User jonah sinick - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T07:03:39Z http://mathoverflow.net/feeds/user/683 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/126340/possible-reasons-why-the-image-of-0-in-the-modular-parametrization-would-always-b/126349#126349 Answer by Jonah Sinick for Possible reasons why the image of 0 in the modular parametrization would always be O for a family of quadratic twists of elliptic curves Jonah Sinick 2013-04-03T05:17:50Z 2013-04-03T07:11:52Z <p>I only know this material in a vague and impressionistic way, but I think that it the fact that $0$ goes to the identity on the elliptic curve follows from Eichler-Shimura theory. </p> <p>If $E/\mathbb{Q}$ is parametrized by a modular curve, then it's a quotient of the Jacobian of the modular curve. In the composite map $$X_{0}(N) \to J_{0}(N) \to E$$ each arrow sends $0$ to $0$.</p> <p>[Edit: Looking at <a href="http://mathoverflow.net/questions/125694/computing-the-order-of-the-image-of-0-under-the-modular-parametrization-map-for-a" rel="nofollow">this question</a> I see that apparently 0 need not go to the identity in general. So now I'm puzzled. I think that my confusion might come from not knowing the definition of "0" on the modular curve in this context, which I was taking to be the preimage of the identity on the Jacobian.]</p> http://mathoverflow.net/questions/125659/the-existence-of-meromorphic-functions-on-riemann-surfaces The existence of meromorphic functions on Riemann surfaces Jonah Sinick 2013-03-26T20:36:14Z 2013-03-27T03:46:54Z <p>In Miranda's book on algebraic curves and Riemann surfaces, Miranda writes:</p> <blockquote> <p>It is a basic and highly nontrivial result that a compact Riemann surface has nonconstant meromorphic functions on it [...] The theory involved in producing meromorphic functions for an unknown compact Riemann surface is rather technical analysis and functional analysis. After one has access to meromorphic functions, however, the theory is completely algebraic, or at least can be made to be so.</p> </blockquote> <p>I've seen this claim a number of other places as well. It seems unnatural to use real analysis to prove a theorem about Riemann surfaces, which are geometric/algebraic objects. Is there genuinely no purely geometric/algebraic way to realize an abstract Riemann surface as a branched cover of the Riemann sphere?</p> http://mathoverflow.net/questions/7318/etale-cohomology-and-l-adic-tate-modules Etale cohomology and l-adic Tate modules Jonah Sinick 2009-11-30T23:35:32Z 2013-03-08T02:14:10Z <p>$\newcommand{\bb}{\mathbb}\DeclareMathOperator{\gal}{Gal}$ Before stating my question I should remark that I know almost nothing about etale cohomology - all that I know, I've gleaned from hearing off hand remarks and reading encyclopedia type articles. So I'm looking for an answer that will have some meaning to an etale cohomology naif. I welcome corrections to any evident misconceptions below.</p> <p>Let $E/\bb Q$ be an elliptic curve the rational numbers $\bb Q$: then to $E/\bb Q$, for each prime $\ell$, we can associate a representation $\gal(\bar{\bb Q}/\bb Q) \to GL(2n, \bb Z_\ell)$ coming from the $\ell$-adic Tate module $T_\ell(E/\bb Q)$ of $E/\bb Q$ (that is, the inverse limit of the system of $\ell^k$ torsion points on $E$ as $k\to \infty$). People say that the etale cohomology group $H^1(E/\bb Q, \bb Z_\ell)$ is dual to $T_\ell(E/\bb Q)$ (presumably as a $\bb Z_\ell$ module) and the action of $\gal(\bar{\bb Q}/\bb Q)$ on $H^1(E/\bb Q, \bb Z_\ell)$ is is the same as the action induced by the action of $\gal(\bar{\bb Q}/\bb Q)$ induced on $T_\ell(E/\bb Q)$.</p> <p>Concerning this coincidence, I could imagine two possible situations:</p> <p>(a) When one takes the definition of etale cohomology and carefully unpackages it, one sees that the coincidence described is tautological, present by definition.</p> <p>(b) The definition of etale cohomology (in the case of an elliptic curve variety) and the action of $\gal(\bar{\bb Q}/\bb Q)$ that it carries is conceptually different from that of the dual of the $\ell$-adic Tate module and the action of $\gal(\bar{\bb Q}/\bb Q)$ that it carries. The coincidence is a theorem of some substance.</p> <p>Is the situation closer to (a) or to (b)? </p> <p>Aside from the action $\gal(\bar{\bb Q}/\bb Q)$ on $T_\ell(E/\bb Q)$, are there other instances where one has a similarly "concrete" description of representation of etale cohomology groups of varieties over number fields and the actions of the absolute Galois group on them?</p> <p>Though I haven't seen this stated explicitly, I imagine that one has the analogy [$\gal(\bar{\bb Q}/\bb Q)$ acts on $T_\ell(E/\bb Q)$: $\gal(\bar{\bb Q}/\bb Q)$ acts on $H^1(E/\bb Q; \bb Z_\ell)$]::[$\gal(\bar{\bb Q}/\bb Q)$ acts on $T_\ell(A/K)$: $\gal(\bar{\bb Q}/\bb Q)$ acts on $H^1(A/K; \bb Z_\ell)$] where $A$ is an abelian variety of dimension $n$ and $K$ is a number field: in asking the last question I am looking for something more substantively different and/or more general than this.</p> <p>I've also inferred that if one has a projective <i>curve</i> $C/\bb Q$, then $H^1(C/\bb Q; \bb Z_\ell)$ is the same as $H^1(J/\bb Q; \bb Z_\ell)$ where $J/\bb Q$ is the Jacobian variety of $C$ and which, by my above inference I assume to be dual to $T_\ell(J/\bb Q)$, with the Galois actions passing through functorially. If this is the case, I'm looking for something more general or substantially different from this as well.</p> <p>The underlying question that I have is: where (in concrete terms, not using a reference to etale cohomology as a black box) do Galois representations come from aside from torsion points on abelian varieties?</p> <hr> <p>[Edit (12/09/12): A sharper, closely related question is as follows. Let $V/\bb Q$ be a (smooth) projective algebraic variety defined over $\bb Q$, and though it may not be necessary let's take $V/\bb Q$ to have good reduction at $p = 5$. Then $V/\bb Q$ is supposed to have an attached 5-adic Galois representation to it (via etale cohomology) and therefore has an attached (mod 5) Galois representation. If $V$ is an elliptic curve, this Galois representation has a number field $K/\bb Q$ attached to it given by adjoining to $\bb Q$ the coordinates of the 5-torsion points of $V$ under the group law, and one can in fact write down a polynomial over $\bb Q$ with splitting field $K$. The field $K/\bb Q$ is Galois and the representation $\gal(\bar{\bb Q}/\bb Q)\to GL(2, \bb F_5)$ comes from a representation $\gal(K/\bb Q) \to GL(2, \bb F_5)$. (I'm aware of the possibility that knowing $K$ does not suffice to recover the representation.)</p> <p>Now, remove the restriction that $V/\bb Q$ is an elliptic curve, so that $V/\bb Q$ is again an arbitrary smooth projective algebraic variety defined over $\bb Q$. Does the (mod 5) Galois representation attached to $V/\bb Q$ have an associated number field $K/\bb Q$ analogous to the (mod 5) Galois representation attached to an elliptic curve does? If so, where does this number field come from? If $V/\bb Q$ is specified by explicit polynomial equations is it possible to write down a polynomial with splitting field $K/\bb Q$ explicitly? If so, is a detailed computation of this type worked out anywhere? </p> <p>I'm posting a bounty for a good answer to the questions succeeding the "Edit" heading.</p> http://mathoverflow.net/questions/123718/counting-smooth-structures-on-manifolds Counting smooth structures on manifolds Jonah Sinick 2013-03-06T07:43:12Z 2013-03-06T07:43:12Z <p>Kervaire and Milnor found a formula for the number of smooth structures on the $4n - 1$ sphere (see, e.g. the last part of <a href="http://mathoverflow.net/questions/61252/why-do-bernoulli-numbers-arise-everywhere/61269#61269" rel="nofollow">this MO answer</a>). It is relatively easy to compute the number of smooth structures on the $k$ sphere for other values of $k$ (aside from $k = 4$).</p> <p>Has there been work finding formulas for the number of smooth structures on elements of other infinite classes of manifolds? </p> http://mathoverflow.net/questions/122442/functional-equations-of-zeta-functions-over-global-fields Functional equations of zeta functions over global fields Jonah Sinick 2013-02-20T19:36:39Z 2013-02-20T20:11:27Z <p>The functional equations for <a href="http://en.wikipedia.org/wiki/Dedekind_zeta_function" rel="nofollow">Dedekind zeta functions</a> (zeta functions attached to rings of integers in algebraic number fields) come from functional equations of theta functions like $\sum_{n \in \mathbb{Z}} q^{n^2}$, where $q = e^{2 \pi i z}$, which in turn come from the <a href="http://en.wikipedia.org/wiki/Poisson_summation_formula#Generalizations" rel="nofollow">Poisson summation formula for lattices in Euclidean spaces</a>. </p> <p>The functional equations for zeta functions attached to curves over finite fields come from the Riemann Roch theorem and Serre Duality (see, e.g. Chapter 2 of <a href="http://math.uchicago.edu/~mitya/beilinson/SamREU07.pdf" rel="nofollow">Sam Raskin's write-up</a>).</p> <p>Rings of integers of algebraic number fields and curves over finite fields are analogous, (see, e.g. Jordan Ellenberg's discussion of <a href="http://quomodocumque.wordpress.com/2009/12/07/the-braid-group-analytic-number-theory-and-weils-three-columns/" rel="nofollow">Weil's three columns</a>).</p> <blockquote> <p>Is there a known uniform proof of the functional equations that covers both cases?</p> </blockquote> http://mathoverflow.net/questions/107500/counting-higher-dimensional-abelian-varieties-of-a-given-conductor Counting higher dimensional abelian varieties of a given conductor Jonah Sinick 2012-09-18T18:24:25Z 2013-02-09T11:35:37Z <p>This question is a follow up to <a href="http://mathoverflow.net/questions/107287/finiteness-of-elliptic-curves-of-a-given-conductor" rel="nofollow">an earlier question</a> of mine on enumerating elliptic curves of a given conductor. </p> <p>I've heard people say that studying higher dimensional varieties via explicit defining equations often leads to hopelessly unmanageable complexity. As apparent evidence of this, on page 65 of Cassels' and Flynn's book titled <a href="http://www.amazon.com/Prolegomena-Middlebrow-Arithmetic-Mathematical-Society/dp/0521483700" rel="nofollow">Prolegomena to a Middlebrow Arithmetic of Curves of Genus $2$</a>, the authors state that the defining equations that they find for Jacobian varieties of genus $2$ curves consist of $72$ quadratic equations in $\mathbb P^{15}$. People say that rather than studying higher dimensional algebraic varieties as solution sets to explicit equations, one typically studies such varieties in a more abstract and geometric way.</p> <p>This makes sense. Yet I wonder how one can get look at concrete examples without defining equations. I know that there are some varieties such as moduli spaces which provide examples. But suppose, say, you want to prove that there are finitely abelian varieties of an arbitrary fixed dimension $d$ over a fixed arbitrary number field $K$ with a fixed conductor $N$ without looking at the automorphic side of things. </p> <blockquote> <p>Is there a (conjectural) method of proving finiteness without writing down explicit defining equations?</p> </blockquote> <p>[Edit: As Barinder Banwait points out, this follows immediately from the Shafarevich conjecture, which was proved by Faltings.]</p> <p>Pushing the envelop further, </p> <blockquote> <p>Is there a (conjectural) algorithm for enumerating these objects?</p> </blockquote> <p>Pushing the envelop still further, </p> <blockquote> <p>Suppose beyond enumerating such varieties, you want to determine, e.g., how many of them have surjective (mod $5$) Galois representation (say, attached to $H^1$) - is there a conjectural algorithm for doing so?</p> </blockquote> http://mathoverflow.net/questions/120067/what-do-theta-functions-have-to-do-with-quadratic-reciprocity/120071#120071 Answer by Jonah Sinick for What do theta functions have to do with quadratic reciprocity? Jonah Sinick 2013-01-28T01:24:36Z 2013-01-28T01:39:27Z <p>Hecke generalizing the argument that you mention to prove quadratic reciprocity relative to any given number field $K$ (see, e.g. his <a href="http://books.google.com/books/about/Lectures_on_the_Theory_of_Algebraic_Numb.html?id=QEocn1kbfxAC" rel="nofollow">Lectures on the Theory of Algebraic Numbers</a>).</p> <p>In <a href="http://www.amazon.com/Fourier-Analytic-Proof-Quadratic-Reciprocity/dp/0471358304" rel="nofollow">The Fourier-Analytic Proof of Quadratic Reciprocity</a> Michael C. Berg describes the subsequent development of this line of research. Quoting from the book's summary:</p> <blockquote> <p>The relative quadratic case was first settled by Hecke in 1923, then recast by Weil in 1964 into the language of unitary group representations. The analytic proof of the general n-th order case is still an open problem today, going back to the end of Hecke's famous treatise of 1923.</p> </blockquote> http://mathoverflow.net/questions/119329/the-riemann-hypothesis-and-the-langlands-program/119356#119356 Answer by Jonah Sinick for The Riemann Hypothesis and the Langlands program Jonah Sinick 2013-01-19T19:03:37Z 2013-01-19T19:03:37Z <p>This doesn't really address the question, but it seems worth mentioning that in order for the Riemann hypothesis to be well formulated for a motivic L-function, one has to know that the L-function is also an automorphic L-function. For example, <em>a priori</em> all one knows about the L-function of an elliptic curve is that it converges for $Re(s) > 3/2$ (by Hasse's bound). One needs automorphicity to extend the L-function to $Re(s) = 1$, and the Riemann hypothesis for the L-function of an elliptic curve states that all (nontrivial?) zeros of the L-function lie on the line $Re(s) = 1$. </p> http://mathoverflow.net/questions/117690/algorithm-for-determining-whether-two-polynomials-have-the-same-splitting-field Algorithm for determining whether two polynomials have the same splitting field Jonah Sinick 2012-12-31T07:09:59Z 2013-01-02T18:24:04Z <p><a href="http://math.stackexchange.com/questions/34522/cubic-polynomials-that-generate-the-same-extension" rel="nofollow">This question</a> asks how to tell whether two cubic polynomials with coefficients in $\mathbb{Q}$ have the same splitting field. There are several answers to the question, but they don't include proofs. Also, it's not clear how the results generalize to higher degree polynomials. Is there an algorithm for determining whether two polynomials in in $\mathbb{Q}[x]$ have the same splitting field? If so, what is it, and why does it work?</p> <p>(Reposted <a href="http://math.stackexchange.com/questions/267873/algorithm-for-determining-whether-two-polynomials-have-the-same-splitting-field" rel="nofollow">from Math Stack Exchange</a>.)</p> http://mathoverflow.net/questions/116244/the-natural-generalization-of-eulers-derivation-of-the-basel-sum The natural generalization of Euler's derivation of the Basel sum Jonah Sinick 2012-12-13T05:37:28Z 2012-12-31T07:21:30Z <p>Euler proved that $$\sum_{n=0}^\infty \frac{1}{n^2} = \frac{{\pi}^2}{6}$$ by comparing the $z^3$ term in the power series expression of $\sin(z)$ given by </p> <p>$$\sin(z) = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} \cdots$$</p> <p>with $z^3$ term obtained by multiplying out the Hadamard product expansion for $\sin(z)$:</p> <p>$$\sin(z) = z\left(1 - \frac{z^2}{\pi^2} \right)\left(1 - \frac{z^2}{4\pi^2} \right)\left(1 - \frac{z^2}{9\pi^2} \right)\cdots$$</p> <p>and more generally, showed that the numbers</p> <p>$$\sum_{n=0}^\infty \frac{1}{n^{2k}}$$</p> <p>are rational multiples of $\pi^{2k}$, giving an algorithm to compute the rational number, by comparing the coefficients of $z^{2j + 1}$ in the sum and the expansion of the product, for each $j \leq k$, together with <a href="http://en.wikipedia.org/wiki/Newton%27s_identities" rel="nofollow">Newton's identities</a>. Euler's results can be generalized to results about special values of $L$-functions outside of the critical strip, for example, Siegel and Klingen gave such a generalization for special values of Dedekind zeta functions of totally real number fields. However, the method of Siegel and Klingen is different from Euler's.</p> <blockquote> <p>What is the natural generalization of Euler's <em>method of proof</em> of the results mentioned above, and what sorts of results does one obtain from it?</p> </blockquote> <p>The sine function is periodic and one could look at doubly periodic functions on the complex plane and try to do the same things, but such functions are not holomorphic and so don't have Hadamard product expansions. One could look at functions of several complex variables that are periodic with respect to each variable, but then it's not immediately clear what such functions look like, or whether one would get anything new. </p> http://mathoverflow.net/questions/117394/applications-of-iwasawa-theory Applications of Iwasawa Theory Jonah Sinick 2012-12-28T07:23:36Z 2012-12-28T15:46:24Z <p>Iwasawa theory gives a formula for the power of $p$ dividing the class group of the $\mathbb{Q}(\zeta_{p^n})$ (where $\zeta_{p^n}$ is a primitive root of unity of exact order $p^n$) for sufficiently large $n$. (See, e.g., Theorem 2 of <a href="http://www.math.harvard.edu/~chaoli/tutorial2012/Lecture14.pdf" rel="nofollow">these notes</a>.) More generally, one gets a similar result for arbitrary $\mathbb{Z}_p$ extensions of number fields. Mazur and Mazur and Rubin have studied the variation of the $p$ part of the Tate-Shafarevich group of an elliptic curve in $\mathbb{Z}_p$ extensions and the variation of the rank of an elliptic curve in $\mathbb{Z}_p$ extensions.</p> <p>I find this somehow unsatisfying. The restriction to the study of the $p$ part of the ideal class group &amp; Tate Shafarevich, and the restriction to the study of $\mathbb{Z}_p$ extensions seem quite strong. </p> <p>Yet I've gotten the impression that Iwasawa theory is considered to be fundamental in number theory. So I feel as though I'm missing perspective on why Iwaswa theory is important.</p> <blockquote> <p>What are some important applications of Iwasawa theory?</p> </blockquote> <p>I'd also be happy with high-level philosophical comments.</p> http://mathoverflow.net/questions/117296/characterizing-primes-that-split-completely-vs-primes-with-a-given-splitting-beh Characterizing primes that split completely vs. primes with a given splitting behavior Jonah Sinick 2012-12-27T01:48:55Z 2012-12-27T05:09:01Z <p>Given a finite abelian extension of number fields $L/K$, the prime ideals $\mathfrak{p}$ in $O_K$ split into primes $\mathfrak{P}$ in $O_L$. The number of primes $\mathfrak{p}$ splits into is necessarily a divisor $d$ of $[L:K]$. One can ask:</p> <p><strong>Question (1):</strong> Which primes in $O_K$ split completely in $O_L$?</p> <p><strong>Question (2):</strong> Given a divisor $d$ of $[L :K]$, which primes $\mathfrak{p}$ split into $d$ primes in $O_L$?</p> <p>For $K = \mathbb{Q}$ and $L$ a cyclotomic extension of $\mathbb{Q}$, Question (2) seems genuinely more difficult than Question (1). A friend pointed this out to me in response to some notes that I put together titled <a href="http://www.mathisbeauty.org/preludereciprocitylaws.pdf" rel="nofollow">A Prelude to the Study of Reciprocity Laws</a>. </p> <p>Let $p$ be a prime and let $$\Phi_{p}(x) = x^{p-1} + x^{p-2} + \ldots + x^2 + x + 1.$$</p> <p>As I discuss in the notes, a prime $\ell$ divides a number of the form $\Phi_{p}(n)$ if and only if $\ell = p$ or $\ell \equiv 1 \pmod p$. In Section 7 of the notes, I show how this follows immediately from the basic theorems of modular arithmetic.</p> <p>Let $\zeta_p$ be a root of the polynomial $\Phi_{p}(x)$. Then a prime $\ell \in \mathbb{Z}$, $\ell \neq p$ divides a number of the form $\Phi_{p}(n)$ if and only if $\ell$ splits completely in $\mathbb{Z}[\zeta_p]$. So the above theorem answers Question (1) in this case where $K = \mathbb{Q}$ and $L = \mathbb{Q}(\zeta_p)$ .</p> <p>In my notes I (apparently) erroneously remark that the answer to Question (1) in this case implies quadratic reciprocity. In fact, quadratic reciprocity follows the answer to Question (2). I knew this, but thought that the answer to Question (2) follows formally from the answer to Question (1). My friend questioned whether this is true, and after spending a few hours on it, I realized that I can't see a way to use the answer to Question (1) to derive the answer to Question (2).</p> <blockquote> <p>Is there a way to do this?</p> </blockquote> <p>The proofs of the theorems of class field theory proceed by answering Question (2) directly, which is consistent with the above paragraph, but I always thought that the reason for this was that the only way to answer Question (1) in general was to answer Question (2). For this reason, I saw the questions as being of comparable difficulty. So I was surprised to see a case in which this doesn't seem to be true.</p> <blockquote> <p>Aside from cyclotomic extensions of $\mathbb{Q}$, are there finite extensions of number fields $L/K$ for which it's easier to answer Question (1) than Question (2)?</p> </blockquote> <p>A few potentially relevant thoughts: </p> <ol> <li><p>The imaginary quadratic case is similar to the cyclotomic case in the sense that one has an explicit construction of the number fields in question via complex multiplication, so it's natural to try to answer the latter question by looking at, e.g., the polynomials over $\mathbb{Z}[i]$ that generate the ray class fields of $\mathbb{Q}(i)$ and try to see if the prime divisors of their values at the Gaussian integers can be determined in a direct way analogous to as in the cyclotomic case.</p></li> <li><p>In the case where $[L : K]$ is a prime, Questions (1) and (2) are the same, because a prime splits completely if and only if it splits at all.</p></li> <li><p>The set of primes that splits completely in a number field uniquely determines the number field and so in principle determines the splitting behavior of the other primes. </p></li> </ol> http://mathoverflow.net/questions/116896/liouvilles-theorem-with-your-bare-hands Liouville's theorem with your bare hands Jonah Sinick 2012-12-20T19:34:27Z 2012-12-22T00:05:51Z <p>Liouville's theorem from complex analysis states that a holomorphic function $f(z)$ on the plane that is bounded in magnitude is constant. The usual proof uses the Cauchy integral formula. But this has always struck me as indirect and unilluminating. There is a proof via harmonic function theory, but this also seems to involve an unnecessarily large amount of prior buildup. So one might seek a more direct proof as below.</p> <p>Assume that $f(z)$ is nonconstant. The fact that $f(z)$ is holomorphic at every point implies that at any given point, there is a direction such that moving in that direction makes $|f(z)|$ larger. But this doesn't prove that $|f(z)|$ is unbounded, because <em>a priori</em> its magnitude could behave like $5 - \frac{1}{|z|}$ or some such thing. </p> <p>In the case of $f(z) = \frac{1}{P(z)}$ where $P(z)$ is a polynomial, one knows that $|f(z)|$ tends toward $0$ as $|z| \to \infty$ so that there's some closed disk such that if $|f(z)|$ is bounded, then it has a maximum in the interior of the disk, which contradicts the fact that one can always make $f(z)$ larger by moving in a suitable direction. But for general $f(z)$, one doesn't have this argument.</p> <p>One can try to reason based on the power series expansion of a holomorphic function $f(z)$ that is not a polynomial. Because polynomials are unbounded as $|z| \to \infty$ and grow in magnitude in a way that's proportional to their degree, one might think that a power series, which can be regarded as an infinite degree polynomial, would also be unbounded as $|z| \to \infty$. This is of course false: take $f(z) = \sin(z)$, then as $|z| \to \infty$ along the real axis, $f(z)$ remains bounded. The point is that the dominant term in the partial sums of the power series varies with $|z|$, and that the relevant coefficients change, alternating in sign and tending toward zero rapidly, so that the gain in size corresponding to moving to the next power of $z$ is counterbalanced by the change in coefficient. But there's <em>some</em> direction that one can move in for which $f(z)$ is unbounded: in particular, for $f(z) = \sin(z)$, $f(z)$ is unbounded along the imaginary axis.</p> <p>This suggests that we write $a_n = s_{n}e^{i \theta_n}$ for the coefficient of $z^n$ in the power series expansion of $f(z)$ and write $z = re^{i \theta}$ (where $s, r > 0$) so that</p> <p>$$f(z) = \sum_{n = 0}^{\infty} {a_n}z^n = \sum_{n = 0}^{\infty} sr^n e^{n\theta + \theta_n}$$</p> <p>and try to find a function $\theta = g(r)$ such that $f(z)$ is unbounded as $r \to \infty$ if one takes $\theta = g(r)$. </p> <p>But I don't know what to do next. Any ideas? Any ideas for <em>other</em> strategies of proving Liouville's theorem that are more direct than the ones using Cauchy's theorem?</p> http://mathoverflow.net/questions/111519/why-might-andre-weil-have-named-carl-ludwig-siegel-the-greatest-mathematician-of Why might André Weil have named Carl Ludwig Siegel the greatest mathematician of the 20th century? Jonah Sinick 2012-11-05T03:57:07Z 2012-12-07T13:28:33Z <p>According to Steven Krantz's <em>Mathematical Apocrypha</em> (<a href="http://books.google.com/books?id=HxpQFHGMfNIC&amp;lpg=PA213&amp;ots=qLx5UTwsd5&amp;dq=mathematical%2520apocrypha%2520siegel&amp;pg=PA186#v=onepage&amp;q=mathematical%2520apocrypha%2520siegel&amp;f=false" rel="nofollow">pg. 186</a>):</p> <blockquote> <p>As was custom, Weil often attended tea at [Princeton] University . Graduate student Steven Weintrab one day went about the room asking various famous mathematicians who was the greatest mathematician of the twentieth century. When he asked Weil, the answer (without hesitation) was "Carl Ludwig Siegel (1896-1981)."</p> </blockquote> <p>As the title of Krantz's book suggests, the anecdote may be apocryphal. However, there are other better grounded accounts of great mathematicians expressing the highest admiration for Siegel:</p> <p>(A) In <em>The Map of My Life</em> Shimura wrote:</p> <blockquote> <p>I always thought that few people really understood my work. I knew that Chevalley, Eichler, Siegel, and Weil understood my work, and that was enough for me [...] Of course [Siegel] established himself as one of the giants in the history of mathematics long ago [...] Among his contemporaries, [Weil] thought highly of Siegel [...]</p> </blockquote> <p>(B) In <a href="http://publications.ias.edu/sites/default/files/DOCdev1.pdf" rel="nofollow">an published interview</a> (pg. 30) Selberg said</p> <blockquote> <p>[Siegel] was in some ways, perhaps, the most impressive mathematician I have met. I would say, in a way, devestatingly so. The things that Siegel tended to do were usually things that seemed impossible. Also after they were done, they seemed still almost impossible.</p> </blockquote> <p>Why might Weil, Shimura and Selberg have been so impressed by Siegel? I should emphasize that I'm <em>not</em> trying to precipitate a debate about the relative standing of historical mathematicians - rather - I'm hoping to learn about aspects of Siegel's work that I might otherwise overlook. I'm also not looking for, e.g. quotations from the Wikipedia article on him, but rather, less familiar material.</p> http://mathoverflow.net/questions/111339/simple-tamagawa-number-calculations Simple Tamagawa number calculations Jonah Sinick 2012-11-03T03:44:54Z 2012-12-05T19:50:48Z <p>As is well known, Euler proved the Basel identity $\displaystyle\sum\limits_{i=0}^{\infty} \frac{1}{n^2} = \frac{{\pi}^2}{6}$. By far the most illuminating explanation of this fact that I've seen is as follows:</p> <blockquote> <p>By another theorem of Euler we can rewrite the identity as $\displaystyle \frac{{\pi}^2}{6}\cdot \prod_{p} \left(1 - \frac{1}{p^2}\right) = 1$, where $p$ ranges over all primes. The first term is the normalized volume of</p> <p>$\displaystyle SL(2,\mathbb{R})/SL(2, \mathbb{Z})$ </p> <p>and the term corresponding to $p$ in the product is the normalized volume of </p> <p>$SL(2, \mathbb{Z}_p)$. </p> <p>With these replacements, the right hand side can be written as the normalized volume of </p> <p>$\displaystyle SL(2,\mathbb{\mathbb{A}_{\mathbb{Q}}})/SL(2, \mathbb{Q})$. </p> <p>where $\mathbb{A}_{\mathbb{Q}}$ denotes the adeles of $\mathbb{Q}$ But this last volume is equal to 1: this is a special case of the <a href="http://en.wikipedia.org/wiki/Weil_conjecture_on_Tamagawa_numbers" rel="nofollow">Weil Conjecture on Tamagawa Numbers</a>.</p> </blockquote> <p>I've been fascinated by this result for years, but have never been able to understand a proof of it (<em>from the adelic perspective</em>) even in cases as simple the one above. In this way, my question contrasts with <a href="http://mathoverflow.net/questions/44184/why-are-tamagawa-numbers-equal-to-pic-sha" rel="nofollow">that of Ben Weiland</a> who asked about the theorem in more general settings. </p> <p>I tried reading André Weil "Adeles and algebraic groups" with a view toward learning a proof but found the book unintelliglbe. I gathered that the idea of the proof is to show that the nonzero volume of some object is equal to the Tamagawa number multiplied by the original volume but beyond that understood nothing. My impression is that Marie-France Vigneras' book titled <a href="http://www.amazon.com/Arithmetique-algebres-quaternions-Lecture-Mathematics/dp/3540099832" rel="nofollow">Arithmetique des algebres de quaternions</a> has this material, but I don't read French.</p> <blockquote> <p>What are some lucid sources that you would recommend for learning proofs of the some of the first few cases (including the case above) of the Weil Tamagawa Number conjecture from an adelic perspective?</p> </blockquote> <hr> <p><a href="http://en.wikipedia.org/wiki/Weil_conjecture_on_Tamagawa_numbers" rel="nofollow">1</a> I learned this material from Yuri Manin's "Reflections on Arithmetical Physics" and Maclachlan and Reid's "The Arithmetic of Hyperbolic 3-Manifolds"</p> http://mathoverflow.net/questions/114284/applications-of-the-class-number-formula-etc Applications of the class number formula, etc. Jonah Sinick 2012-11-23T20:32:38Z 2012-11-23T22:36:28Z <p>This is a big list of applications of the class number formula and its generalizations. I'll start:</p> <ol> <li>The solution to <a href="http://en.wikipedia.org/wiki/Class_number_problem" rel="nofollow">Gauss's class number problem</a> for imaginary quadratic fields, and more generally the <a href="http://en.wikipedia.org/wiki/Brauer%25E2%2580%2593Siegel_theorem" rel="nofollow">Brauer-Siegel theorem</a>.</li> <li>The comparison of the class number of a field with the class numbers of its subfields</li> <li>The application of the Beilinson-Bloch conjecture to the arithmetic Bogolomov-Miyoaka-Yau inequality. See <a href="http://terrytao.wordpress.com/2007/05/04/distinguished-lecture-series-iii-shou-wu-zhang-%25E2%2580%259Ctriple-l-series-and-effective-mordell-conjecture%25E2%2580%259D/" rel="nofollow">here</a>.</li> <li>The explicit construction of class fields of totally real fields via <a href="http://en.wikipedia.org/wiki/Stark_conjectures" rel="nofollow">Stark's conjectures</a>. </li> </ol> <p>What else?</p> http://mathoverflow.net/questions/112699/quotations-about-the-power-of-simple-ideas Quotations about the power of simple ideas Jonah Sinick 2012-11-17T17:45:36Z 2012-11-18T17:05:00Z <p>I'm looking for quotations about how very simple mathematical ideas can be very powerful. I know of a few, but they're not quite what I'm looking for insofar as they contain criticism of other mathematicians, and I'm looking for quotations that are more unambiguously affirmative. </p> <p>Two that are in the direction of what I'm looking for are:</p> <ol> <li><p><em>The very notion of a scheme has a child-like simplicity - so simple, so humble in fact that no one before me had the audacity to take it seriously. So ”infantile” in fact, that for many years afterwards, and in spite of all the evidence, for so many of my ”learned” colleagues, it was treated as a triviality.</em> – Alexander Grothendieck </p></li> <li><p><em>It is the snobbishness of the young to suppose that a theorem is trivial because the proof is trivial.</em> – John Whitehead </p></li> </ol> <p>Any better examples?</p> http://mathoverflow.net/questions/112513/quotations-about-the-class-number-formula-etc Quotations about the class number formula, etc. Jonah Sinick 2012-11-15T19:41:01Z 2012-11-15T19:41:01Z <p>I'm looking for interesting and/or expressive quotations from mathematicians about the class number formula. I'm interested both in quotations from historical mathematicians and from modern mathematicians. I'm also interested in quotations about generalizations of the class number formula.</p> <p>I'll start off by giving one, paraphrasing from Henri Darmon and Claude Levesque's article titled <a href="http://www.math.mcgill.ca/darmon/pub/Articles/Surveys/1.Levis/englishpaper.pdf" rel="nofollow">Infinite sums, diophantine equations and Fermat's Last Theorem</a> (pages 4-6):</p> <blockquote> <p>Let $N_p$ be the number of solutions to $x^2 + y^2 = 1$ over $\mathbb{F_p}$, let $N_{\mathbb{Z}}$ be the number of solutions over $\mathbb{Z}$ and let $N_\mathbb{R}$ be the circumference of the circle. From quadratic reciprocity, Leibniz's formula, and the Euler product formula we deduce $\displaystyle \prod_{p} \frac{N_p}{p} = \frac{4}{\pi}$. We conclude that $\displaystyle \prod_{p} \frac{N_p}{p} \cdot N_{\mathbb{R}} = 2N_{\mathbb{Z}}$. This magical formula shows that the numbers $N_p$ "know" the behavior of the equation over the real numbers. Fundamentally, this is only a simple reinterpretation of Leibniz's formula, but in fact this is quite a fruitful one.</p> </blockquote> http://mathoverflow.net/questions/112062/geometric-physical-probabilistic-interpretations-of-riemann-zetan1/112132#112132 Answer by Jonah Sinick for Geometric / physical / probabilistic interpretations of Riemann zeta(n>1)? Jonah Sinick 2012-11-11T23:47:17Z 2012-11-11T23:47:17Z <p>I wrote an article about this very subject titled <a href="http://mathisbeauty.org/ZetaValuesinGeometryandTopology1016.pdf" rel="nofollow">Zeta Values in Geometry and Topology</a> three years ago. My thinking on the points in the article has evolved, in particular, I'm fairly convinced that Questions 0.1-0.4 aren't fruitful lines of inquiry. Still, the material therein is fascinating to me.</p> http://mathoverflow.net/questions/112015/the-history-of-the-geometrization-of-closed-surfaces The history of the geometrization of closed surfaces Jonah Sinick 2012-11-10T19:38:47Z 2012-11-11T14:25:06Z <ol> <li><p>Who first recognized that the torus supports a flat structure? </p></li> <li><p>Who first characterized the moduli space of flat structures on the torus?</p></li> <li><p>Who first recognized that the closed, orientable genus 2 supports a hyperbolic structure?</p></li> <li><p>Who first thought of a geometrized surface in terms of the property that for any two points $A$ and $B$ there exists an isometry of the surface that takes $A$ to $B$?</p></li> </ol> <p>[Reposted <a href="http://math.stackexchange.com/questions/234034/who-first-discovered-that-the-torus-supports-a-flat-structure" rel="nofollow">from Math Stack Exchange</a>.]</p> http://mathoverflow.net/questions/111851/explicit-examples-of-algebraic-hecke-characters-with-infinite-image Explicit examples of algebraic Hecke characters with infinite image? Jonah Sinick 2012-11-08T21:16:38Z 2012-11-09T15:47:06Z <p>Jerry Shurman has a <a href="http://people.reed.edu/~jerry/361/lectures/heckechar.pdf" rel="nofollow">lovely set of notes</a> explaining the classical definition of Hecke characters, the idelic definition of Hecke characters, their relationship, and the classification of algebraic Hecke characters for $\mathbb{Q}$ as Dirichlet characters. He also gives a single family of examples of algebraic Hecke characters with infinite order, namely $\displaystyle \chi: \mathbb{Z}[i] \to \mathbb{C}^{\times}$ given by $\displaystyle \chi(z) = \left(\frac{z}{|z|}\right)^{4n}$ for integers $n$. </p> <p>It's clear that one has essentially the same family for imaginary quadratic number fields with class number $1$. But what about imaginary quadratic fields with higher class number? I imagine that one has one family analogous to the one above for each ideal class, but I don't know what they should look like...</p> <p>What do infinite image algebraic Hecke characters for real quadratic fields look like? Because the unit group is infinite, one can't kill the unit group as above, by putting a $4$ in the exponent...</p> http://mathoverflow.net/questions/111622/background-for-hejhals-the-selberg-trace-formula-for-psl2-mathbbr Background for Hejhal's "The Selberg Trace Formula for $PSL(2, \mathbb{R})$ Jonah Sinick 2012-11-06T06:06:04Z 2012-11-09T15:13:58Z <p>Reposted <a href="http://math.stackexchange.com/questions/229417/background-for-hejhals-the-selberg-trace-formula-for-psl2-mathbbr" rel="nofollow">from math.stackexchange</a> where my question received only five views and no answers...</p> <p>I'm trying to learn the Selberg trace formula, but have very little background in harmonic analysis. I was referred to Dennis Hejhal's <em>The Selberg Trace Formula for $PSL(2, \mathbb{R})$</em> but just got the book and was dismayed to learn that that the author assumes familiarity with Selberg's original paper (which I don't have access to - would welcome a pointer to an online copy). </p> <p>There's much in the first few pages that I don't know. For example, the author states without proof that the spectrum of the Laplacian on a compact hyperbolic surface is discrete. He gives a reference to a 1912 book by Hilbert, but aside from the fact that I don't read German, it's not clear to me that this is the best place to learn from (in light of the fact that Hejhal's book is from 1976 and many books have been written since). </p> <p>Does anyone have a suggestion as to what to read before Hejhal's book?</p> http://mathoverflow.net/questions/111760/a-natural-way-of-thinking-of-the-definition-of-an-artin-l-function A natural way of thinking of the definition of an Artin $L$-function? Jonah Sinick 2012-11-07T22:06:07Z 2012-11-08T15:51:21Z <p>Emil Artin knew that given a finite extension of $L/\mathbb{Q}$, the local factor of the zeta function $\zeta_{L/\mathbb{Q}}$ at the prime $p$ should be $\displaystyle\prod_{\mathfrak{p}|p}\frac{1}{1 - N(\mathfrak{p})^{-s}}$. He also knew that if $L/K$ is a class field then $\displaystyle\prod_{\mathfrak{P}|\mathfrak{p}}\frac{1}{1 - N(\mathfrak{P})^{-s}} = \displaystyle\prod_{\chi}\frac{1}{1 - \chi{(Frob{_\mathfrak{p})}}\cdot N(\mathfrak{p})^{-s}}$ where $\mathfrak{P}$ runs over all primes in $L$ lying above $\mathfrak{p}$ and $\chi$ runs over all characters of $Gal(L/K)$. </p> <p>It's natural then to </p> <ol> <li>Define $L$-series attached to characters on $Gal(L/K)$.</li> <li>Recognize that the definition makes sense whether or not $L/K$ is a class field.</li> <li>In light of the fact that characters are $1$-dimensional representations of $Gal(L/K)$, ask whether there's a good definition of the $L$-series attached to a higher dimensional representation of non-abelian $Gal(L/K)$. </li> </ol> <p>But having come this far, how does one then arrive at the definition of the local factor of an $L$-series attached to a representation $\rho: Gal(L/K) \to GL_{n}(\mathbb{C})$ at a prime $\mathfrak{p}$ unramified in $K$ as </p> <p>$\displaystyle \frac{1}{\det(Id - \rho(Frob_\mathfrak{p})N(\mathfrak{p})^{-s})}$</p> <p>?</p> <p>To be sure</p> <ol> <li>It specializes to the definition of the $L$-series attached to a character on $Gal(L/K)$.</li> <li>It's well-defined (independent of which member of the conjugacy class $Frob_\mathfrak{p}$ one chooses).</li> <li>One has the theorem $\zeta_{L/\mathbb{Q}} = \prod_{\rho} L(\rho, s)$ where $\rho$ ranges over irreducible representations of $Gal(L/\mathbb{Q})$, generalizing the analogous fact for characters on Galois groups of class fields.</li> </ol> <p>And perhaps the three properties listed above are sufficient to uniquely determine the definition. (Maybe one needs more than the above three, I would have to think about it it.) Maybe this is how Artin discovered the definition. This line of thinking is similar to <a href="http://books.google.com/books?id=ZCYtwHFVZHgC&amp;lpg=PA89&amp;ots=oSBxWClxR0&amp;dq=feynman%20heron%27s%20formula&amp;pg=PA89#v=onepage&amp;q=feynman%20heron%27s%20formula&amp;f=false" rel="nofollow">Feynmann's heuristic derivation of Heron's formula</a>. But I somehow feel as though this doesn't get at the essence of things. Is there a way of thinking about the definition of an Artin L-series that gives it more of a sense of inevitability and canonicity?</p> <p>[Reposted <a href="http://math.stackexchange.com/questions/231964/a-natural-way-of-thinking-of-the-definition-of-an-artin-l-function" rel="nofollow">from mathstackexchange</a>.]</p> http://mathoverflow.net/questions/2791/understanding-gal-bar-q-q "Understanding" Gal(\bar Q/Q) Jonah Sinick 2009-10-27T08:10:01Z 2012-11-06T17:24:02Z <p>I have heard people say that a major goal of number theory is to understand the absolute Galois group of the rational numbers G = Gal(Q bar/Q). What do people mean when they say this? The Kronecker-Weber theorem gives a good idea of what the abelianization of the G looks like. But in one of Richard Taylor's MSRI talks, Taylor said that he's never heard of anyone proposing a similar direct description of G and that to understand G one studies the representations of G.</p> <p>I know that there is a strong interest in showing the Langlands reciprocity conjecture [Edit: What I had in mind in writing this is evidently Clozel's conjecture, not the Langlands reciprocity conjecture - see Kevin Buzzard's post below] - that L-functions attached to l-adic Galois representations coincide with L-functions attached to certain automorphic representations. And I've heard people refer to the Tannakian philosophy which I understand as (roughly speaking) asserting that G is determined by all of its finite dimensional representations. </p> <p>Here is a representation of G understood not to be a representation of G as an abstract group but as a group together with a labeling of some of the conjugacy classes of G by rational primes (the Frobenius elements)?</p> <p>When people talk about "understanding G" do they mean proving [Edit: Clozel's conjecture] (in view of the Tannakian philosophy)? If not, what do they mean? If so, this conceptualization seems quite abstract to me. Is this what people mean when they say "understand G"? Can [Edit: Clozel's conjecture] be used to give more tangible statements about G? </p> <p>Something that I have in mind as I write this is the inverse Galois problem (does every finite group occur as a Galois group of a normal extension of Q?) and Gross' conjecture (mostly proven by now) that for each prime p there exists a nonsolvable extension of Q ramified only at p. But I am open to and interested in other senses and respects in which one might "understand" G</p> http://mathoverflow.net/questions/111522/sets-with-zeta-functions-that-are-not-the-primes/111527#111527 Answer by Jonah Sinick for Sets with zeta functions that are not the primes Jonah Sinick 2012-11-05T06:01:46Z 2012-11-05T20:58:38Z <p>You're asking about how special the prime numbers are as a subset of the integers. One can equally well ask how special the sequence $a_k = 1$ is when viewed as the sequence of coefficients of a Dirichlet series. I don't have anything to offer on your original question, but have read a few things about the latter question that may be of some interest:</p> <ol> <li><p>According to Section 8 of David Farmer's article titled <a href="http://arxiv.org/pdf/math/0408154" rel="nofollow">Basic Analytic Number Theory</a>, if $f(s) = \displaystyle \sum_{k = 1}^{\infty} \frac{a_k}{k^s}$ where the $a_k$ are integers, then a sufficient condition for $f(s)$ to admit a meromorphic continuation <em>to $\Re{(s)} = 0$</em> is that:</p> <p>(1) $a_k$ is of subpolynomial growth</p> <p>(2) $a_k$ is multiplicative</p> <p>(3) If $p$ is prime then $a_{p^m}$ is independent of $m$</p> <p>and may or may not have a natural boundary there.</p></li> <li><p>According to section 9.5 of Titchmarsh's <a href="http://www.amazon.com/Theory-Riemann-Zeta-Function-Science-Publications/dp/0198533691" rel="nofollow">The Theory of the Riemann Zeta-function</a>, if $a_k = 0$ when $k$ is composite and $a_{k} = 1$ when $k$ is prime then $f(s)$ (provably) has a natural boundary at $\Re{(s)} = 0$ </p></li> <li><p>I've also heard of results of the type "a Dirichlet series with $a_k$ chosen at random uniformly from $[-1, 1]$ has natural boundary $\Re{(s)} = 0$ with probability $1$," but don't know a precise statement or a reference.</p></li> </ol> http://mathoverflow.net/questions/110931/what-can-we-do-to-raise-awareness-of-reciprocity-laws What can we do to raise awareness of reciprocity laws? Jonah Sinick 2012-10-28T19:02:28Z 2012-10-28T19:32:20Z <p>The study of reciprocity laws is a centerpiece of modern mathematics. Of the last ten Fields Medalists, two of them (Ngô Bảo Châu and Laurent Lafforgue) were awarded Fields Medals for their work on reciprocity laws. Andrew Wiles proved Fermat's Last Theorem by establishing a reciprocity law. Edward Witten works on reciprocity laws.</p> <p>Not only are reciprocity laws popular today: they have a very distinguished history. Fermat, Euler, Lagrange, Legendre and Gauss were very interested in and spent a lot of time thinking about quadratic reciprocity. As is well known, Gauss called the theorem <em>Aureum Theorema</em> and was sufficiently motivated to understand it that he found eight proofs. The aforementioned mathematicians are not only celebrated for their work in number theory: they were also outstanding mathematician physicists and made outstanding advances in a number areas of math. (In Weil's history of number theory, he comments that while few mathematicians were interested in number theory in the early modern history of mathematics, those who were were of the highest quality.) So their strong appreciation of quadratic reciprocity is an indication that the phenomenon points toward some of the deepest and most interesting math.</p> <p>Aside from being very fertile, quadratic reciprocity is in principle very easy to teach. The fact that</p> <blockquote> <p>If $f(x) = x^2 -5$ then the prime divisors of members of the sequence $f(3)$, $f(4)$, $f(5)$, $f(6)$... are $2$, $5$, and those primes that have final digit $1$ or $9$</p> </blockquote> <p>can be exhibited to middle schoolers. I would guess that people who have a solid understanding of two years of high school algebra can learn the full statement of quadratic reciprocity, its interpretation as a statement about which prime numbers factor further in quadratic number rings, its connection with cyclotomy and a hint as to how the phenomenon generalizes in $20$ hours or less. This statement needs qualification:</p> <ul> <li><p>Here I don't mean understanding the <em>proofs</em> of <em>all</em> of the statements involved, but understanding the big picture using certain theorems from algebra and algebraic number theory as black boxes). </p></li> <li><p>I would also emphasize that the material would have to be taught in a carefully constructed and streamlined fashion.</p></li> </ul> <p>But with these qualifications, I think that my statement is true.</p> <p>Despite all of this, very few math majors ever understand quadratic reciprocity. I would guess that the percentage that do is smaller than 1%. It's even uncommon for mathematicans to understand the theorem (I would guess that the percentage who understand is fewer than 50%, and maybe more like 20%). One reason for this is that most math majors aren't required to take a course in which they see quadratic reciprocity. Another reason for this is that courses in elementary number theory (where quadratic reciprocity is presented) don't present the theorem in a motivated way. In his lectures on <a href="http://www.math.duke.edu/langlands/Three.pdf" rel="nofollow">The Practice of Mathematics</a> (pg. 14 of the pdf) Robert Langlands (one of the major contributors to the study of reciprocity laws) wrote:</p> <blockquote> <p>"I confess that, as a student unaware of the history of the subject and unaware of the connection with cyclotomy, I did not find the law...appealing. I suppose, although I would not have – and could not have – expressed myself in this way that I saw it as little more than a mathematical curiosity, fit more for amateurs than for the attention of the serious mathematician that I then hoped to become. It was only in Hermann Weyl's book on the algebraic theory of numbers that I appreciated it as anything more."</p> </blockquote> <p>I had the same initially reaction as Langlands did, and it was only four years after I first learned the statement of the theorem that I understood it.</p> <blockquote> <p>What can we do as members of the mathematical community to raise awareness of reciprocity laws?</p> </blockquote> <p>Can we change the algebra or elementary number theory course syllabi to address this issue? Can we push for the creation of a Nova or BBC series about reciprocity laws? Any other ideas?</p> http://mathoverflow.net/questions/110882/the-existence-of-an-elliptic-curve-with-a-specific-galois-representation-induced The existence of an elliptic curve with a specific Galois representation induced by a character Jonah Sinick 2012-10-28T06:40:22Z 2012-10-28T10:38:29Z <p>In Kevin Buzzard's <a href="http://www2.imperial.ac.uk/~buzzard/maths/research/papers/survey.pdf" rel="nofollow">survey article on potential modularity</a> Buzzard writes:</p> <blockquote> <p>Let us say that we have an elliptic curve $E$ over a totally real ﬁeld $F$, and we want to prove that $E$ is potentially modular (that is, that $E$ becomes modular over a ﬁnite extension ﬁeld $F^{′}$ of $F$, also assumed totally real). Here is a strategy. Say $p$ is a large prime such that $E[p]$ is irreducible. Let us write down a random odd $2$-dimensional mod $ℓ$ Galois representation $\rho_{ℓ} : Gal(\overline{F}/F) → GL(2,\mathbf{F}_ℓ )$ which is induced from a character; because this representation is induced it is known to be modular. Now let us consider the moduli space parametrising elliptic curves $A$ equipped with </p> <ol> <li>An isomorphism $A[p] \cong E[p]$</li> <li>An isomorphism $A[ℓ]\cong ρ_ℓ$</li> </ol> <p>This moduli problem will be represented by some modular curve, whose connected components will be twists of $X(pℓ)$ and hence, if $p$ and $ℓ$ are large, will typically have large genus. However, such a curve may well still have lots of rational points, as long as I am allowed to look for such things over an arbitrary ﬁnite extension $F^{′}$ of $F$ !</p> </blockquote> <p>It's not immediately obvious to me that there's an elliptic curve $A$ over some $F^{′}$ satisfying the second condition alone (never mind satisfying both conditions simultaneously). Is there a simple explanation for why there should be such an $A$? Did Professor Buzzard mean "consider the set of A such that $A[ℓ]\cong ρ_ℓ$ for <em>some</em> representation induced by a character" (as opposed to a particular one)?</p> http://mathoverflow.net/questions/110737/why-doesnt-functoriality-immediately-imply-the-modularity-theorem Why doesn't functoriality immediately imply the modularity theorem? Jonah Sinick 2012-10-26T08:24:18Z 2012-10-27T21:25:39Z <p>Let $E/\mathbb{Q}$ be an elliptic curve. By the modularity theorem, the prime indexed coefficients of its $L$-function agree with those of a weight $2$ cusp eigenform $f$ with integer coefficients. This immediately imply that the coefficients are congruent (mod $p^k$) for every $k > 0$. However, the converse is also true: if the coefficients of the $L$-series of $E$ and that of $f$ are congruent (mod $p^k$) for every prime $k > 0$, then the $L$-series agree.</p> <p>The work of Langlands and Tunnell can be used to show that if the elliptic curve has irreducible (mod $3$) Galois representation, then coefficients of the $L$-function agree with those of a weight $2$ cusp form with coefficients in some algebraic number field $K$ (mod $v$), where $v$ is a prime above $3$ in $K$. This was the starting point of Wiles' proof of the modularity theorem for semi-stable elliptic curves over $\mathbb{Q}$.</p> <p>One could try to get congruence of coefficients (mod $p^k$) for some value of $p$ and for larger values of $k$ by a method analogous to the method via Langlands and Tunnell rather than as a consequence of modularity lifting theorems. One immediately runs into a stumbling block because when $k$ or $p$ is big enough the group is nonsolvable and the methods of Langlands and Tunnell can't be applied (in a known way) to prove relevant cases of the strong Artin conjecture.</p> <p>Nevertheless, there exists an $n$ such that there is an injective representation $\rho: GL(2, \mathbb{Z}/p^k\mathbb{Z}) \to GL(n, \mathbb{C})$. If one can take this representation to be irreducible, then according to the strong Artin conjecture, its $L$-function should be automorphic. Even assuming that this is the case, it's not at all immediately clear (at least, without knowing the modularity theorem) that the $L$-function of the corresponding automorphic representation is related to that a weight $2$ holomorphic cusp eigenform for $GL(2)$. </p> <p>But functoriality can sometimes be used to relate $L$-functions for automorphic representations on one group to $L$-functions of automorphic representations on another group. The arrows only go one way, and in this case it looks like the wrong way, but sometimes one can characterize the arrows' images. Given that we know <em>ex post</em> that there <em>is</em> a relationship between the (mod $p^k$) Galois representation attached to an elliptic curve and that of $f$ (uniform over $k$!), one can ask whether one can see the relationship directly'' from functoriality, without passing through the modularity lifting theorems.</p> <p>Moreover, if one could do this for infinitely many $k$, perhaps one could show that the coefficients of the $L$-function of the elliptic curve $E$ match up with those of the associated modular form $f$ in characteristic $0$ so as to obtain a different proof of the modularity theorem (conditional, of course, on functoriality).</p> <p>The picture that I've sketched above is full of holes like Swiss cheese and it will take me years to understand the precise statements that I allude to above (never mind the proofs!). Nevertheless, I feel that there's a kernel of a well-defined question in what I write. I assume that the strategy that I allude to breaks down somewhere, because otherwise I would have heard about it. I would be grateful to anybody who would be willing to enlighten me as to what goes wrong.</p> <p>[Edited 10/27/12 to incorporate my last paragraph.] One could also attempt to get a result as $p$ varies rather than $k$ if there are technical problems that come up with varying $k$ but not with varying $p$.</p> http://mathoverflow.net/questions/109417/what-is-the-status-of-the-equidistribution-root-numbers-of-elliptic-curves-l-fun What is the status of the equidistribution root numbers of elliptic curves' L-functions Jonah Sinick 2012-10-11T21:20:47Z 2012-10-11T21:20:47Z <p>In Section 7 of Alice Silverberg's <a href="http://math.uci.edu/~asilverb/connectionstalk.pdf" rel="nofollow">Rank "Cheat Sheet"</a>, Silverberg stated</p> <blockquote> <p>The Bhargava Conjecture: For each $n > 1$ the average size of $S_{n}(E/\mathbb{Q})$ is $\displaystyle\sum\limits_{d|n} d$.</p> </blockquote> <p>Here $S_{n}(E/\mathbb{Q})$ is the $n$-Selmer group of $E/\mathbb{Q}$. Silverberg remarks that assuming the Bhargava conjecture for infinitely many $n$, the <a href="http://mathoverflow.net/questions/71609/the-parity-conjecture" rel="nofollow">parity conjecture</a>, and the equidistribution of root numbers of $L$-functions of elliptic curves over $\mathbb{Q}$, it follows that $50$% of elliptic curves have rank $0$ and $50$% of elliptic curves have rank $1$ (this is called the Rank Distribution Conjecture). </p> <blockquote> <p>Does anyone have a conjectural strategy for proving the equidistribution of root numbers?</p> </blockquote> <p>Silverberg mentions the Poonen-Rains conjecture together with the parity conjecture implying the Rank Distribution Conjecture (which in turn implies equidistribution of root number), so it could be that trying to prove the Poonen-Rains conjecture offers a possible approach, but it seems to me that one in fact needs the equidistribution conjecture as a hypothesis in the latter statement of Silverberg... </p> http://mathoverflow.net/questions/109330/did-gauss-know-dirichlets-class-number-formula-in-1801 Did Gauss know Dirichlet's class number formula in 1801? Jonah Sinick 2012-10-10T20:38:28Z 2012-10-11T21:02:17Z <p>Let $h_d$ be the number of $SL_{2}(\mathbb{Z})$ classes of primitive binary quadratic forms of discriminant $d$. It's natural to impose the hypothesis that $d$ is not at square, as we do below. </p> <p>In Carl Ludwig Siegel's paper titled <em>The Average Measure of Quadratic Forms With Given Discriminant and Signature</em> Siegel cites two formulae given by Gauss in <em>Disquisitiones Arithmeticae</em>: </p> <p>(a) $\displaystyle\sum\limits_{d= -N }^1 h_d \sim \frac{\pi}{18 \zeta(3)}N^{3/2}$ </p> <p>(b) $\displaystyle\sum\limits_{d = 1}^N h_d \log{\epsilon}_d \sim \frac{{\pi}^2}{18 \zeta(3)}N^{3/2}$ </p> <p>Where $N > 0$ and $\epsilon_{d} = \frac{1}{2}(t + u \sqrt{d})$ where $(t,u)$ is the smallest positive solution to $t^2 - ud^2 = 4$.</p> <p>(Actually, Gauss restricts to consideration to binary quadratic forms with even middle coefficient correspondingly arrives at different formulae, but they're essentially the same as those above).</p> <p>Siegel gives two proofs of these formulae: one proceeding from Dirichlet's class number formula together with character sum estimates due to Polya and Landau, and one via a direct lattice point counting argument. </p> <p>In light of the facts that (i) I haven't heard anyone say that Gauss's was the one to discover the class number formula and (ii) the character sum estimates seem outside of the scope of Gauss's work, I imagine that his argument was via lattice point counting. Do we have any evidence otherwise? (I checked Gauss's book and he doesn't describe his methods there.)</p> http://mathoverflow.net/questions/129759/modern-mathematical-achievements-accessible-to-undergraduates Comment by Jonah Sinick Jonah Sinick 2013-05-06T03:57:37Z 2013-05-06T03:57:37Z Is it fair to characterize the use of the Jones Polynomial to distinguish knot as tremendous progress in mathematics? My impression is that its primary source of interest is in its deeper significance. http://mathoverflow.net/questions/129059/grothendieck-s-question-any-update Comment by Jonah Sinick Jonah Sinick 2013-04-29T04:28:12Z 2013-04-29T04:28:12Z I believe that this question was answered in the affirmative by Jean-Pierre Serre, but no longer remember where I read that, and may be misremembering. As for the corresponding question for cohomology groups, see pg. 6 of <a href="http://arxiv.org/pdf/math/0210327v1.pdf" rel="nofollow">arxiv.org/pdf/math/0210327v1.pdf</a> http://mathoverflow.net/questions/126474/new-geometric-methods-in-number-theory-and-automorphic-forms Comment by Jonah Sinick Jonah Sinick 2013-04-04T07:47:54Z 2013-04-04T07:47:54Z The description doesn't explain what breakthroughs it's referring to. This makes it hard to know what the authors have in mind. http://mathoverflow.net/questions/126123/how-did-takagi-prove-kroneckers-jugendtraum-for-qi Comment by Jonah Sinick Jonah Sinick 2013-03-31T23:07:52Z 2013-03-31T23:07:52Z ...because the class number is 1, and the unit group is finite. http://mathoverflow.net/questions/126123/how-did-takagi-prove-kroneckers-jugendtraum-for-qi Comment by Jonah Sinick Jonah Sinick 2013-03-31T21:19:55Z 2013-03-31T21:19:55Z I would imagine that you can mimic the proof of the Kronecker-Weber theorem. http://mathoverflow.net/questions/125189/have-we-ever-proved-any-non-solvable-case-of-reciprocity-without-the-langlands-pr/125807#125807 Comment by Jonah Sinick Jonah Sinick 2013-03-28T20:14:01Z 2013-03-28T20:14:01Z If I understand correctly, Eichler proved the theorem for X_0(11) specifically rather than X_0(N). (The former is an elliptic curve, whereas in the higher genus case one has to look at the Jacobian...) http://mathoverflow.net/questions/125659/the-existence-of-meromorphic-functions-on-riemann-surfaces/125669#125669 Comment by Jonah Sinick Jonah Sinick 2013-03-27T00:18:21Z 2013-03-27T00:18:21Z @Ian: I guess what I want to do is (i) show that the collection of genus g Riemann surfaces with a meromorphic map f to the sphere is dense in the moduli space of genus g Riemann surfaces and then (ii) show that an infinitesimal deformation X' of a Riemann surface X with a meromorphic map to the sphere itself has a meromorphic map to the sphere by infinitesimally deforming f. Presumably this is what the analytic proofs implicitly do: I'll have to think about how they correspond to this mental model. http://mathoverflow.net/questions/125659/the-existence-of-meromorphic-functions-on-riemann-surfaces Comment by Jonah Sinick Jonah Sinick 2013-03-26T23:36:40Z 2013-03-26T23:36:40Z wwcanard —&#160;Ok, so I agree that algebraic arguments won't be relevant. Still, one can hope for a more geometric argument. http://mathoverflow.net/questions/125659/the-existence-of-meromorphic-functions-on-riemann-surfaces/125669#125669 Comment by Jonah Sinick Jonah Sinick 2013-03-26T21:47:30Z 2013-03-26T21:47:30Z I guess one could argue that even the proof in the genus 1 case requires hard analysis (in that one has to verify the convergence of infinite series), but somehow it's still quite conceptual. http://mathoverflow.net/questions/125659/the-existence-of-meromorphic-functions-on-riemann-surfaces/125669#125669 Comment by Jonah Sinick Jonah Sinick 2013-03-26T21:44:25Z 2013-03-26T21:44:25Z Thanks Ian. If I understand correctly, the fact that 1-forms have holomorphic representatives requires the Hodge Theorem, which in turn is usually proved via the theory of elliptic operators, though I found this question <a href="http://mathoverflow.net/questions/28265/proving-hodge-decomposition-without-using-the-theory-of-elliptic-operators" rel="nofollow" title="proving hodge decomposition without using the theory of elliptic operators">mathoverflow.net/questions/28265/&hellip;</a> (which will take some time to digest). The only proofs of the uniformization theorem that I've seen use hard analysis. Maybe the use of hard analysis is inevitable. http://mathoverflow.net/questions/125659/the-existence-of-meromorphic-functions-on-riemann-surfaces Comment by Jonah Sinick Jonah Sinick 2013-03-26T21:18:13Z 2013-03-26T21:18:13Z They're not, but you can still consider the moduli space of algebraic curves of given genus and the moduli space of Riemann surfaces of a given genus and to prove that they're the same. The spaces have the same number of dimensions and the former lies in the latter. Maybe you could use some sort of continuity argument to prove that the former can't be a proper subset of the latter. http://mathoverflow.net/questions/123161/must-read-papers-on-analytic-number-theory Comment by Jonah Sinick Jonah Sinick 2013-02-28T01:36:37Z 2013-02-28T01:36:37Z Are you interested in additive number theory? Multiplicative number theory? Nonabelian harmonic analysis? http://mathoverflow.net/questions/123161/must-read-papers-on-analytic-number-theory Comment by Jonah Sinick Jonah Sinick 2013-02-28T01:33:39Z 2013-02-28T01:33:39Z The proof of the Brauer–Siegel theorem (e.g. as presented in Lang's book), though probably that's something that you'll do regardless / have already done. I know some people who like Soundararajan and Ono's paper titled &quot;Ramanujan's ternary quadratic form.&quot; http://mathoverflow.net/questions/119329/the-riemann-hypothesis-and-the-langlands-program/119532#119532 Comment by Jonah Sinick Jonah Sinick 2013-01-23T04:04:35Z 2013-01-23T04:04:35Z The first paragraph of Langlands' 1978 ICM lecture &quot;L-functions and Automorphic Representations&quot; <a href="http://publications.ias.edu/rpl/paper/65" rel="nofollow">publications.ias.edu/rpl/paper/65</a> reads &quot;Introduction. There are at least three different problems with which one is confronted in the study of L-functions: the analytic continuation and functional equation; the location of the zeroes; and in some cases, the determination of the values at special points. The ﬁrst may bethe easiest. It is certainly the only one with which I have been closely involved.&quot; http://mathoverflow.net/questions/119329/the-riemann-hypothesis-and-the-langlands-program/119356#119356 Comment by Jonah Sinick Jonah Sinick 2013-01-19T20:22:48Z 2013-01-19T20:22:48Z @ GH –&#160;Yes, I'm aware of this, I glossed over this for brevity. Feel free to edit my answer if you'd like.