# Ultraproducts of groups with a right inverse

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.

• 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? – Eric Wofsey Jul 24 '19 at 20:33
• @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 $H_i$. – Z. A. K. Jul 24 '19 at 21:02

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 $$H_p := \mathbb Z/p$$.