Let $G$ be a topological group, and let $F$ be the Markov free topological group over $G$. We define an action of $G$ on $F$ as follows: $G\times F\rightarrow F$, $^{g}(g_{1}^{\epsilon_{1}}...g_{n}^{\epsilon_{n}})$=$(^{g}g_{1})^{\epsilon_{1}}...(^{g}g_{n})^{\epsilon_{n}}$. Question: Is the abovedefined action, a contiuous action? Why? (Note that, here $G$ acts on itself by conjugation.)

$\begingroup$ Nice question ! $\endgroup$ – David Roberts Feb 25 '13 at 23:37

$\begingroup$ I suspect the answer to the following, more general, question may be obvious and standard to anybody who knows about Markov free topological groups (which I don't ... I had to look up the definition), but if a topological group acts continuously on a topological space $X$, then is the induced action on $F(X)$ also continuous? $\endgroup$ – Jeremy Rickard Feb 26 '13 at 15:40

$\begingroup$ The last section of Pestov's paper topology.auburn.edu/tp/reprints/v24/tp24221.pdf may be helpful. $\endgroup$ – Jeremy Brazas Mar 1 '13 at 12:57
I don't know the answer to the general question but the answer seems to be "yes" when $G$ is compact. Here is a straightforward argument.
The corresponding action of $G$ on the free topological monoid $M=\coprod_{n\geq 0}(G\sqcup G^{1})^n$ is certainly continuous. When $G$ is compact $F(G)$ is the topological quotient of $M$ via word reduction $r:M\to F(G)$. When the action of $G$ on itself is conjugation, we have $^{g}(r(g_{1}^{\epsilon_1}...g_{n}^{\epsilon_n}))=r(^{g}(g_{1}^{\epsilon_1}...g_{n}^{\epsilon_n}))$ for any $g_{1}^{\epsilon_1}...g_{n}^{\epsilon_n}\in M$. The action is continuous since $id\times r:G\times M\to G\times F(G)$ is a topological quotient map (compactness of $G$ helps out here too).
This argument should also work when $G$ is a $k_{\omega}$space (inductive limit of nested compact subspaces) but doesn't extend to the general case since $r:M\to F(G)$ is not quotient when $G=\mathbb{Q}$ is the rationals.

$\begingroup$ The answer is very grateful Dr. Brazas. Yes I agree with you. You are right, absolutly. In fact, I was in search of answer for the general case. I put this question to a simpler form by replacing a topological group $G$ with conjugation action instead of a topological $G$space $X$. I guess at least the answer to this question is yes, when X is a topological group. I hope to reach a definitive answer. And if the answer is no, I find a counterexample. $\endgroup$ – Hossein E. Koshkoshi Feb 28 '13 at 21:25

$\begingroup$ I am interested to know the answer to your general question but remain skeptical. I would be surprised if $\mathbb{Q}$ were not a counterexample. You might try to contact someone who works with free topological groups regularly to see if it is known. $\endgroup$ – Jeremy Brazas Mar 1 '13 at 0:11

$\begingroup$ Obviously, $G=\mathbb{Q}$ is not a conterexample. Since $\mathbb{Q}$ is an abelian group, hence the induced action of $\mathbb{Q}$ on $F(\mathbb{Q})$ is trivial and so is continuous. $\endgroup$ – Hossein E. Koshkoshi Mar 1 '13 at 12:20

$\begingroup$ Oh, of course you are right! That was ridiculous to suggest. I suppose what I was going for was a nonlocally compact group. $\endgroup$ – Jeremy Brazas Mar 1 '13 at 12:42