Timeline for Topology on space of hyperfunctions
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 22 at 7:01 | vote | accept | Peter Scholze | ||
| May 21 at 6:34 | answer | added | crow | timeline score: 7 | |
| May 13 at 13:04 | comment | added | Jochen Wengenroth | (continued) Let $X$ be a (nice locally convex) vector space (like $\mathscr D'(\Omega)$ or $\mathcal O(\mathcal C\setminus\mathbb R)$), $L$ a subspace (possibly dense like $\mathscr E(\Omega)$ or $\mathcal O(\mathcal C)$) and $T:X\to X$ a (continuous linear) operator with $T(L)\subseteq L$. Then $T$ induces a linear map $\tilde T$ on $X/L$ which is surjective if and only if $S:X\times L\to X$, $(x,\ell)\mapsto T(x)+\ell$ is surjective. The point is that $S$ is an operator between nice spaces so that surjectivity can be investigated by functional analytic tools (for Fréchet spaces and beyond). | |
| May 13 at 12:56 | comment | added | Jochen Wengenroth | $\mathcal B(\mathbb R)$ is a function space, one wants to do functional analysis with it (like solving differential equations)... : There is a theorem of Hörmander about solving (constant coefficient linear) PDEs in $\mathscr D'(\Omega)/\mathscr E(\Omega)$ where solvability is characterized by $P$-convexity for singular supports. As for hyperfunctions, this space does not seem to have any reasonnable good topology since $\mathscr E(\Omega)$ is dense in $\mathscr D'(\Omega)$. Nevertheless, one can do functional analysis by a quite trivial trick... | |
| May 13 at 5:59 | history | edited | Peter Scholze | CC BY-SA 4.0 |
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| May 11 at 9:33 | comment | added | Peter Scholze | @PedroLauridsenRibeiro Thanks!! Just to be clear, I also don't have a proposal for a nontrivial topology, but rather for a generalization of "topology" that does apply to $\mathcal B(\mathbb R)$. It does go in the direction of considering stacks, but the resulting object is not actually stacky (no nontrivial automorphisms). | |
| May 11 at 5:55 | comment | added | Pedro Lauridsen Ribeiro | (continued) It's not clear to me how to recover a global topology from the local pieces (which, as you've pointed out, do have a nontrivial topology) through that machinery. Maybe some stack-theoretical reasoning could help here, I don't know if this has ever been tried but the rough idea seems natural, considering that stacks also deal well with quotients. | |
| May 11 at 5:43 | comment | added | Pedro Lauridsen Ribeiro | One thing that really gets in the way is that in order to identify distributions with hyperfunctions one first needs to decompose the former as a locally finite sum of distributions with compact support using a partition of unity, which then may be identified with analytic functionals and hence with hyperfunctions. This first step has to be replaced with a Mayer-Vietoris-type sheaf cohomological argument in order to localize a hyperfunction into its compactly supported pieces because there are no real analytic partitions of unity. | |
| May 11 at 5:35 | comment | added | Pedro Lauridsen Ribeiro | Up to as recently as 2018 there seems to be no proposal of a nontrivial topology for spaces of hyperfunctions. J. Bonet and M. Langenbruch in their report on P. Domanski's work in functional analysis (Functiones et Approximatio 59 (2018) 7-39) still claim that "one technical problem (...) is the fact that hyperfunctions do not have a useful topology." There is the approach by Hörmander and Komatsu that treats hyperfunctions on open subsets of $\mathbb{R}^n$ as Dirichlet boundary values of harmonic functions on certain open subsets of $\mathbb{R}^{n+1}$, but it suffers from the same problem. | |
| May 10 at 23:43 | comment | added | Pedro Lauridsen Ribeiro | Ah, OK. Figured it couldn't possibly be the same thing, just wanted to be sure what you meant by it. Thanks! | |
| May 10 at 23:35 | comment | added | Peter Scholze | (And all other $\mathrm{Ext}$ groups vanish.) | |
| May 10 at 23:26 | comment | added | Peter Scholze | @PedroLauridsenRibeiro Of course there are no compactly supported real-analytic functions. But one can still take the compactly supported cohomology of the sheaf of real-analytic functions, and it will have cohomology only in degree $1$, where one will find a nonzero $H^1_c(\mathbb R^n,\mathcal A)$ (with bad topological properties, but it's possible to consider it as condensed $\mathbb R$-vector space). Then $\mathcal B(\mathbb R^n)$ is given by $\mathrm{Ext}^1(H^1_c(\mathbb R^n,\mathcal A),\mathbb R)$. | |
| May 10 at 19:16 | comment | added | Pedro Lauridsen Ribeiro | I'm not familiar with condensed mathematics, so I don't follow your last comment. There are no compactly supported real analytic functions on $\mathbb{R}^n$ besides $0$ - more precisely, because the Taylor coefficients of any smooth, compactly supported function $f$ around any $x_0\in\partial(\text{supp} f)$ are all zero. Since the Taylor series around $x_0$ converges in a neighborhood of $x_0$ by the definition of real analyticity, we should have $f\equiv 0$. What am I missing here? Is the definition of support in condensed mathematics different from the usual one employed in real analysis? | |
| May 10 at 10:19 | comment | added | Peter Scholze | By the way, in the condensed formalism, $\mathcal B(\mathbb R)$ is, as expected, the dual space of the space of compactly supported real-analytic functions $\mathcal A_c(\mathbb R)$, if the latter is understood to be a complex (in this case, supported in degree $1$). | |
| May 10 at 10:17 | comment | added | Peter Scholze | @C.F.G This is a good reference! But it effectively discusses the topology of point 1 in my answer, and then in Note 4.1 remarks that it does not seem possible to put any topology on $\mathcal B(\mathbb R)$: "Not only do we not have, at present, a reasonable way of defining a topology for the space of hyperfunctions on an open set, but it is the common understanding that there exists no topology in the ordinary sense." My question is whether someone found a non-ordinary sense. | |
| May 10 at 9:58 | comment | added | C.F.G | Take a look at page 151 of Introduction to hyperfunctions By A. Kaneko , it might help. | |
| May 10 at 6:30 | comment | added | Peter Scholze | @FedorPetrov Use the cover of $\mathbb C$ by $\mathbb C\setminus \mathbb R$ and $U$, and that the corresponding Cech complex is exact as $\mathbb C$ is Stein. | |
| May 10 at 2:20 | comment | added | Fedor Petrov | Is it clear that $\mathcal O(U\setminus \mathbb R)/\mathcal O(U)$ does not depend on $U$? | |
| May 9 at 19:46 | history | asked | Peter Scholze | CC BY-SA 4.0 |