Locally compact nilpotent group has an open subgroup isomorphic to $mathbb{R}^ntimes K$
$begingroup$
My question is about a possible generalization of the following structure theorem of locally compact abelian groups.
Theorem: Let $G$ be a locally compact abelian group. Then here exists a compact subgroup $K$ and a non-negative number $ninmathbb{N}$ such that $mathbb{R}^ntimes K$ is isomorphic to an open subgroup of $G$.
I wonder whether it is possible to generalize the above for non-abelian but nilpotent groups.
The most simple case is when $G$ is nilpotent of order $3$. In this case $[G,G]$ is an abelian group, hence $[G,G]$ satisfies the Theorem. Moreover $G/[G,G]$ is always an abelian group and so satisfies the Theorem.
Can we deduce that $G$ also satisfies the theorem?
topological-groups locally-compact-groups nilpotent-groups
asked Oct 20 '18 at 12:14
YankoYanko
6,3511527
$endgroup$
add a comment |
$begingroup$
My question is about a possible generalization of the following structure theorem of locally compact abelian groups.
Theorem: Let $G$ be a locally compact abelian group. Then here exists a compact subgroup $K$ and a non-negative number $ninmathbb{N}$ such that $mathbb{R}^ntimes K$ is isomorphic to an open subgroup of $G$.
I wonder whether it is possible to generalize the above for non-abelian but nilpotent groups.
The most simple case is when $G$ is nilpotent of order $3$. In this case $[G,G]$ is an abelian group, hence $[G,G]$ satisfies the Theorem. Moreover $G/[G,G]$ is always an abelian group and so satisfies the Theorem.
Can we deduce that $G$ also satisfies the theorem?
topological-groups locally-compact-groups nilpotent-groups
asked Oct 20 '18 at 12:14
YankoYanko
6,3511527
$endgroup$
add a comment |
$begingroup$
My question is about a possible generalization of the following structure theorem of locally compact abelian groups.
Theorem: Let $G$ be a locally compact abelian group. Then here exists a compact subgroup $K$ and a non-negative number $ninmathbb{N}$ such that $mathbb{R}^ntimes K$ is isomorphic to an open subgroup of $G$.
I wonder whether it is possible to generalize the above for non-abelian but nilpotent groups.
The most simple case is when $G$ is nilpotent of order $3$. In this case $[G,G]$ is an abelian group, hence $[G,G]$ satisfies the Theorem. Moreover $G/[G,G]$ is always an abelian group and so satisfies the Theorem.
Can we deduce that $G$ also satisfies the theorem?
topological-groups locally-compact-groups nilpotent-groups
asked Oct 20 '18 at 12:14
YankoYanko
6,3511527
$endgroup$
My question is about a possible generalization of the following structure theorem of locally compact abelian groups.
Theorem: Let $G$ be a locally compact abelian group. Then here exists a compact subgroup $K$ and a non-negative number $ninmathbb{N}$ such that $mathbb{R}^ntimes K$ is isomorphic to an open subgroup of $G$.
I wonder whether it is possible to generalize the above for non-abelian but nilpotent groups.
The most simple case is when $G$ is nilpotent of order $3$. In this case $[G,G]$ is an abelian group, hence $[G,G]$ satisfies the Theorem. Moreover $G/[G,G]$ is always an abelian group and so satisfies the Theorem.
Can we deduce that $G$ also satisfies the theorem?
topological-groups locally-compact-groups nilpotent-groups
topological-groups locally-compact-groups nilpotent-groups
asked Oct 20 '18 at 12:14
YankoYanko
6,3511527
asked Oct 20 '18 at 12:14
YankoYanko
6,3511527
asked Oct 20 '18 at 12:14
YankoYanko
6,3511527
asked Oct 20 '18 at 12:14
YankoYanko
6,3511527
asked Oct 20 '18 at 12:14
YankoYanko
6,3511527
6,3511527
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, this isn't necessarily true for nilpotent groups. Take the Heisenberg group $$H_3(Bbb{R})=left{begin{pmatrix}
1 & a & c\
0 & 1 & b\
0 & 0 & 1\
end{pmatrix}middle|a,b,cright}.$$ It is a nilpotent group which is diffeomorphic to $Bbb{R}^3$. In particular it is connected, so its only open subgroup is itself. However, it is not isomorphic as a group to $Bbb{R}^3$ since it's not abelian.
answered Dec 19 '18 at 22:12
CronusCronus
1,088517
$endgroup$
$begingroup$
Thanks for that!
$endgroup$
– Yanko
Dec 20 '18 at 18:22
add a 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%2f2963208%2flocally-compact-nilpotent-group-has-an-open-subgroup-isomorphic-to-mathbbrn%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, this isn't necessarily true for nilpotent groups. Take the Heisenberg group $$H_3(Bbb{R})=left{begin{pmatrix}
1 & a & c\
0 & 1 & b\
0 & 0 & 1\
end{pmatrix}middle|a,b,cright}.$$ It is a nilpotent group which is diffeomorphic to $Bbb{R}^3$. In particular it is connected, so its only open subgroup is itself. However, it is not isomorphic as a group to $Bbb{R}^3$ since it's not abelian.
answered Dec 19 '18 at 22:12
CronusCronus
1,088517
$endgroup$
$begingroup$
Thanks for that!
$endgroup$
– Yanko
Dec 20 '18 at 18:22
add a comment |
$begingroup$
No, this isn't necessarily true for nilpotent groups. Take the Heisenberg group $$H_3(Bbb{R})=left{begin{pmatrix}
1 & a & c\
0 & 1 & b\
0 & 0 & 1\
end{pmatrix}middle|a,b,cright}.$$ It is a nilpotent group which is diffeomorphic to $Bbb{R}^3$. In particular it is connected, so its only open subgroup is itself. However, it is not isomorphic as a group to $Bbb{R}^3$ since it's not abelian.
answered Dec 19 '18 at 22:12
CronusCronus
1,088517
$endgroup$
$begingroup$
Thanks for that!
$endgroup$
– Yanko
Dec 20 '18 at 18:22
add a comment |
$begingroup$
No, this isn't necessarily true for nilpotent groups. Take the Heisenberg group $$H_3(Bbb{R})=left{begin{pmatrix}
1 & a & c\
0 & 1 & b\
0 & 0 & 1\
end{pmatrix}middle|a,b,cright}.$$ It is a nilpotent group which is diffeomorphic to $Bbb{R}^3$. In particular it is connected, so its only open subgroup is itself. However, it is not isomorphic as a group to $Bbb{R}^3$ since it's not abelian.
answered Dec 19 '18 at 22:12
CronusCronus
1,088517
$endgroup$
No, this isn't necessarily true for nilpotent groups. Take the Heisenberg group $$H_3(Bbb{R})=left{begin{pmatrix}
1 & a & c\
0 & 1 & b\
0 & 0 & 1\
end{pmatrix}middle|a,b,cright}.$$ It is a nilpotent group which is diffeomorphic to $Bbb{R}^3$. In particular it is connected, so its only open subgroup is itself. However, it is not isomorphic as a group to $Bbb{R}^3$ since it's not abelian.
answered Dec 19 '18 at 22:12
CronusCronus
1,088517
answered Dec 19 '18 at 22:12
CronusCronus
1,088517
answered Dec 19 '18 at 22:12
CronusCronus
1,088517
answered Dec 19 '18 at 22:12
CronusCronus
1,088517
1,088517
$begingroup$
Thanks for that!
$endgroup$
– Yanko
Dec 20 '18 at 18:22
add a comment |
$begingroup$
Thanks for that!
$endgroup$
– Yanko
Dec 20 '18 at 18:22
$begingroup$
Thanks for that!
$endgroup$
– Yanko
Dec 20 '18 at 18:22
$begingroup$
Thanks for that!
$endgroup$
– Yanko
Dec 20 '18 at 18:22
add a 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%2f2963208%2flocally-compact-nilpotent-group-has-an-open-subgroup-isomorphic-to-mathbbrn%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
