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Let $P$ and $Q$ both be sets of $n$ points on a connected manifold $M$. Does there exist an action by a free group $F$ on $M$ such that for every $i$ we have a $g\in F$ with $g\cdot P_i = Q_i$?

I'm not specifically asking for a proof, but I've been trying for hours to find an answer so I would appreciate any hints. (For a start it would be good to know if the answer is yes or no - I am lacking intuition here and I am having real trouble finding out what I even want to show...)

Definition: A free group on $M$ is a group $F$ and a map $f:M\to F$ with the universal property: for any map $\phi:M\to G$ ($G$ group) there exists a unique group homomorphism $\varphi:F\to G$ such that $\phi = \varphi\circ f$.

My attempts:

If the answer to the above question is yes, then as far as I can tell there are not many options for groups to consider. Namely the free groups $\langle P\rangle, \langle Q\rangle$ and $\langle P\rangle * \langle Q\rangle$ come to mind.

  1. Trying any of the 3 groups as $F$ leaves me clueless about how to define $f$ (the obvious map $A\to\langle A\rangle$ for any $A$ yields a free group but that doesn't fit here).

  2. Since $M$ is connected, it is path-connected and the action I want to define just takes $P_i$ to the endpoint of the path that connects to $Q_i$. But that doesn't work out-of-the-box since I have to define the action on the whole $M$. I guess moving only $P$ to $Q$ and leaving the rest as it is is not really continuous... (Plus I still need to define $f$ and $F$ properly.)

If the answer to the above question is no, then:

  1. I've been experimenting with drawing commutative diagrams (since the universal property suggests that kind of) but the only nice thing to notice is that since $|P|=|Q|$, I know that $\langle P\rangle$ and $\langle Q\rangle$ are isomorphic. Taking those two as two $G$-s in the definition above (to get a better understanding if this is plausible) yields a diagram that looks nice somehow yet seems useless to me.

  2. I get that one of the main properties of the free groups above is that they aren't abelian by construction - yet every action defines some kind of symmetry. That strikes me as odd. But this thought is just a product of brain storming that is far too abstract to be of concrete use to me. It's just me trying to get some intuition.

  3. Finding a counterexample feels hard since the only combination of a manifold and a free-ish group that I've come across is the Klein bottle where the fundamental group is generated by two deck transformations $a$ and $b$ of the universal cover. (See Wikipedia.)

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  • $\begingroup$ Just a note: there's nothing wrong with non-commutativity and symmetries. The word symmetry here is not about the group operation. $\endgroup$ – lisyarus Dec 16 '16 at 7:34
  • $\begingroup$ I don't understand what you call "a free group on $M$". Does $F$ have some kind of topology? Is the map assumed to be continuous $f$, or is it just a set map? How is the action ($g \cdot P_i$) related to $f$? $\endgroup$ – Najib Idrissi Dec 16 '16 at 9:08
  • $\begingroup$ @Najib Damn, I forgot to mention that we had the general assumption that throughout that lecture "all maps are continuous". I'm sorry, I'll add that. As far as I can tell $F$ doesn't come with a topology a priori. (For $f$ being cont. we obviously need one...) I guess that "how the action is related to $f$" is a big part of this question. I can try to provide more context, but there is not much more - I didn't leave anything out in the quotes. $\endgroup$ – Piwi Dec 16 '16 at 10:15

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