Example of a parallelizable smooth manifold which is not a Lie Group
All the examples I know of manifolds which are parallelizable are Lie Groups. Can anyone point out an easy example of a parallelizable smooth manifold which is not a Lie Group? Are there conditions on a parallelizable smooth manifold which forces it to be a lie group?
differential-geometry lie-groups smooth-manifolds tangent-bundle
asked Dec 19 '18 at 4:39
PSG
3419
add a comment |
All the examples I know of manifolds which are parallelizable are Lie Groups. Can anyone point out an easy example of a parallelizable smooth manifold which is not a Lie Group? Are there conditions on a parallelizable smooth manifold which forces it to be a lie group?
differential-geometry lie-groups smooth-manifolds tangent-bundle
asked Dec 19 '18 at 4:39
PSG
3419
In the opposite direction, there's a result due to Wolf that $S^7$ is the only exception in the simply-connected case: specifically, for a $X$ compact and simply-connected manifold, $X$ is parallelizable only if it's the product of Lie groups and copies of $S^7$.
– anomaly
18 hours ago
@anomaly Thanks a lot. I was looking for some result like that. I am not experienced to find this results. Can you please mention where can I find this or may be share a link if possible.
– PSG
17 hours ago
I don't have a reference to the original paper, but it's mentioned after Proposition 5.15 in Lee's "Introduction to Smooth Manifolds."
– anomaly
16 hours ago
Noted with thanks. I will definitely look into it.
– PSG
16 hours ago
add a comment |
All the examples I know of manifolds which are parallelizable are Lie Groups. Can anyone point out an easy example of a parallelizable smooth manifold which is not a Lie Group? Are there conditions on a parallelizable smooth manifold which forces it to be a lie group?
differential-geometry lie-groups smooth-manifolds tangent-bundle
asked Dec 19 '18 at 4:39
PSG
3419
All the examples I know of manifolds which are parallelizable are Lie Groups. Can anyone point out an easy example of a parallelizable smooth manifold which is not a Lie Group? Are there conditions on a parallelizable smooth manifold which forces it to be a lie group?
differential-geometry lie-groups smooth-manifolds tangent-bundle
differential-geometry lie-groups smooth-manifolds tangent-bundle
asked Dec 19 '18 at 4:39
PSG
3419
asked Dec 19 '18 at 4:39
PSG
3419
asked Dec 19 '18 at 4:39
PSG
3419
asked Dec 19 '18 at 4:39
PSG
3419
asked Dec 19 '18 at 4:39
PSG
3419
3419
In the opposite direction, there's a result due to Wolf that $S^7$ is the only exception in the simply-connected case: specifically, for a $X$ compact and simply-connected manifold, $X$ is parallelizable only if it's the product of Lie groups and copies of $S^7$.
– anomaly
18 hours ago
@anomaly Thanks a lot. I was looking for some result like that. I am not experienced to find this results. Can you please mention where can I find this or may be share a link if possible.
– PSG
17 hours ago
I don't have a reference to the original paper, but it's mentioned after Proposition 5.15 in Lee's "Introduction to Smooth Manifolds."
– anomaly
16 hours ago
Noted with thanks. I will definitely look into it.
– PSG
16 hours ago
add a comment |
In the opposite direction, there's a result due to Wolf that $S^7$ is the only exception in the simply-connected case: specifically, for a $X$ compact and simply-connected manifold, $X$ is parallelizable only if it's the product of Lie groups and copies of $S^7$.
– anomaly
18 hours ago
@anomaly Thanks a lot. I was looking for some result like that. I am not experienced to find this results. Can you please mention where can I find this or may be share a link if possible.
– PSG
17 hours ago
I don't have a reference to the original paper, but it's mentioned after Proposition 5.15 in Lee's "Introduction to Smooth Manifolds."
– anomaly
16 hours ago
Noted with thanks. I will definitely look into it.
– PSG
16 hours ago
In the opposite direction, there's a result due to Wolf that $S^7$ is the only exception in the simply-connected case: specifically, for a $X$ compact and simply-connected manifold, $X$ is parallelizable only if it's the product of Lie groups and copies of $S^7$.
– anomaly
18 hours ago
In the opposite direction, there's a result due to Wolf that $S^7$ is the only exception in the simply-connected case: specifically, for a $X$ compact and simply-connected manifold, $X$ is parallelizable only if it's the product of Lie groups and copies of $S^7$.
– anomaly
18 hours ago
@anomaly Thanks a lot. I was looking for some result like that. I am not experienced to find this results. Can you please mention where can I find this or may be share a link if possible.
– PSG
17 hours ago
@anomaly Thanks a lot. I was looking for some result like that. I am not experienced to find this results. Can you please mention where can I find this or may be share a link if possible.
– PSG
17 hours ago
I don't have a reference to the original paper, but it's mentioned after Proposition 5.15 in Lee's "Introduction to Smooth Manifolds."
– anomaly
16 hours ago
I don't have a reference to the original paper, but it's mentioned after Proposition 5.15 in Lee's "Introduction to Smooth Manifolds."
– anomaly
16 hours ago
Noted with thanks. I will definitely look into it.
– PSG
16 hours ago
Noted with thanks. I will definitely look into it.
– PSG
16 hours ago
add a comment |
2 Answers
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Any closed, orientable $3$-manifold $X$ is parallelizable. On the other hand, Lie groups have $pi_2$ trivial, and there are a lot of such $X$ that don't. (It's not trivial to find them, though. Note that any such $X$ must have $pi_1 Xnot = 0$ by Poincare duality, and aspherical (e.g., hyperbolic) $X$ have $pi_2 X = 0$ as well. $3$-manifolds are weird.)
(The easiest way to prove the first assertion is to note that the only obstructions to the triviality of $TX$ are the Stiefel-Whitney classes; and $w_1 = 0$ by orientability and $w_2 = 0$ by the Wu formula, since we're conveniently in low dimension. The second is practically a folklore theorem, but it's not that hard to prove directly either via Morse theory or minimal models.)
answered Dec 19 '18 at 4:57
anomaly
17.3k42663
2
You beat me to this answer! Also, we can readily construct examples of oriented, compact $3$-manifolds that do not admit a Lie group structure: Any Lie group $G$ has abelian fundamental group $pi_1(G)$, but the product of $S^1$ and an orientable, compact surface $Sigma$ with genus $> 1$ has nonabelian fundamental group $pi_1(S^1 times Sigma) cong pi_1(S^1) times pi_1(Sigma) cong Bbb Z times pi_1(Sigma)$. (And +1, by the way.)
– Travis
Dec 19 '18 at 5:09
1
Good point! I ignored the hyperbolic case because of $pi_2$, but $pi_1$ also presents an obstruction to Lie-groupness.
– anomaly
Dec 19 '18 at 5:25
add a comment |
The classical example is $S^7$. This comes from the octonions, which are non-associative, so the unit octonions don't form a Lie group.
answered Dec 19 '18 at 4:44
Lord Shark the Unknown
101k958132
Thanks, I was wondering that $S^7$ still has a group structure which is non-associative, are there examples which can not be given a group structure at all.
– PSG
Dec 19 '18 at 4:50
1
Group operations are by definition associative. The restriction of octonionic multiplication to the unit sphere in the octonions still, however, forms a loop (an operation $S times S to S$ with identity and inverses), and the multiplication and inversion maps are smooth with respect to the usual smooth structure.
– Travis
Dec 19 '18 at 4:55
1
Also, this is the only example of parallelizable sphere that admits no Lie group structure.
– Travis
Dec 19 '18 at 4:56
add a comment |
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2 Answers
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Any closed, orientable $3$-manifold $X$ is parallelizable. On the other hand, Lie groups have $pi_2$ trivial, and there are a lot of such $X$ that don't. (It's not trivial to find them, though. Note that any such $X$ must have $pi_1 Xnot = 0$ by Poincare duality, and aspherical (e.g., hyperbolic) $X$ have $pi_2 X = 0$ as well. $3$-manifolds are weird.)
(The easiest way to prove the first assertion is to note that the only obstructions to the triviality of $TX$ are the Stiefel-Whitney classes; and $w_1 = 0$ by orientability and $w_2 = 0$ by the Wu formula, since we're conveniently in low dimension. The second is practically a folklore theorem, but it's not that hard to prove directly either via Morse theory or minimal models.)
answered Dec 19 '18 at 4:57
anomaly
17.3k42663
2
You beat me to this answer! Also, we can readily construct examples of oriented, compact $3$-manifolds that do not admit a Lie group structure: Any Lie group $G$ has abelian fundamental group $pi_1(G)$, but the product of $S^1$ and an orientable, compact surface $Sigma$ with genus $> 1$ has nonabelian fundamental group $pi_1(S^1 times Sigma) cong pi_1(S^1) times pi_1(Sigma) cong Bbb Z times pi_1(Sigma)$. (And +1, by the way.)
– Travis
Dec 19 '18 at 5:09
1
Good point! I ignored the hyperbolic case because of $pi_2$, but $pi_1$ also presents an obstruction to Lie-groupness.
– anomaly
Dec 19 '18 at 5:25
add a comment |
Any closed, orientable $3$-manifold $X$ is parallelizable. On the other hand, Lie groups have $pi_2$ trivial, and there are a lot of such $X$ that don't. (It's not trivial to find them, though. Note that any such $X$ must have $pi_1 Xnot = 0$ by Poincare duality, and aspherical (e.g., hyperbolic) $X$ have $pi_2 X = 0$ as well. $3$-manifolds are weird.)
(The easiest way to prove the first assertion is to note that the only obstructions to the triviality of $TX$ are the Stiefel-Whitney classes; and $w_1 = 0$ by orientability and $w_2 = 0$ by the Wu formula, since we're conveniently in low dimension. The second is practically a folklore theorem, but it's not that hard to prove directly either via Morse theory or minimal models.)
answered Dec 19 '18 at 4:57
anomaly
17.3k42663
2
You beat me to this answer! Also, we can readily construct examples of oriented, compact $3$-manifolds that do not admit a Lie group structure: Any Lie group $G$ has abelian fundamental group $pi_1(G)$, but the product of $S^1$ and an orientable, compact surface $Sigma$ with genus $> 1$ has nonabelian fundamental group $pi_1(S^1 times Sigma) cong pi_1(S^1) times pi_1(Sigma) cong Bbb Z times pi_1(Sigma)$. (And +1, by the way.)
– Travis
Dec 19 '18 at 5:09
1
Good point! I ignored the hyperbolic case because of $pi_2$, but $pi_1$ also presents an obstruction to Lie-groupness.
– anomaly
Dec 19 '18 at 5:25
add a comment |
Any closed, orientable $3$-manifold $X$ is parallelizable. On the other hand, Lie groups have $pi_2$ trivial, and there are a lot of such $X$ that don't. (It's not trivial to find them, though. Note that any such $X$ must have $pi_1 Xnot = 0$ by Poincare duality, and aspherical (e.g., hyperbolic) $X$ have $pi_2 X = 0$ as well. $3$-manifolds are weird.)
(The easiest way to prove the first assertion is to note that the only obstructions to the triviality of $TX$ are the Stiefel-Whitney classes; and $w_1 = 0$ by orientability and $w_2 = 0$ by the Wu formula, since we're conveniently in low dimension. The second is practically a folklore theorem, but it's not that hard to prove directly either via Morse theory or minimal models.)
answered Dec 19 '18 at 4:57
anomaly
17.3k42663
Any closed, orientable $3$-manifold $X$ is parallelizable. On the other hand, Lie groups have $pi_2$ trivial, and there are a lot of such $X$ that don't. (It's not trivial to find them, though. Note that any such $X$ must have $pi_1 Xnot = 0$ by Poincare duality, and aspherical (e.g., hyperbolic) $X$ have $pi_2 X = 0$ as well. $3$-manifolds are weird.)
(The easiest way to prove the first assertion is to note that the only obstructions to the triviality of $TX$ are the Stiefel-Whitney classes; and $w_1 = 0$ by orientability and $w_2 = 0$ by the Wu formula, since we're conveniently in low dimension. The second is practically a folklore theorem, but it's not that hard to prove directly either via Morse theory or minimal models.)
answered Dec 19 '18 at 4:57
anomaly
17.3k42663
answered Dec 19 '18 at 4:57
anomaly
17.3k42663
answered Dec 19 '18 at 4:57
anomaly
17.3k42663
answered Dec 19 '18 at 4:57
anomaly
17.3k42663
17.3k42663
2
You beat me to this answer! Also, we can readily construct examples of oriented, compact $3$-manifolds that do not admit a Lie group structure: Any Lie group $G$ has abelian fundamental group $pi_1(G)$, but the product of $S^1$ and an orientable, compact surface $Sigma$ with genus $> 1$ has nonabelian fundamental group $pi_1(S^1 times Sigma) cong pi_1(S^1) times pi_1(Sigma) cong Bbb Z times pi_1(Sigma)$. (And +1, by the way.)
– Travis
Dec 19 '18 at 5:09
1
Good point! I ignored the hyperbolic case because of $pi_2$, but $pi_1$ also presents an obstruction to Lie-groupness.
– anomaly
Dec 19 '18 at 5:25
add a comment |
2
You beat me to this answer! Also, we can readily construct examples of oriented, compact $3$-manifolds that do not admit a Lie group structure: Any Lie group $G$ has abelian fundamental group $pi_1(G)$, but the product of $S^1$ and an orientable, compact surface $Sigma$ with genus $> 1$ has nonabelian fundamental group $pi_1(S^1 times Sigma) cong pi_1(S^1) times pi_1(Sigma) cong Bbb Z times pi_1(Sigma)$. (And +1, by the way.)
– Travis
Dec 19 '18 at 5:09
1
Good point! I ignored the hyperbolic case because of $pi_2$, but $pi_1$ also presents an obstruction to Lie-groupness.
– anomaly
Dec 19 '18 at 5:25
2
2
You beat me to this answer! Also, we can readily construct examples of oriented, compact $3$-manifolds that do not admit a Lie group structure: Any Lie group $G$ has abelian fundamental group $pi_1(G)$, but the product of $S^1$ and an orientable, compact surface $Sigma$ with genus $> 1$ has nonabelian fundamental group $pi_1(S^1 times Sigma) cong pi_1(S^1) times pi_1(Sigma) cong Bbb Z times pi_1(Sigma)$. (And +1, by the way.)
– Travis
Dec 19 '18 at 5:09
You beat me to this answer! Also, we can readily construct examples of oriented, compact $3$-manifolds that do not admit a Lie group structure: Any Lie group $G$ has abelian fundamental group $pi_1(G)$, but the product of $S^1$ and an orientable, compact surface $Sigma$ with genus $> 1$ has nonabelian fundamental group $pi_1(S^1 times Sigma) cong pi_1(S^1) times pi_1(Sigma) cong Bbb Z times pi_1(Sigma)$. (And +1, by the way.)
– Travis
Dec 19 '18 at 5:09
1
1
Good point! I ignored the hyperbolic case because of $pi_2$, but $pi_1$ also presents an obstruction to Lie-groupness.
– anomaly
Dec 19 '18 at 5:25
Good point! I ignored the hyperbolic case because of $pi_2$, but $pi_1$ also presents an obstruction to Lie-groupness.
– anomaly
Dec 19 '18 at 5:25
add a comment |
The classical example is $S^7$. This comes from the octonions, which are non-associative, so the unit octonions don't form a Lie group.
answered Dec 19 '18 at 4:44
Lord Shark the Unknown
101k958132
Thanks, I was wondering that $S^7$ still has a group structure which is non-associative, are there examples which can not be given a group structure at all.
– PSG
Dec 19 '18 at 4:50
1
Group operations are by definition associative. The restriction of octonionic multiplication to the unit sphere in the octonions still, however, forms a loop (an operation $S times S to S$ with identity and inverses), and the multiplication and inversion maps are smooth with respect to the usual smooth structure.
– Travis
Dec 19 '18 at 4:55
1
Also, this is the only example of parallelizable sphere that admits no Lie group structure.
– Travis
Dec 19 '18 at 4:56
add a comment |
The classical example is $S^7$. This comes from the octonions, which are non-associative, so the unit octonions don't form a Lie group.
answered Dec 19 '18 at 4:44
Lord Shark the Unknown
101k958132
Thanks, I was wondering that $S^7$ still has a group structure which is non-associative, are there examples which can not be given a group structure at all.
– PSG
Dec 19 '18 at 4:50
1
Group operations are by definition associative. The restriction of octonionic multiplication to the unit sphere in the octonions still, however, forms a loop (an operation $S times S to S$ with identity and inverses), and the multiplication and inversion maps are smooth with respect to the usual smooth structure.
– Travis
Dec 19 '18 at 4:55
1
Also, this is the only example of parallelizable sphere that admits no Lie group structure.
– Travis
Dec 19 '18 at 4:56
add a comment |
The classical example is $S^7$. This comes from the octonions, which are non-associative, so the unit octonions don't form a Lie group.
answered Dec 19 '18 at 4:44
Lord Shark the Unknown
101k958132
The classical example is $S^7$. This comes from the octonions, which are non-associative, so the unit octonions don't form a Lie group.
answered Dec 19 '18 at 4:44
Lord Shark the Unknown
101k958132
answered Dec 19 '18 at 4:44
Lord Shark the Unknown
101k958132
answered Dec 19 '18 at 4:44
Lord Shark the Unknown
101k958132
answered Dec 19 '18 at 4:44
Lord Shark the Unknown
101k958132
101k958132
Thanks, I was wondering that $S^7$ still has a group structure which is non-associative, are there examples which can not be given a group structure at all.
– PSG
Dec 19 '18 at 4:50
1
Group operations are by definition associative. The restriction of octonionic multiplication to the unit sphere in the octonions still, however, forms a loop (an operation $S times S to S$ with identity and inverses), and the multiplication and inversion maps are smooth with respect to the usual smooth structure.
– Travis
Dec 19 '18 at 4:55
1
Also, this is the only example of parallelizable sphere that admits no Lie group structure.
– Travis
Dec 19 '18 at 4:56
add a comment |
Thanks, I was wondering that $S^7$ still has a group structure which is non-associative, are there examples which can not be given a group structure at all.
– PSG
Dec 19 '18 at 4:50
1
Group operations are by definition associative. The restriction of octonionic multiplication to the unit sphere in the octonions still, however, forms a loop (an operation $S times S to S$ with identity and inverses), and the multiplication and inversion maps are smooth with respect to the usual smooth structure.
– Travis
Dec 19 '18 at 4:55
1
Also, this is the only example of parallelizable sphere that admits no Lie group structure.
– Travis
Dec 19 '18 at 4:56
Thanks, I was wondering that $S^7$ still has a group structure which is non-associative, are there examples which can not be given a group structure at all.
– PSG
Dec 19 '18 at 4:50
Thanks, I was wondering that $S^7$ still has a group structure which is non-associative, are there examples which can not be given a group structure at all.
– PSG
Dec 19 '18 at 4:50
1
1
Group operations are by definition associative. The restriction of octonionic multiplication to the unit sphere in the octonions still, however, forms a loop (an operation $S times S to S$ with identity and inverses), and the multiplication and inversion maps are smooth with respect to the usual smooth structure.
– Travis
Dec 19 '18 at 4:55
Group operations are by definition associative. The restriction of octonionic multiplication to the unit sphere in the octonions still, however, forms a loop (an operation $S times S to S$ with identity and inverses), and the multiplication and inversion maps are smooth with respect to the usual smooth structure.
– Travis
Dec 19 '18 at 4:55
1
1
Also, this is the only example of parallelizable sphere that admits no Lie group structure.
– Travis
Dec 19 '18 at 4:56
Also, this is the only example of parallelizable sphere that admits no Lie group structure.
– Travis
Dec 19 '18 at 4:56
add a comment |
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In the opposite direction, there's a result due to Wolf that $S^7$ is the only exception in the simply-connected case: specifically, for a $X$ compact and simply-connected manifold, $X$ is parallelizable only if it's the product of Lie groups and copies of $S^7$.
– anomaly
18 hours ago
@anomaly Thanks a lot. I was looking for some result like that. I am not experienced to find this results. Can you please mention where can I find this or may be share a link if possible.
– PSG
17 hours ago
I don't have a reference to the original paper, but it's mentioned after Proposition 5.15 in Lee's "Introduction to Smooth Manifolds."
– anomaly
16 hours ago
Noted with thanks. I will definitely look into it.
– PSG
16 hours ago