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I would greatly appreciate any help with the following problem. If there are existing references related to this, kindly provide them. If not, any help in this matter would be highly valued.

Problem: Let the profinite group $G$ be the inverse limit of the family of finite groups $\{G_i:i\in \mathcal{I}\}$, i.e., $G=\varprojlim G_i$. Let $C(X)$ denote the space of continuous complex-valued functions on $X$. Is it valid to conclude that $C(G)$ is the direct limit of $\{C(G_i):i\in \mathcal{I}\}$, i.e., $C(G)=\varinjlim C(G_i)$?

John
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  • See if this is useful https://math.stackexchange.com/questions/3751271/pontryagin-dual-of-an-inverse-limit –  Aug 29 '23 at 20:14
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    The answers you've received show that it matters what category you consider the $C(G)$ to live in – FShrike Aug 30 '23 at 16:47

2 Answers2

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Yes, a version of this follows from Gelfand duality. Recall that the assignment $X \mapsto C(X)$ is a contravariant equivalence of categories between the category of compact Hausdorff spaces and the category of commutative (unital) $C^{\ast}$-algebras. Profinite sets embed fully faithfully into compact Hausdorff spaces (as the compact Hausdorff totally disconnected spaces, or Stone spaces), and they are the cofiltered limits of finite discrete spaces in this category (the group structure is irrelevant here). A contravariant equivalence of categories converts cofiltered limits into filtered colimits so we get the conclusion, in the category of commutative (unital) $C^{\ast}$-algebras.

How useful this is is unclear; personally I don't know what colimits look like in this category. Already it's not clear to me what finite coproducts look like. Maybe colimits here are the norm completion of the naive colimits?

Qiaochu Yuan
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  • @I am not very familiar with the language of Category theory. Could you please provide a reference or elaborate on how to prove "A contravariant equivalence of categories converts cofiltered limits into filtered colimits so we get the conclusion, in the category of commutative (unital) $C^*$-algebras."? – John Aug 30 '23 at 01:45
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    @John: equivalences of categories preserve all limits and colimits, and contravariant ones swap them. This follows more or less directly from the definitions once you understand the universal properties of limits and colimits. – Qiaochu Yuan Aug 30 '23 at 02:19
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Let $G=\Bbb Z_p$ and let $f:G \to [0,1]$ be a continuous surjection. (This exists as $G$ is topologically a Cantor set.) Consider $f$ as an element of $C(G)$. Suppose that $f$ is in the image of the canonical map from $\varinjlim C(G_i)$. Then by an explicit construction of a filtered colimit, $f$ must be in the image of $C(G_i) \to C(G)$ for some $i$. But then one easily sees that any function in the image of $C(G_i) \to C(G)$ has finite image, which $f$ does not.

The choice of $\Bbb Z_p$ is somewhat arbitrary here, I just needed a profinite group that admits continuous function with non-finite image to $\Bbb C$ and the fact that $\Bbb Z_p$ is a Cantor set is convenient for that.

Lukas Heger
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  • It looks like you're taking the filtered colimit in rings. I agree that this is not the right colimit, e.g. there's no reason to expect it to be complete wrt the $C^{\ast}$-norm. Maybe the correct colimit (in $C^{\ast}$-algebras) is just the norm completion of this one, though. – Qiaochu Yuan Aug 29 '23 at 19:28
  • @QiaochuYuan yeah, I'm taking a colimit of rings/sets/algebras/vector spaces. I don't know how colimits of $C^*$ algebras work, but I agree that the naive collimit has no reason to be complete. – Lukas Heger Aug 29 '23 at 19:30