Is the spatial metric that you get by writing the Minkowski Metric in a rotating coordinate system curved or not? Suppose I am watching a rotating ring. My question is will its radius and circumference get contracted due to length contraction.
My logic to solve the problem:
Each small differential length of proper length $dl_0$ should “appear” to be having a length of
$dl = \frac{ dl_0}{\sqrt{1-\omega^2 R^2/c^2}}$
since the velocity of each  element is tangential.
So the observed circumference should be
$C = \frac{ 2\pi r_0}{\sqrt{1-\omega^2 R^2/c^2}}$
Since the circumference has reduced so the radius observed in the inertial frame should shorten too. **But the radius doesn’t shorten because it’s perpendicular to the velocity ( so no length contraction and thus the ratio of circumference to radius is not $2\pi$)
Questions:
The Minkowski Metric in rotating coordinate system is
\begin{equation}
ds^2=(-c^2\bbox[yellow]{+\omega^2(x'^2+y'^2)})dt'^2
+dx'^2+dy'^2+dz'^2
\bbox[yellow]{+2\omega(x'dy'-y'dx')dt'}.
\end{equation}
which is flat because it’s Riemann curvature tensor vanishes identically. But as shown above the spatial metric’s geometry shouldn’t be Euclidean and hence be curved
If I take a constant $t$ slice and put $dt=0$ is the above metric I just get back the Euclidean space metric which I know is actually flat.
But the space metric should have been curved, why isn’t it curved?
 A: There is a good deal of subtlety buried in this question, tracing back to the idea of the synchronization of clocks.  The beginning of this answer may appear to have very little relevance, but bear with it - an understanding of synchronization is an absolute necessity in answering your question.
Synchronicity
To review, in (3+1)-dimensional Minkowski spacetime we can imagine space being spanned by a 3D rectilinear grid of ideal, identical measuring sticks with a clock at each grid point.  Every event can be labeled by grid coordinates $(x,y,z)$ and the reading $t$ of the clock at that point.  Furthermore, each grid point follows a timelike geodesic.
Being different clocks, the reading on the clock at $(x,y,z)$ need not have anything to do with the reading on the clock at $(x',y',z')$.  To resolve this ambiguity, we introduce the Einstein synchronization procedure.  At time $t_1$, we send a light signal from the clock at $(x,y,z)$ to the clock at $(x',y',z')$.  It is received at time $t_2$ (read off the second clock), immediately reflected backward, and received back at $(x,y,z)$ at time $t_3$.  The synchronization procedure says that the two clocks are synchronized if $t_2 = (t_1+t_3)/2$. If we construct the coordinate system as described above with a network of globally synchronized clocks, the infinitesimal line element takes the familiar form $\mathrm ds^2 = -\mathrm dt^2 + \mathrm dx^2 + \mathrm dy^2 + \mathrm dz^2$.
We can see what would happen if we used clocks that weren't synchronized by introducing an offset into neighboring clocks.  Define a new time coordinate $t' = t+ax$, where $a<1$.  In this new coordinate system, $\mathrm ds^2 = -\mathrm dt'^2 + 2a\mathrm dt' \mathrm dx + (1-a^2)\mathrm dx^2 + \mathrm dy^2 + \mathrm dz^2$.  This metric looks very different - in particular, it has a time/space cross term - but this is a reflection of our choice to use clocks which are not synchronized.
Let's examine the synchronization (or lack thereof) of two clocks - one at the origin, and the other at the point $(x_0,0,0)$.  For a light ray moving along the $x$-axis,
$$\mathrm ds^2 = 0 \implies 1 = 2a\dot x + (1-a^2) \dot x^2 \implies \dot x = \frac{-a \pm 1}{1-a^2}= \begin{cases}\frac{1}{1+a} \\ \frac{-1}{1-a}\end{cases}$$
where $\dot x = \frac{dx}{dt}$ is the velocity of the ray.  We see that in these coordinates, the speed of light is anisotropic - the ray travels faster to the left than it does to the right.  Accordingly, if we implement our synchronization procedure (setting $t_1=0$ for convenience), we find
$$t_2 =  x_0(1+a) \qquad t_3 = t_2 + x_0(1-a) = 2x_0$$
$$\implies \frac{t_1+t_3}{2} = x_0 \neq t_2$$
which means that our clocks aren't synchronized.  To correct for this, we would need to subtract an offset
$$\delta t = t_2 - \frac{t_1 + t_3}{2} = ax_0$$
from $t'$, which happily brings us back to our original time coordinate.

Rest Length
If you look at the spacetime diagram for the motion of an extended object, it traces our a worldsheet, not just a worldline. In its rest frame, the sheet looks like this.

To compute its rest length, we can take a spacelike slice of spacetime which represents a set of simultaneous events.  This slice is a spatial submanifold which inherits a Riemannian metric, and the rest length of the object can be computed along that slice. So how do we choose a slice of simultaneity?

The obvious choice is the surface $t=\mathrm{const}$, but that's only true if your clocks are synchronized. In the previous example, the appropriate choice would be $t'=ax+\mathrm{const}$!

Rotating Frames
At last we can address your question. In inertial polar coordinates (sticking to 2+1 dimensions for simplicity) the line element takes the form $\mathrm ds^2 = -\mathrm dt^2 + \mathrm dr^2 + r^2 \mathrm d\theta^2$.  If we define a rotating coordinate $\psi = \theta-\omega t$, we obtain $\mathrm ds^2 = -(1-r^2\omega^2)\mathrm dt^2 + 2\omega r^2 \mathrm dt \mathrm d\psi + \mathrm dr^2 + r^2 \mathrm d\psi^2$.
Notice that these coordinates have a cross term - just like we obtained in our earlier toy model. Just as before, we find that adjacent clocks are not synchronized with one another.  Two clocks at $(r,\psi)$ and $(r+\delta r,\psi)$ are in fact synchronized with one another - but clocks at $(r,\psi)$ and $(r,\psi + \delta \psi)$ are not.  Operationally, this means we can compute the radius of a disk as usual, but the circumference requires more work.
Repeating our previous exercise, we find that light rays traveling around the circumference of the disk obey
$$\dot \psi = -\omega \pm \frac{1}{r} = \begin{cases}\frac{1}{r}-\omega \\ -\left(\frac{1}{r}+\omega\right)\end{cases}$$
and
$$t_2 =  \frac{r\delta \psi}{1-\omega r} \qquad t_3 = t_2 + \frac{r\delta \psi}{1+\omega r} = \frac{2r \delta \psi}{1-\omega^2 r^2}$$
$$\implies \frac{t_1+t_3}{2} = \frac{r\delta \psi}{1-\omega^2r^2} \neq t_2$$
In order to synchronize these clocks, we must adjust the second by subtracting
$$\delta t = t_2 - \frac{t_1+t_3}{2} = \frac{\omega r^2 \delta \psi}{1-\omega^2 r^2}$$
Alternatively, we can measure the length of the rod by taking a slice along which $\mathrm dt \neq 0$ (because again, the clocks aren't synchronized) but rather $\mathrm dt = \frac{\omega r^2 \mathrm d\psi}{1-\omega^2 r^2}$, which yields
$$\mathrm ds^2 = \frac{r^2 d\psi^2}{1-\omega^2 r^2} \implies \mathrm ds = \frac{r \mathrm d\psi}{\sqrt{1-\omega^2 r^2}}$$
which we recognize as the appropriately length-contracted distance.  Repeating this for measuring rods along the entire circumference, we find that $C = \frac{2\pi R}{\sqrt{1-\omega^2 R^2}}$ as anticipated.
Finally, there is a critical difference here from our original toy model.  We might imagine that we could adjust our clocks by defining a new time coordinate of the form $T = t + \frac{\omega r^2 \psi}{1-\omega^2 r^2}$, thereby obtaining a global network of synchronized clocks. This is not so.  Even if you work hard to eliminate cross terms (you'll need to add another counterterm to compensate for the $r$-dependence, but you can work that out as an exercise), this new time coordinate cannot possibly cover the whole manifold, because it doesn't return to itself after $\psi \rightarrow \psi + 2\pi$.  In other words, if you walk around the circumference of a disk synchronizing clocks, you'll get back to your starting point and find that your final clock disagrees with your first one.  For this reason, there is no way to establish a global network of synchronized clocks in a rotating frame.
If we restrict our attention to extremely small regions, it is possible to obtain a small network of synchronized clocks; if you adopt these coordinates in this small region, you can take a time slice to obtain the spatial metric $\mathrm ds^2 = \mathrm dr^2 + \frac{r^2 d\psi^2}{1-\omega^2 r^2}$, yielding non-Riemannian geometry.  However, it must be emphasized that this chart covers only a tiny portion of the plane, and any attempt to extend it to a large area will result in a failure of global synchronization.
