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
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$omegarightarrow pi^*omega$ 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
60.4k11446
$endgroup$
$begingroup$
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 |
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$begingroup$
$omegarightarrow pi^*omega$ 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
60.4k11446
$endgroup$
$begingroup$
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
60.4k11446
$endgroup$
$begingroup$
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
60.4k11446
$endgroup$
$omegarightarrow pi^*omega$ 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
60.4k11446
edited Jan 7 at 0:58
answered Jan 7 at 0:18
Tsemo AristideTsemo Aristide
60.4k11446
answered Jan 7 at 0:18
Tsemo AristideTsemo Aristide
60.4k11446
answered Jan 7 at 0:18
Tsemo AristideTsemo Aristide
60.4k11446
60.4k11446
$begingroup$
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 |
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