User hewhohungers - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T05:22:37Z http://mathoverflow.net/feeds/user/16504 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/122920/does-the-ech-cohomology-always-yield-long-exact-sequences-from-short-ones Does the Čech cohomology always yield long exact sequences from short ones? HeWhoHungers 2013-02-25T19:47:29Z 2013-02-26T01:45:36Z <p>Does the Čech cohomology always give rise to a long exact sequences given a short exact sequence of sheaves?</p> <p>Clearly that cannot occur for sheaves on a paracomact (perhaps also Hausdorff, I'm not sure about that) space, but the argument one uses there relies on the topological assumptions in a crucial way and cannot be generalized to an arbitrary space, not in a straight-forward manner anyway.</p> <p>Thank you for shedding any light on this issue!</p> http://mathoverflow.net/questions/101888/hsiung-on-the-complex-structure-of-s6 Hsiung on the Complex Structure of $S^6$ HeWhoHungers 2012-07-10T20:30:30Z 2012-07-12T00:51:19Z <p>In 1986 C. C. Hsiung published a paper "Nonexistence of a Complex Structure on the Six-Sphere" and in 1995 he even wrote a monograph "Almost Complex and Complex Structures" to further elaborate on his proof. Yet answers to the 2009 <a href="http://mathoverflow.net/questions/1973/is-there-a-complex-structure-on-the-6-sphere" rel="nofollow">question on this site</a> all agree that the existence of complex structures on $S^6$ is still an open problem. Some recent preprints answering the question with opposite answers are also cited there. I would like to know if there are any known mistakes in Hsiung's approach and if so I would appreciate some reference to a paper that points them out.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99764#99764 Answer by HeWhoHungers for Blackbox Theorems HeWhoHungers 2012-06-16T01:40:19Z 2012-06-16T01:40:19Z <p>In several complex variables it is often desired to be able to <strong>solve the $\bar\partial$ equation</strong>. The standard tool for that is <strong>Hörmander's $L^2$ method</strong> and though I suppose that most who use it have at some point read at least a sketch of the proof, most probably aren't familiar with the tedious details that go into the proof. </p> http://mathoverflow.net/questions/99756/sum-of-univalent-functions Sum of univalent functions HeWhoHungers 2012-06-15T23:05:34Z 2012-06-15T23:05:34Z <p>Let $f,g$ be a pair of univalent functions on a proper subdomain $\Omega$ of $\mathbb C$. Does their sum $f+g$ necessarily omit any complex value? Similarly, can all holomorphic function be written as sums of univalent functions?</p> http://mathoverflow.net/questions/82750/counting-rectilinear-polygons Counting rectilinear polygons HeWhoHungers 2011-12-05T23:14:24Z 2011-12-06T02:37:32Z <p>This is likely a very easy counting question inspired by some elementary geometry:</p> <p>Consider a simple rectilinear polygon embedded in a plane in such a way that each of its edges is parallel to one of the coordinate axis. Two such polygons are considered <em>distinct</em> if they are not related by some composition of translation, scalar multiplication and squeeze mapping.</p> <p>I would like to asses the number of such distinct simple rectilinear polygons which have $2n$ horizontal (equivalently vertical) edges for any chosen $n\in\mathbb N$.</p> <p>Thank you.</p> http://mathoverflow.net/questions/70515/n-times-iterated-cauchy-riemann-operator n-times iterated Cauchy-Riemann operator HeWhoHungers 2011-07-16T18:48:06Z 2011-07-16T19:05:02Z <p>Are there an results on functions annihilated by the n-times iterated Cauchy-Riemann operator ${\partial\over\partial\bar z}$, aka functions $f$ that for some $n\in\mathbb{N}$ satisfy the following equation? $${\partial^n f\over\partial\bar z^n}=0$$ EDIT: I have posted the same questions only a very short period of time ago, when my web browser suddenly froze and I didn't believe the question got posted successively, therefor posting it again here. If such repetition goes against any site rules or regulations, as I would suspect it does, may somebody in power to do so please remove one of the posts. Thank you!</p> http://mathoverflow.net/questions/122920/does-the-ech-cohomology-always-yield-long-exact-sequences-from-short-ones Comment by HeWhoHungers HeWhoHungers 2013-02-25T20:32:02Z 2013-02-25T20:32:02Z Thank you both for your hasty corrections - indeed it was Čech cohomology that I had in mind and it was little but stupidity that led me to write otherwise. http://mathoverflow.net/questions/101888/hsiung-on-the-complex-structure-of-s6 Comment by HeWhoHungers HeWhoHungers 2012-07-11T00:15:00Z 2012-07-11T00:15:00Z @Will Jagy: You really shouldn't think too deeply on the pseudonym I use. It is actually a slight alteration of the biblical quote &quot;blessed are those who hunger and thirst for righteousness, for they shall be satisfied&quot; which I had heard shortly before first posting a question on this site and randomly happened to think of when I had to type a name in. The reference you have found appears to be an amusing coincidence, though it's quite possible that it originates from the said citation also. http://mathoverflow.net/questions/99506/blackbox-theorems/99515#99515 Comment by HeWhoHungers HeWhoHungers 2012-06-19T23:05:31Z 2012-06-19T23:05:31Z Is Brouwer's fixed point theorem not thought in nearly every introductory algebraic geometry course as a rather simple application of homology? http://mathoverflow.net/questions/61882/meromorphic-1-form-and-picards-theorem Comment by HeWhoHungers HeWhoHungers 2012-06-16T01:43:41Z 2012-06-16T01:43:41Z Any particular reason this doesn't have the open problem tag? http://mathoverflow.net/questions/99506/blackbox-theorems/99544#99544 Comment by HeWhoHungers HeWhoHungers 2012-06-16T01:24:41Z 2012-06-16T01:24:41Z The standard proof of the Uniformization theorem with the Green's function, while rather involved, shouldn't really surpass the ability of most who come across it. There also exists a short and elegant proof that uses certain rather more advanced tools: the Mayer-Vietoris sequence and the celebrated Newlander-Nirenberg theorem. But NN for surfaces is just the existence of isothermal coordinates, which is much simpler to prove. This proof can be found in Demailly's &quot;Complex Analytic and Differential Geometry&quot; (available at <a href="http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf" rel="nofollow">www-fourier.ujf-grenoble.fr/~demailly/manuscripts/&hellip;</a>). http://mathoverflow.net/questions/82750/counting-rectilinear-polygons/82763#82763 Comment by HeWhoHungers HeWhoHungers 2011-12-06T09:11:36Z 2011-12-06T09:11:36Z In my innitial question I indeed only considered boundaries composed of 2n vertical and 2n horizontal line segments. But that restriction steams only from the geometric background in which the question originaly arised and there is no particular reason why L-shapes and simmilar objects couldn't be considered as well. Anyway thank you for the answer! http://mathoverflow.net/questions/82750/counting-rectilinear-polygons Comment by HeWhoHungers HeWhoHungers 2011-12-06T09:01:32Z 2011-12-06T09:01:32Z Joseph O'Rourke and Gerry Myerson correctly realized the intended meaning of the perhaps too ambiguous term &quot;squeeze mapping&quot;. As Gerry noted, it is true that under the defined identification all rectangles (that is $n=1$ simple rectilear polygons) are equivalent. I hope this clears up some interpretational issues. http://mathoverflow.net/questions/70515/n-times-iterated-cauchy-riemann-operator/70516#70516 Comment by HeWhoHungers HeWhoHungers 2011-07-16T19:19:46Z 2011-07-16T19:19:46Z I suppose any further properties can be deduced from the one you gave without greater problems so yes, that's just about what I was looking for. Thanks!