$ds^2$ is not a scalar.

A symmetric rank-2 covariant tensor.

@mmeent it is not, the indexes are contracted! do you know anything about the subject?

The $dx^\mu$ are not components of a vector, they are basis covectors.

in curved spacetime light moves locally at c

@mmeent they have contravariant indexes and contract with the covariant ones. if s^2 is a second rank tensor, where are the indexes?

are you just troling me?

If it had indices it would represent the components of a tensor. $ds^2$ is an abstract representation of a tensor.

@mmeent you got it all wrong

No, I did not. Please go back to your introduction text differential geometry. You can come back and apologize after.

@Wolphramjonny The components of a vector have an upstairs index, but a vector itself has downstairs indices. $\partial_\mu$ is a set of basis vectors for the tangent space. Then the vector with components $A^\mu$ is given by $A^\mu \partial_\mu$. Similarly, a set of basis one-forms is given by ${\rm d} x^\mu$. The expression ${\rm d} s^2=g_{\mu\nu} {\rm d} x^\mu {\rm d} x^\nu$ is a basis-independent representation of the metric.

I think we are speaking different languages here. Let me read this again tomorrow

@Wolphramjonny A good free reference is Sean Carroll's lecture notes on GR, arxiv.org/abs/gr-qc/9712019. See the paragraph after Eq 2.32. It's also worth reading this entire chapter, actually.

@Andrew I think you are confusing the common use of the term metric tensor ($g$) with the interval ($ds^2$) which is a scalar and which is what the OP is refers to. It is not a second rank tensor as $ds^2$ has no components except the scalar interval value making it a zero rank tensor. The argument in Carroll's notes explicitly states it is not a one-form. He states it is a "convenient shorthand for the metric tensor", but not that it is the metric tensor (which is rank 2, not rank 0). No one AFAIK treats $ds^2$ as a rank two tensor or a one-form.

@StephenG There are different conventions and people have different languages, so I am not saying anyone is wrong. But it is a consistent (and common) usage to say that $({\rm dx})^2$ is a tensor product of two one-forms, and to expand the metric as components contracted with tensor products of two one-forms as $g_{\mu\nu} {\rm d} x^\mu {\rm d} x^\nu$, and to denote this combination by $ds^2$. We may be using different interpretations, but I am not confused and my interpretation is not idiosyncratic.

@Andrew It might help if you have a reference to something online and accessible where the use of $ds^2$ in that way is shown unambiguously. Always open to expanding my mind. :-)

@StephenG Some random examples just using famous people (a) Eqs 2.2 and 2.3 of arxiv.org/abs/hep-th/9711200 (b) Eq 1 of arxiv.org/pdf/1910.04670.pdf (c) Eq 2.1 of link.springer.com/article/10.1007/BF02345020 I also think the corresponding discussion in Carroll's book is more explicit than his notes, but I don't have that on hand because of COVID. Anyway it's not really my responsibility to justify something that's explained in most good GR textbooks.

Or see the first equation under "Local coordinates and matrix representations" en.wikipedia.org/wiki/Metric_tensor_(general_relativity). And yes I understand that $ds^2$ appears below that, but people are lazy with notation and you will find sources that use $ds^2$ to refer to the actual metric (what wikipedia calls $g$). The sources I linked above all say "the metric is" and then give $ds^2=...$.

@StephenG Wald is very explicit in his book (pg 23): " Sometimes the notation $ds^2$ is used in place of $g$ to represent the metric tensor, ..."

Or try box 3.2 in MTW.

@mmeent This is exactly the point. None of the authors claim it is the metric. They claim it is often be used as shorthand or representation of the metric (a rank two tensor). But no one actually claims that $ds^2$ is the metric tensor. One of the most annoying things learning tensor calculus for GR is that such careless language is used for what are actually precisely defined mathematical constructions. You are actually claiming that $ds^2$ is a second rank tensor and it most certainly is not. Authors being careless will not change that.

@StephenG If it were a scalar field as you claim you should be able to evaluate it at a point and get a number. $ds^2$ is a precisely defined mathematically object and that object is a symmetric rank 2 tensor field.

@mmeent The metric tensor ($g$) is a second rank tensor. But the quantity $ds^2$ is not a second rank tensor - it is a scalar. Explicitly write out $ds^2$ and $g$ in full and compare them.

Then what number do you get when you evaluate $ds^2$ for Schwarzschild at the point $t=1$, $r=10$ $\phi=2$, and $\theta=3$? @StephenG

@mmeent Don't be silly.

What a pointless conversation.