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Is the following statement true: "Every smooth manifold $M$, which is a ring in the category of manifolds, must be diffeomorphic to $\mathbb{R}^n$."? (Actually, homeomorphic would suffice.) I assume manifolds to be Hausdorff, second-countable and positive-dimensional, to exclude finite rings.

I have strong feelings that this must be the case. Is there a "simple" proof of it? I know next to nothing about the theory of Lie groups, so any argument using these would have to be simple for me to understand it. On the other hand, I feel quite comfortable with "standard" algebraic topology (elementary homotopy theory, homology, cohomology rings and so on...).

Edit: I am very sorry for not making this clear the first time, but I assume all manifolds to be without boundary.

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$\mathbb Z$ can be thought of as a $0$-manifold, but obviously you want to rule that out somehow. Maybe you want to assume it has dimension $n>0$. Anyway, nice question. – Grumpy Parsnip Oct 14 '11 at 18:30
(Let us suppose the connectedness) In particular, your $M$ will be an abelian group in the category of manifolds. Standard Lie theory shows that it must be of the form $\mathbb R^n\times (S^1)^m$ (group structure and all) for some $n,m\geq1$. – Mariano Suárez-Alvarez Oct 14 '11 at 18:45
I actually assumed manifolds to be positive dimensional. What I did not write out explicitly is that I assume manifolds to be without boundary. – Piotr Pstrągowski Oct 14 '11 at 19:04
@Piotr: since your $M$ will be a group, it will be homopgeneous and therefore will automatically without boundary. – Mariano Suárez-Alvarez Oct 14 '11 at 19:20
Ah, I guess connectedness is the right hypothesis since you could realize $\mathbb R[x]$ topologically as a disjoint union of countably many copies of $\mathbb R$, for example. – Grumpy Parsnip Oct 14 '11 at 21:29
up vote 8 down vote accepted

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 $R$.

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Nice!$\hspace{1em}$ – Grumpy Parsnip Oct 14 '11 at 20:54
Right! :D ${}{}$ – Mariano Suárez-Alvarez Oct 15 '11 at 0:09

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 $0\times(S^1)^m$.

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