Lasse Rempe-Gillen
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Registered User
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Professor of Pure Mathematics at the University of Liverpool
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Feb 19 |
answered | Asymptotic behavior of entire functions |
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Feb 17 |
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13 months and not even one report. what would you do? I hope you have had or will have some success with getting feedback through your friend. |
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Feb 15 |
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Finding invariant Borel Probability Measures for a contraction map This looks like homework ... |
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Feb 12 |
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Mathematicians whose works were criticized by contemporaries but became widely accepted later The majority of mathematicians will never switch to constructive mathematics, at least not until a contradiction is discovered in ZFC. :) Even in the (rather unlikely) event that this happens, it is far more likely that some weaker, still non-constructive, set of axioms is going to be used. The mathematical world we work in is too powerful and convenient to give up without good reason. Of course this doesn't mean that the study of intuitionistic logic isn't interesting from a foundational point of view. |
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Feb 11 |
awarded | ● Necromancer |
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Feb 11 |
answered | Applications of discrete-time dynamics |
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Feb 11 |
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Functions holomorphic on a region minus a Cantor set As I mention below, the question becomes quite different when asking about "conformal" removability; i.e. removability for conformal homeomorphisms. Of course holomorphic removability in the sense here implies conformal removability, but the converse is far from true. For example, you can easily have conformally removable sets of Hausdorff dimension 2 (but zero Lebesgue measure). |
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Feb 11 |
revised |
Functions holomorphic on a region minus a Cantor set added 376 characters in body |
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Feb 7 |
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Analytic curve on Riemann surface @Zoltan: Since your complex structure is smooth, it would seem that the identity is automatically a diffeomorphism between your surface S with the original structure, and with the new complex structure? Also, you have not answered my question: what do you mean by the curve being analytic? Let me make it more precise. The unit circle is an analytic curve. Take a diffeomorphism of the sphere that maps the unit circle to itself, but not in a real-analytic manner. Pull back the usual complex structure using this diffeomorphism. Does this satisfy your condition or not? |
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Jan 31 |
answered | Functions holomorphic on a region minus a Cantor set |
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Jan 31 |
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Analytic curve on Riemann surface @Zoltan: You should probably clarify what exactly you mean by complex structures here: just a collection of charts? The structure given by a Beltrami differential? Etc. More crucially: what does "$C$ remains analytic" mean? That the map $\gamma$ is analytic, or that the curve has some analytic parametrization? In the former case, the answer is essentially trivial, as noted by Aleksey: the identity map should restrict to be analytic on the curve $C$. In the latter case, I doubt there's a good general answer you can expect - e.g. any homeomorphism preserving $C$ will give such a structure. |
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Jan 30 |
awarded | ● Yearling |
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Jan 27 |
awarded | ● Nice Question |
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Jan 21 |
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13 months and not even one report. what would you do? If you know some of the other editors, it seems like it would be perfectly reasonable to contact them. Make sure to be diplomatic, because you don't want to come across as impatient or unreasonable, but you do have a right to expect a response to your communications from the editor, even if it is just a short "we are still waiting to hear from the referee". Good luck! |
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Jan 21 |
answered | “Explicit” examples of Irrational numbers very well approximated by rationnal numbers |
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Jan 17 |
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13 months and not even one report. what would you do? To not have had contact from the editor seems irregular. It sounds like a tricky situation. I'm not sure what your situation is - if you are a recent PhD student, have you spoken to your supervisor? They might be able to contact someone at the journal on your behalf. |
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Jan 17 |
awarded | ● Nice Answer |
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Jan 14 |
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Examples In Ergodic Theory and Topological Dynamics This being a question that doesn't have a 'right' answer, and one that is quite open-ended, should it perhaps be Community Wiki? |
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Jan 14 |
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Belyi functions on non-compact surfaces; or: Building Riemann surfaces from equilateral triangles Thanks Loic - I wasn't aware of Kevin's work on vector fields.As you say, this isn't quite what I was asking for, but I will take a look at this paper. |
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Jan 12 |
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Belyi functions on non-compact surfaces; or: Building Riemann surfaces from equilateral triangles typo fixed |
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Jan 11 |
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Belyi functions on non-compact surfaces; or: Building Riemann surfaces from equilateral triangles clarified "building from triangles" in the noncompact case, as pointed out by Misha |
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Jan 11 |
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Belyi functions on non-compact surfaces; or: Building Riemann surfaces from equilateral triangles @Misha, I think I see what you mean - for noncompact surfaces, being built from triangles is indeed (formally) weaker than having a Belyi function on it. To have a Belyi function, every corner should be adjacent to only finitely many triangles. I will add the question to clarify this. (Our theorem establishes the stronger property for all non-compact surfaces, and hence the weaker property also holds.) |
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Jan 11 |
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Belyi functions on non-compact surfaces; or: Building Riemann surfaces from equilateral triangles Perhaps I should have clarified that, when we build the Riemann surface from infinitely many triangles, we only include those corner points that are contained in only finitely many triangles. Near these, we define a Riemann surface structure in the obvious way. (Near the others, it isn't at all clear how we would define a Riemann surface structure.) |
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Jan 11 |
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Belyi functions on non-compact surfaces; or: Building Riemann surfaces from equilateral triangles I am not sure that I understand your comment correctly. Any holomorphic map is a "homeomorphism away from branch points". By "branched cover", I mean precisely the definition given in the second paragraph, which is stronger than both of your definitions. (The exponential map is a covering to its image on the plane, but it is not a Belyi function in my sense.) |
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Jan 10 |
asked | Belyi functions on non-compact surfaces; or: Building Riemann surfaces from equilateral triangles |
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Dec 28 |
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Terminology question in dynamical systems Hi Alex, I think Mahdi is right - for the exponential map, $f^{-1}(\mathbb{C})=\mathbb{C}$ ... indeed, $f(Y)\subset Y$ is equivalent to $Y\subset f^{-1}(Y)$. |
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Dec 26 |
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Is this a Julia set (and if so, for which function family is it the Julia set)? @Aaron - do you have any further questions on the topic, or might you consider accepting an answer? |
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Dec 9 |
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Is this a Julia set (and if so, for which function family is it the Julia set)? @Aaron, the fact that $\partial K$ is contained in the bifurcation locus is clear (since the family in question cannot be normal near a point of the boundary). The other direction follows from Montel's theorem. I am not sure what you mean about the neutral cycles. There are many parameters in the bifurcation locus without neutral cycles (e.g. parameters for which the free critical point is eventually mapped to a repelling periodic cycle). |
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Dec 9 |
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Parabolic cylinder functions - explicit estimates? Can you give a link that explains the concepts and notations that you use in your post? |
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Dec 4 |
answered | What are good methods for detecting parabolic components and Siegel disk components in the Fatou set of a rational function? |
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Dec 4 |
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Is this a Julia set (and if so, for which function family is it the Julia set)? @Rodrigo: I think you mean that the similarity will occur near the critical value, not the critical point? |
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Dec 4 |
answered | Is this a Julia set (and if so, for which function family is it the Julia set)? |
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Dec 3 |
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Is this a Julia set (and if so, for which function family is it the Julia set)? On the other hand, there are well-known examples where (distorted) copies of actual Julia sets do appear in parameter spaces. But these will be subsets of the corresponding bifurcation loci. |
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Dec 3 |
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Is this a Julia set (and if so, for which function family is it the Julia set)? What do you mean by "a function family"? Certainly it is not true that your set is the Julia set of a rational function. (If you care enough, you should be able to fashion a formal proof, e.g. from the fact that the complement contains some domains bounded by analytic curves, in the small Mandelbrot copies.) On the other hand, a one-dimensional bifurcation locus can usually be described as the locus of non-normality of a suitable family of analytic functions, formed by taking the iterates of the free critical point. |
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Nov 28 |
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Transcendentality of all irrationals in the Cantor set You may also be interested in Bugeaud's article, "Diophantine approximation and Cantor sets", Mathematische Annalen, Volume 341, Number 3, July 2008 , pp. 677-684(8). This solves the other problem that Mahler states regarding the Cantor set. |
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Nov 28 |
accepted | Transcendentality of all irrationals in the Cantor set |
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Nov 28 |
answered | Transcendentality of all irrationals in the Cantor set |
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Nov 28 |
awarded | ● Organizer |
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Nov 28 |
revised |
Transcendentality of all irrationals in the Cantor set edited tags |
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Nov 28 |
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Integrating Powers without much Calculus To be honest, I do not see the point of withholding your answer. You're much more experienced than I am on MathOverflow, so I'm really puzzled by this; perhaps I am missing something. In my opinion, you ought to post your method, and ask whether people know of a reference to it, and/or ask for other methods that people might know. Why ask a question where people may spend time writing an answer that you already know, but for some reason chose to withhold? |
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Nov 26 |
answered | Proving a least prime factor |
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Nov 25 |
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13 months and not even one report. what would you do? @Andy, that's really unacceptable in my view. When a journal asks you to make revisions to the paper, you ought to be able to expect that they are likely to accept it if these are made to the referee's satisfaction (barring anything unforeseen happening). |
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Nov 25 |
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Zero-dimensional space Ali, You must have misunderstood the answer that Ramiro links to. It gives an example of a subset in $\mathbb{R}^2$ that contains an interval and also a dense set of isolated points. (Take a line segment $I$ in $\mathbb{R}^2$, and add a sequence of points in $\mathbb{R}^2\setminus I$ that accumulates everywhere on $I$.) |
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Nov 24 |
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13 months and not even one report. what would you do? Hi Tom, I certainly didn't mean to suggest that a positive referee's report is a guarantee of acceptance. I don't think I have had a paper rejected by e.g. the main LMS or AMS journals (excluding JAMS of course, which is one of the very top mathematics journals) when the referee clearly recommended publication (or have recommended publication of a paper myself and then heard that it was rejected). But perhaps that experience isn't typical! |
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Nov 24 |
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13 months and not even one report. what would you do? In my experience, most journals except the very top ones (Annals, Acta, Inventiones and a few others) tend to follow the referee's recommendation. A reasonable referee will reject a paper quickly if they are going to reject it, so I hope for you that they are simply very busy and just need to be prodded a bit by the editor. |
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Nov 24 |
accepted | Hausdorff dimension of Julia sets of quadratics not in the Mandelbrot set. |
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Nov 24 |
answered | 13 months and not even one report. what would you do? |
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Nov 24 |
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Functions holomorphic on a region minus a Cantor set - pt.2: Iterated function systems. This question was posed a long time ago, so you may already have found a solution for it yourself. In the case where the boundaries of your domains are disjoint, the resulting limit set will indeed be holomorphically removable. This is because it has absolute area zero. You may be interested in looking at Jeremy Kahn's PhD thesis (Holomorphic Removability of Julia Sets, available on the arXiv) for more background on holomorphic removability. |
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Nov 24 |
answered | Hausdorff dimension of Julia sets of quadratics not in the Mandelbrot set. |
