In (2D) Cartesian coordinates, the Euclidean metric…

$$\begin{bmatrix}

1 & 0 \\

0 & 1

\end{bmatrix}$$

…is flat space. If the diagonal elements are exchanged for other real numbers greater or less than zero, would this still count as flat space even though it is no longer Euclidean? I think the answer is yes but just want to make sure.

Also, when a physicist says that the local tangent space on a manifold is flat, do they imply that the metric is locally Euclidean, or can the diagonal elements be any non-zero real number in that local tangent space?

## Best Answer

1) OP is asking about the use of the word flat metric. It means a pseudo-Riemannian metric (of arbitrary signature) whose corresponding Levi-Civita Riemann curvature tensor vanishes.

2) However, the word

Euclidean spacemay potentially cause confusion among mathematicians and physicists. For a mathematician an Euclidean space is always an affine space, while a physicists often use it as just another word for Riemannian manifold, which is not necessarily affine.In short, the word

Euclideanrefers for a physicist to a positive signature (typically as opposed toMinkowskisignature), while a mathematician use the wordEuclideanto refer to an affine structure.A mathematician calls a pseudo-Riemannian manifold with Minkowski signature for a Lorentzian manifold.