ASKSAGE: Sage Q&A Forum - Individual question feedhttp://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Thu, 22 Feb 2018 06:41:31 -0600Calculations in quotient of a free algebrahttp://ask.sagemath.org/question/41219/calculations-in-quotient-of-a-free-algebra/I want to define (the algebra part of) Sweedler's four-dimensional Hopf algebra, which is freely generated by $x,y$ and subject to the relations
$$
x^2 = 1, \qquad y^2 = 0, \qquad x\cdot y = - y\cdot x~ ,
$$
but I don't see how to do it.
I have tried the following:
sage: A.<x,y> = FreeAlgebra(QQbar)
sage: I = A*[x*x - 1, y*y, x*y + y*x]*A
sage: H.<x,y> = A.quo(I)
sage: H
Quotient of Free Algebra on 2 generators (x, y) over Algebraic Field by the ideal (-1 + x^2, y^2, x*y + y*x)
But then I get
sage: H.one() == H(x*x)
False
So is this currently possibly using a different method?
ThanksWed, 21 Feb 2018 07:44:18 -0600http://ask.sagemath.org/question/41219/calculations-in-quotient-of-a-free-algebra/Answer by Jo Be for <p>I want to define (the algebra part of) Sweedler's four-dimensional Hopf algebra, which is freely generated by $x,y$ and subject to the relations
$$
x^2 = 1, \qquad y^2 = 0, \qquad x\cdot y = - y\cdot x~ ,
$$
but I don't see how to do it.</p>
<p>I have tried the following:</p>
<pre><code>sage: A.<x,y> = FreeAlgebra(QQbar)
sage: I = A*[x*x - 1, y*y, x*y + y*x]*A
sage: H.<x,y> = A.quo(I)
sage: H
Quotient of Free Algebra on 2 generators (x, y) over Algebraic Field by the ideal (-1 + x^2, y^2, x*y + y*x)
</code></pre>
<p>But then I get</p>
<pre><code>sage: H.one() == H(x*x)
False
</code></pre>
<p>So is this currently possibly using a different method?</p>
<p>Thanks</p>
http://ask.sagemath.org/question/41219/calculations-in-quotient-of-a-free-algebra/?answer=41231#post-id-41231I know now that we can do the following:
sage: A = FreeAlgebra(QQbar, 3, "a")
sage: F = A.monoid()
sage: M = MatrixSpace(QQbar,4)
sage: x, y, xy = F.gens()
sage: mons = [F(1), x, y, xy]
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, 0, 0, 0, 0, 0, 0, 0, 0]),
....: M([ 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0 ])
....: ]
sage: H.<x,y,xy> = FreeAlgebraQuotient(A,mons,mats); H
Free algebra quotient on 3 generators ('x', 'y', 'xy') and dimension 4 over Algebraic Field
And we can verify
sage: x^2
1
sage: y^2
0
sage: x*y
xy
sage: y*x
-xy
Note: We cannot take tensor products, but that is not what I originally asked for, so.Thu, 22 Feb 2018 06:41:31 -0600http://ask.sagemath.org/question/41219/calculations-in-quotient-of-a-free-algebra/?answer=41231#post-id-41231Answer by tmonteil for <p>I want to define (the algebra part of) Sweedler's four-dimensional Hopf algebra, which is freely generated by $x,y$ and subject to the relations
$$
x^2 = 1, \qquad y^2 = 0, \qquad x\cdot y = - y\cdot x~ ,
$$
but I don't see how to do it.</p>
<p>I have tried the following:</p>
<pre><code>sage: A.<x,y> = FreeAlgebra(QQbar)
sage: I = A*[x*x - 1, y*y, x*y + y*x]*A
sage: H.<x,y> = A.quo(I)
sage: H
Quotient of Free Algebra on 2 generators (x, y) over Algebraic Field by the ideal (-1 + x^2, y^2, x*y + y*x)
</code></pre>
<p>But then I get</p>
<pre><code>sage: H.one() == H(x*x)
False
</code></pre>
<p>So is this currently possibly using a different method?</p>
<p>Thanks</p>
http://ask.sagemath.org/question/41219/calculations-in-quotient-of-a-free-algebra/?answer=41221#post-id-41221This is definitely a bug, thanks for reporting ! This is now [trac ticket 24808](https://trac.sagemath.org/ticket/24808).
Wed, 21 Feb 2018 09:33:26 -0600http://ask.sagemath.org/question/41219/calculations-in-quotient-of-a-free-algebra/?answer=41221#post-id-41221Comment by Jo Be for <p>This is definitely a bug, thanks for reporting ! This is now <a href="https://trac.sagemath.org/ticket/24808">trac ticket 24808</a>.</p>
http://ask.sagemath.org/question/41219/calculations-in-quotient-of-a-free-algebra/?comment=41222#post-id-41222Thank you for adding the ticket!Wed, 21 Feb 2018 09:37:03 -0600http://ask.sagemath.org/question/41219/calculations-in-quotient-of-a-free-algebra/?comment=41222#post-id-41222