User gene s. kopp - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T12:01:54Z http://mathoverflow.net/feeds/user/8410 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102386/is-a-random-subset-of-the-real-numbers-non-measurable-is-the-set-of-measurable Is a random subset of the real numbers non-measurable? Is the set of measurable sets measurable? Gene S. Kopp 2012-07-16T20:21:25Z 2012-11-29T22:24:54Z <p>One might say, "a random subset of $\mathbb{R}$ is not Lebesgue measurable" without really thinking about it. But if we unpack the standard definitions of all those terms (and work in ZFC), it's not so clear.</p> <p>Let $\Sigma \subset 2^\mathbb{R}$ be the sigma-algebra of all Lebesgue measurable sets. Give $2^\mathbb{R}$ the product measure. (It's a product of continuum many copies of the two-point set.) We want to say that $\Sigma$ is a null set in $2^\mathbb{R}$...but is $\Sigma$ even measurable?</p> <p>Laci Babai posed this question casually several years ago, and no one present knew how to go about it, but it might be easy for a set theorist.</p> <p>Also, a related question: Think of $2^\mathbb{R}$ as a vector space over the field with two elements and $\Sigma$ as a subspace. (Addition is xor, that is, symmetric set difference.) What is $\dim\left(2^\mathbb{R}/\Sigma\right)$?</p> <p>It's not hard to see that $\dim\left(2^\mathbb{R}/\Sigma\right)$ is at least countable, so if $\Sigma$ were measurable, it would be a null set. But that's as far as I made it.</p> http://mathoverflow.net/questions/83027/what-is-ricardo-perez-marcos-ene-product-does-it-explain-his-statistical-resul What is Ricardo Pérez-Marco's eñe product? Does it explain his statistical results on differences of zeta zeros? Gene S. Kopp 2011-12-09T03:53:54Z 2012-11-29T22:04:09Z <p>The number theory community here at University of Michigan is abuzz with talk of this <a href="http://arxiv.org/abs/1112.0346" rel="nofollow">paper</a> recently posted to the arxiv. If you haven't seen it already, the punch line is that the global differences of imaginary parts of zeta zeros show an unusual tendency not to be close to imaginary parts of zeta zeros. "Riemann zeros repel their deltas."</p> <p>The people I have talked to have said that this simple observation appears to be completely new and potentially groundbreaking. Here are my questions:</p> <p>1) Has anyone independently verified these statistics? (See my histograms below, and also the excellent graphs in Jonathan Bober's answer.)</p> <p>2) The author discusses the "eñe product" as a source of a mathematical explanation for his observation, and he refers to his own unpublished manuscripts in which he introduces this product. Are these manuscripts available? Does anyone have any reference which defines the eñe product?</p> <p>Finally, any insights into Marco's results would be much appreciated.</p> <p>The histograms (produced in Mathematica) are of the deltas of the first 10000 zeros of $\zeta(s)$ falling in $[100, 110]$ and $[190, 200]$, respectively. The purple bars show the locations of the zeros.</p> <p><img src="http://s17.postimage.org/e1t4h0jvx/marco1.jpg" alt=""> <img src="http://s7.postimage.org/kogm27i6x/marco2.jpg" alt=""></p> http://mathoverflow.net/questions/97060/asymptotics-of-the-growth-rate-of-a-group/97805#97805 Answer by Gene S. Kopp for Asymptotics of the growth rate of a group Gene S. Kopp 2012-05-24T00:46:25Z 2012-05-24T00:46:25Z <p>There is never a (finite) generating set with that property.</p> <p>Consider a generating set <code>$S=\{x_1,\ldots,x_{\ell}\}$</code> of cardinality $\ell$. Let $B_k := B_k(S)$, $S_k := B_k \setminus B_{k-1}$, and $g := gr(S)$. Let $b_k := |B_k|$ and $s_k: = |S_k|$. Assume for simplicity that $L := \lim_{k \to \infty} \frac{b_k}{g^k}$ exists (although it shouldn't be difficult to get general liminf bounds). Also set $$L' := \lim_{k \to \infty} \frac{s_k}{g^k} = \lim_{k \to \infty} \frac{b_k-b_{k-1}}{g^k} = \left(1-\frac{1}{g}\right)L.$$</p> <p>For the spheres, there's a trivial inequality $s_{m+n} \leq s_m s_n$. (Any word of minimal length $m+n$ in the generators may be written at least one way as a product of two words of minimal length $m$ and $n$.) This is already enough to give $L' \geq 1$, and thus $$L \geq \frac{1}{1-\frac{1}{g}} > 1.$$</p> <p>This is unsatisfying and still leaves the possibility that $L'=1$. Thus, I eliminate that possibility as well, by improving the trivial bound $s_{2k} \leq s_k^2$ to $s_{2k} \leq \left(1-\frac{1}{2\ell}\right)s_k^2$. This improvement (unlike the trivial improvement above) does not hold for monoids, so the extra cancellation comes (unsurprisingly) from the existance of inverses.</p> <p>Specifically, fix $k$, let $E_i$ be the set of all elements of $S_k$ which may be written as a word of length $k$ ending in $x_i$, and let $E_{\ell+i}$ be the set of all elements of $S_k$ which may be written as a word of length $k$ ending in $x_i^{-1}$. Set $$F_i = E_i \setminus \bigcup_{j &lt; i} E_j,$$ so the $F_i$ are disjoint. Let $F_i'$ be the image of $F_i$ under the inversion map, and let $n_i = |F_i|$. The product of an element of $F_i$ and an element of $F_i'$ may be written with $2k-2$ letters (because the $x_i$ and $x_i^{-1}$ cancel), so those $n_i^2$ products are not in $S_{2k}$. Thus, we have $$s_{2k} \leq s_k^2 - \sum_{i=1}^{2\ell} n_i^2.$$ By the Hölder inequality, we have $$s_k = \sum_{i=1}^{2\ell} n_i \leq \sqrt{\sum_{i=1}^{2\ell} 1^2} \sqrt{\sum_{i=1}^{2\ell} n_i^2}$$ $$\frac{s_k^2}{2\ell} \leq \sum_{i=1}^{2\ell} n_i^2.$$ Combining the inequalities gives $$s_{2k} \leq \left(1-\frac{1}{2\ell}\right)s_k^2.$$ Dividing by $g^{2k}$ and sending $k \to \infty$ gives $$L' \geq \frac{1}{1-\frac{1}{2\ell}} > 1.$$ For balls, we obtain the bound $$L \geq \frac{1}{\left(1-\frac{1}{g}\right)\left(1-\frac{1}{2\ell}\right)}.$$ This bound is optimal for free generating sets of free groups.</p> <p>I'd be interested in seeing what Kate's limit says (or don't say) about the underlying group. Can the lower bound I just gave be achieved for non-free groups? Can the limit generally be made arbitrarily close to $1$ by increasing the number of generators appropriately (as Misha speculated in the comments)? I'm far from an expert in combinatorial/asymptotic group theory, so I don't have good intuition for what intrinsic information this value holds.</p> http://mathoverflow.net/questions/94551/are-sets-with-similar-asymptotic-behavior-as-the-primes-necessarily-finite-additi/94977#94977 Answer by Gene S. Kopp for Are sets with similar asymptotic behavior as the primes necessarily finite additive bases? Gene S. Kopp 2012-04-23T20:35:53Z 2012-04-24T05:02:29Z <p>Let <code>$A_n = \{a : a \equiv 1 \mod 2^n \mbox{ and } 2^{2^{n-1}} \leq a &lt; 2^{2^{n}} - 2^{2^{n-1}}\}$</code>, and let $\displaystyle A = \bigcup_{n=1}^\infty A_n$.</p> <p>Then, $A(x) >> \frac{x}{\log x}$, the gap sizes are $&lt;&lt; \sqrt{x}$, and $A$ contains infinitely many non-multiples of $m$ for every $m>1$.</p> <p>However, $2^{2^n}$ cannot be written as a sum of fewer than $n-\log_2 n$ elements of $A$. To prove this, suppose the contrary; say $2^{2^{n}} = a_1 + \cdots + a_k$ for $k &lt; n-\log_2 n$, $a_1 \leq \cdots \leq a_k$, $a_i \in A$. We know that $a_k \in A_n$, because if not, then all the $a_i &lt; 2^{2^{n-1}}$, so, summing, $2^{2^n} &lt; k \cdot 2^{2^{n-1}}$, implying $2^{2^{n-1}} &lt; n$, which is false for positive $n$. One applies a similar argument to each of the partial sums of the $a_i$, in turn from largest to smallest, to show that $\displaystyle a_j \in \bigcup_{i=n-k+j}^n A_i$. In particular, each $a_j \equiv 1 \mod 2^{n-k+1}$, so the sum $2^{2^n} \equiv k \mod 2^{n-k+1}$. Thus, $2^{n-k+1} \mid k$; in particular, $2^{n-k+1} \leq k$. Using the bound $k &lt; n-\log_2 n$ on both sides, we may deduce $2n &lt; n-\log_2 n$, a contradiction.</p> <p>For a possible fix, I would assume that $A$ is well-distributed modulo $m$ (for every $m$) in an appropriate sense. (I'm being purposefully vague here, and you might need something rather strong, like an analogue of <a href="http://en.wikipedia.org/wiki/Bombieri%E2%80%93Vinogradov_theorem" rel="nofollow">Bombieri-Vinogradov</a>). Good luck!</p> http://mathoverflow.net/questions/89324/are-all-zeros-of-gammas-pm-gamma1-s-on-a-line-with-real-part-frac12/92114#92114 Answer by Gene S. Kopp for Are all zeros of $\Gamma(s) \pm \Gamma(1-s)$ on a line with real part = $\frac12$ ? Gene S. Kopp 2012-03-24T22:55:58Z 2012-03-25T20:16:20Z <p>I would like to expand on Guild of Pepperers's answer by noting that the zeros are essentially uniformly spaced and may easily be approximated to a high degree of accuracy. Using Stirling approximation, I obtained the formula $$\Gamma\left(\frac12+it\right) = \sqrt{\frac{2\pi}{1+e^{-2\pi|t|}}}\exp\left(-\frac\pi2|t|+i(t\log|t|-t+\varepsilon(t))\right),$$ valid for real $t$, where the error $\varepsilon(t)$ is an odd, bounded, real-valued function asymptotically equal to $\frac{1}{24t}$. (Indeed, $\varepsilon(t)$ has asymptotic and convergent expansions coming from the asymptotic and convergent versions of Stirling approximation, respectively.) We then have, for $s = \frac12+it$ on the critical line, $$\Gamma(s)+\Gamma(1-s) = 2\sqrt{\frac{2\pi}{1+e^{-2\pi|t|}}}e^{-\frac\pi2|t|}\cos\left(t\log|t|-t+\varepsilon(t)\right),$$ $$\Gamma(s)-\Gamma(1-s) = 2\sqrt{\frac{2\pi}{1+e^{-2\pi|t|}}}e^{-\frac\pi2|t|}\sin\left(t\log|t|-t+\varepsilon(t)\right).$$ One may show by means fair or foul that $t\log|t|-t+\varepsilon(t)$ is monotonically increasing for $|t|\geq1.05$, is bounded between $-0.96$ and $0.96$ for $|t|&lt;1.05$, and is only zero when t = 0. Therefore, the zeros of $\Gamma(s)+\Gamma(1-s)$ on the critical line occur, with multiplicity one, very near those $t$ for which $t\log|t|-t$ is an odd integer multiple of $\frac{\pi}{2}$, and similarly for $\Gamma(s)-\Gamma(1-s)$ and the even integer multiples of $\frac{\pi}{2}$.</p> <p>It's interesting that the number of zeros up to a given height $T$ is of the same order of magnitude, $T \log(T)$, as for the Riemann zeta function, but that these zeros have (essentially) uniform spacings rather than GUE spacings.</p> http://mathoverflow.net/questions/35320/perron-frobenius-inverse-eigenvalue-problem Perron-Frobenius "inverse eigenvalue problem" Gene S. Kopp 2010-08-12T06:36:26Z 2011-11-23T04:26:02Z <p>The Perron-Frobenius theorem says that the largest eigenvalue of a positive real matrix (all entries positive) is real. Moreover, that eigenvalue has a positive eigenvector, and it is the only eigenvalue having a positive eigenvector.</p> <p>Now suppose we want to construct a positive rational matrix with a particular Perron-Frobenius eigenvalue. Specifically, consider a positive real algebraic number $\lambda$ which is greater in absolute value than all of its Galois conjugates. Does there exist a positive rational matrix $A$ with $\lambda$ as its Perron-Frobenius eigenvalue?</p> http://mathoverflow.net/questions/24958/showing-e-is-transcendental-using-its-continued-fraction-expansion/39967#39967 Answer by Gene S. Kopp for Showing e is transcendental using its continued fraction expansion Gene S. Kopp 2010-09-25T19:02:42Z 2011-08-24T19:06:27Z <p>To use the continued fraction for a number to prove it's transcendental, one usually shows that the rational approximations it affords are "too good" for an algebraic number. The traditional tool here is Liouville's theorem, but this has been improved to the more powerful Roth's theorem:</p> <p>If $\alpha$ is algebraic, then for every $\varepsilon > 0$ there are only finitely many rational numbers $p/q$ satisfying $$\left|\alpha - \frac{p}{q}\right| &lt; \frac{1}{q^{2+\epsilon}}.$$</p> <p>Unfortunately, $e$ also satisfies this. Indeed, rational approximations to $e$ are uncommonly <em>bad</em>. As Wadim mentions, there are only finitely many $p/q$ satisfying the following inequality. $$\left|e - \frac{p}{q}\right| &lt; \frac{\log \log q}{3 q^2 \log q}.$$</p> <p>But Khinchin proved that, for almost all $\alpha$, $$\left|\alpha - \frac{p}{q}\right| &lt; \frac{1}{q} \phi(q)$$ has infinitely many solutions if and only if $\sum_q \phi(q)$ diverges.</p> <p>Additionally, Khinchin showed that the geometric means of the entries of the continued fraction expansion of a real number almost always converge to a universal constant. The geometric means of the entries of the continued fraction expansion of $e$ diverge.</p> <p>If either of Khinchin's conditions hold for nonquadratic algebraic numbers, the transcendentality of $e$ would follow, but a proof is probably out of reach.</p> <p>Finally, the continued fraction expansion of $e$ provides an immediate proof of its <em>irrationality</em>, because rational numbers have finite continued fraction expansions. We also have that $e$ is not a quadratic irrational, because those have (eventually) periodic continued fraction expansions.</p> http://mathoverflow.net/questions/48671/examples-of-non-rigorous-but-efficient-mathematical-methods-in-physics/48895#48895 Answer by Gene S. Kopp for Examples of non-rigorous but efficient mathematical methods in physics Gene S. Kopp 2010-12-10T08:11:05Z 2010-12-10T08:11:05Z <p>The use of <a href="http://en.wikipedia.org/wiki/Random_matrix" rel="nofollow">random matrix theory</a> to model energy levels of heavy nuclei and other physical systems. See also the following <a href="http://www.williams.edu/go/math/sjmiller/public_html/ntrmt10/handouts/general/Hayes_spectrum_riemannium.pdf" rel="nofollow">historical piece</a> and the pictures therein: There is striking statistical evidence that the eigenvalues of large random self-adjoint matrices, the energy levels of heavy nuclei, and the normalized zeros of $L$-functions (!) are all spaced about the same.</p> http://mathoverflow.net/questions/48718/is-every-compact-topological-ring-a-profinite-ring Is every compact topological ring a profinite ring? Gene S. Kopp 2010-12-09T05:43:20Z 2010-12-09T09:33:22Z <p>There are a lot of compact (Hausdorff) groups, whereas every compact field is finite. What about rings? Is there a classification theorem for compact rings? If you take a cofiltered limit of finite rings, you get a compact ring; for example, the $p$-adic integers $\mathbb{Z}_p$ are obtained as a limit of $$\cdots \twoheadrightarrow \mathbb{Z}/p^{n+1}\mathbb{Z} \twoheadrightarrow \mathbb{Z}/p^n\mathbb{Z}\twoheadrightarrow \cdots \twoheadrightarrow \mathbb{Z}/p\mathbb{Z}\twoheadrightarrow 0.$$ Can every compact ring be obtained as a cofiltered limit of finite rings?</p> <p>For a counterexample, a compact ring that is not totally disconnected would suffice. In the other direction, proving that such a ring has to be totally disconnected wouldn't suffice <em>a priori</em>: It would show the the additive <em>group</em> is profinite, but not that the ring is a cofiltered limit <em>of rings</em>.</p> <p>Remark: By "compact," I consistently mean "compact Hausdorff."</p> http://mathoverflow.net/questions/2144/a-single-paper-everyone-should-read/46338#46338 Answer by Gene S. Kopp for A single paper everyone should read? Gene S. Kopp 2010-11-17T08:42:58Z 2010-11-17T08:56:03Z <p>Not technically a paper but a lecture (in pdf form) full of pretty pictures and cool ideas:</p> <p><a href="http://math.ucr.edu/home/baez/counting/counting.pdf" rel="nofollow">The Mysteries of Counting: Euler Characteristic versus Homotopy Cardinality</a> by John Baez.</p> <blockquote> <p>We all know what it means for a set to have 6 elements, but what sort of thing has -1 elements, or 5/2? Believe it or not, these questions have nice answers. The Euler characteristic of a space is a generalization of cardinality that admits negative integer values, while the homotopy cardinality is a generalization that admits positive real values. These concepts shed new light on basic mathematics. For example, the space of finite sets turns out to have homotopy cardinality e, and this explains the key properties of the exponential function. Euler characteristic and homotopy cardinality share many properties, but it's hard to tell if they are the same, because there are very few spaces for which both are well-defined. However, in many cases where one is well-defined, the other may be computed by dubious manipulations involving divergent series---and the two then agree! The challenge of unifying them remains open.</p> </blockquote> http://mathoverflow.net/questions/45951/sexy-vacuity/46077#46077 Answer by Gene S. Kopp for Sexy vacuity .... Gene S. Kopp 2010-11-14T22:34:07Z 2010-11-14T22:34:07Z <p>One is not a prime number, but zero is! It's different than the other prime numbers in $\mathbb{Z}$, though, because it has height zero rather than height one.</p> <p>Zero is prime in any integral domain. Remember that the trivial ring is <em>not</em> an integral domain.</p> http://mathoverflow.net/questions/45951/sexy-vacuity/46076#46076 Answer by Gene S. Kopp for Sexy vacuity .... Gene S. Kopp 2010-11-14T22:24:17Z 2010-11-14T22:24:17Z <p>Zero is a limit ordinal, because it is the union of its elements.</p> <p>Transfinite induction has two canonical statements. The "strong" statement, $$(\forall \alpha,\beta)((\beta&lt;\alpha) \wedge P(\beta) \rightarrow P(\alpha))\rightarrow (\forall \alpha)P(\alpha),$$ doesn't split anything into cases. The version used most frequently in proofs says that any property preserved under unions and successors holds for all ordinals. Zero should rarely be a special case.</p> <p>Also, "limit ordinals" should totally be called "colimit ordinals". The term "limit ordinal" refers to limit points in the order topology, thus excluding zero, but this is silly.</p> http://mathoverflow.net/questions/43900/density-of-first-order-definable-sets-in-a-directed-union-of-finite-groups Density of first-order definable sets in a directed union of finite groups Gene S. Kopp 2010-10-27T23:10:17Z 2010-11-04T05:56:47Z <p>This is a generalization of the following <a href="http://mathoverflow.net/questions/39798/in-an-inductive-family-of-groups-does-the-probability-that-a-particular-word-is" rel="nofollow">question</a> by John Wiltshire-Gordon.</p> <p>Consider an inductive family of finite groups: $$G_0 \hookrightarrow G_1 \hookrightarrow \ldots \hookrightarrow G_i \hookrightarrow \ldots$$</p> <p>We may view each group as a subgroup of the next. Let $G$ be the directed union of the $G_i$: $$G := \bigcup_{i=0}^\infty \ G_i$$</p> <p>Now, $G$ is a countable group. Suppose we're asked a question like, "What proportion of the elements of $G$ are commutators?", or "What proportion of pairs $(g,h) \in G^2$ satisfy $g^2h^2=1$ but $g^2 \neq 1$?" We try to make sense of such questions in terms of <em>density</em> on $G^n$, which is not defined for all subsets. For $E \subseteq G^n$, let $$d(E) := \lim_{i \to \infty} \frac{|E \cap G_i^n|}{|G_i^n|},$$ if the limit exists.</p> <p>My question is: If $E$ is a subset of $G^n$ that is first-order definable in the language of groups, does the density of $E$ necessarily exist? John's <a href="http://mathoverflow.net/questions/39798/in-an-inductive-family-of-groups-does-the-probability-that-a-particular-word-is" rel="nofollow">question</a> covers the (still unresolved) special case of when $E$ is defined by an atomic formula.</p> <p>A first-order definable subset $E \subseteq G^n$ is the set of $n$-tuples of group elements where a particular first-order formula in $n$ free variables is true. Such a formula is a finite string of symbols and may involve multiplication, inversion, the identity element, equality, logical connectives ($\wedge$, $\vee$, $\neg$, $\implies$), quatifiers ($\exists$ and $\forall$), and parentheses. For example, the following first-order formula defines the set of commutators $g \in G$:</p> <p>$$(\exists x)(\exists y)(g = xyx^{-1}y^{-1}).$$</p> <p>On the other hand, if you try to define the commutator subgroup by such a formula, you run into trouble. You might want to say, "$g$ is a commutator or $g$ is a product of two commutators or $g$ is a product of three commutators or ...," but infinite disjunctions are not allowed.</p> <p>Please note that the counterexample to a similar question in a comment by Vipul Naik <a href="http://mathoverflow.net/questions/39798/in-an-inductive-family-of-groups-does-the-probability-that-a-particular-word-is" rel="nofollow">here</a> is <em>not</em> a counterexample to this question. In the inductive family $$A_3 \hookrightarrow S_3 \hookrightarrow \ldots \hookrightarrow A_{2i-1} \hookrightarrow S_{2i-1} \hookrightarrow A_{2i+1} \hookrightarrow S_{2i+1} \hookrightarrow \ldots,$$ the groups alternate between having all the elements as commutators and half the elements as commutators. However, in the directed union, which is $A_\infty$, all the elements are commutators. The upshot is that quantifiers in a first-order formula may demand that you "look ahead" in your inductive family.</p> http://mathoverflow.net/questions/42046/efficiently-getting-bits-of-n/42081#42081 Answer by Gene S. Kopp for Efficiently getting bits of N! ? Gene S. Kopp 2010-10-13T23:10:05Z 2010-10-13T23:10:05Z <p>This isn't a complete answer, but I can do it polynomially in $\log N$ and $M$. In other words, the problem isn't hard if $M$ is small.</p> <p>Stirling's formula approximating $N!$ gives the first few terms of a <a href="http://en.wikipedia.org/wiki/Stirling%27s_approximation#A_convergent_version_of_Stirling.27s_formula" rel="nofollow">convergent series</a> for $\log(N!)$. Using this series, we may approximate $\log(N!)$ to any fixed degree of accuracy in polynomial time (in $\log N$). Indeed, if you work it out, you'll see that you can approximate $\log(N!)$ within $2^{-M}$ in $O(M^2 \log N \log\log N)$ time. (Recall $\log N \log\log N$ is the cost of multiplication with a fast Fourier transform; if you multiplied naively, it would be $(\log N)^2$.) Approximating $\log(N!)$ within $2^{-M}$ is exactly what it takes to get the first $M$ digits.</p> <p>Your question is actually equivalent to: Can you quickly pull out <em>any</em> digit of $N!$? There are only about $N \log N$ digits, so $M$ is bounded by a polynomial in $N$. By the above, we can get digits near the beginning. Digits near the end will also be easy: The problem then reduces to modular arithmetic and divisibility considerations. But I have no idea whether there's a fast way to pull out digits in the middle.</p> http://mathoverflow.net/questions/35140/interesting-applications-in-pure-mathematics-of-first-year-calculus/35427#35427 Answer by Gene S. Kopp for Interesting applications (in pure mathematics) of first-year calculus Gene S. Kopp 2010-08-13T02:33:18Z 2010-08-13T02:33:18Z <p>A cool example: The intermediate value theorem may be used to prove the following theorem about continued fractions:</p> <p>Let $\alpha>1$, and suppose that $$\left|\alpha-\frac{p}{q}\right|&lt;\frac{1}{2q^2}.$$ Then, $\dfrac{p}{q}$ is one of the convergents (truncated continued fractions) of $\alpha$.</p> http://mathoverflow.net/questions/126828/irreducible-degrees-and-the-order-of-a-finite-group/126866#126866 Comment by Gene S. Kopp Gene S. Kopp 2013-04-10T23:27:51Z 2013-04-10T23:27:51Z Another interesting point about the Bump and Ginzburg paper is that it provides a combinatorial interpretation for the $a_g$ assuming the existence of an involution on $G$ with certain properties. http://mathoverflow.net/questions/126828/irreducible-degrees-and-the-order-of-a-finite-group/126866#126866 Comment by Gene S. Kopp Gene S. Kopp 2013-04-10T23:24:51Z 2013-04-10T23:24:51Z Bump and Ginzburg remark on page 4 of their paper &quot;Generalized Frobenius-Schur Numbers&quot; that the Mathieu group $M_{11}$ provide another example where some of the $a_g$ are negative (and that the observation goes back to Solomon and Thompson). @David, your example is smaller (order 1920 versus 7920), so maybe it is not well-known. http://mathoverflow.net/questions/122765/exercise-in-milnes-cft-notes Comment by Gene S. Kopp Gene S. Kopp 2013-02-27T05:36:07Z 2013-02-27T05:36:07Z From a practical point of view, an easy way to check if you are right or wrong is to appeal to CM theory. In Mathematica, RootApproximant[KleinInvariantJ[Sqrt[-6]]] gives $1399+988\sqrt{2}$. The j-invariant generates the HCF, so Milne is right. http://mathoverflow.net/questions/102386/is-a-random-subset-of-the-real-numbers-non-measurable-is-the-set-of-measurable/102388#102388 Comment by Gene S. Kopp Gene S. Kopp 2012-07-16T23:47:54Z 2012-07-16T23:47:54Z Nice argument. Thanks for the answer! http://mathoverflow.net/questions/102386/is-a-random-subset-of-the-real-numbers-non-measurable-is-the-set-of-measurable/102390#102390 Comment by Gene S. Kopp Gene S. Kopp 2012-07-16T23:46:57Z 2012-07-16T23:46:57Z +1. Ah, of course. The product sigma-algebra contains only sets $\mathcal{S}$ of sets of real numbers in which membership $S \in \mathcal{S}$ is determined by membership $s \in S$ for countably many real numbers $s$. Thanks! http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration/52731#52731 Comment by Gene S. Kopp Gene S. Kopp 2012-05-25T00:07:19Z 2012-05-25T00:07:19Z @Allen &amp; Dan: My analysis professor saying, &quot;turn your head 90 degrees&quot; was quite helpful the first time I learned Lebesgue integration. In retrospect, I usually choose to think of the two the theories as &quot;approximation by step functions&quot; and &quot;approximation by simple functions,&quot; so both are adding up rows. http://mathoverflow.net/questions/52708/why-should-one-still-teach-riemann-integration Comment by Gene S. Kopp Gene S. Kopp 2012-05-24T23:57:05Z 2012-05-24T23:57:05Z @Harry: I used to think more or less the same thing, but while teaching calculus I found that: (1) Many students had the false belief that the Riemann integral was defined as an anti-derivative. (2) Many students seemed genuinely interested in understanding the geometry of Riemann integration even though they hadn't been interested in other non-computational topics. Even for mathematicians, the geometric intuition of integration as signed area is more important/fundamental than any of its many definitions. Thus, conveying that intuition should be the first goal of teaching integration. http://mathoverflow.net/questions/83027/what-is-ricardo-perez-marcos-ene-product-does-it-explain-his-statistical-resul Comment by Gene S. Kopp Gene S. Kopp 2011-12-23T20:50:37Z 2011-12-23T20:50:37Z @JSE: I would be very interested to see if Marco's observation has a function field analogue. http://mathoverflow.net/questions/83027/what-is-ricardo-perez-marcos-ene-product-does-it-explain-his-statistical-resul Comment by Gene S. Kopp Gene S. Kopp 2011-12-23T20:48:14Z 2011-12-23T20:48:14Z @JSE: I don't think this should be thought of as a fact about the distribution of γ1−γ2−γ3. Marco observes that the deltas of zeros of Dirichlet L-functions repel the zeros of ζ(s)...NOT their own zeros! So, for this phenomenon to be visible on the random matrix side would mean that the eigenvalues of a random Hermitian operator will predict the locations of zeta zeros. I doubt that; instead, Marco's observation deals with the finer properties of $L$-function zeros that are not shared with random matrices. http://mathoverflow.net/questions/83027/what-is-ricardo-perez-marcos-ene-product-does-it-explain-his-statistical-resul Comment by Gene S. Kopp Gene S. Kopp 2011-12-09T04:43:15Z 2011-12-09T04:43:15Z @Harry: Thanks, I fixed it. http://mathoverflow.net/questions/63300/is-every-poset-the-poset-of-prime-ideals-of-a-ring Comment by Gene S. Kopp Gene S. Kopp 2011-04-29T07:51:07Z 2011-04-29T07:51:07Z J&#243;zsef Pelik&#225;n told me that every <i>finite</i> poset can be realized as the spectrum of a ring (although I don't have a reference). A crucial point is that you can obtain closed and open subspaces by ring quotients and localizations, respectively. So the real challenge is building a ring with a big, complicated spectrum with the posets you want as subspaces. I enjoyed constructing examples by hand for some small posets; it's a nice exercise. http://mathoverflow.net/questions/52913/free-subgroups-vs-law Comment by Gene S. Kopp Gene S. Kopp 2011-01-24T06:17:25Z 2011-01-24T06:17:25Z Denis, I think you mean &quot;e.g., the direct sum of all finite groups,&quot; not the direct product. The direct product of all finite groups is a profinite group, is not torsion, and satisfies both (1) and (2). http://mathoverflow.net/questions/48718/is-every-compact-topological-ring-a-profinite-ring/48735#48735 Comment by Gene S. Kopp Gene S. Kopp 2010-12-09T10:27:04Z 2010-12-09T10:27:04Z Yes, this appears to be the argument in Stroppel's book as well (see comments). Thanks for the reference! http://mathoverflow.net/questions/48718/is-every-compact-topological-ring-a-profinite-ring Comment by Gene S. Kopp Gene S. Kopp 2010-12-09T09:55:40Z 2010-12-09T09:55:40Z Gjergji: Thanks for the reference! http://mathoverflow.net/questions/45951/sexy-vacuity/46076#46076 Comment by Gene S. Kopp Gene S. Kopp 2010-12-03T11:15:09Z 2010-12-03T11:15:09Z True :) I just think things should have names and definitions that reflect the way they're most often used.