# A vector field corresponding to the complement of the tangent bundle

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?

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.