Showing that a given vector bundle with connection is not trivial
Given the following exercise:
where d is the trivial connection.
We defined an isomorphism between two vector bundles with connection in the following way:
I'm not sure what I have to show. Can I use that any isomorphism between E and $M times mathbb{R}^2$ must be the identity and then show that the identity is not parallel? So that there exist $X_0 in Gamma (TM), psi_0 in Gamma(E)$ such that: $d_{X_0} psi_0 = nabla_{X_0}psi_0$?
If that's the case could I use the matrix of 1-forms and the standard frame $(e_1, e_2)$ on E to write out the connection $d - nabla$ and then choose as $psi_0$ the section $e_1$, which would yield a term that would not be equal to 0 for some $p in M$.
For the second part of the exercise, we have shown that a vector bundle with connection is trivial if and only if there exists a parallel frame field. In the proof for this, it was shown how to construct an isomorphism given the parallel frame field. Can someone give a hint on how to construct a parallel frame field for $(E, tilde nabla)$?
differential-geometry vector-bundles connections
asked Dec 3 '18 at 15:31
eager2learn
1,23211430
add a comment |
Given the following exercise:
where d is the trivial connection.
We defined an isomorphism between two vector bundles with connection in the following way:
I'm not sure what I have to show. Can I use that any isomorphism between E and $M times mathbb{R}^2$ must be the identity and then show that the identity is not parallel? So that there exist $X_0 in Gamma (TM), psi_0 in Gamma(E)$ such that: $d_{X_0} psi_0 = nabla_{X_0}psi_0$?
If that's the case could I use the matrix of 1-forms and the standard frame $(e_1, e_2)$ on E to write out the connection $d - nabla$ and then choose as $psi_0$ the section $e_1$, which would yield a term that would not be equal to 0 for some $p in M$.
For the second part of the exercise, we have shown that a vector bundle with connection is trivial if and only if there exists a parallel frame field. In the proof for this, it was shown how to construct an isomorphism given the parallel frame field. Can someone give a hint on how to construct a parallel frame field for $(E, tilde nabla)$?
differential-geometry vector-bundles connections
asked Dec 3 '18 at 15:31
eager2learn
1,23211430
Do you know about curvature?
– Ted Shifrin
Dec 3 '18 at 21:59
add a comment |
Given the following exercise:
where d is the trivial connection.
We defined an isomorphism between two vector bundles with connection in the following way:
I'm not sure what I have to show. Can I use that any isomorphism between E and $M times mathbb{R}^2$ must be the identity and then show that the identity is not parallel? So that there exist $X_0 in Gamma (TM), psi_0 in Gamma(E)$ such that: $d_{X_0} psi_0 = nabla_{X_0}psi_0$?
If that's the case could I use the matrix of 1-forms and the standard frame $(e_1, e_2)$ on E to write out the connection $d - nabla$ and then choose as $psi_0$ the section $e_1$, which would yield a term that would not be equal to 0 for some $p in M$.
For the second part of the exercise, we have shown that a vector bundle with connection is trivial if and only if there exists a parallel frame field. In the proof for this, it was shown how to construct an isomorphism given the parallel frame field. Can someone give a hint on how to construct a parallel frame field for $(E, tilde nabla)$?
differential-geometry vector-bundles connections
asked Dec 3 '18 at 15:31
eager2learn
1,23211430
Given the following exercise:
where d is the trivial connection.
We defined an isomorphism between two vector bundles with connection in the following way:
I'm not sure what I have to show. Can I use that any isomorphism between E and $M times mathbb{R}^2$ must be the identity and then show that the identity is not parallel? So that there exist $X_0 in Gamma (TM), psi_0 in Gamma(E)$ such that: $d_{X_0} psi_0 = nabla_{X_0}psi_0$?
If that's the case could I use the matrix of 1-forms and the standard frame $(e_1, e_2)$ on E to write out the connection $d - nabla$ and then choose as $psi_0$ the section $e_1$, which would yield a term that would not be equal to 0 for some $p in M$.
For the second part of the exercise, we have shown that a vector bundle with connection is trivial if and only if there exists a parallel frame field. In the proof for this, it was shown how to construct an isomorphism given the parallel frame field. Can someone give a hint on how to construct a parallel frame field for $(E, tilde nabla)$?
differential-geometry vector-bundles connections
differential-geometry vector-bundles connections
asked Dec 3 '18 at 15:31
eager2learn
1,23211430
asked Dec 3 '18 at 15:31
eager2learn
1,23211430
asked Dec 3 '18 at 15:31
eager2learn
1,23211430
asked Dec 3 '18 at 15:31
eager2learn
1,23211430
asked Dec 3 '18 at 15:31
eager2learn
1,23211430
1,23211430
Do you know about curvature?
– Ted Shifrin
Dec 3 '18 at 21:59
add a comment |
Do you know about curvature?
– Ted Shifrin
Dec 3 '18 at 21:59
Do you know about curvature?
– Ted Shifrin
Dec 3 '18 at 21:59
Do you know about curvature?
– Ted Shifrin
Dec 3 '18 at 21:59
add a comment |
1 Answer
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Here's a hint: If you have an orthonormal frame field $e_1,e_2$ for the bundle with connection form $omega_{12}$ and rotate the frame field by considering $e_1' = costheta e_1+sintheta e_2$, $e_2'=-sintheta e_1+costheta e_2$ for some function $theta$, then $omega_{12}' = omega_{12}+dtheta$. If $omega_{12}=df$, then of course you can make $omega_{12}' = 0$ by choosing $theta=-f$. Then $e_1', e_2'$ will be parallel frame fields.
answered Dec 3 '18 at 22:23
Ted Shifrin
62.9k44489
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
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oldest
votes
active
oldest
votes
Here's a hint: If you have an orthonormal frame field $e_1,e_2$ for the bundle with connection form $omega_{12}$ and rotate the frame field by considering $e_1' = costheta e_1+sintheta e_2$, $e_2'=-sintheta e_1+costheta e_2$ for some function $theta$, then $omega_{12}' = omega_{12}+dtheta$. If $omega_{12}=df$, then of course you can make $omega_{12}' = 0$ by choosing $theta=-f$. Then $e_1', e_2'$ will be parallel frame fields.
answered Dec 3 '18 at 22:23
Ted Shifrin
62.9k44489
add a comment |
Here's a hint: If you have an orthonormal frame field $e_1,e_2$ for the bundle with connection form $omega_{12}$ and rotate the frame field by considering $e_1' = costheta e_1+sintheta e_2$, $e_2'=-sintheta e_1+costheta e_2$ for some function $theta$, then $omega_{12}' = omega_{12}+dtheta$. If $omega_{12}=df$, then of course you can make $omega_{12}' = 0$ by choosing $theta=-f$. Then $e_1', e_2'$ will be parallel frame fields.
answered Dec 3 '18 at 22:23
Ted Shifrin
62.9k44489
add a comment |
Here's a hint: If you have an orthonormal frame field $e_1,e_2$ for the bundle with connection form $omega_{12}$ and rotate the frame field by considering $e_1' = costheta e_1+sintheta e_2$, $e_2'=-sintheta e_1+costheta e_2$ for some function $theta$, then $omega_{12}' = omega_{12}+dtheta$. If $omega_{12}=df$, then of course you can make $omega_{12}' = 0$ by choosing $theta=-f$. Then $e_1', e_2'$ will be parallel frame fields.
answered Dec 3 '18 at 22:23
Ted Shifrin
62.9k44489
Here's a hint: If you have an orthonormal frame field $e_1,e_2$ for the bundle with connection form $omega_{12}$ and rotate the frame field by considering $e_1' = costheta e_1+sintheta e_2$, $e_2'=-sintheta e_1+costheta e_2$ for some function $theta$, then $omega_{12}' = omega_{12}+dtheta$. If $omega_{12}=df$, then of course you can make $omega_{12}' = 0$ by choosing $theta=-f$. Then $e_1', e_2'$ will be parallel frame fields.
answered Dec 3 '18 at 22:23
Ted Shifrin
62.9k44489
edited Dec 3 '18 at 23:30
answered Dec 3 '18 at 22:23
Ted Shifrin
62.9k44489
answered Dec 3 '18 at 22:23
Ted Shifrin
62.9k44489
answered Dec 3 '18 at 22:23
Ted Shifrin
62.9k44489
62.9k44489
add a comment |
add a comment |
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Do you know about curvature?
– Ted Shifrin
Dec 3 '18 at 21:59