This is an obvious continuation of [an MO question](https://mathoverflow.net/q/431642). 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](https://mathoverflow.net/a/431644/) applies to the situation that one of $r$ and $s$ is $\infty$.



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


$\square$


