Talk:Fredholm kernel

message: So what if the spectrum is λn = 1/√n?  Then the operator is 4-summable, and in fact the inf of all p for which the sum exists is 2.  I would like to say that this operator is of order 2, but the article claims that I only consider the inf of 0≤p≤1.  In this case, there are none, what's going on here? -lethe talk [ +] 19:09, 7 March 2006 (UTC)
replies: text: I'm not sure of what you meant by the square root. If \lambda_n\sim 1/n^2 then its summable for any p greater than 1/2. I think it follows from the definition that its always summable for p=1, but I doubt I could reproduce this proof suficiently well to get a passing grade.:) (see section "properties": evry Fredholm kernel has ... such that sum_n lambda_n < infty".) linas 02:29, 8 March 2006 (UTC)
replies: text: Oops, I've put my radical in the wrong place.  There, I've fixed it now.  OK, so anyway, we're in agreement that 1/n2 is p-summable for p>1.  So why does the article restrict only to things which are summable for 0<p<1?  That's my question.  I know lots of series for which the minimum p is greater than 1.  -lethe talk [ +] 02:38, 8 March 2006 (UTC)
replies: text: Yes, but the article claims "ever fredholm kernel can be written as \sum_n \lambda_n e_n e^*_n such that \sum_n |\lambda_n| < \infty" which sounds to me like a theorem that says its at least 1-summable. So p=1 is an upper limit, and maybe one can go lower. Should we try to prove this theorem? See how far we can get? linas 02:22, 10 March 2006 (UTC)
replies: text: Oh, of course you're absolutely right.  In fact, I guess we don't even have to prove it: it's part of the definition.  If the diagonal is not at least l-1 summable, it's not Fredholm. -lethe talk [ +] 05:08, 10 March 2006 (UTC)
replies: text: No, it still needs proof. The definition asks one to take the "inf". One has to prove that the result of taking that inf is that every element of the product space has a representation that is summable. This is rather powerfully counter-intuitive. (Or I'm confused). linas 05:24, 10 March 2006 (UTC)

text: I'm adding this to my list of things to do. There's something just so very odd about this. Or the article may be wrong. Maybe summability is part of the definition. In which case I feel silly. linas 15:16, 10 March 2006 (UTC)
replies: text: I don't find Fredholm kernel in any of my texts, so I don't really know where to start.  Should I be thinking of a Fredholm kernel as an integral kernal whose integral operator is a Fredholm operator? -lethe talk [ +] 16:56, 10 March 2006 (UTC)
replies: text: I wrote a general overview article at Fredholm theory which anchors all of the various concepts into place; and jammed all of the related articles into :Category:Fredholm theory. Unfortunately, this article is far and away the most abstract of the entire collection, and the overview does not make the final leap to the beastie discussed here. The only reference I currently have to the topic of this article is a book on ergodic theory, edited by Caroline Series. Unfortunately, it is equally brief in its treatment, and focuses mostly on the transfer operator.  I would need to hit the library to dig up more info. linas 21:44, 12 March 2006 (UTC)

text: Also, look at the "Fredholm kernel" on Springer link that I gave up above. It won't tell you much, but will be an independent source for linking these concepts together. Also, go to the index "F" on springerlink, there will be a bunch more articles on interconnected Fredholm topics. linas 21:55, 12 March 2006 (UTC)
replies: text: You're right that this article is rather abstract, but actually, I think this is probably the wrong place for such abstraction; this article is supposed to be about a class of integral kernels, not operators between Banach spaces.  That's what it claims in the intro, and that's what the name of the article would imply (Fredholm kernel), while the more abstract treatment should be elsewhere (perhaps Fredholm operator?).  Or at least, that's the way it appears to me, but I don't know for sure. -lethe talk [ +] 00:06, 13 March 2006 (UTC)
replies: text: Yes, well, the less-abstract articles dealing with the same subject can be found at nuclear operator, trace-class and Fredholm operator. I tried to untangle these as best as I can; that's why the intro points back to Fredholm theory for the "big picture". I suppose this article could be renamed as "Grothendieck's theory of Fredholm kernels" but I don't like that title  linas 04:45, 13 March 2006 (UTC)
replies: text: That title would be equally inappropriate for an article in which integral kernels are not mentioned a single time. -lethe talk [ +] 14:20, 13 March 2006 (UTC)

text: I don't have a problem with Fredholm theory being an abstract framework.  Here is my problem: this article says that a Fredholm kernel is an integral kernel in the first sentence.  Integral kernels only make sense on spaces of integrable functions.  Then later they are defined as elements of the completion of a tensor product of an arbitrary Banach space.  This has to be a mistake.  Now surely it's possible that Fredholm theory can be about an abstract class of Banach spaces of which these integral kernels form an important subclass; hell, maybe the two classes are equal, but as the article stands now, it's confusing at best, misleading at worst.  As soon as I figure this stuff out, I'll fix it, but unfortunately none of my functional analysis books (Rudin, Reed and Simon) nor my EDM2 encyclopedia treats Fredholm kernels. -lethe talk [ +] 02:32, 14 March 2006 (UTC)
replies: text: I'm a little frustrated. This conversation keeps going in circles. Yes, this article could be better. I just did a major expansion of the whole collection of fredholm theory articles in order to satisfy your requests, and you're still not happy :(. If you are at all near a univeristy library, go prowl the shelves containing books on operator theory and topological vector spaces. I myself am at the limit of what I know about these topics, and couldn't proceed further without reading a book on it. There's not more that I can do right now. linas 02:03, 16 March 2006 (UTC)
thread_title:  why is p<1 in p-summability? 

message: OK, while day-dreaming, I came up with an utterly hand-waving connection between the topics. 
:1) Let D be a differential operator. We wish to solve Dg(x)=f(x) given a function f(x).
:2) Use the method of Green's functions, namely solve instead DK(x,y)=δ(x-y) where δ is the Dirac delta function and K(x,y) is the Green's function.
:3) But the green's function is nothing but the Fredholm integral equation, viz. our desired solution is \int f(x) K(x,y) dx = g(y).
:4) Obtain the eigenvalues \omega_n and eigenvectors \psi_n(x) of D that is, solve D\psi_n=\omega_n\psi_n.
:5) The Green's function may then be written as K(x,y)=\sum_n \frac {\psi_n^*(y)\psi_n(x)}{\omega_n}
:6) The above is then recognizable as the Fredholm representation, where \lambda_n = 1/\omega_n.
:7) Prove that \lambda_n = 1/\omega_n is 1-summable. Arghhhh. I'm irritated that neither the harmonic oscillator nor the hydrogen atom are summable. (where I take D to be the Hamiltonian H of each system, respectively). So much for a "proof". :-( 
Anyway, that's sort-of the general idea. Ideally, there's an expanded cluster of articles that state this correctly, without the hand-waving.  linas 05:00, 10 March 2006 (UTC)
replies: text: About the hamiltonian not being trace-class, well I sort of expect that to be a general feature of quantum theories; all the infinitesimal actions (generators of time translation, spatial translation, rotation, boost, scaling, etc etc) tend to be unbounded, because they obey canonical commutation relations.  The exponentiations tend to be bounded, but I'm not sure whether they're compact or trace class.  And of course there is an isomorphism between the (bound states of the) Kepler potential and the harmonic oscillator, so you if one is not summable, neither is the other. -lethe talk [ +] 17:48, 10 March 2006 (UTC)
replies: text: Yes, well, figuring out which is which, and figuring out what's safe and what isn't, is one of the major confusions I am struggling with. I see the trees, I'm having trouble seeing the forest. linas 21:48, 12 March 2006 (UTC)

text: The above was meant to be illustrative handwaving, to illustrate the general idea, and not a rigorous development. linas 01:55, 16 March 2006 (UTC)
replies: text: But you added a dubious claim based on this handwaving argument to the article Green's function.  In fact, Fredholm theory studies operators of a more general form than that of the Green's function in that particular case, and furthermore, there are Green's functions which do not admit representations of that form.  So where are you going with this wandwaving analogy? -lethe talk [ +] 02:56, 16 March 2006 (UTC)
thread_title: overview

message: In a vague, day-dreamy way, I can't help noting a certain resemblance to the renormalization problem in perturbative quantum field theory. This article makes the claim that any element of the product space, even one that is superficially divergent, may be re-written as one where the \lambda_i go to zero.  This may seem shallow, except for the fact that the perturbation theory is a theory of integral kernels aka "Green's functions". There's a further analogy: some of the divergences are known to be "anomolous", for example, the famous axial current anomaly in the weak interaction. I can't help thinking that this anomaly is nothing more than an operator that is not "trace class" in the language of this article. linas 04:31, 10 March 2006 (UTC)
thread_title: QFT

message: Not Banach. The space of holomorphic functions on a domain of C^n is never a Banach space. 
In fact, it is an infinite dimensional Frechet space which is also Montel.82.58.202.84 21:03, 3 March 2007 (UTC)
replies: text: I agree that there is a problem here. And as I can see, no reaction for two long years!!! --Bdmy (talk) 11:31, 10 March 2009 (UTC)
thread_title: Examples