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Suppose $\Bbb{K}$ is a topological field and $G$ is a topological group. Recall that $\Bbb{K}[G]$ denotes the group ring of $G$ over $\Bbb{K}$, which consists of sums of the form $\sum_{g \in G} a_g g$ with at most finitely many of the $a_g \in \Bbb{K}$ nonzero. The group ring has a natural vector space structure over $\Bbb{K}$ and multiplication can be defined on it; and, as the name suggests, it is a ring. My question is,

Given the topological structures on $\Bbb{K}$ and $G$, is there natural way of topologizing $\Bbb{K}[G]$ given these structures? Does it make $\Bbb{K}[G]$ into a topological ring? Are there any other interesting features/properties of the topology on $\Bbb{k}[G]$?

EDIT

How about this. Another way of thinking about $\Bbb{K}[G]$ is that it consists of all functions $f : G \to \Bbb{K}$ of finite support. If $G$ is a topological group, what if we instead required that it consists of all $f : G \to \Bbb{K}$ with compact support? If $G$ is a discrete group, then we recover the "usual" group ring. Will this work? Does this result in an "interesting" object?

user193319
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  • This is interesting question... In defining $K[G]$, we did not use any topological structre.. So, to talk about topological structure on $K[G]$ may be we should change the definition slightly?? I do not know.. –  Aug 02 '19 at 15:31
  • @PraphullaKoushik I am open to the possibility. However, I was thinking that the product topology might be helpful. I believe there is another way of thinking about $K[G]$, namely that it is a subspace of the sequence space, consisting of all those sequences with "finite support" ; i.e., at most finitely many of the coordinates are nonzero, where the coordinates are indexed by the group $G$. Not sure if that helps... – user193319 Aug 02 '19 at 15:36
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    For $K=\Bbb C$ and a locally compact group $G$ you can think about $\Bbb CG\subseteq \mathcal C_C(G)\subseteq C^\ast(G)$, the group C*-algebra is isomorphic to $\mathcal C(\hat G)$, so you can topologize this space of functions and let the group algebra inherit the subspace topology Edit: nevermind, this only works for discrete groups – Alessandro Codenotti Aug 02 '19 at 15:42
  • Do you mean to say finitely many $a_g \in \Bbb K$ are nonzero? Because that's not what you wrote. – Robert Shore Aug 02 '19 at 15:44
  • @RobertShore Whoops! Correcting it now. – user193319 Aug 02 '19 at 15:45
  • This seem to be problematic even in finite dimension. If $\mathbb{K}$ is $\mathbb{C}$ or $\mathbb{R}$ then it is well know that every finitely dimensional topological space over $\mathbb{K}$ is a product of an anti-discrete space and the standard $\mathbb{K}^n$. And so there doesn't seem to be a way to make the topology on $\mathbb{K}G$ dependent on the topology on $G$. – freakish Aug 02 '19 at 20:00
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    ...because for any $n$ there are only $n+1$ nonisomorphic t.v.s. of dimension $n$ over $\mathbb{K}$ as above. But for some specific $n$ (with lots of divisors, like factorials) there are many, many more topological groups of order $n$. On the other hand for prime $p$ there are only two topological groups of order $p$. So to me it looks like topologizing $\mathbb{K}G$ based on topology of both $\mathbb{K}$ and $G$ is doomed to fail. Although it may have more sense if we assume Hausdorff everywhere. – freakish Aug 02 '19 at 20:13
  • @freakish How about this. Another way of thinking about $\Bbb{K}[G]$ is that it consists of all functions $f : G \to \Bbb{K}$ of finite support. If $G$ is a topological group, what if we instead required that it consists of all $f : G \to \Bbb{K}$ with compact support? If $G$ is a discrete group, then we recover the "usual" group ring. Will this work? Does this result in an "interesting" object? – user193319 Aug 05 '19 at 12:00
  • @user193319 Well, how do you define multiplication on $C_c(G,\mathbb{K})$ in such a way that it is a ring and $G$ embeds into its group of units? If you don't require this then yes, this is an interesting object. – freakish Aug 05 '19 at 12:41

1 Answers1

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This is not really an answer to your question, since it only talks about $\Bbb K=\Bbb C$, but based on the comments it might be interesting to you nonetheless and it's way too long to fit into a comment.

As pointed out in the comments there's little hope if $G$ isn't discrete, so I'll assume that it is. I'll write something about locally compact groups further down. $\Bbb CG$ is a complex algebra, so in particular a vector space, and the usual way to topologize a vector space is to give it a norm. Note that there's a natural involution that we can define on $\Bbb CG$, namely $$a^\ast(g)=\overline{a(g^{-1})},$$ for $a\in\Bbb CG=\mathcal C_C(G)$ and $g\in G$ so we can actually aim to get a $C^\ast$-algebra out of this norm.

There are two well studied choices at this point, the first is $$\|a\|=\sup\{\|\pi(a)\|\mid\pi\text{ is a representation of }\Bbb CG\},$$ where by representation I mean representation in the unitary operators on a Hilbert space. The completion of $\Bbb CG$ with respect to this norm is denoted by $C^\ast(G)$ and called the (full) group $C^\ast$-algebra, it contains $\Bbb CG$ as a dense subalgebra and this also gives a topology to the group algebra.

The second choice is to set $$\|a\|=\|\pi_\lambda(a)\|,$$ where $\pi_\lambda$ is the left regular representation of $G$ on $\ell^2(G)$ (extended in the usual way to a representation of $\Bbb CG$), the completion of $\Bbb CG$ with respect to this norm is denoted by $C_\lambda^\ast(G)$ and called the reduced group $C^\ast$-algebra. This is the same as taking the $C^\ast$-subalgebra of $B(\ell^2(G))$ generated by the left regular representation.

Note that while there always is a canonical morphism $C^\ast(G)\to C_\lambda^\ast(G)$ those two objects can be wildly different.

In the special case in which $G$ is discrete abelian we have that $C^\ast(G)$ is isomorphic to the reduced group $C^\ast$-algebra, and it is also commutative, so by Gelfand-Naimark it must be isomorphic to a $C^\ast$-algebra of functions on a locally compact Hausdorff space. Turns out that this space has a nice description, it is just $\hat{G}$, the dual group:$$\hat{G}=\{f\colon G\to S^1\mid f \text{ is a group homomorphism}\},$$ with a basis for its topology given by $$U(\hat{f}_0,F,\varepsilon)=\{\hat{f}\in\hat{G}\mid|\hat{f}(g)-\hat{f}_0(g)|<\varepsilon, g\in F\},$$ where $\hat{f}_0\in\hat{G}$, $F\subseteq G$ is finite and $\varepsilon>0$.

The isomorphism $C^\ast(G)\to\mathcal C(\hat{G})$ takes $g\in\Bbb CG$ to the function $\hat{f}\mapsto \hat{f}(g)$ for $\hat{f}\in\hat{G}$.

In the case of a locally compact $G$ we don't have $\Bbb CG=\mathcal C_C(G)$ anymore, but the latter is still a very interesting object, which can be turned into a $C^\ast$-algebra in the same two ways (with some care in the definition of a unitary representation and replacing $\ell^2(G)$ by $L^2(G)$ with respect to the Haar measure), where the product is convolution with respect to the Haar measure, while the involution is defined as follows, where $\Delta\colon G\to\Bbb R$ is the modular function of $G$:

$$a^\ast(g)=\frac1{\Delta(g)}\overline{a(g^{-1})},$$ note that discrete groups are unimodular so this agrees with the previous definition in the discrete case.

Once again if $G$ is abelian we get an isomorphism of $C^\ast(G)$ and $\mathcal C(\hat{G})$, where $\hat{G}$ is defined as above, but with continuous group homomorphisms instead.

Alessandro Codenotti
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