Group Rings of Topological Groups and Fields 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?
 A: 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.
