Example of a parallelizable smooth manifold which is not a Lie Group












7














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?






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    7














    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







    share|cite|improve this question

























      7












      7








      7


      3





      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







      share|cite|improve this question













      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






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      asked Dec 19 at 4:39









      PSG

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          6














          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.)






          share|cite|improve this answer

















          • 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 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 at 5:25



















          5














          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.






          share|cite|improve this answer





















          • 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 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 at 4:55






          • 1




            Also, this is the only example of parallelizable sphere that admits no Lie group structure.
            – Travis
            Dec 19 at 4:56











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          6














          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.)






          share|cite|improve this answer

















          • 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 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 at 5:25
















          6














          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.)






          share|cite|improve this answer

















          • 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 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 at 5:25














          6












          6








          6






          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.)






          share|cite|improve this answer












          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.)







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 19 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 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 at 5:25














          • 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 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 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 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 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 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 at 5:25











          5














          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.






          share|cite|improve this answer





















          • 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 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 at 4:55






          • 1




            Also, this is the only example of parallelizable sphere that admits no Lie group structure.
            – Travis
            Dec 19 at 4:56
















          5














          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.






          share|cite|improve this answer





















          • 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 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 at 4:55






          • 1




            Also, this is the only example of parallelizable sphere that admits no Lie group structure.
            – Travis
            Dec 19 at 4:56














          5












          5








          5






          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 19 at 4:44









          Lord Shark the Unknown

          100k958131




          100k958131












          • 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 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 at 4:55






          • 1




            Also, this is the only example of parallelizable sphere that admits no Lie group structure.
            – Travis
            Dec 19 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 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 at 4:55






          • 1




            Also, this is the only example of parallelizable sphere that admits no Lie group structure.
            – Travis
            Dec 19 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 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 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 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 at 4:55




          1




          1




          Also, this is the only example of parallelizable sphere that admits no Lie group structure.
          – Travis
          Dec 19 at 4:56




          Also, this is the only example of parallelizable sphere that admits no Lie group structure.
          – Travis
          Dec 19 at 4:56


















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