Let $X$ is a strip between two different parallel lines $a$ and $b$ on a plane ($a,b\subset X$) and $h(x)=\min\limits_{l\in \{a,b\}}\{d(x,l)\}$. Let $(X,*)$ be a topological group with the following property: $$h(xy)\leq \max\{h(x),h(y) \}.$$ It is a locally compact, connected, simply connected, Hausdorff group. I think it might be a Lie group, but which one I don't know. This situation can be transferred to higher dimensions and there I am also interested in the next question. Does such a group exist?
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3$\begingroup$ Do you have a specific definition for a group structure on $X$ or are you considering any group structure (provided it exists...)? $\endgroup$– QfwfqSep 15, 2021 at 23:37
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1$\begingroup$ Does the statement "It is a [list of conditions] group" mean that you want additionally (to the condition on $h$) to impose these conditions, or that they follow from the condition on $h$? $\endgroup$– LSpiceSep 15, 2021 at 23:46
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$\begingroup$ @Qfwfq, I am considering the standard definition of a topological group, this is just a special case of interest to me. $\endgroup$– Ben TomSep 16, 2021 at 1:29
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1$\begingroup$ If I understand well, $X$ would be homeomorphic to $\mathbb{R}^2$ (I assume you're considering an open strip, otherwise it's not homogeneous so it can't be a topological group). Do you have any explicit examples of such objects? $\endgroup$– QfwfqSep 16, 2021 at 16:32
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1$\begingroup$ (Or, if the topology you're putting on $X$ is not the one inherited from the plane, could you specify how you define it?) $\endgroup$– QfwfqSep 16, 2021 at 16:40
1 Answer
Such group does not exist. To derive a contradiction, assume that the strip $X=\mathbb R\times(-1,1)$ admits a continuous group operation $X\times X\to X$, $(x,y)\mapsto xy$, such that $h(xy)\le\max\{h(x),h(y)\}$ for all $x,y\in X$.
Let $c$ be any point on the central line $L=\mathbb R\times\{0\}$ and $f:X\to X$ be the homeomorphism defined by $f(x)=x^{-1}c$. Observe that $xf(x)=xx^{-1}c=c\in L$ for every $x\in X$.
Since the line $L$ is nowhere dense in $X$, the set $L\cup f^{-1}[L]$ is nowhere dense in $X$ and hence we can choose an element $x\in X\setminus (L\cup f^{-1}[L])$. Then for the points $x\notin L$ and $y=f(x)=x^{-1}c\notin L$ we have $$1=h(c)=h(xy)>\max\{h(x),h(x^{-1}c)\},$$ which contradicts the choice of $h$.
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$\begingroup$ I guess here $a$ and $b$ are $\mathbb R \times \{1\}$ and $\mathbb R \times \{-1\}$? $\endgroup$– LSpiceSep 19, 2021 at 17:31
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$\begingroup$ @LSpice Yes, I had in mind the strip $\mathbb R\times(-1,1)$. $\endgroup$ Sep 19, 2021 at 20:47