Hi all,
Thanks you Simon for the time you spent to give answers. Thanks you also
Vincent for the suggestion. I have now a trac number :
http://trac.sagemath.org/ticket/15160
Feel free to rephrase the ticket description or modify anything (still
my English...). This is far from final but partial results may interest
some people in the combinat community. The suggestion and advises can
now go to the trac ticket. I am still motivated to do the job but alone,
It will remains too hard for me so thanks you Simon for having describe
how attack this problem.
I just uploaded a drafty patch with some tests but no documentation.
Before this small patch, once cannot multiply a scalar and a matrix if
the coefficient didn't inherit from RingElement. Now, here is a small
log of what can be done :
*
sage: from sage.matrix.matrix_space import test_special_ring
sage: test_special_ring(verbose=True)
Test with : An example of an algebra with basis: the free algebra on the
generators ('a', 'b', 'c') over Rational Field
Let M be an invertible matrix:
[ B[word: ] B[word: b] B[word: a]]
[ 0 B[word: ] B[word: ab]]
[ 0 0 B[word: ]]
Here is the invert computed:
[ B[word: ]-B[word: b] -B[word: a] +
B[word: bab]]
[ 0 B[word: ] -B[word: ab]]
[ 0 0 B[word: ]]
inversion OK
Test with : Symmetric Functions over Rational Field in the Schur basis
Let M be an invertible matrix:
[ s[] s[2] s[]]
[ 0 s[] s[1]]
[ 00 s[]]
Here is the invert computed:
[ s[] -s[2] -s[] + s[2, 1] + s[3]]
[0 s[] -s[1]]
[0 0 s[]]
inversion OK
Test with : An example of Hopf algebra with basis: the group algebra of
the Dihedral group of order 6 as a permutation group over Rational Field
Let M be an invertible matrix:
[ B[()] B[(1,2,3)] B[(1,3,2)]]
[ 0 B[()] B[(1,3)]]
[ 0 0 B[()]]
Here is the invert computed:
[B[()] -B[(1,2,3)] B[(1,2)] - B[(1,3,2)]]
[0 B[()] -B[(1,3)]]
[0 0 B[()]]
inversion OK
Test with : Group algebra of Symmetric group of order 4! as a
permutation group over Rational Field
Let M be an invertible matrix:
[ B[()] 0 0]
[B[(2,3,4)] B[()] 0]
[B[(2,4,3)] B[(2,4)] B[()]]
Here is the invert computed:
[B[()] 0 0]
[ -B[(2,3,4)] B[()] 0]
[B[(3,4)] - B[(2,4,3)] -B[(2,4)] B[()]]
inversion OK
Test with : Non-Commutative Symmetric Functions over the Rational Field
in the Phi basis
Let M be an invertible matrix:
[ Phi[] Phi[2, 1, 2] Phi[3]]
[ 0Phi[] Phi[1, 0, 1]]
[ 00Phi[]]
Here is the invert computed:
[ Phi[] -Phi[2, 1, 2] Phi[2, 1,
2, 1, 0, 1] - Phi[3]]
[ 0 Phi[] -Phi[1, 0, 1]]
[ 0 0 Phi[]]
inversion OK
*
Inversion OK means that the result was the right answer. Note also that
the first and last example are in the non-commutative world.
Feel free to comments (ok on the trac or on this list if you prefer)
what I already done to allows me to continue in a good way.
Cheers,
Nicolas B.
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