Mixing
Newman's “proof” that surface groups are LERF?
$begingroup$
In trying to find alternatives to Scott's paper, I came across Tretkoff Covering Spaces, Subgroup Separability, and the Generalized M. Hall Property, which references this paper of Newman to show surface groups are LERF.
But I don't see how that paper proves it. The main pertinent result of Newman's paper is that for a closed orientable surface of genus $2$, $pi_1(S)$ embeds in $SL(8,mathbb{Z})$.
Newman says the results of Scott's paper follow from the following theorem (which I have paraphrased):
Let $F={a_1,ldots,a_k}subset pi_1(S)$, and suppose $binpi_1(S)$ is such that $ba_ineq a_ib$ for $1le ile k$. Then there is a finite index subgroup $Hle pi_1(S)$ with $bin H$ but $Fcap H=emptyset$.
But how does this show $pi_1(S)$ is LERF? Isn't the conclusion in the wrong direction? By that I mean that LERF would follow if $Fsubset H$ and $bnotin H$. The result as stated seems to only show self-centralizing cyclic subgroups are separable.
group-theory surfaces geometric-group-theory
asked Jan 2 at 16:07
HempeliciousHempelicious
142110
$endgroup$
|
show 1 more comment
$begingroup$
In trying to find alternatives to Scott's paper, I came across Tretkoff Covering Spaces, Subgroup Separability, and the Generalized M. Hall Property, which references this paper of Newman to show surface groups are LERF.
But I don't see how that paper proves it. The main pertinent result of Newman's paper is that for a closed orientable surface of genus $2$, $pi_1(S)$ embeds in $SL(8,mathbb{Z})$.
Newman says the results of Scott's paper follow from the following theorem (which I have paraphrased):
Let $F={a_1,ldots,a_k}subset pi_1(S)$, and suppose $binpi_1(S)$ is such that $ba_ineq a_ib$ for $1le ile k$. Then there is a finite index subgroup $Hle pi_1(S)$ with $bin H$ but $Fcap H=emptyset$.
But how does this show $pi_1(S)$ is LERF? Isn't the conclusion in the wrong direction? By that I mean that LERF would follow if $Fsubset H$ and $bnotin H$. The result as stated seems to only show self-centralizing cyclic subgroups are separable.
group-theory surfaces geometric-group-theory
asked Jan 2 at 16:07
HempeliciousHempelicious
142110
$endgroup$
$begingroup$
I don't see any such claim in Newman's paper. I'd need to see Tretkoff's paper...
$endgroup$
– YCor
Jan 2 at 23:44
$begingroup$
@YCor: Newman's paper contains the line "Scott contains a proof of theorem 2 for surface groups". Since theorem 2 says that self-centralizing subgroups are separable, and Scott doesn't explicitly prove that, it does seem weird to state it that way. But Tretkoff's paper just says surface groups are LERF, and points to Scott, Newman, or Burns,Brunner,Solitar for a proof.
$endgroup$
– Hempelicious
Jan 3 at 1:43
$begingroup$
@YCor: I should add that it is entirely possible Newman does not prove this, but I just wanted to make sure I wasnt missing something.
$endgroup$
– Hempelicious
Jan 3 at 1:44
$begingroup$
If "self-centralizing" means "equal to its own centralizer", it is trivial that every self-centralizing subgroup of a residually finite group is separable (= closed in the profinite topology). Because the centralizer of any subset in a Hausdorff topological group, is closed.
$endgroup$
– YCor
Jan 3 at 17:06
$begingroup$
@YCor: but then what exactly is Newman's paper showing? That these groups are linear?
$endgroup$
– Hempelicious
Jan 4 at 1:29
|
show 1 more comment
$begingroup$
In trying to find alternatives to Scott's paper, I came across Tretkoff Covering Spaces, Subgroup Separability, and the Generalized M. Hall Property, which references this paper of Newman to show surface groups are LERF.
But I don't see how that paper proves it. The main pertinent result of Newman's paper is that for a closed orientable surface of genus $2$, $pi_1(S)$ embeds in $SL(8,mathbb{Z})$.
Newman says the results of Scott's paper follow from the following theorem (which I have paraphrased):
Let $F={a_1,ldots,a_k}subset pi_1(S)$, and suppose $binpi_1(S)$ is such that $ba_ineq a_ib$ for $1le ile k$. Then there is a finite index subgroup $Hle pi_1(S)$ with $bin H$ but $Fcap H=emptyset$.
But how does this show $pi_1(S)$ is LERF? Isn't the conclusion in the wrong direction? By that I mean that LERF would follow if $Fsubset H$ and $bnotin H$. The result as stated seems to only show self-centralizing cyclic subgroups are separable.
group-theory surfaces geometric-group-theory
asked Jan 2 at 16:07
HempeliciousHempelicious
142110
$endgroup$
In trying to find alternatives to Scott's paper, I came across Tretkoff Covering Spaces, Subgroup Separability, and the Generalized M. Hall Property, which references this paper of Newman to show surface groups are LERF.
But I don't see how that paper proves it. The main pertinent result of Newman's paper is that for a closed orientable surface of genus $2$, $pi_1(S)$ embeds in $SL(8,mathbb{Z})$.
Newman says the results of Scott's paper follow from the following theorem (which I have paraphrased):
Let $F={a_1,ldots,a_k}subset pi_1(S)$, and suppose $binpi_1(S)$ is such that $ba_ineq a_ib$ for $1le ile k$. Then there is a finite index subgroup $Hle pi_1(S)$ with $bin H$ but $Fcap H=emptyset$.
But how does this show $pi_1(S)$ is LERF? Isn't the conclusion in the wrong direction? By that I mean that LERF would follow if $Fsubset H$ and $bnotin H$. The result as stated seems to only show self-centralizing cyclic subgroups are separable.
group-theory surfaces geometric-group-theory
group-theory surfaces geometric-group-theory
asked Jan 2 at 16:07
HempeliciousHempelicious
142110
asked Jan 2 at 16:07
HempeliciousHempelicious
142110
asked Jan 2 at 16:07
HempeliciousHempelicious
142110
asked Jan 2 at 16:07
HempeliciousHempelicious
142110
asked Jan 2 at 16:07
HempeliciousHempelicious
142110
142110
$begingroup$
I don't see any such claim in Newman's paper. I'd need to see Tretkoff's paper...
$endgroup$
– YCor
Jan 2 at 23:44
$begingroup$
@YCor: Newman's paper contains the line "Scott contains a proof of theorem 2 for surface groups". Since theorem 2 says that self-centralizing subgroups are separable, and Scott doesn't explicitly prove that, it does seem weird to state it that way. But Tretkoff's paper just says surface groups are LERF, and points to Scott, Newman, or Burns,Brunner,Solitar for a proof.
$endgroup$
– Hempelicious
Jan 3 at 1:43
$begingroup$
@YCor: I should add that it is entirely possible Newman does not prove this, but I just wanted to make sure I wasnt missing something.
$endgroup$
– Hempelicious
Jan 3 at 1:44
$begingroup$
If "self-centralizing" means "equal to its own centralizer", it is trivial that every self-centralizing subgroup of a residually finite group is separable (= closed in the profinite topology). Because the centralizer of any subset in a Hausdorff topological group, is closed.
$endgroup$
– YCor
Jan 3 at 17:06
$begingroup$
@YCor: but then what exactly is Newman's paper showing? That these groups are linear?
$endgroup$
– Hempelicious
Jan 4 at 1:29
|
show 1 more comment
$begingroup$
I don't see any such claim in Newman's paper. I'd need to see Tretkoff's paper...
$endgroup$
– YCor
Jan 2 at 23:44
$begingroup$
@YCor: Newman's paper contains the line "Scott contains a proof of theorem 2 for surface groups". Since theorem 2 says that self-centralizing subgroups are separable, and Scott doesn't explicitly prove that, it does seem weird to state it that way. But Tretkoff's paper just says surface groups are LERF, and points to Scott, Newman, or Burns,Brunner,Solitar for a proof.
$endgroup$
– Hempelicious
Jan 3 at 1:43
$begingroup$
@YCor: I should add that it is entirely possible Newman does not prove this, but I just wanted to make sure I wasnt missing something.
$endgroup$
– Hempelicious
Jan 3 at 1:44
$begingroup$
If "self-centralizing" means "equal to its own centralizer", it is trivial that every self-centralizing subgroup of a residually finite group is separable (= closed in the profinite topology). Because the centralizer of any subset in a Hausdorff topological group, is closed.
$endgroup$
– YCor
Jan 3 at 17:06
$begingroup$
@YCor: but then what exactly is Newman's paper showing? That these groups are linear?
$endgroup$
– Hempelicious
Jan 4 at 1:29
$begingroup$
I don't see any such claim in Newman's paper. I'd need to see Tretkoff's paper...
$endgroup$
– YCor
Jan 2 at 23:44
$begingroup$
I don't see any such claim in Newman's paper. I'd need to see Tretkoff's paper...
$endgroup$
– YCor
Jan 2 at 23:44
$begingroup$
@YCor: Newman's paper contains the line "Scott contains a proof of theorem 2 for surface groups". Since theorem 2 says that self-centralizing subgroups are separable, and Scott doesn't explicitly prove that, it does seem weird to state it that way. But Tretkoff's paper just says surface groups are LERF, and points to Scott, Newman, or Burns,Brunner,Solitar for a proof.
$endgroup$
– Hempelicious
Jan 3 at 1:43
$begingroup$
@YCor: Newman's paper contains the line "Scott contains a proof of theorem 2 for surface groups". Since theorem 2 says that self-centralizing subgroups are separable, and Scott doesn't explicitly prove that, it does seem weird to state it that way. But Tretkoff's paper just says surface groups are LERF, and points to Scott, Newman, or Burns,Brunner,Solitar for a proof.
$endgroup$
– Hempelicious
Jan 3 at 1:43
$begingroup$
@YCor: I should add that it is entirely possible Newman does not prove this, but I just wanted to make sure I wasnt missing something.
$endgroup$
– Hempelicious
Jan 3 at 1:44
$begingroup$
@YCor: I should add that it is entirely possible Newman does not prove this, but I just wanted to make sure I wasnt missing something.
$endgroup$
– Hempelicious
Jan 3 at 1:44
$begingroup$
If "self-centralizing" means "equal to its own centralizer", it is trivial that every self-centralizing subgroup of a residually finite group is separable (= closed in the profinite topology). Because the centralizer of any subset in a Hausdorff topological group, is closed.
$endgroup$
– YCor
Jan 3 at 17:06
$begingroup$
If "self-centralizing" means "equal to its own centralizer", it is trivial that every self-centralizing subgroup of a residually finite group is separable (= closed in the profinite topology). Because the centralizer of any subset in a Hausdorff topological group, is closed.
$endgroup$
– YCor
Jan 3 at 17:06
$begingroup$
@YCor: but then what exactly is Newman's paper showing? That these groups are linear?
$endgroup$
– Hempelicious
Jan 4 at 1:29
$begingroup$
@YCor: but then what exactly is Newman's paper showing? That these groups are linear?
$endgroup$
– Hempelicious
Jan 4 at 1:29
|
show 1 more comment
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059651%2fnewmans-proof-that-surface-groups-are-lerf%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059651%2fnewmans-proof-that-surface-groups-are-lerf%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
I don't see any such claim in Newman's paper. I'd need to see Tretkoff's paper...
$endgroup$
– YCor
Jan 2 at 23:44
$begingroup$
@YCor: Newman's paper contains the line "Scott contains a proof of theorem 2 for surface groups". Since theorem 2 says that self-centralizing subgroups are separable, and Scott doesn't explicitly prove that, it does seem weird to state it that way. But Tretkoff's paper just says surface groups are LERF, and points to Scott, Newman, or Burns,Brunner,Solitar for a proof.
$endgroup$
– Hempelicious
Jan 3 at 1:43
$begingroup$
@YCor: I should add that it is entirely possible Newman does not prove this, but I just wanted to make sure I wasnt missing something.
$endgroup$
– Hempelicious
Jan 3 at 1:44
$begingroup$
If "self-centralizing" means "equal to its own centralizer", it is trivial that every self-centralizing subgroup of a residually finite group is separable (= closed in the profinite topology). Because the centralizer of any subset in a Hausdorff topological group, is closed.
$endgroup$
– YCor
Jan 3 at 17:06
$begingroup$
@YCor: but then what exactly is Newman's paper showing? That these groups are linear?
$endgroup$
– Hempelicious
Jan 4 at 1:29