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The groups ${\mathbb Z}_N$ may be viewed as a series, $N=1,2,3,\ldots$, which in the limit $N\to\infty$ approaches $U(1)$. I realize this is a bit hand waving but I'm pretty sure it can be made precise.

What other examples are there? For instance can $S_N$, the permutation groups, be viewed as approaching a Lie group in the $N\to\infty$ limit? If not, why not?

Is there an example with a non-abelian Lie group as the limit?

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    $\begingroup$ I guess you can say that $\underset{\longrightarrow}{\operatorname{lim}} \mathbb{Z}_N \cong \mathbb{Q}/\mathbb{Z}$ embeds in $S^1$. In the same way, you can form the direct limit $\underset{\longrightarrow}{\operatorname{lim}} S_N$ which we might call $S_{(\infty)}$. I don't know whether this embeds in some Lie group. $\endgroup$
    – Smiley1000
    Commented Aug 21 at 12:25
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    $\begingroup$ You might be interested in the concept of “Gromov-Hausdorff” convergence of (pointed) metric spaces. This applies to Cayley graphs, so can be used to find limits of sequences of groups, equipped with generating sets and a “rescaling” function. $\endgroup$
    – Sam Nead
    Commented Aug 21 at 12:32
  • $\begingroup$ The notion of limit of finitely generated groups was refined in the work of van den Dries and Wilkie. See sciencedirect.com/science/article/pii/0021869384902230. To apply this in the case of $S_N$ you would need to choose generators. $\endgroup$
    – Sean Eberhard
    Commented Aug 21 at 12:35
  • $\begingroup$ @Smiley Infinite symmetric group has no finite dimensional representations by Jordan's theorem: every finite subgroup of GL(N) is abelian-by-(finite of bounded order); and because all Lie groups are abelian-by-linear, there's no such embedding. $\endgroup$
    – Denis T
    Commented Aug 21 at 13:04

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It was shown by Alan Turing that (in a certain precise sense) the only connected Lie groups approximable by finite groups are the compact abelian Lie groups, i.e. $U(1)^n$. See Theorem 2 of this paper.

If you allow infinite discrete subgroups then an approximable connected Lie group must be nilpotent and further any simply-connected nilpotent Lie group is approximable. See this paper.

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    $\begingroup$ The first link is a link to a file on your computer. $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Aug 21 at 13:44
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There's a way to formulate an easy version of this question, which I could partially answer. (I'll make it CW, so people who know more about lattices could complete it.)

Question. Which Lie groups have a dense subgroup, such that this dense subgroup is a (sequential) filtered colimit of finitely generated lattices?

If Lie group is nilpotent, then there's such a sequence of subgroups if and only if its Lie algebra admits a basis with rational structural constants. This follows from the works of Malcev; having any lattice at all is equivalent to being Q-algebraic nilpotent group.

If Lie group is semisimple, then (as far as I remember) every lattice is contained in a maximal lattice, so there's no such 'approximation'.

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    $\begingroup$ For the last statement: you need "with finite center"; still the conclusion holds (for every nontrivial semisimple Lie group). $\endgroup$
    – YCor
    Commented Aug 21 at 14:49
  • $\begingroup$ I guess that every connected Lie group with this condition is indeed nilpotent. (By the way, the iff condition for nilpotent you stated is valid only for simply connected nilpotent Lie groups.) $\endgroup$
    – YCor
    Commented Aug 22 at 8:35

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