As an example, consider polar coordinates. We have \begin{align} r^2 &= x^2 + y^2 \\ \tan\phi &= \frac{y}{x} \end{align} so that \begin{align} r\,dr &= x\,dx + y\,dy \\ \frac{r^2}{x^2}\,d\phi = (1+\tan^2\phi) \,d\phi &= \frac{x\,dy-y\,dx}{x^2} \end{align} Thus, \begin{align} g(r\,dr,r\,dr) &= g(x\,dx+y\,dy,x\,dx+y\,dy) = x^2 + y^2 = r^2\\ g(r^2\,d\phi,r^2\,d\phi) &= g(x\,dy-y\,dx,x\,dy-y\,dx) = x^2 + y^2 = r^2\\ g(r\,dr,r^2\,d\phi) &= g(x\,dx+y\,dy,x\,dy-y\,dx) = yx - xy = 0 \end{align} and we conclude that $\{dr,r\,d\phi\}$ is also an orthonormal basis.

This is not surprising! Recall that infinitesimal arclength is given by \begin{equation} ds^2 = dx^2 + dy^2 = dr^2 + r^2\,d\phi^2 \end{equation} Thus, the line element tells us directly which basis of 1-forms is orthonormal.

We can take this analogy even further by considering the vector differential \begin{equation} d\rr = dx\,\xhat + dy\,\yhat = dr\,\rhat + r\,d\phi\,\phat \end{equation} which we now interpret as a vector-valued 1-form. As usual, we have \begin{equation} d\rr\cdot d\rr = ds^2 \end{equation} so we can think of $d\rr$ as the “square root” of $ds^2$. Again, $d\rr$ tells us directly which 1-forms are orthonormal.

We will develop these ideas further in Chapter 3.