Is $W$ a finite-dimensional vector space in Proposition 3.13?
$begingroup$
Is $W$ a finite-dimensional vector space in Proposition $3.13$?
manifolds smooth-manifolds
edited Dec 13 '18 at 11:31
amWhy
1
asked Dec 13 '18 at 11:13
Born to be proudBorn to be proud
854510
$endgroup$
add a comment |
$begingroup$
Is $W$ a finite-dimensional vector space in Proposition $3.13$?
manifolds smooth-manifolds
edited Dec 13 '18 at 11:31
amWhy
1
asked Dec 13 '18 at 11:13
Born to be proudBorn to be proud
854510
$endgroup$
$begingroup$
@JackLee It seems there isn't enough information about $W$ in $textbf{ Proposition 3.13}$.
$endgroup$
– Born to be proud
Dec 13 '18 at 12:50
$begingroup$
The tangent space was only defined for finite-dimensional smooth manifolds, so there is no need to explicitely state this.
$endgroup$
– TheGeekGreek
Dec 13 '18 at 14:55
add a comment |
$begingroup$
Is $W$ a finite-dimensional vector space in Proposition $3.13$?
manifolds smooth-manifolds
edited Dec 13 '18 at 11:31
amWhy
1
asked Dec 13 '18 at 11:13
Born to be proudBorn to be proud
854510
$endgroup$
Is $W$ a finite-dimensional vector space in Proposition $3.13$?
manifolds smooth-manifolds
manifolds smooth-manifolds
edited Dec 13 '18 at 11:31
amWhy
1
asked Dec 13 '18 at 11:13
Born to be proudBorn to be proud
854510
edited Dec 13 '18 at 11:31
amWhy
1
asked Dec 13 '18 at 11:13
Born to be proudBorn to be proud
854510
edited Dec 13 '18 at 11:31
amWhy
1
edited Dec 13 '18 at 11:31
amWhy
1
edited Dec 13 '18 at 11:31
amWhy
1
1
asked Dec 13 '18 at 11:13
Born to be proudBorn to be proud
854510
asked Dec 13 '18 at 11:13
Born to be proudBorn to be proud
854510
asked Dec 13 '18 at 11:13
Born to be proudBorn to be proud
854510
854510
$begingroup$
@JackLee It seems there isn't enough information about $W$ in $textbf{ Proposition 3.13}$.
$endgroup$
– Born to be proud
Dec 13 '18 at 12:50
$begingroup$
The tangent space was only defined for finite-dimensional smooth manifolds, so there is no need to explicitely state this.
$endgroup$
– TheGeekGreek
Dec 13 '18 at 14:55
add a comment |
$begingroup$
@JackLee It seems there isn't enough information about $W$ in $textbf{ Proposition 3.13}$.
$endgroup$
– Born to be proud
Dec 13 '18 at 12:50
$begingroup$
The tangent space was only defined for finite-dimensional smooth manifolds, so there is no need to explicitely state this.
$endgroup$
– TheGeekGreek
Dec 13 '18 at 14:55
$begingroup$
@JackLee It seems there isn't enough information about $W$ in $textbf{ Proposition 3.13}$.
$endgroup$
– Born to be proud
Dec 13 '18 at 12:50
$begingroup$
@JackLee It seems there isn't enough information about $W$ in $textbf{ Proposition 3.13}$.
$endgroup$
– Born to be proud
Dec 13 '18 at 12:50
$begingroup$
The tangent space was only defined for finite-dimensional smooth manifolds, so there is no need to explicitely state this.
$endgroup$
– TheGeekGreek
Dec 13 '18 at 14:55
$begingroup$
The tangent space was only defined for finite-dimensional smooth manifolds, so there is no need to explicitely state this.
$endgroup$
– TheGeekGreek
Dec 13 '18 at 14:55
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The author seems to assume that $W$ is a vector space and also a smooth manifold. So yes, it is finite-dimensional.
answered Dec 13 '18 at 11:25
ChristophChristoph
12.4k1642
$endgroup$
add a comment |
$begingroup$
Yes it is, and there is really no need to write down that explicitly.
The theorem states that there is a canonical isomorphism $Vcong T_aV$ when $V$ is finite-dimensional, so in writing $Wcong T_{La}W$ it is implicitly assumed that $W$ is finite-dimensional as well as it is using this canonical isomorphism for $W.$
answered Dec 13 '18 at 14:27
positrón0802positrón0802
4,468520
$endgroup$
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%2f3037877%2fis-w-a-finite-dimensional-vector-space-in-proposition-3-13%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
$begingroup$
The author seems to assume that $W$ is a vector space and also a smooth manifold. So yes, it is finite-dimensional.
answered Dec 13 '18 at 11:25
ChristophChristoph
12.4k1642
$endgroup$
add a comment |
$begingroup$
The author seems to assume that $W$ is a vector space and also a smooth manifold. So yes, it is finite-dimensional.
answered Dec 13 '18 at 11:25
ChristophChristoph
12.4k1642
$endgroup$
add a comment |
$begingroup$
The author seems to assume that $W$ is a vector space and also a smooth manifold. So yes, it is finite-dimensional.
answered Dec 13 '18 at 11:25
ChristophChristoph
12.4k1642
$endgroup$
The author seems to assume that $W$ is a vector space and also a smooth manifold. So yes, it is finite-dimensional.
answered Dec 13 '18 at 11:25
ChristophChristoph
12.4k1642
answered Dec 13 '18 at 11:25
ChristophChristoph
12.4k1642
answered Dec 13 '18 at 11:25
ChristophChristoph
12.4k1642
answered Dec 13 '18 at 11:25
ChristophChristoph
12.4k1642
12.4k1642
add a comment |
add a comment |
$begingroup$
Yes it is, and there is really no need to write down that explicitly.
The theorem states that there is a canonical isomorphism $Vcong T_aV$ when $V$ is finite-dimensional, so in writing $Wcong T_{La}W$ it is implicitly assumed that $W$ is finite-dimensional as well as it is using this canonical isomorphism for $W.$
answered Dec 13 '18 at 14:27
positrón0802positrón0802
4,468520
$endgroup$
add a comment |
$begingroup$
Yes it is, and there is really no need to write down that explicitly.
The theorem states that there is a canonical isomorphism $Vcong T_aV$ when $V$ is finite-dimensional, so in writing $Wcong T_{La}W$ it is implicitly assumed that $W$ is finite-dimensional as well as it is using this canonical isomorphism for $W.$
answered Dec 13 '18 at 14:27
positrón0802positrón0802
4,468520
$endgroup$
add a comment |
$begingroup$
Yes it is, and there is really no need to write down that explicitly.
The theorem states that there is a canonical isomorphism $Vcong T_aV$ when $V$ is finite-dimensional, so in writing $Wcong T_{La}W$ it is implicitly assumed that $W$ is finite-dimensional as well as it is using this canonical isomorphism for $W.$
answered Dec 13 '18 at 14:27
positrón0802positrón0802
4,468520
$endgroup$
Yes it is, and there is really no need to write down that explicitly.
The theorem states that there is a canonical isomorphism $Vcong T_aV$ when $V$ is finite-dimensional, so in writing $Wcong T_{La}W$ it is implicitly assumed that $W$ is finite-dimensional as well as it is using this canonical isomorphism for $W.$
answered Dec 13 '18 at 14:27
positrón0802positrón0802
4,468520
answered Dec 13 '18 at 14:27
positrón0802positrón0802
4,468520
answered Dec 13 '18 at 14:27
positrón0802positrón0802
4,468520
answered Dec 13 '18 at 14:27
positrón0802positrón0802
4,468520
4,468520
add a comment |
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%2f3037877%2fis-w-a-finite-dimensional-vector-space-in-proposition-3-13%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
$begingroup$
@JackLee It seems there isn't enough information about $W$ in $textbf{ Proposition 3.13}$.
$endgroup$
– Born to be proud
Dec 13 '18 at 12:50
$begingroup$
The tangent space was only defined for finite-dimensional smooth manifolds, so there is no need to explicitely state this.
$endgroup$
– TheGeekGreek
Dec 13 '18 at 14:55