We have just seen that if $\alpha,\beta\in\bigwedge^1$ then \begin{align} g(\alpha\wedge\beta,\alpha\wedge\beta) &= \left| \begin{matrix} g(\alpha,\alpha)& g(\alpha,\beta)\cr g(\beta,\alpha)& g(\beta,\beta)\cr \end{matrix} \right| \nonumber\\ &= g(\alpha,\alpha)g(\beta,\beta) - g(\alpha,\beta)^2 \end{align} But we also know that over $\RR^2$ we must have \begin{equation} \alpha\wedge\beta = f\,dx\wedge dy \end{equation} for some function $f$ (which was computed explicitly in the preceding section), so that 1) \begin{align} g(\alpha\wedge\beta,\alpha\wedge\beta) &= f^2 \,g(dx\wedge dy,dx\wedge dy) \nonumber\\ &= f^2 \,g(dx,dx) \,g(dy,dy) \nonumber\\ &= f^2 \ge 0 \end{align} Putting these two computations together, we obtain the Schwarz inequality, which says that, in Euclidean space, \begin{equation} g(\alpha,\beta)^2 \le g(\alpha,\alpha) \,g(\beta,\beta) \end{equation}

Consider the same argument in Minkowski space. The first equation is unchanged, but we now have \begin{equation} \alpha\wedge\beta = f\,dx\wedge dt \end{equation} and the last computation becomes: \begin{align} g(\alpha\wedge\beta,\alpha\wedge\beta) &= f^2 \,g(dx\wedge dt,dx\wedge dt) \nonumber\\ &= f^2 \,g(dx,dx) \,g(dt,dt) \nonumber\\ &= -f^2 \le 0 \end{align} which yields the reverse Schwarz inequality, namely \begin{equation} g(\alpha,\beta)^2 \ge g(\alpha,\alpha) \,g(\beta,\beta) \end{equation}