Integral curves are horizontal
$begingroup$
Suppose we're given an Ehresmann connection on a submersion (or a fiber bundle, but I don't think it's needed) $begin{smallmatrix}X\
downarrow\
Y
end{smallmatrix}$. Given a curve $gamma$ in the base $Y$, consider the pullback of the submersion and its connection along the curve. The horizontal bundle of the pulled-back bundle $begin{smallmatrix}gamma ^{ast}X\
downarrow\
I
end{smallmatrix}$ is a line subbundle of the tangent bundle $begin{smallmatrix}mathrm Tgamma^{ast}X\
downarrow\
gamma^ast X
end{smallmatrix}$. Thus we locally have integral curves in $gamma^ast X$ for the horizontal bundle. These integral curves are also transverse to the fibers of the bundle $gamma^ast Xto I$.
Write $c_{(s,x)}(t)in gamma^ast X$ for the value of at $t$ of the integral curve $c$ based at $(s,x)in gamma^ast X$. How can I prove the first coordinate of $c_{(s,x)}(t)$ is always $s+t$? I drew a picture, which tells me I should use transversality, but I don't know how to formulate a proof.
(The context: I want to prove that flow along the horizontal bundle defines parallel transport, and for that I need to know the horizontal position of an integral curve at each instance.)
differential-geometry fiber-bundles connections
asked Dec 20 '18 at 20:19
ArrowArrow
5,13731446
$endgroup$
add a comment |
$begingroup$
Suppose we're given an Ehresmann connection on a submersion (or a fiber bundle, but I don't think it's needed) $begin{smallmatrix}X\
downarrow\
Y
end{smallmatrix}$. Given a curve $gamma$ in the base $Y$, consider the pullback of the submersion and its connection along the curve. The horizontal bundle of the pulled-back bundle $begin{smallmatrix}gamma ^{ast}X\
downarrow\
I
end{smallmatrix}$ is a line subbundle of the tangent bundle $begin{smallmatrix}mathrm Tgamma^{ast}X\
downarrow\
gamma^ast X
end{smallmatrix}$. Thus we locally have integral curves in $gamma^ast X$ for the horizontal bundle. These integral curves are also transverse to the fibers of the bundle $gamma^ast Xto I$.
Write $c_{(s,x)}(t)in gamma^ast X$ for the value of at $t$ of the integral curve $c$ based at $(s,x)in gamma^ast X$. How can I prove the first coordinate of $c_{(s,x)}(t)$ is always $s+t$? I drew a picture, which tells me I should use transversality, but I don't know how to formulate a proof.
(The context: I want to prove that flow along the horizontal bundle defines parallel transport, and for that I need to know the horizontal position of an integral curve at each instance.)
differential-geometry fiber-bundles connections
asked Dec 20 '18 at 20:19
ArrowArrow
5,13731446
$endgroup$
$begingroup$
The horizontal curves are the trajectories of a horizontal vector field which you seem to ignore. This vector field is characterized by two properties. The first is horizontality, as you explain in your question. The second property is that it projects to the vector field $partial/partial t$ on $I$.
$endgroup$
– Amitai Yuval
Dec 20 '18 at 20:41
$begingroup$
Dear @AmitaiYuval, thank you for your comment. I posted an answer in an attempt to make sure I understood you well. Would appreciate if you could take a look and point out any errors!
$endgroup$
– Arrow
Dec 20 '18 at 22:21
add a comment |
$begingroup$
Suppose we're given an Ehresmann connection on a submersion (or a fiber bundle, but I don't think it's needed) $begin{smallmatrix}X\
downarrow\
Y
end{smallmatrix}$. Given a curve $gamma$ in the base $Y$, consider the pullback of the submersion and its connection along the curve. The horizontal bundle of the pulled-back bundle $begin{smallmatrix}gamma ^{ast}X\
downarrow\
I
end{smallmatrix}$ is a line subbundle of the tangent bundle $begin{smallmatrix}mathrm Tgamma^{ast}X\
downarrow\
gamma^ast X
end{smallmatrix}$. Thus we locally have integral curves in $gamma^ast X$ for the horizontal bundle. These integral curves are also transverse to the fibers of the bundle $gamma^ast Xto I$.
Write $c_{(s,x)}(t)in gamma^ast X$ for the value of at $t$ of the integral curve $c$ based at $(s,x)in gamma^ast X$. How can I prove the first coordinate of $c_{(s,x)}(t)$ is always $s+t$? I drew a picture, which tells me I should use transversality, but I don't know how to formulate a proof.
(The context: I want to prove that flow along the horizontal bundle defines parallel transport, and for that I need to know the horizontal position of an integral curve at each instance.)
differential-geometry fiber-bundles connections
asked Dec 20 '18 at 20:19
ArrowArrow
5,13731446
$endgroup$
Suppose we're given an Ehresmann connection on a submersion (or a fiber bundle, but I don't think it's needed) $begin{smallmatrix}X\
downarrow\
Y
end{smallmatrix}$. Given a curve $gamma$ in the base $Y$, consider the pullback of the submersion and its connection along the curve. The horizontal bundle of the pulled-back bundle $begin{smallmatrix}gamma ^{ast}X\
downarrow\
I
end{smallmatrix}$ is a line subbundle of the tangent bundle $begin{smallmatrix}mathrm Tgamma^{ast}X\
downarrow\
gamma^ast X
end{smallmatrix}$. Thus we locally have integral curves in $gamma^ast X$ for the horizontal bundle. These integral curves are also transverse to the fibers of the bundle $gamma^ast Xto I$.
Write $c_{(s,x)}(t)in gamma^ast X$ for the value of at $t$ of the integral curve $c$ based at $(s,x)in gamma^ast X$. How can I prove the first coordinate of $c_{(s,x)}(t)$ is always $s+t$? I drew a picture, which tells me I should use transversality, but I don't know how to formulate a proof.
(The context: I want to prove that flow along the horizontal bundle defines parallel transport, and for that I need to know the horizontal position of an integral curve at each instance.)
differential-geometry fiber-bundles connections
differential-geometry fiber-bundles connections
asked Dec 20 '18 at 20:19
ArrowArrow
5,13731446
asked Dec 20 '18 at 20:19
ArrowArrow
5,13731446
asked Dec 20 '18 at 20:19
ArrowArrow
5,13731446
asked Dec 20 '18 at 20:19
ArrowArrow
5,13731446
asked Dec 20 '18 at 20:19
ArrowArrow
5,13731446
5,13731446
$begingroup$
The horizontal curves are the trajectories of a horizontal vector field which you seem to ignore. This vector field is characterized by two properties. The first is horizontality, as you explain in your question. The second property is that it projects to the vector field $partial/partial t$ on $I$.
$endgroup$
– Amitai Yuval
Dec 20 '18 at 20:41
$begingroup$
Dear @AmitaiYuval, thank you for your comment. I posted an answer in an attempt to make sure I understood you well. Would appreciate if you could take a look and point out any errors!
$endgroup$
– Arrow
Dec 20 '18 at 22:21
add a comment |
$begingroup$
The horizontal curves are the trajectories of a horizontal vector field which you seem to ignore. This vector field is characterized by two properties. The first is horizontality, as you explain in your question. The second property is that it projects to the vector field $partial/partial t$ on $I$.
$endgroup$
– Amitai Yuval
Dec 20 '18 at 20:41
$begingroup$
Dear @AmitaiYuval, thank you for your comment. I posted an answer in an attempt to make sure I understood you well. Would appreciate if you could take a look and point out any errors!
$endgroup$
– Arrow
Dec 20 '18 at 22:21
$begingroup$
The horizontal curves are the trajectories of a horizontal vector field which you seem to ignore. This vector field is characterized by two properties. The first is horizontality, as you explain in your question. The second property is that it projects to the vector field $partial/partial t$ on $I$.
$endgroup$
– Amitai Yuval
Dec 20 '18 at 20:41
$begingroup$
The horizontal curves are the trajectories of a horizontal vector field which you seem to ignore. This vector field is characterized by two properties. The first is horizontality, as you explain in your question. The second property is that it projects to the vector field $partial/partial t$ on $I$.
$endgroup$
– Amitai Yuval
Dec 20 '18 at 20:41
$begingroup$
Dear @AmitaiYuval, thank you for your comment. I posted an answer in an attempt to make sure I understood you well. Would appreciate if you could take a look and point out any errors!
$endgroup$
– Arrow
Dec 20 '18 at 22:21
$begingroup$
Dear @AmitaiYuval, thank you for your comment. I posted an answer in an attempt to make sure I understood you well. Would appreciate if you could take a look and point out any errors!
$endgroup$
– Arrow
Dec 20 '18 at 22:21
add a comment |
1 Answer
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$begingroup$
This is just a slight expansion of Amitai Yuval's comment.
The problem is that integral curves are not defined with respect to a line subbundle of the tangent bundle, but with respect to a vector field. (Only the images of the integral curves are "known" to the horizontal subbundle, much like only integral submanifolds - without any parametrization - are "known" to general subbundles of the tangent bundle.) Geometrically this says the horizontal bundle does not determine how quickly to flow along its integral submanifolds (the images of the integral curves).
The horizontal subbundle amounts to a multiple of a global horizontal vector field by a scalar function. Of these it is naturally best to choose a lift of the global vector field $partial/partial t$ on $I$. With respect to this horizontal vector field the integral curves are now defined.
That this field lifts $partial/partial t$ implies by the chain rule that $frac{partial}{partial t}(picirc c_{(s,x)})equiv 1$. But $picirc c_{(s,x)}$ is precisely the map $f$ giving the first coordinate of the integral curve $c$. By assumption $f(0)=s$ and $f^prime equiv 1$. By uniqueness of solutions to ODE it must then by given by $f(t)=s+t$, as desired.
answered Dec 20 '18 at 22:18
ArrowArrow
5,13731446
$endgroup$
$begingroup$
Seems fine to me.
$endgroup$
– Amitai Yuval
Dec 20 '18 at 22:40
$begingroup$
@AmitaiYuval great, thanks again.
$endgroup$
– Arrow
Dec 20 '18 at 22:41
add a comment |
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$begingroup$
This is just a slight expansion of Amitai Yuval's comment.
The problem is that integral curves are not defined with respect to a line subbundle of the tangent bundle, but with respect to a vector field. (Only the images of the integral curves are "known" to the horizontal subbundle, much like only integral submanifolds - without any parametrization - are "known" to general subbundles of the tangent bundle.) Geometrically this says the horizontal bundle does not determine how quickly to flow along its integral submanifolds (the images of the integral curves).
The horizontal subbundle amounts to a multiple of a global horizontal vector field by a scalar function. Of these it is naturally best to choose a lift of the global vector field $partial/partial t$ on $I$. With respect to this horizontal vector field the integral curves are now defined.
That this field lifts $partial/partial t$ implies by the chain rule that $frac{partial}{partial t}(picirc c_{(s,x)})equiv 1$. But $picirc c_{(s,x)}$ is precisely the map $f$ giving the first coordinate of the integral curve $c$. By assumption $f(0)=s$ and $f^prime equiv 1$. By uniqueness of solutions to ODE it must then by given by $f(t)=s+t$, as desired.
answered Dec 20 '18 at 22:18
ArrowArrow
5,13731446
$endgroup$
$begingroup$
Seems fine to me.
$endgroup$
– Amitai Yuval
Dec 20 '18 at 22:40
$begingroup$
@AmitaiYuval great, thanks again.
$endgroup$
– Arrow
Dec 20 '18 at 22:41
add a comment |
$begingroup$
This is just a slight expansion of Amitai Yuval's comment.
The problem is that integral curves are not defined with respect to a line subbundle of the tangent bundle, but with respect to a vector field. (Only the images of the integral curves are "known" to the horizontal subbundle, much like only integral submanifolds - without any parametrization - are "known" to general subbundles of the tangent bundle.) Geometrically this says the horizontal bundle does not determine how quickly to flow along its integral submanifolds (the images of the integral curves).
The horizontal subbundle amounts to a multiple of a global horizontal vector field by a scalar function. Of these it is naturally best to choose a lift of the global vector field $partial/partial t$ on $I$. With respect to this horizontal vector field the integral curves are now defined.
That this field lifts $partial/partial t$ implies by the chain rule that $frac{partial}{partial t}(picirc c_{(s,x)})equiv 1$. But $picirc c_{(s,x)}$ is precisely the map $f$ giving the first coordinate of the integral curve $c$. By assumption $f(0)=s$ and $f^prime equiv 1$. By uniqueness of solutions to ODE it must then by given by $f(t)=s+t$, as desired.
answered Dec 20 '18 at 22:18
ArrowArrow
5,13731446
$endgroup$
$begingroup$
Seems fine to me.
$endgroup$
– Amitai Yuval
Dec 20 '18 at 22:40
$begingroup$
@AmitaiYuval great, thanks again.
$endgroup$
– Arrow
Dec 20 '18 at 22:41
add a comment |
$begingroup$
This is just a slight expansion of Amitai Yuval's comment.
The problem is that integral curves are not defined with respect to a line subbundle of the tangent bundle, but with respect to a vector field. (Only the images of the integral curves are "known" to the horizontal subbundle, much like only integral submanifolds - without any parametrization - are "known" to general subbundles of the tangent bundle.) Geometrically this says the horizontal bundle does not determine how quickly to flow along its integral submanifolds (the images of the integral curves).
The horizontal subbundle amounts to a multiple of a global horizontal vector field by a scalar function. Of these it is naturally best to choose a lift of the global vector field $partial/partial t$ on $I$. With respect to this horizontal vector field the integral curves are now defined.
That this field lifts $partial/partial t$ implies by the chain rule that $frac{partial}{partial t}(picirc c_{(s,x)})equiv 1$. But $picirc c_{(s,x)}$ is precisely the map $f$ giving the first coordinate of the integral curve $c$. By assumption $f(0)=s$ and $f^prime equiv 1$. By uniqueness of solutions to ODE it must then by given by $f(t)=s+t$, as desired.
answered Dec 20 '18 at 22:18
ArrowArrow
5,13731446
$endgroup$
This is just a slight expansion of Amitai Yuval's comment.
The problem is that integral curves are not defined with respect to a line subbundle of the tangent bundle, but with respect to a vector field. (Only the images of the integral curves are "known" to the horizontal subbundle, much like only integral submanifolds - without any parametrization - are "known" to general subbundles of the tangent bundle.) Geometrically this says the horizontal bundle does not determine how quickly to flow along its integral submanifolds (the images of the integral curves).
The horizontal subbundle amounts to a multiple of a global horizontal vector field by a scalar function. Of these it is naturally best to choose a lift of the global vector field $partial/partial t$ on $I$. With respect to this horizontal vector field the integral curves are now defined.
That this field lifts $partial/partial t$ implies by the chain rule that $frac{partial}{partial t}(picirc c_{(s,x)})equiv 1$. But $picirc c_{(s,x)}$ is precisely the map $f$ giving the first coordinate of the integral curve $c$. By assumption $f(0)=s$ and $f^prime equiv 1$. By uniqueness of solutions to ODE it must then by given by $f(t)=s+t$, as desired.
answered Dec 20 '18 at 22:18
ArrowArrow
5,13731446
edited Dec 20 '18 at 22:27
answered Dec 20 '18 at 22:18
ArrowArrow
5,13731446
answered Dec 20 '18 at 22:18
ArrowArrow
5,13731446
answered Dec 20 '18 at 22:18
ArrowArrow
5,13731446
5,13731446
$begingroup$
Seems fine to me.
$endgroup$
– Amitai Yuval
Dec 20 '18 at 22:40
$begingroup$
@AmitaiYuval great, thanks again.
$endgroup$
– Arrow
Dec 20 '18 at 22:41
add a comment |
$begingroup$
Seems fine to me.
$endgroup$
– Amitai Yuval
Dec 20 '18 at 22:40
$begingroup$
@AmitaiYuval great, thanks again.
$endgroup$
– Arrow
Dec 20 '18 at 22:41
$begingroup$
Seems fine to me.
$endgroup$
– Amitai Yuval
Dec 20 '18 at 22:40
$begingroup$
Seems fine to me.
$endgroup$
– Amitai Yuval
Dec 20 '18 at 22:40
$begingroup$
@AmitaiYuval great, thanks again.
$endgroup$
– Arrow
Dec 20 '18 at 22:41
$begingroup$
@AmitaiYuval great, thanks again.
$endgroup$
– Arrow
Dec 20 '18 at 22:41
add a comment |
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$begingroup$
The horizontal curves are the trajectories of a horizontal vector field which you seem to ignore. This vector field is characterized by two properties. The first is horizontality, as you explain in your question. The second property is that it projects to the vector field $partial/partial t$ on $I$.
$endgroup$
– Amitai Yuval
Dec 20 '18 at 20:41
$begingroup$
Dear @AmitaiYuval, thank you for your comment. I posted an answer in an attempt to make sure I understood you well. Would appreciate if you could take a look and point out any errors!
$endgroup$
– Arrow
Dec 20 '18 at 22:21