# Since not all compact subspaces of $\mathbb R^n$ are manifolds with boundaries, how can we this apply index of a vector field formula?

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?

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

• 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 at 4:05
• 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 at 4:07
• @RyleeLyman Oh I just edited the post. Thanks. Anyway, I took that off already. – Selene Auckland May 13 at 4:07
• @RyleeLyman Thanks! I guess this isn't a technical difference issue, then. – Selene Auckland May 13 at 4:07
• 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 at 10:29