Timeline for Integrals from a non-analytic point of view
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50 events
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| May 10, 2020 at 2:49 | history | edited | Dmitri Pavlov | CC BY-SA 4.0 |
added 1261 characters in body
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Nov 12, 2016 at 2:56 | comment | added | Dmitri Pavlov | @Arrow: Additional details appear here in the comments: mathoverflow.net/questions/43681/motivating-the-de-rham-theorem/… | |
| Nov 12, 2016 at 0:28 | comment | added | Arrow | @DmitriPavlov have you by any chance written up the details of an integration-free construction? I (and surely many others) would love to learn it. | |
| Apr 21, 2015 at 18:03 | comment | added | Dmitri Pavlov | @YemonChoi: Many thanks for your efforts. I very much like TeX formatting, but not if it takes several minutes to process… | |
| Apr 21, 2015 at 0:52 | comment | added | Yemon Choi | and rolled back again ... | |
| Apr 21, 2015 at 0:52 | history | rollback | Yemon Choi |
Rollback to Revision 3
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| S Apr 21, 2015 at 0:50 | history | suggested | user5794 | CC BY-SA 3.0 |
Basic LaTex
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| Apr 21, 2015 at 0:04 | review | Suggested edits | |||
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| Apr 20, 2015 at 23:28 | comment | added | Yemon Choi | [admin note: I just rolled back a "Let's put LaTeX in something four years old" edit, because IIRC Dmitri very deliberately avoids MathJax. If I misremembered, Dmitri, and you would like "LaTeX" formatting, then please let us know] | |
| Apr 20, 2015 at 23:26 | history | rollback | Yemon Choi |
Rollback to Revision 1
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| S Apr 20, 2015 at 22:46 | history | suggested | Arrow | CC BY-SA 3.0 |
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| Apr 20, 2015 at 22:35 | review | Suggested edits | |||
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| Dec 15, 2010 at 21:14 | comment | added | Yemon Choi | @Dmitri: OK, thanks, especially for the last comment. (For what it's worth I learned a large chunk of my functional analysis without any measure theory, by dealing with completions of various normed spaces; but I had to learn the rudiments of measure theory during a course in (rigorous) probability theory. So I have an ambivalent attitude towards it myself.) | |
| Dec 15, 2010 at 20:42 | comment | added | Dmitri Pavlov | @Yemon: I do not think that the word “strident” justly describes the situation here. The question asked for the “correct natural definition of the integral from a non-analytic point of view” and I suggested one. In general, every mathematical theory should be developed in the right context. Theories like integration on smooth manifolds, general topology, and measure theory were first developed as a part of analysis but now we realize that the proper context for them is different. | |
| Dec 15, 2010 at 20:39 | comment | added | Dmitri Pavlov | @Yemon: This stuff is too elementary to be included in my thesis, but I might write it up at some point, perhaps as a blog post. I agree that addtional arguments are required to justify the claim. | |
| Dec 15, 2010 at 19:03 | comment | added | Yemon Choi | Finally: I think it is far from obvious that this "construction of the Lebesgue integral" is non-circular. I am not saying that it is, but the various dualities and completions you rely on as you move through the argument in your answer and in some of your comments need independent justification; this is why I would be interested to see a fuller account written up (at some future date). | |
| Dec 15, 2010 at 18:58 | comment | added | Yemon Choi | Afterthought: I really think it would help if you were a little less strident in your promotion of (NC)geometric ideas, topics and toolkits above all else. Not everything has to be geared towards Doing Better Geometry, surely? | |
| Dec 15, 2010 at 18:52 | comment | added | Yemon Choi | Just seen this. Dmitri, are you planning to write up your answer and your various claims in this comment thread in a more formal form (either as an expository article or as part of your thesis)? I would be interested to read this, although currently I share some of Johannes' qualms - if you don't know a priori that a chain complex of Frechet spaces is exact (when regarded as a complex of vector spaces) then it is not true that the "spaces of boundaries" are always closed in "the spaces of cycles", at least not without additional justification/hypotheses. | |
| Nov 8, 2010 at 12:25 | comment | added | Dmitri Pavlov | @Johannes: What I meant is that you should use the fourth criterion in the above list. However, the proof seems to be more technical than Schapiro's proof combined with the proof of surjectivity of differentiation via Lagrange's theorem, therefore I suggest we stick to the latter proof. | |
| Nov 3, 2010 at 18:42 | comment | added | Johannes Ebert | absurd conclusion that Im(u) is complete, hence closed in F. You can apply the open mapping theorem to show that Im(u) is closed in F if you know that the quotient and subspace topology are equivalent. | |
| Nov 3, 2010 at 18:37 | comment | added | Johannes Ebert | @Dmitri: but you need to read this with care. I do not have the English edition here, but the word "strict" means something, probably to the effect that u is a quotient map tom its image. Consider a map of Banach spaces u: E \to F, injective, dense image, but not surjective. Examples abund. Im(u) has two norms, the quotient norm (in which it is complete since E is complete) and the restriction of the norm of F (in which Im(u) is not complete since it is not closed in F). u is strict if you consider the first norm, but it is not strict in the second norm, because that would lead to the. | |
| Nov 3, 2010 at 17:36 | comment | added | Dmitri Pavlov | @Johannes: Perhaps you are referring to one of the older editions of Bourbaki. I use the 1987 English translation, which says “Let E and F be two metrisable vector spaces over a non-discrete valued division ring K, and let u be a continuous linear mapping of E in F. Suppose that E is complete. Then the following conditions are equivalent: (i) u is a strict surjective morphism. (ii) F is complete and u is surjective. (iii) The image of u is not meagre in F (GT, IX, § 5.2). (iv) For every neighbourhood V of 0 in E, the set \overline{u(V)} is a neighbourhood of 0 in F.” | |
| Nov 3, 2010 at 17:04 | comment | added | Johannes Ebert | @Dmitri: Bourbaki, loc. cit.: "Soient E et F deux espaces vectoriels metrisables et complets sur un corps value non discret K. Toute application lineaire continue u de E sur F est un homomorphisme". "homomorphisme"=open linear continuous map, "sur": surjective, "complet"=complete. Otherwise, any operator from a complete source would have a complete image, which is absurd. | |
| Nov 3, 2010 at 13:48 | comment | added | Dmitri Pavlov | @Johannes: No, the open mapping theorem requires only the domain of the map to be complete, the completeness of the codomain is implied by the theorem. Reference: Theorem I.3.1 in Bourbaki's Topological Vector Spaces. | |
| Nov 2, 2010 at 19:46 | comment | added | Johannes Ebert | @Dmitri: the open mapping theorem has an assumption: completeness of the target. | |
| Nov 2, 2010 at 19:40 | comment | added | Johannes Ebert | @Dmitri: "You can easily derive Hodge theory from the elliptic regularity theorem". This is how it is done. Elliptic regularity, however, is the problem. "However, I prefer to develop integration first, then prove the elliptic regularity theorem": so do I, but I do not think this is a matter of opinion. | |
| Nov 2, 2010 at 18:16 | comment | added | Dmitri Pavlov | @Johannes: The nuclear space of forms is complete, therefore the closure of the subspace of exact forms is isomorphic to its completion. Thus the space of exact forms is closed if and only if it is complete. The space of (k+1)-exact forms is isomorphic to the space of k-forms modulo closed k-forms as a topological vector space (a consequence of the open mapping theorem). Closed k-forms form a closed subspace of the complete nuclear space of k-forms, therefore the factor space is complete. | |
| Nov 2, 2010 at 11:56 | comment | added | Dmitri Pavlov | @Johannes: You can easily derive Hodge theory from the elliptic regularity theorem. If you prove the latter without integration, you obtain an integration-free exposition of Hodge theory. However, I prefer to develop integration first, then prove the elliptic regularity theorem. | |
| Oct 30, 2010 at 13:14 | comment | added | Johannes Ebert | @Dmitri: "The subspaces of exact forms and closed forms are closed in this topology". Both statements are true, one proof is trivial, the other one isn't. How do I see that the space of exact forms is closed? | |
| Oct 30, 2010 at 13:05 | comment | added | Johannes Ebert | @Dmititri: well, in terms of logic, you are of course right. But the fact is that ALL textbooks I know (around 20 texts) make heavy use of Lebesgue integration theory (or quote the main technical result, or do not say anything specific about the theory). Where, exactly, can I read about an "integration-free" approach to the basic result of Hodge theory, i.e. the statement that each cohomology class (under the known assumptions) has a unique harmonic representative?? | |
| Oct 27, 2010 at 15:22 | comment | added | Dmitri Pavlov | @Johannes: The fact that some textbooks use integration to develop Hodge theory does not imply that you cannot develop Hodge theory without integration. | |
| Oct 26, 2010 at 18:15 | comment | added | Johannes Ebert | @Paul Siegel: "use Hodge theory (which I think is integration-free)" Nope, take a look at an arbitrary textbook. | |
| Sep 14, 2010 at 9:13 | comment | added | Dmitri Pavlov | @Victor: The space of polynomial functions that are equal to their own derivative is not dense in the space of all functions that are equal to their own derivative. | |
| Sep 14, 2010 at 6:31 | comment | added | Victor Protsak | Hmm... There are some issues with this argument. Consider the following "lemma": only constant function is equal to its own derivative. Observe that the lemma is true for polynomial functions. Now consider the complete nuclear space of smooth functions equipped with the natural topology. The subspace of functions equal to its own derivatives and zero subspace are closed in this topology. Every polynomial function that is equal to its own derivative is zero, hence zero subspace contains the closure of the subspace of polynomial functions equal to its own derivative. Have we proved the lemma? | |
| Sep 13, 2010 at 16:32 | comment | added | Dmitri Pavlov | Here is one way to prove Poincaré's lemma without integration for a finite-dimensional real inner product space V. Observe that the lemma is true for polynomial forms. Now consider the complete nuclear space of all smooth k-forms equipped with the natural topology. The subspaces of exact forms and closed forms are closed in this topology. The subspace of all polynomial closed forms is dense in the subspace of all closed forms. Every polynomial closed form is also exact, hence the subspace of exact forms contains the closure of the subspace of polynomial closed forms. | |
| Sep 13, 2010 at 16:15 | comment | added | Dmitri Pavlov | @Victor: Measure theory = integration theory. @Robin: There is no “anti-measure theory ideology” in my answer. I just point out that measure theory for smooth manifolds can be developed much more easily (modulo existing tools) than in the standard approach (modulo existing tools). | |
| Sep 13, 2010 at 5:29 | comment | added | Per Vognsen | Count me in as interested in a integration-free proof of Poincare's lemma for de Rham cohomology. Integration exhibits a particular form in terms of the smooth contraction map. An alternative existence proof would have to be non-constructive. Maybe you can rephrase it as an existence problem for an associated Laplacian when the set is geodesically convex, cf. Paul's suggestion of Hodge theory? | |
| Sep 12, 2010 at 22:02 | comment | added | Yemon Choi | What does this proof or strategy do in the case where M=(0,1) or [0,1] or the circle? | |
| Sep 12, 2010 at 17:50 | comment | added | Paul Siegel | An integration-free proof of Poincare duality can be found in section 3.1 here: math.uni-muenster.de/u/lueck/publ/lueck/ictp.pdf The idea is to first prove Poincare duality abstractly (i.e. for topological manifolds) and then use Hodge theory (which I think is integration-free) to bring differential forms into the picture. It may seem a little perverse to define integration via Hodge theory, but on the other hand surgery theory seems to tell us the Hodge theory is deeply relevant to smooth manifold theory (via the signature theorem). | |
| Sep 12, 2010 at 17:42 | comment | added | Robin Chapman | Me too, Victor :-) I don't understand this anti-measure theory ideology either. I'm the first to want to avoid lengthy and technical proofs ('cos I can't understand them), but I prefer not to replace them with even more lengthy and even more technical proofs. :-) | |
| Sep 12, 2010 at 17:34 | comment | added | Victor Protsak | I don't understand the emphasis on measure theory: suppose we stick to differential forms version of the de Rham theory, then the corresponding notion of integral is "Riemann integral on manifolds". What I'd like to know is what proof of the Poincaré lemma doesn't use integration! | |
| Sep 12, 2010 at 16:58 | comment | added | Dmitri Pavlov | @Robin: Also, I do not agree with this “in favor of“ ideology. Proofs in measure theory can hardly be useful for anything else. On the other hand, Poincaré duality and de Rham theorem have independent value. | |
| Sep 12, 2010 at 16:55 | comment | added | Dmitri Pavlov | @Robin: Of course, integration is one way to prove Poincaré's lemma, but certainly not the only one. | |
| Sep 12, 2010 at 16:50 | comment | added | Robin Chapman | So I should amend my original statement to "So, you are discarding the lengthy and technical proofs in measure theory in favour of Poincare duality and the de Rham theorem". (It is fortunate that these have short and non-technical proofs :-)). Alas, the proof of de Rham's theorem I have come closest to understanding is that in Warner's book. He relies on Poincare's Lemma (for the proof that the sheafified de Rham complex is exact) and for that he uses integration :-( | |
| Sep 12, 2010 at 16:16 | comment | added | Dmitri Pavlov | @Robin: As you might know, the de Rham cohomology is canonically isomorphic to the singular cohomology with real coefficients. | |
| Sep 12, 2010 at 16:09 | comment | added | Robin Chapman | Neither does Hatcher's proof use de Rham cohomology. | |
| Sep 12, 2010 at 15:49 | comment | added | Dmitri Pavlov | @Robin: For example, see the proof of Poincaré duality in Hatcher's textbook (Section 3.3). It does not use integration. | |
| Sep 12, 2010 at 15:25 | comment | added | Robin Chapman | So you are discarding the lengthy and technical proofs in measure theory in favour of Poincare duality. So, how does one construct the duality isomorphism $H^n_c(M,\mathbb{Or}(M))\to H_0(M)$ (presumably with real coefficients) without integration? I'd appreciate any hints/references (even if just for the case $M=\mathbb{R}^n$). | |
| Sep 12, 2010 at 14:58 | history | answered | Dmitri Pavlov | CC BY-SA 2.5 |