'Complex dimensions' in a metric In Special Relativity the metric is (with $\eta=\text{diag}(1, -1, -1, -1)$)
$$
\text{d}s^2 = \text{d}t^2 - \text{d}\mathbf{x}^2.
$$
What sets time apart from space in this equation is the "$-$" in front of the space differential. Is there ever any context in which a dimension has a complex number in front of its differential? e.g.
$$
\text{d}s^2 = \text{d}t^2 - \text{d}\mathbf{x}^2 +i\text{d}q^2
$$
for some dimension $q$. If so, what does this correspond to physically?
 A: You can certainly have a metric like $\operatorname{diag}(7i,-7i,-7i,-7i)$, and the physical interpretation is no different than in the case of a standard real-valued metric. Multiplicative factors in the metric are unobservable.
If the relative phases of the different components are nontrivial, then you would have to have to say what those phases meant. For example, in AC circuits, an impedance of $i$ ohms means that the voltage leads the current by 90 degrees, while an impedance of $-i$ means that the voltage trails. Because the complex number system is isomorphic under an interchange of $i$ and $-i$, any physical application is going to need to provide a physical interpretation that would distinguish them. I don't see what that would be here.
If $ds^2$ is real, then $ds$ has two values that differ by a sign, and we know what that sign means -- it's just a matter of choosing a time orientation. But if $ds^2$ is complex, then the two values of $ds$ do not necessarily differ by just a sign. They could differ by some nontrivial phase. You would have to impute some meaning to that.
I guess if you had in mind quantum gravity, you could imagine that the metric would become an operator in a Hilbert space. But the metric is not an observable, only the curvature is. The curvature emerges only from the second derivatives of the metric.
