Non-flat connection on trivial bundle? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Why do people care about principal bundles?Connections on principal bundles and vector bundlesMetric non symmetric connection in the tangent bundle?Parallel vector fields imply a flat connection?Frobenius theorem to prove that flat connection admits a local basis of flat sectionsReferences: Equivalence between local systems and vector bundles (with flat connections)Recovering a principal connection from its monodromyDifferential geometry of line bundlesShowing that a given vector bundle with connection is not trivialConstructing the flat vector bundle associated to a given linear representation of the fundamental group
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Non-flat connection on trivial bundle?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Why do people care about principal bundles?Connections on principal bundles and vector bundlesMetric non symmetric connection in the tangent bundle?Parallel vector fields imply a flat connection?Frobenius theorem to prove that flat connection admits a local basis of flat sectionsReferences: Equivalence between local systems and vector bundles (with flat connections)Recovering a principal connection from its monodromyDifferential geometry of line bundlesShowing that a given vector bundle with connection is not trivialConstructing the flat vector bundle associated to a given linear representation of the fundamental group
$begingroup$
From what I have read it seems like there exists non-flat connections on trivial vector/principal bundles. However I can't find any notes on it or examples.
Can anyone confirm that such connections exist and perhaps provide an example?
differential-geometry vector-bundles curvature connections principal-bundles
asked Apr 2 at 17:24
jojojojo
6417
$endgroup$
add a comment |
$begingroup$
From what I have read it seems like there exists non-flat connections on trivial vector/principal bundles. However I can't find any notes on it or examples.
Can anyone confirm that such connections exist and perhaps provide an example?
differential-geometry vector-bundles curvature connections principal-bundles
asked Apr 2 at 17:24
jojojojo
6417
$endgroup$
$begingroup$
Yes, they do exist. But you have to exclude 1-d base.
$endgroup$
– Moishe Kohan
Apr 2 at 23:42
$begingroup$
Is it correct that on the trivial complex line bundles, no flat-connection exists? This is because then applying Weil-Chern theory we get that the first chern class cannot be trivial and thus the bundle cannot be trivial which is a contradiction?
$endgroup$
– jojo
Apr 3 at 6:41
add a comment |
$begingroup$
From what I have read it seems like there exists non-flat connections on trivial vector/principal bundles. However I can't find any notes on it or examples.
Can anyone confirm that such connections exist and perhaps provide an example?
differential-geometry vector-bundles curvature connections principal-bundles
asked Apr 2 at 17:24
jojojojo
6417
$endgroup$
From what I have read it seems like there exists non-flat connections on trivial vector/principal bundles. However I can't find any notes on it or examples.
Can anyone confirm that such connections exist and perhaps provide an example?
differential-geometry vector-bundles curvature connections principal-bundles
differential-geometry vector-bundles curvature connections principal-bundles
asked Apr 2 at 17:24
jojojojo
6417
asked Apr 2 at 17:24
jojojojo
6417
edited Apr 3 at 6:18
jojo
asked Apr 2 at 17:24
jojojojo
6417
asked Apr 2 at 17:24
jojojojo
6417
asked Apr 2 at 17:24
jojojojo
6417
6417
$begingroup$
Yes, they do exist. But you have to exclude 1-d base.
$endgroup$
– Moishe Kohan
Apr 2 at 23:42
$begingroup$
Is it correct that on the trivial complex line bundles, no flat-connection exists? This is because then applying Weil-Chern theory we get that the first chern class cannot be trivial and thus the bundle cannot be trivial which is a contradiction?
$endgroup$
– jojo
Apr 3 at 6:41
add a comment |
$begingroup$
Yes, they do exist. But you have to exclude 1-d base.
$endgroup$
– Moishe Kohan
Apr 2 at 23:42
$begingroup$
Is it correct that on the trivial complex line bundles, no flat-connection exists? This is because then applying Weil-Chern theory we get that the first chern class cannot be trivial and thus the bundle cannot be trivial which is a contradiction?
$endgroup$
– jojo
Apr 3 at 6:41
$begingroup$
Yes, they do exist. But you have to exclude 1-d base.
$endgroup$
– Moishe Kohan
Apr 2 at 23:42
$begingroup$
Yes, they do exist. But you have to exclude 1-d base.
$endgroup$
– Moishe Kohan
Apr 2 at 23:42
$begingroup$
Is it correct that on the trivial complex line bundles, no flat-connection exists? This is because then applying Weil-Chern theory we get that the first chern class cannot be trivial and thus the bundle cannot be trivial which is a contradiction?
$endgroup$
– jojo
Apr 3 at 6:41
$begingroup$
Is it correct that on the trivial complex line bundles, no flat-connection exists? This is because then applying Weil-Chern theory we get that the first chern class cannot be trivial and thus the bundle cannot be trivial which is a contradiction?
$endgroup$
– jojo
Apr 3 at 6:41
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You are confusing the notion of flatness for connections with characteristic classes. A connection is flat if and only if it has (identically) zero curvature.
Thus, to construct a non-flat connection on a trivial bundle $p: Eto M$, it suffices to do so locally: Construct a non-flat connection form with compact support on $p^-1(U)$ for some small open subset in the base $M$ and then extend by zero to the rest of the manifold (this makes no sense for general bundles but is well-defined for the trivial bundle). I will do so for complex vector bundles (of rank $ge 1$) since it appears that this is what you are interested in. Furthermore, it suffices to do so for complex line bundles $p: Lto M$ (since our trivial vector bundle is a direct sum of trivial line bundles). Then, a connection can be identified with a complex-valued 1-form $omega$ on $M$ (the actual connection will equal $d+omega$). The curvature of the connection equals
$$
domega+ omegawedge omega=domega$$
in our situation. Thus, all what you need is to construct a compactly supported complex-valued (actually, even real-valued is enough!) 1-form $omega$ on the open ball $B$ in $R^n$, $nge 2$, such that $domegane 0$. For instance, take a smooth compactly supported function $eta(x)$ on $B$ which is not identically zero and let $omega= eta(x) dx_1$. I will leave it to you to verify that $domega$ is not identically zero.
answered Apr 3 at 16:06
Moishe KohanMoishe Kohan
49k345111
$endgroup$
add a comment |
$begingroup$
They do exist (provided that the base has dimension bigger than one) and any non-flat connection provides an example.They point here is that curvature is a local invariant of a connection, while triviality of a bundle is a global condition. So whenever you have an example of a connection and a point in which the curvature of the connection is non-trivial, then the bundle is trivial on a neighbourhood of that point, an the restriction of the connection to that trivial bundle still is non-flat.
Implications can only be obtained in the other direction, say that certain bundles do not admit (global) flat connections.
answered Apr 3 at 7:49
Andreas CapAndreas Cap
11.5k923
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
You are confusing the notion of flatness for connections with characteristic classes. A connection is flat if and only if it has (identically) zero curvature.
Thus, to construct a non-flat connection on a trivial bundle $p: Eto M$, it suffices to do so locally: Construct a non-flat connection form with compact support on $p^-1(U)$ for some small open subset in the base $M$ and then extend by zero to the rest of the manifold (this makes no sense for general bundles but is well-defined for the trivial bundle). I will do so for complex vector bundles (of rank $ge 1$) since it appears that this is what you are interested in. Furthermore, it suffices to do so for complex line bundles $p: Lto M$ (since our trivial vector bundle is a direct sum of trivial line bundles). Then, a connection can be identified with a complex-valued 1-form $omega$ on $M$ (the actual connection will equal $d+omega$). The curvature of the connection equals
$$
domega+ omegawedge omega=domega$$
in our situation. Thus, all what you need is to construct a compactly supported complex-valued (actually, even real-valued is enough!) 1-form $omega$ on the open ball $B$ in $R^n$, $nge 2$, such that $domegane 0$. For instance, take a smooth compactly supported function $eta(x)$ on $B$ which is not identically zero and let $omega= eta(x) dx_1$. I will leave it to you to verify that $domega$ is not identically zero.
answered Apr 3 at 16:06
Moishe KohanMoishe Kohan
49k345111
$endgroup$
add a comment |
$begingroup$
You are confusing the notion of flatness for connections with characteristic classes. A connection is flat if and only if it has (identically) zero curvature.
Thus, to construct a non-flat connection on a trivial bundle $p: Eto M$, it suffices to do so locally: Construct a non-flat connection form with compact support on $p^-1(U)$ for some small open subset in the base $M$ and then extend by zero to the rest of the manifold (this makes no sense for general bundles but is well-defined for the trivial bundle). I will do so for complex vector bundles (of rank $ge 1$) since it appears that this is what you are interested in. Furthermore, it suffices to do so for complex line bundles $p: Lto M$ (since our trivial vector bundle is a direct sum of trivial line bundles). Then, a connection can be identified with a complex-valued 1-form $omega$ on $M$ (the actual connection will equal $d+omega$). The curvature of the connection equals
$$
domega+ omegawedge omega=domega$$
in our situation. Thus, all what you need is to construct a compactly supported complex-valued (actually, even real-valued is enough!) 1-form $omega$ on the open ball $B$ in $R^n$, $nge 2$, such that $domegane 0$. For instance, take a smooth compactly supported function $eta(x)$ on $B$ which is not identically zero and let $omega= eta(x) dx_1$. I will leave it to you to verify that $domega$ is not identically zero.
answered Apr 3 at 16:06
Moishe KohanMoishe Kohan
49k345111
$endgroup$
add a comment |
$begingroup$
You are confusing the notion of flatness for connections with characteristic classes. A connection is flat if and only if it has (identically) zero curvature.
Thus, to construct a non-flat connection on a trivial bundle $p: Eto M$, it suffices to do so locally: Construct a non-flat connection form with compact support on $p^-1(U)$ for some small open subset in the base $M$ and then extend by zero to the rest of the manifold (this makes no sense for general bundles but is well-defined for the trivial bundle). I will do so for complex vector bundles (of rank $ge 1$) since it appears that this is what you are interested in. Furthermore, it suffices to do so for complex line bundles $p: Lto M$ (since our trivial vector bundle is a direct sum of trivial line bundles). Then, a connection can be identified with a complex-valued 1-form $omega$ on $M$ (the actual connection will equal $d+omega$). The curvature of the connection equals
$$
domega+ omegawedge omega=domega$$
in our situation. Thus, all what you need is to construct a compactly supported complex-valued (actually, even real-valued is enough!) 1-form $omega$ on the open ball $B$ in $R^n$, $nge 2$, such that $domegane 0$. For instance, take a smooth compactly supported function $eta(x)$ on $B$ which is not identically zero and let $omega= eta(x) dx_1$. I will leave it to you to verify that $domega$ is not identically zero.
answered Apr 3 at 16:06
Moishe KohanMoishe Kohan
49k345111
$endgroup$
You are confusing the notion of flatness for connections with characteristic classes. A connection is flat if and only if it has (identically) zero curvature.
Thus, to construct a non-flat connection on a trivial bundle $p: Eto M$, it suffices to do so locally: Construct a non-flat connection form with compact support on $p^-1(U)$ for some small open subset in the base $M$ and then extend by zero to the rest of the manifold (this makes no sense for general bundles but is well-defined for the trivial bundle). I will do so for complex vector bundles (of rank $ge 1$) since it appears that this is what you are interested in. Furthermore, it suffices to do so for complex line bundles $p: Lto M$ (since our trivial vector bundle is a direct sum of trivial line bundles). Then, a connection can be identified with a complex-valued 1-form $omega$ on $M$ (the actual connection will equal $d+omega$). The curvature of the connection equals
$$
domega+ omegawedge omega=domega$$
in our situation. Thus, all what you need is to construct a compactly supported complex-valued (actually, even real-valued is enough!) 1-form $omega$ on the open ball $B$ in $R^n$, $nge 2$, such that $domegane 0$. For instance, take a smooth compactly supported function $eta(x)$ on $B$ which is not identically zero and let $omega= eta(x) dx_1$. I will leave it to you to verify that $domega$ is not identically zero.
answered Apr 3 at 16:06
Moishe KohanMoishe Kohan
49k345111
answered Apr 3 at 16:06
Moishe KohanMoishe Kohan
49k345111
answered Apr 3 at 16:06
Moishe KohanMoishe Kohan
49k345111
answered Apr 3 at 16:06
Moishe KohanMoishe Kohan
49k345111
49k345111
add a comment |
add a comment |
$begingroup$
They do exist (provided that the base has dimension bigger than one) and any non-flat connection provides an example.They point here is that curvature is a local invariant of a connection, while triviality of a bundle is a global condition. So whenever you have an example of a connection and a point in which the curvature of the connection is non-trivial, then the bundle is trivial on a neighbourhood of that point, an the restriction of the connection to that trivial bundle still is non-flat.
Implications can only be obtained in the other direction, say that certain bundles do not admit (global) flat connections.
answered Apr 3 at 7:49
Andreas CapAndreas Cap
11.5k923
$endgroup$
add a comment |
$begingroup$
They do exist (provided that the base has dimension bigger than one) and any non-flat connection provides an example.They point here is that curvature is a local invariant of a connection, while triviality of a bundle is a global condition. So whenever you have an example of a connection and a point in which the curvature of the connection is non-trivial, then the bundle is trivial on a neighbourhood of that point, an the restriction of the connection to that trivial bundle still is non-flat.
Implications can only be obtained in the other direction, say that certain bundles do not admit (global) flat connections.
answered Apr 3 at 7:49
Andreas CapAndreas Cap
11.5k923
$endgroup$
add a comment |
$begingroup$
They do exist (provided that the base has dimension bigger than one) and any non-flat connection provides an example.They point here is that curvature is a local invariant of a connection, while triviality of a bundle is a global condition. So whenever you have an example of a connection and a point in which the curvature of the connection is non-trivial, then the bundle is trivial on a neighbourhood of that point, an the restriction of the connection to that trivial bundle still is non-flat.
Implications can only be obtained in the other direction, say that certain bundles do not admit (global) flat connections.
answered Apr 3 at 7:49
Andreas CapAndreas Cap
11.5k923
$endgroup$
They do exist (provided that the base has dimension bigger than one) and any non-flat connection provides an example.They point here is that curvature is a local invariant of a connection, while triviality of a bundle is a global condition. So whenever you have an example of a connection and a point in which the curvature of the connection is non-trivial, then the bundle is trivial on a neighbourhood of that point, an the restriction of the connection to that trivial bundle still is non-flat.
Implications can only be obtained in the other direction, say that certain bundles do not admit (global) flat connections.
answered Apr 3 at 7:49
Andreas CapAndreas Cap
11.5k923
answered Apr 3 at 7:49
Andreas CapAndreas Cap
11.5k923
answered Apr 3 at 7:49
Andreas CapAndreas Cap
11.5k923
answered Apr 3 at 7:49
Andreas CapAndreas Cap
11.5k923
11.5k923
add a comment |
add a comment |
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$begingroup$
Yes, they do exist. But you have to exclude 1-d base.
$endgroup$
– Moishe Kohan
Apr 2 at 23:42
$begingroup$
Is it correct that on the trivial complex line bundles, no flat-connection exists? This is because then applying Weil-Chern theory we get that the first chern class cannot be trivial and thus the bundle cannot be trivial which is a contradiction?
$endgroup$
– jojo
Apr 3 at 6:41