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











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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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    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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      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







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      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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      PSG

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          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.






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          • 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
            32 mins ago










          • 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
            27 mins ago






          • 1




            Also, this is the only example of parallelizable sphere that admits no Lie group structure.
            – Travis
            26 mins ago


















          up vote
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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.)






          share|cite|improve this answer





















          • 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
            12 mins ago











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          up vote
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          accepted










          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
            32 mins ago










          • 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
            27 mins ago






          • 1




            Also, this is the only example of parallelizable sphere that admits no Lie group structure.
            – Travis
            26 mins ago















          up vote
          1
          down vote



          accepted










          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
            32 mins ago










          • 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
            27 mins ago






          • 1




            Also, this is the only example of parallelizable sphere that admits no Lie group structure.
            – Travis
            26 mins ago













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          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 38 mins ago









          Lord Shark the Unknown

          99.8k958131




          99.8k958131












          • 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
            32 mins ago










          • 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
            27 mins ago






          • 1




            Also, this is the only example of parallelizable sphere that admits no Lie group structure.
            – Travis
            26 mins ago


















          • 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
            32 mins ago










          • 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
            27 mins ago






          • 1




            Also, this is the only example of parallelizable sphere that admits no Lie group structure.
            – Travis
            26 mins ago
















          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
          32 mins ago




          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
          32 mins ago












          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
          27 mins ago




          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
          27 mins ago




          1




          1




          Also, this is the only example of parallelizable sphere that admits no Lie group structure.
          – Travis
          26 mins ago




          Also, this is the only example of parallelizable sphere that admits no Lie group structure.
          – Travis
          26 mins ago










          up vote
          4
          down vote













          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





















          • 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
            12 mins ago















          up vote
          4
          down vote













          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





















          • 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
            12 mins ago













          up vote
          4
          down vote










          up vote
          4
          down vote









          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 25 mins ago









          anomaly

          17.3k42662




          17.3k42662












          • 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
            12 mins ago


















          • 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
            12 mins ago
















          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
          12 mins ago




          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
          12 mins ago


















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