User marius overholt - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T09:14:50Z http://mathoverflow.net/feeds/user/13218 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/72474/axioms-for-zeta-functions Axioms for zeta functions Marius Overholt 2011-08-09T14:17:40Z 2011-08-09T14:17:40Z <p>The Selberg class is an axiomatization of arithmetically significant zeta functions (a.k.a. L-functions) by a few analytic properties (functional equation etc.) However there do exist other zeta functions that do not seem to come from arithmetic, but instead from geometry, for example. Some of these are known to have zeros on the real axis to the right of the critical line, but are otherwise expected to satisfy an analogue of the Riemann Hypothesis. Have any attempt been made to write down axioms for such more general zeta functions that are expected to satisfy an appropriately modified Riemann Hypothesis?</p> http://mathoverflow.net/questions/68546/teaching-a-pedagogy-course/68592#68592 Answer by Marius Overholt for Teaching a pedagogy course Marius Overholt 2011-06-23T07:04:25Z 2011-06-23T07:04:25Z <p>Among possible sources of readings, I would mention the booklet How to Teach Mathematics by Steven G. Krantz. It is quite practical, and it costs very little. </p> <p>By the way, one of the other responders mentioned an unpleasant issue. I would like to mention another issue that is potentially much more unpleasant: The TA should keep the door to the office open when a student comes to office hour, most particularly if the student is of the opposite sex. Very often the student closes the door when entering the office, even if the door is open. In such cases, I tell the student to keep the door open "in case someone else comes along." The unpleasant developments that this precaution prevents are extremely rare, but when they occur, they cause quite exceptional amounts of trouble.</p> http://mathoverflow.net/questions/58839/does-de-brangess-theorem-extend-to-several-variables/58981#58981 Answer by Marius Overholt for Does de Branges's theorem extend to several variables? Marius Overholt 2011-03-20T16:25:48Z 2011-06-08T11:15:09Z <p>Such a result would have to be quite different in several variables, because the holomorphic automorphism group of $\mathbb{C}^n$ is very big when $n \geq 2$. For injectivity, we need to look at equidimensional mappings $F$ from the domain (whatever it may be), and into $\mathbb{C}^n$ say. For simplicity, choose $n = 2$, which already contains the main features of the general case. Any mapping $\psi$ of the form $(z,w) \mapsto (z,w + h(z))$ (a "shear") with $h$ an arbitrary entire function is holomorphic everywhere on $\mathbb{C}^2$ and injective. Thus $\psi \circ F$ is also an injective holomorphic mapping from the domain into $\mathbb{C}^2$. Assuming the domain contains the origin and that $F$ is normalized (taking the origin to the origin and having the identity as derivative at the origin), we can ensure that $\psi \circ F$ is also normalized by choosing the entire function $h$ to have a double zero at the origin. The freedom of choosing $h$ implies that the power series coefficients of the mappings in the normalized family of injective holomorphic mappings have no universal bounds over the family; unlike in the one-dimensional case, the normalized family is not compact.</p> <p>The key difficulty is this: In one variable the automorphism group of the target $\mathbb{C}^1$ consists of mappings of the form $z \mapsto az + b$, and we can mod out this automorphism group simply by fixing $f(0)$ and $f^{\prime}(0)$, and then it turns out that injectivity implies compactness. When $n \geq 2$ the automorphism group is enormous (a dense subgroup of it was determined explicitly by E. Andersén and L. Lempert, so it is known in a sense), thus a linear normalization is very inadequate, and we should normalize by the whole automorphism group. But it is not clear what it means in practice to mod out by the automorphism group, and whether injectivity would imply compactness after moding out when $n \geq 2$. If we obtain a compact family, there will exist sharp bounds on all coefficients, of course, though we might not be able to establish them.</p> <p>A precise question (though in an unexplicit form) would be: When can we write the family of all injective holomorphic mappings from a domain $\Omega$ into $\mathbb{C}^n$ by composing $\mathrm{Aut}(\mathbb{C}^n)$ with a compact family? Trivially we can do it when $\Omega = \mathbb{C}^n$ by choosing the compact family to consist of the identity map alone.</p> <p>If one is willing to impose strong geometric conditions (e. g. starlikeness) on the injective mappings, one can get compactness. See the book Geometric function theory in one and several variables by I. Graham and G. Kohr for example.</p> <p>ADDED: Having thought about this a little more, I believe it unlikely that for $n \geq 2$ compactness can hold for $F : \mathbb{B}_n \rightarrow \mathbb{C}^n$ even after moding out by the automorphism group of $\mathbb{C}^n$. For generic domains have only the identity automorphism, and so we could replace $\mathbb{C}^n$ by a slightly smaller domain $D$ with only the identity automorphism and look at injective holomorphic mappings $F : \mathbb{B}_n \rightarrow D$. Since $D$ is nearly as "roomy" as $\mathbb{C}^n$ the family should be noncompact, but obviously cannot be made compact by moding out by automorphisms. So moding out by automorphisms seems to be an "accidental" device, so to speak, and unlikely to work when $n \geq 2$. I would support this conclusion as follows:</p> <p>One way to see that the family of injective holomorphic functions $f : \mathbb{D} \rightarrow \mathbb{C}$ with $f(0) = 0$ and $f'(0) = 1$ is compact is to apply the Schwarz Lemma to the inverse $f^{-1}$ to see that $f$ omits a point on the circle $|w| = 1$. Since the omitted set of $f$ on the unit sphere is a continuum containing $\infty$, $f$ also omits a point on the circle $|w| = 2$. A well known theorem states that if a family of holomorphic functions omits three points on the Riemann sphere, then it is precompact ("normal" in complex analysis parlance). Here we have three points, but two are "movable" (on the circles) and the third is fixed (at infinity) and a routine extension of the theorem mentioned shows that the family is still precompact, because the three omitted points stay uniformly away from each other. Since the normalized family of injective holomorphic functions $f$ is closed by a theorem of Hurwitz, the family is compact.</p> <p>Now let us see what this yields for injective holomorphic mappings $F : \mathbb{B}_n \rightarrow D$ with $F(0) = 0$ and $F'(0) = I$ (we assume $0 \in D$) for $n \geq 2$. We can still apply the Schwarz Lemma (in several variables) to the inverse $F^{-1}$ to conclude that $F$ omits a point on the sphere $|w| = 1$, so that the omitted set of $F$ is an unbounded continuum with a point on that sphere. But this information is hardly strong enough to conclude that the family is precompact. The higher-dimensional analogue of the theorem on three omitted points is that if we remove $2n+1$ hyperplanes in general position from $\mathbb{C}^n$ then any family of holomorphic mappings into what remains is precompact. So it seems that we have removed much too little when $n \geq 2$.</p> http://mathoverflow.net/questions/67102/is-anyone-aware-of-a-good-exposition-of-the-gauss-kramer-model-of-integers/67122#67122 Answer by Marius Overholt for Is anyone aware of a good exposition of the Gauss-Kramer model of Integers? Marius Overholt 2011-06-07T12:21:48Z 2011-06-07T12:21:48Z <p>Chapter 3 of The Prime Numbers and Their Distribution by Gérald Tenenbaum and Michel Mendès France has a nice exposition of the model, including modifications indicated by Maier's discovery.</p> http://mathoverflow.net/questions/59741/are-there-infinitely-many-triples-of-consecutive-square-free-integers/59817#59817 Answer by Marius Overholt for are there infinitely many triples of consecutive square-free integers? Marius Overholt 2011-03-28T10:32:10Z 2011-03-28T10:32:10Z <p>The question of the number of positive integers $n \leq x$ for which all members of an associated fixed pattern are squarefree (or r-free) was studied by Leon Mirsky:</p> <p>L. Mirsky, Note on an asymptotic formula connected with r-free integers. Quart. J. Math., Oxford Ser. 18 (1947), 178-182. </p> <p>L. Mirsky, Arithmetical pattern problems relating to divisibility by rth powers. Proc. London Math. Soc. (2) 50 (1949), 497–508.</p> <p>As I remember it, Mirsky proved that the number is $cx + O(x^{2/3})$ for patterns of squarefrees, where $c$ is a constant depending on the pattern, and is positive if the pattern is not excluded by certain necessary congruential conditions.</p> http://mathoverflow.net/questions/56736/euler-product-and-stieltjes-convergence-theorem Euler Product and Stieltjes convergence theorem. Marius Overholt 2011-02-26T13:43:00Z 2011-02-26T14:30:58Z <p>Scratched this question, because there was a serious problem with it.</p> http://mathoverflow.net/questions/56534/surface-automorphisms-and-conformal-automorphisms Surface automorphisms and conformal automorphisms Marius Overholt 2011-02-24T16:59:42Z 2011-02-24T19:44:04Z <p>Given a closed orientable surface $S$ and a topological automorphism $\sigma$ of $S$, it is not in general possible to find a conformal structure $\Sigma$ on $S$ so that $\sigma$ is isotopic to a conformal automorphism of the Riemann surface $(S,\Sigma)$. For example by the theorem of Hurwitz that the conformal automorphism group is finite, while $\sigma$ on the other hand may be of infinite order in the mapping class group. But by a theorem of Colin Maclachlan in "Modulus space is simply connected", Proc. Amer. Math. Soc. 29 (1971), 85–86, every surface automorphism is isotopic to the composition of finitely many conformal automorphisms (for varying complex structures on $S$). For being isotopic to a conformal automorphism is equivalent to being isotopic to an topological automorphism of finite order (one direction by Hurwitz, the other by averaging a metric). Maclachlan proved that the mapping class group is generated by elements of finite order.</p> <p>I am interested in the minimal number $m(\sigma)$ of conformal structures required, especially for the torus, where the mapping classes have a nice explicit description. Unfortunately when I tried to use this explicit description, it translated into some obscure number theory with a Diophantine flavor. I could not even show that for the torus in general arbitrarily many conformal structures would be needed, i.e. that $m(\sigma)$ is unbounded for $T^2$. This is my question. An upper bound for $m(\sigma)$ for $T^2$ in terms of the explicit description of $\sigma$ by a 2 by 2 integer matrix would also be interesting. Perhaps the higher genus case could be worth looking at after the torus case. I got stuck on $T^2$ and gave up quite a long time ago. But now that Math Overflow is here, I can ask this as a question.</p> <p>CLARIFICATION: For the case of a topological torus I am concretely asking how many conformal automorphisms (relative to various complex structures on the topological torus) I need to compose together to have enough freedom to represent a topological automorphism up to isotopy. I tend to agree with Sam Nead that for every positive integer $K$ there will be some topological automorphism $\sigma$ that will require at least $K$ conformal automorphisms in order to be so represented. But I don't know how to prove this, though Sam Nead's comment on proving it seems reasonable.</p> http://mathoverflow.net/questions/72474/axioms-for-zeta-functions Comment by Marius Overholt Marius Overholt 2011-08-10T06:20:27Z 2011-08-10T06:20:27Z @GH: I checked on the web, and there is a translation into English at <a href="http://www.ams.org/journals/bull/2008-45-04/S0273-0979-08-01223-8/home.html" rel="nofollow">ams.org/journals/bull/2008-45-04/&hellip;</a>. However there are a few differences between the English and the Norwegian versions concerning which questions were included. But the above question and answer is also included in the version in English. http://mathoverflow.net/questions/72474/axioms-for-zeta-functions Comment by Marius Overholt Marius Overholt 2011-08-10T06:11:55Z 2011-08-10T06:11:55Z If you look at a compact Riemannian surface with the hyperbolic metric, and consider the closed geodesics, you can say that their lengths correspond to the logarithms of the primes. In the compact case one knows that the Riemann Hypothesis essentially holds, except that in some particular cases we have some zeros between 1/2 and 1 on the real axis, which I do not think can happen with those functions that we usually consider in number theory. Though I know that some have believed that there may be quadratic L-functions with zeros between 1/2 and 1. http://mathoverflow.net/questions/72474/axioms-for-zeta-functions Comment by Marius Overholt Marius Overholt 2011-08-10T06:03:53Z 2011-08-10T06:03:53Z @GH: It was the way I thought I remembered it. Selberg made the remark in part 3 of the interview, about the Riemann Hypothesis and the trace formula. The interviewers were Nils A. Baas and Christian F. Skau. I translate their question from Norwegian into English: Have you considered whether there exists any kind of geometrical analogies to the primes in a fundamental sense? Selberg's answer translated from the Norwegian: http://mathoverflow.net/questions/72474/axioms-for-zeta-functions Comment by Marius Overholt Marius Overholt 2011-08-09T14:55:13Z 2011-08-09T14:55:13Z @GH: Actually, I was quoting Atle Selberg from a long interview that he gave about a year before he died. I have to scratch around for the interview, but suspect that he was thinking of the Selberg zeta function for a generic compact Riemann surface. I will report back. http://mathoverflow.net/questions/71453/what-should-be-learned-in-an-introductory-analytic-number-theory-course/71458#71458 Comment by Marius Overholt Marius Overholt 2011-07-30T10:14:21Z 2011-07-30T10:14:21Z @Igor: Well, there is a brief chapter on arithmetic functions, without any estimates. But in the section of problems following this chapter, there is a selection of estimates to prove. It certainly is not a big chunk of the book. Maybe we are not speaking of the same book? I am only familiar with the translated edition, and perhaps there was more material in the Russian edition. http://mathoverflow.net/questions/71453/what-should-be-learned-in-an-introductory-analytic-number-theory-course/71458#71458 Comment by Marius Overholt Marius Overholt 2011-07-29T08:42:24Z 2011-07-29T08:42:24Z I didn't downvote it, but I am familiar with the book. It is one of the nicer of the traditional introductory treatments of elementary number theory, up to quadratic reciprocity and primitive roots. The strong point of the book is a large number of both theoretical problems and computational exercises, with about 90 pages of solutions and answers at the back. This book may be the finest resource available for someone who wants to learn elementary number theory on their own. A weak point is that there is no index. I imagine the answer was downvoted because there is no analysis there. http://mathoverflow.net/questions/71453/what-should-be-learned-in-an-introductory-analytic-number-theory-course/71462#71462 Comment by Marius Overholt Marius Overholt 2011-07-29T08:26:02Z 2011-07-29T08:26:02Z @David: For the Hecke formula and its consequences, I wrote out full details of an approach outlined without details in the expository paper The Analytic Theory of Algebraic Numbers by H. M. Stark (An invited address at the 1973 AMS Summer meeting in Missoula). It came to 14 pages in all after making modifications. (In Stark's paper it takes 3 pages). For the Ikehara theorem, I rearranged and expanded a bit the argument in The Wiener-Ikehara theorem by Complex Analysis by Korevaar, making it easier to read for students without much analysis background, after which it came to 6 pages. http://mathoverflow.net/questions/71453/what-should-be-learned-in-an-introductory-analytic-number-theory-course/71462#71462 Comment by Marius Overholt Marius Overholt 2011-07-28T19:51:18Z 2011-07-28T19:51:18Z Another worthy goal is the Prime Ideal Theorem of Landau. Specializing this to $\mathbb{Q}$ yields the Prime Number Theorem in the form $\pi(x) \sim x/\log(x)$. A complete proof of the Prime Ideal Theorem by means of Ikehara's Tauberian theorem takes only a couple of pages. And the road to Ikehara's theorem is easier than it used to be, after Korevaar found a proof by means of Newman's contour integration method. With full details, this proof takes six pages, and does not require any Fourier analysis (but if you want the strongest form, you need the Riemann-Lebesgue Lemma.) http://mathoverflow.net/questions/71453/what-should-be-learned-in-an-introductory-analytic-number-theory-course/71462#71462 Comment by Marius Overholt Marius Overholt 2011-07-28T19:50:51Z 2011-07-28T19:50:51Z But do the students have a little background in alg. num. thry? If they have, you can prove the Hecke theta formula (assuming Poisson summation) and use it to prove (1) the Dirichlet Unit Theorem, (2) finiteness of the class number, (3) the functional equation of the Dedekind zeta function, (4) the analytic class number formula, (5) the Dirichlet density of primes that split completely in a normal extension K of $\mathbb{Q}$ is $1/n_K$, and (6) the rational primes that split completely in a normal extension of $\mathbb{Q}$ determine it, in 14 pages with full proofs. http://mathoverflow.net/questions/70036/a-symbol-to-denote-the-set-of-prime-numbers/70038#70038 Comment by Marius Overholt Marius Overholt 2011-07-12T06:46:47Z 2011-07-12T06:46:47Z @Michael: Isn't P more common for probabilities than $\mathbb{P}$? I have only a couple of books about probability; both use P. http://mathoverflow.net/questions/70036/a-symbol-to-denote-the-set-of-prime-numbers/70038#70038 Comment by Marius Overholt Marius Overholt 2011-07-12T06:43:53Z 2011-07-12T06:43:53Z @Seva: In complex analysis $\mathbb{D}$ denotes the open unit disk and $\mathbb{H}$ the open upper half plane. The latter notation is also used in number theory (modular forms). I think that $\mathbb{P}$ is quite reasonable, because it is very unlikely to be confused with $\mathbb{P}^n$ (projective space) from algebraic geometry. http://mathoverflow.net/questions/69543/good-books-on-dirichlets-class-number-formula Comment by Marius Overholt Marius Overholt 2011-07-07T10:33:18Z 2011-07-07T10:33:18Z There is also Number Theory by Borevich and Shafarevich. It certainly counts as a good book! http://mathoverflow.net/questions/67102/is-anyone-aware-of-a-good-exposition-of-the-gauss-kramer-model-of-integers/67122#67122 Comment by Marius Overholt Marius Overholt 2011-06-07T16:48:28Z 2011-06-07T16:48:28Z I don't think there has been much hope of making it rigorous, because the conclusions it yields are so far beyond what can be proved today. For example, if $p_k$ denotes the k-th prime, it is known that $p_{k+1} = p_k + O(p_k^{\alpha + \varepsilon})$ with $\alpha = 0.525$. The Riemann Hypothesis would allow $\alpha = 0.5$. But the Cram&#233;r model would yield $p_{k+1} = p_k + O(\log^2(p_k))$. http://mathoverflow.net/questions/66735/bounded-open-set-with-frontier-of-nonzero-lebesgue-measure Comment by Marius Overholt Marius Overholt 2011-06-02T18:15:26Z 2011-06-02T18:15:26Z There exists an example of a Jordan curve that has positive planar Lebesgue measure. The construction dates to 1903 and is due to William Fogg Osgood. It proceeds iteratively, by threading a curve through nested boxes. His paper is available on the web at <a href="http://www.jstor.org/stable/1986455" rel="nofollow">jstor.org/stable/1986455</a>. http://mathoverflow.net/questions/56736/euler-product-and-stieltjes-convergence-theorem Comment by Marius Overholt Marius Overholt 2011-02-26T14:32:14Z 2011-02-26T14:32:14Z @George: You are right, of course. I was thinking in terms of series rather than meromorphic functions.