ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Fri, 23 Mar 2012 03:12:08 +0100Non-commutative ring with inverseshttps://ask.sagemath.org/question/8808/non-commutative-ring-with-inverses/Hello Sage,
I would like to make a ring over $\mathbb{Q}$ with $n$ variables which are non-commutative and also include their inverses.
So I want to generate the free algebra over $\mathbb{Q}$ with generates $x_1, x_2, ..., x_n$ and $x_1^{-1}, x_2^{-1}, ..., x_n^{-1}$. How can I do this?
Best regards,
NoudWed, 21 Mar 2012 08:26:28 +0100https://ask.sagemath.org/question/8808/non-commutative-ring-with-inverses/Comment by Noud for <p>Hello Sage,</p>
<p>I would like to make a ring over $\mathbb{Q}$ with $n$ variables which are non-commutative and also include their inverses.
So I want to generate the free algebra over $\mathbb{Q}$ with generates $x_1, x_2, ..., x_n$ and $x_1^{-1}, x_2^{-1}, ..., x_n^{-1}$. How can I do this?</p>
<p>Best regards,
Noud</p>
https://ask.sagemath.org/question/8808/non-commutative-ring-with-inverses/?comment=20079#post-id-20079Thank you Simon! I'm looking forward to an answer on the sage-combinat-devel mailinglist.Fri, 23 Mar 2012 03:12:08 +0100https://ask.sagemath.org/question/8808/non-commutative-ring-with-inverses/?comment=20079#post-id-20079Comment by Simon King for <p>Hello Sage,</p>
<p>I would like to make a ring over $\mathbb{Q}$ with $n$ variables which are non-commutative and also include their inverses.
So I want to generate the free algebra over $\mathbb{Q}$ with generates $x_1, x_2, ..., x_n$ and $x_1^{-1}, x_2^{-1}, ..., x_n^{-1}$. How can I do this?</p>
<p>Best regards,
Noud</p>
https://ask.sagemath.org/question/8808/non-commutative-ring-with-inverses/?comment=20084#post-id-20084I think that sage-combinat-devel would be even better than sage-devel or sage-algebra (though the latter would make sense as well). Actually, yesterday I did post on sage-combinat-devel, in order to point people to this question. After all, Noud had already asked on sage-support, but didn't get a good answer (sorry for that).Thu, 22 Mar 2012 13:44:58 +0100https://ask.sagemath.org/question/8808/non-commutative-ring-with-inverses/?comment=20084#post-id-20084Comment by niles for <p>Hello Sage,</p>
<p>I would like to make a ring over $\mathbb{Q}$ with $n$ variables which are non-commutative and also include their inverses.
So I want to generate the free algebra over $\mathbb{Q}$ with generates $x_1, x_2, ..., x_n$ and $x_1^{-1}, x_2^{-1}, ..., x_n^{-1}$. How can I do this?</p>
<p>Best regards,
Noud</p>
https://ask.sagemath.org/question/8808/non-commutative-ring-with-inverses/?comment=20094#post-id-20094you could try making a quotient of a free algebra: http://www.sagemath.org/doc/reference/sage/algebras/free_algebra_quotient.html I'm not sure if the kind of quotient you need is implemented though.Wed, 21 Mar 2012 16:50:17 +0100https://ask.sagemath.org/question/8808/non-commutative-ring-with-inverses/?comment=20094#post-id-20094Comment by niles for <p>Hello Sage,</p>
<p>I would like to make a ring over $\mathbb{Q}$ with $n$ variables which are non-commutative and also include their inverses.
So I want to generate the free algebra over $\mathbb{Q}$ with generates $x_1, x_2, ..., x_n$ and $x_1^{-1}, x_2^{-1}, ..., x_n^{-1}$. How can I do this?</p>
<p>Best regards,
Noud</p>
https://ask.sagemath.org/question/8808/non-commutative-ring-with-inverses/?comment=20087#post-id-20087yes, I see your problem; I think this means that such algebras are not implemented yet for Sage. They certainly could be implemented, either by wrapping functionality for free groups from GAP, or by building on the CombinatorialFreeModule class, as was done for the Steenrod Algebra. If you're interested in either of these, I would ask about it on the sage developer email list to see if anyone else has thought about this (or has a better idea!).Thu, 22 Mar 2012 08:30:59 +0100https://ask.sagemath.org/question/8808/non-commutative-ring-with-inverses/?comment=20087#post-id-20087Comment by Noud for <p>Hello Sage,</p>
<p>I would like to make a ring over $\mathbb{Q}$ with $n$ variables which are non-commutative and also include their inverses.
So I want to generate the free algebra over $\mathbb{Q}$ with generates $x_1, x_2, ..., x_n$ and $x_1^{-1}, x_2^{-1}, ..., x_n^{-1}$. How can I do this?</p>
<p>Best regards,
Noud</p>
https://ask.sagemath.org/question/8808/non-commutative-ring-with-inverses/?comment=20088#post-id-20088I tried to do this, but I do not know how to make the generators of infinite order. I define `FreeAlgebra(QQ, 2n, 'x')` and define with `FreeAlgebraQuotient` relations $x_i x_{2n-i} = 1 = x_{2n-i} x_i$. But how do you define that $x_i^k = x_i^k$ (this looks a bit ambiguous)? In the examples in this reference everything has finite order.Thu, 22 Mar 2012 07:06:24 +0100https://ask.sagemath.org/question/8808/non-commutative-ring-with-inverses/?comment=20088#post-id-20088