Where does the minus sign appear from in the metric tensor? Trying to understand Schutz's AFCIGR, where does the minus sign appear from in the metric tensor?

I understand that this expresses the invariance of the spacetime interval. Schutz says (I think) that the metric is a (0,2) tensor. I assume that means it is the product of two one-forms, so presumably one of these one-forms has a -ve time component. What does that mean? Why don't both one-forms have a -ve time component? Looking at a Minkowski diagram what is a straightforward way to understand/visualise those one-forms? At my level, two things have been multiplied together to give a 4x4 matrix which has -1 in the top left corner. What are those two things and why do they give a -1 time component?
 A: The metric is a symmetric bilinear form
$$
ds^2 = \eta_{ij} dx^i dx^j = -dt^2 + dx^2 + dy^2 + dz^2
$$
Hence the minus sign is not a property of the one-forms $dx^i$ but of the coefficients $\eta_{ij}$
A: Although one can think of the metric as about infinitesimal distances, it can also be thought of in a dual way, as about differential equations. Thus, with your sign convention for the metric, the wave equation is
$$g^{ij}\frac{\partial^2\phi(x)}{\partial x^i\partial x^j}=-\frac{\partial^2\phi(x)}{\partial t\partial t}+\frac{\partial^2\phi(x)}{\partial x\partial x}+\frac{\partial^2\phi(x)}{\partial y\partial y}+\frac{\partial^2\phi(x)}{\partial z\partial z}=0.$$
In this POV, the metric determines the differential equations that are satisfied by the fields that are introduced in the space-time. The Laplace equation has very different properties than does the wave equation.
There are formalisms in which d$t$ is replaced by id$t$, so that the metric can be uniformly positive. However most people keep this as a mathematical trick up one's sleeve, not as a fundamental part of the construction of a theory. There was a period in the development of GR when imaginary time was much more commonly used than it is now.
