04/26/2022, 01:37 AM
(This post was last modified: 07/14/2022, 07:35 PM by MphLee.
Edit Reason: FOUND ERROR
)
2022, april 23 - actions, iterations and generalized elements (reboot of old notes)
I try to link this to the previous so to make the whole more reasonable.
Proposition 1: Fix a monoid \((A,+_A)\). Every pair of functions \(f,g\) s.t. \(f:A\to Y\) and \(g:Y\to A\) inverts \(f\) on the left, i.e. \(gf={\rm id}_A\) , defines a semigroup (!) action \(\gamma_{f:g}:A\times Y\to Y\) on the set \(Y\). Not only it is a semigroup action, it also make \(f\) into an \(A\)-equivariant map of semigroup actions.
The last line means that the naturally defined semigroup action \(\gamma_{f:g}\) is homomorphic to the natural translation action of \(A\) on itself via the map \(f\).
Proof: define the binary function \(\gamma_{f:g}:A\times Y\to Y\) as \(\gamma_{f:g,a}(y):=f(a+g(y))\).
This map is an action, i.e. \(\gamma_{f:g,a+b}(y)=\gamma_{f:g,a}(\gamma_{f:g,b}(y))\). Unwrap the definitions to obtain \((f\lambda_ag)(f\lambda_bg) =f\lambda_a\lambda_bg=f\lambda_{a+b}g\).
To conclude the proof we have to show that \(f\in {Hom}_{A{\rm Act}}(A^{\circlearrowleft \lambda},Y^{\circlearrowleft \gamma_{f:g}})\), i.e.
\[f(a+b)=\gamma_{f:g,a}(f(b)) \] Unwrap the definitions again \(f\lambda_a=f\lambda_a(gf)=(f\lambda_ag)f\). \(\square\)
Corollary the map \(\gamma_{f:g}:A\to End(Y)\) that takes \(a\in A\) and produces endomaps \(\gamma_{f:g,a}:Y\to Y\) is a semigroup homomorphism. It sends \(+_A\) into composition over \(Y\) .
Observation 1 The map \(\gamma_{f:g}:A\to End(Y)\) is also a monoid morphism precisely when it sends the element \(0_A\) to the identity. This happens precisely when \(g\) is also a right inverse of \(f\). When this happens \(f\) must be invertible and \(g=f^{-1}\).
Observation 2 (\(A\)-time) Every monoid \(A\)-action over a set \(Y\) defines a monoid morphism \(A\to {\rm End}Y\) (from the "monoid of time" to the "monoid of endofunctions over Y"). We can interpret the latter as a kind of \(A\)-time exponentiation of something that acts on \(Y\) or an algebraic flow over \(Y\). In symbols we can express this as a one-to-one correspondence
![[Image: image.png]](https://i.ibb.co/BCB6QgS/image.png)
In the special case of the integers \(\mathbb Z\) we have something peculiar that unlocks a whole kind of heuristic/philosophy around this correspondence.
Observation 2 (integer-time) If we set \(A=\mathbb Z\), the monoid morphism \(\mathbb Z\to A\), induced by a chosen \(\mathbb Z\)-action, is just the iteration/exponential map \(n\to f^{\circ n}\) and it is a group homomorphism sending \(+_A\) to composition of bijections. In this special case, the initiality of the integers as a group (CORRECTION: the initiality used here is of the pointed group \((\mathbb Z,1)\) in the cat. of pointed groups), i.e. thanks to the recursion theorem, we have that group morphisms \(\mathbb Z\to {\rm Bij}Y\) are the same as, in one-to-one correspondence with, bijective maps \(f:Y \to Y\).
![[Image: image.png]](https://i.ibb.co/NVJ5VQ9/image.png)
Philosophy What can we take from the previous one-to-one correspondences? If \(\mathbb Z\)-actions over \(Y\) are the same as iteration/discrete flow/exponential map of a bijection, and if choosing a map of that kind is the same as choosing a bijection over \(Y\) then we can can adopt the following philosophy:
Given a group \(A\) we can see \(A\)-actions over a set \(Y\) as if they are \(A\)-time iterations of something that we can think as a (generalized) bijection over \(Y\). In particular we can call that thing a generalized element of the group \({\rm Bij}Y\) (see Generalized element).
Key problem of iteration given an exponential map \(f\) we recover the base "it is exponential of" just by evaluating it at \(1\in \mathbb R\). The same for functions and integer iterations. We go from a bijection \(f:Y\to Y\) to its iteration \({\rm ite}_f:\mathbb Z\to {\rm Bij Y}\) easily, but also given an integer-time action \(\alpha(n,y)\) we know it is the iteration of some function, and we can recover it just by evaluating it at one \(\alpha (1,y)\).
The key problem is: someone gives us an \(A\)-action over \(Y\), or equivalently a group homomorphism \({\rm ite}_f:A\to {\rm Bij Y}\) how can we recover the function \(f:Y\to Y\) that we believe it is iteration? Every choice of an element \(u\in A\) extracts from \(\alpha\) a different bijection \(\alpha_u\) over \(Y\).
Proposition 2 Every choice of an element \(u\in A\) defines a process \(u^*: A{\rm - Act}(Y)\to\) of restriction of an \(A\)-Action over \(Y\) to a \(Z\)-action over \(Y\).
Proof: We define \(u^*f:=f\circ u\) Consider an \(A\)-Action, write it as a morphism \(f:\mathbb Z\to {\rm Bij Y}\), precompose it with the iterations of \(u\), i.e. the map \(u:\mathbb Z\to A;\,\, n\mapsto nu\) \[(u^*f)^n(y):=f_{u(n)}(y)\]
![[Image: image.png]](https://i.ibb.co/H4kSGRm/image.png)
A composition of morphism is still a morphism hence it induces an action. \(\square\)
Example (Sanity check): Consider a \((\mathbb C,+)\)-action over the complex numbers, e.g. an exponential map \(\alpha(t,z):=b^t\cdot z\). Every choice of a (additive) group morphism \(\mathbb Z\to\mathbb C\), or equivalently every choice of an element \(u\in \mathbb C\) defines a way to restrict the exponential action to the integers:
\[u^*\alpha(n,z):=(b^u)^n\cdot z\]
but in this case we are lucky since among all the maps \(\mathbb Z\to\mathbb C\) there is a special one: the inclusion of the integers \(1:\mathbb Z\hookrightarrow \mathbb C\), that is precisely the choice of \(u=1\). In this case we say conclude by regarding \( \alpha(t,z):=b^t\cdot z\) as THE (one of the) extension(s) of \(1^*\alpha(n,z):=b^n\cdot z\), an the latter is the iteration of \(b\cdot z\).
Observation 3: For each way \(u\) to send \(\mathbb Z\) to the group \(A\), we get a machine that restricts \(A\)-actions into \(\mathbb Z\)-actions. In other words, for every morphism \(\mathbb Z\to A\) we obtain a way to restrict \(A\)-iterations to integer-iterations. This process is functorial. Much more is true. Every group morphism \(A\to B\) defines a functor that mechanically turns \(B\)-iterations into \(A\) iterations.
Question 3: when we can invert the above construction? In general it is not invertible... there can be two reasons:
Question 4 All of the above discussion adopts a kind of action-point of view, or exp-like. Because it goes from the time to what is to be iterated. It is a contravariant approach that reminds me of constructions like homotopy and homology. But one should study the dual point-of-view too:.
Instead fo the maps \(A\to ...\) we could study maps \(...\to A\), i.e. the measure-like point of view, or log-like (\(L(fg)=L(f)+_AL(g)\)). In this approach we look for Abel functions, the various sets inherit the additive structure. Is this second p-o-v related with character theory and with cohomology?
I try to link this to the previous so to make the whole more reasonable.
Proposition 1: Fix a monoid \((A,+_A)\). Every pair of functions \(f,g\) s.t. \(f:A\to Y\) and \(g:Y\to A\) inverts \(f\) on the left, i.e. \(gf={\rm id}_A\) , defines a semigroup (!) action \(\gamma_{f:g}:A\times Y\to Y\) on the set \(Y\). Not only it is a semigroup action, it also make \(f\) into an \(A\)-equivariant map of semigroup actions.
The last line means that the naturally defined semigroup action \(\gamma_{f:g}\) is homomorphic to the natural translation action of \(A\) on itself via the map \(f\).
Proof: define the binary function \(\gamma_{f:g}:A\times Y\to Y\) as \(\gamma_{f:g,a}(y):=f(a+g(y))\).
This map is an action, i.e. \(\gamma_{f:g,a+b}(y)=\gamma_{f:g,a}(\gamma_{f:g,b}(y))\). Unwrap the definitions to obtain \((f\lambda_ag)(f\lambda_bg) =f\lambda_a\lambda_bg=f\lambda_{a+b}g\).
To conclude the proof we have to show that \(f\in {Hom}_{A{\rm Act}}(A^{\circlearrowleft \lambda},Y^{\circlearrowleft \gamma_{f:g}})\), i.e.
\[f(a+b)=\gamma_{f:g,a}(f(b)) \] Unwrap the definitions again \(f\lambda_a=f\lambda_a(gf)=(f\lambda_ag)f\). \(\square\)
Corollary the map \(\gamma_{f:g}:A\to End(Y)\) that takes \(a\in A\) and produces endomaps \(\gamma_{f:g,a}:Y\to Y\) is a semigroup homomorphism. It sends \(+_A\) into composition over \(Y\) .
Observation 1 The map \(\gamma_{f:g}:A\to End(Y)\) is also a monoid morphism precisely when it sends the element \(0_A\) to the identity. This happens precisely when \(g\) is also a right inverse of \(f\). When this happens \(f\) must be invertible and \(g=f^{-1}\).
Observation 2 (\(A\)-time) Every monoid \(A\)-action over a set \(Y\) defines a monoid morphism \(A\to {\rm End}Y\) (from the "monoid of time" to the "monoid of endofunctions over Y"). We can interpret the latter as a kind of \(A\)-time exponentiation of something that acts on \(Y\) or an algebraic flow over \(Y\). In symbols we can express this as a one-to-one correspondence
Observation 2 (integer-time) If we set \(A=\mathbb Z\), the monoid morphism \(\mathbb Z\to A\), induced by a chosen \(\mathbb Z\)-action, is just the iteration/exponential map \(n\to f^{\circ n}\) and it is a group homomorphism sending \(+_A\) to composition of bijections. In this special case, the initiality of the integers as a group (CORRECTION: the initiality used here is of the pointed group \((\mathbb Z,1)\) in the cat. of pointed groups), i.e. thanks to the recursion theorem, we have that group morphisms \(\mathbb Z\to {\rm Bij}Y\) are the same as, in one-to-one correspondence with, bijective maps \(f:Y \to Y\).
Philosophy What can we take from the previous one-to-one correspondences? If \(\mathbb Z\)-actions over \(Y\) are the same as iteration/discrete flow/exponential map of a bijection, and if choosing a map of that kind is the same as choosing a bijection over \(Y\) then we can can adopt the following philosophy:
Given a group \(A\) we can see \(A\)-actions over a set \(Y\) as if they are \(A\)-time iterations of something that we can think as a (generalized) bijection over \(Y\). In particular we can call that thing a generalized element of the group \({\rm Bij}Y\) (see Generalized element).
Key problem of iteration given an exponential map \(f\) we recover the base "it is exponential of" just by evaluating it at \(1\in \mathbb R\). The same for functions and integer iterations. We go from a bijection \(f:Y\to Y\) to its iteration \({\rm ite}_f:\mathbb Z\to {\rm Bij Y}\) easily, but also given an integer-time action \(\alpha(n,y)\) we know it is the iteration of some function, and we can recover it just by evaluating it at one \(\alpha (1,y)\).
The key problem is: someone gives us an \(A\)-action over \(Y\), or equivalently a group homomorphism \({\rm ite}_f:A\to {\rm Bij Y}\) how can we recover the function \(f:Y\to Y\) that we believe it is iteration? Every choice of an element \(u\in A\) extracts from \(\alpha\) a different bijection \(\alpha_u\) over \(Y\).
Proposition 2 Every choice of an element \(u\in A\) defines a process \(u^*: A{\rm - Act}(Y)\to\) of restriction of an \(A\)-Action over \(Y\) to a \(Z\)-action over \(Y\).
Proof: We define \(u^*f:=f\circ u\) Consider an \(A\)-Action, write it as a morphism \(f:\mathbb Z\to {\rm Bij Y}\), precompose it with the iterations of \(u\), i.e. the map \(u:\mathbb Z\to A;\,\, n\mapsto nu\) \[(u^*f)^n(y):=f_{u(n)}(y)\]
Example (Sanity check): Consider a \((\mathbb C,+)\)-action over the complex numbers, e.g. an exponential map \(\alpha(t,z):=b^t\cdot z\). Every choice of a (additive) group morphism \(\mathbb Z\to\mathbb C\), or equivalently every choice of an element \(u\in \mathbb C\) defines a way to restrict the exponential action to the integers:
\[u^*\alpha(n,z):=(b^u)^n\cdot z\]
but in this case we are lucky since among all the maps \(\mathbb Z\to\mathbb C\) there is a special one: the inclusion of the integers \(1:\mathbb Z\hookrightarrow \mathbb C\), that is precisely the choice of \(u=1\). In this case we say conclude by regarding \( \alpha(t,z):=b^t\cdot z\) as THE (one of the) extension(s) of \(1^*\alpha(n,z):=b^n\cdot z\), an the latter is the iteration of \(b\cdot z\).
Observation 3: For each way \(u\) to send \(\mathbb Z\) to the group \(A\), we get a machine that restricts \(A\)-actions into \(\mathbb Z\)-actions. In other words, for every morphism \(\mathbb Z\to A\) we obtain a way to restrict \(A\)-iterations to integer-iterations. This process is functorial. Much more is true. Every group morphism \(A\to B\) defines a functor that mechanically turns \(B\)-iterations into \(A\) iterations.
Question 3: when we can invert the above construction? In general it is not invertible... there can be two reasons:
- (loss of information/non-injectivity) many \(A\)-iterations could be restricted to the same integer iteration. For example, there are multiple extension of integer tetration.
- (non-surjectivity) this obstruction happens only if there exists an integer iteration that cannot be extended to \(A\)-time exponent, i.e. there are integer iterations not in the image of \(u^*\).
Question 4 All of the above discussion adopts a kind of action-point of view, or exp-like. Because it goes from the time to what is to be iterated. It is a contravariant approach that reminds me of constructions like homotopy and homology. But one should study the dual point-of-view too:.
Instead fo the maps \(A\to ...\) we could study maps \(...\to A\), i.e. the measure-like point of view, or log-like (\(L(fg)=L(f)+_AL(g)\)). In this approach we look for Abel functions, the various sets inherit the additive structure. Is this second p-o-v related with character theory and with cohomology?
Mother Law \(\sigma^+\circ 0=\sigma \circ \sigma^+ \)
\({\rm Grp}_{\rm pt} ({\rm RK}J,G)\cong \mathbb N{\rm Set}_{\rm pt} (J, \Sigma^G)\)