6

Take a profinite group $G=\varprojlim G_\alpha$. We know that the inner automorphism group $\text{Inn}(G)$ of $G$ is profinite since $\text{Inn}(G)=G/Z(G)$, and the quotient of a profinite group by a closed normal subgroup (which $Z(G)$ must be as the intersection of centralizers) is profinite. I'm considering another object $\widehat{\text{Inn}}(G)=\varprojlim\text{Inn}(G_\alpha)$. Now I would like to show that $\text{Inn}(G)=\widehat{\text{Inn}}(G)$. A priori there may be things in $\widehat{\text{Inn}}(G)$ that are not in $\text{Inn}(G)$, so I believe that showing this is a worthwhile goal. There is an embedding $\text{Inn}(G)\hookrightarrow\widehat{\text{Inn}}(G)$ and I've managed to show that $\text{Inn}(G)$ is dense in $\widehat{\text{Inn}}(G)$. It remains to show that $\text{Inn}(G)$ is closed in $\widehat{\text{Inn}}(G)$. I believe that one way to go about this is to take a limit point $\varphi$ of $\text{Inn}(G)$ and show that any open subset of $\widehat{\text{Inn}}(G)$ that contains $\varphi$ must intersect $\text{Inn}(G)$. Can someone help me with this argument?

Another possible way involves considering what the basic closed sets look like in a Stone topological space presented as an inverse limit. I know what the basic open sets look like, but what do basic closed sets look like?

Edit: I want to assume all the maps $f:G_\alpha\to G_\beta$ are surjective.

Alex Byard
  • 516
  • As I pointed out in a comment to your previous question, $\DeclareMathOperator{\Inn}{Inn}\varprojlim \Inn(G_\alpha)$ may not make sense, since $\Inn$ is not a functor on the category of (finite) groups. If $f\colon G_\alpha\to G_\beta$ is an arrow in the diagram, we only get a canonical arrow $\Inn(G_\alpha)\to \Inn(G_\beta)$ if $f(Z(G_\alpha))\subseteq Z(G_\beta)$. Fortunately, this is always true when $f$ is surjective. Do you want to assume all the maps in the diagram $(G_\alpha)$ are surjective? – Alex Kruckman Jul 04 '23 at 02:44
  • @AlexKruckman Yes I do want to assume that. – Alex Byard Jul 04 '23 at 14:48

1 Answers1

1

The image of a compact set under a continuous map is compact, and compact subsets of Hausdorff spaces are closed, so the image of $\mathrm{Inn}(G)$ in $\widehat{\mathrm{Inn}}(G)$ is closed.

It follows (assuming you've correctly proved that the canonical map $\mathrm{Inn}(G)\to \widehat{\mathrm{Inn}}(G)$ is injective and has dense image) that this map is an isomorphism of profinite groups.

To answer your second question: What counts as a "basic open" or "basic closed" set in a topological space depends on the basis you've chosen (e.g. if we take the entire topology as a basis, every open set is basic open). Now if we have a Stone space $X$ presented as an inverse limit of finite sets, $X = \varprojlim (X_\alpha)$ and we write $\rho_\alpha\colon X\to X_\alpha$ for the projection maps, then a standard choice of basis is $\{\rho_\alpha^{-1}(Y)\mid Y\subseteq X_\alpha\}$. Note that if $Z = \rho_\alpha^{-1}(Y)$ is basic open, then its complement $X\setminus Z = X\setminus \rho_\alpha^{-1}(Y) = \rho_\alpha^{-1}(X_\alpha\setminus Y)$ is also a basic open set. So this basis $\mathcal{B}$ of open sets is closed under complement, and the corresponding basis of closed sets is again $\mathcal{B}$.

Alex Kruckman
  • 86,811
  • Could you explain why $\text{Inn}(G)$ is compact? – Alex Byard Jul 04 '23 at 23:53
  • 1
    @AlexByard Because it's profinite. – Alex Kruckman Jul 05 '23 at 00:36
  • Yes, you're correct I overlooked that. Do you know of an argument supporting the closedness of $\text{Inn}(G)$ that doesn't reference the fact that we know $\text{Inn}(G)$ is profinite? I am trying to generalize such things to objects for which the notion of "center" does not exist and "quotient" is a little more mysterious. Really, I am trying to use this to draw out the fact that $\text{Inn}(G)$ is profinite, where $G$ isn't necessarily a group. – Alex Byard Jul 05 '23 at 00:40
  • 1
    Sorry, I don't think I have anything more useful to say on that front. – Alex Kruckman Jul 05 '23 at 00:49
  • Can you say why $X\setminus\rho_\alpha^{-1}(Y)=\rho_\alpha^{-1}(X_\alpha\setminus Y)$? This does not immediately follow for me. – Alex Byard Jul 07 '23 at 18:34
  • Also, does ${\rho_\alpha^{-1}(Y):Y\subseteq X_\alpha}$ form a basis or a subbasis? – Alex Byard Jul 07 '23 at 19:37
  • 1
    It's an elementary fact of set theory that preimages under arbitrary functions preserve intersections, unions, and complements. Check it! This answers your first question. – Alex Kruckman Jul 07 '23 at 19:57
  • Just worked it out. For the second question, I would have thought that a basis of opens would be when we restrict our elements in finitely many coordinates, and that ${\rho_\alpha^{-1}(Y):Y\subseteq X_\alpha}$ forms a subbasis, but perhaps I'm wrong to view it this way. – Alex Byard Jul 07 '23 at 19:59
  • 1
    For your second: yes, sets of this form are closed under finite intersections, so they quality as a basis for a topology. To see this, suppose we have $Y\subseteq X_\alpha$ and $Z\subseteq X_\beta$. Now find $X_\gamma$ in the diagram that maps to both $X_\alpha$ and $X_\beta$ by maps $f$ and $g$. Now check that $\rho_\alpha^{-1}(Y)\cap \rho_\beta^{-1}(X) = \rho_\gamma^{-1}(f^{-1}(X)\cap g^{-1}(Y))$. – Alex Kruckman Jul 07 '23 at 20:02
  • Thank you for the argument on the basis. Is there a reason these generate the Stone topology? – Alex Byard Jul 07 '23 at 20:05
  • 1
    @AlexByard there are two ways to interpret your question. Which do you mean? (1) Given a filtered diagram of finite groups, these sets form the basis for a topology on the limit. Why is this topology a Stone topology? (2) Given a topological group with a Stone topology, we can form the filtered diagram of all finite quotients by open normal subgroups and recover the original group as the limit. Why is the topology arising from this basis equal to the original topology? – Alex Kruckman Jul 07 '23 at 20:21
  • I think I mean in the sense of (1). If I take a limit of a filtered diagram of finite discrete things, then the obtained profinite thing has a Stone topology. I agree that the sets $\rho_\alpha^{-1}(Y)$ form a basis for a topology on the profinite thing, indeed they satisfy (B1) and (B2). I want to know why the topology they generate is necessarily the Stone topology we already have on the profinite thing. – Alex Byard Jul 07 '23 at 20:31
  • Ok... so what is your definition of the Stone topology on the profinite thing? To me, it is the topology generated by this basis. – Alex Kruckman Jul 07 '23 at 20:36
  • @AlexByard This topology is, after all, the limit topology in the category of topological spaces: the coarsest topology that makes all the maps $\rho_\alpha$ continuous when we equip the finite groups with the discrete topologies. – Alex Kruckman Jul 07 '23 at 20:49
  • Okay, I'm just used to thinking of Stone topology very abstractly: Some topology that makes the profinite thing compact, Hausdorff, totally disconnected. If this is simply the topology generated by the mentioned basis, and these properties follow, then I am clear on the situation at hand. – Alex Byard Jul 07 '23 at 20:53
  • Also, would taking preimages of singletons in each $X_\alpha$ be sufficient to construct a basis? – Alex Byard Jul 07 '23 at 20:57
  • 1
    @AlexByard Yes, because each set in the "all preimages" basis is a (finite) union of sets in the "just preimages of singletons" basis. – Alex Kruckman Jul 08 '23 at 03:23