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Note: this question is really a subquestion of this one, but I decided to ask it separately since it seems it might be attacked first.

Let $B$ be a topological space and $G$ a topological group acting on a space $F$. Then the space $G^B$ of continuous maps from $B$ to $G$ has a group structure given by pointwise multiplication. Moreover, for nice enough spaces the compact-open topology gives $G^B$ the structure of a topological group. This group acts on the space $F^B$ (again endowed with the compact open-topology) in the obvious pointwise way and the action is continuous.

My question is the following: assume $G$ acts transitively; can we infer the same holds for $G^B$? If not, under what assumptions is transitivity preserved?

Notice that freeness and faithfulness are (separately) preserved (it is enough to think pointwise).

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johndoe
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No. Consider the case $G=(\Bbb R,+),~B=F=S^1$, and $G$ acts on $F$ via $$t\cdot z=\exp(it)\cdot z$$ Then there is no map $g:B\to G$ such that $g\cdot 1=\mathrm{id}$, where $1$ is the constant map $B\to F, z\mapsto 1$. Indeed any map $g: S^1\to\Bbb R$ is homotopic to a constant map because $\Bbb R$ is contractible, however $1$ and $\mathrm{id}$ are not homotopic as maps $S^1\to S^1$.

More precisely, in this example we have that two maps $B\to F$ are in the same $G^B$ orbit iff they are homotopic as maps $S^1\to S^1$.

Olivier Bégassat
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  • Do you want me to expand on this answer? Is it clear enough? – Olivier Bégassat May 20 '13 at 20:00
  • Very nice counterexample. Basically each $g\in G^B$ does not change the winding number of any section, so sections with different winding numbers cannot be in the same orbit. Am I right? I notice that this action is not faithful. Do you think it is possible to tweak your example a little in order to get the same behaviour with a faithful action? – johndoe May 20 '13 at 21:35
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    @johndoe if the action is faithful and transitive, then upon fixing a point $\mathrm{pt}\in F$, we get an identification $\psi:G\stackrel{\approx}{\rightarrow} F,~g\mapsto g\cdot\mathrm{pt}$. This is only a continuous bijection, not necessarily a homeomorphism, but in case it is a homeomorphism (for instance if $G$ is compact and $F$ is Hausdorff), any function $f:B\to F$ defines a continuous map $\tilde{f}:B\to G$ with $\tilde{f}=\psi^{-1}\circ f$ such that $f=\tilde{f}\cdot\mathrm{pt}$ where this time $\mathrm{pt}$ stands for the constant map $B\to F$ associated to $\mathrm{pt}\in F$. – Olivier Bégassat May 20 '13 at 21:53
  • And in this case the above behavior is ruled out. – Olivier Bégassat May 20 '13 at 21:55
  • So here you use the fact that $G$ is abelian to derive that the action is free, hence basically recovering the case of a trivial principal $G$-bundle. Interesting. So up to now the answer to my second question is: for $G$ acting freely. I wonder whether there are different conditions we can impose, for instance on $B$. Maybe something involving the homotopy groups of $B$ and $F$ and obstruction theory? – johndoe May 20 '13 at 22:52
  • @johndoe I made a mistake, I thought the action would be free, but it won't be in general! What I wrote works if we assume the action is free instead of merely faithful, sorry. – Olivier Bégassat May 20 '13 at 22:59
  • No, I think you're right in your comment about faithfulness. In your example $G$ is abelian, so assuming faithfulness together with transitivity we obtain freeness automatically. There is this subtlety that a free transitive action might not give a homeomorphism $G\to F$ (see this question), as you actually already pointed out. So modulo this difficulty we can safely get that for $G$ abelian, faithfulness makes $G^B$ transitive. Do you agree? – johndoe May 21 '13 at 08:37