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Our original plane has curvature $K$, which can be any real, and may not be constant, though as a function of location it will be smooth. Assume we have some Euclidean projected plane which is a conformal projection of the original plane. At every point in the projected plane we define a scale $S$, where if we start moving from the corresponding point in the original plane at a speed of $1$, we will be moving at a speed of $S$ in the projection. Since the projection is conformal, the scale will be the same regardless of what direction we move in.

We know $S$ as a function of location and $P$ the point of interest in the projected plane. We can get other information like the derivative of $S$ if needed. We do not know $K$, or what point $P$ corresponds to in the original plane. Can we solve for $K$ at that point?


For constant $K$: Consider a reference curve in the original plane, the signed distance along the reference curve $u$ and the signed distance from the reference curve $v$. Then we can name $P$ alternatively with coordinates $(u,v)$. The reference curve is chosen so any two points with the same $u$ produce the same $S$. The actual reference curve can be anything because we only care about local changes to $S$. We can equivalently frame $S$ in terms of $v$ - if $f(v)$ is the signed distance from the reference curve in the projected plane, then $S(P)=f'(f^{-1}(P_v))$ ignoring sign.

We can scale the projected plane, making the substitution $f(x)=cg(x)$. This results in $S(P)=cg'(c(\frac{1}{c}g^{-1}(P_v)))=cg'(g^{-1}(P_v))$. Thus any constant factor of $S$ is equivalent.

$K=0$ with identity mapping, with a straight line as the reference curve: $f(v)=v$. Then $S(P)=1$.

$K=0$ projected using circle inversion with the unit circle, with the unit circle as the reference curve with $v$ offset by $1$: $f(v)=\frac{1}{v}$. Then $S(P)=r^2=x^2+y^2$. This shows a non-constant $S$ can still result in $K=0$.

$K=1$ modelled by the unit sphere, projected using stereographic projection from $(0,0,-1)$ to $z=1$, with $(0,0,1)$ as the reference curve: $f(v)=\tan(\frac{v}{2})$. Then $S(P)=\frac{1+r^2}{2}=\frac{1+x^2+y^2}{2}$.

$K=1$ using Mercator projection, with the equator as the reference curve: $f(v)=\int{\sec(v)}=\ln(\tan(\frac{v}{2}+\frac{\pi}{4}))$. Then $S(P)=\cosh(y)$.

$K=-1$ projected using Poincaré disk, with $(0,0)$ in the projected plane as the reference curve: $f(v)=\tanh(\frac{v}{2})$. Then $S(P)=\frac{1-r^2}{2}=\frac{1-x^2-y^2}{2}$.

$K=-1$ projected using Poincaré half-plane, with the horocycle corresponding to $y=1$ in the projected plane as the reference curve: $f(v)=e^v$. Then $S(P)=y$.

Across these examples, there isn't much similarity, so there's probably some reframing needed before an answer will pop out.

This is a big question, so please excuse any small mistakes or bad notation in the examples, unless the result is wrong.


We can also scale the entire original plane and change $K$ accordingly, so the same point in the projected plane now has a different corresponding $K$. However, the relative values are preserved, as $K\propto R^{-2}$ and $R$ scales uniformly, thus for two points $P$ and $Q$, $\ln(\left|\frac{K(P)}{K(Q)}\right|)$ is preserved. The directions are also preserved, as Euclidean remains Euclidean, hyperbolic remains hyperbolic, and elliptic remains elliptic. We could say then that the value of $K$ is not normalized, as multiplying all $K$ by $c$ for some $c>0$ produces an equivalent result. It doesn't invalidate the original question, we just need to be able to derive some $K(P)$ which is consistent.

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