As before, let $\{\sigma^i\}$ be an orthonormal basis of $\bigwedge^1(M)$, and consider $\bigwedge^n(M)$. First of all, this space is 1-dimensional, with basis \begin{equation} \omega = \sigma^1 \wedge \,…\, \wedge \sigma^n \end{equation} But $\omega$ is a unit $n$-form, since \begin{equation} g(\omega,\omega) = g(\sigma^1,\sigma^1) … g(\sigma^n,\sigma^n) = (-1)^s \label{orientation} \end{equation} Furthermore, there are precisely two unit $n$-forms, namely $\pm\omega$, a statement which does not, of course, depend on the choice of basis.

An orientation $\omega$ on $M$ is a choice $\omega$ of one of the two unit $n$-forms. If an orthonormal basis is given explicitly, as above, we will always assume that $\omega$ is precisely as given in $(\ref{orientation})$, that is, that the basis is ordered to match (or define) the orientation.