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2024-07-18 15:07:13 +0200 commented question Alternative to changing the __add__ of Integers class (need $0_{\mathbb Z} + x = x$, $x$ belonging to custom class)

Thanks, I'm taking a look.

2024-07-18 12:58:21 +0200 edited question Alternative to changing the __add__ of Integers class (need $0_{\mathbb Z} + x = x$, $x$ belonging to custom class)

Alternative to changing the __add__ of Integers class (need $0_{\mathbb Z} + x = x$, $x$ belonging to custom class) I ha

2024-07-18 12:49:07 +0200 edited question Alternative to changing the __add__ of Integers class (need $0_{\mathbb Z} + x = x$, $x$ belonging to custom class)

Alternative to changing the __add__ of Integers class (need $0 + x = x$, $x$ belonging to custom class) I have a class c

2024-07-18 11:24:36 +0200 edited question Alternative to changing the __add__ of Integers class (need $0_{\mathbb Z} + x = x$, $x$ belonging to custom class)

Alternative to changing the __add__ of Integers class (need $0 + x = x$, $x$ belonging to custom class) I have a class c

2024-07-18 10:16:23 +0200 edited question Alternative to changing the __add__ of Integers class (need $0_{\mathbb Z} + x = x$, $x$ belonging to custom class)

Alternative to changing the __add__ of Integers class I have a class called phi defined on certain ring $R$, which moral

2024-07-18 10:16:20 +0200 edited question Alternative to changing the __add__ of Integers class (need $0_{\mathbb Z} + x = x$, $x$ belonging to custom class)

Alternative to changing the __add__ of Integers class I have a class called phi defined on certain ring $R$, which moral

2024-07-18 08:36:31 +0200 edited question Alternative to changing the __add__ of Integers class (need $0_{\mathbb Z} + x = x$, $x$ belonging to custom class)

Alternative to changing the __add__ of Integers class I have a class called phi defined on certain ring $R$, which moral

2024-07-18 08:22:52 +0200 edited question Alternative to changing the __add__ of Integers class (need $0_{\mathbb Z} + x = x$, $x$ belonging to custom class)

Alternative to changing the __add__ of Integers class I have a class called phi defined on certain ring $R$, which moral

2024-07-18 08:22:03 +0200 edited question Alternative to changing the __add__ of Integers class (need $0_{\mathbb Z} + x = x$, $x$ belonging to custom class)

Alternative to changing the __add__ of Integers class I have a class called phi defined on certain ring which morally sh

2024-07-18 08:19:39 +0200 edited question Alternative to changing the __add__ of Integers class (need $0_{\mathbb Z} + x = x$, $x$ belonging to custom class)

Alternative to changing the __add__ of Integers class I have a class called phi defined on certain ring which morally sh

2024-07-18 08:19:20 +0200 asked a question Alternative to changing the __add__ of Integers class (need $0_{\mathbb Z} + x = x$, $x$ belonging to custom class)

Alternative to changing the __add__ of Integers class I have a class called phi defined on certain ring which morally sh

2024-07-18 08:09:31 +0200 answered a question Tensor product of Clifford algebra with another (non-Clifford)

The main question was already answered in the comments. Of course, both tensor factors should be a module of the same ri

2024-07-18 08:07:06 +0200 edited question Tensor product of Clifford algebra with another (non-Clifford)

Tensor product of Clifford algebra with another (non-Clifford) For a given quadratic form $Q$, I'd wish to symbolically

2024-07-18 08:05:28 +0200 commented question Tensor product of Clifford algebra with another (non-Clifford)

Thanks for the link

2024-07-12 17:52:31 +0200 commented question Tensor product of Clifford algebra with another (non-Clifford)

Many thanks. I should say that forming a triple product sometimes work, sometimes it doesn't (by running the same code)

2024-07-12 12:36:56 +0200 edited question Tensor product of Clifford algebra with another (non-Clifford)

Tensor product of Clifford algebra with another (non-Clifford) over a multivariable polynomial ring For a given quadrati

2024-07-12 12:36:52 +0200 edited question Tensor product of Clifford algebra with another (non-Clifford)

Tensor product of Clifford algebra with another (non-Clifford) over a multivariable polynomial ring For a given quadrati

2024-07-12 10:36:23 +0200 edited question Tensor product of Clifford algebra with another (non-Clifford)

Tensor product of Clifford algebra with another (non-Clifford) For a given quadratic form $Q$, I'd wish to symbolically

2024-07-12 08:05:19 +0200 commented question Tensor product of Clifford algebra with another (non-Clifford)

Many thanks, (moral: I should think in the math first for next time). I should say that forming a triple product sometim

2024-07-12 08:04:48 +0200 commented question Tensor product of Clifford algebra with another (non-Clifford)

Many thanks, (moral: I should think in the math first for next time). I should say that forming a triple product sometim

2024-07-12 08:04:33 +0200 commented question Tensor product of Clifford algebra with another (non-Clifford)

Many thanks, (moral: I should think in the math first for next time). I should say that forming a triple product sometim

2024-07-12 08:00:05 +0200 commented question Comparing two symbolic expressions without sympy

It surely does, thanks.

2024-07-11 23:33:31 +0200 asked a question Comparing two symbolic expressions without sympy

Comparing two symbolic expressions without sympy From diff(sin(x),x) == cos(x) one gets just trivial equality cos(x)

2024-07-10 09:39:32 +0200 commented question Tensor product of Clifford algebra with another (non-Clifford)

Many thanks, (moral: I should think in the math first for next time).

2024-07-10 09:33:56 +0200 edited answer complex expand

Hi, welcome. Don't know how far the next works, but while nobody answers, consider: var('x,y',domain=RR) def csimp(comp

2024-07-10 08:39:44 +0200 answered a question complex expand

Hi, welcome. Don't know how far the next goes, but if nobody answers consider: var('x,y',domain=RR) def csimp(complex_p

2024-07-10 08:01:19 +0200 edited question Radical numerical expressions not handled by simplify_full ?

Rational numerical expressions not handled by simplify_full ? To simplify rational sqrt(21)*sqrt(6) to, 3*sq

2024-07-10 08:01:15 +0200 edited question Radical numerical expressions not handled by simplify_full ?

Rational numerical expressions not handled by simplify_full ? To simplify rational sqrt(21)*sqrt(6) to, 3*sq

2024-07-09 23:45:13 +0200 edited answer partial derivatives

The next: x,y,q = SR.var('x,y,q') y = function('y')(x) f(y,z) = sqrt(1+z^2) z = function('z')(x) (((f(y,q).diff(q)).s

2024-07-09 23:44:53 +0200 edited answer partial derivatives

The next: x,y,q = SR.var('x,y,q') y = function('y')(x) f(y,z) = sqrt(1+z^2) z = function('z')(x) (((f(y,q).diff(q)).s

2024-07-09 23:44:33 +0200 edited answer partial derivatives

x,y,q = SR.var('x,y,q') y = function('y')(x) f(y,z) = sqrt(1+z^2) z = function('z')(x) (((f(y,q).diff(

2024-07-09 23:44:19 +0200 edited answer partial derivatives

x,y,q = SR.var('x,y,q') y = function('y')(x) f(y,z) = sqrt(1+z^2) z = function('z')(x) (((f(y,q).diff(

2024-07-09 23:43:41 +0200 answered a question partial derivatives

x,y,q = SR.var('x,y,q') y = function('y')(x) f(y,z) = sqrt(1+z^2) fy1 = f.diff(z) z = function('z')(x) (((f(y,q).diff(q

2024-07-09 21:58:30 +0200 edited question Tensor product of Clifford algebra with another (non-Clifford)

Tensor product of Clifford algebra with another (non-Clifford) For a given quadratic form $Q$, I'd wish to symbolically

2024-07-09 21:54:21 +0200 edited question Tensor product of Clifford algebra with another (non-Clifford)

Tensor product of Clifford algebra with another (non-Clifford) For a given quadratic form $Q$, I'd wish to symbolically

2024-07-09 21:53:19 +0200 asked a question Tensor product of Clifford algebra with another (non-Clifford)

Tensor product of Clifford algebra with another (non-Clifford) For a given quadratic form $Q$, I'd wish to symbolically

2024-06-21 17:43:08 +0200 commented answer Question on tensor products (of the free algebra)

sure I forgot to specify $F\otimes F$ has non-homogeneous elements, was a little bit sloppy of me. Thanks

2024-06-21 17:40:14 +0200 edited question Question on tensor products (of the free algebra)

Question on tensor products (of the free algebra) If $F$ is the free algebra over the rational numbers with generators

2024-06-21 17:36:42 +0200 marked best answer Question on tensor products (of the free algebra)

If $F$ is the free algebra over the rational numbers with generators ${a,b,c,\ldots}$, then $F\otimes F$ is implemented

F<a,b,c>=FreeAlgebra(QQ)
Fsquare=F.tensor_square()

Typical element are (sums of)

F(a) # F(1)
F(a) # F(b^2)
(1/2)*F(1) # F(a* b*a* c^2)

A first issue I have is that -- despite the typing alternative e.g. a.tensor(b) for a#b -- output happens in the hashtag format, which, if copied, is recognised only as a comment after that sign. Is there a way to make output useful [or rendered as tensor(a,b)] ?

And more importantly, how read off each factor of this type of expressions? That is, how to map

(1/2)*F(1) # F(a* b*a* c^2)

to, say,

(1/2)  ,   a*b*a*c^2

or

 1,  a*b*a*c^2 /2

(who gets the numerical factor I don't care) ?

2024-06-21 14:59:15 +0200 asked a question Free algebra 'inside the trace'

Free algebra 'inside the trace' In the free algebra of generators $a,b,c,\ldots$, is there a way to do additive simplifi

2024-06-21 12:03:55 +0200 edited question Question on tensor products (of the free algebra)

Question on tensor products (of the free algebra) If $F$ is the free algebra over the rational numbers with generators

2024-06-21 12:01:50 +0200 edited question Question on tensor products (of the free algebra)

Question on tensor products (of the free algebra) If $F$ is the free algebra over the rational numbers with generators

2024-06-21 12:01:01 +0200 edited question Question on tensor products (of the free algebra)

Question on tensor products (notation) If $F$ is the free algebra over the rational numbers with generators ${a,b,c,\ld

2024-06-21 11:59:18 +0200 edited question Question on tensor products (of the free algebra)

Question on tensor products (notation) If $F$ is the free algebra over the rational numbers with generators ${a,b,c,\ld

2024-06-21 11:58:40 +0200 edited question Question on tensor products (of the free algebra)

Question on tensor products (notation) If $F$ is the free algebra over the rational numbers with generators ${a,b,c,\ld

2024-06-21 11:56:41 +0200 asked a question Question on tensor products (of the free algebra)

Question on tensor products (notation) If $F$ is the free algebra over the rational numbers with generators ${a,b,c,\ld

2024-06-19 23:08:16 +0200 marked best answer Free algebra identity (empty word) and involution

I didn't find the way to define an involution on the free algebra (say in two generators) yet. Specifically I want unitary elements. Thus, I added two variables, $y$ and $z$ which serve as $w^\star,x^\star$ and force a relation so that $w y=1$, the empty word, and similar relation $x z=1$ . But then typing

> A.<w,x,y,z> = FreeAlgebra(QQ,4)  
> G=A.g_algebra({w*y: 1})    
(x,y,z) =
> G.gens()

yields an error (#This is dirty, coercion is broken). That is: Question: which is the free algebra empty word here?

Of course, I tried to add a symbol $E$ which then satisfies $gE=g=Eg$ for all the generators $g$, But then

> G=A.g_algebra({w*y: E})

yields also an error.

> /usr/lib/python3/dist-packages/sage/algebras/free_algebra.py
> in g_algebra(self, relations, names,    
> order, check)    
>     873                     d_poly = commuted - self(c) * self(m)     
>     874                     break    
> --> 875             assert c_coef is not None, list(m)   
>     876             v2_ind = self.gens().index(v2)   
>     877             v1_ind = self.gens().index(v1)   
> 
> AssertionError: [(E, 1)]

Second question: does somebody know an easier way to to implement the involution here?

2024-04-07 22:04:51 +0200 received badge  Notable Question (source)
2023-09-06 21:10:07 +0200 commented question Free algebra identity (empty word) and involution

It doesn't but thanks anyway, good to know.