Positive orientation of a manifold

Question

Let $S$ be the boundary of a 3-dim manifold $\Omega$. I know that for given $v_1,v_2$, $\{v_1,v_2\}$ is a positive orientation of $T_p(S)$ if $\{n(p),v_1,v_2\}$ is a positive orientation of $T_p(\Omega)$ where $n(p)$ is the unit outward normal. May I know how to check if $\{n(p),v_1,v_2\}$ is a positive orientation of $T_p(\Omega)$? Should I take the cross-product of $v_1$ and $v_2$ and check if it is in the direction of $n(p)$? Thanks.

Answer 1

Short answer is yes. Somewhat more elaborately: an orientation of an $n$-manifold is given by a so-called orientation $n$-form $\omega$, with the idea being that the sign of $\omega(v_1,\ldots,v_n)$ determines whether $\{v_1,\ldots,v_n\}$ is positively or negatively oriented. In this case, the normal vector $n(p)$ gives rise to an orientation $2$-form by defining $$ \omega(v_1,v_2):=n(p)\cdot (v_1\times v_2), $$ also known as the scalar triple product. It gives the signed volume of the parallelepiped spanned by $n(p)$, $v_1$, and $v_2$. It is equivalent to what you suggested, since $n(p)$ and $v_1\times v_2$ will both be perpendicular to $T_p(S)$ and the sign of the dot product simply determines whether the two vectors point in the same or opposite directions.