Is Zero Curvature Space Equivalent to Flat Space in General Relativity?

[Orodruin]

Perhaps I can not express my opnion clearly, what I stressed is just that: to a vector field $v(x)$, and we can express $v(x)$ using basis $e^\mu(x)$.then
$$v(x)=v_\mu(x)e^\mu(x)$$
also we can express $v(x)$ using basis $e'^\mu(x)$, and
$$v(x)=v_\mu(x)e^\mu(x)=v'_\mu(x)e'^\mu(x)$$
then because all the basis $e^\mu(x)$ are coordinate basis, so sometimes when we express $v(x)$ from one basis to another, then the changes of components can not be equivalantly described as change under coordinate transformation

I am sorry, it is impossible to deduce what you are trying to say here. If anything, this post made your argument murkier, not clearer.

 

[Orodruin]

In 2D just extend @Orodruin 's example of the 1D circle to a 2D cylinder surface

Actually, as I hinted to in a later post (#17), the cylinder is not as easy to get as the circle as the cylinder does admit a global coordinate system (being homeomorphic to $\mathbb R^2 \setminus \{0\}$). The problem becomes showing that there is no way to arrange such a global coordinate system such that the metric becomes diagonal with the diagonal entries equal to one.

Edit: Just to be a little more specific. The global coordinate system on the cylinder $x^2+y^2=r^2_0$ as a submanifold of $\mathbb R^3$ using coordinates $\xi$ and $\eta$ on $\mathbb R^2 \setminus \{0\}$ can be constructed as
$$
x = \frac{r_0 \xi}{\sqrt{\xi^2 + \eta^2}}, \quad
y = \frac{r_0 \eta}{\sqrt{\xi^2 + \eta^2}}, \quad
z = \frac{r_0}{2} \ln\left(\frac{\xi^2 + \eta^2}{r_0^2}\right).
$$
Of course, in these coordinates, the induced metric from the embedding in $\mathbb R^3$ does not take the $\delta$ form, but it shows that a global coordinate system exists, making the argument a bit muddier than for the circle.

 

[Jianbing_Shao]

What do you mean by "it can change" as it goes around the loop? To what are you comparing it?

I know where I was wrong, thanks ! everyone

 

[Orodruin]

I know where I was wrong, thanks ! everyone

Please tell us. We would like to make sure that you have gotten it right. (It happens more than you would think that people tell us they have gotten it, only to return with similar misconceptions a few days later.)

 

[Jianbing_Shao]

I am sorry, it is impossible to deduce what you are trying to say here. If anything, this post made your argument murkier, not clearer.

Perhaps it can be stated that not all metric $g_{\mu\nu}(x)$ can globally transform to $\eta_{\mu\nu}$ under coordinate transformation.

 

[romsofia]

Another way to start classifying random metrics, and if they're flat, is to look into Petrov classifications if you have not OP.