Principal fiber bundles and invariant differential forms
$begingroup$
Let $G$ be a real Lie group with a right action on a smooth manifold $X$. Assume the action is free and proper. This implies there is a unique manifold structure on the quotient space $G backslash X$ such that the quotient map $pi$ is a submersion, and that $pi$ is a principal fiber bundle with $G$ is a fiber.
In other words, locally, the quotient map looks like $U times G rightarrow U, (u,g) mapsto u$ for sufficiently small open sets $U subset G backslash X$, with $G$ acting by $(u,x).g = (u,xg)$.
Let $omega in Omega^k(X backslash G)$ be a smooth differential $k$-form on $X backslash G$. Then $omega$ pulls back to a differential $k$-form $pi^{ast}(omega)$ on $X$ which is $G$-invariant.
1 . Is $omega mapsto pi^{ast}(omega)$ an injective map $Omega^k(G backslash X) rightarrow Omega^k(X)$?
2 . What is the image of $pi^{ast}$? Does it consist of all $G$-invariant differential $k$-forms on $X$?
If $X$ is a trivial principal fiber bundle, i.e. $X = Y times G$ for a smooth manifold $Y$, then all this seems obvious, but I'm not sure if there are some complications that can arise if $X$ is merely covered by such things.
differential-geometry differential-forms
asked Jan 7 at 0:09
D_SD_S
14.2k61754
$endgroup$
add a comment |
$begingroup$
Let $G$ be a real Lie group with a right action on a smooth manifold $X$. Assume the action is free and proper. This implies there is a unique manifold structure on the quotient space $G backslash X$ such that the quotient map $pi$ is a submersion, and that $pi$ is a principal fiber bundle with $G$ is a fiber.
In other words, locally, the quotient map looks like $U times G rightarrow U, (u,g) mapsto u$ for sufficiently small open sets $U subset G backslash X$, with $G$ acting by $(u,x).g = (u,xg)$.
Let $omega in Omega^k(X backslash G)$ be a smooth differential $k$-form on $X backslash G$. Then $omega$ pulls back to a differential $k$-form $pi^{ast}(omega)$ on $X$ which is $G$-invariant.
1 . Is $omega mapsto pi^{ast}(omega)$ an injective map $Omega^k(G backslash X) rightarrow Omega^k(X)$?
2 . What is the image of $pi^{ast}$? Does it consist of all $G$-invariant differential $k$-forms on $X$?
If $X$ is a trivial principal fiber bundle, i.e. $X = Y times G$ for a smooth manifold $Y$, then all this seems obvious, but I'm not sure if there are some complications that can arise if $X$ is merely covered by such things.
differential-geometry differential-forms
asked Jan 7 at 0:09
D_SD_S
14.2k61754
$endgroup$
add a comment |
$begingroup$
Let $G$ be a real Lie group with a right action on a smooth manifold $X$. Assume the action is free and proper. This implies there is a unique manifold structure on the quotient space $G backslash X$ such that the quotient map $pi$ is a submersion, and that $pi$ is a principal fiber bundle with $G$ is a fiber.
In other words, locally, the quotient map looks like $U times G rightarrow U, (u,g) mapsto u$ for sufficiently small open sets $U subset G backslash X$, with $G$ acting by $(u,x).g = (u,xg)$.
Let $omega in Omega^k(X backslash G)$ be a smooth differential $k$-form on $X backslash G$. Then $omega$ pulls back to a differential $k$-form $pi^{ast}(omega)$ on $X$ which is $G$-invariant.
1 . Is $omega mapsto pi^{ast}(omega)$ an injective map $Omega^k(G backslash X) rightarrow Omega^k(X)$?
2 . What is the image of $pi^{ast}$? Does it consist of all $G$-invariant differential $k$-forms on $X$?
If $X$ is a trivial principal fiber bundle, i.e. $X = Y times G$ for a smooth manifold $Y$, then all this seems obvious, but I'm not sure if there are some complications that can arise if $X$ is merely covered by such things.
differential-geometry differential-forms
asked Jan 7 at 0:09
D_SD_S
14.2k61754
$endgroup$
Let $G$ be a real Lie group with a right action on a smooth manifold $X$. Assume the action is free and proper. This implies there is a unique manifold structure on the quotient space $G backslash X$ such that the quotient map $pi$ is a submersion, and that $pi$ is a principal fiber bundle with $G$ is a fiber.
In other words, locally, the quotient map looks like $U times G rightarrow U, (u,g) mapsto u$ for sufficiently small open sets $U subset G backslash X$, with $G$ acting by $(u,x).g = (u,xg)$.
Let $omega in Omega^k(X backslash G)$ be a smooth differential $k$-form on $X backslash G$. Then $omega$ pulls back to a differential $k$-form $pi^{ast}(omega)$ on $X$ which is $G$-invariant.
1 . Is $omega mapsto pi^{ast}(omega)$ an injective map $Omega^k(G backslash X) rightarrow Omega^k(X)$?
2 . What is the image of $pi^{ast}$? Does it consist of all $G$-invariant differential $k$-forms on $X$?
If $X$ is a trivial principal fiber bundle, i.e. $X = Y times G$ for a smooth manifold $Y$, then all this seems obvious, but I'm not sure if there are some complications that can arise if $X$ is merely covered by such things.
differential-geometry differential-forms
differential-geometry differential-forms
asked Jan 7 at 0:09
D_SD_S
14.2k61754
asked Jan 7 at 0:09
D_SD_S
14.2k61754
edited Jan 7 at 0:16
D_S
asked Jan 7 at 0:09
D_SD_S
14.2k61754
asked Jan 7 at 0:09
D_SD_S
14.2k61754
asked Jan 7 at 0:09
D_SD_S
14.2k61754
14.2k61754
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$omegarightarrow pi^*omega$ is linear. Suppose that $pi^*omega=0$, since $pi$ is a submersion, for every $xin G/X, u_1,...,u_kin T_x(G/X)$ and $yinpi^{-1}(x)$, there exists $vin T_yX$ such that $dpi_y(v_i)=u_i$, $pi^*omega_y(v_1,...,v_k)=omega_x(u_1,...,u_k)=0$ implies that $omega=0$ and $omegarightarrow pi^*omega$ is injective.
Consider $Xtimes G$ and take any non zero $1$-form $beta$ invariant on $G$ by the right translations. Write $alpha_{(x,g)}(u,v)=beta_g(v)$ is a form invariant by $G$. You cannot have $alpha=pi^*omega$ since $pi^*omega$ vanishes on the fibre.
answered Jan 7 at 0:18
Tsemo AristideTsemo Aristide
60.5k11446
$endgroup$
$begingroup$
Okay, thanks. So the injectivity is not a big deal.
$endgroup$
– D_S
Jan 7 at 0:19
$begingroup$
The images of $pi^*$ consists of all $G$-invariant forms which are horizontal in the sense that they vanish upon insertion of a single vertical vector field.
$endgroup$
– Andreas Cap
Jan 7 at 11:30
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%2f3064547%2fprincipal-fiber-bundles-and-invariant-differential-forms%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
$begingroup$
$omegarightarrow pi^*omega$ is linear. Suppose that $pi^*omega=0$, since $pi$ is a submersion, for every $xin G/X, u_1,...,u_kin T_x(G/X)$ and $yinpi^{-1}(x)$, there exists $vin T_yX$ such that $dpi_y(v_i)=u_i$, $pi^*omega_y(v_1,...,v_k)=omega_x(u_1,...,u_k)=0$ implies that $omega=0$ and $omegarightarrow pi^*omega$ is injective.
Consider $Xtimes G$ and take any non zero $1$-form $beta$ invariant on $G$ by the right translations. Write $alpha_{(x,g)}(u,v)=beta_g(v)$ is a form invariant by $G$. You cannot have $alpha=pi^*omega$ since $pi^*omega$ vanishes on the fibre.
answered Jan 7 at 0:18
Tsemo AristideTsemo Aristide
60.5k11446
$endgroup$
$begingroup$
Okay, thanks. So the injectivity is not a big deal.
$endgroup$
– D_S
Jan 7 at 0:19
$begingroup$
The images of $pi^*$ consists of all $G$-invariant forms which are horizontal in the sense that they vanish upon insertion of a single vertical vector field.
$endgroup$
– Andreas Cap
Jan 7 at 11:30
add a comment |
$begingroup$
$omegarightarrow pi^*omega$ is linear. Suppose that $pi^*omega=0$, since $pi$ is a submersion, for every $xin G/X, u_1,...,u_kin T_x(G/X)$ and $yinpi^{-1}(x)$, there exists $vin T_yX$ such that $dpi_y(v_i)=u_i$, $pi^*omega_y(v_1,...,v_k)=omega_x(u_1,...,u_k)=0$ implies that $omega=0$ and $omegarightarrow pi^*omega$ is injective.
Consider $Xtimes G$ and take any non zero $1$-form $beta$ invariant on $G$ by the right translations. Write $alpha_{(x,g)}(u,v)=beta_g(v)$ is a form invariant by $G$. You cannot have $alpha=pi^*omega$ since $pi^*omega$ vanishes on the fibre.
answered Jan 7 at 0:18
Tsemo AristideTsemo Aristide
60.5k11446
$endgroup$
$begingroup$
Okay, thanks. So the injectivity is not a big deal.
$endgroup$
– D_S
Jan 7 at 0:19
$begingroup$
The images of $pi^*$ consists of all $G$-invariant forms which are horizontal in the sense that they vanish upon insertion of a single vertical vector field.
$endgroup$
– Andreas Cap
Jan 7 at 11:30
add a comment |
$begingroup$
$omegarightarrow pi^*omega$ is linear. Suppose that $pi^*omega=0$, since $pi$ is a submersion, for every $xin G/X, u_1,...,u_kin T_x(G/X)$ and $yinpi^{-1}(x)$, there exists $vin T_yX$ such that $dpi_y(v_i)=u_i$, $pi^*omega_y(v_1,...,v_k)=omega_x(u_1,...,u_k)=0$ implies that $omega=0$ and $omegarightarrow pi^*omega$ is injective.
Consider $Xtimes G$ and take any non zero $1$-form $beta$ invariant on $G$ by the right translations. Write $alpha_{(x,g)}(u,v)=beta_g(v)$ is a form invariant by $G$. You cannot have $alpha=pi^*omega$ since $pi^*omega$ vanishes on the fibre.
answered Jan 7 at 0:18
Tsemo AristideTsemo Aristide
60.5k11446
$endgroup$
$omegarightarrow pi^*omega$ is linear. Suppose that $pi^*omega=0$, since $pi$ is a submersion, for every $xin G/X, u_1,...,u_kin T_x(G/X)$ and $yinpi^{-1}(x)$, there exists $vin T_yX$ such that $dpi_y(v_i)=u_i$, $pi^*omega_y(v_1,...,v_k)=omega_x(u_1,...,u_k)=0$ implies that $omega=0$ and $omegarightarrow pi^*omega$ is injective.
Consider $Xtimes G$ and take any non zero $1$-form $beta$ invariant on $G$ by the right translations. Write $alpha_{(x,g)}(u,v)=beta_g(v)$ is a form invariant by $G$. You cannot have $alpha=pi^*omega$ since $pi^*omega$ vanishes on the fibre.
answered Jan 7 at 0:18
Tsemo AristideTsemo Aristide
60.5k11446
edited Jan 7 at 0:58
answered Jan 7 at 0:18
Tsemo AristideTsemo Aristide
60.5k11446
answered Jan 7 at 0:18
Tsemo AristideTsemo Aristide
60.5k11446
answered Jan 7 at 0:18
Tsemo AristideTsemo Aristide
60.5k11446
60.5k11446
$begingroup$
Okay, thanks. So the injectivity is not a big deal.
$endgroup$
– D_S
Jan 7 at 0:19
$begingroup$
The images of $pi^*$ consists of all $G$-invariant forms which are horizontal in the sense that they vanish upon insertion of a single vertical vector field.
$endgroup$
– Andreas Cap
Jan 7 at 11:30
add a comment |
$begingroup$
Okay, thanks. So the injectivity is not a big deal.
$endgroup$
– D_S
Jan 7 at 0:19
$begingroup$
The images of $pi^*$ consists of all $G$-invariant forms which are horizontal in the sense that they vanish upon insertion of a single vertical vector field.
$endgroup$
– Andreas Cap
Jan 7 at 11:30
$begingroup$
Okay, thanks. So the injectivity is not a big deal.
$endgroup$
– D_S
Jan 7 at 0:19
$begingroup$
Okay, thanks. So the injectivity is not a big deal.
$endgroup$
– D_S
Jan 7 at 0:19
$begingroup$
The images of $pi^*$ consists of all $G$-invariant forms which are horizontal in the sense that they vanish upon insertion of a single vertical vector field.
$endgroup$
– Andreas Cap
Jan 7 at 11:30
$begingroup$
The images of $pi^*$ consists of all $G$-invariant forms which are horizontal in the sense that they vanish upon insertion of a single vertical vector field.
$endgroup$
– Andreas Cap
Jan 7 at 11:30
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%2f3064547%2fprincipal-fiber-bundles-and-invariant-differential-forms%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
