If $G$ is an infinite group, is there necessarily an unbounded left-invariant metric on $G$?
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$\begingroup$ If $G$ is countable, then tehre exists even a proper invariant metric, as can be seen here: math.uni-bielefeld.de/~amanouss/papers/jlms1.pdf $\endgroup$– user1688Commented May 4, 2013 at 14:48
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$\begingroup$ Do you mean $\textit{countable}$ group? Otherwise just take the uncountable product of a discrete group. The result is reasonably nice (ie it's a compact topological group). However, the underlying space is not metrizable. $\endgroup$– Owen SizemoreCommented May 4, 2013 at 18:02
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1$\begingroup$ @Owen Sizemore: there is no compatibility with any given topology asked in the question. $\endgroup$– Benoît KloecknerCommented May 4, 2013 at 18:35
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$\begingroup$ No compatibility? Then the 0-1 metrics would be an answer. $\endgroup$– Włodzimierz HolsztyńskiCommented May 4, 2013 at 21:26
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$\begingroup$ @Wlodzimierz: how do you prove that the 0-1 metric is unbounded? :) $\endgroup$– YCorCommented May 4, 2013 at 21:28
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1 Answer
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The answer is no. See Thm 1.2 of http://homepages.math.uic.edu/~rosendal/PapersWebsite/Property(OB)10.pdf. There is a property discussed in the intro of this paper which is equivalent to all left invariant metrics are bounded. It is known that certain large permutation groups have this property.
answered May 4, 2013 at 19:52
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2$\begingroup$ This is actually originally due to Bergman (Bull London Math Soc) arxiv.org/abs/math/0401304, Rosendal considers the version for a group endowed with a topology. $\endgroup$– YCorCommented May 4, 2013 at 20:25
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$\begingroup$ Thanks Yves. I realized only the intro talks about this but I didn't have a link to Bergman's paper. $\endgroup$ Commented May 4, 2013 at 21:15
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$\begingroup$ Very interesting answers. Thank you. $\endgroup$– H-HookCommented May 5, 2013 at 21:34