Let $k$ be a cardinal (possibly infinite). Denote by $F_k$ the free group of rank $k$ and consider its profinite completion $G := \widehat{F_k}$.

What is the (topological) abelianization $G'_k := G / \overline{[G,G]}$ of $G$ ?

It can be seen that the abelianization of $F_k$ is the free abelian group $\Bbb Z^{(k)}$. But it is not expected that abelianization commutes with profinite completion. This question seems to claim that the abstract abelianization of $G$ is $\widehat{\Bbb Z}^2$ (without reference nor proof).

Thank you!

  • $\begingroup$ I think that an answer to this question would be very helpful to math.stackexchange.com/questions/2836916 $\endgroup$ – Watson Jul 1 '18 at 13:55
  • $\begingroup$ [By the way, I don't really know how profinite completion behaves w.r.t. exact sequences, products, coproducts of groups…] $\endgroup$ – Watson Jul 1 '18 at 13:55

This is not a complete answer but it's too long to be a comment.

Profinite completion is a functor $C:\mathbf{Grp}\to \mathbf{Prof}$ (the latter being the category of profinite groups and continuous group morphisms) that has a very nice property: it is left adjoint to the forgetful functor $\mathbf{Prof}\to \mathbf{Grp}$.

Therefore, as all left adjoints do, it commutes with colimits (in particular, this tells you how it behaves wrt exact sequences and coproducts: it is right exact, and commutes with coproducts -note however that this last bit is subtle : it takes coproducts in $\mathbf{Grp}$ to coproducts in $\mathbf{Prof}$ which may not look like coproducts in $\mathbf{Grp}$ !)

Taking a quotient is a colimit, hence $C(\mathbb{Z}^{(k)})=C(F_k/[F_k,F_k])=C(F_k)/C([F_k,F_k])=G/C([F_k,F_k])$ (this quotient is meant as "the coequalizer of the diagram $C([F_k,F_k]) \to C(F_k)$" where the two arrows are $C($inclusion$[F_k,F_k]\to F_k)$ and $C($trivial morphism$) =$trivial morphism.

It suffices now to see what the image of $C([F_k,F_k])$ in $G$ looks like, and what its normal closure looks like. If it happened to be $\overline{[G,G]}$, that would be great because it would imply that $G/\overline{[G,G]} = C(\mathbb{Z}^{(k)})$

Another way to look at this would be to say that abelianization is also a functor, here we're looking at profinite abelianization, so it's a functor $^{abProf}:\mathbf{Prof}\to \mathbf{AbProf}$ (the category of abelian profinite groups). This one is left adjoint to the inclusion $\mathbf{AbProf}\to \mathbf{Prof}$, so again it commutes with colimits.

Therefore $G^{abProf} = C(F_k)^{abProf}= C(\displaystyle\coprod_{i\in k}\mathbb{Z})^{abProf} = (\displaystyle\coprod_{i\in k}C(\mathbb{Z}))^{abProf}$ (because $C$ commutes with coproducts -note that the coproducts are to be understood in the correct category; i.e. they need not be the same !), so $G^{abProf} = \displaystyle\coprod_{i\in k}C(\mathbb{Z})^{abProf}$. Now $C(\mathbb{Z})=\widehat{\mathbb{Z}}$ is already abelian, so $G^{abProf} = \displaystyle\coprod_{i\in k}C(\mathbb{Z})=\displaystyle\coprod_{i\in k}\widehat{\mathbb{Z}}$, where this coproduct is to be taken in $\mathbf{AbProf}$.

Now what does the coproduct in $\mathbf{AbProf}$ look like?

To answer this let's make a detour through $\mathbf{Ab}$: we have another profinite completion functor $K: \mathbf{Ab}\to \mathbf{AbProf}$ and it's again left adjoint to the forgetful functor, so it commutes with coproducts. Hence in $\mathbf{AbProf}$, $\displaystyle\coprod_{i\in k}\widehat{\mathbb{Z}}= \displaystyle\coprod_{i\in k}K(\mathbb{Z})=K(\mathbb{Z}^{(k)})$.

Therefore $G^{abProf} = $ the profinite completion of $\mathbb{Z}^{(k)}$.

For finite $k$ I expect this should be $\widehat{\mathbb{Z}}^{(k)}$.

Hence computing what you call $G'_k$ is reduced to one of the following tasks :

-computing the topological closure of the normal closure of the image of $C([F_k,F_k]) $ in $G$ (under the induced morphism) and finding that this is $\overline{[G,G]}$ ( I have very little intuition/knowledge about profinite groups; but this seems like it should be true ? )

-Or describe the coproduct in the category $\mathbf{AbProf}$. I expect this shouldn't be complicated (I'm just not seeing it right now for some reason). If you have this, then $G'_k$ is the coproduct in this category of $k$ copies of $\widehat{\mathbb{Z}}$.

-Or compute the profinite completion of $\mathbb{Z}^{(k)}$. For finite $k$ this should be easy, for infinite $k$ I don't know. If you have this, then $G'_k$ is the profinite completion of $\mathbb{Z}^{(k)}$.

Edit: I can confirm that it works for a finite $k$. Indeed, the finite coproduct of profinite abelian groups (in $\mathbf{AbProf}$ !) is their product: if $G_1,..,G_n$ are such groups then their product is Hausdorff, compact, totally disconnected, and it's a topological group so it's an abelian profinite group, and then the universal property is immediate. So for finite $k$, we do have $G_k' = \widehat{\mathbb{Z}}^k$.

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  • $\begingroup$ Just to be sure, reading your last edit: should there be a topological closure to the commutator subgroup in this question? Or is it true that the abstract abelianization of $\widehat{F_2}$ (as discrete group) is $\widehat{\Bbb Z}^2$? $\endgroup$ – Watson Jul 1 '18 at 15:41
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    $\begingroup$ I don't know. Adding a closure definitely makes it work ($^{abProf}$ is defined with a closure for it to be Hausdorff) but I wouldn't be surprised if $[\widehat{F_2}, \widehat{F_2}]$ were closed (after all it's generated by a compact set) - but again I don't know enough about profinite groups to know if it's closed (if it's not closed it can't be true since it would be $f^{-1}(0)$ for some continuous $f$) $\endgroup$ – Maxime Ramzi Jul 1 '18 at 15:50

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