TITLE: Is the scalar curvature the only isometric invariant of a Riemannian 2-manifold?
QUESTION [11 upvotes]: Given two Riemannian Manifolds of dimension 2, and a point on each. If the scalar curvatures are isomorphic (as functions) in some neighbourhoods of these points, are then the manifolds necessarily locally isometric?

REPLY [16 votes]: The answer is 'no' for simple reasons:  If the two Riemannian surfaces $(M_1,g_1)$ and $(M_2,g_2)$ have Gauss curvatures $K_1:M_1\to\mathbb{R}$ and $K_2:M_2\to\mathbb{R}$ respectively and there happen to be points $p_i\in M_i$ such that $K_1(p_1) = K_2(p_2)$ and such that $dK_i$ is nonvanishing at $p_i$, then there will always be $p_i$-neighborhoods $U_i\subset M_i$ and a local diffeomorphism $\phi:U_1\to U_2$ with $\phi(p_1)=p_2$ such that $K_1 = K_2\circ\phi$ on $U_1$.  This is just basic calculus.  Of course, there is no reason for $\phi$ to be a local isometry.
However, now set $L_i = |dK_i|^2_{g_i}$ and suppose, in addition, that $L_1(p_1)=L_2(p_2)$ and that $dK_i\wedge dL_i$ is nonvanishing in a neighborhood of $p_i$ for $i=1,2$.  Then there will be $p_i$-neighborhoods $U_i\subset M_i$ and a unique local diffeomorphism $\phi:U_1\to U_2$ with $\phi(p_1)=p_2$ such that $K_1 = K_2\circ\phi$ and $L_1 = L_2\circ\phi$ on $U_1$.  Using $\phi$, you can now compare $g_1$ with $\phi^*(g_2)$.  If these are equal on $U_1$, then the two metrics are (obviously) locally isometric.  If these are not equal, then there is no local isometry between the two metrics that carries $p_1$ to $p_2$.  
If you happen to have the bad luck that $dK_i\wedge dL_i$ vanishes identically on a neighborhood of $p_i$ (while, still $dK_i$ is nonvanishing), then, locally, one can write $L_i = f_i\circ K_i$ for some (essentially unique) functions $f_i$, and these will have to be equal in a neighborhood of $K_i(p_i)$ or there is no isometry.  However, even if this does hold, there might still be no isometry, you have to go to higher order.
For example, if $K_1(p_1) = K_2(p_2)$, each $dK_i$ is nonvanishing on a neighborhood of $p_i$, and there exists a function $f$ on a neighborhood of $K_i(p_1)$ such that $L_i = f\circ K_i$ on some neighborhood of $p_i$, then you might consider the function $J_i = \Delta K_i$ and hope that this is independent of $K_i$, and, if so, ask whether the map $\phi$ that satisfies $(K_1,J_1) = (K_2\circ\phi, J_2\circ\phi)$ is an isometry.  Etc.
The point is, though, that, after a finite number of tests of this kind (assuming that some mild nondegeneracy conditions hold), you will be able to completely determine whether the metrics are locally isometric in neighborhoods of the points $p_i$.