Metric properties

Question

Let $f: \Omega \rightarrow \mathbb{R}^3$ be a submanifold in $\mathbb{R}^3$ and also $f' : \Omega' \rightarrow \mathbb{R}^3$ another one. Now if $f(\Omega) \cap f' ( \Omega')$ is a regular curve $c: I \rightarrow f(\Omega) \cap f' ( \Omega')$ and we have $T_{c(t)}f = T_{c'(t)}f'.$ Does this mean that along $c$ the metric or the Christoffel symbols of the two surfaces are the same?

Answer 1

For convenience I'll write $N = f(\Omega)$ and $N' = f'(\Omega')$.

It's evident that at any point along $c$, say $c(t)$, the inner product on $T_{c(t)}N$ and $T_{c(t)}N'$ agree, since both are the restriction of the ambient inner product on $\mathbb{R}^3$.

I'm not sure what it would mean for the Christoffel symbols to "agree", since there doesn't seem to be a coherent way to choose coordinates on $N$ and $N'$ in a way that "matches up" to check for agreement. However, one could ask about the covariant derivative $\nabla_{c'(t)} X$ of a vector field $X$ along $c$ (with values in $T_{c'(t)}N = T_{c'(t)}N'$), which makes sense taking this derivative both in $N$ and in $N'$. These covariant derivatives agree. The proof is to note that, for example, the connection $\nabla^N$ on $N$ is given by $$ \nabla^N_Y X = \text{proj}_{TN} \big(\nabla^{\mathbb{R}^3}_Y X \big), $$ whenever $X$ and $Y$ are such that this makes sense; and of course a similar formula holds for $N'$. If $Y = c'(t)$ and $X$ is a vector field along $c$ with values in $T_{c(t)} N = T_{c(t)}N'$ then this formula makes sense for both $N$ and $N'$, so that $$ \nabla^N_{c'(t)} X = \nabla^{N'}_{c'(t)} X $$ as claimed.

On the other hand, the sectional curvatures of $N$ and $N'$ need not agree along $c$. For example, let $f : \mathbb{S}^2 \to \mathbb{R}^3$ be the standard embedding of the sphere of radius 1 centered at the origin, and let $f': \mathbb{S}^1 \times \mathbb{R} \to \mathbb{R}^3$ be the infinite cylinder of radius 1 with, say, the x-axis as its axis. The submanifolds $f$ and $f'$ intersect in the circle $x = 0, y^2 + z^2 = 1$, and the tangent spaces agree on this circle, but $\mathbb{S}^2$ has positive curvature, whereas the cylinder is flat.