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In the post Adapted Orthonormal frame on Riemann manifold, the poster offered the outline of a proof for IRM Proposition 2.14. I can fill in the details to make that proof work when the embedded submanifold $M$ has empty boundary. But the proposition allows $M$ to have a boundary. Here is the text of the proposition:

Proposition 2.14 (Existence of Adapted Orthonormal Frames). Let $(\tilde{M},\tilde{g})$ be a Riemannian manifold (without boundary), and let $M\subseteq\tilde{M}$ be an embedded smooth manifold with or without boundary. Given $p\in M$, there exist a neighborhood $\tilde{U}$ of $p$ in $\tilde{M}$ and a smooth orthonormal frame for $\tilde{M}$ on $\tilde{U}$ that is adapted to $M$.

And here is the definition for adapted:

a local frame $(E_1,\dots,E_m)$ for $\tilde{M}$ on an open subset $\tilde{U}\subseteq\tilde{M}$ is said to be adapted to $M$ if the first $n$ vector fields $(E_1,\dots,E_n)$ are tangent to $M$. [$n=\mathrm{dim}\,M,\,m=\mathrm{dim}\,\tilde{M}$]

I can even complete the proof when $M$ has a boundary and $p\in\mathrm{Int}(M)$. But I cannot prove the proposition when $p\in\partial M$. The orthonormal part is not an issue; it's the adapted part that's giving me trouble. I have been trying to use the idea in the proof of Introduction to Smooth Manifolds (ISM) Proposition 5.37, which for a particular choice of smooth charts for $M$ and for $\tilde{M}$ (called $S$ and $M$, respectively, in that proof) shows that the first $n$ coordinates of the coordinate basis for $T_p\tilde{M}$ are a basis for $di_p(T_pM)$ where $i\colon M\to\tilde{M}$ is the inclusion map.

As far as I can tell, this only works because the combination of ISM Theorem 5.8 and ISM Theorem 5.31 has been used to generate a particular smooth chart for $M$. ISM Theorem 5.8 has two parts. The first starts with an embedded submanifold (without boundary) and gets it to satisfy the local $k$-slice condition. The second part starts with a subset satisfying the local $k$-slice condition and provides a smooth atlas with respect to which the subset is an embedded submanifold. If you start with an embedded submanifold, you get a potentially new smooth structure with respect to which it is still an embedded submanifold. ISM Theorem 5.31 says that it's the same smooth structure.

But for the present problem, we have to use ISM Theorem 5.51 which is the embedded submanifold with boundary version of ISM Theorem 5.8. As I mentioned in my post Uniqueness of smooth structures on submanifolds with boundary, I don't know of a uniqueness result (corresponding to ISM Theorem 5.31) for the smooth structure generated by ISM Theorem 5.51 when used in place of ISM Theorem 5.8. This prevents me from creating the necessary smooth chart for $M$ needed to complete use of the idea from ISM Proposition 5.37.

So my question is: how do you prove IRM Proposition 2.14 for the case when $M$ has a nonempty boundary and $p\in\partial M$?

  • It might not be a satisfying answer, but my conclusion about this issue is that we simply have to explicitly choose the smooth structure on $M$ to be the one given by Proposition 5.51. Perhaps there are other smooth structures that make it into an embedded submanifold with boundary. Then we just have to assume all embedded submanifolds with boundary are given the smooth structure compatible with slice charts, and consider the other possibilities as "exotic". – Tob Ernack Aug 14 '23 at 01:56
  • @TobErnack Wow! I wonder if there is a standard name for that kind of embedded submanifold with boundary (sliceable?), and if we have to make the same assumption in other places in IRM (or even ISM), for example IRM Proposition 2.16 (The Normal Bundle). – Jeff Rubin Aug 17 '23 at 15:24
  • If my answer to this question is correct, then actually the idea in Proposition 5.37 should work with no further assumptions in the case of an embedded submanifold with boundary. – Tob Ernack Aug 29 '23 at 20:01

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