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A while ago I managed to convince myself that every finite group is isomorphic to a subgroup of a finite dimensional Euclidean space. But my little simplex construction doesn't extend to infinite groups. Is it the case that every infinite group is isomorphic to an isometry subgroup of a finite dimensional space?

I don't mind if groups are only subgroups of an isometry group, and I don't mind if the space is non-Euclidean.

My naive thought is that non-Euclidean spaces are necessary to deal with groups like $GL({\mathbb R}^n)$ that explicitly include scale transformations.

Luke
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  • By infinite do you mean countable? – azarel Sep 17 '14 at 02:56
  • I mean any infinite group, countable or uncountable, but restricted answers are welcome. I know that many Lie groups arise naturally as symmetry groups and are uncountable but finite-dimensional. – Luke Sep 17 '14 at 03:00
  • Presumably you mean "isomorphic to a subgroup of the isometry group of...." The answer is no; take, for example, any group with cardinality larger than $\mathbb{R}$. – Qiaochu Yuan Sep 17 '14 at 03:44
  • "isometry group of a finite-dimensional space": you should explain what you mean by space so that isometries make sense: metric spaces? connected Riemannian manifolds? cell complexes?... and if you deal with metric spaces you should specify what you call "finite-dimensional". – YCor Sep 17 '14 at 22:34

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Start by taking a finitely generated infinite simple group $G$ (see this MSE question for references), say, $G$ is a Higman group. Then $G$ (despite being countable) does not embed as a subgroup in any matrix group $GL(n, F)$ where $F$ is any field. This follows from a theorem of A.Malcev, which states that every finitely generated subgroup $G$ of $GL(n, F)$ is residually finite, i.e., the intersection of all finite index subgroups of $G$ is trivial. Malcev's theorem implies that $G$ (if infinite) contains proper finite index subgroups, hence, proper normal finite index subgroups, hence, cannot be simple. See a proof of Malcev's theorem here in the case $F={\mathbb C}$.

On the other hand, if you do not mind infinite-dimensional vector spaces, then every group $G$ embeds into the isometry group of some Banach space (a complete normed vector space). Namely, let $V$ denote the space of bounded functions on $G$ with sup-norm. Then $G$ acts isometrically, linearly (and faithfully) on $V$ by the rule: $$ g\cdot f(x)= f(xg) $$ where $f: G\to {\mathbb R}$ is in $V$.

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Moishe Kohan
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