# A version of the group ring using direct product rather than direct sum?

Let $G$ be an infinite group. It's (integral) group ring $\mathbb{Z}[G]$ has as its elements the finite formal linear combinations $$m_1g_1 + m_2g_2 + \cdots + m_ng_n,\qquad n\in\mathbb{N},\quad m_i\in\mathbb{Z},\quad g_i\in G,$$ and these are added and multiplied in the obvious way such that the usual ring axioms are satisfied. Thus it has underlying abelian group isomorphic to a direct sum of copies of $\mathbb{Z}$, with one copy for each group element $g\in G$.

I wonder what happens if we replace direct sum with direct product in the above construction? Can we make the direct product $\prod_{g\in G} \mathbb{Z}$ in the category of abelian groups into a ring by simply defining $$(\sum m_gg)(\sum n_h h) = \sum m_g n_h (gh)$$ and not worrying about whether the sums converge?

And if so, is this ring considered anywhere in the literature? Is it some sort of "completed group ring"?

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Ignoring convergence could not be as useful as one might think because it robs us of the opportunity to use methods from functional analysis etc. For example the space $\ell^2(G)$ is very useful (despite not being an algebra). If we use $\ell^1(G)$ we should get a true algebra, if I'm not mistaken. – Johannes Hahn Jan 18 '13 at 12:49
It looks like you will have trouble with infinite sums on the right hand side. For example, consider $x=\Sigma g$, the sum of all elements of $G$. What is the coefficient of the identity element in the expansion of $x^2$? Looks like it is the order of $G$, which is infinite. – Gregory Arone Jan 18 '13 at 12:50
It seems to me however that I've seen a tiny amount of literature on the formal algebraic calculus of the Dirac distribution (which is exactly Gregory's sum in the case $G = \mathbb{Z}$). As usual, you get into trouble when you try to multiplicatively square the Dirac distribution. However, the formal calculus of $\delta$ and its derivatives can be interesting. See for example the book by Frankel, Lepowsky, and Meurman, Vertex Operator Algebras and the Monster. Thus, these direct product analogues might be best seen by pursuing analogies with distributions. – Todd Trimble Jan 18 '13 at 14:05
@Gregory: Thanks, I had an inkling this might be a stupid question, but couldn't see why! If you make your comment into an answer, I will accept it. @Everyone else: Thanks for making the question seem less stupid. I'll look into these references. – Mark Grant Jan 19 '13 at 11:45
This can be made to work for semigroups in which each element has only finitely many factorizations of length 2, eg, the free monoid. – Benjamin Steinberg Jan 19 '13 at 15:27