2
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $G_k' = \widehat{\mathbb{Z}}^k$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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 $f$) $\endgroup$ – Max Jul 1 '18 at 15:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.