User adam sheffer - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-22T00:24:03Zhttp://mathoverflow.net/feeds/user/17509http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/126507/two-questions-about-combinatorics-journalsTwo questions about combinatorics journalsAdam Sheffer2013-04-04T12:53:53Z2013-04-04T18:34:18Z
<p>Hello,</p>
<p>I have two questions regarding combinatorics journals. I hope that this is the right place for such questions.</p>
<ol>
<li><p>Which combinatorics/DM journals would you consider as the "top tier"?
I tried to look for an answer online, and found these two links:
<a href="http://www.scimagojr.com/journalrank.php?category=2607" rel="nofollow">http://www.scimagojr.com/journalrank.php?category=2607</a> and
<a href="http://mathoverflow.net/questions/3512/top-specialized-journals" rel="nofollow">http://mathoverflow.net/questions/3512/top-specialized-journals</a> .
These somewhat contradict each other (especially regarding EJC), and I assume that the SJR ranking might not be identical to the general public opinion.</p></li>
<li><p>What exactly is the difference between Journal of Combinatorial Theory series A and Journal of Combinatorial Theory series B? Wikipedia states that "Series A is concerned primarily with structures, designs, and applications of combinatorics. Series B is concerned primarily with graph and matroid theory.", but this seems a bit vague. For example, JCTA does contain many papers concerning graph theory. I also heard that the journal split due to a disagreement between its founders (or editors?). Can this disagreement shed some light on the difference?</p></li>
</ol>
<p>Many thanks,
Adam </p>
http://mathoverflow.net/questions/114171/irregular-hexagons-with-equal-anglesIrregular hexagons with equal angles Adam Sheffer2012-11-22T15:20:53Z2012-11-22T16:52:43Z
<p>Hello,</p>
<p>I have an irregular hexagon whose interior angles are all $2\pi/3$. That is, every pair of opposite edges are parallel. I have the distance between every pair of opposite edges, and I wish to find the perimeter of the hexagon. Intuitively, this seems possible, but I'm not sure how to do it.
Also, is there a name for such a hexagon?</p>
<p>If that helps, I can also assume that every pair of opposite edges have the same length.</p>
<p>Many thanks,
Adam</p>
http://mathoverflow.net/questions/109389/proofs-for-doubly-ruled-surfacesProofs for doubly ruled surfacesAdam Sheffer2012-10-11T16:08:52Z2012-10-13T15:51:41Z
<p>Hello,</p>
<p>I am interested in proofs for why the only irreducible doubly ruled surfaces in ${\mathbb R}^3$ are the one sheeted hyperboloid and the hyperbolic paraboloid. While many books and papers state that this is "well known", I could hardly find any sources that give more details. I only found the following two:</p>
<ol>
<li><p>In the book "Mathematical Omnibus: Thirty Lectures on Classic Mathematics" by Fuchs and Tabachnikov there is a proof relying on rather unusual tools. The proof heavily relies on the property that the neighborhood of every (non-singular) point behaves similarly to a plane.</p></li>
<li><p>Various places state that we can take three lines from one generating family, and these should intersect every line of the second family. I am not sure how to prove such a claim, and couldn't find a reference with more details (it does seem much simpler in the complex projective space, where one could rely on plucker coordinates).</p></li>
</ol>
<p>Could anyone provide references to proofs of this property? Or describe a proof different from the one I mentioned in item 1?</p>
<p>Many thanks!
Adam </p>
http://mathoverflow.net/questions/74165/is-a-point-that-is-incident-to-several-circles-not-on-the-same-sphere-necessarilyIs a point that is incident to several circles not on the same sphere necessarily singular?Adam Sheffer2011-08-31T14:47:26Z2011-08-31T14:47:26Z
<p>I have an irreducible polynomial $f \in R[x,y,z]$, and a point $p$ that is in the zero-set $Z$ of $f$. My question is, given the following properties of $p$, is it necessarily a singular point of $f$. There is a set of (one dimensional) circles, all passing through $p$ and fully contained in $Z$. Moreover, all of these circles, except for a single circle $c$, are fully contained in the same sphere $\sigma$. In case that it matters, I am only interested in the case where all of the circles that are on $\sigma$ also have a second point $b$ in common. Intuitively, it seems to me that $a$ has to be singular because of the additional circle $c$. How can I verify or contradict this? </p>
<p>Perhaps one way to answer this is by answering the follow-up question: If the zero set (of an irreducible $f$) contains several circles on the same sphere, all passing through two common points $a,b$, must the sphere be contained in the zero-set? I think that the answer is positive if the number of circles is sufficiently large, at least proportional to the degree of $f$, but I would like to know the answer when the number of circles is at least some (large) constant.</p>
<p>I apologize in advance in case this question is trivial. My knowledge in algebraic geometry is somewhat limited, and in fact, this problem arose while studying a combinatorial problem. Many thanks. </p>
http://mathoverflow.net/questions/119173/every-rigid-motion-in-the-plane-has-a-simple-description/119179#119179Comment by Adam ShefferAdam Sheffer2013-01-17T16:39:59Z2013-01-17T16:39:59ZThanks Benoît Kloeckner.
If the question is out of place I can remove it myself. I'm not sure how to spot when a question is indeed research-level. I require this question for research purposes (this claim is used in the recent solution to the Erdos distinct distances problem by Guth and Katz), and I don't recall studying such a topic in a course.
http://mathoverflow.net/questions/119173/every-rigid-motion-in-the-plane-has-a-simple-descriptionComment by Adam ShefferAdam Sheffer2013-01-17T16:27:11Z2013-01-17T16:27:11ZMany thanks for the references!
Deane, I'm afraid I don't see the problem. The claim is that a motion involving both rotations and translations can always be considered as a single rotation or translation.
http://mathoverflow.net/questions/119173/every-rigid-motion-in-the-plane-has-a-simple-descriptionComment by Adam ShefferAdam Sheffer2013-01-17T15:12:45Z2013-01-17T15:12:45ZMy mistake! I rephrased the question. Thanks.http://mathoverflow.net/questions/114171/irregular-hexagons-with-equal-anglesComment by Adam ShefferAdam Sheffer2012-11-22T16:51:18Z2012-11-22T16:51:18ZIndeed. Thanks for the correction.
An answer to either case would be helpful (either assuming that opposite edges have the same length or not).http://mathoverflow.net/questions/109389/proofs-for-doubly-ruled-surfacesComment by Adam ShefferAdam Sheffer2012-10-12T15:30:57Z2012-10-12T15:30:57ZThanks Robert. I am interested in any kind of proof for the claim.http://mathoverflow.net/questions/109389/proofs-for-doubly-ruled-surfaces/109408#109408Comment by Adam ShefferAdam Sheffer2012-10-12T03:04:36Z2012-10-12T03:04:36ZMany thanks Sue! I looked throughout the book and could not find a proof for this claim. Could you please specify where it is exactly? If I understand correctly, Guth and Katz refer to this book for issues regarding flecnode polynomials, and not for the above claim. In the introduction of their paper, they do state this claim, but provide the same brief explanation that I stated in my second itemhttp://mathoverflow.net/questions/74165/is-a-point-that-is-incident-to-several-circles-not-on-the-same-sphere-necessarilyComment by Adam ShefferAdam Sheffer2011-09-04T16:56:07Z2011-09-04T16:56:07ZThanks. Indeed, this case is clear to me, but I wish to know if the point is always singular under the above conditions. That is, including when the additional circle has the same tangent plane as the sphere.
Also, it seems that I have a counterexample for the first part of the follow-up question. That is, any constant number of circles can be fully contained in $f$ while the sphere that contains them isn't in $f$. But I don't have an example for the case where the additional circle is also in $f$.