If $TM$ is trivial, then $Lambda^n(M)$ is also trivial and $M$ is orientable












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Suppose that $M$ is a smooth $n-$manifold. Suppose $$ TM=coprod_{pin M}T_pM $$ is the tangent bundle of $M$. And let $$ Lambda^n(M)=coprod_{pin M}Lambda^n(T_pM) $$ where $Lambda^n(T_pM)$ is alternating covariant $n-$tensor product. i.e.



$$Lambda^n(T_pM)=T^*_pMwedge T^*_pM wedge cdots wedge T^*_pM.$$



and



$$dim(Lambda^n(T_pM))=begin{bmatrix} dim(T_pM)\ n end{bmatrix}$$




If $TM$ is trivial, then $Lambda^n(M)$ is also trivial and $M$ is orientable.




I am trying to understand but seems difficult. I would be very appreciate any help. Thanks in advance.






orientation tangent-bundle







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    0












    $begingroup$


    Suppose that $M$ is a smooth $n-$manifold. Suppose $$ TM=coprod_{pin M}T_pM $$ is the tangent bundle of $M$. And let $$ Lambda^n(M)=coprod_{pin M}Lambda^n(T_pM) $$ where $Lambda^n(T_pM)$ is alternating covariant $n-$tensor product. i.e.



    $$Lambda^n(T_pM)=T^*_pMwedge T^*_pM wedge cdots wedge T^*_pM.$$



    and



    $$dim(Lambda^n(T_pM))=begin{bmatrix} dim(T_pM)\ n end{bmatrix}$$




    If $TM$ is trivial, then $Lambda^n(M)$ is also trivial and $M$ is orientable.




    I am trying to understand but seems difficult. I would be very appreciate any help. Thanks in advance.






    orientation tangent-bundle







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Suppose that $M$ is a smooth $n-$manifold. Suppose $$ TM=coprod_{pin M}T_pM $$ is the tangent bundle of $M$. And let $$ Lambda^n(M)=coprod_{pin M}Lambda^n(T_pM) $$ where $Lambda^n(T_pM)$ is alternating covariant $n-$tensor product. i.e.



      $$Lambda^n(T_pM)=T^*_pMwedge T^*_pM wedge cdots wedge T^*_pM.$$



      and



      $$dim(Lambda^n(T_pM))=begin{bmatrix} dim(T_pM)\ n end{bmatrix}$$




      If $TM$ is trivial, then $Lambda^n(M)$ is also trivial and $M$ is orientable.




      I am trying to understand but seems difficult. I would be very appreciate any help. Thanks in advance.






      orientation tangent-bundle







      share|cite|improve this question











      $endgroup$




      Suppose that $M$ is a smooth $n-$manifold. Suppose $$ TM=coprod_{pin M}T_pM $$ is the tangent bundle of $M$. And let $$ Lambda^n(M)=coprod_{pin M}Lambda^n(T_pM) $$ where $Lambda^n(T_pM)$ is alternating covariant $n-$tensor product. i.e.



      $$Lambda^n(T_pM)=T^*_pMwedge T^*_pM wedge cdots wedge T^*_pM.$$



      and



      $$dim(Lambda^n(T_pM))=begin{bmatrix} dim(T_pM)\ n end{bmatrix}$$




      If $TM$ is trivial, then $Lambda^n(M)$ is also trivial and $M$ is orientable.




      I am trying to understand but seems difficult. I would be very appreciate any help. Thanks in advance.






      orientation tangent-bundle




      orientation tangent-bundle






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 3 '18 at 14:00







      Lev Ban

















      asked Nov 30 '18 at 16:52









      Lev BanLev Ban

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      1,0701317






















          1 Answer
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          $begingroup$

          If $M$ is $n$-dimensional, $TM$ is trivial is equivalent to saying that there exists $n$-vector fields $X_1,...,X_n$ such that for every $xin M$, $X_i(x)neq 0$. Take a differentiable metric on $M$ and define the $1$-form $f_i(x)(u)=langle X_i(x),urangle$ where $uin T_xM$, $Lambda^n(M)_x$ is generated by $f_1(x)wedge...wedge f_n(x)$ which is also a volume form on $M$. This implies that $M$ is orientable.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your answer. But could you let me know why $Lambda^n(M)$ is also trivial?
            $endgroup$
            – Lev Ban
            Nov 30 '18 at 17:09










          • $begingroup$
            because it is a dim 1 and you have the volume form
            $endgroup$
            – Tsemo Aristide
            Nov 30 '18 at 17:21











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          $begingroup$

          If $M$ is $n$-dimensional, $TM$ is trivial is equivalent to saying that there exists $n$-vector fields $X_1,...,X_n$ such that for every $xin M$, $X_i(x)neq 0$. Take a differentiable metric on $M$ and define the $1$-form $f_i(x)(u)=langle X_i(x),urangle$ where $uin T_xM$, $Lambda^n(M)_x$ is generated by $f_1(x)wedge...wedge f_n(x)$ which is also a volume form on $M$. This implies that $M$ is orientable.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your answer. But could you let me know why $Lambda^n(M)$ is also trivial?
            $endgroup$
            – Lev Ban
            Nov 30 '18 at 17:09










          • $begingroup$
            because it is a dim 1 and you have the volume form
            $endgroup$
            – Tsemo Aristide
            Nov 30 '18 at 17:21
















          1












          $begingroup$

          If $M$ is $n$-dimensional, $TM$ is trivial is equivalent to saying that there exists $n$-vector fields $X_1,...,X_n$ such that for every $xin M$, $X_i(x)neq 0$. Take a differentiable metric on $M$ and define the $1$-form $f_i(x)(u)=langle X_i(x),urangle$ where $uin T_xM$, $Lambda^n(M)_x$ is generated by $f_1(x)wedge...wedge f_n(x)$ which is also a volume form on $M$. This implies that $M$ is orientable.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your answer. But could you let me know why $Lambda^n(M)$ is also trivial?
            $endgroup$
            – Lev Ban
            Nov 30 '18 at 17:09










          • $begingroup$
            because it is a dim 1 and you have the volume form
            $endgroup$
            – Tsemo Aristide
            Nov 30 '18 at 17:21














          1












          1








          1





          $begingroup$

          If $M$ is $n$-dimensional, $TM$ is trivial is equivalent to saying that there exists $n$-vector fields $X_1,...,X_n$ such that for every $xin M$, $X_i(x)neq 0$. Take a differentiable metric on $M$ and define the $1$-form $f_i(x)(u)=langle X_i(x),urangle$ where $uin T_xM$, $Lambda^n(M)_x$ is generated by $f_1(x)wedge...wedge f_n(x)$ which is also a volume form on $M$. This implies that $M$ is orientable.






          share|cite|improve this answer









          $endgroup$



          If $M$ is $n$-dimensional, $TM$ is trivial is equivalent to saying that there exists $n$-vector fields $X_1,...,X_n$ such that for every $xin M$, $X_i(x)neq 0$. Take a differentiable metric on $M$ and define the $1$-form $f_i(x)(u)=langle X_i(x),urangle$ where $uin T_xM$, $Lambda^n(M)_x$ is generated by $f_1(x)wedge...wedge f_n(x)$ which is also a volume form on $M$. This implies that $M$ is orientable.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 30 '18 at 17:01









          Tsemo AristideTsemo Aristide

          58k11445




          58k11445












          • $begingroup$
            Thank you for your answer. But could you let me know why $Lambda^n(M)$ is also trivial?
            $endgroup$
            – Lev Ban
            Nov 30 '18 at 17:09










          • $begingroup$
            because it is a dim 1 and you have the volume form
            $endgroup$
            – Tsemo Aristide
            Nov 30 '18 at 17:21


















          • $begingroup$
            Thank you for your answer. But could you let me know why $Lambda^n(M)$ is also trivial?
            $endgroup$
            – Lev Ban
            Nov 30 '18 at 17:09










          • $begingroup$
            because it is a dim 1 and you have the volume form
            $endgroup$
            – Tsemo Aristide
            Nov 30 '18 at 17:21
















          $begingroup$
          Thank you for your answer. But could you let me know why $Lambda^n(M)$ is also trivial?
          $endgroup$
          – Lev Ban
          Nov 30 '18 at 17:09




          $begingroup$
          Thank you for your answer. But could you let me know why $Lambda^n(M)$ is also trivial?
          $endgroup$
          – Lev Ban
          Nov 30 '18 at 17:09












          $begingroup$
          because it is a dim 1 and you have the volume form
          $endgroup$
          – Tsemo Aristide
          Nov 30 '18 at 17:21




          $begingroup$
          because it is a dim 1 and you have the volume form
          $endgroup$
          – Tsemo Aristide
          Nov 30 '18 at 17:21


















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