Given a (topological) abelian group $G$ and a (bicontinuous) $G$-bilinear map $\mu: G \times G \to G$, clearly $G$ becomes a (topological) ring by specifying $$ x y := \mu(x, y) \quad \forall x, y \in G $$ Of course, we want associativity, so we also insist that $\mu(x, \mu(y, z)) = \mu(\mu(x, y), z)$.
But in general such a $\mu$ may be hard to find, and indeed is just the definition of the ring multiplication - so, in a sense, requires you to know ahead of time what you want your ring to be.
Can I get away with less initial data? What is the minimum amount of multiplicative structure that I can specify and still obtain a consistent ring structure? Can I give an extremely local description of how I want the multiplication to behave and still construct a sensible ring?
For instance, suppose I have a continuous function $\sigma: G \to G$ which fixes the identity; $\sigma(0_G) = 0_G$. We interpret (or wish to interpret) $\sigma(x)$ as $x^2$, and seek to define a ring structure compatible with this.
This is equivalent to having some kind of process that can take $\sigma$ and generate a collection of continuous $G$-homomorphims $$ \{\sigma_x : G \to G\}_{x \in G} $$ which also satisfy $$ \sigma_{x+y} = \sigma_{x} + \sigma_y \quad \forall x, y \in G $$ and $$ \sigma_x(x) = \sigma(x) $$ so that we may take $\sigma_x$ as `multiplication by $x$'; $$ xy := \sigma_x(y). $$ Associativity follows from that of function composition.
When can we actually describe such a process for finding the $\sigma_x$? When $G$ has additional structure such as smoothness, it seems like this should be (at least locally) possible. If the $\sigma_x$s cannot be extended globally, then is there some appropriate weakened ring structure (some `locally defined' multiplication?) in which they have a sensible interpretation?
Another equivalent formulation is that we define a $\sigma_\Delta$ on the diagonal subgroup $$ \Delta G = \{(x, x)|x \in G\}\subset G \times G $$ mapping into $G$ via $$ \sigma_\Delta(x, x) := \sigma(x) $$ and attempt to find a $G$-bicontinuous extension of $\sigma_\Delta$ to all of $G\times G$ which satisfies associativity. Are there existence conditions for such a thing?
If it is possible to answer the above, then for a given group $G$, how many essentially inequivalent $\sigma$s are there, in the sense that they give non-isomorphic ring structures? Is each ring structure uniquely determined by its $\sigma$, or can there be more than one compatible ring structure with each? Of course, all the above holds mutando mutandis for semirings, near-rings, Lie rings, and other algebraic structures.
