Topological rings which are manifolds

Solution 1:

If your ring $R$ is a (path-)connected topological manifold, then it is contractible. Those posting previously have mentioned already that your manifold must be of the form $\mathbb{R}^n \times (S^1)^m$, and contractibility excludes any factors of $S^1$.

To show this, choose a path $\gamma$ from $1$ to $0$ in $R$. Then the map $$ H(r,t) = r \cdot \gamma(t) : R \times [0,1] \to R $$ is a contraction of $R$.

Solution 2:

A couple of notes. Let us suppose $M$ is connected.

$M$ will be an abelian group in the category of manifolds. Since it act on itself by translation homogeneously, it is an homogeneous manifold and therefore has no boundary. Standard Lie theory shows that it must be of the form $\mathbb R^n\times (S^1)^m$ for some $n,m\geq0$.

The set of elements $x\in M$ such that the additive subgroup $(x)$ has compact closure is an ideal.

The set of torsion elements is an ideal, and it is proper and non-zero. It follows that the closure is also an ideal: this is $0\times(S^1)^m$.