I'm searching for a weak condition that makes us able to perform iteration in right divisible magams with an easy method. If this holds for some special structures this can solve the problem of the fractional hyperoperations.
I would like to know if someone here know the answer:
The question is already on MathSE
http://math.stackexchange.com/questions/...-divisible
The question is about wich is the weakest condition for a general algebraic structure \( (G,*) \) with one binary operation \( * \) that makes the set of integer iterations of the left (right) translation \( iter(a):=\{L_a^{\circ n}:n \in \mathbb{N}\} \) for a fixed \( a \)a commutative subsemigroup of the semigroup of all the left (right) translations under function composition \( left(G,*):=\{L_b:b \in G\} \).
Where a left (right) traslation is defined as usual \( L_a(x)=a*x \) and \( R_a(x)=x*a \)
This should be equivalent to the fact that every iteration of a left traslation \( L_a^{\circ n}(x)=a_1*(...(a_{n-1}*(a_{n}*x))) \) by a fixed \( a \) is still a left traslation by another element \( \alpha \) such that \( L_a^{\circ n}(x)=\alpha * x \)and this will make us able to perform the left(or right) iteration on such structures that are not associative (like addition/multiplication)
For associative structures this is trivially true and the function is the nth-power in that semi/group/monoid. But if the associativity implies that exist such injection I don't know if everytime the bijection exists then the operation is associative.
I should note that, for example, this injection doesn't exist for the reals structured with the exponentiation as binary operation.
If it was true we should have that for every \( b \) exist a \( \beta \) such that \( b^{b^x}=\beta^x \) .
I would like to know if someone here know the answer:
The question is already on MathSE
http://math.stackexchange.com/questions/...-divisible
The question is about wich is the weakest condition for a general algebraic structure \( (G,*) \) with one binary operation \( * \) that makes the set of integer iterations of the left (right) translation \( iter(a):=\{L_a^{\circ n}:n \in \mathbb{N}\} \) for a fixed \( a \)a commutative subsemigroup of the semigroup of all the left (right) translations under function composition \( left(G,*):=\{L_b:b \in G\} \).
Where a left (right) traslation is defined as usual \( L_a(x)=a*x \) and \( R_a(x)=x*a \)
This should be equivalent to the fact that every iteration of a left traslation \( L_a^{\circ n}(x)=a_1*(...(a_{n-1}*(a_{n}*x))) \) by a fixed \( a \) is still a left traslation by another element \( \alpha \) such that \( L_a^{\circ n}(x)=\alpha * x \)and this will make us able to perform the left(or right) iteration on such structures that are not associative (like addition/multiplication)
For associative structures this is trivially true and the function is the nth-power in that semi/group/monoid. But if the associativity implies that exist such injection I don't know if everytime the bijection exists then the operation is associative.
I should note that, for example, this injection doesn't exist for the reals structured with the exponentiation as binary operation.
If it was true we should have that for every \( b \) exist a \( \beta \) such that \( b^{b^x}=\beta^x \) .
Quote:APPLICATIONS
But if this weak condition is satisfied by non associative structures and to be more precise for every conjugation quandle or LD system then it should holds for the conjugation in a group of functions under composition...and so the fractional iteration of conjugation could be easier easier and we can obtain a solution for the left traslations of the fractional-rank hyperoperations. Or some related (weaker) result if we use the LD systems.
Another interesting point is that this make us able to reduce the existence of a sequence o hyperoperation with some wanted properties to an abstract algebra problem (even if I fear that uniqueness will be only settled by an additional structure on the algebraic structure).
MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
