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I have the following definition (from Hubbard's vector calculus book) for an oriented boundary of piece with boundary of an oriented manifold:

Let $M$ be a $k$ dimensional manifold oriented by $\Omega$ and $P$ a piece with boundary of $M$. Let $x$ be a point of the smooth boundary $\partial^{ \ S}_MP$ and let $\vec{V}_{\text{out}}\in T_xM$ be an outward pointing bector. Then the function $\Omega^\partial : \mathcal{B}(T_x\partial P)\to\left\{+1,-1\right\}$ given by $$ \Omega_x^\partial(\vec{v}_1,...,\vec{v}_{k-1}) = \Omega_x(\vec{V}_{\text{out}},\vec{v}_1,...,\vec{v}_{k-1}) $$ defines an orientation on the smooth boundary $\partial_M^{ \ S}P,$ where $\vec{v}_1,...,\vec{v}_{k-1}$ is an ordered basis of $T_x\partial_M^{ \ S}P$.

I'm working on a problem that asks me to find a basis for the $T_x\partial P$ that is direct to a certain orientation (given by an elementary 3-form). My question is this:

When I choose a basis for $T_x\partial P$, does this basis also need to lie in $T_xM$? Also, are there restrictions to how I should choose $\vec{V}_{\text{out}}$? In other words, does $\vec{V}_{\text{out}}$ need only lie in $T_xM$ and not in $T_x\partial P$?

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up vote 2 down vote accepted

Yes, $T_x\partial P$ is a hyperplane in $T_xM$. $\vec V_{\text{out}}$ is by definition the outward-pointing normal to $\partial P$. This means that if $\vec v_1,\dots,\vec v_{k-1}$ are chosen as a basis for $T_x\partial P$, then $\vec V_{\text{out}},\vec v_1,\dots,\vec v_{k-1}$ will give you a basis for $T_xM$. The whole point of this orientation stuff is that when you pick $\vec v_1,\dots,\vec v_{k-1}$ so that $\vec V_{\text{out}},\vec v_1,\dots,\vec v_{k-1}$ gives you a positively-oriented basis for $T_xM$, then you win: You have achieved a positively-oriented basis for $T_x\partial P$.

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