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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 $U\ni p$ such that $X_i\rvert_p = e_i$ for each $i$.

But can we further stipulate that $(X_1\rvert_q, \ldots, X_k\rvert_q)$ is oriented for each $q\in U$?

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.

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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 $T_qM$.

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  • $\begingroup$ 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
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    $\begingroup$ "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
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    $\begingroup$ 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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