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For an inverse system {$G_i$} of finite groups, and a fixed field $\mathbb{k}$, one can consider the corresponding group algebras $\mathbb{k}[G_i]$. The latter form an inverse system of $\mathbb{k}$-algebras {$\mathbb{k}[G_i]$} (unless I miss something obvious). Is it true that the inverse limit of {$\mathbb{k}[G_i]$} is the group algebra $\mathbb{k}[G]$, for $G=\lim\limits_\leftarrow${$G_i$} ?

In my case $\mathbb{k}=\mathbb{Z}/p\mathbb{Z}$, and $G_i$ are abelian $p$-groups, if this helps.

Added: I see that the answer is much less trivial than I expected. What about the simplest case, perhaps, when $G_i=\mathbb{Z}/p^i\mathbb{Z}$, for $i\geq 1$, and thus $G$ is the additive group of $\mathbb{Z}_p$? Are there fields for which the complete group algebra $[[\mathbb{k}G]]$ is easy to describe (particularly interesting for me would be the case $\mathbb{k}=\mathbb{Z}/p\mathbb{Z}$.) ? [This is answered by Simon Wadsley in the comment below.]

P.S. This is a spill-over of an innocently looking final year project on invertible circulant matrices of an undergraduate student of mine---I am unfamiliar with profinite things...

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    $\begingroup$ Did you try $G_i$ to be a vector space over the 2-element field, $p=2$, the homomorphisms kill the last coordinate? It seems like the inverse limit of group algebras is much bigger than the group algebra of inverse limits. $\endgroup$
    – user6976
    Commented Dec 5, 2011 at 6:36
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    $\begingroup$ In the case you just added the ring is isomorphic to the formal polynomial ring in one variable $k[[T]]$. The element $T$ corresponds to an element $g-1$ with $g$ a generator of $\mathbb{Z}_p$. The references you have already been given prove this. $\endgroup$
    – Simon Wadsley
    Commented Dec 5, 2011 at 15:04
  • $\begingroup$ Sorry. I just noticed that I typed polynomial when I meant power series. $\endgroup$
    – Simon Wadsley
    Commented Dec 5, 2011 at 16:55
  • $\begingroup$ Simon, thanks! Perhaps you can also confirm my suspicion that the inverse limit of the system of groups $\mathbb{k}[G_i]^*$, i.e. groups of invertible elements of $\mathbb{k}[G_i]$, is the group $[[\mathbb{k}G]]^*$ ? $\endgroup$
    – Dima Pasechnik
    Commented Dec 6, 2011 at 4:22

2 Answers 2

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See section 5.3 in "Profinite Groups" by Ribes and Zalesskii. The inverse limit of group algebras you are referring to is called the complete group algebra and it is the completion of the ordinary group algebra $\mathbb k[G]$ with its natural profinite topology. In other words $\mathbb k[G]$ is densely embedded in this algebra.

This works for any profinite ring $\mathbb k$. What I mean by natural profinite topology on $\mathbb k[G]$ is the one given by the fundamental system of neighbourhoods of $0$ consisting of the ideals $$\text{Ker}(\mathbb k[G]\to (\mathbb k/I)[G/U])$$ where $I$ ranges over the open ideals of $\mathbb k$ and $U$ over the open normal subgroups of $G$.

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    $\begingroup$ Maybe one sould mention that the notation for the complete group algebra is $k[[G]]$. $\endgroup$
    – Chandan Singh Dalawat
    Commented Dec 5, 2011 at 8:06
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    $\begingroup$ Perhaps more accurately the notation for the complete group algebra is sometimes k[[G]]. In the case the OP is interested in I have also seen $[[kG]]$, $\Omega(G)$ or $\Omega_G$. As it is the most natural algebraic object to attach to the pair $(k,G)$ I would normally use $kG$ myself. $\endgroup$
    – Simon Wadsley
    Commented Dec 5, 2011 at 12:27
  • $\begingroup$ I wonder if the particular case I just added to the question can be described explicitly... $\endgroup$
    – Dima Pasechnik
    Commented Dec 5, 2011 at 13:02
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Although the reference to Ribes and Zalesski (given in another answer) is excellent, another very good starting point for this area is:

A. Brumer, Pseudocompact algebras, profinite groups and class formations , J. Alg, 4, (1966), 442–470.

The key point is the pseudocompactness of the result. Brumer goes into the homological algebra of these algebras. Of course, in the 45 years since that was published there have been a lot of advances, but the basic theory is very well explained there.

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  • $\begingroup$ Thanks, looks quite useful! I can only accept one answer though :) $\endgroup$
    – Dima Pasechnik
    Commented Dec 6, 2011 at 4:07
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    $\begingroup$ @Dima Brumer is useful and, I find, very well written. My answer was written to be of help, not to gain `brownie points' so don't worry about the other point. Another interesting reference for you may be 'Analytic pro-p groups', by Dixon, du Sautoy, Mann, and Segal, Cambridge studies in advanced mathematics no 61. This looks at special cases which are possibly nearer to what you want. $\endgroup$
    – Tim Porter
    Commented Dec 6, 2011 at 6:32

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