work
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
On Feb 17, 8:53 am, Nick Alexander ncalexan...@gmail.com 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
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 combinat group
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.
I
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
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]
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]
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 =
On Feb 17, 7:57 am, javier vengor...@gmail.com 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
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