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Start with an $I$-indexed family of vector spaces $V_i$. Take an ultrafilter $\mathcal{D}$ over $I$. We have a linear transformation $$[-]: \prod_{i \in I} V_i \rightarrow \prod_{i \in I} V_i / \mathcal{D} $$ and we can always find another linear transformation $$ r: \prod_{i \in I} V_i/\mathcal{D} \rightarrow \prod_{i \in I} V_i$$ such that $[ r([f]) ] = [f]$ holds for every $f$.

This observation does not tell us anything specific about ultraproducts: we can find a right inverse to every surjective linear transformation. However, the analogous property fails badly for groups, e.g. the obvious group homomorphism $\mathbb{Z}/4\mathbb{Z} \rightarrow \mathbb{Z}/2\mathbb{Z}$ lacks a right inverse.

If we have an $I$-indexed family of groups $H_i$, and $\mathcal{D}$ is a principal ultrafilter, then the quotient map $[-]_\mathcal{D}$ still admits a right inverse. What can we say when the ultrafilter $\mathcal{D}$ is non-principal? Do we always have a right inverse? If not, can we characterize the pairs of families and ultrafilters for which a right inverse exists?

edit As people have pointed out, a general characterization is unlikely to say the least.

However, if all the $H_i$ are finite and there is a section, then the ultraproduct embeds into a direct product of finite groups, and is therefore residually finite. I'd accept as an answer any partial result which puts nintrivial restrictive conditions on the group $G$ without assuming the finiteness of the $H_i$.

I'd also like to know which residually finite groups arise this way for finite $H_i$ (the answer is not all of them, since $SO(3)$ is trivially the quotient of a residually finite group, but I know it's not a quotient of any product of finite groups), but perhaps this is best asked as a separate question.

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  • $\begingroup$ This can depend on the ultrafilter in a serious way--for instance, if the $V_i$ are all $\mathbb{Z}$, then there is a section iff $\mathcal{D}$ is countably complete. I doubt there is any nice answer in full generality, though. What sort of answer are you looking for? $\endgroup$ – Eric Wofsey Jul 24 '19 at 20:33
  • $\begingroup$ @EricWofsey I added a clarification for the kind of answer I'm looking for. If the the $H_i$ are finite groups, and there is a section, then the ultraproduct is residually finite. I'd accept as an answer any partial result which puts similarly restrictive conditions on the group $G$ for general $H_i$. $\endgroup$ – Z. A. K. Jul 24 '19 at 21:02
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This is not a full answer and only adresses the first part of the question. To that, the answer is that there isn't always a section.

For an example take $I$ to be the set of prime numbers and for $p\in I$, $H_p := \mathbb Z[1/p!]$.

Then the class of $(1,1,1,...)$ in $\prod_p H_p/\mathcal D$ is $n$-divisible for all integers $n$ (because for each $n$, $\{p\in I\mid 1$ is $n$ divisible in $H_p\}$ is cofinite hence in $\mathcal D$ ($\mathcal D$ is non principal), and then just apply Los's theorem) and nonzero , but there is no such nonzero element in $\prod_p H_p$, so any section would send $(1,...)$ to $0$, which is absurd.

You can do a similar thing with $H_p := \mathbb Z/p$.

If I had to guess, I would say that for "most pairs" there is no section, but that would be hard to quantify and I can't really base it on anything but my intuition. However, I think perhaps you should clarify in your question what kind of characterization you're looking for.

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  • $\begingroup$ Thanks for the answer. I just realized that I already knew the answer to the "is there always a section" part: if all the H_i are finite and there is a section, then the ultraproduct embeds into a direct product of finite groups, and is therefore residually finite. This is essentially your argument (you construct a divisible group as an ultraproduct). I don't know precisely what characterization I'm looking for, but I imagine it'd start with "which residually finite groups arise this way?". I'll add this to the question. $\endgroup$ – Z. A. K. Jul 24 '19 at 20:25
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    $\begingroup$ Ah ! so I answered precisely the uninteresting part of your question ;) I won't delete my answer but I do hope to see the second part answered ! (I have no clue how to start looking) $\endgroup$ – Maxime Ramzi Jul 24 '19 at 21:03

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