Proposition. Let $V$ be a vector space of positive dimension $n$. Then every nonzero alternating $n$-tensor $\omega$ on $V$ determines an orientation of $V$ by collecting all the bases $\{v_1,\ldots,v_n\}$ for $V$ whose images under $\omega$ exhibit the same sign.
As far as I know, there are exactly two orientations on $V$. Now the proposition seems to indicate that there are in fact three of them. The first one is obtained by collecting all the bases $\{v_1,\ldots,v_n\}$ with $\omega(v_1,\ldots,v_n)>0$. The second one is obtained by collecting all the bases $\{v_1,\ldots,v_n\}$ with $\omega(v_1,\ldots,v_n)<0$. The third on is obtained by collecting all the bases $\{v_1,\ldots,v_n\}$ with $\omega(v_1,\ldots,v_n)=0$. These three orientations seems to violate what I already learned about orientations of vector spaces. Why is that? Thank you.
Added: A nonzero alternating $n$-tensor $\omega$ on $V$ is supposed to mean a function on $V^n$ that is not identically zero; thus, there exists $w_1,\ldots,w_n\in V$ such that $\omega(w_1,\ldots,w_n)\neq 0$. I know what to do with those bases mapped to a nonzero value by $\omega$, but I don't know what to do with those bases mapped to zero by $\omega$. Is the latter case not supposed to happen? Or is the proposition simply inconclusive when it comes to bases $\beta$ mapped to zero? In this case, do we determine which orientation $\beta$ belongs to in the old-fashioned way, that is, by calculating the transition matrix between $\beta$ and $\{v_1,\ldots,v_n\}$? Thank you.
If there is a basis $\{v_1,\dots, v_n\}$ such that $\omega(v_1,\dots, v_n)=0$, then $\omega=0$. This is a consequence of multilinearity, and the alternating nature of $\omega$.