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Let $M$ be a compact topological n-manifold. Suppose we are given a locally flat embedding $M \subset \mathbb{R}^{n+k}$. This induces a metric on $M$ by restriction. Is it true that for $\epsilon$ small enough, the close $B(x,\epsilon)$ is homeomorphic to $D^n$ for all $x \in M$?

My thinking is that the point of the locally flat conditions is that small neigborhoods of points look like $(\mathbb{R}^{n+k},\mathbb{R}^n)$ so small balls should have a decomposition like $(D^{n+k},D^n)$. But perhaps the boundary of $D^n$ causes issues.

If not, is there some metric on $M$ with this property?

I am interested in using this to show that $F(M,k)$, configurations of k points in $M$, for any compact topological manifold $M$ can be compactified by adding a boundary. This is known for smooth $M$; one may simply remove a regular neighborhood of the fat diagonal. However, I am sure that treating regular neighborhoods of the fat diagonal must be delicate in low dimensions for topological manifolds.

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  • $\begingroup$ When $M$ is compact, $M^k$ is a compactification of $F(M,k)$. I think you need to be a little more specific about what you might want "compactification" to mean. The closure of $F(M,k)$ in $M^k$ is $M^k$, and this is a compact space, so it certainly is a compactification. $\endgroup$
    – Ryan Budney
    Apr 21, 2022 at 3:21
  • $\begingroup$ @RyanBudney Thanks, I've clarified to mean that the compactification should just be adding boundary points. $\endgroup$
    – Connor Malin
    Apr 21, 2022 at 3:22
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    $\begingroup$ An embedding doesn't give you such a metric. Take the embedding of the line into the plane given by $x\mapsto x\sin(1/x)$ and modify it in a way that the waves ``tip over'' towards the origin. You can do this in a way that all balls around zero become disconnected. $\endgroup$
    – user473423
    Apr 21, 2022 at 7:07
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    $\begingroup$ @Echo Is that embedding locally flat? $\endgroup$
    – Connor Malin
    Apr 21, 2022 at 12:59
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    $\begingroup$ Yes, as the neighbourhood, that you need for local flatness can be chosen not to be a ball itself. Already the construction tells you that it is, as the deformation can be reversed. $\endgroup$
    – user473423
    Apr 22, 2022 at 7:12

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