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Benjamin Steinberg
Benjamin Steinberg
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Now the issue is that if a proper subset $S$ generates $G$, it is rarely the case that $X$$S$ generates $M(G)$ under the universal dual prehomomorphism $\theta$ above. Maybe it is never the case if $G$ is non-trivial. It is not the case for free groups. So you have some partial actions that are not determined on generators.

Now the issue is that if $S$ generates $G$, it is rarely the case that $X$ generates $M(G)$ under the universal dual prehomomorphism $\theta$ above. Maybe it is never the case if $G$ is non-trivial. It is not the case for free groups. So you have some partial actions that are not determined on generators.

Now the issue is that if a proper subset $S$ generates $G$, it is rarely the case that $S$ generates $M(G)$ under the universal dual prehomomorphism $\theta$ above. Maybe it is never the case if $G$ is non-trivial. It is not the case for free groups. So you have some partial actions that are not determined on generators.

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Benjamin Steinberg
Benjamin Steinberg
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There is a way to replace dual prehomomorphisms by inverse semigroup homomorphism using the so-called Birget-Rhodes expansion of a group $G$ (rediscovered by Exel). It is the inverse monoid $M(G)$ whose element consist of pairs $(A,g)$ where $A$ is a finite subset of $G$ containing $g$$1,g$ and you multiply by the rule $(A,g)(B,h) = (A\cup gB,h)$$(A,g)(B,h) = (A\cup gB,gh)$. There is a natural homomorphism $\pi\colon M(G)\to G$ that projects to the second coordinate and it has the property the the preimage of each element of $g$ has a maximum element in the natural partial order (namely $(\{g\},g)$$(\{1,g\},g)$) and so $M(G)$ is what is called an $F$-inverse monoid. There is a dual prehomomorphism $\theta\colon G\to M(G)$ given by $\theta(g) =(\{g\},g)$$\theta(g) =(\{1,g\},g)$ and it is universal in the sense that any dual prehomomorphism from $G$ into an inverse monoid factors through this one. In particular, partial actions of $G$ on a set $X$ correspond to genuine inverse semigroup actions of $M(G)$ on $X$ via a homomorphism $M(G)\to I_X$.

Now the issue is that if $S$ generates $G$, it is rarely the case that $X$ generates $M(G)$ under the universal dual prehomomorphism $\theta$ above. Maybe it is never the case if $G$ is non-trivial. It is not the case for free groups.. So you have some partial actions that are not determined on generators.

There is a way to replace dual prehomomorphisms by inverse semigroup homomorphism using the so-called Birget-Rhodes expansion of a group $G$ (rediscovered by Exel). It is the inverse monoid $M(G)$ whose element consist of pairs $(A,g)$ where $A$ is a finite subset of $G$ containing $g$ and you multiply by the rule $(A,g)(B,h) = (A\cup gB,h)$. There is a natural homomorphism $\pi\colon M(G)\to G$ that projects to the second coordinate and it has the property the the preimage of each element of $g$ has a maximum element in the natural partial order (namely $(\{g\},g)$) and so $M(G)$ is what is called an $F$-inverse monoid. There is a dual prehomomorphism $\theta\colon G\to M(G)$ given by $\theta(g) =(\{g\},g)$ and it is universal in the sense that any dual prehomomorphism from $G$ into an inverse monoid factors through this one. In particular, partial actions of $G$ on a set $X$ correspond to genuine inverse semigroup actions of $M(G)$ on $X$ via a homomorphism $M(G)\to I_X$.

Now the issue is that if $S$ generates $G$, it is rarely the case that $X$ generates $M(G)$ under the universal dual prehomomorphism $\theta$ above. Maybe it is never the case if $G$ is non-trivial. It is not the case for free groups.. So you have some partial actions that are not determined on generators.

There is a way to replace dual prehomomorphisms by inverse semigroup homomorphism using the so-called Birget-Rhodes expansion of a group $G$ (rediscovered by Exel). It is the inverse monoid $M(G)$ whose element consist of pairs $(A,g)$ where $A$ is a finite subset of $G$ containing $1,g$ and you multiply by the rule $(A,g)(B,h) = (A\cup gB,gh)$. There is a natural homomorphism $\pi\colon M(G)\to G$ that projects to the second coordinate and it has the property the the preimage of each element of $g$ has a maximum element in the natural partial order (namely $(\{1,g\},g)$) and so $M(G)$ is what is called an $F$-inverse monoid. There is a dual prehomomorphism $\theta\colon G\to M(G)$ given by $\theta(g) =(\{1,g\},g)$ and it is universal in the sense that any dual prehomomorphism from $G$ into an inverse monoid factors through this one. In particular, partial actions of $G$ on a set $X$ correspond to genuine inverse semigroup actions of $M(G)$ on $X$ via a homomorphism $M(G)\to I_X$.

Now the issue is that if $S$ generates $G$, it is rarely the case that $X$ generates $M(G)$ under the universal dual prehomomorphism $\theta$ above. Maybe it is never the case if $G$ is non-trivial. It is not the case for free groups. So you have some partial actions that are not determined on generators.

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Benjamin Steinberg
Benjamin Steinberg
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This is not a real answer but things are getting too long for a comment. If $X$ is a set, let $I_X$ be the symmetric inverse monoid, the monoid of all partial bijections of $X$. It is naturally ordered by the restriction relation and this partial order is compatible with multiplication and preserved by inversion. It can also be defined by $f\leq g$ if $f=ge$ for some idempotent $e$ and this definition works in any inverse semigroup. Inverse semigroup is the abstract algebraic structure that encodes pseudogroups of transformations.

A partial group action is intuitively what you get when a group $G$ acts on a set $X$ and you have a subset $Y$ and view each $g\in G$ as giving a partial bijection from $Y$ to itself by defining $gy$ only when $gy\in Y$. In fact, it is shown in the Kellendonk-Lawson article I mentioned in the comments that every partial action arises this way. However, if the partial action preserves some extra structure on $Y$, there is no reason that the globalized set $X$ on which $G$ acts satisfies this extra structure.

The formal definition (which was given by Exel in a different but equivalent way) is a partial action of $G$ on a set $X$ is a dual prehomomorphism $\theta\colon G\to I_X$. What does this mean? It means $\theta(1)=1$, $\theta(g)\theta(h)\leq \theta(gh)$ and $\theta(g^{-1}) = \theta(g)^{-1}$. The middle axioms says that first acting by $g$ and then by $h$ is a restriction of the action of $gh$. This makes sense because maybe $hy$ is not in $Y$ but $ghy$ is.

Because $\theta$ is not a genuine homomorphism, there is no reason that $\theta$ should be determined by what it does to generators of $G$ and in fact it isn't.

There is a way to replace dual prehomomorphisms by inverse semigroup homomorphism using the so-called Birget-Rhodes expansion of a group $G$ (rediscovered by Exel). It is the inverse monoid $M(G)$ whose element consist of pairs $(A,g)$ where $A$ is a finite subset of $G$ containing $g$ and you multiply by the rule $(A,g)(B,h) = (A\cup gB,h)$. There is a natural homomorphism $\pi\colon M(G)\to G$ that projects to the second coordinate and it has the property the the preimage of each element of $g$ has a maximum element in the natural partial order (namely $(\{g\},g)$) and so $M(G)$ is what is called an $F$-inverse monoid. There is a dual prehomomorphism $\theta\colon G\to M(G)$ given by $\theta(g) =(\{g\},g)$ and it is universal in the sense that any dual prehomomorphism from $G$ into an inverse monoid factors through this one. In particular, partial actions of $G$ on a set $X$ correspond to genuine inverse semigroup actions of $M(G)$ on $X$ via a homomorphism $M(G)\to I_X$.

Now the issue is that if $S$ generates $G$, it is rarely the case that $X$ generates $M(G)$ under the universal dual prehomomorphism $\theta$ above. Maybe it is never the case if $G$ is non-trivial. It is not the case for free groups.. So you have some partial actions that are not determined on generators.

When a partial action of a free group on a set is determined by generators is understood, but I don't know about the general case. If you have any $F$-inverse cover of $G$, that is an inverse monoid $M$ with a surjective homomorphism $\pi\colon M\to G$ such that each fiber has a maximum element, then you can define a dual prehomomorphism $\theta\colon G\to M$ by $\theta(g)$ is the max element of its fiber. Then any action of $M$ on a set $X$ can be turned into a partial action of $G$.

Now if $S$ is a set, the free inverse monoid $FIM(S)$ is an $F$-inverse cover of the free group $FG(S)$ under the natural projection. The maximum element of each fiber is the element represented by the reduced word in $S$ and its generators mapping to the group element. So there is a dual prehomomorphism $\theta\colon FG(S)\to FIM(S)$ and using the universal property of $FIM(S)$, it is easy to see that any collection of partial bijections of a set $X$ indexed by elements of $S$ give a partial action of $FG(S)$ in which the action is determined by what happens to $S$. Now somebody I think wrote up in some old papers axiomatically which actions of $FG(S)$ come about this way, but I forget where. If you had any partial action of $FG(S)$ generated by $S$ in any reasonable sense, it will have to factor through this one. It is in some sense, that I forget how to make precise, the universal generator preserving dual prehomomorphism. But I don't know for other groups how to do this.