We are speaking about Gauss curvature here, which is, according to the Theorema Egregium, intrinsic.
Yes.
You are speaking about Gaussian curvature, which is a work of abstraction. Since you claim this curvature is observable, I encourage you to present it to me.
Yes, the Theorema Egregium has already been mentioned, but apparently you still don't get it, as you keep babbling about embedding space
The only times I have used the words "embed" and "space" was when momentia forced me to, and they were usually in the context of "I'm not 'embedding' anything here".
and asking "relative to what" the curvature is defined.
Because that's a very crucial question. By the definition momentia referenced, and which he confirmed to be valid, Theorema Ergegium works regardlessly of
how we embed the surface
"in the ambient 3-dimensional Euclidean space". The fact that we're not operating in "the ambient 3-dimensional Euclidean space" in the first place makes it somewhat useless.
Like it or not, Gauss curvature is intrinsic and thus independant from embedding space.
Like it or not, I'm not talking about unobservable abstract claims of the same category as the square root of -1. I don't care about imagination, what-ifs, hypothetical solutions of problems that don't exist, and the like. What I'm talking about is
measurable curvature, keeping in mind the important distinction between optical and mechanical measurements.
If it's not zero, then the earth is not flat, as flat surfaces have Gauss curvature zero.
In "the ambient 3-dimensional Euclidean space", yes. This is not the case.
If it's zero, then the earth is locally euclidean, hence it has an euclidean map.
In "the ambient 3-dimensional Euclidean space", yes. This is not the case.
Please provide such a map.
Same old assumption, same old response coming through. As usual, I agree that
if the Earth is round, it's not flat. However, it's not round.