# What is different between coordinates that locally see a Minkowski metric and those that see a Euclidiean metric?

The text I'm reading claims that in an infinitesimally small region of the manifold, we can always find observers that locally see the metric as either Minkowski, $$\eta_{\mu\nu}$$, or Euclidean, $$\delta_{\mu\nu}$$. The observer seeing a Minkowski metric makes sense to me, if we look at a small enough region the curvature is small enough that the metric reduces locally to that of flat space, at least that's what I think. But finding the metric to be Euclidean implies $$diag(1,1,1,1)$$, unless I am wrong about that. My question is in what physical situation could an observer see a Euclidean metric?

• Which text? The first "or" should be an "xor": The manifold is Euclidean xor Lorentzian. Not both. Possible duplicate: physics.stackexchange.com/q/129187/2451 Mar 31, 2020 at 15:15
• Hmmm. What's the text? I guess they may mean that if all velocities are relatively small, then we get Galilean relativity. Mar 31, 2020 at 15:16
• The notes for my GR course, perhaps text was a poor choice of wording, the exact quote is "For this, let us assume that we start with an infinitesimally small region of our manifold. In this region one can find a set of ‘privileged’ observers that locally see the metric as trivial (the LIF for GR). This means that in this coordinate system the metric is diagonal, with +1 or −1 components (two examples are δµν or ηµν) and has vanishing first derivatives" Mar 31, 2020 at 15:17
• I think that the text could be referring to that the spacetime metric is locally Minkowskian $diag(1,-1,-1,-1)$ and then the projected 3D space-type for an instant $t=t_0$ is euclidean $(1,1,1)$ Mar 31, 2020 at 15:22
• Maybe, the notes have consistently used the (-,+,+,+) signature though. Also the indices on the Kronecker delta metric are spacetime indices which unless the notation has been confused suggests the entire metric is the identity. Mar 31, 2020 at 15:26