partial converse of existence of covering spaces
$begingroup$
suppose $X$ and $Y$ are two spaces and let $Z$ be a covering spaces for both $X$ and $Y$, then is it true that there is some space $W$ such that $X$ and $Y$ is a covering space of $W$???
(sorry but I didn't get any answer... my intuition says that it might be correct or rather I need to put some more little conditions)
and sorry if somebody thinks that the title is not appropriate.
general-topology algebraic-topology
edited Feb 1 '15 at 16:33
Stefan Hamcke
21.7k42879
asked Feb 1 '15 at 14:57
Anubhav MukherjeeAnubhav Mukherjee
4,9861826
$endgroup$
add a comment |
$begingroup$
suppose $X$ and $Y$ are two spaces and let $Z$ be a covering spaces for both $X$ and $Y$, then is it true that there is some space $W$ such that $X$ and $Y$ is a covering space of $W$???
(sorry but I didn't get any answer... my intuition says that it might be correct or rather I need to put some more little conditions)
and sorry if somebody thinks that the title is not appropriate.
general-topology algebraic-topology
edited Feb 1 '15 at 16:33
Stefan Hamcke
21.7k42879
asked Feb 1 '15 at 14:57
Anubhav MukherjeeAnubhav Mukherjee
4,9861826
$endgroup$
add a comment |
$begingroup$
suppose $X$ and $Y$ are two spaces and let $Z$ be a covering spaces for both $X$ and $Y$, then is it true that there is some space $W$ such that $X$ and $Y$ is a covering space of $W$???
(sorry but I didn't get any answer... my intuition says that it might be correct or rather I need to put some more little conditions)
and sorry if somebody thinks that the title is not appropriate.
general-topology algebraic-topology
edited Feb 1 '15 at 16:33
Stefan Hamcke
21.7k42879
asked Feb 1 '15 at 14:57
Anubhav MukherjeeAnubhav Mukherjee
4,9861826
$endgroup$
suppose $X$ and $Y$ are two spaces and let $Z$ be a covering spaces for both $X$ and $Y$, then is it true that there is some space $W$ such that $X$ and $Y$ is a covering space of $W$???
(sorry but I didn't get any answer... my intuition says that it might be correct or rather I need to put some more little conditions)
and sorry if somebody thinks that the title is not appropriate.
general-topology algebraic-topology
general-topology algebraic-topology
edited Feb 1 '15 at 16:33
Stefan Hamcke
21.7k42879
asked Feb 1 '15 at 14:57
Anubhav MukherjeeAnubhav Mukherjee
4,9861826
edited Feb 1 '15 at 16:33
Stefan Hamcke
21.7k42879
asked Feb 1 '15 at 14:57
Anubhav MukherjeeAnubhav Mukherjee
4,9861826
edited Feb 1 '15 at 16:33
Stefan Hamcke
21.7k42879
edited Feb 1 '15 at 16:33
Stefan Hamcke
21.7k42879
edited Feb 1 '15 at 16:33
Stefan Hamcke
21.7k42879
21.7k42879
asked Feb 1 '15 at 14:57
Anubhav MukherjeeAnubhav Mukherjee
4,9861826
asked Feb 1 '15 at 14:57
Anubhav MukherjeeAnubhav Mukherjee
4,9861826
asked Feb 1 '15 at 14:57
Anubhav MukherjeeAnubhav Mukherjee
4,9861826
4,9861826
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $X$ and $Y$ be the two graphs having two vertices and three edges. There is a common two-sheeted covering space $Z$ of $X$ and $Y$ which is a graph with four vertices and six edges. Exercise: Find $Z$ and show that $X$ and $Y$ are not covering spaces of any other spaces (besides themselves, of course).
Interesting side note: The covering spaces $Zto X$ and $Zto Y$ are defined by free actions of ${mathbb Z}_2$ on $Z$. These generate a ${mathbb Z}_2times{mathbb Z}_2$ action, but it is not a free action since the product of the two generators has a fixed point.
answered Feb 11 '15 at 22:36
Allen HatcherAllen Hatcher
85666
$endgroup$
2
$begingroup$
wonderful!! it is just too beautiful... Z is tow circles connecting their noth poles by an edge and south poles by an edge...and if W exists then it has to be wedge of 3 circles then both X and Y be two sheeted covering of W but any two sheeted covering of W contain 6 edges...so contradiction
$endgroup$
– Anubhav Mukherjee
Feb 11 '15 at 23:20
add a comment |
$begingroup$
No, it is not true.
The open 2-dimensional disc $Z=D^2$ is a covering space of the genus 1 closed surface $X=F_1 = S^1 times S^1$ and of the genus 2 closed surface $Y=F_2 = F_1 # F_1$ (that symbol $#$ means connected sum). One sees this best using geometry. The torus $F_1$ has a Euclidean structure and so there is a locally isometric universal covering map from the Euclidean plane $mathbb{R}^2 to F_1$. And the surface $F_2$ has a hyperbolic structure and so there is a locally isometric universal covering map from the hyperbolic plane $mathbb{H}^2 to F_2$. And, of course, each of $mathbb{R}^2$, $mathbb{H}^2$ is homeomorphic to $D^2$.
But $F_1$ and $F_2$ cannot cover the same space $W$. For suppose they did, compactness of $F_1$ and $F_2$ would imply that the covering maps $F_1 mapsto W$ and $F_2 mapsto W$ are each of finite degree. The fundamental group $pi_1 W$ could therefore contain finite index subgroups $A_1,A_2$ isomorphic to $pi_1F_1$, $pi_1F_2$ respectively. Since $pi_1F_1 = mathbb{Z} oplus mathbb{Z}$ is abelian, $A_1 cap A_2$ is a finite index abelian subgroup of $A_2$. But $pi_1F_2$ has no finite index abelian subgroup, in fact its only abelian subgroups are trivial or infinite cyclic of infinite index.
answered Feb 1 '15 at 16:28
Lee MosherLee Mosher
49.1k33685
$endgroup$
$begingroup$
$pi_1(F_2) = <a,b,c,d/ aba^-1b^-1cdc^-1d^-1>$ but from here how do I conclude the last line that it has no finite index ableian subgroup ?? can you plese help me?
$endgroup$
– Anubhav Mukherjee
Feb 1 '15 at 18:33
$begingroup$
@AnubhaV: that is also best proved using geometry: every discrete abelian subgroup of the group of isometries of $mathbb{H}^2$ fixes some line in $mathbb{H}^2$, but the action of the deck transformation group on $mathbb{H}^2$ has no finite index subgroup fixing a line.
$endgroup$
– Lee Mosher
Feb 1 '15 at 21:06
$begingroup$
here if I restricted Z as a finite sheeted covering space...Now can you give me an counter examle...thanks for helping
$endgroup$
– Anubhav Mukherjee
Feb 6 '15 at 14:07
$begingroup$
@AnubhaV: That's a strong restriction which really makes it a different question. So you should post it as a different question (with a link to this one and an explanation of why you have added the restriction to finite sheeted covering spaces).
$endgroup$
– Lee Mosher
Feb 6 '15 at 20:51
$begingroup$
thanks...I am doing so now...and sir please help me to post properly the question...I dont know how to properly edit questions
$endgroup$
– Anubhav Mukherjee
Feb 6 '15 at 21:56
|
show 1 more comment
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%2f1129047%2fpartial-converse-of-existence-of-covering-spaces%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $X$ and $Y$ be the two graphs having two vertices and three edges. There is a common two-sheeted covering space $Z$ of $X$ and $Y$ which is a graph with four vertices and six edges. Exercise: Find $Z$ and show that $X$ and $Y$ are not covering spaces of any other spaces (besides themselves, of course).
Interesting side note: The covering spaces $Zto X$ and $Zto Y$ are defined by free actions of ${mathbb Z}_2$ on $Z$. These generate a ${mathbb Z}_2times{mathbb Z}_2$ action, but it is not a free action since the product of the two generators has a fixed point.
answered Feb 11 '15 at 22:36
Allen HatcherAllen Hatcher
85666
$endgroup$
2
$begingroup$
wonderful!! it is just too beautiful... Z is tow circles connecting their noth poles by an edge and south poles by an edge...and if W exists then it has to be wedge of 3 circles then both X and Y be two sheeted covering of W but any two sheeted covering of W contain 6 edges...so contradiction
$endgroup$
– Anubhav Mukherjee
Feb 11 '15 at 23:20
add a comment |
$begingroup$
Let $X$ and $Y$ be the two graphs having two vertices and three edges. There is a common two-sheeted covering space $Z$ of $X$ and $Y$ which is a graph with four vertices and six edges. Exercise: Find $Z$ and show that $X$ and $Y$ are not covering spaces of any other spaces (besides themselves, of course).
Interesting side note: The covering spaces $Zto X$ and $Zto Y$ are defined by free actions of ${mathbb Z}_2$ on $Z$. These generate a ${mathbb Z}_2times{mathbb Z}_2$ action, but it is not a free action since the product of the two generators has a fixed point.
answered Feb 11 '15 at 22:36
Allen HatcherAllen Hatcher
85666
$endgroup$
2
$begingroup$
wonderful!! it is just too beautiful... Z is tow circles connecting their noth poles by an edge and south poles by an edge...and if W exists then it has to be wedge of 3 circles then both X and Y be two sheeted covering of W but any two sheeted covering of W contain 6 edges...so contradiction
$endgroup$
– Anubhav Mukherjee
Feb 11 '15 at 23:20
add a comment |
$begingroup$
Let $X$ and $Y$ be the two graphs having two vertices and three edges. There is a common two-sheeted covering space $Z$ of $X$ and $Y$ which is a graph with four vertices and six edges. Exercise: Find $Z$ and show that $X$ and $Y$ are not covering spaces of any other spaces (besides themselves, of course).
Interesting side note: The covering spaces $Zto X$ and $Zto Y$ are defined by free actions of ${mathbb Z}_2$ on $Z$. These generate a ${mathbb Z}_2times{mathbb Z}_2$ action, but it is not a free action since the product of the two generators has a fixed point.
answered Feb 11 '15 at 22:36
Allen HatcherAllen Hatcher
85666
$endgroup$
Let $X$ and $Y$ be the two graphs having two vertices and three edges. There is a common two-sheeted covering space $Z$ of $X$ and $Y$ which is a graph with four vertices and six edges. Exercise: Find $Z$ and show that $X$ and $Y$ are not covering spaces of any other spaces (besides themselves, of course).
Interesting side note: The covering spaces $Zto X$ and $Zto Y$ are defined by free actions of ${mathbb Z}_2$ on $Z$. These generate a ${mathbb Z}_2times{mathbb Z}_2$ action, but it is not a free action since the product of the two generators has a fixed point.
answered Feb 11 '15 at 22:36
Allen HatcherAllen Hatcher
85666
answered Feb 11 '15 at 22:36
Allen HatcherAllen Hatcher
85666
answered Feb 11 '15 at 22:36
Allen HatcherAllen Hatcher
85666
answered Feb 11 '15 at 22:36
Allen HatcherAllen Hatcher
85666
85666
2
$begingroup$
wonderful!! it is just too beautiful... Z is tow circles connecting their noth poles by an edge and south poles by an edge...and if W exists then it has to be wedge of 3 circles then both X and Y be two sheeted covering of W but any two sheeted covering of W contain 6 edges...so contradiction
$endgroup$
– Anubhav Mukherjee
Feb 11 '15 at 23:20
add a comment |
2
$begingroup$
wonderful!! it is just too beautiful... Z is tow circles connecting their noth poles by an edge and south poles by an edge...and if W exists then it has to be wedge of 3 circles then both X and Y be two sheeted covering of W but any two sheeted covering of W contain 6 edges...so contradiction
$endgroup$
– Anubhav Mukherjee
Feb 11 '15 at 23:20
2
2
$begingroup$
wonderful!! it is just too beautiful... Z is tow circles connecting their noth poles by an edge and south poles by an edge...and if W exists then it has to be wedge of 3 circles then both X and Y be two sheeted covering of W but any two sheeted covering of W contain 6 edges...so contradiction
$endgroup$
– Anubhav Mukherjee
Feb 11 '15 at 23:20
$begingroup$
wonderful!! it is just too beautiful... Z is tow circles connecting their noth poles by an edge and south poles by an edge...and if W exists then it has to be wedge of 3 circles then both X and Y be two sheeted covering of W but any two sheeted covering of W contain 6 edges...so contradiction
$endgroup$
– Anubhav Mukherjee
Feb 11 '15 at 23:20
add a comment |
$begingroup$
No, it is not true.
The open 2-dimensional disc $Z=D^2$ is a covering space of the genus 1 closed surface $X=F_1 = S^1 times S^1$ and of the genus 2 closed surface $Y=F_2 = F_1 # F_1$ (that symbol $#$ means connected sum). One sees this best using geometry. The torus $F_1$ has a Euclidean structure and so there is a locally isometric universal covering map from the Euclidean plane $mathbb{R}^2 to F_1$. And the surface $F_2$ has a hyperbolic structure and so there is a locally isometric universal covering map from the hyperbolic plane $mathbb{H}^2 to F_2$. And, of course, each of $mathbb{R}^2$, $mathbb{H}^2$ is homeomorphic to $D^2$.
But $F_1$ and $F_2$ cannot cover the same space $W$. For suppose they did, compactness of $F_1$ and $F_2$ would imply that the covering maps $F_1 mapsto W$ and $F_2 mapsto W$ are each of finite degree. The fundamental group $pi_1 W$ could therefore contain finite index subgroups $A_1,A_2$ isomorphic to $pi_1F_1$, $pi_1F_2$ respectively. Since $pi_1F_1 = mathbb{Z} oplus mathbb{Z}$ is abelian, $A_1 cap A_2$ is a finite index abelian subgroup of $A_2$. But $pi_1F_2$ has no finite index abelian subgroup, in fact its only abelian subgroups are trivial or infinite cyclic of infinite index.
answered Feb 1 '15 at 16:28
Lee MosherLee Mosher
49.1k33685
$endgroup$
$begingroup$
$pi_1(F_2) = <a,b,c,d/ aba^-1b^-1cdc^-1d^-1>$ but from here how do I conclude the last line that it has no finite index ableian subgroup ?? can you plese help me?
$endgroup$
– Anubhav Mukherjee
Feb 1 '15 at 18:33
$begingroup$
@AnubhaV: that is also best proved using geometry: every discrete abelian subgroup of the group of isometries of $mathbb{H}^2$ fixes some line in $mathbb{H}^2$, but the action of the deck transformation group on $mathbb{H}^2$ has no finite index subgroup fixing a line.
$endgroup$
– Lee Mosher
Feb 1 '15 at 21:06
$begingroup$
here if I restricted Z as a finite sheeted covering space...Now can you give me an counter examle...thanks for helping
$endgroup$
– Anubhav Mukherjee
Feb 6 '15 at 14:07
$begingroup$
@AnubhaV: That's a strong restriction which really makes it a different question. So you should post it as a different question (with a link to this one and an explanation of why you have added the restriction to finite sheeted covering spaces).
$endgroup$
– Lee Mosher
Feb 6 '15 at 20:51
$begingroup$
thanks...I am doing so now...and sir please help me to post properly the question...I dont know how to properly edit questions
$endgroup$
– Anubhav Mukherjee
Feb 6 '15 at 21:56
|
show 1 more comment
$begingroup$
No, it is not true.
The open 2-dimensional disc $Z=D^2$ is a covering space of the genus 1 closed surface $X=F_1 = S^1 times S^1$ and of the genus 2 closed surface $Y=F_2 = F_1 # F_1$ (that symbol $#$ means connected sum). One sees this best using geometry. The torus $F_1$ has a Euclidean structure and so there is a locally isometric universal covering map from the Euclidean plane $mathbb{R}^2 to F_1$. And the surface $F_2$ has a hyperbolic structure and so there is a locally isometric universal covering map from the hyperbolic plane $mathbb{H}^2 to F_2$. And, of course, each of $mathbb{R}^2$, $mathbb{H}^2$ is homeomorphic to $D^2$.
But $F_1$ and $F_2$ cannot cover the same space $W$. For suppose they did, compactness of $F_1$ and $F_2$ would imply that the covering maps $F_1 mapsto W$ and $F_2 mapsto W$ are each of finite degree. The fundamental group $pi_1 W$ could therefore contain finite index subgroups $A_1,A_2$ isomorphic to $pi_1F_1$, $pi_1F_2$ respectively. Since $pi_1F_1 = mathbb{Z} oplus mathbb{Z}$ is abelian, $A_1 cap A_2$ is a finite index abelian subgroup of $A_2$. But $pi_1F_2$ has no finite index abelian subgroup, in fact its only abelian subgroups are trivial or infinite cyclic of infinite index.
answered Feb 1 '15 at 16:28
Lee MosherLee Mosher
49.1k33685
$endgroup$
$begingroup$
$pi_1(F_2) = <a,b,c,d/ aba^-1b^-1cdc^-1d^-1>$ but from here how do I conclude the last line that it has no finite index ableian subgroup ?? can you plese help me?
$endgroup$
– Anubhav Mukherjee
Feb 1 '15 at 18:33
$begingroup$
@AnubhaV: that is also best proved using geometry: every discrete abelian subgroup of the group of isometries of $mathbb{H}^2$ fixes some line in $mathbb{H}^2$, but the action of the deck transformation group on $mathbb{H}^2$ has no finite index subgroup fixing a line.
$endgroup$
– Lee Mosher
Feb 1 '15 at 21:06
$begingroup$
here if I restricted Z as a finite sheeted covering space...Now can you give me an counter examle...thanks for helping
$endgroup$
– Anubhav Mukherjee
Feb 6 '15 at 14:07
$begingroup$
@AnubhaV: That's a strong restriction which really makes it a different question. So you should post it as a different question (with a link to this one and an explanation of why you have added the restriction to finite sheeted covering spaces).
$endgroup$
– Lee Mosher
Feb 6 '15 at 20:51
$begingroup$
thanks...I am doing so now...and sir please help me to post properly the question...I dont know how to properly edit questions
$endgroup$
– Anubhav Mukherjee
Feb 6 '15 at 21:56
|
show 1 more comment
$begingroup$
No, it is not true.
The open 2-dimensional disc $Z=D^2$ is a covering space of the genus 1 closed surface $X=F_1 = S^1 times S^1$ and of the genus 2 closed surface $Y=F_2 = F_1 # F_1$ (that symbol $#$ means connected sum). One sees this best using geometry. The torus $F_1$ has a Euclidean structure and so there is a locally isometric universal covering map from the Euclidean plane $mathbb{R}^2 to F_1$. And the surface $F_2$ has a hyperbolic structure and so there is a locally isometric universal covering map from the hyperbolic plane $mathbb{H}^2 to F_2$. And, of course, each of $mathbb{R}^2$, $mathbb{H}^2$ is homeomorphic to $D^2$.
But $F_1$ and $F_2$ cannot cover the same space $W$. For suppose they did, compactness of $F_1$ and $F_2$ would imply that the covering maps $F_1 mapsto W$ and $F_2 mapsto W$ are each of finite degree. The fundamental group $pi_1 W$ could therefore contain finite index subgroups $A_1,A_2$ isomorphic to $pi_1F_1$, $pi_1F_2$ respectively. Since $pi_1F_1 = mathbb{Z} oplus mathbb{Z}$ is abelian, $A_1 cap A_2$ is a finite index abelian subgroup of $A_2$. But $pi_1F_2$ has no finite index abelian subgroup, in fact its only abelian subgroups are trivial or infinite cyclic of infinite index.
answered Feb 1 '15 at 16:28
Lee MosherLee Mosher
49.1k33685
$endgroup$
No, it is not true.
The open 2-dimensional disc $Z=D^2$ is a covering space of the genus 1 closed surface $X=F_1 = S^1 times S^1$ and of the genus 2 closed surface $Y=F_2 = F_1 # F_1$ (that symbol $#$ means connected sum). One sees this best using geometry. The torus $F_1$ has a Euclidean structure and so there is a locally isometric universal covering map from the Euclidean plane $mathbb{R}^2 to F_1$. And the surface $F_2$ has a hyperbolic structure and so there is a locally isometric universal covering map from the hyperbolic plane $mathbb{H}^2 to F_2$. And, of course, each of $mathbb{R}^2$, $mathbb{H}^2$ is homeomorphic to $D^2$.
But $F_1$ and $F_2$ cannot cover the same space $W$. For suppose they did, compactness of $F_1$ and $F_2$ would imply that the covering maps $F_1 mapsto W$ and $F_2 mapsto W$ are each of finite degree. The fundamental group $pi_1 W$ could therefore contain finite index subgroups $A_1,A_2$ isomorphic to $pi_1F_1$, $pi_1F_2$ respectively. Since $pi_1F_1 = mathbb{Z} oplus mathbb{Z}$ is abelian, $A_1 cap A_2$ is a finite index abelian subgroup of $A_2$. But $pi_1F_2$ has no finite index abelian subgroup, in fact its only abelian subgroups are trivial or infinite cyclic of infinite index.
answered Feb 1 '15 at 16:28
Lee MosherLee Mosher
49.1k33685
edited Feb 1 '15 at 16:35
answered Feb 1 '15 at 16:28
Lee MosherLee Mosher
49.1k33685
answered Feb 1 '15 at 16:28
Lee MosherLee Mosher
49.1k33685
answered Feb 1 '15 at 16:28
Lee MosherLee Mosher
49.1k33685
49.1k33685
$begingroup$
$pi_1(F_2) = <a,b,c,d/ aba^-1b^-1cdc^-1d^-1>$ but from here how do I conclude the last line that it has no finite index ableian subgroup ?? can you plese help me?
$endgroup$
– Anubhav Mukherjee
Feb 1 '15 at 18:33
$begingroup$
@AnubhaV: that is also best proved using geometry: every discrete abelian subgroup of the group of isometries of $mathbb{H}^2$ fixes some line in $mathbb{H}^2$, but the action of the deck transformation group on $mathbb{H}^2$ has no finite index subgroup fixing a line.
$endgroup$
– Lee Mosher
Feb 1 '15 at 21:06
$begingroup$
here if I restricted Z as a finite sheeted covering space...Now can you give me an counter examle...thanks for helping
$endgroup$
– Anubhav Mukherjee
Feb 6 '15 at 14:07
$begingroup$
@AnubhaV: That's a strong restriction which really makes it a different question. So you should post it as a different question (with a link to this one and an explanation of why you have added the restriction to finite sheeted covering spaces).
$endgroup$
– Lee Mosher
Feb 6 '15 at 20:51
$begingroup$
thanks...I am doing so now...and sir please help me to post properly the question...I dont know how to properly edit questions
$endgroup$
– Anubhav Mukherjee
Feb 6 '15 at 21:56
|
show 1 more comment
$begingroup$
$pi_1(F_2) = <a,b,c,d/ aba^-1b^-1cdc^-1d^-1>$ but from here how do I conclude the last line that it has no finite index ableian subgroup ?? can you plese help me?
$endgroup$
– Anubhav Mukherjee
Feb 1 '15 at 18:33
$begingroup$
@AnubhaV: that is also best proved using geometry: every discrete abelian subgroup of the group of isometries of $mathbb{H}^2$ fixes some line in $mathbb{H}^2$, but the action of the deck transformation group on $mathbb{H}^2$ has no finite index subgroup fixing a line.
$endgroup$
– Lee Mosher
Feb 1 '15 at 21:06
$begingroup$
here if I restricted Z as a finite sheeted covering space...Now can you give me an counter examle...thanks for helping
$endgroup$
– Anubhav Mukherjee
Feb 6 '15 at 14:07
$begingroup$
@AnubhaV: That's a strong restriction which really makes it a different question. So you should post it as a different question (with a link to this one and an explanation of why you have added the restriction to finite sheeted covering spaces).
$endgroup$
– Lee Mosher
Feb 6 '15 at 20:51
$begingroup$
thanks...I am doing so now...and sir please help me to post properly the question...I dont know how to properly edit questions
$endgroup$
– Anubhav Mukherjee
Feb 6 '15 at 21:56
$begingroup$
$pi_1(F_2) = <a,b,c,d/ aba^-1b^-1cdc^-1d^-1>$ but from here how do I conclude the last line that it has no finite index ableian subgroup ?? can you plese help me?
$endgroup$
– Anubhav Mukherjee
Feb 1 '15 at 18:33
$begingroup$
$pi_1(F_2) = <a,b,c,d/ aba^-1b^-1cdc^-1d^-1>$ but from here how do I conclude the last line that it has no finite index ableian subgroup ?? can you plese help me?
$endgroup$
– Anubhav Mukherjee
Feb 1 '15 at 18:33
$begingroup$
@AnubhaV: that is also best proved using geometry: every discrete abelian subgroup of the group of isometries of $mathbb{H}^2$ fixes some line in $mathbb{H}^2$, but the action of the deck transformation group on $mathbb{H}^2$ has no finite index subgroup fixing a line.
$endgroup$
– Lee Mosher
Feb 1 '15 at 21:06
$begingroup$
@AnubhaV: that is also best proved using geometry: every discrete abelian subgroup of the group of isometries of $mathbb{H}^2$ fixes some line in $mathbb{H}^2$, but the action of the deck transformation group on $mathbb{H}^2$ has no finite index subgroup fixing a line.
$endgroup$
– Lee Mosher
Feb 1 '15 at 21:06
$begingroup$
here if I restricted Z as a finite sheeted covering space...Now can you give me an counter examle...thanks for helping
$endgroup$
– Anubhav Mukherjee
Feb 6 '15 at 14:07
$begingroup$
here if I restricted Z as a finite sheeted covering space...Now can you give me an counter examle...thanks for helping
$endgroup$
– Anubhav Mukherjee
Feb 6 '15 at 14:07
$begingroup$
@AnubhaV: That's a strong restriction which really makes it a different question. So you should post it as a different question (with a link to this one and an explanation of why you have added the restriction to finite sheeted covering spaces).
$endgroup$
– Lee Mosher
Feb 6 '15 at 20:51
$begingroup$
@AnubhaV: That's a strong restriction which really makes it a different question. So you should post it as a different question (with a link to this one and an explanation of why you have added the restriction to finite sheeted covering spaces).
$endgroup$
– Lee Mosher
Feb 6 '15 at 20:51
$begingroup$
thanks...I am doing so now...and sir please help me to post properly the question...I dont know how to properly edit questions
$endgroup$
– Anubhav Mukherjee
Feb 6 '15 at 21:56
$begingroup$
thanks...I am doing so now...and sir please help me to post properly the question...I dont know how to properly edit questions
$endgroup$
– Anubhav Mukherjee
Feb 6 '15 at 21:56
|
show 1 more comment
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%2f1129047%2fpartial-converse-of-existence-of-covering-spaces%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
