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?
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.