Is $C^r(M)$ non-isomorphic to $C^s(N)$ for $r\neq s$ and nontrivial manifolds $M,N$?

Question

This is an obvious continuation of an MO question. Let $r,s\in\mathbb N\cup\{\infty\}$ with $r\neq s$, and $M,N$ two connected manifolds of positive dimension (which roots out the trivial case of a single point). I wonder whether the rings $C^r(M)$ and $C^s(N)$ of real-valued continuously $r$-th (resp. $s$-th) differentiable functions are non-isomorphic?

More precisely, we have the following (combinations of) variants of conditions on $M$ and $N$, although they might not affect the result nontrivially.

1. Smoothness:

• weakest: $M$ is a $C^r$-manifold and $N$ is a $C^s$-manifold;
• strongest: both $M$ and $N$ are $C^\infty$.

2. Compactness:

• weakest: both are paracompact Hausdorff (usually present in the definition of manifolds);
• strongest: both are compact Hausdorff.

There is also a variant of manifolds with boundaries, but I am not looking at this generality in this question.

It seems that, in this case, Morse's lemma might not hold, thus the structure of vanishing ideals $\mathfrak m_x=\ker(C^r(M)\to\mathbb R)$ at a point $x\in M$ might be more complicated (say, maybe not regular). However, it seems possible that Tom Goodwillie's answer applies to the situation that one of $r$ and $s$ is $\infty$.

Answer 1

If $r\neq s$, then the rings $C_r(M),C_s(N)$ cannot be isomorphic nor can they be elementarily equivalent to each other. I claim that for all $r<\infty$, there is a first order formula $\phi$ where $C_r(M)\models\phi$ if and only if $s=r$.

If $p/q$ is a reduced rational number where $q$ is odd, and $f$ is a real valued function (or real number), then we shall write $f^{p/q}$ for the unique function such that $(f^{p/q})^q=f^p$. While the function $f^{p/q}$ is unique, the function $f^{p/q}$ may no longer be contained in a ring of $r$-times continuously differentiable functions.

If $\alpha=p/q$ is a reduced rational number, then let $\phi_\alpha$ denote the first statement "For each $f$, the object $f^\alpha$ exists and is unique." More precisely, $\phi_\alpha$ is the statement "For each $f$, there is a unique $g$ with $g^q=f^p$."

Now, suppose that $\alpha=p/q$ is a reduced rational number with $q$ odd and that $\alpha$ is not an integer. Then $\frac{d^r}{dx^r}x^{\alpha}=\frac{\alpha!}{(\alpha-r)!}x^{\alpha-r}$ which is continuous and defined for every real number $x$ if and only if $\alpha>r$.

For non-logicians, the following proposition says that for $\alpha=p/q$ reduced, $q>1$, $q$ odd, the ring $C_r(M)$ is closed under the function $f\mapsto f^\alpha$ if and only if $\alpha>r$.

Proposition: Suppose that $\alpha=p/q$ is a reduced rational number with $q$ odd and where $\alpha$ is not an integer. Then $C_r(M)\models\phi_\alpha$ if and only if $\alpha>r$.

Proof: Suppose $\alpha<r$. Let $U\subseteq M$ and let $V\subseteq\mathbb{R}^n$ where both $U$ and $V$ are open. Let $\phi:U\rightarrow V$ be a chart. We can assume that $\mathbf{0}\in V$ since we may translate the set $V$ if necessarily. Now suppose that $g:V\rightarrow\mathbb{R}$ be a smooth function with compact support where $g|_W=1$ for some neighborhood $W$ of $\mathbf{0}$. Let $h:V\rightarrow\mathbb{R}$ be the function defined by letting $h(x_1,\dots,x_n)=x_1\cdot g(x_1,\dots,x_n).$ Let $f:M\rightarrow\mathbb{R}$ be the function defined by letting $f(\mathbf{x})=0$ for $\mathbf{x}\not\in U$ and $f(\mathbf{x})=h(\phi(\mathbf{x}))$ for $\mathbf{x}\in U$. Then $f\in C^r(M)$. On the other hand, $h(x_1,\dots,x_n)^\alpha=x_1^\alpha$ in a neighborhood of $\mathbf{0}$ which is not a $C^r$ function. Therefore, since $f(\mathbf{x})^\alpha=h(\phi(\mathbf{x}))^\alpha$ for $\mathbf{x}\in U$, the function $f^\alpha$ is not in $C^r(M).$

Now assume that $r<\alpha$ and $f\in C^r(M)$. Then $f^\alpha\in C^r(M)$ since $f^\alpha$ is the composition of the two $C^r$ functions, namely $f$ and $x\mapsto x^\alpha.$

$\square$

Answer 2

I claim that if $M,N$ are possibly non-paracompact manifolds and $C^r(M)$ is isomorphic as a ring to $C^s(N)$, then $r=s$ and $M$ is diffeomorphic to $N$. We shall prove this by establishing that the differentiable manifold $M$ is always an invariant of the ring $C^r(M)$, and we shall prove a more general result for a wider class of unital subalgebras of the form $\mathcal{A}\subseteq\mathbb{R}^X$ for some set $X$.

A function $f\in C^r(M)$ is said to be bounded if and only if there is some natural number $n$ where $\sqrt{n-f^2}$ exists. Therefore, the ring of all bounded functions in $C^r(M)$ is an invariant of the ring $C^r(M)$. We shall therefore recover the differentiable manifold $M$ from the ring $C^{r,*}(M)$ of bounded functions in $C^r(M)$. In fact, we shall prove a much more general result.

Let $X$ be a set, and let $\mathcal{A}\subseteq\mathbb{R}^X$ be a unital $\mathbb{R}$-subalgebra. We observe that the $\mathbb{R}$-algebra structure on $\mathcal{A}$ can be obtained from the ring structure on $\mathcal{A}$. We say that $\mathcal{A}$ separates points if whenever $x,y\in X,x\neq y$, there is some $f\in\mathcal{A}$ with $f(x)\neq f(y)$. We shall assume that all algebras separate points. We say that $\mathcal{A}$ is bounded if each $f\in\mathcal{A}$ is bounded. Suppose now that $\mathcal{A}$ is bounded. Let $e:X\rightarrow\mathbb{R}^{\mathcal{A}}$ be the mapping where $e(x)=(f(x))_{f\in\mathcal{A}}$ for each $f\in\mathcal{A}$. Then let $Y=\overline{e[X]}$. By associated $X$ with $e[X]$, we may assume that $X\subseteq Y$ and we may give $X,Y$ the subspace topology. In this case, each $f\in\mathcal{A}$ is continuous and extends to a unique $\overline{f}:Y\rightarrow\mathbb{R}$. Let $\mathcal{A}^\sharp=\{\overline{f}:f\in\mathcal{A}\}$.

We say that $\mathcal{A}$ is closed under bounded inversion if whenever $f\in\mathcal{A}$ and $1/f$ is bounded, then $1/f\in\mathcal{A}$ as well. If $y_0\in Y$, then we shall write $I_{y_0}$ for the ideal $\{f\in\mathcal{A}^\sharp:f(y_0)=0\}$.

Proposition: If $\mathcal{A}$ is bounded and closed under bounded inversion, then the maximal ideals in $\mathcal{A}^\sharp$ are precisely the mappings of the form $I_{y_0}$.

Proof: If $I\subseteq\mathcal{A}^\sharp$ is an ideal with $I\not\subseteq I_{y_0}$ for each $y_0\in Y$, then whenever $y\in Y$, there is some $f_y\in\mathcal{A}^\sharp$ with $f_y\geq 0$ and where $f_y(y)>0$. By compactness, there are $y_1,\dots,y_n\in Y$ where $(f_{y_1}+\dots+f_{y_n})(y)>0$ for all $y\in Y$. Therefore, we have $1/(f_{y_1}+\dots+f_{y_n})\in\mathcal{A}^\sharp$, so since $I$ is an ideal, we have $1=(f_{y_1}+\dots+f_{y_n})\cdot 1/(f_{y_1}+\dots+f_{y_n})\in I$. Therefore, the maximal ideals are precisely the ideals of the form $I_{y_0}$ for $y_0\in Y$. $\square$

In particular, if $\mathcal{A}$ is bounded and closed under bounded inversion, then whenever $M$ is a maximal ideal in $\mathcal{A}^\sharp$, then $\mathcal{A}^\sharp/M\simeq\mathbb{R}$. Therefore, the unital homomorphisms $\phi:\mathcal{A}^\sharp\rightarrow\mathbb{R}$, the maximal ideals in $\mathcal{A}^\sharp$ and the points in $Y$ are all in a one-to-one correspondence with each other, and the topology on $Y$ is an invariant of the ring $\mathcal{A}$. We now just need to be able to distinguish the subset $X\subseteq Y$ only using the ring structure of $\mathcal{A}$.

We say that $\mathcal{A}$ has the splitting property if whenever $(U_n)_{n\in\omega}$ is a locally finite and pairwise disjoint collection of non-empty open sets, there is some $f\in\mathcal{A}$ along with infinite subsets $B,C\subseteq\omega$ where $B\cap C=\emptyset$ and where if $n\in B$, then $f(x)>1$ for some $x\in U_n$ and if $n\in A$, then $f(x)<0$ for some $x\in U_n$.

For example, if $M$ is a manifold, then $C^r(M)$ always satisfies the splitting property.

Proposition: If $\mathcal{A}$ is a bounded algebra that satisfies the splitting property, then no point in $Y\setminus X$ has a countable local basis with respect to the space $Y$.

Proof: Suppose for the sake of contradiction that a point $y\in Y\setminus X$ does have a countable local basis $(O_n)_{n\in\omega}$ where $\overline{O_{n+1}}\subseteq O_n$ for $n\geq 0$ and where each $\overline{O_{n+1}}\subseteq O_n$ is nonempty. Now, set $U_n=O_{2n}\setminus\overline{O_{2n+1}}$. Then $(U_n\cap X)_{n\in\omega}$ is our locally finite family of pairwise disjoint open subsets of $X$. Therefore, there is some $f\in\mathcal{A}$ and $(x_n)_{n\in\omega}$ where $x_n\in U_n\cap X$ for all $n$ but where $f(x_n)>1$ for even $n$ and $f(x_n)<0$ for odd $n$. In this case, $x_n\rightarrow y$, but the sequence $(\overline{f}(x_n))_{f\in\mathcal{A}}$ does not converge. This contradicts the continuity of $\overline{f}$. $\square$.

If $X$ is locally compact and first countable and $\mathcal{A}$ satisfies the splitting property, then $X$ is the collection of all points in $Y$ that have a countable local basis with respect to $Y$. We therefore have the following result which allows us to recover up-to-isomorphism the set $X$ and the embedding of $\mathcal{A}$ into $\mathbb{R}^X$ from only the unital $\mathbb{R}$-algebra structure of $\mathcal{A}$.

Proposition: Suppose that $(X_i,\mathcal{A}_i)$ is a pair for $i\in\{0,1\}$ where $\mathcal{A}_i\subseteq\mathbb{R}^{X_i}$ is a bounded unital $\mathbb{R}$-subalgebra where each $X_i$ is first countable, locally compact, and where each $\mathcal{A}_i$ is closed under bounded inversion and satisfies the splitting property.

Then $\mathcal{A}_0$ is isomorphic to the ring $\mathcal{A}_1$ if and only if there is a bijection $L:X_0\rightarrow X_1$ where if $f:X_1\rightarrow\mathbb{R}$, then $f\in\mathcal{A}_1$ if and only if $f\circ L\in\mathcal{A}_0.$ In particular, if each $X_i$ is a manifold and $\mathcal{A}_i=C^{r_i,*}(X_i)$ for each $i$, then $\mathcal{A}_0$ is isomorphic to the ring $\mathcal{A}_1$ if and only if $r=s$ and the manifolds $X_0,X_1$ are diffeomorphic.