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Calculations in quotient of a free algebra

asked 2018-02-21 07:44:18 -0500

Jo Be gravatar image

updated 2018-02-21 09:34:03 -0500

tmonteil gravatar image

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?

Thanks

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answered 2018-02-21 09:33:26 -0500

tmonteil gravatar image

This is definitely a bug, thanks for reporting ! This is now trac ticket 24808.

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Thank you for adding the ticket!

Jo Be gravatar imageJo Be ( 2018-02-21 09:37:03 -0500 )edit
0

answered 2018-02-22 06:41:31 -0500

Jo Be gravatar image

updated 2018-02-22 06:45:16 -0500

I 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.

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Asked: 2018-02-21 07:44:18 -0500

Seen: 57 times

Last updated: Feb 22