7
$\begingroup$

Consider the Fréchet spaces $C^\infty(\mathbb{R},\mathbb{R})$ and $\mathbb{R}^\infty$, and the continuous linear map $$ J\colon C^\infty(\mathbb{R},\mathbb{R}) \to \mathbb{R}^\infty $$ returning the infinite jet at 0, which is a surjection by Borel's lemma. Here $\mathbb{R}^\infty$ is the set of all sequences of real numbers, with the family of seminorms induced by the truncations to the first $n$-coordinates. The map $J$ does not have a continuous linear section, but it has a continuous non-linear section. What I'd like to know is if $J$ has a smooth nonlinear section, even if just in a neighbourhood of $0$. Here smoothness is taken in the sense of Michal–Bastiani.

I recently learned that smooth maps $\mathbb{R}^n \to \mathbb{R}^\infty$ lift smoothly to $C^\infty(\mathbb{R},\mathbb{R})$ (via Enxin Wu's paper Homological Algebra for Diffeological Vector Spaces), which makes this projection maps a subduction of the associated diffeological spaces, making our friend $J$ above a diffeological principal bundle (which, I remind you, are not assumed locally trivial!). But I do wonder if it's a bundle in the traditional sense, in the category of Fréchet manifolds. Perhaps a continuous section could be smoothed, but I really am grasping at straws.

$\endgroup$
6
  • $\begingroup$ Your question is full of jargon and hardly accessible to most of Overflow users. For example, why didn't you explain what $\mathbb{R}^\infty$ is? $\endgroup$ Oct 5, 2022 at 4:04
  • 1
    $\begingroup$ Hi Piotr. Thanks for the feedback. I'm sorry it's confusing, but there is explanation at the links. I'm not sure why this question is somehow more aimed at specialists and/or particularly egregious, compared to many others I could point to. Here's another reference that contains everything one needs to understand the first paragraph, which is all that is needed to answer the question (modulo sufficient expertise in functional analysis): math.stackexchange.com/a/63062/3835 The 2nd paragraph is set-dressing. $\endgroup$
    – David Roberts
    Oct 5, 2022 at 4:11
  • $\begingroup$ I still feel you could make the problem more accessible to a wider audience. I have known Borel's lemma for more than 30 years and I don't find your question appealing for for the lack of basic explanations. $\endgroup$ Oct 5, 2022 at 4:22
  • $\begingroup$ @Piotr that's ok. As long as the person who knows the answer feels like sharing their ideas, I'll be happy. $\endgroup$
    – David Roberts
    Oct 5, 2022 at 4:31
  • $\begingroup$ Borel's lemma is in fact a theorem of Peano who proved it a decade before Borel. Here is a link to a note of A. Besenyei in the Amer. Math. Monthly researchgate.net/publication/… $\endgroup$ Oct 5, 2022 at 6:50

1 Answer 1

6
$\begingroup$

No, there is not even any $C^1$ (in the Michal−Bastiani, i.e. Keller $C_c$ sence) map $\mathbb R^\infty=G\sqsupseteq{\rm dom}\,f\to E=C^\infty(\mathbb R)=C^\infty(\mathbb R,\mathbb R)$ with ${\rm dom}\,f$ a zero neighbourhood and $J\circ f={\rm id}$ on ${\rm dom}\,f$. The argument goes as follows. Supposing there is, for $j={\rm D}\,f(0)$ we have $J\circ j={\rm id}$ on $G$. Then for $F$ the subspace of $E$ formed by functions that are infinitely flat at $0$, and for $\rho:E\to F$ given by $x\mapsto x-j\circ J\,x$ we have $\rho$ the identity on $F$ which is shown to be impossible in Corollary 7.1.3 on page 206 in the Frölicher−Kriegl book Linear Spaces and Differentiation Theory. Note that since all spaces here are Fréchet, there is only one reasonable concept of smoothness that (in this restricted case) also equals that in the FK book. A $\underline{\rm Con}$−morphism (there and here; in this case) just means a continuous linear map.

$\endgroup$
4
  • $\begingroup$ Oh, excellent! This will take me a little to digest, so I can accept this after a little delay. $\endgroup$
    – David Roberts
    Oct 7, 2022 at 12:16
  • $\begingroup$ Can I ask: is this proof reducing the issue to that of a linear section of $J$ by looking at the derivative of the local section? And then proving that a linear section cannot exist, since we cannot have a retraction to the kernel? If that's the case, then I totally get it. $\endgroup$
    – David Roberts
    Oct 8, 2022 at 2:09
  • $\begingroup$ Yes, it is exactly so. $\endgroup$
    – TaQ
    Oct 8, 2022 at 7:30
  • $\begingroup$ Oh cool. I was worried that the tangent spaces wouldn't work out, but given that they do, this is an excellent argument. Thanks! $\endgroup$
    – David Roberts
    Oct 8, 2022 at 8:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.