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Given a infinite family of groups $(G_i)$ for $i\in I$. Is there a ring theoretic construction, that produces $R[\prod_{i\in I} G_i]$ using only the rings $(R[G_i])_{i\in I}$ ?

For the case of a finite family, we have $R[G\times H]\cong R[G][H]$ and for commutative $R$ we have $R[G\times H]\cong R[G]\otimes R[H]$. Neither of those constructions generalizes to the infinite case, e.g.

The map $R[\prod_i G_i]\rightarrow \mbox{invlim}_{I'\subset I, |I'|<\infty}R[\prod G_i]$ is not surjective (This product runs over $i\in I'$). The same holds for the map into the infinite tensor product (assuming that $R$ is commutative).

So I am hoping, that there is a better contruction in a more elaborate category (like $R$-Algebras with an augmentation), that produces $R[\prod_{i\in I} G_i]$ out of the group rings $(R[G_i])_{i\in I}$ .

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(See for what Henrik means by the infinite tensor product here. There $g_1 \otimes g_2 \otimes ...$ makes sense, in contrast to the usual infinite tensor product of algebras.) – Martin Brandenburg Nov 22 '10 at 14:37
How do we see that the product to the infinite tensor product is not surjective (or better say bijective)? – Thomas Nikolaus Nov 22 '10 at 15:05
let all groups be trivial and let $R$ be the rationals. Then $2\otimes 2\otimes 2...$ is not in the image, which is generated by $1\otimes 1\otimes ...$ and these two elements are linear independent: Consider the multilinear form $f(x_1,....):=\prod x_i$, if almost all $x_i$ are $1$,$0$ otherwise. – HenrikRüping Nov 22 '10 at 15:19

I'm guessing that $R[\prod_i G_i]$ might be obtained as the inverse limit you wrote in the category of Hopf algebras over $R$. Here the forgetful functor

$$U: HopfAlg_R \to Coalg_R$$

preserves and reflects limits, so it suffices to check the claim in the category of (cocommutative) coalgebras. The guess then is that, by some application of the principle that every coalgebra is the filtered colimit of its finite-dimensional subcoalgebras, that the limit in $Coalg_R$ picks up only functions $\prod_i G_i \to R$ of finite support.

See these slides for some hints on calculating limits of Hopf algebras.

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So this would mean especially, that a inverse limit in the category of hopf algebras (over a cofiltered system) is not constructed by taking the inverse limits of the underlying sets and defining the operations componentwise (otherwise the map would still be not surjective). I cannot (yet) really imagine how the construction looks like. – HenrikRüping Nov 22 '10 at 16:22
Well, that's right. For example, the product of two cocommutative coalgebras (over a commutative ring, which is actually the case I was considering) is the tensor product of the underlying modules, and same for cocommutative Hopf algebras. I'll see if I can improve my answer; a few months back I was thinking some about the many amazing categorical properties of the category of coalgebras over a commutative ring. – Todd Trimble Nov 22 '10 at 17:11

I started a work in 2009 concerning infinite tensor products of complex vector spaces, $^*$-algebras and Hilbert spaces. The resulting article has just been accepted for publication and I put it in the arxiv: In the article, it is shown in Example 3.1 that the infinite tensor product of $\mathbb{C}[G_i]$ is big enough to contain $\mathbb{C}[\prod_{i\in I} G_i]$. But in order to identify $\mathbb{C}[\prod_{i\in I} G_i]$, I still need to consider the group $\prod_{i\in I} G_i$.

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After some thoughts, I think your question has a positive answer if $R=\mathbb{C}$ (or $\mathbb{R}$) and one equips $\mathbb{C}[G_i]$ with the extra structure of the $\ell^1$-norm. By the way, your construction of an inverse limit also requires the extra structure of a fixed character on each $R[G_i]$ (in order to map $\bigotimes_{i\in I'} R[G_i]$ to $\bigotimes_{i\in I''} R[G_i]$, when $I''\subseteq I'$).

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Could you explain this a bit more. So you equip the group ring with the $l^1$-norm and take the inverse limit in which category? Or are you considering its completions w.r.t. this norm. Concerning your second remark: Yes I need the augmentation. But this is just the functor "group ring" applied to the group homomorphism $G\rightarrow 1$. – HenrikRüping Dec 20 '11 at 12:50

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