11
$\begingroup$

The Klee Trick allows one to find an $\mathbb{R}^m$ where two embeddings of same compact metric space have homeomorphic complements. More precisely, given two embeddings of a compact metric space $K$ into $\mathbb{R}^n$, $f_{n,1}$, $f_{n,2}$, we can construct two embeddings of $K$ into $\mathbb{R}^{2n}$ such that the images of $K$ are equivalent under a homeomorphism of $\mathbb{R}^{2n}$. (The trick itself produces an isotopy of the the two embeddings in $\mathbb{R}^{2n}$.)

However, I was wondering if there is an example of a compact metric space $K$ and a pair of embeddings $f_{n,1},f_{n,2}:K \rightarrow \mathbb{R}^n$ such that the embeddings are not equivalent under a homeomorphism of $\mathbb{R}^{n+m}=\mathbb{R}^n \times\mathbb{R}^m$ for all $m < n$.

To clarify with a non-example, any two (tamely) embedded knots in $\mathbb{R}^3$ will be isotopic in $\mathbb{R}^4=\mathbb{R}^3 \times\mathbb{R}$, and so for embeddings of $S^1$ into $\mathbb{R}^3$, the isotopy obtained from the Klee trick is not optimal.

$\endgroup$
4
+50
$\begingroup$

Are the embeddings required to be isometric embeddings? If not, then what about including three points into $\mathbb{R}$ in two ways, so that the middle point of the three changes? More explicitly, let $K=\{a,b,c\}$ and define $f(a)=0$, $f(b)=1$, $f(c)=2$, whereas $g(a)=1$, $g(b)=0$, $g(c)=2$. There isn't any self-homeomorphism of $\mathbb{R}$ whose composition with $f$ is equal to $g$.

$\endgroup$
  • $\begingroup$ Thanks! This is probably the simplest example of what I was looking for. Its also fairly satisfying to see the isotopy in $\mathbb{R}^2$ of the two embeddings $(f(x),0)$ and $(g(x),0)$. However, it would be nice to have an example of where $n\ne 1$ so $\mathbb{R}^{2n}\ne \mathbb{R}^n \times \mathbb{R}$. $\endgroup$ – Neil Hoffman Jan 17 at 15:50
  • $\begingroup$ Mea Culpa: I misunderstood your answer and I voted it down. Now, the only way to remove my vote down and vote it up was to edit your answer so I had to do it. I just replaced whereas with where and then where back with whereas. Sorry for that. I hope you appreciate my effort and honesty :) $\endgroup$ – Piotr Hajlasz Jan 17 at 23:45
  • $\begingroup$ I couldn't think of a way to make any higher-dimensional examples, but my guess is that the place to look is when $K$ is either $n$ or $n-1$-dimensional. $\endgroup$ – IJL Jan 18 at 16:04
  • $\begingroup$ Piotr: thanks for your message and correcting yourself with an edit. $\endgroup$ – IJL Jan 18 at 16:08

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.