Extending to a local frame that agrees with given orientation












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Suppose that $(e_1, ldots, e_k)$ is an oriented basis for $T_pM$ where $M$ is an oriented Riemannian manifold. In general, we know that we can extend to a smooth local frame $(X_1, ldots, X_k)$ on $Uni p$ such that $X_irvert_p = e_i$ for each $i$.



But can we further stipulate that $(X_1rvert_q, ldots, X_krvert_q)$ is oriented for each $qin U$?



I have tried thinking of ways to shrink $U$ in a suitable way. I have played around with using the orientation form $omega$ on $M$ and with first applying Gram-Schmidt to $(X_1, ldots, X_k)$, but no progress.



Any help is much appreciated.






linear-algebra differential-geometry differential-topology riemannian-geometry smooth-manifolds







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$endgroup$

















    0












    $begingroup$


    Suppose that $(e_1, ldots, e_k)$ is an oriented basis for $T_pM$ where $M$ is an oriented Riemannian manifold. In general, we know that we can extend to a smooth local frame $(X_1, ldots, X_k)$ on $Uni p$ such that $X_irvert_p = e_i$ for each $i$.



    But can we further stipulate that $(X_1rvert_q, ldots, X_krvert_q)$ is oriented for each $qin U$?



    I have tried thinking of ways to shrink $U$ in a suitable way. I have played around with using the orientation form $omega$ on $M$ and with first applying Gram-Schmidt to $(X_1, ldots, X_k)$, but no progress.



    Any help is much appreciated.






    linear-algebra differential-geometry differential-topology riemannian-geometry smooth-manifolds







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Suppose that $(e_1, ldots, e_k)$ is an oriented basis for $T_pM$ where $M$ is an oriented Riemannian manifold. In general, we know that we can extend to a smooth local frame $(X_1, ldots, X_k)$ on $Uni p$ such that $X_irvert_p = e_i$ for each $i$.



      But can we further stipulate that $(X_1rvert_q, ldots, X_krvert_q)$ is oriented for each $qin U$?



      I have tried thinking of ways to shrink $U$ in a suitable way. I have played around with using the orientation form $omega$ on $M$ and with first applying Gram-Schmidt to $(X_1, ldots, X_k)$, but no progress.



      Any help is much appreciated.






      linear-algebra differential-geometry differential-topology riemannian-geometry smooth-manifolds







      share|cite|improve this question











      $endgroup$




      Suppose that $(e_1, ldots, e_k)$ is an oriented basis for $T_pM$ where $M$ is an oriented Riemannian manifold. In general, we know that we can extend to a smooth local frame $(X_1, ldots, X_k)$ on $Uni p$ such that $X_irvert_p = e_i$ for each $i$.



      But can we further stipulate that $(X_1rvert_q, ldots, X_krvert_q)$ is oriented for each $qin U$?



      I have tried thinking of ways to shrink $U$ in a suitable way. I have played around with using the orientation form $omega$ on $M$ and with first applying Gram-Schmidt to $(X_1, ldots, X_k)$, but no progress.



      Any help is much appreciated.






      linear-algebra differential-geometry differential-topology riemannian-geometry smooth-manifolds




      linear-algebra differential-geometry differential-topology riemannian-geometry smooth-manifolds






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 4 '18 at 18:25







      CuriousKid7

















      asked Dec 4 '18 at 1:01









      CuriousKid7CuriousKid7

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          1 Answer
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          $begingroup$

          The local frame $(X_i)$ will agree with the chosen orientation form: if $omega(e_1,dots,e_k)>0$, then $omega(X_1,dots,X_k)>0$ in $U$, because $omega(X_1,dots,X_k)|_q=0$ would imply that the $X_i's$ do not form a basis of $T_qM$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This solution shows how to restrict $U$ appropriately, but I don't see why vanishing at $omega rvert_q$ would imply that the $X_i$ don't form a basis. It's possible that an alternating tensor vanishes at a linearly independent tuple, right?
            $endgroup$
            – CuriousKid7
            Dec 5 '18 at 5:01












          • $begingroup$
            You don't need to restrict $U$ in any way. $M$ is orientable and you have fixed an orientation form $omega$. What I'm saying is that if $(X_i)$ is a frame inside $U$, then $omega(X_1,dots,X_k)$ doesn't change sign inside $U$, because if it does there is a point at which it must vanish, and at that point $(X_i)$ would not be a frame
            $endgroup$
            – Federico
            Dec 5 '18 at 13:51






          • 1




            $begingroup$
            "It's possible that an alternating tensor vanishes at a linearly independent tuple, right?" No, unless $omega=0$ at that point. If it vanishes at a linearly independent tuple, it vanishes on any tuple.
            $endgroup$
            – Federico
            Dec 5 '18 at 13:52






          • 1




            $begingroup$
            Assume $omega(e_1,dots,e_k)=0$ for some basis $(e_i)$. Take any $k$-uple $v_j=a^i_je_i$. Then $omega(v_1,dots,v_k)=det(A)omega(e_1,dots,e_k)=0$, where $A$ is the matrix with entries $a^i_j$.
            $endgroup$
            – Federico
            Dec 5 '18 at 13:54










          • $begingroup$
            Got it, thanks for clarifying!
            $endgroup$
            – CuriousKid7
            Dec 5 '18 at 16:27











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          1 Answer
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          2












          $begingroup$

          The local frame $(X_i)$ will agree with the chosen orientation form: if $omega(e_1,dots,e_k)>0$, then $omega(X_1,dots,X_k)>0$ in $U$, because $omega(X_1,dots,X_k)|_q=0$ would imply that the $X_i's$ do not form a basis of $T_qM$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This solution shows how to restrict $U$ appropriately, but I don't see why vanishing at $omega rvert_q$ would imply that the $X_i$ don't form a basis. It's possible that an alternating tensor vanishes at a linearly independent tuple, right?
            $endgroup$
            – CuriousKid7
            Dec 5 '18 at 5:01












          • $begingroup$
            You don't need to restrict $U$ in any way. $M$ is orientable and you have fixed an orientation form $omega$. What I'm saying is that if $(X_i)$ is a frame inside $U$, then $omega(X_1,dots,X_k)$ doesn't change sign inside $U$, because if it does there is a point at which it must vanish, and at that point $(X_i)$ would not be a frame
            $endgroup$
            – Federico
            Dec 5 '18 at 13:51






          • 1




            $begingroup$
            "It's possible that an alternating tensor vanishes at a linearly independent tuple, right?" No, unless $omega=0$ at that point. If it vanishes at a linearly independent tuple, it vanishes on any tuple.
            $endgroup$
            – Federico
            Dec 5 '18 at 13:52






          • 1




            $begingroup$
            Assume $omega(e_1,dots,e_k)=0$ for some basis $(e_i)$. Take any $k$-uple $v_j=a^i_je_i$. Then $omega(v_1,dots,v_k)=det(A)omega(e_1,dots,e_k)=0$, where $A$ is the matrix with entries $a^i_j$.
            $endgroup$
            – Federico
            Dec 5 '18 at 13:54










          • $begingroup$
            Got it, thanks for clarifying!
            $endgroup$
            – CuriousKid7
            Dec 5 '18 at 16:27
















          2












          $begingroup$

          The local frame $(X_i)$ will agree with the chosen orientation form: if $omega(e_1,dots,e_k)>0$, then $omega(X_1,dots,X_k)>0$ in $U$, because $omega(X_1,dots,X_k)|_q=0$ would imply that the $X_i's$ do not form a basis of $T_qM$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This solution shows how to restrict $U$ appropriately, but I don't see why vanishing at $omega rvert_q$ would imply that the $X_i$ don't form a basis. It's possible that an alternating tensor vanishes at a linearly independent tuple, right?
            $endgroup$
            – CuriousKid7
            Dec 5 '18 at 5:01












          • $begingroup$
            You don't need to restrict $U$ in any way. $M$ is orientable and you have fixed an orientation form $omega$. What I'm saying is that if $(X_i)$ is a frame inside $U$, then $omega(X_1,dots,X_k)$ doesn't change sign inside $U$, because if it does there is a point at which it must vanish, and at that point $(X_i)$ would not be a frame
            $endgroup$
            – Federico
            Dec 5 '18 at 13:51






          • 1




            $begingroup$
            "It's possible that an alternating tensor vanishes at a linearly independent tuple, right?" No, unless $omega=0$ at that point. If it vanishes at a linearly independent tuple, it vanishes on any tuple.
            $endgroup$
            – Federico
            Dec 5 '18 at 13:52






          • 1




            $begingroup$
            Assume $omega(e_1,dots,e_k)=0$ for some basis $(e_i)$. Take any $k$-uple $v_j=a^i_je_i$. Then $omega(v_1,dots,v_k)=det(A)omega(e_1,dots,e_k)=0$, where $A$ is the matrix with entries $a^i_j$.
            $endgroup$
            – Federico
            Dec 5 '18 at 13:54










          • $begingroup$
            Got it, thanks for clarifying!
            $endgroup$
            – CuriousKid7
            Dec 5 '18 at 16:27














          2












          2








          2





          $begingroup$

          The local frame $(X_i)$ will agree with the chosen orientation form: if $omega(e_1,dots,e_k)>0$, then $omega(X_1,dots,X_k)>0$ in $U$, because $omega(X_1,dots,X_k)|_q=0$ would imply that the $X_i's$ do not form a basis of $T_qM$.






          share|cite|improve this answer









          $endgroup$



          The local frame $(X_i)$ will agree with the chosen orientation form: if $omega(e_1,dots,e_k)>0$, then $omega(X_1,dots,X_k)>0$ in $U$, because $omega(X_1,dots,X_k)|_q=0$ would imply that the $X_i's$ do not form a basis of $T_qM$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 4 '18 at 18:47









          FedericoFederico

          5,014514




          5,014514












          • $begingroup$
            This solution shows how to restrict $U$ appropriately, but I don't see why vanishing at $omega rvert_q$ would imply that the $X_i$ don't form a basis. It's possible that an alternating tensor vanishes at a linearly independent tuple, right?
            $endgroup$
            – CuriousKid7
            Dec 5 '18 at 5:01












          • $begingroup$
            You don't need to restrict $U$ in any way. $M$ is orientable and you have fixed an orientation form $omega$. What I'm saying is that if $(X_i)$ is a frame inside $U$, then $omega(X_1,dots,X_k)$ doesn't change sign inside $U$, because if it does there is a point at which it must vanish, and at that point $(X_i)$ would not be a frame
            $endgroup$
            – Federico
            Dec 5 '18 at 13:51






          • 1




            $begingroup$
            "It's possible that an alternating tensor vanishes at a linearly independent tuple, right?" No, unless $omega=0$ at that point. If it vanishes at a linearly independent tuple, it vanishes on any tuple.
            $endgroup$
            – Federico
            Dec 5 '18 at 13:52






          • 1




            $begingroup$
            Assume $omega(e_1,dots,e_k)=0$ for some basis $(e_i)$. Take any $k$-uple $v_j=a^i_je_i$. Then $omega(v_1,dots,v_k)=det(A)omega(e_1,dots,e_k)=0$, where $A$ is the matrix with entries $a^i_j$.
            $endgroup$
            – Federico
            Dec 5 '18 at 13:54










          • $begingroup$
            Got it, thanks for clarifying!
            $endgroup$
            – CuriousKid7
            Dec 5 '18 at 16:27


















          • $begingroup$
            This solution shows how to restrict $U$ appropriately, but I don't see why vanishing at $omega rvert_q$ would imply that the $X_i$ don't form a basis. It's possible that an alternating tensor vanishes at a linearly independent tuple, right?
            $endgroup$
            – CuriousKid7
            Dec 5 '18 at 5:01












          • $begingroup$
            You don't need to restrict $U$ in any way. $M$ is orientable and you have fixed an orientation form $omega$. What I'm saying is that if $(X_i)$ is a frame inside $U$, then $omega(X_1,dots,X_k)$ doesn't change sign inside $U$, because if it does there is a point at which it must vanish, and at that point $(X_i)$ would not be a frame
            $endgroup$
            – Federico
            Dec 5 '18 at 13:51






          • 1




            $begingroup$
            "It's possible that an alternating tensor vanishes at a linearly independent tuple, right?" No, unless $omega=0$ at that point. If it vanishes at a linearly independent tuple, it vanishes on any tuple.
            $endgroup$
            – Federico
            Dec 5 '18 at 13:52






          • 1




            $begingroup$
            Assume $omega(e_1,dots,e_k)=0$ for some basis $(e_i)$. Take any $k$-uple $v_j=a^i_je_i$. Then $omega(v_1,dots,v_k)=det(A)omega(e_1,dots,e_k)=0$, where $A$ is the matrix with entries $a^i_j$.
            $endgroup$
            – Federico
            Dec 5 '18 at 13:54










          • $begingroup$
            Got it, thanks for clarifying!
            $endgroup$
            – CuriousKid7
            Dec 5 '18 at 16:27
















          $begingroup$
          This solution shows how to restrict $U$ appropriately, but I don't see why vanishing at $omega rvert_q$ would imply that the $X_i$ don't form a basis. It's possible that an alternating tensor vanishes at a linearly independent tuple, right?
          $endgroup$
          – CuriousKid7
          Dec 5 '18 at 5:01






          $begingroup$
          This solution shows how to restrict $U$ appropriately, but I don't see why vanishing at $omega rvert_q$ would imply that the $X_i$ don't form a basis. It's possible that an alternating tensor vanishes at a linearly independent tuple, right?
          $endgroup$
          – CuriousKid7
          Dec 5 '18 at 5:01














          $begingroup$
          You don't need to restrict $U$ in any way. $M$ is orientable and you have fixed an orientation form $omega$. What I'm saying is that if $(X_i)$ is a frame inside $U$, then $omega(X_1,dots,X_k)$ doesn't change sign inside $U$, because if it does there is a point at which it must vanish, and at that point $(X_i)$ would not be a frame
          $endgroup$
          – Federico
          Dec 5 '18 at 13:51




          $begingroup$
          You don't need to restrict $U$ in any way. $M$ is orientable and you have fixed an orientation form $omega$. What I'm saying is that if $(X_i)$ is a frame inside $U$, then $omega(X_1,dots,X_k)$ doesn't change sign inside $U$, because if it does there is a point at which it must vanish, and at that point $(X_i)$ would not be a frame
          $endgroup$
          – Federico
          Dec 5 '18 at 13:51




          1




          1




          $begingroup$
          "It's possible that an alternating tensor vanishes at a linearly independent tuple, right?" No, unless $omega=0$ at that point. If it vanishes at a linearly independent tuple, it vanishes on any tuple.
          $endgroup$
          – Federico
          Dec 5 '18 at 13:52




          $begingroup$
          "It's possible that an alternating tensor vanishes at a linearly independent tuple, right?" No, unless $omega=0$ at that point. If it vanishes at a linearly independent tuple, it vanishes on any tuple.
          $endgroup$
          – Federico
          Dec 5 '18 at 13:52




          1




          1




          $begingroup$
          Assume $omega(e_1,dots,e_k)=0$ for some basis $(e_i)$. Take any $k$-uple $v_j=a^i_je_i$. Then $omega(v_1,dots,v_k)=det(A)omega(e_1,dots,e_k)=0$, where $A$ is the matrix with entries $a^i_j$.
          $endgroup$
          – Federico
          Dec 5 '18 at 13:54




          $begingroup$
          Assume $omega(e_1,dots,e_k)=0$ for some basis $(e_i)$. Take any $k$-uple $v_j=a^i_je_i$. Then $omega(v_1,dots,v_k)=det(A)omega(e_1,dots,e_k)=0$, where $A$ is the matrix with entries $a^i_j$.
          $endgroup$
          – Federico
          Dec 5 '18 at 13:54












          $begingroup$
          Got it, thanks for clarifying!
          $endgroup$
          – CuriousKid7
          Dec 5 '18 at 16:27




          $begingroup$
          Got it, thanks for clarifying!
          $endgroup$
          – CuriousKid7
          Dec 5 '18 at 16:27


















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