Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
@DmitryK: No, the description I gave above very much uses the specific parametrization to simplify the conditions for an infinitesimal deformation. I don't see how to generalize this to an arbitary surface.
Note that your $C(n)$ is also the smallest number such that $\mathbb{R}^n$ has a bounded, smooth isometric embedding into $\mathbb{R}^{C(n)}$: As soon as you have such an isometric embedding into a ball of some radius $R>0$, you can construct one into a ball of radius $\epsilon$ for any $\epsilon>0$.
Because the system is invariant under the solvable $2$-parameter Lie group of transformations of the form $$F(x,y,t) = (x/p,y/p,pt+q)$$ for constants $q$ and $p\not=0$, it is (parametrically) solvable by quadrature, so there will be explicit formulae, and they won't be very complicated in form, but they may not be elementary functions.
@FriederLadisch: You ask a good question, as this is not obvious. I remember that that argument was tedious, but straightforward. One has to prove that, if $B^O_PB^P_QB^Q_O$ fixes a ray, then it must be the identity, so one assumes that it fixes a ray and uses that ray to show that it fixes all the rays. I don't remember the argument now (that was 5 years ago), but I'll try to reconstruct it.
@dennis: I can, but it would take a while to set up the notation properly. The equation is quadratic in the first derivatives of $g$, and it also involves the first derivatives of $\mathcal{R}$. It is the unique new first order equation that shows up among the derivatives of the 9 equations on $g$ already listed. I never published my calculations, but I shared them with DeTurck and Yang. The equation is described more or less explicitly in their paper that I referenced above, so I recommend that you have a look at that paper.
@SamBlitz: No, there's no nondegeneracy condition in his Prop 2.1, but his assumption is that you have found a normal vector $A$ satisfying $A\cdot I\!I = 0$. This is not a condition purely on the differential invariants of the submanifold. For example, for curves (i.e., $d=3$ in your convention), this is not the condition that the torsion $\tau$ vanish, it's the condition that there exists a normal vector to the curve that is perpendicular to all of the principal normals of the curve. This implies $\tau=0$, of course, but it's not equivalent to it, as the example I gave shows.
Have a look at the following paper and its references: R. B. Gardner, New viewpoints in the geometry of submanifolds of $\mathbb{R}^n$, Bulletin of the AMS 83 (1977), 1–35. I think you'll find it helpful. (R. B. Gardner was my thesis advisor.). Certainly, if $M^{d-2}\subset\mathbb{R}^d$ lies in a hyperplane, the rank of the second fundamental form is at most one, and, if it is $1$ everywhere, then there will be a normal vector field $n$ such that $n\cdot{I\!I}=0$, but that's not sufficient if the nullity of ${I\!I}$ is $d{-}3$, as the case $d=3$ shows.
@CrashBandicoot: Oh...OK. I think that it is probably true that the 'generic' smooth curve in $\mathbb{R}^n$ is not contained in any analytic submanifold of $\mathbb{R}^n$, but I don't know any literature about this.
@RomainGicquaud: Actually, there are many (incomplete) Riemannian manifolds $(M^n,g)$ that admit a function $f$ that satisfies $\mathrm{Hess}(f) = g$. For example, let $(N^{n-1},h)$ be any Riemannian manifold, let $M= \mathbb{R}^+\times N$, and let $g = \mathrm{d}r^2 + r^2\,h$ be the usual cone metric for $(N,h)$. Then $f = \tfrac12 r^2$ has $\mathrm{Hess}(f) = g$. However, $(M,g)$ won't be complete, and including a point for $r=0$ gives a complete smooth metric only if $(N,g)$ is a unit sphere in $\mathbb{R}^n$.
I think you are essentially asking whether every smooth Riemannian manifold be locally conformally harmonic, which would imply the weaker statement that every smooth Riemannian manifold be conformally Einstein. This is already not true for dimension $n=3$, since, in this dimension, conformally Einstein implies conformally flat, and not all Riemannian $3$-manifolds are conformally flat.
A simpler formula for the map of $V_{2n}\times V_{2n}\to V_{4n-2}$ is this: You just send $F\wedge G$ to $\mathrm{d}F\wedge\mathrm{d}G$ which is a $2$-form in the variables $x_1,x_2$ and then take the coefficient of the basis $2$-form $\mathrm{d}x_1\wedge\mathrm{d}x_2$. In other words, you send $F\wedge G$ to the Poisson bracket $\{F,G\}$.
