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Assume $M$ is an embedded submanifold of $\mathbb R^n$ with codimension 1, by a specific coordinate of $\mathbb R^n$ $\{x_1,\dots,x_n\}$ in an area $U$, the $M\cap U$ has coordinate $\{x_1,\dots,x_{n-1}\}$. Under this coordinate,the orientation of $\mathbb R^n$ can be determined by the standard volume form $dx_1\wedge dx_2\wedge\dots\wedge dx_n$, so the standard volume form of $dx_1\wedge dx_2\wedge\dots\wedge dx_{n-1}$, thus the submanifold of $\mathbb R^n$ is orientable? Am I right?

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  • $\begingroup$ I don't think that suffices. You've shown that $M\cap U$ is orientable, but you need to show the form you produced extends to a nowhere vanishing form on all of $M$ in order to conclude that $M$ is orientable. $\endgroup$ – user4571 Dec 9 '15 at 7:24
  • $\begingroup$ In fact, depending on the definition of embedded submanifold, your claim might be false. Consider embedding the Mobius band in $\mathbb R^3$, for example. $\endgroup$ – user4571 Dec 9 '15 at 7:26
  • $\begingroup$ @ Patrick, Thanks, I think you are right, actually, I arise this question from proving the preimage of regular value of a smooth function from $R^n$ to R is orientable. Could you give me some ideas about this question? $\endgroup$ – 6666 Dec 9 '15 at 7:32
  • $\begingroup$ See: math.stackexchange.com/questions/384700/… $\endgroup$ – user4571 Dec 9 '15 at 7:34
  • $\begingroup$ @Patrick, I guess the result also holds for the case that the function is on a manifold M, right? $\endgroup$ – 6666 Dec 9 '15 at 16:46

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