If $TM$ is trivial, then $Lambda^n(M)$ is also trivial and $M$ is orientable
$begingroup$
Suppose that $M$ is a smooth $n-$manifold. Suppose $$ TM=coprod_{pin M}T_pM $$ is the tangent bundle of $M$. And let $$ Lambda^n(M)=coprod_{pin M}Lambda^n(T_pM) $$ where $Lambda^n(T_pM)$ is alternating covariant $n-$tensor product. i.e.
$$Lambda^n(T_pM)=T^*_pMwedge T^*_pM wedge cdots wedge T^*_pM.$$
and
$$dim(Lambda^n(T_pM))=begin{bmatrix} dim(T_pM)\ n end{bmatrix}$$
If $TM$ is trivial, then $Lambda^n(M)$ is also trivial and $M$ is orientable.
I am trying to understand but seems difficult. I would be very appreciate any help. Thanks in advance.
orientation tangent-bundle
asked Nov 30 '18 at 16:52
Lev BanLev Ban
1,0701317
$endgroup$
add a comment |
$begingroup$
Suppose that $M$ is a smooth $n-$manifold. Suppose $$ TM=coprod_{pin M}T_pM $$ is the tangent bundle of $M$. And let $$ Lambda^n(M)=coprod_{pin M}Lambda^n(T_pM) $$ where $Lambda^n(T_pM)$ is alternating covariant $n-$tensor product. i.e.
$$Lambda^n(T_pM)=T^*_pMwedge T^*_pM wedge cdots wedge T^*_pM.$$
and
$$dim(Lambda^n(T_pM))=begin{bmatrix} dim(T_pM)\ n end{bmatrix}$$
If $TM$ is trivial, then $Lambda^n(M)$ is also trivial and $M$ is orientable.
I am trying to understand but seems difficult. I would be very appreciate any help. Thanks in advance.
orientation tangent-bundle
asked Nov 30 '18 at 16:52
Lev BanLev Ban
1,0701317
$endgroup$
add a comment |
$begingroup$
Suppose that $M$ is a smooth $n-$manifold. Suppose $$ TM=coprod_{pin M}T_pM $$ is the tangent bundle of $M$. And let $$ Lambda^n(M)=coprod_{pin M}Lambda^n(T_pM) $$ where $Lambda^n(T_pM)$ is alternating covariant $n-$tensor product. i.e.
$$Lambda^n(T_pM)=T^*_pMwedge T^*_pM wedge cdots wedge T^*_pM.$$
and
$$dim(Lambda^n(T_pM))=begin{bmatrix} dim(T_pM)\ n end{bmatrix}$$
If $TM$ is trivial, then $Lambda^n(M)$ is also trivial and $M$ is orientable.
I am trying to understand but seems difficult. I would be very appreciate any help. Thanks in advance.
orientation tangent-bundle
asked Nov 30 '18 at 16:52
Lev BanLev Ban
1,0701317
$endgroup$
Suppose that $M$ is a smooth $n-$manifold. Suppose $$ TM=coprod_{pin M}T_pM $$ is the tangent bundle of $M$. And let $$ Lambda^n(M)=coprod_{pin M}Lambda^n(T_pM) $$ where $Lambda^n(T_pM)$ is alternating covariant $n-$tensor product. i.e.
$$Lambda^n(T_pM)=T^*_pMwedge T^*_pM wedge cdots wedge T^*_pM.$$
and
$$dim(Lambda^n(T_pM))=begin{bmatrix} dim(T_pM)\ n end{bmatrix}$$
If $TM$ is trivial, then $Lambda^n(M)$ is also trivial and $M$ is orientable.
I am trying to understand but seems difficult. I would be very appreciate any help. Thanks in advance.
orientation tangent-bundle
orientation tangent-bundle
asked Nov 30 '18 at 16:52
Lev BanLev Ban
1,0701317
asked Nov 30 '18 at 16:52
Lev BanLev Ban
1,0701317
edited Dec 3 '18 at 14:00
Lev Ban
asked Nov 30 '18 at 16:52
Lev BanLev Ban
1,0701317
asked Nov 30 '18 at 16:52
Lev BanLev Ban
1,0701317
asked Nov 30 '18 at 16:52
Lev BanLev Ban
1,0701317
1,0701317
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $M$ is $n$-dimensional, $TM$ is trivial is equivalent to saying that there exists $n$-vector fields $X_1,...,X_n$ such that for every $xin M$, $X_i(x)neq 0$. Take a differentiable metric on $M$ and define the $1$-form $f_i(x)(u)=langle X_i(x),urangle$ where $uin T_xM$, $Lambda^n(M)_x$ is generated by $f_1(x)wedge...wedge f_n(x)$ which is also a volume form on $M$. This implies that $M$ is orientable.
answered Nov 30 '18 at 17:01
Tsemo AristideTsemo Aristide
58k11445
$endgroup$
$begingroup$
Thank you for your answer. But could you let me know why $Lambda^n(M)$ is also trivial?
$endgroup$
– Lev Ban
Nov 30 '18 at 17:09
$begingroup$
because it is a dim 1 and you have the volume form
$endgroup$
– Tsemo Aristide
Nov 30 '18 at 17:21
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%2f3020313%2fif-tm-is-trivial-then-lambdanm-is-also-trivial-and-m-is-orientable%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
$begingroup$
If $M$ is $n$-dimensional, $TM$ is trivial is equivalent to saying that there exists $n$-vector fields $X_1,...,X_n$ such that for every $xin M$, $X_i(x)neq 0$. Take a differentiable metric on $M$ and define the $1$-form $f_i(x)(u)=langle X_i(x),urangle$ where $uin T_xM$, $Lambda^n(M)_x$ is generated by $f_1(x)wedge...wedge f_n(x)$ which is also a volume form on $M$. This implies that $M$ is orientable.
answered Nov 30 '18 at 17:01
Tsemo AristideTsemo Aristide
58k11445
$endgroup$
$begingroup$
Thank you for your answer. But could you let me know why $Lambda^n(M)$ is also trivial?
$endgroup$
– Lev Ban
Nov 30 '18 at 17:09
$begingroup$
because it is a dim 1 and you have the volume form
$endgroup$
– Tsemo Aristide
Nov 30 '18 at 17:21
add a comment |
$begingroup$
If $M$ is $n$-dimensional, $TM$ is trivial is equivalent to saying that there exists $n$-vector fields $X_1,...,X_n$ such that for every $xin M$, $X_i(x)neq 0$. Take a differentiable metric on $M$ and define the $1$-form $f_i(x)(u)=langle X_i(x),urangle$ where $uin T_xM$, $Lambda^n(M)_x$ is generated by $f_1(x)wedge...wedge f_n(x)$ which is also a volume form on $M$. This implies that $M$ is orientable.
answered Nov 30 '18 at 17:01
Tsemo AristideTsemo Aristide
58k11445
$endgroup$
$begingroup$
Thank you for your answer. But could you let me know why $Lambda^n(M)$ is also trivial?
$endgroup$
– Lev Ban
Nov 30 '18 at 17:09
$begingroup$
because it is a dim 1 and you have the volume form
$endgroup$
– Tsemo Aristide
Nov 30 '18 at 17:21
add a comment |
$begingroup$
If $M$ is $n$-dimensional, $TM$ is trivial is equivalent to saying that there exists $n$-vector fields $X_1,...,X_n$ such that for every $xin M$, $X_i(x)neq 0$. Take a differentiable metric on $M$ and define the $1$-form $f_i(x)(u)=langle X_i(x),urangle$ where $uin T_xM$, $Lambda^n(M)_x$ is generated by $f_1(x)wedge...wedge f_n(x)$ which is also a volume form on $M$. This implies that $M$ is orientable.
answered Nov 30 '18 at 17:01
Tsemo AristideTsemo Aristide
58k11445
$endgroup$
If $M$ is $n$-dimensional, $TM$ is trivial is equivalent to saying that there exists $n$-vector fields $X_1,...,X_n$ such that for every $xin M$, $X_i(x)neq 0$. Take a differentiable metric on $M$ and define the $1$-form $f_i(x)(u)=langle X_i(x),urangle$ where $uin T_xM$, $Lambda^n(M)_x$ is generated by $f_1(x)wedge...wedge f_n(x)$ which is also a volume form on $M$. This implies that $M$ is orientable.
answered Nov 30 '18 at 17:01
Tsemo AristideTsemo Aristide
58k11445
answered Nov 30 '18 at 17:01
Tsemo AristideTsemo Aristide
58k11445
answered Nov 30 '18 at 17:01
Tsemo AristideTsemo Aristide
58k11445
answered Nov 30 '18 at 17:01
Tsemo AristideTsemo Aristide
58k11445
58k11445
$begingroup$
Thank you for your answer. But could you let me know why $Lambda^n(M)$ is also trivial?
$endgroup$
– Lev Ban
Nov 30 '18 at 17:09
$begingroup$
because it is a dim 1 and you have the volume form
$endgroup$
– Tsemo Aristide
Nov 30 '18 at 17:21
add a comment |
$begingroup$
Thank you for your answer. But could you let me know why $Lambda^n(M)$ is also trivial?
$endgroup$
– Lev Ban
Nov 30 '18 at 17:09
$begingroup$
because it is a dim 1 and you have the volume form
$endgroup$
– Tsemo Aristide
Nov 30 '18 at 17:21
$begingroup$
Thank you for your answer. But could you let me know why $Lambda^n(M)$ is also trivial?
$endgroup$
– Lev Ban
Nov 30 '18 at 17:09
$begingroup$
Thank you for your answer. But could you let me know why $Lambda^n(M)$ is also trivial?
$endgroup$
– Lev Ban
Nov 30 '18 at 17:09
$begingroup$
because it is a dim 1 and you have the volume form
$endgroup$
– Tsemo Aristide
Nov 30 '18 at 17:21
$begingroup$
because it is a dim 1 and you have the volume form
$endgroup$
– Tsemo Aristide
Nov 30 '18 at 17:21
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%2f3020313%2fif-tm-is-trivial-then-lambdanm-is-also-trivial-and-m-is-orientable%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
