# Can one cancel $\mathbb R$ in a bi-Lipschitz embedding?

Let $X$ be a metric space. Suppose that the product $X\times\mathbb R$ admits a bi-Lipschitz embedding into $\mathbb R^{n}$. Does it follow that $X$ admits a bi-Lipschitz embedding into $\mathbb R^{n-1}$?

1. A map $F\colon (Y,\rho)\to\mathbb R^n$ is a bi-Lipschitz embedding if there exists a constant $L$ such that $L^{-1}\rho(a,b)\le |F(a)-F(b)|\le L\rho(a,b)$ for all $a,b\in Y$.
2. The choice of a product metric on $X\times \mathbb R$ makes no difference. For definiteness, let the distance between $(x,t)$ and $(x',t')$ be $d_X(x,x')+|t-t'|$.
3. The cancellation fails if bi-Lipschitz is replaced by topological or smooth embedding. For example, $S^1\times \mathbb R$ embeds into $\mathbb R^2$ smoothly, but $S^1$ does not embed into $\mathbb R$ in any sense.
4. The cancellation is possible when $n=1$, according to $\mathbb R^0=\{0\}$. :) I'm pretty sure it's also possible when $n=2$, but don't have a proof. For $n\ge 3$ I don't have a clue.
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I think your example in 3 should work in the Lipschitz case: You can extend a smooth embedding $S^1\times (0,1)\to \mathbb{R}^2$ to a $C^1$ diffeomorphism of open sets in $\mathbb{R}^3$, where this sets have Lipschitz boundary, then this diff. is bi-Lipschitz, and then so is the restriction. – Jose27 Jun 13 '12 at 2:48
@Jose27 You cannot replace $\mathbb R$ with $(0,1)$ -- these are different animals in the bi-Lipschitz zoo. $S^1\times \mathbb R$ does not have a bi-Lipschitz embedding to a plane, since the images of some half-lines $(x,t)$ would be trapped inside the image of $S^1$, forcing them to be bounded. – user31373 Jun 13 '12 at 13:49
@Thursday, do you got any progress on this problem? – Tomás Sep 13 '14 at 11:59

I think the answer to your question does not depend at all on the target space, but only on the domain of your function and how you define the metric on it. If you use the metric you suggested, then the answer is positive for whichever target space $Y$. For, suppose $F\colon X\times\mathbb{R}\to Y$ is a bi-Lipschitz embedding into any metric space $Y$. Define a new map $\hat F\colon X\to Y$ by setting $\hat F(x)=F(x,0)$. Then you get: $d_{Y}(\hat F(x),\hat F(y))=d_Y(F(x,0),F(y,0))\le L d_{X\times\mathbb{R}}((x,0),(y,0))=d_X(x,y)$ and similarly for the other inequality. I hope this helps!
The point is that if $X\times \mathbb{R}$ embeds into $\mathbb{R}^n$ then $X$ should embed into $\mathbb{R}^{n-1}$, not just into $\mathbb{R}^n$. (Since I'm commenting, I also wish to say that this is a great question which I'd like to know the answer to.) – GCD Sep 24 '13 at 3:58