User gordon craig - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-21T21:01:22Zhttp://mathoverflow.net/feeds/user/2497http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/75462/applications-of-rademachers-theoremApplications of Rademacher's TheoremGordon Craig2011-09-15T00:10:51Z2013-03-21T17:47:59Z
<p>Rademacher's Theorem(that every Lipschitz function on $\mathbb{R}^{n}$ is almost everywhere differentiable) is a remarkable result on the structure of the space of Lipschitz functions, but I was wondering whether it has any interesting applications. All of the "useful" results(or maybe "applicable") that I know of about weak versions of differentiability involve estimates(eg Sobolev embedding, Lebesgue differentiation theorem.)</p>
http://mathoverflow.net/questions/110468/examples-of-separable-ordinary-differential-equations-in-economicsExamples of separable ordinary differential equations in economicsGordon Craig2012-10-23T20:34:09Z2012-10-25T02:13:47Z
<p>I'm currently teaching an integral calculus course for business students, and we're just about to discuss differential equations. They've worked hard, and I'd like to reward them with some economic applications of ODEs, but they can only handle simple separable equations. </p>
<p>I'm going to frame exponential growth in terms of economic growth(among other things,) and then I'm currently planning on looking at which demand functions have constant elasticity and looking at the logistic model of a population. I might be asking for too much, but I was wondering whether anyone could suggest a separable equation that arises from a simple model(they've all taken an introduction to economics, but no more.)</p>
http://mathoverflow.net/questions/60685/reference-for-mathematical-economicsReference for Mathematical EconomicsGordon Craig2011-04-05T13:31:05Z2011-04-06T06:33:59Z
<p>I'm looking for a good introduction to basic economics from a mathematically solid(or, even better, rigorous) perspective. I know just about nothing about economics, but I've picked up bits and pieces in the course of teaching Calculus for business and social sciences, and I'd like to know more, both for my personal culture and to incorporate into my courses. My Platonic ideal of such a book would be along the lines of T W Körner's Naive Decision Making, but I'll take what I can get.</p>
http://mathoverflow.net/questions/53302/is-there-a-chern-gauss-bonnet-theorem-for-orbifoldsIs there a Chern-Gauss-Bonnet theorem for orbifolds?Gordon Craig2011-01-26T00:39:08Z2011-02-03T17:08:14Z
<p>There's a Gauss-Bonnet theorem for compact 2-orbifolds(due to Satake, I think), which gives a relation between the curvature of a Riemannian orbifold and the orbifold topology(i.e. taking into account not just the structure of O as a Hausdorff topological space, but also the structure of the singular points.) The only proof that I've seen was very much like the classical one of the Gauss-Bonnet theorem, with geodesic triangulations, angle defects and so on. This approach doesn't generalise to higher dimensions to prove the Chern-Gauss-Bonnet theorem, and I haven't even found a conjectured Chern-Gauss-Bonnet formula. I'm especially interested in the four-dimensional case, where the integrand is amenable. One reason I'm interested is just to get a feel for how much more complicated orbifolds are than manifolds; on the one hand, many basic definitions seem to go through from the manifold world to the orbifold world, but the generalisation leads to significant complications in practice. On the other hand, Kleiner and Lott posted a paper on the arxiv in which they use Ricci flow to geometrise 3-orbifolds, so orbifolds certainly seem like a good arena in which to generalise differential geometry. Chern-Gauss-Bonnet is a bit of a benchmark for higher-dimensional Riemannian geometry, and I'd like to know the state of the art is in this case. </p>
http://mathoverflow.net/questions/36824/applications-of-measure-integration-and-banach-spaces-to-combinatoricsApplications of Measure, Integration and Banach Spaces to CombinatoricsGordon Craig2010-08-27T00:30:12Z2010-08-28T22:28:52Z
<p>I'm going to be teaching a Master's level analysis course(measure theory, Lebesgue integration, Banach and Hilbert spaces, and if there's time, some spectral or PDE stuff) in the fall. My problem is that while roughly half of the students will actually be using analysis in their further work, the rest of them are going to specialize in combinatorics, and while I want to convince them that they should know this stuff as part of their general mathematical culture, I'd also like to try to connect it to what they'll be working on. So far, I've found survey articles on applications of Ramsey theory to Banach spaces, and applications of harmonic analysis to additive number theory, but I was wondering whether anyone had some suggestions of references for applications of classical analysis to old-fashioned, classical combinatorics. (I realise that this is a pretty tall order, as on many levels, these two fields are at antipodes.) </p>
<p>PS: I'm planning on talking about probability measures on discrete spaces, but I don't think that will convince the combinatorics people that hacking through the construction of the Lebesgue integral could have a practical payoff someday for them.</p>
http://mathoverflow.net/questions/13538/hausdorff-measure-questionHausdorff measure questionGordon Craig2010-01-31T04:14:24Z2010-04-24T21:06:45Z
<p>Say we have some compact metrisable topological space $X$ with a measure $\mu$ defined on the Borel sets of $X$. Then is there some way to determine whether $\mu$ is the Hausdorff measure associated to some metric $d$ compatible with the topology of $X$? And if so, is there some process to recover a metric from the measure? I'd imagine that there would have to be some conditions placed on the space $X$, eg. that it's connected, and it might even be necessary to assume that it's some nice space such as a manifold, with a "gauge" metric $d_{0}$ relative to whose Hausdorff measure $\mu$ is absolutely continuous, but I'd like to ask the question in the greatest generality possible, in the hope that there is an answer out there.</p>
http://mathoverflow.net/questions/18844/does-every-smooth-manifold-of-infinite-topological-type-admit-a-complete-riemanniDoes every smooth manifold of infinite topological type admit a complete Riemannian metric?Gordon Craig2010-03-20T15:19:59Z2010-03-20T16:03:09Z
<p>To elaborate a bit, I should say that the question of the existence of a complete metric is only of interest in the case of manifolds of infinite topological type; if a manifold is compact, any metric is complete, and if a noncompact manifold has finite topological type(ie is diffeomorphic to the interior of a compact manifold with boundary,) one can contruct a complete metric on the manifold with boundary via a partition of unity, and then divide by the square of a defining function to get an complete asymptotically metric on the interior.</p>
<p>I have absolutely no intuition for how "wild" these manifolds can be. The only examples I can think of are infinite connected sums and quotients of negatively curved symmetric spaces by sufficiently complicated groups, but I'd imagine that one can construct some pathological examples by limiting arguments.</p>
http://mathoverflow.net/questions/9401/number-theory-and-geometry-several-complex-variablesNumber Theory and Geometry/Several Complex VariablesGordon Craig2009-12-20T03:28:49Z2010-01-15T00:24:34Z
<p>This is a question for all you number theorists out there...based on my skimming of number theory textbooks and survey articles, it seems like most of the applications of geometry and complex variables to number theory are restricted to surfaces and the theory of a single complex variable. My questions are</p>
<p>1) Is this impression indeed accurate?</p>
<p>and, if so</p>
<p>2) Why is this? Is it because the theories of surfaces and of a single complex variable are "easier"(in the sense that they're simple enough to have a single unified theory,) or is there some deeper reason? And if the former is the case, are there some deep conjectures in number theory that could be solved using higher-dimensional geometry/complex variables?</p>
http://mathoverflow.net/questions/9990/cone-angles-for-riemannian-metrics-in-polar-coordinatesCone angles for Riemannian metrics in polar coordinatesGordon Craig2009-12-28T23:49:34Z2009-12-29T11:11:27Z
<p>This is the simplest case of a question that's been bugging me for a while: say we have a Riemannian metric in polar coordinates on a (2-d) surface:
g=dr<sup>2</sup>+f<sup>2</sup>(r, θ)dθ<sup>2</sup>, such that the θ parameter runs from 0 to 2π. Assume that f is a smooth function on (0,∞)X S<sup>1</sup> such that f(0, θ)=0. </p>
<p>Define the cone angle at the pole to be
$ C=\lim_{r\rightarrow 0^{+}} \frac{L(\partial B(r))}{r} $, where B(r) is the geodesic disc of radius r centered at the origin. Then it's fairly easy to see(by switching into Cartesian coordinates) that a necessary condition for the metric to be smooth is that C=2π. If C<2π, there is a cone point at the origin. One can write out a cone metric, and show that the triangle inequality holds, so there is a singular metric, but which still induces a metric space structure.</p>
<p>Now, if C>2π, it seems pretty clear that we'll end up with a space which violates the triangle inequality; it will be shorter to take a broken segment through the origin than to follow the shortest geodesic(in the sense of a curve γ(t) such that D<sub>γ'</sub>γ'=0.) One can show this directly for some simple cases, eg a flat metric with a cone angle greater than 2π. </p>
<p>But there must be an elementary proof of the general case! I can't seem to find one though, and I spent the afternoon playing around with the Topogonov and Rauch comparison estimates to no avail. The basic problem I'm having is that the cone angle condition is essentially a condition on metric balls, but we expect a violation of the triangle inequality, which is a condition on distances. </p>
<p>This is not really related to anything I'm working on, but it's driving me crazy, so I'd appreciate any insight.</p>
http://mathoverflow.net/questions/8513/characterization-of-riemannian-metricsCharacterization of Riemannian metricsGordon Craig2009-12-11T01:45:17Z2009-12-11T06:10:14Z
<p>This is probably an insanely hard question, but given an abstract metric space, is there some way to determine whether it's a manifold with a Riemannian, or more generally a Finslerian, metric? If that's too hard, one could start off by assuming that the underlying space is a manifold. The example that got me thinking about this was the induced metric on the 2-sphere embedded in $R^3$...the underlying space is obviously a smooth manifold, and the metric should be smooth(the geodesics would even be great circles, as they are for the standard metric on $S^2$,) but I don't see how you could prove the triangle inequality pointwise, as you'd have to to show that it's Finslerian, let alone Riemannian. This example is already incredibly simple, since it's a homogeneous space with a the metric induced by being a subspace of another homogeneous space.</p>
http://mathoverflow.net/questions/97775/n-elements-t-are-black-choose-k-probability-of-choosing-m-tsComment by Gordon CraigGordon Craig2012-05-23T17:00:01Z2012-05-23T17:00:01ZThis seems like a homework problem.http://mathoverflow.net/questions/75462/applications-of-rademachers-theorem/75478#75478Comment by Gordon CraigGordon Craig2011-11-16T18:37:13Z2011-11-16T18:37:13ZThanks. It should have occured to me that it would be natural to be able to derive nonexistence results from Rademacher.http://mathoverflow.net/questions/60685/reference-for-mathematical-economics/60686#60686Comment by Gordon CraigGordon Craig2011-05-18T20:00:56Z2011-05-18T20:00:56ZThanks very muchhttp://mathoverflow.net/questions/60685/reference-for-mathematical-economics/60690#60690Comment by Gordon CraigGordon Craig2011-05-18T20:00:34Z2011-05-18T20:00:34Z@Steven: Thanks for the recommendation. I'll look at the textbook on Price Theory, but I read the first few chapters of the Armchair Economist, and in all honesty, I was uncomfortable with some of the implicit assumptions. I'd be happier with something where the axioms are laid out explicitly, which is one of the reason that I'm interested in a mathematical treatment.http://mathoverflow.net/questions/60685/reference-for-mathematical-economics/60709#60709Comment by Gordon CraigGordon Craig2011-05-18T19:52:00Z2011-05-18T19:52:00ZThanks very much!http://mathoverflow.net/questions/60685/reference-for-mathematical-economics/60702#60702Comment by Gordon CraigGordon Craig2011-05-18T19:51:37Z2011-05-18T19:51:37Z@Gil: Thanks very much. And I second Thierry's comments.http://mathoverflow.net/questions/60685/reference-for-mathematical-economics/60738#60738Comment by Gordon CraigGordon Craig2011-05-18T19:50:50Z2011-05-18T19:50:50ZThanks! I'll investigate both, although I'm more interested in the economic theory than the Finance aspect.http://mathoverflow.net/questions/60685/reference-for-mathematical-economics/60772#60772Comment by Gordon CraigGordon Craig2011-05-18T19:49:45Z2011-05-18T19:49:45ZThank very much. I'll check those out.http://mathoverflow.net/questions/60685/reference-for-mathematical-economicsComment by Gordon CraigGordon Craig2011-05-18T19:49:15Z2011-05-18T19:49:15Z@James O: I heard about that book a long time ago, and I was meaning to read it. Thanks for reminding me of it!http://mathoverflow.net/questions/60685/reference-for-mathematical-economicsComment by Gordon CraigGordon Craig2011-05-18T19:48:12Z2011-05-18T19:48:12Z@Gil: I'll have to get back to you in a few years rather than a few months, since this is more of a weekend and holiday project than everything else. (Somewhat) in defense of the content of Zen's post, I don't find his comments offensive(although they sounds a bit far-fetched, and may well violate MO protocol) and his comment about Nobel's descendants is accurate(I heard one say it on the radio,) although the relevance what someone's great-grandson has to say is questionable.http://mathoverflow.net/questions/53302/is-there-a-chern-gauss-bonnet-theorem-for-orbifolds/54145#54145Comment by Gordon CraigGordon Craig2011-02-18T19:04:59Z2011-02-18T19:04:59ZThanks very much!http://mathoverflow.net/questions/53302/is-there-a-chern-gauss-bonnet-theorem-for-orbifolds/53351#53351Comment by Gordon CraigGordon Craig2011-02-18T18:48:46Z2011-02-18T18:48:46ZThanks very much. The this gives me a second, more abstract way of looking at the problem.http://mathoverflow.net/questions/53302/is-there-a-chern-gauss-bonnet-theorem-for-orbifoldsComment by Gordon CraigGordon Craig2011-01-26T20:59:19Z2011-01-26T20:59:19Z@Theo: Thanks! I'll get in touch with him.http://mathoverflow.net/questions/53302/is-there-a-chern-gauss-bonnet-theorem-for-orbifoldsComment by Gordon CraigGordon Craig2011-01-26T01:50:53Z2011-01-26T01:50:53Z@Ryan: That's a good question, and the observation is definitely correct for good orbifolds. But for bad orbifolds(ones which are not global quotients of a Riemannian manifold of the same dimension,) I'm unsure how you'd come up with the right correction term to take into account the isotropy of the singular strata. http://mathoverflow.net/questions/48045/why-are-matrices-ubiquitous-but-hypermatrices-rareComment by Gordon CraigGordon Craig2010-12-02T13:25:47Z2010-12-02T13:25:47ZJust a guess, but it may have to do with the difficulty of defining a (canonical) product of hypermatrices; you can't view them naturally as linear maps between vector spaces, and define a product via composition.