Ask Your Question

Group algebra/matrix space homomorphism

asked 2013-11-01 23:49:02 -0500

vuur gravatar image

updated 2013-11-02 10:20:43 -0500

I would like to know if it is possible to define a homomorphism (of algebras) between a group algebra over the rationals and a matrix space over the rationals.

I have

sage: S3 = SymmetricGroup(3)  
sage: QG = GroupAlgebra(S3, QQ)  
sage: QG.gens()  
Finite family {(1,2,3): (1,2,3), (1,2): (1,2)}  
sage: M_3 = MatrixSpace(QQ, 2)  
sage: ma = M_3([[0,-1],[1,-1]])  
sage: mb = M_3([[1,-1],[0,-1]])

Now I would like to map each generator of QG to a matrix. In this case I would like to map (1,2,3) to ma and (1,2) to mb and use that to define a homomorphism. But everything I tried didn't work.

I've encountered another issue, though I can work around it. So this is only for information. The elements of G.gens() don't coincide with the reduced words of length 1.

sage: S3 = SymmetricGroup(3)
sage: S3.gens()
[(1,2,3), (1,2)]
sage: {g: g.reduced_word() for g in S3}
{(): [],
 (2,3): [2],
 (1,2): [1],
 (1,2,3): [2, 1],
 (1,3,2): [1, 2],
 (1,3): [1, 2, 1]}
edit retag flag offensive close merge delete


Looks like a bug... I modified ma = matrix(...) to ma = M_3(...) and similarly for mb.

vdelecroix gravatar imagevdelecroix ( 2013-11-02 04:23:50 -0500 )edit

It even seems impossible to get a group morphism from S3 to MatrixGroup([ma,mb])... which looks like another bug!

vdelecroix gravatar imagevdelecroix ( 2013-11-02 04:26:33 -0500 )edit

yes, I used M_3 instead of matrix, I posted the wrong code here, sorry for that

vuur gravatar imagevuur ( 2013-11-02 06:49:39 -0500 )edit

about your edit: I would not characterize it as a bug. This is well documented, the reduced_word method returns a word over the simple reflections (1,2), (2,3), (3,4), ...

vdelecroix gravatar imagevdelecroix ( 2013-11-02 18:14:05 -0500 )edit

2 answers

Sort by » oldest newest most voted

answered 2013-11-02 05:19:05 -0500

updated 2013-11-02 05:55:36 -0500


it seems to me that R.module_morphism for R a parent inheriting from CombinatorialFreeModule does not support a definition on_gens (probably some code by Robert Bradshaw do that but I do not remember where it is in Sage...). So the only current way to define your morphism imply to do some math to get a definition on_basis instead on_gens.

For example, this work but is ugly :

sage: S3 = SymmetricGroup(3)
sage: QG = GroupAlgebra(S3, QQ)
sage: QG.gens()
Finite family {(1,2,3): (1,2,3), (1,2): (1,2)}
sage: M_3 = MatrixSpace(QQ, 2)
sage: ma = M_3([[0,-1],[1,-1]])
sage: mb = M_3([[1,-1],[0,-1]])
sage: def img_of_perm(p, gens_mat):
    w = p.reduced_word()
    R = gens_mat[0].parent()
    return[gens_mat[i-1] for i in w])
sage: f = QG.module_morphism(on_basis=lambda x: img_of_perm(x, [ma,mb]), 
codomain=ma.parent(), category=Algebras(QQ))
sage: f(QG.an_element())    
[ 2 -5]
[ 2 -3]

Another bug : if you replace category=Algebras(QQ) by category=AlgebrasWithBasis(QQ), you get the error : TypeError: Full MatrixSpace of 2 by 2 dense matrices over Rational Field is not in Category of algebras with basis over Rational Field

In a perfect world, the category should be at least FiniteDimensionnalAlgebraWithBasis(QQ) (because the parents should lies in it if I don't say something false)

The other way is trying to find stuff from Robert to define maps between parents on generators... I am not sure but I think it won't work either because it assume that elements of the two parents are not based on RingElement (old category stuff) and your group algebra is a CombinatorialFreeModule in data structure.

anyway, generally, define an algebra homomorphism between a group algebra and a matrix space implies there exist an algorithm to decompose each element of the domain in a polynomial over the algebra generators (I am not very sure it is a prerequisite but I think so...). For the symmetric group, we know how to do that with reduced word but a lot of combinatorial free module in Sage have a method algebra_generators() but do not know how to decompose an element along these algebra generators...

GAP do probably that very well but I also never tried (GAP knows how to decompose a group element along group generators...).

I hope that help a little...


Nicolas B.

edit flag offensive delete link more


I don´t mind using (or calculating) the basis instead of gens, Thanks for answering. It helps alot :-)

vuur gravatar imagevuur ( 2013-11-02 06:58:08 -0500 )edit

answered 2013-11-02 06:57:17 -0500

updated 2013-11-02 06:58:33 -0500


I reproduce here the answer I got from Travis Scrimshaw on the mailing list sage-combinat devel.

First method

sage: S3 = SymmetricGroup(3)
sage: SGA = GroupAlgebra(S3, QQ)
sage: MS = MatrixSpace(QQ,2)
sage: mtx = [MS([[0,-1],[1,-1]]), MS([[1,-1],[0,-1]])]
sage: f = lambda x:[i-1] for i in x.reduced_word())
sage: phi = SGA.module_morphism(f, codomain=MS)

Where the -1 is needed because of indexing. Now you can check:

sage: elt = SGA.gen(0) + SGA.gen(1); elt
(1,2) + (1,2,3)
sage: phi(elt)
[-1 -1]
[ 0  0]

Second method An alternative method is to build a group morphism and then build the associate algebra morphism. There is a subtle issue as there is currently no method to express an element of a finite group in terms of generators. In particular, a morphism can not be built by providing only the image of the generators.

sage: MG = MatrixGroup(mtx)    # take some time because it launches GAP
sage: H = Hom(S3, MG)
sage: psi = H(lambda x: for i in x.reduced_word()))
sage: kappa = SGA.module_morphism(lambda x: MS(psi(x)), codomain=MS)

You can again check

sage: elt = SGA.gen(0) + SGA.gen(1); elt
(1,2) + (1,2,3)
sage: kappa(elt)
[-1 -1]
[ 0  0]

Alternative to the second method The following shows how in general, given a finite group, one may express a given element in terms of generators using the Cayley graph:

sage: G = S3.cayley_graph(generators=S3.gens())
sage: path = G.shortest_path(, prod(S3.gens())); path
[(), (1,2,3), (2,3)]
sage: labels = [G.edge_label(path[i], path[i+1]) for i in range(len(path)-1)]; labels
[(1,2,3), (1,2)]
sage: map(lambda x: list(S3.gens()).index(x), labels)
[0, 1]
edit flag offensive delete link more


Thanks for answering. I will try out what works best for me.

vuur gravatar imagevuur ( 2013-11-02 07:05:41 -0500 )edit

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

Question Tools


Asked: 2013-11-01 23:49:02 -0500

Seen: 320 times

Last updated: Nov 02 '13