Consider a 2-dimensional rectangular plane sheet of paper $S \subset \mathbb{R}^2$, then push both small sides towards the middle to form a Gaussian-like surface, that we call $D$, see the figure below for only one wave. Unless I am mistaken, this is a developable surface since its Gaussian curvature is everywhere zero.
We don't push the ends to close to each other, so that $D$ is also the graph in $\mathbb{R}^3$ of a function $f(u)$. Denoting by $U$ the parameter space $(u,v)$, the map $F$ associated the graph is: $$ F : U \to \mathbb{R}^3 : (u,v) \mapsto (u,v,f(u)). $$ There is also a map $G$ such that: $$ G : S \to U : (x,y) \mapsto (u(x,y),v(x,y)), $$ and $$ F \circ G : S \to \mathbb{R}^3 : (x,y) \mapsto (u(x,y),v(x,y), f(u(x,y)). $$ The induced Riemannian metric $g_D = \iota^*\bar{g} $ on $D$ is the Euclidean metric of $\mathbb{R}^3$ restricted to vectors tangent to $D$, and the pullback of this metric on the parameter space is given in coordinates by: $$ (F^*g_D)_{ij} = \begin{pmatrix} 1 + (f'(u))^2 & 0\\0 & 1 \end{pmatrix}. $$ Thus, we have 3 Riemannian manifolds with each their Levi-Civita connection: $(S,\bar{g},\nabla_S)$ is the plane sheet with the Euclidean metric, $(U,F^*g_D,\nabla_U)$ is the plane parameter space equipped with the pullback metric of the graph, and $(D,g_D,\nabla_D)$ is the graph with the induced metric from $\mathbb{R}^3$. I am trying to check which of these manifolds are isometric, and I am also trying to verify it by checking that the pullback of the Levi-Civita on each manifold matches the Levi-Civita of another. I am doing these checks in coordinates, using the Christoffel symbols.
My reasoning so far:
$(S,\bar{g})$ and $(D,g_D)$ are isometric if the Jacobian matrix of G is of the form: $$ J_G = \frac{\partial G^i}{\partial x^j} = \begin{pmatrix}\frac{\displaystyle1}{\sqrt{\displaystyle 1 + (f'(u))^2}} & 0\\ 0 & 1\end{pmatrix}. $$ The Jacobian depends on $u$, so this defines it implcitely. In particular, $u$ must be solution of the nonlinear ODE: $$u'(x) = \frac{\displaystyle1}{\sqrt{\displaystyle 1 + (f'(u))^2}}$$ for a given $f(u)$. Thus $u = u(x)$ and $v = y$, and we then have: $$ J_{F \circ G} = \frac{\partial (F \circ G)^i}{\partial x^j} = \begin{pmatrix}u'(x) & 0\\ 0 & 1\\ \displaystyle\frac{d}{dx}f(u(x)) & 0\end{pmatrix} = \begin{pmatrix}\frac{\displaystyle1}{\sqrt{\displaystyle 1 + (f'(u))^2}} & 0\\ 0 & 1\\ \displaystyle\frac{f'(u)}{\sqrt{\displaystyle 1 + (f'(u))^2}} & 0\end{pmatrix}, $$ so that $J_{F \circ G}^T I^{3 \times 3}J_{F \circ G} = I^{2 \times 2}$, i.e., the pullback of $g_D$ matches the Euclidean metric, where $I^{n \times n}$ is the identity matrix in $n$ dimensions. This was expected because developable surfaces are flat manifolds, thus isometric to the Euclidean space. Question 1: How can we show that the pullback of the Levi-Civita connection on $(D,g_D)$ matches the one of $(S,\bar{g})$, which has zero coefficients everywhere? I believe the connection on $(D,g_D)$ is the tangential connection, but I'm not sure how to manipulate it.
Regarding $(S,\bar{g})$ and $(U,F^*g_D)$: For the choice of Jacobian matrix for $G$, we have $J_G^T [F^*g_D] J_G = I^{2 \times 2}$, where $[F^*g_D]$ is the matrix of coefficients of $F^*g_D$, hence $G$ pulls back $F^*g_D$ to the Euclidean metric and is an isometry. However, I cannot show that $G$ pulls back $\nabla_U$ to $\nabla_S$. The Christoffel symbols of the pullback connection are: $$ {}^{G^*\nabla_U}\Gamma^k_{ij} = {}^{\nabla_U}\Gamma^k_{im}\frac{\partial G^m}{\partial x^j}, $$ and to check for metric compatibility with the Euclidean metric, we need to show that: $$ {}^{G^*\nabla_U}\Gamma^l_{ki}\delta_{lj} + {}^{G^*\nabla_U}\Gamma^l_{kj}\delta_{il} = \frac{\partial \delta_{ij}}{\partial x^k} = 0 \Longleftrightarrow {}^{G^*\nabla_U}\Gamma^j_{ki} + {}^{G^*\nabla_U}\Gamma^i_{kj} = 0. $$ Question 2: Only the ${}^{\nabla_U}\Gamma^1_{11}$ term is nonzero and the Jacobian of $G$ is diagonal, so only the ${}^{G^*\nabla_U}\Gamma^1_{11}$ term is nonzero for the pullback connection, so how could the compatibility condition be satisfied?
Sorry for the long post!