Confused about FreeAlgebra quotients

qw
13 ●1 ●1 ●3

I think I'm misunderstanding how quotients of free algebras work. I tried to make a free algebra on two generators x, y and mod out by xy = yx, so you get a polynomial algebra -- but that's not what happened.

sage: R.<x,y> = FreeAlgebra(QQ)
sage: I = R*[x*y-y*x]*R
sage: Q.<a,b> = R.quo(I)
sage: a*b is b*a
False
sage: Q.is_commutative()
False

Relatedly, the documentation for free_module_quotient gives an example (constructing the quaternions as a free quotient):

sage: n = 2
sage: A = FreeAlgebra(QQ,n,'x')
sage: F = A.monoid()
sage: i, j = F.gens()
sage: mons = [ F(1), i, j, i*j ]
sage: r = len(mons)
sage: M = MatrixSpace(QQ,r)
sage: mats = [M([0,1,0,0, -1,0,0,0, 0,0,0,-1, 0,0,1,0]), M([0,0,1,0, 0,0,0,1, -1,0,0,0, 0,-1,0,0]) ]
sage: H2.<i,j> = A.quotient(mons,mats)

but I'm confused exactly how the matrices in the penultimate line describe the (multiplication) action? I.e. if I have some relation, say x^2 = 0, that I want to mod out by, how do I accomplish that using matrices?

Comments

small remark: you should not use "is" for testing equality as it does something else

sage: (3*2) is (3*2)
False

vdelecroix ( 9 years ago )

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letterplace is a completely different implementation of FreeAlgebras which emulates noncommutative homogeneous polynomials as commutative polynomials using Singular. It sure works well, but it has some limitations, for instance the elements of the FreeAlgebra aren't callable and hence substitutions won't work. On the other hand, if the described behaviour of the non letterplace implementation is not a bug, then one might ast what are such quotients good for? In any case, it is desirable that both implementations maintained a good compatibility level with each other.

Waldeck ( 8 years ago )

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