Oriented vector bundle with odd-dimensional fibersVanishing of Euler classAre there analogous statements for the number of zeros of a section in terms of the Euler class, even when the relevant spaces are not manifolds? When does an even-dimensional manifold fiber over an odd-dimensional manifold?Computing a cobordism group of manifolds endowed with a real vector bundle with constraints on the Stiefel-Whitney classesHow to flow submanifolds?Analogue of the Euler class of a circle bundle and the global angular formIs every vector bundle over a noncompact finite-dimensional manifold a summand of a trivial bundle?Consequences of the Euler characteristic vanishing mod pVector bundle over an oriented manifold with non-vanishing w_2w_3Lifting sections of a projective bundle to a vector bundle
Oriented vector bundle with odd-dimensional fibers
Vanishing of Euler classAre there analogous statements for the number of zeros of a section in terms of the Euler class, even when the relevant spaces are not manifolds? When does an even-dimensional manifold fiber over an odd-dimensional manifold?Computing a cobordism group of manifolds endowed with a real vector bundle with constraints on the Stiefel-Whitney classesHow to flow submanifolds?Analogue of the Euler class of a circle bundle and the global angular formIs every vector bundle over a noncompact finite-dimensional manifold a summand of a trivial bundle?Consequences of the Euler characteristic vanishing mod pVector bundle over an oriented manifold with non-vanishing w_2w_3Lifting sections of a projective bundle to a vector bundle
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Is it true that for every oriented vector bundle with odd-dimensional fibers, there is always a global section that vanishes nowhere?
at.algebraic-topology differential-topology
edited Jul 22 at 8:40
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Is it true that for every oriented vector bundle with odd-dimensional fibers, there is always a global section that vanishes nowhere?
at.algebraic-topology differential-topology
edited Jul 22 at 8:40
user43326
2,2558 silver badges21 bronze badges
asked Jul 22 at 8:15
FredyFredy
483 bronze badges
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add a comment |
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Is it true that for every oriented vector bundle with odd-dimensional fibers, there is always a global section that vanishes nowhere?
at.algebraic-topology differential-topology
edited Jul 22 at 8:40
user43326
2,2558 silver badges21 bronze badges
asked Jul 22 at 8:15
FredyFredy
483 bronze badges
$endgroup$
Is it true that for every oriented vector bundle with odd-dimensional fibers, there is always a global section that vanishes nowhere?
at.algebraic-topology differential-topology
at.algebraic-topology differential-topology
edited Jul 22 at 8:40
user43326
2,2558 silver badges21 bronze badges
asked Jul 22 at 8:15
FredyFredy
483 bronze badges
edited Jul 22 at 8:40
user43326
2,2558 silver badges21 bronze badges
asked Jul 22 at 8:15
FredyFredy
483 bronze badges
edited Jul 22 at 8:40
user43326
2,2558 silver badges21 bronze badges
edited Jul 22 at 8:40
user43326
2,2558 silver badges21 bronze badges
edited Jul 22 at 8:40
user43326
2,2558 silver badges21 bronze badges
2,2558 silver badges21 bronze badges
asked Jul 22 at 8:15
FredyFredy
483 bronze badges
asked Jul 22 at 8:15
FredyFredy
483 bronze badges
asked Jul 22 at 8:15
FredyFredy
483 bronze badges
483 bronze badges
add a comment |
add a comment |
1 Answer
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No.
Let us consider some vector bundles over $S^4$. Since $S^4$ is simply connected all vector bundles are oriented and rank $k$ vector bundles over $S^4$ are classified by $pi_3(SO(k))$ by the clutching construction (I am glossing over basepoint issues here, but I think it is correct in this setting). Now $pi_3(SO(1))$ and $ pi_3(SO(2)=pi_3(S^1)$ are trivial groups, so there do not exist non-trivial rank $1$ and rank $2$ bundles over $S^4$. What about rank $3$? Well $SO(3)cong mathbb RP^3$ and the long exact sequence of homotopy groups of the fiber bundle
$$
mathbbZ_2rightarrow S^3rightarrow mathbbRP^3
$$
shows that $pi_3(SO(3))congpi_3(mathbbRP^3)cong mathbbZ$. Hence there are $mathbbZ$ different rank $3$ vector bundles over $S^4$.
Only the trivial one admits a non-zero section. If another one admits a section, then the orthogonal complement (a rank $2$ bundle) must be non-trivial, otherwise the bundle is trivial. But our previous computation shows that there are no-nontrivial rank $2$ bundles over $S^4$.
answered Jul 22 at 8:55
Thomas RotThomas Rot
3,6591 gold badge19 silver badges39 bronze badges
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4
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Moreover an important example (often denoted $Lambda^2_+$) of a rank $3$ oriented vector bundle with no non-vanishing section is given by the self-dual $2$-forms on the round sphere $S^4$ (any other conformal structure and orientation would do). The reason is that otherwise $S^4$ would admit an (orthogonal) almost complex structure, hence two, one for each orientation ($S^4$ has an orentation reversing diffeo). Considering their only two (exercise) common tangent complex lines at each point, $TS^4$ would split in two rank $2$ subbundles, which is excluded as in this answer.
$endgroup$
– BS.
Jul 22 at 10:28
$begingroup$
Thanks a lot, it's wonderful! I conjectured it to be true in order to solve a exercise from Bott/Tu's GTM 82, "the Euler class of an oriented sphere bundle with even-dimensional fibers is zero, when the sphere bundle comes from a vector bundle". I hope to know why this statement turn out to be true.
$endgroup$
– Fredy
Jul 22 at 13:37
1
$begingroup$
@Fredy: Note that the euler class is also zero in this example. But this does not imply that there is a non-zero section.
$endgroup$
– Thomas Rot
Jul 22 at 15:06
$begingroup$
@ThomasRot: Thanks. Can you tell me more details about the proof of that exercise? I seem to lack other tools besides looking for a global section...
$endgroup$
– Fredy
Jul 23 at 2:40
add a comment |
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$begingroup$
No.
Let us consider some vector bundles over $S^4$. Since $S^4$ is simply connected all vector bundles are oriented and rank $k$ vector bundles over $S^4$ are classified by $pi_3(SO(k))$ by the clutching construction (I am glossing over basepoint issues here, but I think it is correct in this setting). Now $pi_3(SO(1))$ and $ pi_3(SO(2)=pi_3(S^1)$ are trivial groups, so there do not exist non-trivial rank $1$ and rank $2$ bundles over $S^4$. What about rank $3$? Well $SO(3)cong mathbb RP^3$ and the long exact sequence of homotopy groups of the fiber bundle
$$
mathbbZ_2rightarrow S^3rightarrow mathbbRP^3
$$
shows that $pi_3(SO(3))congpi_3(mathbbRP^3)cong mathbbZ$. Hence there are $mathbbZ$ different rank $3$ vector bundles over $S^4$.
Only the trivial one admits a non-zero section. If another one admits a section, then the orthogonal complement (a rank $2$ bundle) must be non-trivial, otherwise the bundle is trivial. But our previous computation shows that there are no-nontrivial rank $2$ bundles over $S^4$.
answered Jul 22 at 8:55
Thomas RotThomas Rot
3,6591 gold badge19 silver badges39 bronze badges
$endgroup$
4
$begingroup$
Moreover an important example (often denoted $Lambda^2_+$) of a rank $3$ oriented vector bundle with no non-vanishing section is given by the self-dual $2$-forms on the round sphere $S^4$ (any other conformal structure and orientation would do). The reason is that otherwise $S^4$ would admit an (orthogonal) almost complex structure, hence two, one for each orientation ($S^4$ has an orentation reversing diffeo). Considering their only two (exercise) common tangent complex lines at each point, $TS^4$ would split in two rank $2$ subbundles, which is excluded as in this answer.
$endgroup$
– BS.
Jul 22 at 10:28
$begingroup$
Thanks a lot, it's wonderful! I conjectured it to be true in order to solve a exercise from Bott/Tu's GTM 82, "the Euler class of an oriented sphere bundle with even-dimensional fibers is zero, when the sphere bundle comes from a vector bundle". I hope to know why this statement turn out to be true.
$endgroup$
– Fredy
Jul 22 at 13:37
1
$begingroup$
@Fredy: Note that the euler class is also zero in this example. But this does not imply that there is a non-zero section.
$endgroup$
– Thomas Rot
Jul 22 at 15:06
$begingroup$
@ThomasRot: Thanks. Can you tell me more details about the proof of that exercise? I seem to lack other tools besides looking for a global section...
$endgroup$
– Fredy
Jul 23 at 2:40
add a comment |
$begingroup$
No.
Let us consider some vector bundles over $S^4$. Since $S^4$ is simply connected all vector bundles are oriented and rank $k$ vector bundles over $S^4$ are classified by $pi_3(SO(k))$ by the clutching construction (I am glossing over basepoint issues here, but I think it is correct in this setting). Now $pi_3(SO(1))$ and $ pi_3(SO(2)=pi_3(S^1)$ are trivial groups, so there do not exist non-trivial rank $1$ and rank $2$ bundles over $S^4$. What about rank $3$? Well $SO(3)cong mathbb RP^3$ and the long exact sequence of homotopy groups of the fiber bundle
$$
mathbbZ_2rightarrow S^3rightarrow mathbbRP^3
$$
shows that $pi_3(SO(3))congpi_3(mathbbRP^3)cong mathbbZ$. Hence there are $mathbbZ$ different rank $3$ vector bundles over $S^4$.
Only the trivial one admits a non-zero section. If another one admits a section, then the orthogonal complement (a rank $2$ bundle) must be non-trivial, otherwise the bundle is trivial. But our previous computation shows that there are no-nontrivial rank $2$ bundles over $S^4$.
answered Jul 22 at 8:55
Thomas RotThomas Rot
3,6591 gold badge19 silver badges39 bronze badges
$endgroup$
4
$begingroup$
Moreover an important example (often denoted $Lambda^2_+$) of a rank $3$ oriented vector bundle with no non-vanishing section is given by the self-dual $2$-forms on the round sphere $S^4$ (any other conformal structure and orientation would do). The reason is that otherwise $S^4$ would admit an (orthogonal) almost complex structure, hence two, one for each orientation ($S^4$ has an orentation reversing diffeo). Considering their only two (exercise) common tangent complex lines at each point, $TS^4$ would split in two rank $2$ subbundles, which is excluded as in this answer.
$endgroup$
– BS.
Jul 22 at 10:28
$begingroup$
Thanks a lot, it's wonderful! I conjectured it to be true in order to solve a exercise from Bott/Tu's GTM 82, "the Euler class of an oriented sphere bundle with even-dimensional fibers is zero, when the sphere bundle comes from a vector bundle". I hope to know why this statement turn out to be true.
$endgroup$
– Fredy
Jul 22 at 13:37
1
$begingroup$
@Fredy: Note that the euler class is also zero in this example. But this does not imply that there is a non-zero section.
$endgroup$
– Thomas Rot
Jul 22 at 15:06
$begingroup$
@ThomasRot: Thanks. Can you tell me more details about the proof of that exercise? I seem to lack other tools besides looking for a global section...
$endgroup$
– Fredy
Jul 23 at 2:40
add a comment |
$begingroup$
No.
Let us consider some vector bundles over $S^4$. Since $S^4$ is simply connected all vector bundles are oriented and rank $k$ vector bundles over $S^4$ are classified by $pi_3(SO(k))$ by the clutching construction (I am glossing over basepoint issues here, but I think it is correct in this setting). Now $pi_3(SO(1))$ and $ pi_3(SO(2)=pi_3(S^1)$ are trivial groups, so there do not exist non-trivial rank $1$ and rank $2$ bundles over $S^4$. What about rank $3$? Well $SO(3)cong mathbb RP^3$ and the long exact sequence of homotopy groups of the fiber bundle
$$
mathbbZ_2rightarrow S^3rightarrow mathbbRP^3
$$
shows that $pi_3(SO(3))congpi_3(mathbbRP^3)cong mathbbZ$. Hence there are $mathbbZ$ different rank $3$ vector bundles over $S^4$.
Only the trivial one admits a non-zero section. If another one admits a section, then the orthogonal complement (a rank $2$ bundle) must be non-trivial, otherwise the bundle is trivial. But our previous computation shows that there are no-nontrivial rank $2$ bundles over $S^4$.
answered Jul 22 at 8:55
Thomas RotThomas Rot
3,6591 gold badge19 silver badges39 bronze badges
$endgroup$
No.
Let us consider some vector bundles over $S^4$. Since $S^4$ is simply connected all vector bundles are oriented and rank $k$ vector bundles over $S^4$ are classified by $pi_3(SO(k))$ by the clutching construction (I am glossing over basepoint issues here, but I think it is correct in this setting). Now $pi_3(SO(1))$ and $ pi_3(SO(2)=pi_3(S^1)$ are trivial groups, so there do not exist non-trivial rank $1$ and rank $2$ bundles over $S^4$. What about rank $3$? Well $SO(3)cong mathbb RP^3$ and the long exact sequence of homotopy groups of the fiber bundle
$$
mathbbZ_2rightarrow S^3rightarrow mathbbRP^3
$$
shows that $pi_3(SO(3))congpi_3(mathbbRP^3)cong mathbbZ$. Hence there are $mathbbZ$ different rank $3$ vector bundles over $S^4$.
Only the trivial one admits a non-zero section. If another one admits a section, then the orthogonal complement (a rank $2$ bundle) must be non-trivial, otherwise the bundle is trivial. But our previous computation shows that there are no-nontrivial rank $2$ bundles over $S^4$.
answered Jul 22 at 8:55
Thomas RotThomas Rot
3,6591 gold badge19 silver badges39 bronze badges
edited Jul 22 at 9:02
answered Jul 22 at 8:55
Thomas RotThomas Rot
3,6591 gold badge19 silver badges39 bronze badges
answered Jul 22 at 8:55
Thomas RotThomas Rot
3,6591 gold badge19 silver badges39 bronze badges
answered Jul 22 at 8:55
Thomas RotThomas Rot
3,6591 gold badge19 silver badges39 bronze badges
3,6591 gold badge19 silver badges39 bronze badges
4
$begingroup$
Moreover an important example (often denoted $Lambda^2_+$) of a rank $3$ oriented vector bundle with no non-vanishing section is given by the self-dual $2$-forms on the round sphere $S^4$ (any other conformal structure and orientation would do). The reason is that otherwise $S^4$ would admit an (orthogonal) almost complex structure, hence two, one for each orientation ($S^4$ has an orentation reversing diffeo). Considering their only two (exercise) common tangent complex lines at each point, $TS^4$ would split in two rank $2$ subbundles, which is excluded as in this answer.
$endgroup$
– BS.
Jul 22 at 10:28
$begingroup$
Thanks a lot, it's wonderful! I conjectured it to be true in order to solve a exercise from Bott/Tu's GTM 82, "the Euler class of an oriented sphere bundle with even-dimensional fibers is zero, when the sphere bundle comes from a vector bundle". I hope to know why this statement turn out to be true.
$endgroup$
– Fredy
Jul 22 at 13:37
1
$begingroup$
@Fredy: Note that the euler class is also zero in this example. But this does not imply that there is a non-zero section.
$endgroup$
– Thomas Rot
Jul 22 at 15:06
$begingroup$
@ThomasRot: Thanks. Can you tell me more details about the proof of that exercise? I seem to lack other tools besides looking for a global section...
$endgroup$
– Fredy
Jul 23 at 2:40
add a comment |
4
$begingroup$
Moreover an important example (often denoted $Lambda^2_+$) of a rank $3$ oriented vector bundle with no non-vanishing section is given by the self-dual $2$-forms on the round sphere $S^4$ (any other conformal structure and orientation would do). The reason is that otherwise $S^4$ would admit an (orthogonal) almost complex structure, hence two, one for each orientation ($S^4$ has an orentation reversing diffeo). Considering their only two (exercise) common tangent complex lines at each point, $TS^4$ would split in two rank $2$ subbundles, which is excluded as in this answer.
$endgroup$
– BS.
Jul 22 at 10:28
$begingroup$
Thanks a lot, it's wonderful! I conjectured it to be true in order to solve a exercise from Bott/Tu's GTM 82, "the Euler class of an oriented sphere bundle with even-dimensional fibers is zero, when the sphere bundle comes from a vector bundle". I hope to know why this statement turn out to be true.
$endgroup$
– Fredy
Jul 22 at 13:37
1
$begingroup$
@Fredy: Note that the euler class is also zero in this example. But this does not imply that there is a non-zero section.
$endgroup$
– Thomas Rot
Jul 22 at 15:06
$begingroup$
@ThomasRot: Thanks. Can you tell me more details about the proof of that exercise? I seem to lack other tools besides looking for a global section...
$endgroup$
– Fredy
Jul 23 at 2:40
4
4
$begingroup$
Moreover an important example (often denoted $Lambda^2_+$) of a rank $3$ oriented vector bundle with no non-vanishing section is given by the self-dual $2$-forms on the round sphere $S^4$ (any other conformal structure and orientation would do). The reason is that otherwise $S^4$ would admit an (orthogonal) almost complex structure, hence two, one for each orientation ($S^4$ has an orentation reversing diffeo). Considering their only two (exercise) common tangent complex lines at each point, $TS^4$ would split in two rank $2$ subbundles, which is excluded as in this answer.
$endgroup$
– BS.
Jul 22 at 10:28
$begingroup$
Moreover an important example (often denoted $Lambda^2_+$) of a rank $3$ oriented vector bundle with no non-vanishing section is given by the self-dual $2$-forms on the round sphere $S^4$ (any other conformal structure and orientation would do). The reason is that otherwise $S^4$ would admit an (orthogonal) almost complex structure, hence two, one for each orientation ($S^4$ has an orentation reversing diffeo). Considering their only two (exercise) common tangent complex lines at each point, $TS^4$ would split in two rank $2$ subbundles, which is excluded as in this answer.
$endgroup$
– BS.
Jul 22 at 10:28
$begingroup$
Thanks a lot, it's wonderful! I conjectured it to be true in order to solve a exercise from Bott/Tu's GTM 82, "the Euler class of an oriented sphere bundle with even-dimensional fibers is zero, when the sphere bundle comes from a vector bundle". I hope to know why this statement turn out to be true.
$endgroup$
– Fredy
Jul 22 at 13:37
$begingroup$
Thanks a lot, it's wonderful! I conjectured it to be true in order to solve a exercise from Bott/Tu's GTM 82, "the Euler class of an oriented sphere bundle with even-dimensional fibers is zero, when the sphere bundle comes from a vector bundle". I hope to know why this statement turn out to be true.
$endgroup$
– Fredy
Jul 22 at 13:37
1
1
$begingroup$
@Fredy: Note that the euler class is also zero in this example. But this does not imply that there is a non-zero section.
$endgroup$
– Thomas Rot
Jul 22 at 15:06
$begingroup$
@Fredy: Note that the euler class is also zero in this example. But this does not imply that there is a non-zero section.
$endgroup$
– Thomas Rot
Jul 22 at 15:06
$begingroup$
@ThomasRot: Thanks. Can you tell me more details about the proof of that exercise? I seem to lack other tools besides looking for a global section...
$endgroup$
– Fredy
Jul 23 at 2:40
$begingroup$
@ThomasRot: Thanks. Can you tell me more details about the proof of that exercise? I seem to lack other tools besides looking for a global section...
$endgroup$
– Fredy
Jul 23 at 2:40
add a comment |
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