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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. I'm happy if I can deal with unitary elements. So I add two variables, $y$ and $z$ which serve as $w^\star,x^\star$ and force a relation so that $wy=1=()$, the empty word, and similar relation $zx=1=()$ . 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 $E*g$ 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?

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. I'm happy if I can deal with unitary elements. So I add two variables, $y$ and $z$ which serve as $w^\star,x^\star$ and force a relation so that $wy=1=()$, y=1=( )$, the empty word, and similar relation $zx=1=()$ x=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 $E*g$ 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? here?

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. I'm happy if I can deal with unitary elements. So I add two variables, $y$ and $z$ which serve as $w^\star,x^\star$ and force a relation so that $wy=1=( )$, y=1$, the empty word, and similar relation $zx=1=( )$ x=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 $E*g$ 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?

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. I'm happy if I can deal with unitary elements. So I add two variables, $y$ and $z$ which serve as $w^\star,x^\star$ and force a relation so that $w$w y=1$, the empty word, and similar relation $zx=1$ $x z=1$ . 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 $E*g$ 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?

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. I'm happy if I can deal with unitary elements. So I add 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 $E*g$ $gE=g=Eg$ 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?

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. I'm happy if Specifically I can deal with want unitary elements. So Thus, I add 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?

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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()

> 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}) 

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)]

1)]

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