7
$\begingroup$

Update: The lead paragraph has been changed to reflect the solution to a related question.

I asked the question Is dimension given by the Klee trick ever sharp? and it has been answered in the affirmative using two embeddings of a $0$-dimensional metric spaces in $\mathbb{R}^1$.

In the interests of trying to better understand the Klee Trick, I thought I might ask a slightly more concrete question along those lines.

Given a metric space $K$ that embeds into $\mathbb{R}^n$, define a $Klee_n(K)$ as the minimum value for $m$ such that for any two embeddings $f_1, f_2 : K \rightarrow \mathbb{R}^n$ and any embeddings $g_1,g_2: \mathbb{R}^n \rightarrow \mathbb{R}^m$ there exists with a homeomorphism $h:\mathbb{R}^m \rightarrow \mathbb{R}^m$ such that $$h(g_1(f_1(K)))=g_2(f_2(K)).$$

(The Klee trick says $Klee_n(K) \leq 2n$.)

For example, $Klee_3(\mathbb{S}^1)=4$, since if $g_1, g_2$ correspond to embeddings of distinct (hyperbolic) knots then we see $Klee_3(\mathbb{S}^1)> 3$, but any two embeddings of $\mathbb{S}^1$ into $\mathbb{R}^4$ can be unknotted via an isotopy.

The other question asked if there was a metric space $K$ such that $Klee_n(K)=2n.$ However, for this question, is there a $K$ such that $Klee_3(K)=5$?

$\endgroup$
  • $\begingroup$ Just to be sure, is it true that $Klee_1([0,1]) =1$? $\endgroup$ – Mathieu Baillif Sep 19 '14 at 7:19
  • $\begingroup$ @MathieuBaillif Yes. $Klee_1([0,1])=1$. $\endgroup$ – Neil Hoffman Sep 23 '14 at 1:24

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.

Browse other questions tagged or ask your own question.