Why is the exterior differential independent of the chosen basis?
$begingroup$
Let M be as smooth manifold, W a k-dimensional vector space with basis $(e_1,...,e_k)$ and $Omega^r(M,W)$ be the differential r-form on M with values in W. Let $omega in Omega^r$ and $v_1,..., v_r in T_xM$. Then we can write $omega_x(v_1,...,v_r)$ as a linear combination of the $e_i$ as follows: $w_x(v_1,...,v_r) = omega_x^i(v_1,...,v_r)e_i$, i.e. both $omega_x$ and $omega_x^i$ are multilinear maps and $omega_i$ is a differential r-form on M. Let's now define the exterior differential as a sheaf morphism from $Omega^r(M,W) rightarrow Omega^{r+1}(M,W)$:
$$d(omega^iotimes w_i):=domega_iotimes e_i.$$
My question is now why this is independent of the basis we have chosen for W?
differential-geometry sheaf-theory exterior-derivative
asked Dec 30 '18 at 10:02
QuasarQuasar
1278
$endgroup$
add a comment |
$begingroup$
Let M be as smooth manifold, W a k-dimensional vector space with basis $(e_1,...,e_k)$ and $Omega^r(M,W)$ be the differential r-form on M with values in W. Let $omega in Omega^r$ and $v_1,..., v_r in T_xM$. Then we can write $omega_x(v_1,...,v_r)$ as a linear combination of the $e_i$ as follows: $w_x(v_1,...,v_r) = omega_x^i(v_1,...,v_r)e_i$, i.e. both $omega_x$ and $omega_x^i$ are multilinear maps and $omega_i$ is a differential r-form on M. Let's now define the exterior differential as a sheaf morphism from $Omega^r(M,W) rightarrow Omega^{r+1}(M,W)$:
$$d(omega^iotimes w_i):=domega_iotimes e_i.$$
My question is now why this is independent of the basis we have chosen for W?
differential-geometry sheaf-theory exterior-derivative
asked Dec 30 '18 at 10:02
QuasarQuasar
1278
$endgroup$
add a comment |
$begingroup$
Let M be as smooth manifold, W a k-dimensional vector space with basis $(e_1,...,e_k)$ and $Omega^r(M,W)$ be the differential r-form on M with values in W. Let $omega in Omega^r$ and $v_1,..., v_r in T_xM$. Then we can write $omega_x(v_1,...,v_r)$ as a linear combination of the $e_i$ as follows: $w_x(v_1,...,v_r) = omega_x^i(v_1,...,v_r)e_i$, i.e. both $omega_x$ and $omega_x^i$ are multilinear maps and $omega_i$ is a differential r-form on M. Let's now define the exterior differential as a sheaf morphism from $Omega^r(M,W) rightarrow Omega^{r+1}(M,W)$:
$$d(omega^iotimes w_i):=domega_iotimes e_i.$$
My question is now why this is independent of the basis we have chosen for W?
differential-geometry sheaf-theory exterior-derivative
asked Dec 30 '18 at 10:02
QuasarQuasar
1278
$endgroup$
Let M be as smooth manifold, W a k-dimensional vector space with basis $(e_1,...,e_k)$ and $Omega^r(M,W)$ be the differential r-form on M with values in W. Let $omega in Omega^r$ and $v_1,..., v_r in T_xM$. Then we can write $omega_x(v_1,...,v_r)$ as a linear combination of the $e_i$ as follows: $w_x(v_1,...,v_r) = omega_x^i(v_1,...,v_r)e_i$, i.e. both $omega_x$ and $omega_x^i$ are multilinear maps and $omega_i$ is a differential r-form on M. Let's now define the exterior differential as a sheaf morphism from $Omega^r(M,W) rightarrow Omega^{r+1}(M,W)$:
$$d(omega^iotimes w_i):=domega_iotimes e_i.$$
My question is now why this is independent of the basis we have chosen for W?
differential-geometry sheaf-theory exterior-derivative
differential-geometry sheaf-theory exterior-derivative
asked Dec 30 '18 at 10:02
QuasarQuasar
1278
asked Dec 30 '18 at 10:02
QuasarQuasar
1278
edited Jan 1 at 12:02
Quasar
asked Dec 30 '18 at 10:02
QuasarQuasar
1278
asked Dec 30 '18 at 10:02
QuasarQuasar
1278
asked Dec 30 '18 at 10:02
QuasarQuasar
1278
1278
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The reason is that the change of basis is given by constants. If you take a basis ${f_j}$ then you get an invertible matrix $A=(a_j^i)$ charakterized by the fact that $f_j=sum_i a_j^ie_i$. Then writing $omega$ as $sum_jtildeomega^jotimes f_j$ we see that $omega=sum_iomega^iotimes e_i$, where $omega^i=sum_j a^j_itildeomega^j$. But since the $a^j_i$ are constants, this implies $domega^i=sum_j a^j_idtildeomega^j$ and so $sum_idomega^iotimes e_i=sum_{i,j}a^j_idtildeomega^jotimes e_i$. But then the constants can be brought to the other factor in the tensor product (the right one) and the $dtildeomega^j$ can be taken out of the sum over $i$ to obtain $sum_jdtildeomega^jotimes(sum_ia^i_je_i)=sum_j dtildeomega^jotimes f_j$.
answered Jan 2 at 10:31
Andreas CapAndreas Cap
11.5k923
$endgroup$
add a comment |
Your Answer
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%2f3056675%2fwhy-is-the-exterior-differential-independent-of-the-chosen-basis%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
$begingroup$
The reason is that the change of basis is given by constants. If you take a basis ${f_j}$ then you get an invertible matrix $A=(a_j^i)$ charakterized by the fact that $f_j=sum_i a_j^ie_i$. Then writing $omega$ as $sum_jtildeomega^jotimes f_j$ we see that $omega=sum_iomega^iotimes e_i$, where $omega^i=sum_j a^j_itildeomega^j$. But since the $a^j_i$ are constants, this implies $domega^i=sum_j a^j_idtildeomega^j$ and so $sum_idomega^iotimes e_i=sum_{i,j}a^j_idtildeomega^jotimes e_i$. But then the constants can be brought to the other factor in the tensor product (the right one) and the $dtildeomega^j$ can be taken out of the sum over $i$ to obtain $sum_jdtildeomega^jotimes(sum_ia^i_je_i)=sum_j dtildeomega^jotimes f_j$.
answered Jan 2 at 10:31
Andreas CapAndreas Cap
11.5k923
$endgroup$
add a comment |
$begingroup$
The reason is that the change of basis is given by constants. If you take a basis ${f_j}$ then you get an invertible matrix $A=(a_j^i)$ charakterized by the fact that $f_j=sum_i a_j^ie_i$. Then writing $omega$ as $sum_jtildeomega^jotimes f_j$ we see that $omega=sum_iomega^iotimes e_i$, where $omega^i=sum_j a^j_itildeomega^j$. But since the $a^j_i$ are constants, this implies $domega^i=sum_j a^j_idtildeomega^j$ and so $sum_idomega^iotimes e_i=sum_{i,j}a^j_idtildeomega^jotimes e_i$. But then the constants can be brought to the other factor in the tensor product (the right one) and the $dtildeomega^j$ can be taken out of the sum over $i$ to obtain $sum_jdtildeomega^jotimes(sum_ia^i_je_i)=sum_j dtildeomega^jotimes f_j$.
answered Jan 2 at 10:31
Andreas CapAndreas Cap
11.5k923
$endgroup$
add a comment |
$begingroup$
The reason is that the change of basis is given by constants. If you take a basis ${f_j}$ then you get an invertible matrix $A=(a_j^i)$ charakterized by the fact that $f_j=sum_i a_j^ie_i$. Then writing $omega$ as $sum_jtildeomega^jotimes f_j$ we see that $omega=sum_iomega^iotimes e_i$, where $omega^i=sum_j a^j_itildeomega^j$. But since the $a^j_i$ are constants, this implies $domega^i=sum_j a^j_idtildeomega^j$ and so $sum_idomega^iotimes e_i=sum_{i,j}a^j_idtildeomega^jotimes e_i$. But then the constants can be brought to the other factor in the tensor product (the right one) and the $dtildeomega^j$ can be taken out of the sum over $i$ to obtain $sum_jdtildeomega^jotimes(sum_ia^i_je_i)=sum_j dtildeomega^jotimes f_j$.
answered Jan 2 at 10:31
Andreas CapAndreas Cap
11.5k923
$endgroup$
The reason is that the change of basis is given by constants. If you take a basis ${f_j}$ then you get an invertible matrix $A=(a_j^i)$ charakterized by the fact that $f_j=sum_i a_j^ie_i$. Then writing $omega$ as $sum_jtildeomega^jotimes f_j$ we see that $omega=sum_iomega^iotimes e_i$, where $omega^i=sum_j a^j_itildeomega^j$. But since the $a^j_i$ are constants, this implies $domega^i=sum_j a^j_idtildeomega^j$ and so $sum_idomega^iotimes e_i=sum_{i,j}a^j_idtildeomega^jotimes e_i$. But then the constants can be brought to the other factor in the tensor product (the right one) and the $dtildeomega^j$ can be taken out of the sum over $i$ to obtain $sum_jdtildeomega^jotimes(sum_ia^i_je_i)=sum_j dtildeomega^jotimes f_j$.
answered Jan 2 at 10:31
Andreas CapAndreas Cap
11.5k923
edited Jan 2 at 12:15
answered Jan 2 at 10:31
Andreas CapAndreas Cap
11.5k923
answered Jan 2 at 10:31
Andreas CapAndreas Cap
11.5k923
answered Jan 2 at 10:31
Andreas CapAndreas Cap
11.5k923
11.5k923
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%2f3056675%2fwhy-is-the-exterior-differential-independent-of-the-chosen-basis%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
