Mixing
Splitting of the tangent bundle of a vector bundle and connections
$begingroup$
Let $pi:Eto M$ be a smooth vector bundle. Then we have the following exact sequence of vector bundles over $E$:
$$
0to VExrightarrow{} TExrightarrow{mathrm{d}pi}pi^*TMto 0
$$
Here $VE$ is the vertical bundle, that is kernel of the bunble map $mathrm{d}pi$. This sequence splits, so there is a bundle morphism $sigma:pi^*TMto TE$ such that $mathrm{d}picircsigma=mathrm{Id}$. It can be shown that the data of such a $sigma$ is equivalent to a connection (covariant derivative) $nabla$ on $Eto M$ (see for instance the volume 2 of Greub-Halperin-Vanstone). Also we know that $VEcongpi^*E$ so that we can write the following decomposition:
$$
TEcongpi^*Eopluspi^*TM
$$
My question is the following: considering $E$ as a manifold, I would like to consider differential forms on $E$, that is $Gamma(Lambda T^*E)$; is the De Rham differential linked to $nabla$ in some way?
Thank you for your help. Any reference would be very useful.
differential-geometry vector-bundles exact-sequence connections
asked Jun 18 '14 at 15:45
bc87bc87
1249
$endgroup$
add a comment |
$begingroup$
Let $pi:Eto M$ be a smooth vector bundle. Then we have the following exact sequence of vector bundles over $E$:
$$
0to VExrightarrow{} TExrightarrow{mathrm{d}pi}pi^*TMto 0
$$
Here $VE$ is the vertical bundle, that is kernel of the bunble map $mathrm{d}pi$. This sequence splits, so there is a bundle morphism $sigma:pi^*TMto TE$ such that $mathrm{d}picircsigma=mathrm{Id}$. It can be shown that the data of such a $sigma$ is equivalent to a connection (covariant derivative) $nabla$ on $Eto M$ (see for instance the volume 2 of Greub-Halperin-Vanstone). Also we know that $VEcongpi^*E$ so that we can write the following decomposition:
$$
TEcongpi^*Eopluspi^*TM
$$
My question is the following: considering $E$ as a manifold, I would like to consider differential forms on $E$, that is $Gamma(Lambda T^*E)$; is the De Rham differential linked to $nabla$ in some way?
Thank you for your help. Any reference would be very useful.
differential-geometry vector-bundles exact-sequence connections
asked Jun 18 '14 at 15:45
bc87bc87
1249
$endgroup$
add a comment |
$begingroup$
Let $pi:Eto M$ be a smooth vector bundle. Then we have the following exact sequence of vector bundles over $E$:
$$
0to VExrightarrow{} TExrightarrow{mathrm{d}pi}pi^*TMto 0
$$
Here $VE$ is the vertical bundle, that is kernel of the bunble map $mathrm{d}pi$. This sequence splits, so there is a bundle morphism $sigma:pi^*TMto TE$ such that $mathrm{d}picircsigma=mathrm{Id}$. It can be shown that the data of such a $sigma$ is equivalent to a connection (covariant derivative) $nabla$ on $Eto M$ (see for instance the volume 2 of Greub-Halperin-Vanstone). Also we know that $VEcongpi^*E$ so that we can write the following decomposition:
$$
TEcongpi^*Eopluspi^*TM
$$
My question is the following: considering $E$ as a manifold, I would like to consider differential forms on $E$, that is $Gamma(Lambda T^*E)$; is the De Rham differential linked to $nabla$ in some way?
Thank you for your help. Any reference would be very useful.
differential-geometry vector-bundles exact-sequence connections
asked Jun 18 '14 at 15:45
bc87bc87
1249
$endgroup$
Let $pi:Eto M$ be a smooth vector bundle. Then we have the following exact sequence of vector bundles over $E$:
$$
0to VExrightarrow{} TExrightarrow{mathrm{d}pi}pi^*TMto 0
$$
Here $VE$ is the vertical bundle, that is kernel of the bunble map $mathrm{d}pi$. This sequence splits, so there is a bundle morphism $sigma:pi^*TMto TE$ such that $mathrm{d}picircsigma=mathrm{Id}$. It can be shown that the data of such a $sigma$ is equivalent to a connection (covariant derivative) $nabla$ on $Eto M$ (see for instance the volume 2 of Greub-Halperin-Vanstone). Also we know that $VEcongpi^*E$ so that we can write the following decomposition:
$$
TEcongpi^*Eopluspi^*TM
$$
My question is the following: considering $E$ as a manifold, I would like to consider differential forms on $E$, that is $Gamma(Lambda T^*E)$; is the De Rham differential linked to $nabla$ in some way?
Thank you for your help. Any reference would be very useful.
differential-geometry vector-bundles exact-sequence connections
differential-geometry vector-bundles exact-sequence connections
asked Jun 18 '14 at 15:45
bc87bc87
1249
asked Jun 18 '14 at 15:45
bc87bc87
1249
edited Jun 20 '14 at 8:53
bc87
asked Jun 18 '14 at 15:45
bc87bc87
1249
asked Jun 18 '14 at 15:45
bc87bc87
1249
asked Jun 18 '14 at 15:45
bc87bc87
1249
1249
add a comment |
add a comment |
1 Answer
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$begingroup$
The covariant derivative $nabla$ is a map $nabla: Gamma(TM)times Gamma(E)longrightarrow Gamma(E)$ which is $mathbb R$-bilinear and satisfies $$nabla_{fcdot X} s=fcdot nabla_X squad textrm{and}quad nabla_alpha (fcdot s)=fcdot nabla_X s+mathcal{L}_{X}(f)cdot s,$$ for every $fin C^infty(M)$, $XinGamma(TM)$ and $sin Gamma(E)$. We can then define $$d_nabla: Gamma(Lambda^p T^*Motimes E)longrightarrow Gamma(Lambda^{p+1} T^*Motimes E),$$ setting $$begin{eqnarray*}
d_nabla varepsilon(X_1, ldots, X_{p+1})&&:=sum_{j=1}^{p+1}(-1)^{j+1} nabla_{X_j} varepsilon(X_1, ldots, widehat{X_j}, ldots, X_j)\
&&+sum_{i<j} (-1)^{i+j} varepsilon([X_i, X_j], X_1, ldots, widehat{X_i}, ldots, widehat{X_j}, ldots, X_{p+1}).end{eqnarray*}$$ To define this I'm using the isomorphism:
$$Gamma(Lambda^p T^* Motimes E)simeq mathsf{Hom}_{C^infty(M)} (Lambda^p Gamma(TM), Gamma(E))simeq mathsf{Alt}^p_{C^infty(M)}(Gamma(TM), Gamma(E)).$$
If I'm not wrong, $d_nabla^2=0$ if and only if $nabla$ is a flat connection.
Up to this point everything is ok. Now, my guess is that you have a product:
$$cdot:Gamma(Lambda^p T^*M)times Gamma(Lambda^q T^* Motimes E)longrightarrow Gamma(Lambda^{p+q} T^*M otimes E), $$ defined in the obvious way and it holds:
$$d_nabla (alphacdot omega)=d_{mathsf{dR}}alphacdot omega+(-1)^{|alpha|} alpha cdot d_nabla omega. $$
Hope it helps.
P.s: If you want to know where those things came from look for Lie algebroid representations. If I'm not wrong you can find something here.
answered Dec 25 '17 at 18:21
PtFPtF
4,04821734
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add a comment |
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$begingroup$
The covariant derivative $nabla$ is a map $nabla: Gamma(TM)times Gamma(E)longrightarrow Gamma(E)$ which is $mathbb R$-bilinear and satisfies $$nabla_{fcdot X} s=fcdot nabla_X squad textrm{and}quad nabla_alpha (fcdot s)=fcdot nabla_X s+mathcal{L}_{X}(f)cdot s,$$ for every $fin C^infty(M)$, $XinGamma(TM)$ and $sin Gamma(E)$. We can then define $$d_nabla: Gamma(Lambda^p T^*Motimes E)longrightarrow Gamma(Lambda^{p+1} T^*Motimes E),$$ setting $$begin{eqnarray*}
d_nabla varepsilon(X_1, ldots, X_{p+1})&&:=sum_{j=1}^{p+1}(-1)^{j+1} nabla_{X_j} varepsilon(X_1, ldots, widehat{X_j}, ldots, X_j)\
&&+sum_{i<j} (-1)^{i+j} varepsilon([X_i, X_j], X_1, ldots, widehat{X_i}, ldots, widehat{X_j}, ldots, X_{p+1}).end{eqnarray*}$$ To define this I'm using the isomorphism:
$$Gamma(Lambda^p T^* Motimes E)simeq mathsf{Hom}_{C^infty(M)} (Lambda^p Gamma(TM), Gamma(E))simeq mathsf{Alt}^p_{C^infty(M)}(Gamma(TM), Gamma(E)).$$
If I'm not wrong, $d_nabla^2=0$ if and only if $nabla$ is a flat connection.
Up to this point everything is ok. Now, my guess is that you have a product:
$$cdot:Gamma(Lambda^p T^*M)times Gamma(Lambda^q T^* Motimes E)longrightarrow Gamma(Lambda^{p+q} T^*M otimes E), $$ defined in the obvious way and it holds:
$$d_nabla (alphacdot omega)=d_{mathsf{dR}}alphacdot omega+(-1)^{|alpha|} alpha cdot d_nabla omega. $$
Hope it helps.
P.s: If you want to know where those things came from look for Lie algebroid representations. If I'm not wrong you can find something here.
answered Dec 25 '17 at 18:21
PtFPtF
4,04821734
$endgroup$
add a comment |
$begingroup$
The covariant derivative $nabla$ is a map $nabla: Gamma(TM)times Gamma(E)longrightarrow Gamma(E)$ which is $mathbb R$-bilinear and satisfies $$nabla_{fcdot X} s=fcdot nabla_X squad textrm{and}quad nabla_alpha (fcdot s)=fcdot nabla_X s+mathcal{L}_{X}(f)cdot s,$$ for every $fin C^infty(M)$, $XinGamma(TM)$ and $sin Gamma(E)$. We can then define $$d_nabla: Gamma(Lambda^p T^*Motimes E)longrightarrow Gamma(Lambda^{p+1} T^*Motimes E),$$ setting $$begin{eqnarray*}
d_nabla varepsilon(X_1, ldots, X_{p+1})&&:=sum_{j=1}^{p+1}(-1)^{j+1} nabla_{X_j} varepsilon(X_1, ldots, widehat{X_j}, ldots, X_j)\
&&+sum_{i<j} (-1)^{i+j} varepsilon([X_i, X_j], X_1, ldots, widehat{X_i}, ldots, widehat{X_j}, ldots, X_{p+1}).end{eqnarray*}$$ To define this I'm using the isomorphism:
$$Gamma(Lambda^p T^* Motimes E)simeq mathsf{Hom}_{C^infty(M)} (Lambda^p Gamma(TM), Gamma(E))simeq mathsf{Alt}^p_{C^infty(M)}(Gamma(TM), Gamma(E)).$$
If I'm not wrong, $d_nabla^2=0$ if and only if $nabla$ is a flat connection.
Up to this point everything is ok. Now, my guess is that you have a product:
$$cdot:Gamma(Lambda^p T^*M)times Gamma(Lambda^q T^* Motimes E)longrightarrow Gamma(Lambda^{p+q} T^*M otimes E), $$ defined in the obvious way and it holds:
$$d_nabla (alphacdot omega)=d_{mathsf{dR}}alphacdot omega+(-1)^{|alpha|} alpha cdot d_nabla omega. $$
Hope it helps.
P.s: If you want to know where those things came from look for Lie algebroid representations. If I'm not wrong you can find something here.
answered Dec 25 '17 at 18:21
PtFPtF
4,04821734
$endgroup$
add a comment |
$begingroup$
The covariant derivative $nabla$ is a map $nabla: Gamma(TM)times Gamma(E)longrightarrow Gamma(E)$ which is $mathbb R$-bilinear and satisfies $$nabla_{fcdot X} s=fcdot nabla_X squad textrm{and}quad nabla_alpha (fcdot s)=fcdot nabla_X s+mathcal{L}_{X}(f)cdot s,$$ for every $fin C^infty(M)$, $XinGamma(TM)$ and $sin Gamma(E)$. We can then define $$d_nabla: Gamma(Lambda^p T^*Motimes E)longrightarrow Gamma(Lambda^{p+1} T^*Motimes E),$$ setting $$begin{eqnarray*}
d_nabla varepsilon(X_1, ldots, X_{p+1})&&:=sum_{j=1}^{p+1}(-1)^{j+1} nabla_{X_j} varepsilon(X_1, ldots, widehat{X_j}, ldots, X_j)\
&&+sum_{i<j} (-1)^{i+j} varepsilon([X_i, X_j], X_1, ldots, widehat{X_i}, ldots, widehat{X_j}, ldots, X_{p+1}).end{eqnarray*}$$ To define this I'm using the isomorphism:
$$Gamma(Lambda^p T^* Motimes E)simeq mathsf{Hom}_{C^infty(M)} (Lambda^p Gamma(TM), Gamma(E))simeq mathsf{Alt}^p_{C^infty(M)}(Gamma(TM), Gamma(E)).$$
If I'm not wrong, $d_nabla^2=0$ if and only if $nabla$ is a flat connection.
Up to this point everything is ok. Now, my guess is that you have a product:
$$cdot:Gamma(Lambda^p T^*M)times Gamma(Lambda^q T^* Motimes E)longrightarrow Gamma(Lambda^{p+q} T^*M otimes E), $$ defined in the obvious way and it holds:
$$d_nabla (alphacdot omega)=d_{mathsf{dR}}alphacdot omega+(-1)^{|alpha|} alpha cdot d_nabla omega. $$
Hope it helps.
P.s: If you want to know where those things came from look for Lie algebroid representations. If I'm not wrong you can find something here.
answered Dec 25 '17 at 18:21
PtFPtF
4,04821734
$endgroup$
The covariant derivative $nabla$ is a map $nabla: Gamma(TM)times Gamma(E)longrightarrow Gamma(E)$ which is $mathbb R$-bilinear and satisfies $$nabla_{fcdot X} s=fcdot nabla_X squad textrm{and}quad nabla_alpha (fcdot s)=fcdot nabla_X s+mathcal{L}_{X}(f)cdot s,$$ for every $fin C^infty(M)$, $XinGamma(TM)$ and $sin Gamma(E)$. We can then define $$d_nabla: Gamma(Lambda^p T^*Motimes E)longrightarrow Gamma(Lambda^{p+1} T^*Motimes E),$$ setting $$begin{eqnarray*}
d_nabla varepsilon(X_1, ldots, X_{p+1})&&:=sum_{j=1}^{p+1}(-1)^{j+1} nabla_{X_j} varepsilon(X_1, ldots, widehat{X_j}, ldots, X_j)\
&&+sum_{i<j} (-1)^{i+j} varepsilon([X_i, X_j], X_1, ldots, widehat{X_i}, ldots, widehat{X_j}, ldots, X_{p+1}).end{eqnarray*}$$ To define this I'm using the isomorphism:
$$Gamma(Lambda^p T^* Motimes E)simeq mathsf{Hom}_{C^infty(M)} (Lambda^p Gamma(TM), Gamma(E))simeq mathsf{Alt}^p_{C^infty(M)}(Gamma(TM), Gamma(E)).$$
If I'm not wrong, $d_nabla^2=0$ if and only if $nabla$ is a flat connection.
Up to this point everything is ok. Now, my guess is that you have a product:
$$cdot:Gamma(Lambda^p T^*M)times Gamma(Lambda^q T^* Motimes E)longrightarrow Gamma(Lambda^{p+q} T^*M otimes E), $$ defined in the obvious way and it holds:
$$d_nabla (alphacdot omega)=d_{mathsf{dR}}alphacdot omega+(-1)^{|alpha|} alpha cdot d_nabla omega. $$
Hope it helps.
P.s: If you want to know where those things came from look for Lie algebroid representations. If I'm not wrong you can find something here.
answered Dec 25 '17 at 18:21
PtFPtF
4,04821734
edited Dec 25 '17 at 18:27
answered Dec 25 '17 at 18:21
PtFPtF
4,04821734
answered Dec 25 '17 at 18:21
PtFPtF
4,04821734
answered Dec 25 '17 at 18:21
PtFPtF
4,04821734
4,04821734
add a comment |
add a comment |
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