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Let $M$ be a $m$ dimensional orientable manifold, and $N$ a $m-1$ dimensional orientable submanifold in $M$, then we know at each point $x \in N$, $T_{x}M = T_x N \oplus$its complement. I need to produce a vector field $X$ such that $X_x$ is a vector in the complement of $T_{x}N$.

I am trying to exploit the assumption that both manifolds are orientable meaning they all have a non-vanishing top form, but what next?

UPDATE1: Take a slice chart $(x_1, \ldots, x_n)$ of $M$. Let the orientation form on $M$ be $w_M = fdx_1 \wedge \ldots \wedge dx_n$, and the orientation form on $N$ be $w_N = fdx_1 \wedge \ldots \wedge dx_{n-1}$. A possible approach is to find a vector $V$ such that $V \lnot w_{M} = w_N$. The computation is easy. Locally $V$ should be $\frac{g}{f}\frac{\partial}{\partial x_n}$. However, how to show this expression is independent of coordinate thus can be globalized?

UPDATE2: Trying to understand @Tsemo Aristide's answer. Locally, the choice of the sign has to be consistent. so in a neighborhood, it has to be either $u_x$ or $-u_x$ throughout, which proves the smoothness. Is this correct?

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Let $\Omega^M$ be the volume form of $M$ and $\Omega^N$ the volume form of $N$. Consider a differentiable metric defined on $M$. For every $x\in N$, there exists two vectors of norm $1$, $u_x,-u_x$ orthogonal to $T_xN$, If the restriction of $i_{u_x}{\Omega^M}_x$ to $T_xN$ is $c{\Omega^N}_x, c>0$ write $n(x)=u_x$ otherwise, write $n(x)=-u_x$. Show that $n(x)$ is differentiable by using local coordinates.

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  • $\begingroup$ May you be more specific about how to check the differentiability? I cannot really figure out how to write down an explicit formula locally $\endgroup$ – Keith Dec 8 '18 at 2:17

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