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My book is From Calculus to Cohomology by Ib Madsen and Jørgen Tornehave. I didn't study much of the definitions or theorems in Chapters 1 to 10 that might already be found in An Introduction to Manifolds by Loring W. Tu. I mostly assume the concepts are the same until there is evidence otherwise.

I believe I might have come across evidence that disproves one of my assumptions of equivalence of concepts. The assumption I made is:

Let $M$ be a smooth $n$-manifold with dimension. Let $N \subseteq M$. $N$ is what Madsen and Tornehave would call a "domain with smooth boundary" if and only if $N$ is what Tu would call a smooth $n$-manifold with boundary (see context below). $\tag{A}$

Now, this post shows it is not necessarily the case that compact subspaces of $\mathbb R^n$ can be realized as smooth manifolds with boundary. I assume this is not a dimension issue (see context below), so I guess I'll say that not all compact subspaces of $\mathbb R^n$ can be realized as smooth n-manifolds (or k-manifolds) with boundary.


Question: (Definitions given below) Let $K$ be the compact set of Lemma 11.25. How does the proposition (the "(14)") after Lemma 11.25 apply Corollary 11.23 and thus Theorem 11.22 given $K$ is not stated to be a (compact) domain with smooth boundary while Theorem 11.22 assumes a domain with smooth boundary?

Possible answers:

  1. $K$ disproves (A): $K$ is a domain with smooth boundary but not a smooth $n$-manifold with boundary.

  2. $K$ does not disprove (A), but (A) is still false: $K$ is both a domain with smooth boundary and a smooth $n$-manifold with boundary, but, in general, domains with smooth boundary and smooth $n$-manifolds with boundary are not equivalent.

  3. $K = \{||x|| \le 2\}$, and $K$ does not disprove (A), and (A) is true: $K$ is both a domain with smooth boundary and a smooth $n$-manifold with boundary.

  4. $K \ne \{||x|| \le 2\}$, and $K$ does not disprove (A), and (A) is true: $K$ is both a domain with smooth boundary and a smooth $n$-manifold with boundary.

  5. The proposition is incorrect and should have assumed $K$ is a (compact) domain with smooth boundary.


Context:

  • On (A) and on dimensions:

    • Definition 10.5, the definition of "domain with smooth boundary" of Madsen and Tornehave.

    • Tu Definition 22.6 (part 1) and Tu Definition 22.6 (part 2), of $n$-manifold with boundary (and manifold with boundary), where $\mathcal H^n := \{x \in \mathbb R^n | x_n \ge 0\}$.

    • Note that Madsen and Tornehave have $x_1 \le 0$ while Tu has $x_n \ge 0$.

    • Also note that Tu's manifolds are not the same as his $n$-manifolds because not all his manifolds' connected components have the same dimension (see this and this), but for Madsen and Tornehave, I believe their manifolds and domains with smooth boundary have (uniform) dimensions.

  • Definitions of index: Definition 11.16, Definition 11.19, Definition 11.21

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    2.) Is false. Plenty of compact subsets of $\mathbb R^n$ are not manifolds at all, and thus are not domains with smooth boundary. – Rylee Lyman May 13 '19 at 04:05
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    From my cursory reading, the only way that (A) could be false is if there were some technical difference between the author's definitions of submanifold. – Rylee Lyman May 13 '19 at 04:07
  • @RyleeLyman Oh I just edited the post. Thanks. Anyway, I took that off already. –  May 13 '19 at 04:07
  • @RyleeLyman Thanks! I guess this isn't a technical difference issue, then. –  May 13 '19 at 04:07
  • @RyleeLyman Wait the conjecture in the previous edit was that I know not all compact subspaces of $\mathbb R^n$ are $k$-manifolds with or without boundary, but perhaps they are domains with smooth boundary thus disproving the equivalence in $(A)$. Also, if $(A)$ is true, then why does not manifold without boundary imply not a manifold with boundary? I think it should be the converse: Not a manifold with boundary implies not a manifold without boundary. –  May 13 '19 at 04:11
  • @RyleeLyman Wait, is $K ={||x|| \le 2}$? –  May 13 '19 at 04:15
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    this $K$ is both an $n$-manifold with boundary and a smooth domain with boundary. I think it's apparent that it would help you to figure out why. – Rylee Lyman May 13 '19 at 10:29
  • @RyleeLyman So we can actually strengthen Lemma 11.25 from "compact set" to "compact domain with smooth boundary"? Thanks! –  May 13 '19 at 21:51
  • This would be a weakening of Lemma 11.25. Compact domains with smooth boundary in particular have nonempty interior, while arbitrary compact sets do not have to have nonempty interior. :) – Rylee Lyman May 14 '19 at 00:15
  • @RyleeLyman Why am I weakening? I mean that Lemma 11.25 says $F|_K^c = F|_K^c$ for a compact set $K$. If I were to say that $F|_K^c = F|_K^c$ for a compact set $K$ that is also a compact manifold with boundary, then it sounds like I am saying more. Also, are you talking about topological interior for both? –  May 16 '19 at 03:18
  • @RyleeLyman Saying more is strengthening, or not actually? –  May 16 '19 at 07:06

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