Does it show that a coordinate transformation occurrs, and how?
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"a degenerate metric indeed violates one of the axioms of GR, namely the equivalence principle" – bolbteppa Jul 28 '19 at 16:47
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Zero determinant usually means some manner of singularity, either real or coordinate. You should check for scalars there. – Slereah Jul 28 '19 at 16:48
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2Why in the world would someone downvote this perfectly reasonable question? – Jul 28 '19 at 17:30
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If the metric has a zero determinant at some point or surface, then the metric is changing signature at that point, e.g., from $+---$ to $0---$. It's conceivable that spacetime would actually do this. However, it would violate the equivalence principle, and the standard mathematical machinery of GR breaks down in this situation, because everything is predicated on the assumption that we can use the metric to raise and lower indices at will.
If you encounter a spacetime, described in some coordinates, where this is happening, then the more likely situation is that you're looking at it using an inappropriate choice of coordinates. If so, then by Sylvester's law the transformation to the well-behaved coordinates will have to be singular.
I have a worked-out example of this kind of thing in section 6.4 of my GR book, which is free online: http://www.lightandmatter.com/genrel/
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Ben Crowell's answer correctly covers the case in which by "space" you mean the full-dimensional spacetime. However, the induced metric on a submanifold of a perfectly well-behaved spacetime can be singular. For example, as discussed in this answer, the induced metric on a null hypersurface is always singular.
tparker
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