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Let $M$ be a manifold. Then is the ring of smooth functions $C^\infty(M,\mathbb{R})$ formally smooth over $\mathbb{R}$?

If it helps, feel free to assume that $M$ is compact.


(This is not a joke question. And yes, I know about $C^\infty$-rings and topological algebras, but I'm still interested in the above as stated.)

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  • $\begingroup$ This is only a partial answer, maybe someone sees how to finish: for a point $x\in M$, we have the evaluation map $C^\infty(M)\to \mathbb{R}$. Its cotangent complex is really just $T^*_xM[1]$. To see this, we may localise to assume $M=\mathbb{R}^n$ with $x=0$, and then the kernel of the evaluation map is generated by the regular sequence of coordinates $x_1,..,x_n$. So from the conormal sequence we at least get that the base-changes of the cotangent complex $L_{C^\infty(M)/\mathbb{R}} $ along the evaluation maps are projective, given by $T_x^*M$. $\endgroup$ Nov 20, 2022 at 22:15
  • $\begingroup$ (one would hope before basechange it is given by sections of $T^*M$, but I'm currently not sure whether we can detect this at the residue fields) $\endgroup$ Nov 20, 2022 at 22:17
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    $\begingroup$ Is it formally smooth when $M=\mathbb N$, namely, what about the cotangent complex of ${\mathbb R}^{\mathbb N}$ over $\mathbb R$? $\endgroup$
    – Z. M
    Nov 21, 2022 at 9:05
  • $\begingroup$ Am I understanding correctly that these comments are suggesting to employ Quillen's criterion that a ring map $A \to B$ is formally smooth iff $\Omega_{B/A}$ is projective and the cotangent complex is homotopy equivalent to it in degree zero? According to this other question, this also holds without finite presentation, right? $\endgroup$ Nov 21, 2022 at 10:16
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    $\begingroup$ I believe that $k^I$ is not formally smooth over $k$ when $k$ is a field and $I$ is an infinite set. $\endgroup$ Nov 21, 2022 at 12:02

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It seems $C^\infty(M;\mathbb{R})$ is not formally smooth for any positive-dimensional manifold. (The following argument came up in a discussion with Thomas Nikolaus, we later also found it in this MO question):

Let's start by discussing the case $M=\mathbb{R}$. Our strategy will be to show that $\Omega^1_{C^\infty(M;\mathbb{R})/\mathbb{R}}$ is not a projective $C^\infty(M;\mathbb{R})$-module. For a projective module $P$ over some ring $R$, different elements $x,x'\in P$ can always be separated by homomorphisms $P\to R$ (e.g. by embedding $P$ into a free module and using coordinate functions), equivalently the map $P\to P^{\vee\vee}$ into the double dual is injective. So it suffices to exhibit distinct elements of $\Omega^1_{C^\infty(M;\mathbb{R})/\mathbb{R}}$ which cannot be separated by homomorphisms to $C^\infty(M;\mathbb{R})$.

Claim 1: In $\Omega^1_{C^\infty(\mathbb{R};\mathbb{R})/\mathbb{R}}$, $de^x \neq e^x dx$. To detect this, it suffices to find a derivation $\partial: C^\infty(\mathbb{R};\mathbb{R})\to M$ with $\partial e^x \neq e^x \partial x$. Let $K$ be the fraction field of the local ring of stalks of smooth functions on $\mathbb{R}$ at $0$. Then $e^x$ and $x$ are algebraically independent in $K$, so we may find a transcendence basis of $K$ over $\mathbb{R}$ consisting of $e^x$, $x$ and other elements $a_i$. It follows that $\Omega^1_{K/\mathbb{R}}$ is a free module on $de^x$, $dx$ and the $da_i$, in particular we find a derivation $K\to K$ taking $de^x$ to $1$ and $dx$ to $0$. Precomposing with the map $C^\infty(\mathbb{R};\mathbb{R})\to K$ yields the claim.

Claim 2: Every homomorphism $\Omega^1_{C^\infty(\mathbb{R};\mathbb{R})}\to C^\infty(\mathbb{R};\mathbb{R})$ takes $de^x$ and $e^x dx$ to the same element. Indeed, such homomorphisms correspond to derivations $C^\infty(\mathbb{R};\mathbb{R})\to C^\infty(\mathbb{R};\mathbb{R})$, which are given by smooth vector fields. But any smooth vector field $f\cdot \frac{\partial}{\partial x}$ takes $de^x$ to $f\cdot\frac{\partial}{\partial x} e^x$, and $e^x dx$ to $f\cdot e^x \frac{\partial}{\partial x} x$, which agree.

So $\Omega^1_{C^\infty(\mathbb{R};\mathbb{R})/\mathbb{R}}$ cannot be projective, and thus $C^\infty(\mathbb{R};\mathbb{R})$ is not formally smooth.

A zoomed-out version of the above argument is that the $C^\infty(M;\mathbb{R})$-linear dual of $\Omega^1_{C^\infty(M;\mathbb{R})/\mathbb{R}}$ is always vector fields, and so the double dual is always $1$-forms. Hence the above observation tells us that the map from "algebraic $1$-forms" $\Omega^1_{C^\infty(M;\mathbb{R})/\mathbb{R}}$ into its double dual identifies with the map to "smooth $1$-forms" $\Omega^1(M;\mathbb{R})$, and we have used algebraically independent functions with a linear dependence between their derivatives to show that map is not injective.

For $M=\mathbb{R}^n$, the same argument works for $de^{x_1}$ and $e^{x_1}dx_1$ with the coordinate function $x_1$. Finally, on a general manifold, take a coordinate ball around an arbitrary point, and extend the functions $e^{x_1}$, $x_1$ in any way. If $\psi$ is a function supported on our ball (and $1$ in a smaller neighbourhood of $x$), then we still have that $\psi \cdot d\widetilde{e^{x_1}}$ and $\psi\cdot \widetilde{e^{x_1}}d \widetilde{x_1}$ agree as "smooth $1$-forms", but not as "algebraic $1$-forms", as witnessed by going into the fraction field of stalks around our point.

EDIT: For a short proof that the stalks of $e^x$ and $x$ at $0$ are algebraically independent, first observe that since these are analytic functions, any polynomial relation $f(e^x,x)=0$ which holds in a neighbourhood of $0$ holds on all of $\mathbb{R}$, and in fact on all of $\mathbb{C}$. Since every nonzero $a$ is attained infinitely often as value of $e^x$, each of the polynomials $f(a,x)$ for fixed nonzero $a$ has infinitely many zeros in $x$, thus vanishes. So the coefficient polynomials in $f(y,x) = \sum g_i(y) x^i$ each have infinitely many zeros, thus vanish, and so $f=0$ as polynomial, proving algebraic independence.

EDIT2: In the zero-dimensional case, if $M$ is finite, $C^\infty(M;\mathbb{R})=\mathbb{R}^M$ is clearly formally smooth over $\mathbb{R}$. In the infinite case, they shouldn't be (following the comment by Martin Brandenburg). Indeed, a variant of the argument works. We still have that $C^\infty(M;\mathbb{R})$-valued derivations are vector fields, hence trivial. However, there exist nontrivial derivations on $\mathbb{R}^M$. Identify $M=\mathbb{N}$, fix a nonprincipal ultrafilter on $\mathbb{N}$, and let $K$ be the corresponding ultraproduct (which is automatically a field). The function $f: n\mapsto n$ is algebraically independent from $1$ in $K$, and so we find a derivation $K\to K$ taking $df$ to $1$. In particular, $df\neq 0$ in $\Omega^1_{\mathbb{R}^\mathbb{N}/\mathbb{R}}$. (If your manifolds are not second-countable, they still contain $\mathbb{N}$ as retract and you can still pull this nonzero element back.)

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  • $\begingroup$ Thanks, that's beautiful and beautifully elementary! $\endgroup$ Nov 22, 2022 at 9:53
  • $\begingroup$ What about the $0$-dimensional case? :-) $\endgroup$ Nov 22, 2022 at 9:54
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    $\begingroup$ I think a variant of my argument works, replacing the fraction field of stalks by a suitable ultraproduct. I edited my answer with details. $\endgroup$ Nov 22, 2022 at 10:17
  • $\begingroup$ I wonder whether the (algebraic) cotangent complex is concentrated in degree zero? $\endgroup$
    – Z. M
    Nov 22, 2022 at 11:51

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