Appendix: Mathematical Details

### Divergence of the Metric in Two Dimensions

Consider two-dimensional Euclidean space, for which $$d\rr = dx\,\xhat + dy\,\yhat$$ so that $${*}d\rr = dy\,\xhat - dx\,\yhat$$ and clearly $$d{*}d\rr = \zero \label{metricdiv}$$

This computation shows that (\ref{metricdiv}) must in fact hold in flat space in any signature. We check this explicitly in polar coordinates, for which $$d\rr = dr\,\rhat + r\,d\phi\,\phat$$ so that $${*}d\rr = r\,d\phi\,\rhat - dr\,\phat$$ and we compute \begin{align} d{*}d\rr &= dr\wedge d\phi\,\rhat - r\,d\phi\wedge d\rhat + dr\wedge d\phat \nonumber\\ &= dr\wedge d\phi\,\rhat + r\,d\phi\wedge d\phi\,\phat - dr\wedge d\phi\,\rhat \nonumber\\ &= \zero \end{align} as expected.

A similar computation shows that (\ref{metricdiv}) holds in any two-dimensional geometry. We have $$d\rr = \sigma^i \,\ee_i$$ so that $${*}d\rr = {*}\sigma^i \,\ee_i = \epsilon^i{}_j \,\sigma^j \,\ee_i$$ and we compute \begin{align} d{*}d\rr &= \epsilon^i{}_j \,d\sigma^j \,\ee_i - \epsilon^i{}_j \,\sigma^j\wedge d\ee_i \nonumber\\ &= -\epsilon^i{}_j \,\omega^j{}_k\wedge\sigma^k \,\ee_i - \epsilon^i{}_j \,\sigma^j\wedge\omega^k{}_i\,\ee_k \nonumber\\ &= -\sum\limits_{ij} \left( \epsilon^i{}_j \,\omega^j{}_i\wedge\sigma^i \,\ee_i + \epsilon^i{}_j \,\sigma^j\wedge\omega^j{}_i\,\ee_j \right) \nonumber\\ &= -\sum\limits_{ij} \left( \epsilon^i{}_j \,\omega^j{}_i\wedge\sigma^i \,\ee_i + \epsilon^j{}_i \,\sigma^i\wedge\omega^i{}_j\,\ee_i \right) \nonumber\\ &= -\sum\limits_{ij} \left( \epsilon^i{}_j \,\omega^j{}_i\wedge\sigma^i \,\ee_i + \epsilon^i{}_j \,\sigma^i\wedge\omega^j{}_i\,\ee_i \right) \nonumber\\ &= \zero \end{align} where we have used the fact that $$\epsilon^i{}_j\,\omega^j{}_i = \epsilon^j{}_i\,\omega^i{}_j \qquad\hbox{(no sum)}$$ for each value of $i$ and $j$.

A similar computation can be done in any dimension; the divergence of the metric always vanishes, that is, (\ref{metricdiv}) is always true.