I am currently reading this paper https://blms.oxfordjournals.org/content/35/5/624.abstract and I have some difficulties to understand two steps of the proof of the main theorem.

Let $G$ be a non discrete topological group.

First of all, the authors indicate by $w(X_1,...,X_n)$ a free word. They don't say what it means exactly but I suppose the correct definition is the following:

$\textbf {Definition}$ A free word $w(X_1,...,X_n)$ is a function $w:G^n\to G$ such that $w(X_1,...,X_n)=X_{i_1}^{a_1}...X_{i_k}^{a_k}$ where $k\ge 1$, $i_1,...,i_k\in \{1,...,n\}$ and $a_1,...,a_k$ are integers different from $0$.

Now define for every $n\ge 2$: $F_n=\{ (g_1,...,g_n) \in G^n: \{g_1,...,g_n\}$ freely generates a free subgroup of G}, $F_\infty=\{(g_i)_1^\infty \in G^\infty: \{g_i\}_1^\infty $freely generates a free subgroup of G }

$\textbf {Warning}$ The ranks of the free groups generated by the elements of $F_n$ and $F_\infty$ are respectively $\textbf {at most } n$ and $\textbf{at most } \infty$. About this see If I say $\{g_1,\dots,g_n\}$ freely generates a subgroup of $G$, does that mean the elements $g_i$ are all distinct?

Now for every word $w$ define $C(w)=\{(g_1,...,g_n)\in G^n: w(g_1,...,g_n)=1\}$ and for every $n$-uple$ K=(k_1,...,k_n) $, where $ k_i$ are naturals,define $C(w,K)=\{(g_i)_1^\infty \in G^\infty: w(g_{k_1},...,g_{k_n})=1 $and the $g_{k_j} $are all distinct}.

$\textbf {Problem 1} $Prove that $F_n=G^n\setminus\bigcup_w C(w)$

$\textbf {Problem 2}$ Prove that $F_\infty=G^\infty\setminus\bigcup_{w,K} C(w,K)$

I know that the idea is that if the word is $1$ then we can in some way deny the definition of free group but I need your help for the details.

  • 1
    $\begingroup$ If for example $\;g_1=g_2\;$ then the reduced word $\;g_1g_2^{-1}=1\;$ , which means the group fulfills a non-trivial relation on its generatos and it thus cannot be free on them. $\endgroup$ – DonAntonio Jul 19 '16 at 10:53
  • $\begingroup$ About your comment, please see the linked question. $\endgroup$ – Richard Jul 19 '16 at 12:17
  • $\begingroup$ Well, up to some agreement in the objections that were rised there, I'd say it still follows that we're assuming the elements $\;g_i\;$ to be different, otherwise things could be too heavy imo. $\endgroup$ – DonAntonio Jul 19 '16 at 12:20
  • $\begingroup$ Indeed, but I think the authors do not consider the $g_i$ distinct and so I need help exactly because the things are heavy. $\endgroup$ – Richard Jul 19 '16 at 12:23
  • $\begingroup$ The paper is 14 years old. Odds are at least one of the authors is still alive. Have you considered writing them/him? $\endgroup$ – DonAntonio Jul 19 '16 at 12:30

[Warning: I do not have access to the article itself and am basing everything below only on what you have written.]

First, a minor point: your definition of "free word" isn't quite right: you need to also require that $i_j\neq i_{j+1}$ for each $j$.

In context it seems clear that $F_n$ is meant to be defined as the set of tuples $(g_1,\dots,g_n)\in G^n$ such that $\{g_1,\dots,g_n\}$ freely generates a subgroup and the $g_i$ are all distinct (that is, in the language of my answer to the previous question, $(g_1,\dots,g_n)$ freely generates a subgroup). Indeed, with your interpretation the claim is not true: if for instance $g\in G$ generates a free subgroup then $(g,g)$ would be in $F_2$ by your interpretation, but it is in $C(w)$ where $w(X_1,X_2)=X_1X_2^{-1}$.

With the interpretation that the $g_i$ must all be distinct, the claim is easy to prove. Indeed, fix $(g_1,\dots,g_n)\in G^n$ and let $H$ be the free group on generators $x_1,\dots,x_n$. There is then a unique homomorphism $\varphi:H\to G$ that sends $x_i$ to $g_i$. Since $H$ is freely generated by the $x_i$, each free word on the $x_i$ is a non-identity element of $H$ (and conversely each non-identity element of $H$ can be represented uniquely as a free word on the $x_i$). Since $\varphi(w(x_1,\dots,x_n))=w(g_1,\dots,x_n)$, the set of non-identity elements of the kernel of $\varphi$ is in bijection with the set of $w$ such that $(g_1,\dots,g_n)\in C(w)$. But $(g_1,\dots,g_n)$ freely generates a subgroup iff $\varphi$ is injective (and hence an isomorphism between $H$ and the subgroup generated by the $g_i$). Thus $(g_1,\dots,g_n)$ freely generates a subgroup iff it is not in any $C(w)$.

For $F_\infty$, your interpretation does make the equation $F_\infty=G^\infty\setminus\bigcup_{w,K} C(w,K)$ true, since the definition of $C(w,K)$ only considers the case that the $g_{k_j}$ are all distinct. However, it seems rather strange to me that this would be intended. I would guess that the definition of $F_\infty$ is meant to require the $g_i$ to all be distinct, and that in defining $C(w,K)$ you should require the $k_j$ to be distinct rather than the $g_{k_j}$. With this interpretation, the proof of the second claim is nearly identical to the proof of the first claim: you now take $H$ to be free on $x_1,x_2,\dots$, and define $\varphi:H\to G$ sending $x_i$ to $g_i$. For any $(w,K)$ (with the $k_j$ all distinct), the element $w(x_{k_1},\dots,x_{k_n})\in H$ is not the identity (and conversely every non-identity element of $H$ has a unique representation of this form). So there is a nontrivial element of the kernel of $\varphi$ iff there is a $w$ and $K$ such that $(g_i)\in C(w,K)$.

Alternatively, it is plausible that actually your interpretation is the intended one for both the $F_n$ and for $F_\infty$, in which case the definition of $C(w)$ should be changed to require that for all $i$ such that $X_i$ appears in $w$, the $g_i$ are distinct. I can't say for sure without seeing the context in the article.

  • $\begingroup$ I really thank you for your help. I really need it. However I still have some doubts. 1) In order to have uniqueness of representation as free word,is my definition ok or should I add that $X_{i_j}\not=X_{i_{j+1}}$? 2) About the infinity case, if I consider the $g_i$ as not distinct, what is a counterexample? The $\infty$-uple $(g,g,....)$ does not work because of the definition of $C(w,K)$. Moreover note that we can have the situation where some $g_i$ are equal but at the same time the set of all the $g_i$ still generates a free subgroup of infinity rank. $\endgroup$ – Richard Jul 19 '16 at 21:56
  • $\begingroup$ Personally I prefer an edited answer instead of comments if it is possible. Thank you again $\endgroup$ – Richard Jul 19 '16 at 21:58
  • $\begingroup$ The context is about "almost freeness". They also extended a result by Dixon, who studied the finite case and actually he said explicitly that the rank of the elements of $F_n$ must be $n$. I copied the definitions from the article ( also C(w) and C(w,K)). So I'd say to use the interpretations that make the two claims true. We are ok for the finite case.For the infinity, so the intepretation is: the elements of $F_\infty$ generate a free subgroup of rank at most $\infty$, right? So in theory also finite rank can be obtained. I m sorry but I need to ask you how can I prove the second claim? $\endgroup$ – Richard Jul 20 '16 at 7:16
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    $\begingroup$ 5) That's correct, the equation $C(w,K)=\pi_K^{-1}[C(w)\cap G^{(n)}]$ is only true for the original definition of $C(w,K)$. Nevertheless, I still seriously doubt that this is what the authors really meant, since the corresponding definition of $F_\infty$ is quite unnatural and not analogous with the definition of $F_n$ that makes the first claim true. For what it's worth, while I can't access the paper in question, in the paper arxiv.org/pdf/1103.1099.pdf that appears to build on its results I can definitely say that my interpretation is the intended one. $\endgroup$ – Eric Wofsey Jul 20 '16 at 15:17
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    $\begingroup$ If $X_1X_1^{-1}$ counts as a free word, then any tuple is in $C(X_1X_1^{-1})$, so $G^n\setminus\bigcup_w C(w)$ would be empty. My proof breaks down because you need to know that $w(x_1,\dots,x_n)\neq 1$ in $H$ for any $w$ in order to obtain a nontrivial element of the kernel of $\varphi$. $\endgroup$ – Eric Wofsey Jul 20 '16 at 22:20

To Eric Wofsey: this answer needs to me to see if I have understood everything of your ideas. Please tell me if there are some mistakes in proofs or definitions and feel free to edit my answer or to make corrections. I put some questions at the end. $\textbf{The first one is fundamental for me}$. After your answers and corrections I will mark your answer as the official one.

$\textbf{Def 1}$ A free word is a function $w:G^n\to G$ such that $w(X_1,...,X_n)=X_{i_1}^{a_1}...X_{i_k}^{a_k}$ with $k\ge 1, a_1,...,a_k$ are integers different from $0$, $i_1,...,i_k\in \{1,...,n\}$ and $i_j\not=i_{j+1}$ for every $j$

$\textbf{Def 2}$ For every $n\ge 2$ let $F_n=\{ (g_1,...,g_n) \in G^n: \{g_1,...,g_n\}$ freely generates a free subgroup of G and all the $g_i$ are distinct}

$\textbf{Def 3} F_\infty=\{(g_i)_1^\infty \in G^\infty: \{g_i\}_1^\infty $freely generates a free subgroup of G and all the $g_i$ are distinct}

$\textbf{Def 4}$ For every free word $w$ define $C(w)=\{(g_1,...,g_n)\in G^n: w(g_1,...,g_n)=1\}$

$\textbf{Def 5}$ For every $n$-uple $K=(k_1,...,k_n) $, where $k_i$ are strictly positive integers and $\textbf{all distinct},$ define $C(w,K)=\{(g_i)_1^\infty \in G^\infty: w(g_{k_1},...,g_{k_n})=1 $}.

$\textbf{Def 6}$ For every $n$-uple$ K=(k_1,...,k_n) $ with the $k_j$ $\textbf{all distinct}$, define $\pi_K(g_1,g_2,...)=(g_{k_1},...,g_{k_n})$

$\textbf {Prop 1}$ $F_n=G^n\setminus\bigcup_w C(w)$

Proof. Fix $(g_1,...,g_n)\in G^n$. Let $H$ be the free group of rank $n$ and generators $x_1,...,x_n$. Then there exists a unique isomorphism $f:H\to G $ such that it send every $x_i$ in $g_i$.

$\textbf{Notation}$ I will indicate by $w'$ the free words defined on $H^n$ and by $w$ the "same" words defined on $G^n$ ( I mean replacin $h_i$ by $g_i$). I will indicate by $T*$ the set $T$ minus the identity.

By definition of free group, we know that for every element in $H*$ there exists a unique free word $w'_h$ such that $w'_h(x_1,...,x_n)=h$. Note also that $f(h)=f(w'_h(x_1,...,.x_n))=w_h(g_1,...,g_n)$.

$\textbf{Fact 1}$ $\alpha:ker f*\to \{w':(g_1,...,g_n)\in C(w)\}=M$ such that $\alpha(t)=w'_t$ is a bijection

It is easy to prove that $\alpha$ is well defined, namely that $\alpha(t)\in M$

Surjectivity: Let $w'\in M$. We have $w(g_1,...,g_n)=1$. Let $t=w'(x_1,...,x_n)$. By definition of free word and free group, we have that $w'(x_1,...,x_n)\not=1$ and $f(w'(x_1,...,x_n))=w(g_1,...,g_n)=1$, namely $t\in ker f*$. By construction and uniqueness we have $\alpha(t)=\alpha(w'(x_1,...,x_n))=w'.$

Injectivity: $\alpha(t_1)=\alpha(t_2)$ iff $w'_{t_1}=w'_{t_2}$ iff $t_1=t_2$ by the uniqueness of representation of the $t_i$ as free words.

$\textbf{Fact 2}$ $(g_1,...,g_n)\in F_n$ iff $f$ is injective

Indeed: if $(g_1,...,g_n)\in F_n$ then the $g_i$ are all distinct and for every $w$ we have $w(g_1,...,g_n)\not =1$. Hence $ ker f*=\{h\in H*:f(h)=1\}=\{h\in H*: f(w'_h(x_1,...,x_n))=1\}=\{h\in H*:w_h(g_1,...,g_n)=1\}=\emptyset$.

On the other hand, if $f$ is injective, then the $g_i$ are all distinct and $H$ is isomorphic to $Im f$. So $Im f$ is the free group of rank $n$. Moreover generators are sent to generators, so the $g_1$ generate $Im f$. Note that $Im f*=\{w_h(g_1,...,g_n):h=w'_h(x_1,...,x_n)\in H*\}$ and if $h_1\not=h_2$ then $w_{h_1}(g_1,...,g_n)\not=w_{h_2}(g_1,...,g_n)$ since $f$ is injective. So by definition of free group we have that $Im f$ is free on the $g_1$.

Now $(g_1,...,g_n)\in F_n$ iff $f$ is injective iff $ker f *=\emptyset=M$ iff $(g_1,...,g_n)$ is not in $C(w)$ for every free word $w$, namely the thesis.

$\textbf {Prop 2}$ $F_\infty=G^\infty\setminus\bigcup_{w,K}C(w,K)$

Proof. Fix $(g_i)\in G^\infty$. Let $H$ be the free group of rank $|\mathbb N|$ on the generators $x_1,x_2,...$. Then there exists a unique homomorphism $f:H\to G$ such that $f(x_i)=g_i$. Moreover we know that for every $h\in H*$ there exists a unique $(w_h,K_h)$ with the $k_j$ all distinct such that $w'_h(x_{k_1},...,x_{k_n})=h$.

$\textbf {Notation}$ Same of the previous proposition and $(g_i)_1^\infty=(g_i)$.

$\textbf {Fact 3}$ $\alpha:ker f*\to \{(w',K):(g_i)\in C(w,K)\}=N$ such that $t\mapsto (w'_t,K_t)$ is a bijection.

It is easy to prove that $\alpha$ is well defined, namely that $\alpha(t)\in N$

Surjectivity: Let $(w',K)\in N$. Now $w(g_{k_1},...,g_{k_n})=1$ with the $k_j$ distinct. Let $t=w'(x_{k_1},...,x_{k_n})$. We have $t\not=1$ and $f(t)=w(g_{k_1},...,g_{k_n})=1$, namely $t\in ker f*$ and clearly $\alpha(t)=(w',K)$.

Injectivity: $\alpha(t_1)=\alpha(t_2)$ iff $(w'_{t_1},K_{t_1})=(w'_{t_2},K_{t_2})$ iff $K_{t_1}=K_{t_2}$ and $w'_{t_1}=w'_{t_2}$. By the uniqueness of the couples $(w'_{t_i}, K_{t_i})$ we have $t_1=t_2$.

$\textbf {Fact 4}$ $(g_i)\in F_\infty$ iff $f$ is injective

Indeed: if $(g_i)\in F_\infty$ then all the $g_i$ are distinct and for every $(w,K)$ with the $k_j$ all distinct we have $w(g_{k_1},...,g_{k_n})\not =1$. So $ker f*=\{h=w'_h(x_{k_1},...,x_{k_n})\in H*:f(h)=w_h(g_{k_1},...,g_{k_n})=1$ and all the $k_j$ are distinct}$=\emptyset$, which implies $f$ is injective.

On the other hand, if $f$ is injective then all the $g_i$ are distinct and $H$ is isomorphic to $Im f$, so $Im f$ is the free group of rank $|\mathbb N|$. It is also generated by the $g_i$ and note that $Im f*=\{w_h(g_{k_1},...,g_{k_n}):h=w'_h(x_{k_1},...,x_{k_n})\in H*$ with all the $k_j$ distinct} and that if $h_1\not=h_2$ then $w_{h_1}(g_{s_1},...,g_{s_n})\not=w_{h_2}(g_{r_1},...,g_{r_n})$, since $f$ is injective (and with the $s_i$ distinct and $r_j$ distinct). So by definition of free group, we deduce that $Im f$ is free on the $g_i$.

Finally we have that $(g_i)\in F_\infty$ iff $f$ is injective iff $ker f*=\emptyset=N$ iff $(g_i)$ is not in $C(w,K)$ for every free word $w$ and every $K$ with the $k_j$ all distinct.

$\textbf {Prop 3}$ $ C(w,K)=\pi_K^{-1}(C(w))$.

Proof. Consider $(g_i)\in C(w,K)$. Then there exists (w,K) with the $k_j$ distinct such that $w(g_{k_1},...,g_{k_n})=1$. On the other hand $\pi_K(g_i)=(g_{k_1},...,g_{k_n})$ and so we have an inclusion. For the other one, note that if $(g_i)\in \pi_K^{-1}(C(w))$ then $\pi_K(g_i)=(g_{k_1},...,g_{k_n})\in C(w)$ and so $w(g_{k_1},...,g_{k_n})=1$ and we have the thesis since the $k_j$ are distinct.

$\textbf{Questions}$ 1) If I consider the definition of $F_\infty$ of this answer but the original definition of C(w,K), namely all the $g_{k_i}$ are distinct which part of my proof breaks down? Note that $g_{k_i}$ distinct implies $k_i$ distinct. Do you have a counterexample about this? 2) If I want $\pi_K$ to be an open function, is it important to require in its definition that all the $k_j$ are distinct?

  • $\begingroup$ @Eric wofsey please see this answer. Thank you very much $\endgroup$ – Richard Jul 23 '16 at 12:11
  • $\begingroup$ 1) Fact 3 can fail. Indeed, $\alpha(t)$ need not even be an element of $N$ for every $t\in\ker(f*)$. If $g_i=g_j$, then $t=x_ix_j^{-1}$ is in $\ker(f*)$ but $\alpha(t)$ is not in $N$. 2) Yes. If $k_i=k_j$, then the image of $\pi_K$ is contained in the set of points whose $i$th coordinate equals their $j$th coordinate, which has empty interior if $G$ has no isolated points. $\endgroup$ – Eric Wofsey Jul 23 '16 at 14:47
  • $\begingroup$ 1)Why not? Consider $K=(i,j)$ and note that $w'_t=w_t=X_iX_j^{-1}$. We have that $w_t(g_j,g_i)=1$, namely $(w'_t,K)\in N$. 2) In your comments you said that with the original definition prop 3 fails. Were you wrong? 3) do you agree with the proofs and definitions of my answer? $\endgroup$ – Richard Jul 23 '16 at 15:07
  • $\begingroup$ 1) By definition of $C(w,K)$, $(g_i)$ cannot be in $C(w,K)$ if $g_i=g_j$. 2) Prop 3 does indeed fail with the orignal definition. 3) They look correct, though I may have overlooked a small detail or two. $\endgroup$ – Eric Wofsey Jul 23 '16 at 15:17
  • $\begingroup$ 1)sorry why not? You just need that in K the indices are distinct and we actually have $i\not =j$. Moreover the evaluation through $w$ is $1$. Could you give me more details please? 2) which part of the prop 3 fails and why? $\endgroup$ – Richard Jul 23 '16 at 15:26

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