Can I comb unoriented hair on a ball?
up vote
16
down vote
favorite
I know there is no non-vanishing vector field on $S^2$, so I cannot comb the hair on a ball.
(I am treating $S^2$ as a manifold without the ambient space $mathbb R^3$, which amounts to demanding that the vector field is tangential to $S^2$ at every point if you prefer the Euclidean point of view.)
Is this still impossible if the hair is unoriented?
By unoriented hair I mean that instead of looking at a tangent vector at every point, I am looking at a tangent line.
That is, is there a rank one subbundle of $TS^2$ (which is a line bundle on $S^2$)?
Geometrical intuition suggests that there is not, but that is far from proof.
I am mostly interested in $S^2$, but I expect the answer is the same for all even-dimensional spheres just like in the oriented case.
algebraic-topology manifolds vector-fields line-bundles
asked May 2 at 15:01
Joonas Ilmavirta
20.5k84282
|
show 1 more comment
up vote
16
down vote
favorite
I know there is no non-vanishing vector field on $S^2$, so I cannot comb the hair on a ball.
(I am treating $S^2$ as a manifold without the ambient space $mathbb R^3$, which amounts to demanding that the vector field is tangential to $S^2$ at every point if you prefer the Euclidean point of view.)
Is this still impossible if the hair is unoriented?
By unoriented hair I mean that instead of looking at a tangent vector at every point, I am looking at a tangent line.
That is, is there a rank one subbundle of $TS^2$ (which is a line bundle on $S^2$)?
Geometrical intuition suggests that there is not, but that is far from proof.
I am mostly interested in $S^2$, but I expect the answer is the same for all even-dimensional spheres just like in the oriented case.
algebraic-topology manifolds vector-fields line-bundles
asked May 2 at 15:01
Joonas Ilmavirta
20.5k84282
I just found this older question on the topic. Should I close this as duplicate?
– Joonas Ilmavirta
May 2 at 17:55
6
OT: holy semantic overloading batman
– Kuba Ober
May 2 at 18:54
Presumably, you want a continuously varying subbundle. In which case, wouldn't picking a "forward" direction on one tangent line allow one to extend that orientation throughout the sphere?
– Acccumulation
May 2 at 21:06
@Acccumulation Not necessarily. You can always locally orient the line bundle and thus get a non-vanishing vector field as you describe. This does not always happen globally, as the orientations might not match for topological reasons. The simplest example I could think of is considering a line bundle over the Möbius strip with all lines being "across" the strip. No local orientation of this line bundle can be extended to the whole manifold.
– Joonas Ilmavirta
May 2 at 21:11
@JoonasIlmavirta But it does seem like, for orientable surfaces, unoriented hair can be combed only if oriented hair can be.
– Acccumulation
May 2 at 21:29
|
show 1 more comment
up vote
16
down vote
favorite
up vote
16
down vote
favorite
I know there is no non-vanishing vector field on $S^2$, so I cannot comb the hair on a ball.
(I am treating $S^2$ as a manifold without the ambient space $mathbb R^3$, which amounts to demanding that the vector field is tangential to $S^2$ at every point if you prefer the Euclidean point of view.)
Is this still impossible if the hair is unoriented?
By unoriented hair I mean that instead of looking at a tangent vector at every point, I am looking at a tangent line.
That is, is there a rank one subbundle of $TS^2$ (which is a line bundle on $S^2$)?
Geometrical intuition suggests that there is not, but that is far from proof.
I am mostly interested in $S^2$, but I expect the answer is the same for all even-dimensional spheres just like in the oriented case.
algebraic-topology manifolds vector-fields line-bundles
asked May 2 at 15:01
Joonas Ilmavirta
20.5k84282
I know there is no non-vanishing vector field on $S^2$, so I cannot comb the hair on a ball.
(I am treating $S^2$ as a manifold without the ambient space $mathbb R^3$, which amounts to demanding that the vector field is tangential to $S^2$ at every point if you prefer the Euclidean point of view.)
Is this still impossible if the hair is unoriented?
By unoriented hair I mean that instead of looking at a tangent vector at every point, I am looking at a tangent line.
That is, is there a rank one subbundle of $TS^2$ (which is a line bundle on $S^2$)?
Geometrical intuition suggests that there is not, but that is far from proof.
I am mostly interested in $S^2$, but I expect the answer is the same for all even-dimensional spheres just like in the oriented case.
algebraic-topology manifolds vector-fields line-bundles
algebraic-topology manifolds vector-fields line-bundles
asked May 2 at 15:01
Joonas Ilmavirta
20.5k84282
asked May 2 at 15:01
Joonas Ilmavirta
20.5k84282
edited May 2 at 15:39
asked May 2 at 15:01
Joonas Ilmavirta
20.5k84282
asked May 2 at 15:01
Joonas Ilmavirta
20.5k84282
asked May 2 at 15:01
Joonas Ilmavirta
20.5k84282
20.5k84282
I just found this older question on the topic. Should I close this as duplicate?
– Joonas Ilmavirta
May 2 at 17:55
6
OT: holy semantic overloading batman
– Kuba Ober
May 2 at 18:54
Presumably, you want a continuously varying subbundle. In which case, wouldn't picking a "forward" direction on one tangent line allow one to extend that orientation throughout the sphere?
– Acccumulation
May 2 at 21:06
@Acccumulation Not necessarily. You can always locally orient the line bundle and thus get a non-vanishing vector field as you describe. This does not always happen globally, as the orientations might not match for topological reasons. The simplest example I could think of is considering a line bundle over the Möbius strip with all lines being "across" the strip. No local orientation of this line bundle can be extended to the whole manifold.
– Joonas Ilmavirta
May 2 at 21:11
@JoonasIlmavirta But it does seem like, for orientable surfaces, unoriented hair can be combed only if oriented hair can be.
– Acccumulation
May 2 at 21:29
|
show 1 more comment
I just found this older question on the topic. Should I close this as duplicate?
– Joonas Ilmavirta
May 2 at 17:55
6
OT: holy semantic overloading batman
– Kuba Ober
May 2 at 18:54
Presumably, you want a continuously varying subbundle. In which case, wouldn't picking a "forward" direction on one tangent line allow one to extend that orientation throughout the sphere?
– Acccumulation
May 2 at 21:06
@Acccumulation Not necessarily. You can always locally orient the line bundle and thus get a non-vanishing vector field as you describe. This does not always happen globally, as the orientations might not match for topological reasons. The simplest example I could think of is considering a line bundle over the Möbius strip with all lines being "across" the strip. No local orientation of this line bundle can be extended to the whole manifold.
– Joonas Ilmavirta
May 2 at 21:11
@JoonasIlmavirta But it does seem like, for orientable surfaces, unoriented hair can be combed only if oriented hair can be.
– Acccumulation
May 2 at 21:29
I just found this older question on the topic. Should I close this as duplicate?
– Joonas Ilmavirta
May 2 at 17:55
I just found this older question on the topic. Should I close this as duplicate?
– Joonas Ilmavirta
May 2 at 17:55
6
6
OT: holy semantic overloading batman
– Kuba Ober
May 2 at 18:54
OT: holy semantic overloading batman
– Kuba Ober
May 2 at 18:54
Presumably, you want a continuously varying subbundle. In which case, wouldn't picking a "forward" direction on one tangent line allow one to extend that orientation throughout the sphere?
– Acccumulation
May 2 at 21:06
Presumably, you want a continuously varying subbundle. In which case, wouldn't picking a "forward" direction on one tangent line allow one to extend that orientation throughout the sphere?
– Acccumulation
May 2 at 21:06
@Acccumulation Not necessarily. You can always locally orient the line bundle and thus get a non-vanishing vector field as you describe. This does not always happen globally, as the orientations might not match for topological reasons. The simplest example I could think of is considering a line bundle over the Möbius strip with all lines being "across" the strip. No local orientation of this line bundle can be extended to the whole manifold.
– Joonas Ilmavirta
May 2 at 21:11
@Acccumulation Not necessarily. You can always locally orient the line bundle and thus get a non-vanishing vector field as you describe. This does not always happen globally, as the orientations might not match for topological reasons. The simplest example I could think of is considering a line bundle over the Möbius strip with all lines being "across" the strip. No local orientation of this line bundle can be extended to the whole manifold.
– Joonas Ilmavirta
May 2 at 21:11
@JoonasIlmavirta But it does seem like, for orientable surfaces, unoriented hair can be combed only if oriented hair can be.
– Acccumulation
May 2 at 21:29
@JoonasIlmavirta But it does seem like, for orientable surfaces, unoriented hair can be combed only if oriented hair can be.
– Acccumulation
May 2 at 21:29
|
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
14
down vote
accepted
An alternative way to see why this isn't true is to fix a Riemannian metric. If such a subbundle existed, then restricting to those vectors of norm $1$ would give a two-sheeted covering of $S^2$. If it were disconnected, there would be a nowhere zero vector field on $S^2$, a contradiction. If it were connected, $S^2$ would admit a connected two-sheeted covering, a contradiction with it being simply connected.
answered May 2 at 17:16
Aloizio Macedo♦
23.3k23485
add a comment |
up vote
10
down vote
No. Line bundles on $S^2$ are classified by $H^1(S^2;mathbb{Z}/2)=0$ which means that there is only the trivial line bundle $L$ on $S^2$. But then $Loplus L$ is the trivial bundle hence not the tangent bundle of $S^2$.
An interesting question is if $TS^n$ splits as a direct sum $Eoplus F$ of low dimensional bundles. If $n$ is odd the Euler characteristic is zero, so there is a non-zero vector field. So let us assume $n$ is even.
If $E$ and $F$ are required to be orientable bundles this is not possible: The Euler class is multiplicative and $e(TS^n)$ is twice the generator of $H^n(S^n;mathbb{Z})$ if $n$ is even and in particular it is non-zero. But if $E$ and $F$ are proper subbundles the euler classes $e(E)$ and $e(F)$ lie in some $H^i(S^n;mathbb{Z})$ for $0<i<n$ hence $e(E)cup e(F)=0$ which contradicts the fact that $e(E)cup e(F)=e(TS^n)not=0$.
A bundle is orientable iff its determinant bundle has a section, i.e. is a trivial line bundle. But we have seen that the sphere only admits the trivial line bundle. This implies that every bundle on $S^n$ is orientable and the previous argument shows that the tangent bundle cannot split at all.
This discussion btw does not imply that there are no non-trivial bundles on $S^n$ of rank $r<n$, and indeed there are some.
answered May 2 at 15:47
Thomas Rot
6,8231643
And, of course, the same holds in any even dimension.
– anomaly
May 2 at 15:48
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
14
down vote
accepted
An alternative way to see why this isn't true is to fix a Riemannian metric. If such a subbundle existed, then restricting to those vectors of norm $1$ would give a two-sheeted covering of $S^2$. If it were disconnected, there would be a nowhere zero vector field on $S^2$, a contradiction. If it were connected, $S^2$ would admit a connected two-sheeted covering, a contradiction with it being simply connected.
answered May 2 at 17:16
Aloizio Macedo♦
23.3k23485
add a comment |
up vote
14
down vote
accepted
An alternative way to see why this isn't true is to fix a Riemannian metric. If such a subbundle existed, then restricting to those vectors of norm $1$ would give a two-sheeted covering of $S^2$. If it were disconnected, there would be a nowhere zero vector field on $S^2$, a contradiction. If it were connected, $S^2$ would admit a connected two-sheeted covering, a contradiction with it being simply connected.
answered May 2 at 17:16
Aloizio Macedo♦
23.3k23485
add a comment |
up vote
14
down vote
accepted
up vote
14
down vote
accepted
An alternative way to see why this isn't true is to fix a Riemannian metric. If such a subbundle existed, then restricting to those vectors of norm $1$ would give a two-sheeted covering of $S^2$. If it were disconnected, there would be a nowhere zero vector field on $S^2$, a contradiction. If it were connected, $S^2$ would admit a connected two-sheeted covering, a contradiction with it being simply connected.
answered May 2 at 17:16
Aloizio Macedo♦
23.3k23485
An alternative way to see why this isn't true is to fix a Riemannian metric. If such a subbundle existed, then restricting to those vectors of norm $1$ would give a two-sheeted covering of $S^2$. If it were disconnected, there would be a nowhere zero vector field on $S^2$, a contradiction. If it were connected, $S^2$ would admit a connected two-sheeted covering, a contradiction with it being simply connected.
answered May 2 at 17:16
Aloizio Macedo♦
23.3k23485
answered May 2 at 17:16
Aloizio Macedo♦
23.3k23485
answered May 2 at 17:16
Aloizio Macedo♦
23.3k23485
answered May 2 at 17:16
Aloizio Macedo♦
23.3k23485
23.3k23485
add a comment |
add a comment |
up vote
10
down vote
No. Line bundles on $S^2$ are classified by $H^1(S^2;mathbb{Z}/2)=0$ which means that there is only the trivial line bundle $L$ on $S^2$. But then $Loplus L$ is the trivial bundle hence not the tangent bundle of $S^2$.
An interesting question is if $TS^n$ splits as a direct sum $Eoplus F$ of low dimensional bundles. If $n$ is odd the Euler characteristic is zero, so there is a non-zero vector field. So let us assume $n$ is even.
If $E$ and $F$ are required to be orientable bundles this is not possible: The Euler class is multiplicative and $e(TS^n)$ is twice the generator of $H^n(S^n;mathbb{Z})$ if $n$ is even and in particular it is non-zero. But if $E$ and $F$ are proper subbundles the euler classes $e(E)$ and $e(F)$ lie in some $H^i(S^n;mathbb{Z})$ for $0<i<n$ hence $e(E)cup e(F)=0$ which contradicts the fact that $e(E)cup e(F)=e(TS^n)not=0$.
A bundle is orientable iff its determinant bundle has a section, i.e. is a trivial line bundle. But we have seen that the sphere only admits the trivial line bundle. This implies that every bundle on $S^n$ is orientable and the previous argument shows that the tangent bundle cannot split at all.
This discussion btw does not imply that there are no non-trivial bundles on $S^n$ of rank $r<n$, and indeed there are some.
answered May 2 at 15:47
Thomas Rot
6,8231643
And, of course, the same holds in any even dimension.
– anomaly
May 2 at 15:48
add a comment |
up vote
10
down vote
No. Line bundles on $S^2$ are classified by $H^1(S^2;mathbb{Z}/2)=0$ which means that there is only the trivial line bundle $L$ on $S^2$. But then $Loplus L$ is the trivial bundle hence not the tangent bundle of $S^2$.
An interesting question is if $TS^n$ splits as a direct sum $Eoplus F$ of low dimensional bundles. If $n$ is odd the Euler characteristic is zero, so there is a non-zero vector field. So let us assume $n$ is even.
If $E$ and $F$ are required to be orientable bundles this is not possible: The Euler class is multiplicative and $e(TS^n)$ is twice the generator of $H^n(S^n;mathbb{Z})$ if $n$ is even and in particular it is non-zero. But if $E$ and $F$ are proper subbundles the euler classes $e(E)$ and $e(F)$ lie in some $H^i(S^n;mathbb{Z})$ for $0<i<n$ hence $e(E)cup e(F)=0$ which contradicts the fact that $e(E)cup e(F)=e(TS^n)not=0$.
A bundle is orientable iff its determinant bundle has a section, i.e. is a trivial line bundle. But we have seen that the sphere only admits the trivial line bundle. This implies that every bundle on $S^n$ is orientable and the previous argument shows that the tangent bundle cannot split at all.
This discussion btw does not imply that there are no non-trivial bundles on $S^n$ of rank $r<n$, and indeed there are some.
answered May 2 at 15:47
Thomas Rot
6,8231643
And, of course, the same holds in any even dimension.
– anomaly
May 2 at 15:48
add a comment |
up vote
10
down vote
up vote
10
down vote
No. Line bundles on $S^2$ are classified by $H^1(S^2;mathbb{Z}/2)=0$ which means that there is only the trivial line bundle $L$ on $S^2$. But then $Loplus L$ is the trivial bundle hence not the tangent bundle of $S^2$.
An interesting question is if $TS^n$ splits as a direct sum $Eoplus F$ of low dimensional bundles. If $n$ is odd the Euler characteristic is zero, so there is a non-zero vector field. So let us assume $n$ is even.
If $E$ and $F$ are required to be orientable bundles this is not possible: The Euler class is multiplicative and $e(TS^n)$ is twice the generator of $H^n(S^n;mathbb{Z})$ if $n$ is even and in particular it is non-zero. But if $E$ and $F$ are proper subbundles the euler classes $e(E)$ and $e(F)$ lie in some $H^i(S^n;mathbb{Z})$ for $0<i<n$ hence $e(E)cup e(F)=0$ which contradicts the fact that $e(E)cup e(F)=e(TS^n)not=0$.
A bundle is orientable iff its determinant bundle has a section, i.e. is a trivial line bundle. But we have seen that the sphere only admits the trivial line bundle. This implies that every bundle on $S^n$ is orientable and the previous argument shows that the tangent bundle cannot split at all.
This discussion btw does not imply that there are no non-trivial bundles on $S^n$ of rank $r<n$, and indeed there are some.
answered May 2 at 15:47
Thomas Rot
6,8231643
No. Line bundles on $S^2$ are classified by $H^1(S^2;mathbb{Z}/2)=0$ which means that there is only the trivial line bundle $L$ on $S^2$. But then $Loplus L$ is the trivial bundle hence not the tangent bundle of $S^2$.
An interesting question is if $TS^n$ splits as a direct sum $Eoplus F$ of low dimensional bundles. If $n$ is odd the Euler characteristic is zero, so there is a non-zero vector field. So let us assume $n$ is even.
If $E$ and $F$ are required to be orientable bundles this is not possible: The Euler class is multiplicative and $e(TS^n)$ is twice the generator of $H^n(S^n;mathbb{Z})$ if $n$ is even and in particular it is non-zero. But if $E$ and $F$ are proper subbundles the euler classes $e(E)$ and $e(F)$ lie in some $H^i(S^n;mathbb{Z})$ for $0<i<n$ hence $e(E)cup e(F)=0$ which contradicts the fact that $e(E)cup e(F)=e(TS^n)not=0$.
A bundle is orientable iff its determinant bundle has a section, i.e. is a trivial line bundle. But we have seen that the sphere only admits the trivial line bundle. This implies that every bundle on $S^n$ is orientable and the previous argument shows that the tangent bundle cannot split at all.
This discussion btw does not imply that there are no non-trivial bundles on $S^n$ of rank $r<n$, and indeed there are some.
answered May 2 at 15:47
Thomas Rot
6,8231643
edited Nov 21 at 10:18
answered May 2 at 15:47
Thomas Rot
6,8231643
answered May 2 at 15:47
Thomas Rot
6,8231643
answered May 2 at 15:47
Thomas Rot
6,8231643
6,8231643
And, of course, the same holds in any even dimension.
– anomaly
May 2 at 15:48
add a comment |
And, of course, the same holds in any even dimension.
– anomaly
May 2 at 15:48
And, of course, the same holds in any even dimension.
– anomaly
May 2 at 15:48
And, of course, the same holds in any even dimension.
– anomaly
May 2 at 15:48
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2763466%2fcan-i-comb-unoriented-hair-on-a-ball%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
I just found this older question on the topic. Should I close this as duplicate?
– Joonas Ilmavirta
May 2 at 17:55
6
OT: holy semantic overloading batman
– Kuba Ober
May 2 at 18:54
Presumably, you want a continuously varying subbundle. In which case, wouldn't picking a "forward" direction on one tangent line allow one to extend that orientation throughout the sphere?
– Acccumulation
May 2 at 21:06
@Acccumulation Not necessarily. You can always locally orient the line bundle and thus get a non-vanishing vector field as you describe. This does not always happen globally, as the orientations might not match for topological reasons. The simplest example I could think of is considering a line bundle over the Möbius strip with all lines being "across" the strip. No local orientation of this line bundle can be extended to the whole manifold.
– Joonas Ilmavirta
May 2 at 21:11
@JoonasIlmavirta But it does seem like, for orientable surfaces, unoriented hair can be combed only if oriented hair can be.
– Acccumulation
May 2 at 21:29