Orientation on normal bundle












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Let $M subseteq N$ be an embedded submanifold. Then if both $M$, $N$ are oriented, it is claimed, pg87 line +17:




this induces an orientation on the normal bundle of $M$.














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  • $begingroup$
    This follows from the comments at the bottom of page 78 and top of page 79.
    $endgroup$
    – Tyrone
    Jan 3 at 11:01
















0












$begingroup$


Let $M subseteq N$ be an embedded submanifold. Then if both $M$, $N$ are oriented, it is claimed, pg87 line +17:




this induces an orientation on the normal bundle of $M$.














share|cite|improve this question











$endgroup$












  • $begingroup$
    This follows from the comments at the bottom of page 78 and top of page 79.
    $endgroup$
    – Tyrone
    Jan 3 at 11:01














0












0








0





$begingroup$


Let $M subseteq N$ be an embedded submanifold. Then if both $M$, $N$ are oriented, it is claimed, pg87 line +17:




this induces an orientation on the normal bundle of $M$.














share|cite|improve this question











$endgroup$




Let $M subseteq N$ be an embedded submanifold. Then if both $M$, $N$ are oriented, it is claimed, pg87 line +17:




this induces an orientation on the normal bundle of $M$.











differential-geometry algebraic-topology differential-topology vector-bundles






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edited Jan 2 at 10:52









Bernard

122k740116




122k740116










asked Jan 2 at 10:48









CL.CL.

2,2552925




2,2552925












  • $begingroup$
    This follows from the comments at the bottom of page 78 and top of page 79.
    $endgroup$
    – Tyrone
    Jan 3 at 11:01


















  • $begingroup$
    This follows from the comments at the bottom of page 78 and top of page 79.
    $endgroup$
    – Tyrone
    Jan 3 at 11:01
















$begingroup$
This follows from the comments at the bottom of page 78 and top of page 79.
$endgroup$
– Tyrone
Jan 3 at 11:01




$begingroup$
This follows from the comments at the bottom of page 78 and top of page 79.
$endgroup$
– Tyrone
Jan 3 at 11:01










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The obstruction of the orientability is the first Stiefel Whitney class $w_1$. Suppose that $M$ is a submanifold of $N$, we denote by $TN$ the tangent bundle of $N$, $T(N,M)$ the restriction of $TM$ to $N$ and $LN$ the normal bundle of $TN$. Remark that by using a differentiable metric one can write $T(N,M)=TNoplus LN$. This implies that $w_1(T(N,M))=w_1(TN)w_1(LN)$. We deduce that $w_1(T(N,M))=w_1(TN)=1$ implies that $w_1(LN)=1$.






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    $begingroup$

    The obstruction of the orientability is the first Stiefel Whitney class $w_1$. Suppose that $M$ is a submanifold of $N$, we denote by $TN$ the tangent bundle of $N$, $T(N,M)$ the restriction of $TM$ to $N$ and $LN$ the normal bundle of $TN$. Remark that by using a differentiable metric one can write $T(N,M)=TNoplus LN$. This implies that $w_1(T(N,M))=w_1(TN)w_1(LN)$. We deduce that $w_1(T(N,M))=w_1(TN)=1$ implies that $w_1(LN)=1$.






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      0












      $begingroup$

      The obstruction of the orientability is the first Stiefel Whitney class $w_1$. Suppose that $M$ is a submanifold of $N$, we denote by $TN$ the tangent bundle of $N$, $T(N,M)$ the restriction of $TM$ to $N$ and $LN$ the normal bundle of $TN$. Remark that by using a differentiable metric one can write $T(N,M)=TNoplus LN$. This implies that $w_1(T(N,M))=w_1(TN)w_1(LN)$. We deduce that $w_1(T(N,M))=w_1(TN)=1$ implies that $w_1(LN)=1$.






      share|cite|improve this answer









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        $begingroup$

        The obstruction of the orientability is the first Stiefel Whitney class $w_1$. Suppose that $M$ is a submanifold of $N$, we denote by $TN$ the tangent bundle of $N$, $T(N,M)$ the restriction of $TM$ to $N$ and $LN$ the normal bundle of $TN$. Remark that by using a differentiable metric one can write $T(N,M)=TNoplus LN$. This implies that $w_1(T(N,M))=w_1(TN)w_1(LN)$. We deduce that $w_1(T(N,M))=w_1(TN)=1$ implies that $w_1(LN)=1$.






        share|cite|improve this answer









        $endgroup$



        The obstruction of the orientability is the first Stiefel Whitney class $w_1$. Suppose that $M$ is a submanifold of $N$, we denote by $TN$ the tangent bundle of $N$, $T(N,M)$ the restriction of $TM$ to $N$ and $LN$ the normal bundle of $TN$. Remark that by using a differentiable metric one can write $T(N,M)=TNoplus LN$. This implies that $w_1(T(N,M))=w_1(TN)w_1(LN)$. We deduce that $w_1(T(N,M))=w_1(TN)=1$ implies that $w_1(LN)=1$.







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        answered Jan 2 at 11:53









        Tsemo AristideTsemo Aristide

        58.8k11445




        58.8k11445






























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