Mixing
Using connectedness of $SL_n(R)$ How to show that $Gl_n(R)^+$ is connected?
2
$begingroup$
Using connectedness of $SL_n(R)$ How to show that $Gl_n(R)^+$ is connected?
I can prove that Sl_n(R) is connected . How to prove using this fact only connectedness of all matrices with positive determinant ?
I wanted to solve this problem. Please give me hint.
general-topology lie-groups
asked Jan 31 at 16:52
MathLoverMathLover
59210
$endgroup$
add a comment |
2
$begingroup$
Using connectedness of $SL_n(R)$ How to show that $Gl_n(R)^+$ is connected?
I can prove that Sl_n(R) is connected . How to prove using this fact only connectedness of all matrices with positive determinant ?
I wanted to solve this problem. Please give me hint.
general-topology lie-groups
asked Jan 31 at 16:52
MathLoverMathLover
59210
$endgroup$
add a comment |
2
2
2
$begingroup$
Using connectedness of $SL_n(R)$ How to show that $Gl_n(R)^+$ is connected?
I can prove that Sl_n(R) is connected . How to prove using this fact only connectedness of all matrices with positive determinant ?
I wanted to solve this problem. Please give me hint.
general-topology lie-groups
asked Jan 31 at 16:52
MathLoverMathLover
59210
$endgroup$
Using connectedness of $SL_n(R)$ How to show that $Gl_n(R)^+$ is connected?
I can prove that Sl_n(R) is connected . How to prove using this fact only connectedness of all matrices with positive determinant ?
I wanted to solve this problem. Please give me hint.
general-topology lie-groups
general-topology lie-groups
asked Jan 31 at 16:52
MathLoverMathLover
59210
asked Jan 31 at 16:52
MathLoverMathLover
59210
asked Jan 31 at 16:52
MathLoverMathLover
59210
asked Jan 31 at 16:52
MathLoverMathLover
59210
asked Jan 31 at 16:52
MathLoverMathLover
59210
59210
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
2
$begingroup$
Hint: $GL_n(mathbb R)^+ = mathbb R^+ cdot SL_n(mathbb R)$.
answered Jan 31 at 16:54
rabotarabota
14.4k32784
$endgroup$
$begingroup$
Sir, Please check my answer: $Ain Gl_n(R)^+$ $A=xB$ where $Bin Sl_n(R)$ Now there is path which take $x(0)=x,x(t)=1$ and As Sl_n(R) is connected we can path for B to Identity so together with there is path form A to Identity Hence Connected
$endgroup$
– MathLover
Feb 1 at 2:52
$begingroup$
That works if you assume that $SL_n(R)^+$ is path-connected. But connected and path-connected is not the same thing. Instead, you need a different argument. If you want a further hint, ask away.
$endgroup$
– rabota
Feb 1 at 9:58
1
$begingroup$
@barto In case of manifold, Both connected and path-connected are same
$endgroup$
– SRJ
Feb 3 at 13:34
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
});
}
});
draft saved
draft discarded
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%2f3095136%2fusing-connectedness-of-sl-nr-how-to-show-that-gl-nr-is-connected%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
2
$begingroup$
Hint: $GL_n(mathbb R)^+ = mathbb R^+ cdot SL_n(mathbb R)$.
answered Jan 31 at 16:54
rabotarabota
14.4k32784
$endgroup$
$begingroup$
Sir, Please check my answer: $Ain Gl_n(R)^+$ $A=xB$ where $Bin Sl_n(R)$ Now there is path which take $x(0)=x,x(t)=1$ and As Sl_n(R) is connected we can path for B to Identity so together with there is path form A to Identity Hence Connected
$endgroup$
– MathLover
Feb 1 at 2:52
$begingroup$
That works if you assume that $SL_n(R)^+$ is path-connected. But connected and path-connected is not the same thing. Instead, you need a different argument. If you want a further hint, ask away.
$endgroup$
– rabota
Feb 1 at 9:58
1
$begingroup$
@barto In case of manifold, Both connected and path-connected are same
$endgroup$
– SRJ
Feb 3 at 13:34
add a comment |
2
$begingroup$
Hint: $GL_n(mathbb R)^+ = mathbb R^+ cdot SL_n(mathbb R)$.
answered Jan 31 at 16:54
rabotarabota
14.4k32784
$endgroup$
$begingroup$
Sir, Please check my answer: $Ain Gl_n(R)^+$ $A=xB$ where $Bin Sl_n(R)$ Now there is path which take $x(0)=x,x(t)=1$ and As Sl_n(R) is connected we can path for B to Identity so together with there is path form A to Identity Hence Connected
$endgroup$
– MathLover
Feb 1 at 2:52
$begingroup$
That works if you assume that $SL_n(R)^+$ is path-connected. But connected and path-connected is not the same thing. Instead, you need a different argument. If you want a further hint, ask away.
$endgroup$
– rabota
Feb 1 at 9:58
1
$begingroup$
@barto In case of manifold, Both connected and path-connected are same
$endgroup$
– SRJ
Feb 3 at 13:34
add a comment |
2
2
2
$begingroup$
Hint: $GL_n(mathbb R)^+ = mathbb R^+ cdot SL_n(mathbb R)$.
answered Jan 31 at 16:54
rabotarabota
14.4k32784
$endgroup$
Hint: $GL_n(mathbb R)^+ = mathbb R^+ cdot SL_n(mathbb R)$.
answered Jan 31 at 16:54
rabotarabota
14.4k32784
answered Jan 31 at 16:54
rabotarabota
14.4k32784
answered Jan 31 at 16:54
rabotarabota
14.4k32784
answered Jan 31 at 16:54
rabotarabota
14.4k32784
14.4k32784
$begingroup$
Sir, Please check my answer: $Ain Gl_n(R)^+$ $A=xB$ where $Bin Sl_n(R)$ Now there is path which take $x(0)=x,x(t)=1$ and As Sl_n(R) is connected we can path for B to Identity so together with there is path form A to Identity Hence Connected
$endgroup$
– MathLover
Feb 1 at 2:52
$begingroup$
That works if you assume that $SL_n(R)^+$ is path-connected. But connected and path-connected is not the same thing. Instead, you need a different argument. If you want a further hint, ask away.
$endgroup$
– rabota
Feb 1 at 9:58
1
$begingroup$
@barto In case of manifold, Both connected and path-connected are same
$endgroup$
– SRJ
Feb 3 at 13:34
add a comment |
$begingroup$
Sir, Please check my answer: $Ain Gl_n(R)^+$ $A=xB$ where $Bin Sl_n(R)$ Now there is path which take $x(0)=x,x(t)=1$ and As Sl_n(R) is connected we can path for B to Identity so together with there is path form A to Identity Hence Connected
$endgroup$
– MathLover
Feb 1 at 2:52
$begingroup$
That works if you assume that $SL_n(R)^+$ is path-connected. But connected and path-connected is not the same thing. Instead, you need a different argument. If you want a further hint, ask away.
$endgroup$
– rabota
Feb 1 at 9:58
1
$begingroup$
@barto In case of manifold, Both connected and path-connected are same
$endgroup$
– SRJ
Feb 3 at 13:34
$begingroup$
Sir, Please check my answer: $Ain Gl_n(R)^+$ $A=xB$ where $Bin Sl_n(R)$ Now there is path which take $x(0)=x,x(t)=1$ and As Sl_n(R) is connected we can path for B to Identity so together with there is path form A to Identity Hence Connected
$endgroup$
– MathLover
Feb 1 at 2:52
$begingroup$
Sir, Please check my answer: $Ain Gl_n(R)^+$ $A=xB$ where $Bin Sl_n(R)$ Now there is path which take $x(0)=x,x(t)=1$ and As Sl_n(R) is connected we can path for B to Identity so together with there is path form A to Identity Hence Connected
$endgroup$
– MathLover
Feb 1 at 2:52
$begingroup$
That works if you assume that $SL_n(R)^+$ is path-connected. But connected and path-connected is not the same thing. Instead, you need a different argument. If you want a further hint, ask away.
$endgroup$
– rabota
Feb 1 at 9:58
$begingroup$
That works if you assume that $SL_n(R)^+$ is path-connected. But connected and path-connected is not the same thing. Instead, you need a different argument. If you want a further hint, ask away.
$endgroup$
– rabota
Feb 1 at 9:58
1
1
$begingroup$
@barto In case of manifold, Both connected and path-connected are same
$endgroup$
– SRJ
Feb 3 at 13:34
$begingroup$
@barto In case of manifold, Both connected and path-connected are same
$endgroup$
– SRJ
Feb 3 at 13:34
add a comment |
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f3095136%2fusing-connectedness-of-sl-nr-how-to-show-that-gl-nr-is-connected%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
