On the $\mathbb R$-algebra structure on $C^\infty(M)$.

Question

As it is more or less well-know, and as it has come up on MO a couple of times, the $\mathbb R$-algebra $C^\infty(M)$ of smooth functions on a (say) compact manifold contains essentially everything there is to know about $M$ itself.

Does one really need to know the $\mathbb R$-algebra structure, though? Can we manage with $C^\infty(M)$ given as an abstract ring?

For comparison, if I recall correctly one can reconstruct (as an analytic manifold) an open subset $U\subseteq\mathbb C$ from the field of meromorphic functions on it, given as an abstract field. (It follows, of course, that the ring of holomorphic functions also can be used to reconstruct $U$.)

Answer 1

If $M$ is connected then one can determine $\mathbb{R}$ (the constant functions) within $C^\infty(M)$ (the ring of smooth real-valued functions on $M$). One can certainly determine $\mathbb{Q}$ within $C^\infty(M)$. If $f$ is not in $\mathbb{Q}$ then it's a constant function iff $f-a$ is a unit in $C^\infty(M)$ for all $a\in\mathbb{Q}$. The key is that all nonconstant functions must take a rational value.

This doesn't work for non-connected $M$, since a function $f$ may take distinct constant values on a clopen partition on $M$. Maybe one can use idempotents...

Answer 2

You can determine the R-algebra structure of $C^\infty(M)$ purely from its ring structure. As Robin Chapman mentions, the constant function 1_M is uniquely determined by the fact that it is the identity element, and multiplication by rationals is uniquely defined, so the functions equal to a constant rational value are uniquely determined.

Actually, the ring homomorphism $F\colon\mathbb{R}\to C^\infty(M)$ is unique, which also uniquely defines the R-algebra structure.

The positive elements $x\in\mathbb{R}$ are squares, so $F(x)$ must be a square in $C^\infty(M)$, hence nonnegative everywhere. Then, for any $x\in\mathbb{R}$ and rational numbers $a\le x\le b$ we have $F(x)-a1_M=F(x-a)\ge0$ and $b1_M-F(x)=F(b-x)\ge0$, so $F(x)\in C^\infty(M)$ takes values in the interval $[a,b]$. This shows that, in fact, $F(x)=x1_M$.

Thinking about it, this works because $\mathbb{R}/\mathbb{Q}$ has trivial Galois group. You can see this by asking if the C-algebra structure on $A\equiv C^{\infty}(M,\mathbb{C})$ is uniquely determined by its ring structure, for which the answer is no. For any $\sigma\in{\rm Gal}(\mathbb{C}/\mathbb{Q})$ it is not possible to distinguish a constant $f\in A$ from $\sigma(f)$ in terms of ring operations [edit: if $\sigma$ is continuous, that is. So, only considering the identity element and complex conjugation]. Instead, you could ask if it is possible to determine the C-algebra structure up to the action of the Galois group. If the manifold is connected then the answer to this is yes. The constant function taking the value $\pm i$ everywhere is given by $i_M^2+1_M=0$, and the constant functions $f\in A$ are those for which $f-\lambda1_M-\mu i_M$ are units for all but at most one choice of $\lambda,\mu\in\mathbb{Q}$. The constant functions are isomorphic to $\mathbb{C}$, which is determined up to the action of the Galois group. If it is not connected, then we can't even say that much. For any locally constant map $\sigma\colon M\to{\rm Gal}(\mathbb{C}/\mathbb{Q})$, it is not possible to distinguish $f\in A$ from $f_\sigma(P)\equiv\sigma(P)(f(P))$ using ring operations. The C-algebra structure is uniquely determined up to the action of such a locally constant $\sigma$ though, which should still be enough to tell you everything about the manifold. Working over the reals, none of this matters, because of the triviality of the Galois group.