Re: Given the christoffel symbols (in differential geometry), can I solve for the metric

> metrics [1 0;0 1] and [2 0; 0 2] they most
> definitely are *not* isometric.

huh? They clearly are isometric. Trivially so. If

ds^2=dX^2 + dY^2

and

ds^2=2*dx^2 + 2*dy^2

just choose the bijection (X, Y) = sqrt(2) * (x, y), therefore ((1, 0),
(0, 1)) and ((2, 0), (0, 2)) are isometric. (ie- both are flat
euclidean spaces, and it doesn't matter what coordinates I use to
describe them, much like Germany and Deutschland both describe the same
country).



> Consider also the pseudo-Riemannian metric, [1 0;0 -1]---same Christoffel symbols
> again.

That is a valid point that I didn't think of, and definitely answers
the question that I posed. I am curious as to whether restricting to
positive definite spaces would change the answer though. Clearly all
constant positive definite metrics can be transformed to diag(1), so
the trivial counterexample wouldn't work.

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