A nowhere vanishing vector field $X$ tangent to the leaves of the foliation?
I want to know if the following statement is true or false:
For every manifold $M$ and for every foliation on $M$ whose leaves have dimension $1$, there exists a nowhere vanishing vector field $X$ on $M$ tangent to the leaves of the foliation.
I think this is false. I was trying to come up with a counter example, namely the Klein Bottle. See here for a construction of the Klein Bottle.
I would say: draw a vector field vertically downwards (from top to bottom) on the square (parallel to the red lines). Then after construction, this would be a nowhere vanishing vector field on the Klein Bottle. A foliation whose leaves are 1-dimensional would be circles, see e.g. here
I'm not sure if this vector field would be tangent to each circle or not?
differential-geometry manifolds vector-fields
asked yesterday
Kamil
1,98321445
add a comment |
I want to know if the following statement is true or false:
For every manifold $M$ and for every foliation on $M$ whose leaves have dimension $1$, there exists a nowhere vanishing vector field $X$ on $M$ tangent to the leaves of the foliation.
I think this is false. I was trying to come up with a counter example, namely the Klein Bottle. See here for a construction of the Klein Bottle.
I would say: draw a vector field vertically downwards (from top to bottom) on the square (parallel to the red lines). Then after construction, this would be a nowhere vanishing vector field on the Klein Bottle. A foliation whose leaves are 1-dimensional would be circles, see e.g. here
I'm not sure if this vector field would be tangent to each circle or not?
differential-geometry manifolds vector-fields
asked yesterday
Kamil
1,98321445
The Klein bottle carries a smooth circle action so that if $gx = x$, then either $g = 1$ or $g = -1$. In particular, every orbit is a circle. Pushing forward the vector field $partial/partial t$ you obtain a nonvanishing vector field on any manifold with such a circle action, and so have constructed the desired vector field. AN IMPORTANT POINT: Foliations with 1-dimensional leaves do not need to have only circle leaves! Leaves can be noncompact, and often are; this is why foliations are interesting.
– Mike Miller
yesterday
add a comment |
I want to know if the following statement is true or false:
For every manifold $M$ and for every foliation on $M$ whose leaves have dimension $1$, there exists a nowhere vanishing vector field $X$ on $M$ tangent to the leaves of the foliation.
I think this is false. I was trying to come up with a counter example, namely the Klein Bottle. See here for a construction of the Klein Bottle.
I would say: draw a vector field vertically downwards (from top to bottom) on the square (parallel to the red lines). Then after construction, this would be a nowhere vanishing vector field on the Klein Bottle. A foliation whose leaves are 1-dimensional would be circles, see e.g. here
I'm not sure if this vector field would be tangent to each circle or not?
differential-geometry manifolds vector-fields
asked yesterday
Kamil
1,98321445
I want to know if the following statement is true or false:
For every manifold $M$ and for every foliation on $M$ whose leaves have dimension $1$, there exists a nowhere vanishing vector field $X$ on $M$ tangent to the leaves of the foliation.
I think this is false. I was trying to come up with a counter example, namely the Klein Bottle. See here for a construction of the Klein Bottle.
I would say: draw a vector field vertically downwards (from top to bottom) on the square (parallel to the red lines). Then after construction, this would be a nowhere vanishing vector field on the Klein Bottle. A foliation whose leaves are 1-dimensional would be circles, see e.g. here
I'm not sure if this vector field would be tangent to each circle or not?
differential-geometry manifolds vector-fields
differential-geometry manifolds vector-fields
asked yesterday
Kamil
1,98321445
asked yesterday
Kamil
1,98321445
asked yesterday
Kamil
1,98321445
asked yesterday
Kamil
1,98321445
asked yesterday
Kamil
1,98321445
1,98321445
The Klein bottle carries a smooth circle action so that if $gx = x$, then either $g = 1$ or $g = -1$. In particular, every orbit is a circle. Pushing forward the vector field $partial/partial t$ you obtain a nonvanishing vector field on any manifold with such a circle action, and so have constructed the desired vector field. AN IMPORTANT POINT: Foliations with 1-dimensional leaves do not need to have only circle leaves! Leaves can be noncompact, and often are; this is why foliations are interesting.
– Mike Miller
yesterday
add a comment |
The Klein bottle carries a smooth circle action so that if $gx = x$, then either $g = 1$ or $g = -1$. In particular, every orbit is a circle. Pushing forward the vector field $partial/partial t$ you obtain a nonvanishing vector field on any manifold with such a circle action, and so have constructed the desired vector field. AN IMPORTANT POINT: Foliations with 1-dimensional leaves do not need to have only circle leaves! Leaves can be noncompact, and often are; this is why foliations are interesting.
– Mike Miller
yesterday
The Klein bottle carries a smooth circle action so that if $gx = x$, then either $g = 1$ or $g = -1$. In particular, every orbit is a circle. Pushing forward the vector field $partial/partial t$ you obtain a nonvanishing vector field on any manifold with such a circle action, and so have constructed the desired vector field. AN IMPORTANT POINT: Foliations with 1-dimensional leaves do not need to have only circle leaves! Leaves can be noncompact, and often are; this is why foliations are interesting.
– Mike Miller
yesterday
The Klein bottle carries a smooth circle action so that if $gx = x$, then either $g = 1$ or $g = -1$. In particular, every orbit is a circle. Pushing forward the vector field $partial/partial t$ you obtain a nonvanishing vector field on any manifold with such a circle action, and so have constructed the desired vector field. AN IMPORTANT POINT: Foliations with 1-dimensional leaves do not need to have only circle leaves! Leaves can be noncompact, and often are; this is why foliations are interesting.
– Mike Miller
yesterday
add a comment |
1 Answer
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Given a foliated manifold $M$, we have a splitting $TM cong Tmathcal F oplus N mathcal F$. The first term is a line bundle; finding the desired nonvanishing section is precisely asking that this line bundle is trivial. If the ambient manifold is orientable, this is the same as saying that the foliation is so-called "co-orientable" (cut out by a single globally defined 1-form; alternatively, its normal bundle is orientable).
Observe that every line subbundle of $TM$ is integrable to a unique foliation, and so because you are asking for an example where $T mathcal F$ is not orientable, all we actually need to do is find an example of a manifold that has a non-orientable line bundle on it inside of the tangent bunde. The Klein bottle is an example, because you can split off a trivial line bundle from the tangent bundle; your desired non-orientable foliation is essentially the complement (orthogonal, if you like) of this one, at the level of tangent spaces.
Another good example is many oriented 3-manifolds. If $Y$ is an oriented 3-manifold, then $TY$ is trivial; then if $eta$ is a real line bundle on $Y$, $E = eta oplus eta oplus Bbb R$ is a rank 3 vector bundle with $w_1(E) = 0$ and $w_2(E) = w_1(eta)^2$. So if the cup-square map $H^1(Y;Bbb Z/2) to H^2(Y;Bbb Z/2)$ is identically zero, then $E$ has trivial Stiefel-Whitney classes; by the classification of vector bundles over a 3-complex, it is necessarily trivializable, and thus $eta$ is a summand of $TY$ as desired, and non-orientable foliations exist.
In particular, because $H^1(M;Bbb Z/2)$ classifies real line bundles, a manifold with $H^1(M;Bbb Z/2) = 0$ always has a tangent vector field to any 1D foliation.
answered yesterday
Mike Miller
36.4k470137
I'm afraid my background in differential geometry is not sufficient enough to understand your answer. By the way: the question I asked was posed in the context of an introductory class (master's level) in differential geometry which I follow. So I doubt that the answer can be so complicated.
– Kamil
yesterday
1
@Kamil It is not complicated to come up with a counterexample, no. What I explained is that you're just trying to show that $Tmathcal F$ is a non-orientable line bundle, so you should rig a foliation so that's true. The 3-manifold stuff is just to provide many more examples. There is a simple explicit example on the Klein bottle, it's just not the one you wrote down.
– Mike Miller
yesterday
2
@Kamil There is nothing complicated about this answer, and I would argue that a student in a master's level differential geometry class should spend time trying to understand this answer. Any $k$-dimensional foliation of a manifold gives a $k$-dimensional distribution inside the tangent bundle of the manifold. If $k = 1$, having a tangent field to the foliation is equivalent to having a section of this line subbundle. This exists iff the line subbundle is orientable. Easiest example is the foliation on the Moebius strip transverse to it's width, parallel to the boundary.
– Balarka Sen
yesterday
add a comment |
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Given a foliated manifold $M$, we have a splitting $TM cong Tmathcal F oplus N mathcal F$. The first term is a line bundle; finding the desired nonvanishing section is precisely asking that this line bundle is trivial. If the ambient manifold is orientable, this is the same as saying that the foliation is so-called "co-orientable" (cut out by a single globally defined 1-form; alternatively, its normal bundle is orientable).
Observe that every line subbundle of $TM$ is integrable to a unique foliation, and so because you are asking for an example where $T mathcal F$ is not orientable, all we actually need to do is find an example of a manifold that has a non-orientable line bundle on it inside of the tangent bunde. The Klein bottle is an example, because you can split off a trivial line bundle from the tangent bundle; your desired non-orientable foliation is essentially the complement (orthogonal, if you like) of this one, at the level of tangent spaces.
Another good example is many oriented 3-manifolds. If $Y$ is an oriented 3-manifold, then $TY$ is trivial; then if $eta$ is a real line bundle on $Y$, $E = eta oplus eta oplus Bbb R$ is a rank 3 vector bundle with $w_1(E) = 0$ and $w_2(E) = w_1(eta)^2$. So if the cup-square map $H^1(Y;Bbb Z/2) to H^2(Y;Bbb Z/2)$ is identically zero, then $E$ has trivial Stiefel-Whitney classes; by the classification of vector bundles over a 3-complex, it is necessarily trivializable, and thus $eta$ is a summand of $TY$ as desired, and non-orientable foliations exist.
In particular, because $H^1(M;Bbb Z/2)$ classifies real line bundles, a manifold with $H^1(M;Bbb Z/2) = 0$ always has a tangent vector field to any 1D foliation.
answered yesterday
Mike Miller
36.4k470137
I'm afraid my background in differential geometry is not sufficient enough to understand your answer. By the way: the question I asked was posed in the context of an introductory class (master's level) in differential geometry which I follow. So I doubt that the answer can be so complicated.
– Kamil
yesterday
1
@Kamil It is not complicated to come up with a counterexample, no. What I explained is that you're just trying to show that $Tmathcal F$ is a non-orientable line bundle, so you should rig a foliation so that's true. The 3-manifold stuff is just to provide many more examples. There is a simple explicit example on the Klein bottle, it's just not the one you wrote down.
– Mike Miller
yesterday
2
@Kamil There is nothing complicated about this answer, and I would argue that a student in a master's level differential geometry class should spend time trying to understand this answer. Any $k$-dimensional foliation of a manifold gives a $k$-dimensional distribution inside the tangent bundle of the manifold. If $k = 1$, having a tangent field to the foliation is equivalent to having a section of this line subbundle. This exists iff the line subbundle is orientable. Easiest example is the foliation on the Moebius strip transverse to it's width, parallel to the boundary.
– Balarka Sen
yesterday
add a comment |
Given a foliated manifold $M$, we have a splitting $TM cong Tmathcal F oplus N mathcal F$. The first term is a line bundle; finding the desired nonvanishing section is precisely asking that this line bundle is trivial. If the ambient manifold is orientable, this is the same as saying that the foliation is so-called "co-orientable" (cut out by a single globally defined 1-form; alternatively, its normal bundle is orientable).
Observe that every line subbundle of $TM$ is integrable to a unique foliation, and so because you are asking for an example where $T mathcal F$ is not orientable, all we actually need to do is find an example of a manifold that has a non-orientable line bundle on it inside of the tangent bunde. The Klein bottle is an example, because you can split off a trivial line bundle from the tangent bundle; your desired non-orientable foliation is essentially the complement (orthogonal, if you like) of this one, at the level of tangent spaces.
Another good example is many oriented 3-manifolds. If $Y$ is an oriented 3-manifold, then $TY$ is trivial; then if $eta$ is a real line bundle on $Y$, $E = eta oplus eta oplus Bbb R$ is a rank 3 vector bundle with $w_1(E) = 0$ and $w_2(E) = w_1(eta)^2$. So if the cup-square map $H^1(Y;Bbb Z/2) to H^2(Y;Bbb Z/2)$ is identically zero, then $E$ has trivial Stiefel-Whitney classes; by the classification of vector bundles over a 3-complex, it is necessarily trivializable, and thus $eta$ is a summand of $TY$ as desired, and non-orientable foliations exist.
In particular, because $H^1(M;Bbb Z/2)$ classifies real line bundles, a manifold with $H^1(M;Bbb Z/2) = 0$ always has a tangent vector field to any 1D foliation.
answered yesterday
Mike Miller
36.4k470137
I'm afraid my background in differential geometry is not sufficient enough to understand your answer. By the way: the question I asked was posed in the context of an introductory class (master's level) in differential geometry which I follow. So I doubt that the answer can be so complicated.
– Kamil
yesterday
1
@Kamil It is not complicated to come up with a counterexample, no. What I explained is that you're just trying to show that $Tmathcal F$ is a non-orientable line bundle, so you should rig a foliation so that's true. The 3-manifold stuff is just to provide many more examples. There is a simple explicit example on the Klein bottle, it's just not the one you wrote down.
– Mike Miller
yesterday
2
@Kamil There is nothing complicated about this answer, and I would argue that a student in a master's level differential geometry class should spend time trying to understand this answer. Any $k$-dimensional foliation of a manifold gives a $k$-dimensional distribution inside the tangent bundle of the manifold. If $k = 1$, having a tangent field to the foliation is equivalent to having a section of this line subbundle. This exists iff the line subbundle is orientable. Easiest example is the foliation on the Moebius strip transverse to it's width, parallel to the boundary.
– Balarka Sen
yesterday
add a comment |
Given a foliated manifold $M$, we have a splitting $TM cong Tmathcal F oplus N mathcal F$. The first term is a line bundle; finding the desired nonvanishing section is precisely asking that this line bundle is trivial. If the ambient manifold is orientable, this is the same as saying that the foliation is so-called "co-orientable" (cut out by a single globally defined 1-form; alternatively, its normal bundle is orientable).
Observe that every line subbundle of $TM$ is integrable to a unique foliation, and so because you are asking for an example where $T mathcal F$ is not orientable, all we actually need to do is find an example of a manifold that has a non-orientable line bundle on it inside of the tangent bunde. The Klein bottle is an example, because you can split off a trivial line bundle from the tangent bundle; your desired non-orientable foliation is essentially the complement (orthogonal, if you like) of this one, at the level of tangent spaces.
Another good example is many oriented 3-manifolds. If $Y$ is an oriented 3-manifold, then $TY$ is trivial; then if $eta$ is a real line bundle on $Y$, $E = eta oplus eta oplus Bbb R$ is a rank 3 vector bundle with $w_1(E) = 0$ and $w_2(E) = w_1(eta)^2$. So if the cup-square map $H^1(Y;Bbb Z/2) to H^2(Y;Bbb Z/2)$ is identically zero, then $E$ has trivial Stiefel-Whitney classes; by the classification of vector bundles over a 3-complex, it is necessarily trivializable, and thus $eta$ is a summand of $TY$ as desired, and non-orientable foliations exist.
In particular, because $H^1(M;Bbb Z/2)$ classifies real line bundles, a manifold with $H^1(M;Bbb Z/2) = 0$ always has a tangent vector field to any 1D foliation.
answered yesterday
Mike Miller
36.4k470137
Given a foliated manifold $M$, we have a splitting $TM cong Tmathcal F oplus N mathcal F$. The first term is a line bundle; finding the desired nonvanishing section is precisely asking that this line bundle is trivial. If the ambient manifold is orientable, this is the same as saying that the foliation is so-called "co-orientable" (cut out by a single globally defined 1-form; alternatively, its normal bundle is orientable).
Observe that every line subbundle of $TM$ is integrable to a unique foliation, and so because you are asking for an example where $T mathcal F$ is not orientable, all we actually need to do is find an example of a manifold that has a non-orientable line bundle on it inside of the tangent bunde. The Klein bottle is an example, because you can split off a trivial line bundle from the tangent bundle; your desired non-orientable foliation is essentially the complement (orthogonal, if you like) of this one, at the level of tangent spaces.
Another good example is many oriented 3-manifolds. If $Y$ is an oriented 3-manifold, then $TY$ is trivial; then if $eta$ is a real line bundle on $Y$, $E = eta oplus eta oplus Bbb R$ is a rank 3 vector bundle with $w_1(E) = 0$ and $w_2(E) = w_1(eta)^2$. So if the cup-square map $H^1(Y;Bbb Z/2) to H^2(Y;Bbb Z/2)$ is identically zero, then $E$ has trivial Stiefel-Whitney classes; by the classification of vector bundles over a 3-complex, it is necessarily trivializable, and thus $eta$ is a summand of $TY$ as desired, and non-orientable foliations exist.
In particular, because $H^1(M;Bbb Z/2)$ classifies real line bundles, a manifold with $H^1(M;Bbb Z/2) = 0$ always has a tangent vector field to any 1D foliation.
answered yesterday
Mike Miller
36.4k470137
answered yesterday
Mike Miller
36.4k470137
answered yesterday
Mike Miller
36.4k470137
answered yesterday
Mike Miller
36.4k470137
36.4k470137
I'm afraid my background in differential geometry is not sufficient enough to understand your answer. By the way: the question I asked was posed in the context of an introductory class (master's level) in differential geometry which I follow. So I doubt that the answer can be so complicated.
– Kamil
yesterday
1
@Kamil It is not complicated to come up with a counterexample, no. What I explained is that you're just trying to show that $Tmathcal F$ is a non-orientable line bundle, so you should rig a foliation so that's true. The 3-manifold stuff is just to provide many more examples. There is a simple explicit example on the Klein bottle, it's just not the one you wrote down.
– Mike Miller
yesterday
2
@Kamil There is nothing complicated about this answer, and I would argue that a student in a master's level differential geometry class should spend time trying to understand this answer. Any $k$-dimensional foliation of a manifold gives a $k$-dimensional distribution inside the tangent bundle of the manifold. If $k = 1$, having a tangent field to the foliation is equivalent to having a section of this line subbundle. This exists iff the line subbundle is orientable. Easiest example is the foliation on the Moebius strip transverse to it's width, parallel to the boundary.
– Balarka Sen
yesterday
add a comment |
I'm afraid my background in differential geometry is not sufficient enough to understand your answer. By the way: the question I asked was posed in the context of an introductory class (master's level) in differential geometry which I follow. So I doubt that the answer can be so complicated.
– Kamil
yesterday
1
@Kamil It is not complicated to come up with a counterexample, no. What I explained is that you're just trying to show that $Tmathcal F$ is a non-orientable line bundle, so you should rig a foliation so that's true. The 3-manifold stuff is just to provide many more examples. There is a simple explicit example on the Klein bottle, it's just not the one you wrote down.
– Mike Miller
yesterday
2
@Kamil There is nothing complicated about this answer, and I would argue that a student in a master's level differential geometry class should spend time trying to understand this answer. Any $k$-dimensional foliation of a manifold gives a $k$-dimensional distribution inside the tangent bundle of the manifold. If $k = 1$, having a tangent field to the foliation is equivalent to having a section of this line subbundle. This exists iff the line subbundle is orientable. Easiest example is the foliation on the Moebius strip transverse to it's width, parallel to the boundary.
– Balarka Sen
yesterday
I'm afraid my background in differential geometry is not sufficient enough to understand your answer. By the way: the question I asked was posed in the context of an introductory class (master's level) in differential geometry which I follow. So I doubt that the answer can be so complicated.
– Kamil
yesterday
I'm afraid my background in differential geometry is not sufficient enough to understand your answer. By the way: the question I asked was posed in the context of an introductory class (master's level) in differential geometry which I follow. So I doubt that the answer can be so complicated.
– Kamil
yesterday
1
1
@Kamil It is not complicated to come up with a counterexample, no. What I explained is that you're just trying to show that $Tmathcal F$ is a non-orientable line bundle, so you should rig a foliation so that's true. The 3-manifold stuff is just to provide many more examples. There is a simple explicit example on the Klein bottle, it's just not the one you wrote down.
– Mike Miller
yesterday
@Kamil It is not complicated to come up with a counterexample, no. What I explained is that you're just trying to show that $Tmathcal F$ is a non-orientable line bundle, so you should rig a foliation so that's true. The 3-manifold stuff is just to provide many more examples. There is a simple explicit example on the Klein bottle, it's just not the one you wrote down.
– Mike Miller
yesterday
2
2
@Kamil There is nothing complicated about this answer, and I would argue that a student in a master's level differential geometry class should spend time trying to understand this answer. Any $k$-dimensional foliation of a manifold gives a $k$-dimensional distribution inside the tangent bundle of the manifold. If $k = 1$, having a tangent field to the foliation is equivalent to having a section of this line subbundle. This exists iff the line subbundle is orientable. Easiest example is the foliation on the Moebius strip transverse to it's width, parallel to the boundary.
– Balarka Sen
yesterday
@Kamil There is nothing complicated about this answer, and I would argue that a student in a master's level differential geometry class should spend time trying to understand this answer. Any $k$-dimensional foliation of a manifold gives a $k$-dimensional distribution inside the tangent bundle of the manifold. If $k = 1$, having a tangent field to the foliation is equivalent to having a section of this line subbundle. This exists iff the line subbundle is orientable. Easiest example is the foliation on the Moebius strip transverse to it's width, parallel to the boundary.
– Balarka Sen
yesterday
add a comment |
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The Klein bottle carries a smooth circle action so that if $gx = x$, then either $g = 1$ or $g = -1$. In particular, every orbit is a circle. Pushing forward the vector field $partial/partial t$ you obtain a nonvanishing vector field on any manifold with such a circle action, and so have constructed the desired vector field. AN IMPORTANT POINT: Foliations with 1-dimensional leaves do not need to have only circle leaves! Leaves can be noncompact, and often are; this is why foliations are interesting.
– Mike Miller
yesterday