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Group stability considers the question of whether "almost"-homomorphisms are "close to" true homomorphisms. Here, "almost" and "close to" are made rigorous using a group metric.

More precisely, suppose we have a group $\Gamma$, and a family $\{G_n\}$ of groups equipped with a bi-invariant metric $d_n$. An almost (or asymptotic) homomorphism from $\Gamma$ to the family $\{G_n\}$ is a sequence of maps $$\phi_n: \Gamma \to G_n$$ such that for all $x,y \in \Gamma$, $$d_n\left( \phi_n(xy), \phi_n(x)\phi_n(y) \right) \longrightarrow 0$$ as $n \to \infty$. The group $\Gamma$ is said to be stable with respect to $\{G_n\}$ if any such almost homomorphism is close to a sequence of homomorphisms. That is, given any almost homomorphism $\{\phi_n\}$, there exists a sequence $\{\psi_n:\Gamma \to G_n\}$ of homomorphisms such that for all $x \in \Gamma$, $$d_n\left(\psi_n(x),\phi_n(x) \right) \longrightarrow 0$$ as $n \to \infty$.

This phenomenon of stability can also be interpreted as a lifting problem by combining all the groups $G_n$ into a direct product. Consider the direct product $\mathcal{G}=G_1 \times G_2 \times \dots$ and let $N$ be its subset of elements $(g_1,g_2,\dots)$ such that $d_n(g_n, 1) \to 0$ as $n \to \infty$ (that is, $N$ is the subset of elements (or sequences) that converge to the identity element). This $N$ is a normal subgroup of $\mathcal{G}$ , and an almost homomorphism $\{\phi_n\}$ can be interpreted simply as a homomorphism $$\phi: \Gamma \to \mathcal{G}/N$$ The stability question is now whether this homomorphism $\phi$ from $\Gamma$ to the quotient $\mathcal{G}/N$, can be lifted to a homomorphism $\psi: \Gamma \to \mathcal{G}$.

Note that the speed of convergence was never an issue above. What if we consider uniform stability? Now we have an almost homomorphism $$\phi_n: \Gamma \to G_n$$ such that given any $\epsilon>0$, then there exists $N$ such that for all $x,y \in \Gamma$, $$d_n\left( \phi_n(xy), \phi_n(x)\phi_n(y) \right) < \epsilon$$ when $n \geq N$. That is, the rate of convergence to $0$ is uniform over all elements of $\Gamma$. The group $\Gamma$ is said to be uniformly-stable with respect to $\{G_n\}$ if for any such uniformly almost homomorphism, there exists a sequence $\{\psi_n:\Gamma \to G_n\}$ of homomorphisms such that given any $\epsilon>0$, there exists $N$ such that for all $x \in \Gamma$, $$d_n\left(\psi_n(x),\phi_n(x) \right) < \epsilon$$ for $n \geq N$.

Do we have an interpretation of uniform stability in terms of lifting homomorphisms from a quotient group as in the non-uniform case?

The main difficulty is that when quotienting out $N$ from $\mathcal{G}$, the information regarding uniformity is lost. If instead of taking the homomorphism $$\phi: \Gamma \to \mathcal{G}/N$$ we take the homomorphism to be $$\phi: \Gamma \times \Gamma \times \dots \to \mathcal{G}/N$$ would this work?

Relevant References: 1) A survey by Andreas Thom on group approximations: https://arxiv.org/abs/1712.01052

2) Notes/thesis of Christoph Gamm on uniform stability https://www.math.uni-leipzig.de/preprints/p1110.0020.pdf

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    $\begingroup$ This is related to very recent research, maybe you could point out relevant references. $\endgroup$
    – YCor
    Commented Mar 16, 2019 at 15:34
  • $\begingroup$ @YCor Thanks for the suggestion. I have added links to two broad survey articles I found useful. However, I think my specific question is rather simple and general, and does not go into or need the technicalities of the papers. $\endgroup$
    – BharatRam
    Commented Mar 16, 2019 at 17:09
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    $\begingroup$ Sure, these references rather come to give context and motivation, but it's excellent you made the question self-contained. $\endgroup$
    – YCor
    Commented Mar 16, 2019 at 18:34
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    $\begingroup$ I here forward an email comment by Goulnara Arzhantseva: The lifting theorem he mentions is my theorem with Liviu, see Theorem 4.2 (and a comment just after its proof) of G.N. Arzhantseva, L. Paunescu, Almost commuting permutations are near commuting permutations, J. Funct. Anal., 269(3) (2015), 745-757. Thom does not give a correct attribution to this theorem in his survey (if I remember well, have no possibility to check it now). I do not have the MO account, thought probably this information is useful for the users. $\endgroup$
    – YCor
    Commented Mar 16, 2019 at 22:38
  • $\begingroup$ @YCor Thanks for the comment. I did look up the paper of Arzhantseva-Parnescu, though that only discusses the non-uniform model. In the non-uniform case, the lifting idea is easy to see. As for the uniform case, I asked Prof.Lev Glebsky about this, and he did agree that the above method works, and I kind of see it now, though I am not fully satisfied by my own proof. Please do tell me if you think it makes sense. Thanks! $\endgroup$
    – BharatRam
    Commented Mar 23, 2019 at 13:58

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