Can a Riemannian submanifold effectively act as a local tangent space?

Question

I have been studying rudimentary gauge gravity, using the soldering equation to show how local P-translations can be related to global Lorentz and coordinate transformations. My question relates to the tangent space $T_{p}M$ at a point $p$ on a manifold $M$, and the properties of hypersurfaces (submanifolds) of $M$, which I'll denote $X$. Essentially, if $X$ and $M$ share a point $p$, then is the the exponential map $exp:T_{p}M\rightarrow{M}$ equivalent to $exp:T_{p}X\rightarrow{X}$?

Answer 1

I think this is what you are trying to ask: Suppose that $M$ is a Riemannian manifold, $X\subset M$ is a hypersurface equipped with the Riemannian metric induced from $M$. Fix a point $p\in X$. We then have two exponential maps: $\exp_p^M: T_pM\to M$ and $\exp_p^X: T_pX\to X$. (A caveat: The metrics might be incomplete and the exponential maps might be defined only on neighborhoods of $0$ in the respective tangent spaces. I will ignore this issue.) We have the natural inclusion $T_pX\subset T_pM$. Then your question is:

Is it true that $\exp_p^X$ equals the restriction of $\exp_p^M$ to $T_pM$?

Since you do not seem to know what the word "restriction" means, here is a reformulation of the above question:

Is it true that for every vector $v\in T_pM$ (where both $\exp_p^M$ and $\exp_p^X$ are defined) we have $\exp_p^X(v)=\exp_p^M(v)$?

The answer to this question is in general negative. The reason is the failure of $X$ to be totally geodesic in $M$. (A submanifold $X$ of a Riemannian manifold $M$ is called totally geodesic if each geodesic in $X$, with respect to the metric of $X$, is also a geodesic in $M$.) As a concrete example, consider $M={\mathbb R}^n$ with the standard flat Riemannian metric (with the metric tensor $g_{ij}(x)=\delta_{ij}$). Let $X\subset M$ be a round sphere. Then for each point $p\in X$, $v\in T_pX$, $\exp^M_p(v)\ne \exp_p^X(v)$, unless $v=0$.

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Addendum: Much of the trouble that physicists have with explaining themselves to mathematicians comes down to the matter of language, since mathematics is both science and language. Here is a fragment taken from Richard Feynman's book "Surely you are joking, Mr. Feynman." There, Feynman explains that he had invented alternatives to the standard math notation and what was the result:

I thought my symbols were just as good, if not better, than the regular symbols - it doesn't make any difference what symbols you use - but I discovered later that it does make a difference. Once when I was explaining something to another kid in high school, without thinking I started to make these symbols, and he said, "What the hell are those?" I realized then that if I'm going to talk to anybody else, I'll have to use the standard symbols, so I eventually gave up my own symbols.