User va - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T15:58:51Z http://mathoverflow.net/feeds/user/1784 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/10667/euler-maclaurin-formula-and-riemann-roch Euler-Maclaurin formula and Riemann-Roch VA 2010-01-04T05:10:05Z 2012-12-14T02:25:23Z <p>Let $Df$ denote the derivative of a function $f(x)$ and $\bigtriangledown f=f(x)-f(x-1)$ be the discrete derivative. Using the Taylor series expansion for $f(x-1)$, we easily get $\bigtriangledown = 1- e^{-D}$ or, by taking the inverses, $$\frac{1}{\bigtriangledown} = \frac{1}{1-e^{-D}} = \frac{1}{D}\cdot \frac{D}{1-e^{-D}}= \frac{1}{D} + \frac12+ \sum_{k=1}^{\infty} B_{2k}\frac{D^{2k-1}}{(2k)!} ,$$ where $B_{2k}$ are Bernoulli numbers.</p> <p>(<b>Edit:</b> I corrected the signs to adhere to the most common conventions.)</p> <p>Here, $(1/D)g$ is the opposite to the derivative, i.e. the integral; adding the limits this becomes a definite integral $\int_0^n g(x)dx$. And $(1/\bigtriangledown)g$ is the opposite to the discrete derivative, i.e. the sum $\sum_{x=1}^n g(x)$. So the above formula, known as Euler-Maclaurin formula, allows one, sometimes, to compute the discrete sum by using the definite integral and some error terms. </p> <p>Usually, there is a nontrivial remainder in this formula. For example, for $g(x)=1/x$, the remainder is Euler's constant $\gamma\simeq 0.57$. Estimating the remainder and analyzing the convergence of the power series is a long story, which is explained for example in the nice book "Concrete Mathematics" by Graham-Knuth-Patashnik. But the power series becomes finite with zero remainder if $g(x)$ is a polynomial. OK, so far I am just reminding elementary combinatorics.</p> <p>Now, for my question. In the (Hirzebruch/Grothendieck)-Riemann-Roch formula one of the main ingredients is the Todd class which is defined as the product, going over Chern roots $\alpha$, of the expression $\frac{\alpha}{1-e^{-\alpha}}$. This looks so similar to the above, and so suggestive (especially because in the Hirzebruch's version $$\chi(X,F) = h^0(F)-h^1(F)+\dots = \int_X ch(F) Td(T_X)$$ there is also an "integral", at least in the notation) that it makes me wonder: is there a connection?</p> <p>The obvious case to try (which I did) is the case when $X=\mathbb P^n$ and $F=\mathcal O(d)$. But the usual proof in that case is a residue computation which, to my eye, does not look anything like Euler-Maclaurin formula. </p> <p>But is there really a connection?</p> <p><hr></p> <p><b>An edit after many answers:</b> Although the connection with Khovanskii-Pukhlikov's paper and the consequent work, pointed out by Dmitri and others, is undeniable, it is still not obvious to me how the usual Riemann-Roch for $X=\mathbb P^n$ and $F=\mathcal O(d)$ follows from them. It appears that one has to prove the following nontrivial </p> <p><b>Identity:</b> The coefficient of $x^n$ in $Td(x)^{n+1}e^{dx}$ equals $$\frac{1}{n!} Td(\partial /\partial h_0) \dots Td(\partial /\partial h_n) (d+h_0+\dots + h_n)^n |_{h_0=\dots h_n=0}$$</p> <p>A complete answer to my question would include a proof of this identity or a reference to where this is shown. (I did not find it in the cited papers.) I removed the acceptance to encourage a more complete explanation.</p> http://mathoverflow.net/questions/96620/flatness-for-family-of-hypersurfaces/96837#96837 Answer by VA for Flatness for family of hypersurfaces VA 2012-05-13T16:05:56Z 2012-05-14T03:23:44Z <p>This is an example when proving "locally free" instead of merely "flat" is easier and more straightforward, and no Noetherian assumption on the base is needed. The point is that if some coefficient $a$ of a polynomial $f\in R[x_1,\dotsc x_n]$ is nonzero at $p\in Spec R$ (i.e. nonzero in $R/p$) then it is <em>invertible</em> in an open neighborhood $D(a)\ni p$. </p> <p>So let $f\in R[x_1,\dotsc,x_n]$ be a polynomial of degree $d$, $p$ be a point of $Spec R$ (i.e. a prime ideal in $R$) and $k$ be the quotient field of $R/p$. Let $\bar f \in k[x_1,\dotsc,x_n]$ be the reduction of $f$ modulo $p$.</p> <p>Using a change of coordinates in $k[x_1,\dotsc,x_n]$, put $\bar f$ in a Weierstrass form w.r.t. to the variable $x_n$. This means that </p> <p>$$\bar f= \bar a x_n^d + p_{d-1}x_n^{d-1} + \dots + p_0$$</p> <p>for some polynomials $p_j$ in the remaining variables $x_1,\dotsc,x_{n-1}$, and $\bar a\in k$, $\bar a\ne 0$.</p> <p>If $k$ is infinite, this can be done by a linear change of coordinates. If $k$ is finite, there is a little trick.</p> <p>If $r_i/s_i\in k$ are the coefficients involved in the change of coordinates ($r_i,s_i\in R$) then this change of coordinates can be done already in the ring $R'=R[1/a \prod s_i]$, i.e. over the open set $Spec R'= D(a\prod s_i)$ in $Spec R$ containing $[p]$. Further, $a$ is invertible over this set.</p> <p>Now, over $R'$ the quotient $R'[x_1,\dotsc,x_n]/(f)$ is a free $R'[x_1,\dotsc,x_{n-1}]$-module with a basis $1,x_n,\dotsc, x_n^{d-1}$. Hence, it is a free $R'$-module. QED</p> <p>This proves the statement for a family of nonzero hypersurfaces in $\mathbb A^n$. For a family of nonzero hypersurfaces in $\mathbb P^n$, cover $\mathbb P^n$ by $\mathbb A^n$ appropriately.</p> http://mathoverflow.net/questions/93610/detecting-tilings-by-toric-geometry/93625#93625 Answer by VA for Detecting tilings by toric geometry VA 2012-04-10T04:53:26Z 2012-04-10T04:53:26Z <p>A related question (but not exactly the one you asked) is: </p> <blockquote> <p>Can one tell if a convex polytope $P$ and its translations by $\mathbb Z^n$ tile $\mathbb R^n$? Which polytopes $P$ have this property?</p> </blockquote> <p>Fix some positive quadratic form $q$ on $\mathbb R^n$ and the corresponding distance function. Let $P^0$ be the set of points in $\mathbb R^n$ which are closer to $0$ than to any other integral (i.e. in $\mathbb Z^n$) point. The closure $P$ of $P^0$ is called the Voronoi polytope w.r.t. $q$. Then $P$ obviously has the above property.</p> <p>Voronoi's conjectured circa 1907 that the opposite is true, i.e. any such $P$ is a Voronoi polytope w.r.t. some $q$. </p> <p>This conjecture is known for $n\le 4$ due to Delaunay and for zonotopes by Erdahl "Zonotopes, Dicings, and Voronoi Conjecture on Parallelohedra". It is still open in general, I believe.</p> <p>So what is special about the toric variety $X_P$ corresponding to $P$? I am not sure. If you look at the Delaunay tiling which is <em>dual</em> to the Voronoi tiling $P+\mathbb Z^n$, then the polytopes in that tiling and the corresponding toric varieties have a clear geometric meaning: they describe degenerations of principally polarized abelian varieties. But this is a dual picture.</p> <p>Note by the way that Delaunay polytopes have vertices in $\mathbb Z^n$, so they indeed correspond to projective polarized toric varieties. In contrast, the Voronoi polytope for a generic $q$ will have irrational vertices. Also, when you vary $q$ continuously, the Voronoi polytope will vary continuously. But the Delaunay polytopes will jump discretely, and there are only finitely many Delaunay polytopes modulo $GL(n,\mathbb Z)$.</p> <p>One place where the Voronoi tilings appear is tropical geometry. Indeed, a principally polarized tropical abelian variety $A$ is just the real torus $\mathbb R^n / \mathbb Z^n$ together with the positive definite form $q$. Then the $(n-1)$-skeleton of the Voronoi tiling modulo $\mathbb Z^n$ is the theta divisor on $A$. See Mikhalkin-Zharkov <a href="http://arxiv.org/abs/math/0612267" rel="nofollow">http://arxiv.org/abs/math/0612267</a> for more details. </p> http://mathoverflow.net/questions/20696/a-question-regarding-a-claim-of-v-i-arnold/20720#20720 Answer by VA for A question regarding a claim of V. I. Arnold VA 2010-04-08T12:39:40Z 2012-04-09T19:40:03Z <p>Here is a problem which I heard Arnold give in an ODE lecture when I was an undergrad. Arnold indeed talked about Barrow, Newton and Hooke that day, and about how modern mathematicians can not calculate quickly but for Barrow this would be a one-minute exercise. He then dared anybody in the audience to do it in 10 minutes and offered immediate monetary reward, which was not collected. I admit that it took me more than 10 minutes to do this by computing Taylor series.</p> <p>This is consistent with what Angelo is describing. But for all I know, this could have been a lucky guess on Faltings' part, even though he is well known to be very quick and razor sharp. </p> <p>The problem was to find the limit</p> <p>$$\lim_{x\to 0} \frac { \sin(\tan x) - \tan(\sin x) } { \arcsin(\arctan x) - \arctan(\arcsin x) }$$</p> <p>The answer is the same for $$\lim_{x\to 0} \frac { f(x) - g(x) } { g^{-1}(x) - f^{-1}(x) }$$ for any two analytic around 0 functions $f,g$ with $f(0)=g(0)=0$ and $f'(0)=g'(0)=1$, which you can easily prove by looking at the power expansions of $f$ and $f^{-1}$ or, in the case of Barrow, by looking at the graph.</p> <p>End of Apr 8 2010 edit <hr> Beg of Apr 9 2012 edit</p> <p>Here is a computation for the inverse functions. Suppose $$f(x) = x + a_2 x^2 + a_3 x^3 + \dots \quad \text{and} \quad f^{-1}(x) = x + A_2 x^2 + A_3 x^3 + \dots$$</p> <p>Computing recursively, one sees that for $n\ge2$ one has $$A_n = -a_n + P_n(a_2, \dotsc, a_{n-1} )$$ for some universal polynomial $P_n$. </p> <p>Now, let $$g(x) = x + b_2 x^2 + b_3 x^3 + \dots \quad \text{and} \quad g^{-1}(x) = x + B_2 x^2 + B_3 x^3 + \dots$$</p> <p>and suppose that $b_i=a_i$ for $i\le n-1$ but $b_n\ne a_n$. Then by induction one has $B_i=A_i$ for $i\le n-1$, $A_n=-a_n+ P_n(a_2,\dotsc,a_{n-1})$ and $B_n=-b_n+ P_n(a_2,\dotsc,a_{n-1})$.</p> <p>Thus, the power expansion for $f(x)-g(x)$ starts with $(a_n-b_n)x^n$, and the power expansion for $g^{-1}(x)-f^{-1}(x)$ starts with $(B_n-A_n)x^n = (a_n-b_n)x^n$. So the limit is 1.</p> http://mathoverflow.net/questions/22141/how-do-i-see-latex-math-on-any-web-page-and-in-email How do I see LaTeX math on any web page and in email? VA 2010-04-22T02:19:12Z 2010-11-24T18:27:34Z <p>This is a follow up to <a href="http://mathoverflow.net/questions/19679/seeing-math-when-viewing-abstracts-on-arxiv-org-closed" rel="nofollow">this closed question</a>.</p> <p>I open a random page, such as something on arXiv at 8:05 p.m. EST, and I see all these dollar signs, and I sigh and I wish that I could see nicely formatted math formulas instead, just like on MO. Is it possible? Can one write a Greasemonkey script to apply jsMath after the fact even if the page authors did not think of it? A Mozilla Firefox addon?</p> <p>Please share your solutions. Seeing like this is an active community of people with similar interests, I am sure that hundreds or thousands of mathematicians would benefit from a solution.</p> http://mathoverflow.net/questions/22141/how-do-i-see-latex-math-on-any-web-page-and-in-email/27524#27524 Answer by VA for How do I see LaTeX math on any web page and in email? VA 2010-06-08T21:19:54Z 2010-11-24T18:27:34Z <p>I wrote a little program called GmailTeX which adds $\TeX$ capability to Gmail. You can get it <a href="http://alexeev.org/gmailtex.html" rel="nofollow">here</a>. </p> <p><img src="http://alexeev.org/mathoverflow.png" alt="alt text"></p> <p>GmailTeX now includes the "live" fully automatic mode, just like when you type a question / answer on MO. (Actually, faster than the current MO math preview.)</p> <p><strong>Added 9/1/2010:</strong> Thanks to Kristi Tsukida, it is now available as a <a href="http://userscripts.org/scripts/show/84589" rel="nofollow">Firefox greasemonkey script</a> and as a <a href="https://chrome.google.com/extensions/detail/eenhaafocdcfjengcoflbdnoljeaepgb" rel="nofollow">Google Chrome extension</a>, for easy installation.</p> <p><strong>Added 11/24/2010:</strong> Now available as a <a href="https://addons.mozilla.org/en-US/firefox/addon/257869/" rel="nofollow">Firefox add-on</a>.</p> http://mathoverflow.net/questions/46116/a-valued-points-of-projective-space/46175#46175 Answer by VA for A-valued points of projective space VA 2010-11-16T02:20:10Z 2010-11-16T02:20:10Z <p>Yes, you (and BCnrd) are absolutely correct and the quoted statement is wrong.</p> <p>Over any scheme $S$, the $S$-points of $\mathbb P^n$ are the surjections $\mathcal O_S^{\oplus n+1} \to F$ with invertible $\mathcal O_S$-module $F$. More generally, the $S$-points of the grassmannian $Gr(m,k)$ are the surjections $\mathcal O_S^{\oplus m}\to F$ with $F$ locally free of rank $k$. </p> <p>Note: no Noetherian assumptions on $S$ are necessary. This is the first step for Grothendieck's construction of Hilbert schemes, without Noetherian assumption.</p> <p>So, for a ring $A$, the $A$-points of $\mathbb P^n$ are the surjections $A^{n+1}\to P$ with $P$ locally free (equivalently, projective) $A$-modules of rank 1.</p> http://mathoverflow.net/questions/44445/about-isogenies-of-abelian-varieties/44686#44686 Answer by VA for About isogenies of abelian varieties VA 2010-11-03T15:12:25Z 2010-11-03T15:12:25Z <p>This is to fill some prerequisites to BCnrd's comment-answer.</p> <p>First of all, there are several definitions of a polarization on an abelian variety, and the most "coordinate-free" one is that it is a homomorphism $\lambda:A\to A^t = Pic^0(A)$ given by some (non-unique) ample divisor $D$, so that $\lambda(a) = \mathcal O_A( T^*_a D - D)$, where $T_a:A\to A$ is the translation by $a\in A$. A polarization is principal if $\lambda$ is an isomorphism.</p> <p>Then the basic steps, all requiring proof, are:</p> <ol> <li><p>Every abelian variety (over a field) has a polarization.</p></li> <li><p>$\lambda$ is surjective, and $K(\lambda)=\ker\lambda$ is a finite group subscheme of $A$ of length $d^2$. The degree of $\lambda$ is defined to be $d$. Thus, a polarization is principal iff $d=1$.</p></li> <li><p>$K(\lambda)$ comes with a perfect skew symmetric Weil pairing $b: K(\lambda)\times K(\lambda) \to \mathbb G_m$ with the values in the multiplicative group. "Perfect pairing" means that it defines an iso from $K(\lambda)$ to its Cartier dual.</p></li> <li><p>If $H\subset K(\lambda)$ is a subgroup which is isotropic w.r.t. this pairing (i.e. $b$ restricted to $H\times H$ is trivial), then $\lambda$ descends to a polarization on $B=A/H$ of degree $d/|H|$.</p></li> </ol> <p>So then it is enough to find a nontrivial isotropic subgroup $H$, replace $A$ by $A/H$, and continue by induction, until you reach $d=1$.</p> <p>In char 0, this is trivial: just pick any cyclic subgroup in $K(\lambda)$. Since $b$ is skew-symmetric, it is automatically isotropic. </p> <p>In char p, you need to get into the classification of finite group schemes, and learn the difference between $\mathbb Z/p\mathbb Z$, $\mu_p$, and $\alpha_p$. Once you learn that (<em>Introduction to affine group schemes</em> by Waterhouse is a good source), then you are perhaps ready to read BCnrd's comment-answer. </p> http://mathoverflow.net/questions/40624/root-systems-and-sums-of-squares Root systems and sums of squares VA 2010-09-30T15:02:27Z 2010-10-01T06:31:49Z <p>It is easy to see that the quadratic form for the root system $A_n$ is a sum of $n+1$ squares of <em>integral</em> linear forms:</p> <p>$$q_{A_n} = 2 \sum_{i=1}^n x_i^2 - 2 \sum_{i=1}^{n-1} x_i x_{i+1} = x_1^2 + (x_1-x_2)^2 + \dots (x_{n-1}-x_n)^2 + x_n^2$$</p> <p>It is equally easy to see that $q_{D_n}$ is a sum of $n$ squares. It is a little harder to see but I think is true that $q_{E_n}$ ($n=6,7,8$) is <em>not</em> a sum of $\ge n$ squares of integral forms.</p> <p><strong>Question:</strong> is this a standard fact, well-known to experts? Is there a standard reference? (I hate to reinvent a bycicle.) And has this fact been used for something interesting? (I have an interesting application in mind, so I am looking for connections...)</p> http://mathoverflow.net/questions/38105/is-there-an-underlying-explanation-for-the-magical-powers-of-the-schwarzian-deriv/38249#38249 Answer by VA for Is there an underlying explanation for the magical powers of the Schwarzian derivative? VA 2010-09-09T23:53:02Z 2010-09-09T23:53:02Z <p>Nobody pointed out <a href="http://www.ams.org/notices/200901/tx090100034p.pdf" rel="nofollow">What is ... Schwarzian Derivative</a>? (Notices of AMS Jan 2009), which succinctly explains quite a lot.</p> http://mathoverflow.net/questions/37381/integral-positive-definite-quadratic-forms-and-graphs Integral positive definite quadratic forms and graphs VA 2010-09-01T14:18:35Z 2010-09-03T14:04:51Z <p>Let me start with a question for which I know the answer. Consider a symmetric integral $n\times n$ matrix $A=(a_{ij})$ such that $a_{ii}=2$, and for $i\ne j$ one has $a_{ij}=0$ or $-1$. One can encode such a matrix by a graph $G$: it has $n$ vertices, and two vertices $i,j$ are connected by an edge iff $a_{ij}=-1$. </p> <p><strong>Question 1:</strong> Which graphs correspond to positive definite $A$?</p> <p><strong>Answer 1:</strong> (Classical, well-known, and easy) $G$ is a disjoint union of $A_n$, $D_n$, $E_n$. (http://en.wikipedia.org/wiki/Root_system)</p> <p>Now let me put a little twist to this. Let us also allow $a_{ij}=+1$ for $i\ne j$, and encode this by putting a broken edge between $i$ and $j$ (or an edge of different color, if you prefer).</p> <p><strong>Real question:</strong> Which of these graphs correspond to positive definite $A$?</p> <p><hr> Let me add some partial considerations which do not quite go far enough. (Some of these were in the helpful Gjergji's answer.)</p> <p>(1) Consider the set $R$ of shortest vectors in $\mathbb Z^n$; they have square 2. Reflections in elements $r\in R$ send $R$ to itself, and $R$ spans $\mathbb Z^n$ since it contains the standard basis vectors $e_i$. By the standard result about root lattices, $\mathbb Z^n$ is then a direct sum of the $A_n$, $D_n$, $E_n$ root lattices, and one can restrict to the case of a single direct summand.</p> <p>Hence, the question equivalent to the following: what are the graphs corresponding to all possible bases of $\mathbb R^n$ in which the basis vectors are roots?</p> <p>The case of $R=A_n$ is relatively easy. The roots are of the form $f_a-f_b$, with $a,b \in (1,\dots,n+1)$. Every collection $e_i$ corresponds to an oriented spanning tree on the set $(1,\dots,n+1)$. The 2-colored graph is computed from that tree. I don't see a clean description of the graphs obtained via this procedure, but it is something.</p> <p>For $D_n$, similarly, a basis is described by an auxiliary connected graph $S$ on $n$ vertices with $n$ edges whose ends are labeled $+$ or $-$. The graph $G$ is computed from $S$.</p> <p>And for $E_6,E_7,E_8$ there are of course only finitely many cases, but for me the emphasis is on MANY, very many.</p> <p><em>So has anybody done this?</em> Is there a table in some paper or book which contains all the 2-colored graphs obtained this way, or -- better still -- a clean characterization of such graphs?</p> <p>(2) There is a notion of "weakly positive quadratic forms" used in the cluster algebra theory (see for example the first pages of LNM 1099 (Tame Algebras and Integral Quadratic Forms) by Ringel. And there is some kind of classification theory for them. Maybe I am mistaken, but this seems to be quite different: a quadratic form $q$ is "weakly positive" if $q\ge 0$ on the first quadrant $\mathbb Z_{\ge0}^n$. So there is no direct relation to my question, it seems.</p> http://mathoverflow.net/questions/37384/mirror-of-flop/37386#37386 Answer by VA for Mirror of Flop? VA 2010-09-01T14:47:05Z 2010-09-01T14:47:05Z <p>Small contractions are mirrors to degenerations, so: degenerate, then deform out.</p> http://mathoverflow.net/questions/27413/when-are-ehrhart-polynomials-polynomials/27420#27420 Answer by VA for When are Ehrhart polynomials polynomials? VA 2010-06-08T02:28:48Z 2010-06-08T02:28:48Z <p>Just to remark that for a rational polytope whose vertices are <em>not</em> integral, the function $f_P(t)$ could still be a polynomial (and not just a quasipolynomial). A large class of examples is provided by degenerations of flag varieties $G/B$. There are many degenerations, each corresponding to a representation of the longest word $w\in W$ in the Weil group as the shortest product of standard reflections. All of these correspond to rational polytopes. They all have the same Erhart function. Some of them are integral but others are not.</p> <p>For more details, see R. Chiriv`ı, LS algebras and application to Schubert varieties, Transform. Groups 5 (2000), no. 3, 245–264, or Alexeev-Brion <a href="http://arxiv.org/abs/math/0403379" rel="nofollow">Toric degenerations of spherical varieties.</a></p> http://mathoverflow.net/questions/27144/why-didnt-vladimir-arnold-get-the-fields-medal-in-1974/27171#27171 Answer by VA for Why didn't Vladimir Arnold get the Fields Medal in 1974? VA 2010-06-05T17:08:17Z 2010-06-05T19:15:04Z <p>Pontryagin wrote a book "Biography of Lev Semenovich Pontryagin, a mathematician, composed by himself". It is available online at <a href="http://www.ega-math.narod.ru/LSP/book.htm" rel="nofollow">http://www.ega-math.narod.ru/LSP/book.htm</a>, in the original Russian. Google does a fairly good job of translation, although it refuses to translate the individual chapters completely because of their length.</p> <p>In the book, Pontryagin shares a lot about the inner workings of the IMU Executive Board and his own role in holding the Soviet party line there as its vice president. For example, he recounts his version of how France got the IMU presidency in 1974, so that neither the Soviet Union nor the US would dominate. </p> <p>The only relevant mention of Arnold that I could find in that book is in chapter 5. He states that in 1974 Arnold was not allowed to leave the country to lecture abroad, and that there was a conflict about this with the Executive Board of the IMU, who insisted that he should. From this, you could extrapolate the reasons for blocking Arnold's Fields medal, if the story is true.</p> http://mathoverflow.net/questions/25922/is-the-euler-characteristic-a-birational-invariant/25965#25965 Answer by VA for Is the Euler characteristic a birational invariant VA 2010-05-26T04:19:02Z 2010-05-26T16:03:27Z <p>The dimensions $h^i(\mathcal O_X)$ of the cohomology groups of $\mathcal O_X$, and thus the Euler characteristic, are birational invariants of <em>smooth</em> proper varieties <em>in positive characteristic</em> as well, by a recent work of Andre Chatzistamatiou and Kay Rülling. It is not published yet but a <a href="http://arxiv.org/abs/0911.3599" rel="nofollow">preprint is available.</a> </p> http://mathoverflow.net/questions/25603/ample-line-bundles-sections-morphisms-to-projective-space/25622#25622 Answer by VA for Ample line bundles, sections, morphisms to projective space VA 2010-05-23T03:30:26Z 2010-05-24T01:52:25Z <p><em>1. Are there simple examples (say on a curve or surface) of line bundles that are globally generated but not ample, of ample line bundles with no sections, of ample line bundles that are globally generated but not very ample, and of very ample line bundles with higher cohomology?</em></p> <p>On a curve of genus $g$, a general divisor of degree $d \le g-1$ has no sections. Of course, if $d>0$ then it is ample. </p> <p>$K_X$ on a hyperelliptic curve is globally generated but not very ample. </p> <p>Look at $L=\mathcal O(1)$ on a plane curve of genus $d$. Then from $$0\to \mathcal O_{\mathbb P^2}(1-d) \to \mathcal O_{\mathbb P^2}(1) \to \mathcal O_C(1)\to 0$$ </p> <p>you see that $H^1(\mathcal O_C(1))=H^2(\mathcal O_{\mathbb P^2}(1-d))$ which is dual to $H^0(\mathcal O_{\mathbb P^2}(d-4))$. So that's nonzero for $\ge4$.</p> <p><em>2. Given an ample line bundle L, what is the minimal number k so that I can be sure $L^k$ has sections, is globally generated, is very ample? Is k related to the dimension of X?</em></p> <p>Again, just look at the divisor of a degree 1 on a curve of genus $g$. You need $k\ge g$, so you see that there is no bound in terms of the dimension. </p> <p>It turns out that a better right question to ask is about the <em>adjoint</em> line bundles $\omega_X\otimes L^k$ ($K_X+kL$ written additively). Then the basic guiding conjecture is by Fujita, and which says that for $k\ge \dim X+1$ the sheaf is globally generated, and for $k\ge \dim X+2$ it is very ample. This is proved for $\dim X=2$, proved with slightly worse bounds for $\dim X=3$. For higher dimensions the best result is due to Angehrn-Siu who gave a quadratic bound on $k$ instead of linear. There are some small improvements for some special cases.</p> <p><em>3.If L is very ample, I can use it to embed X into some projective space. Then by projecting from points off of XÚN , I can eventually get a finite morphism X!d , where d is the dimension of X. But what if I just know that L is ample and globally generated? Can I also use it to get such a finite morphism to d?</em></p> <p>But of course $L$ gives a morphism $f$, and it follows that $f$ is finite: $f$ contacts no curve so $f$ is quasifinite, and $f$ is projective (since $X$ was assumed to be projective). And quasifinite + proper = finite.</p> http://mathoverflow.net/questions/24711/ways-of-formulating-homological-algebra-without-diagram-chasing/24902#24902 Answer by VA for Ways of formulating homological algebra without diagram chasing VA 2010-05-16T16:59:41Z 2010-05-16T16:59:41Z <p>But isn't doing homological algebra without diagram chasing akin to doing geometry without circles, or doing algebra without multiplication? It's just the nature of the subject, a big part of it (and a fairly pleasant, I might add).</p> <p>A standard solution is to use Freyd's Embedding theorem for abelian categories, which turns on diagram chase for all abelian categories. Taking it on faith and proving separately later is not ideal, but to me it seems better than the alternatives.</p> http://mathoverflow.net/questions/22141/how-do-i-see-latex-math-on-any-web-page-and-in-email/24135#24135 Answer by VA for How do I see LaTeX math on any web page and in email? VA 2010-05-10T19:20:51Z 2010-05-15T22:16:56Z <p>By now, I am aware of three ways to accomplish this:</p> <ol> <li><a href="http://www.mathjax.org/" rel="nofollow">MathJax</a>.</li> <li>display-latex2 described in <a href="http://mathoverflow.net/questions/22141/how-do-i-see-latex-math-on-any-web-page-and-in-email/22378#22378" rel="nofollow">the other answer</a> (written by Steve Cheng, to which I made some modifications).</li> <li><a href="http://www1.chapman.edu/~jipsen/mathml/asciimath.html" rel="nofollow">ASCIIMathML</a>, written by Peter Jipsen.</li> </ol> <p>I think the best choice by far is to use MathJax, an active project with a very professional development team. It is free, open source, and it is backed by the American Mathematical Society, the American Physical Society, and SIAM, among others. That is the way to go, if you are able to install it on your server.</p> <p>(In the previous version of this answer, I mentioned several problems. They were very promptly resolved by the developers, who are truly impressive, and whom I can not praise enough.)</p> <p>I wrote a <a href="http://sites.google.com/site/vaatmo/home" rel="nofollow">very simple Greasemonkey script</a> which allows you to use your local installation of MathJax on any web page, and in Gmail (in "basic HTML" and "print" views).</p> <p><a href="http://sites.google.com/site/vaatmo/home" rel="nofollow">The same web page</a> also contains a Greasemonkey script allowing you to pick and choose between the three locally installed math engines.</p> <p>In the long run, the best way would be for arXiv, MathSciNet, and Gmail to use MathJax on their servers. I made a <a href="http://groups.google.com/group/gmail-labs-suggest-a-labs-feature/browse_thread/thread/4c3c262ca0459385?pli=1" rel="nofollow">Gmail Lab request for this.</a> If more people support it, maybe they will do it, that would be great. </p> <p><hr> Here are detailed instructions. I am hesitant to bump this question too often. So for minor edits I will update <a href="http://sites.google.com/site/vaatmo/home" rel="nofollow">this web site</a> instead. <br><br></p> <h2> Installation instructions </h2> <p>This solution assumes that you have access to a web server and can install Javascript programs on it. </p> <p>First, download and install MathJax and MathJax web fonts. Install MathJax on your web server. Install the MathJax fonts from the MathJax-webfonts(-beta2)/fonts/HTML-CSS/TeX/otf directory.</p> <p>The customization is done by editing the file MathJax/config/MathJax.js. You need to set</p> <pre><code>webFont: null, </code></pre> <p>so that MathJax uses your locally installed MathJax TeX fonts. Next, download mathjaxthispage.user.js and save it to your Desktop. The script assumes that your MathJax installation resides in <a href="http://localhost/MathJax" rel="nofollow">http://localhost/MathJax</a>. If it is different, edit the script accordingly. From the Firefox menu bar, File > Open File, navigate to the downloaded script and open it. Greasemonkey will offer to install; do that. Start surfing.</p> <h3> Muting the TeX errors (optional) </h3> <p>The following settings make for a more pleasant viewing experience when browsing the pages with non-standard TeX macros, for example arXiv.org.</p> <blockquote> <p>extensions: ["tex2jax.js", "TeX/noErrors.js", "TeX/noUndefined.js"], </p> </blockquote> <p>and inside the TeX block, the following code</p> <blockquote> <p>TeX: { </p> <pre><code> noErrors: { inlineDelimiters: ["",""], multiLine: false, style: { "font-family": "serif", "font-size": "120%", "color": "gray", "border": "" } }, noUndefined: { attributes: { mathcolor: "red", mathbackground: "#FFEEEE", mathsize: "100%" } }, </code></pre> <p>// The rest follows...</p> </blockquote> <p>If you don't have the extensions noErrors.js and noUndefined.js in your MathJax/extensions directory, you can get them from a more recent build available at <a href="https://sourceforge.net/projects/mathjax/develop" rel="nofollow">https://sourceforge.net/projects/mathjax/develop</a></p> <h3> Using the native MathMML output (optional) </h3> <p>With Mozilla Firefox, you have an option of using the native MathMML instead of HTML-CSS output, which is faster. For this, you will need to set the following in MathJax/config/MathJax.js :</p> <blockquote> <p>jax: ["input/TeX","output/NativeMML"],</p> </blockquote> <p>Next, add the following to your userContent.css file (see <a href="http://www.mozilla.org/unix/customizing.html" rel="nofollow">Customizing Mozilla</a>):</p> <blockquote> <p>math { font-size: 112% } [mathvariant="double-struck"] {font-family: MathJax_AMS; } [mathvariant="script"] {font-family: MathJax_Script; } [mathvariant="fraktur"] {font-family: MathJax_Fraktur;} [mathvariant="-tex-caligraphic"] {font-family: MathJax_Caligraphic; } [mathvariant="bold-script"] {font-family: MathJax_Script; font-weight: bold;} [mathvariant="bold-fraktur"] {font-family: MathJax_Fraktur; font-weight: bold;} [mathvariant="monospace"]{font-family: monospace} </p> </blockquote> <p>The first line controls the magnification of math output, and you can change it to your liking. The other lines are needed to fix a bug with Mozilla's display (otherwise, Mozilla does not display MathML correctly).</p> <p>For font consistency, you could also type 'about:config' (without the quotes) in the location bar, and change the variable font.mathfont-family to</p> <blockquote> <p>MathJax_Main, MathJax_Math, MathJax_Size1, MathJax_Size2, MathJax_Size3, MathJax_Typewritter, MathJax_AMS, MathJax_Caligraphic, MathJax_Fraktur, MathJax_SansSerif</p> </blockquote> http://mathoverflow.net/questions/24503/contracting-divisors-to-a-point/24516#24516 Answer by VA for Contracting divisors to a point VA 2010-05-13T15:48:06Z 2010-05-15T00:22:17Z <p>For a smooth $Y$, a necessary condition for contractibility is that the conormal line bundle $N_{Y,X}^*$ is ample. It is also sufficient for contracting to an algebraic space. The reference is <em>Algebraization of formal moduli. II. Existence of modifications.</em> by M. Artin. </p> <p>$Y$ can be contracted to a point on an algebraic (projective) variety if in addition $Y=\mathbb P^{n-1}$, $n=\dim X$. You can prove this easily by hands. Start with an ample divisor $H$ and then prove that an appropriate linear combination $|aH+bY|$ is base point free and is zero exactly on $Y$. You will find the argument in Matsuki's book on Mori's program for example.</p> <p>So if $X$ is a surface and $Y=\mathbb P^1$ with $Y^2&lt;0$ then it is contractible to a projective surface. For a reducible divisor $Y=\sum Y_i$ a necessary condition (which is also sufficient in the category of algebraic spaces) is that the matrix $(Y_i.Y_j)$ is negative definite. The strongest elementary sufficient condition for contractibility to a variety is that $\sum Y_i$ is a <em>rational configuration</em> of curves. This is contained in <em>On isolated rational singularities of surfaces</em> by M. Artin. </p> <p>This paper also contains an example of an elliptic curve $Y$ with $Y^2=-1$ which is not contractible to an algebraic surface. The surface $X$ is the blowup of $\mathbb P^2$ at 10 sufficiently general points lying on a smooth cubic, $Y$ is the strict preimage of that cubic.</p> <p>Finally, for an irreducible divisor $Y$ the resulting space $V$ is smooth iff $Y=\mathbb P^{n-1}$ and $N_{Y,X}=\mathcal O(-1)$. Indeed, $X\to V$ has to factor through the blowup of $V$ at a point by the universal property of the blowup. But then $X$ has to coincide with this blowup by Zariski main theorem. And on the blowup at a point the exceptional divisor is $\mathbb P^{n-1}$ with the normal bundle $\mathcal O(-1)$.</p> http://mathoverflow.net/questions/24082/what-are-the-prime-ideals-of-kx-y/24200#24200 Answer by VA for What are the prime ideals of k[[x,y]]? VA 2010-05-11T03:49:27Z 2010-05-11T03:49:27Z <p>You will find the proof of the fact that $k[[x_1,\dots,x_n]]$ is a UFD in Shafarevish, <em>Basic algebraic geometry,</em> Chapter II, 3.1. The proof uses the Weierstrass preparation lemma. The same section contains the proof that a codimension 1 prime ideal is principal. The only codimension prime ideal of codimension 2 is $(x,y)$ of course, since it has to be contained in $(x,y)$. Miles Reid's <em>Undergraduate commutative algebra</em> contains the proof as well, I think.</p> http://mathoverflow.net/questions/24175/why-does-the-euler-characteristic-of-a-toric-variety-equal-the-number-of-vertices/24179#24179 Answer by VA for Why does the Euler characteristic of a toric variety equal the number of vertices in the defining polytope? VA 2010-05-10T23:04:12Z 2010-05-10T23:04:12Z <p>Merely observe that a toric variety is the union of torus orbits $(\mathbb C^*)^r$ for various dimensions $r$, and that the Euler characteristic of $(\mathbb C^*)^r$ is zero if $r>0$ and $1$ if $r=0$. </p> <p>Vertices of a polytope correspond to 0-dimensional orbits, $r$-dimensional faces -- to $r$-dimensional orbits.</p> <p>$\mathbb P^2$ corresponds to a triangle, $\mathbb P^1\times\mathbb P^1$ to a square. It is not very hard to count their vertices.</p> http://mathoverflow.net/questions/24132/what-are-examples-of-mathematical-concepts-named-after-the-wrong-people-stigler/24139#24139 Answer by VA for What are examples of mathematical concepts named after the wrong people? (Stigler's law) VA 2010-05-10T19:42:02Z 2010-05-10T19:42:02Z <p>Chow varieties were invented by Van der Waerden (Chow was his student). And Hilbert schemes were invented by Grothendieck (who called them Hilbert schemes himself, however).</p> http://mathoverflow.net/questions/22141/how-do-i-see-latex-math-on-any-web-page-and-in-email/22378#22378 Answer by VA for How do I see LaTeX math on any web page and in email? VA 2010-04-23T17:53:25Z 2010-05-10T19:24:45Z <p>The <a href="http://www.gold-saucer.org/mathml/greasemonkey/display-latex.user.js" rel="nofollow">Greasemonkey MathML script</a> written by Steve Cheng and linked to in <a href="http://mathoverflow.net/questions/22141/how-do-i-see-latex-math-on-any-web-page/22248#22248" rel="nofollow">Scott Morrison's answer</a> worked only partially for me in Firefox on Windows 7: it did not display many \mathbb, \mathcal, and \mathfrak characters because the corresponding Unicode characters were missing in the fonts. Installing additional <a href="https://developer.mozilla.org/en/Mozilla_MathML_Project/Fonts" rel="nofollow">STIX and Asana Math fonts</a> did not help, in fact it made the display looking worse. </p> <p>So I rewrote the script (a long and tedious job finding the correct Unicode codes and putting them in the right places). I also added arxiv.org, front.math.ucdavis.edu, MathSciNet, and mail.google.com to the sites supported by default, and added miscellaneous characters and TeX commands missing in the original script.</p> <p><strong>Yes, it works with gmail (!)</strong> if you switch to the basic HTML view. So now you can read an email from your collaborator and see typeset math right there. Now tell me you haven't always wished and prayed for this? I know I have.</p> <p>Here are the detailed instructions for the method that produces good results using Mozilla Firefox on Windows 7. I haven't tested on other systems, you are welcome to share your experiences in the comments.</p> <ol> <li><p>Click <a href="https://addons.mozilla.org/en-US/firefox/addon/748" rel="nofollow">here</a> to install the Greasemonkey Firefox extension.</p></li> <li><p>Download a modified Greasemonkey script from <a href="http://sites.google.com/site/vaatmo/home" rel="nofollow">here</a> and save it to your Desktop.</p></li> <li><p>From the Firefox menu bar, File > Open File, navigate to the downloaded script and open it. Greasemonkey will offer to install it. Do that.</p></li> </ol> <p>That should be it. Check how it works by looking at some arXiv abstracts such as <a href="http://arxiv.org/abs/1004.2614" rel="nofollow">this</a>, or <a href="http://arxiv.org/abs/1004.4188" rel="nofollow">this</a>.</p> <p>Even when the authors use custom notations, such as \red or \cE, removing the dollar signs, putting math in a different font, and using sub- and superscripts dramatically increases the readability in my experience. </p> <p><strong>Edit:</strong> I also fixed the displayed formulas with double dollars, which the original script did not handle correctly. So now you can also view <a href="http://arxiv.org/abs/0903.1879" rel="nofollow">this</a> and <a href="http://arxiv.org/abs/1004.4623" rel="nofollow">this</a>.</p> <p>So in the end this was more of a community service than a question. Enjoy the results!</p> http://mathoverflow.net/questions/23394/applications-of-algebraic-geometry-over-a-field-with-one-element/23400#23400 Answer by VA for Applications of algebraic geometry over a field with one element VA 2010-05-04T03:08:25Z 2010-05-04T03:08:25Z <p>Connes et al. motivate their study of geometry over $\mathbb F_1$ in part by an idea that it may help to prove Riemann's Hypothesis. Some people, such as Mochizuki I think, hope that this study may help to prove the $abc$ conjecture. </p> <p>These are all potential, not yet real, applications. But hey, if one of these works...</p> http://mathoverflow.net/questions/23091/is-there-an-obvious-way-for-showing-singularities-are-quotient/23139#23139 Answer by VA for Is there an obvious way for showing singularities are quotient? VA 2010-04-30T19:48:39Z 2010-05-01T12:48:20Z <p>Say $X=\mathbb A^n$ and $D_1,\dots, D_n$, the components of $D$, are the coordinate hyperplanes $x_i=0$, for simplicity. You can assume that WLOG, because your $(X,D)$ is isomorphic to this one in &eacute;tale topology. </p> <p>$\pi_1(X\setminus D) = \mathbb Z^n$. So the cover $V\to U$ corresponds to a finite quotient of $\mathbb Z^n$, which is a finite abelian group $G$. So the ring of regular functions on $V$ is generated by the <em>roots</em> of monomials in $x$. You can write these as $x^m$ for some $m\in \mathbb Q^n$. The lattice $H$ generated by $m_i$ contains $\mathbb Z^n$, and the quotient $H/\mathbb Z^n$ is the dual abelian group $G^{\vee}$.</p> <p>So what is $Y$ now? It is the normalization of the ring $k[x_1,\dots,x_n]$ in the bigger field $k(x^{m_i})$. So it is a toric problem now. The normalization is generated by monomials in the lattice $H$ which lie in the cone $(\mathbb R_{\ge0})^n$.</p> <p>So $Y$ is toric and simplicial, and every such singularity is an abelian quotient singularity.</p> <p>The condition $\pi_1(X\setminus D) = \mathbb Z^n$ fails in char $p$ (indeed, the fundamental group in that case is huge), so this argument and the statement both fail in char $p$.</p> http://mathoverflow.net/questions/23107/what-is-known-about-the-mmp-over-non-algebraically-closed-fields/23141#23141 Answer by VA for What is known about the MMP over non-algebraically closed fields VA 2010-04-30T21:39:47Z 2010-04-30T21:46:42Z <p>Both of these cases follow more or less automatically from the Minimal Model Program over an algebraically closed field $\bar k$. This is well, known, see for example the original Mori's paper <em>"Threefolds whose canonical bundles are not numerically effective"</em> or <a href="http://www.math.princeton.edu/~kollar/FromMyHomePage/rat2.ps" rel="nofollow">Koll&aacute;r's paper</a> on 3-folds over $\mathbb R$. Iskovskikh and Manin did both versions for surfaces in the 1970s, and their arguments still apply.</p> <p>The point is that $K_X$ is invariant under the action of any group $G\subset Aut(X)$ and $Gal(\bar k/k)$. So if $\bar C$ is a curve on $\bar X= X\otimes_k \bar k$ with $K_X . \bar C&lt;0$ then $C= \sum_{g\in G} g.\bar C$ (resp. the sum of the conjugates) also intersects $K_X$ negatively.</p> <p>So you can work with the $G$-invariant (resp. $Gal$-invariant) part of the Mori cone $NE(X)\subset N_1(X)$. If $R$ is an extremal ray of $NE(X)^G$ then the supporting divisor $D$ can be chosen to be $G$-invariant, and it contracts a face of $NE(X)$ (instead of just a ray). It is either a divisorial contraction over $k$, or a flipping contraction. In the latter case there is a flip defined over $k$, since it is an appropriate relative canonical model, and every canonical model is automatically $G$-equivariant, resp. $Gal$-equivariant, by its uniqueness.</p> <p>So you just do the MMP over $k$. Unless $K_X$ is pseudoeffective but not effective, MMP terminates by [BCHM], and you get either a minimal model (with $K_{X_{\rm min}}$ nef) of a Mori-Fano fibration.</p> <p>Finally, if you started with a variety $X$ such that $\bar X$ is covered by rational curves then $K_X$ is not pseudoeffective.</p> http://mathoverflow.net/questions/22907/confusion-about-how-the-first-cohomology-classifies-torsors/22921#22921 Answer by VA for Confusion about how the first cohomology classifies torsors VA 2010-04-29T01:29:09Z 2010-04-29T01:29:09Z <p>The general principle is: if you have some objects which are locally trivial but globally possibly not trivial then the isomorphism classes of such objects are classified by $H^1(X,\underline{Aut})$, where $\underline{Aut}$ is the sheaf of automorphisms of your objects.</p> <p>So, if your objects locally are $U\times \mathbb A^n$ (i.e. vector bundles) or they are $\mathcal O_U^{\oplus n}$ (i.e. locally free sheaf) then either of these are classifed by $H^1(X, GL(n,\mathcal O))$. For $n=1$, you get $GL(1,\mathcal O)=\mathcal O^*$. </p> <p>Now what is $\mathbb C$ an automorphism group of? Certainly not of line bundles (zero has to go to zero).</p> http://mathoverflow.net/questions/22111/extending-vector-bundles-on-a-given-open-subscheme/22124#22124 Answer by VA for Extending vector bundles on a given open subscheme VA 2010-04-21T22:14:11Z 2010-04-22T17:55:22Z <p>This is true if $X$ satisfies Serre's condition $S_2$, i.e. $\mathcal O_X$ is $S_2$. Then a vector bundle is $S_2$ since locally it is isomorphic to $\mathcal O_X^n$.</p> <p>More generally, a coherent sheaf $F$ on a Japanese scheme (for example: $X$ is of finite type over a field) which is $S_2$ has a unique extension from an open subset $U$ with $\operatorname{codim} (X\setminus U)\ge 2$. This follows at once from the cohomological characterization of $S_2$.</p> <p>Thus, another name for the $S_2$-sheaves: they are sheaves which are <em>saturated in codimension 2</em>, and another name for the $S_2$-fication: <em>saturation in codimension 2</em>.</p> <p>P.S. Of course, by Serre's criterion, normal = $S_2+R_1$. So the above statement is true for any normal (e.g. smooth) variety.</p> <p>P.P.S. And of course, Gorenstein implies Cohen-Macaulay implies $S_2$. So the statement is also true for hypersurfaces and complete intersections, which could be very singular and non-reduced.</p> <p><hr> Edit to define some terms:</p> <ol> <li><p>A Japanese (or Nagata) ring is a ring obtained from a ring finitely generated over a field or $\mathbb Z$ by optionally applying localizations and completions. The property used here is that for a Japanese ring $R$, its integral closure (normalization) $\tilde R$ is a finitely generated $R$-module. This is important because the $S_2$-fication $S_2(R)$ lies between $R$ and $\tilde R$.</p></li> <li><p>A coherent sheaf $F$ satisfies $S_n$ if for any point $x\in Supp(F)$, one has $$depth_x (F) \ge \min(\dim_x Supp(F),n)$$ If $F$ locally corresponds to an $R$-module $M$, and $x$ to a prime ideal $p$, then the depth is the length of a maximal regular sequence $(f_1,\dots, f_k)$ of elements of $R_p$ for $M_p$ (so, $f_1$ is a nonzerodivisor in $M_p$, etc.).</p></li> </ol> http://mathoverflow.net/questions/22080/line-bundles-vs-cartier-divisors-on-a-non-integral-scheme/22083#22083 Answer by VA for Line bundles vs. Cartier divisors on a non-integral scheme VA 2010-04-21T17:22:50Z 2010-04-21T17:22:50Z <p>From the exact sequence </p> <p>$$1\to O^*\to K^*\to K^* / O^*\to 1$$</p> <p>you see that, for as long as $H^1(K^*)=0$, the map from $H^0( K^*/ O^*)$ (i.e. Cartier divisors) to $H^1( O^*)$ (i.e. line bundles) is surjective.</p> <p>On a Noetherian scheme without embedded primes (for example, reduced), $\mathcal K^*$ is the direct sum of several constant sheaves on the irreducible components, so it has trivial $H^1$.</p> <p>So the example would have to be a scheme with embedded primes with a tricky nonconstant $K^*$ (the sheaf of nonzero divisors). I've seen it but can't remember right now. So this is just some general observations to narrow the search.</p> http://mathoverflow.net/questions/21232/when-is-a-blow-up-non-singular/21234#21234 Answer by VA for When is a blow-up non-singular? VA 2010-04-13T17:54:25Z 2010-04-13T18:17:13Z <p>There is no general criterion, as far as I know, it is all try and see.</p> <p>Any projective birational morphism $f:X\to Y$ between varieties is the blowup of <em>some</em> sheaf of ideals $I$ on $Y$, so you can see that anything can happen.</p> http://mathoverflow.net/questions/20696/a-question-regarding-a-claim-of-v-i-arnold/64414#64414 Comment by VA VA 2012-04-09T19:04:15Z 2012-04-09T19:04:15Z Dear Robert, let me try to explain myself better. This is what I am saying: Denote the vertex of the right triangle by $P$ (it lies on the line $y=x$). Then it is easy to draw a triangle in which the slopes of AB and DC are very close to 1, but AD/BC=2, <b>*for as long as*</b> AD and BC are much much smaller than AP and BP. But this is indeed the case here. http://mathoverflow.net/questions/20696/a-question-regarding-a-claim-of-v-i-arnold/64414#64414 Comment by VA VA 2012-04-09T16:41:43Z 2012-04-09T16:41:43Z &quot;BD&quot; should read &quot;BC&quot; everywhere. http://mathoverflow.net/questions/20696/a-question-regarding-a-claim-of-v-i-arnold/64414#64414 Comment by VA VA 2012-04-09T16:17:36Z 2012-04-09T16:17:36Z At first, this looks like a great geometric explanation. Unfortunately, I do not understand it. The problem is that AD and BC are much smaller than the catheti, i.e. the sides of the triangle (in the original example, AD and BD are on the order of $x^7$ and the sides are on the order of $x^3$). So you can easily draw a similar triangle like, with slopes very close to 1, in which AD/BD=2 or anything else you like. http://mathoverflow.net/questions/46/what-is-the-universal-property-of-normalization/56#56 Comment by VA VA 2010-11-18T03:48:35Z 2010-11-18T03:48:35Z Since you speak of <i>associated points</i> and not just <i>irreducible components</i>, you allow embedded primes, right? So what if $X$ is a nodal curve with an embedded prime at the node, $Z$ is a point, and $f$ maps $Z$ to the node? http://mathoverflow.net/questions/46116/a-valued-points-of-projective-space/46235#46235 Comment by VA VA 2010-11-16T15:04:36Z 2010-11-16T15:04:36Z ... and &quot;multiplication by a nonzero element&quot;? For $A=\mathbb Z$, (1,1)=(2,2) in $\mathbb P^1$. http://mathoverflow.net/questions/46116/a-valued-points-of-projective-space/46235#46235 Comment by VA VA 2010-11-16T14:07:14Z 2010-11-16T14:07:14Z ... $\mathcal O_K$, the ring of integers in a number field $K$. http://mathoverflow.net/questions/40624/root-systems-and-sums-of-squares/40628#40628 Comment by VA VA 2010-10-01T15:22:21Z 2010-10-01T15:22:21Z Robin, thank you for your answer. I think I can prove all my statements quite easily by a direct combinatorial argument, working with the Dynkin graph. But you give a nice argument. http://mathoverflow.net/questions/40624/root-systems-and-sums-of-squares/40649#40649 Comment by VA VA 2010-09-30T20:09:24Z 2010-09-30T20:09:24Z Although the paper of Ellenberg-Venkatesh you quoted is really interesting, I don't see how to easily apply it: it only applies to forms with large minima, plus there are local conditions to check. I actually think it does not apply at all, since neither of $A_n,D_n,E_n$ is a sum of $\ge n+2$ squares. In any case, I can prove the statement easily, so my questions were (1) standard reference or a 2-line proof? and (2) did this simple characterization of $A_n,D_n,E_n$ show up in relation with something interesting? http://mathoverflow.net/questions/40624/root-systems-and-sums-of-squares/40628#40628 Comment by VA VA 2010-09-30T16:09:37Z 2010-09-30T16:09:37Z This does not rule out that $E_n$ is a sum of $&gt;n$ squares. But it's a good observation. http://mathoverflow.net/questions/40624/root-systems-and-sums-of-squares Comment by VA VA 2010-09-30T16:05:26Z 2010-09-30T16:05:26Z In fact, I think $E_n$ is not a sum of ANY number of squares of integral linear forms. http://mathoverflow.net/questions/37381/integral-positive-definite-quadratic-forms-and-graphs/37590#37590 Comment by VA VA 2010-09-23T07:41:06Z 2010-09-23T07:41:06Z They are linearly dependent, so the quadratic form is not positive definite. This may work for the problem you investigated but not for my question. Indeed, I think that there is no positive definite form for this graph, for any way of putting solid and broken edges. http://mathoverflow.net/questions/37381/integral-positive-definite-quadratic-forms-and-graphs/37590#37590 Comment by VA VA 2010-09-20T12:52:15Z 2010-09-20T12:52:15Z Do you really get all graphs with ≤8 vertices, as you say? For example, consider the graph on 5 vertices with the edges 12,13,14,15,23,24,25. How do you get that? What is the root system? http://mathoverflow.net/questions/38891/is-there-a-connected-k-group-scheme-g-such-that-g-red-is-not-a-subgroup Comment by VA VA 2010-09-15T23:00:48Z 2010-09-15T23:00:48Z A disconnected example can be found in <a href="http://staff.science.uva.nl/~bmoonen/boek/BookAV.html" rel="nofollow">staff.science.uva.nl/~bmoonen/boek/BookAV.html</a>, Ex.3.2. But you already knew that. http://mathoverflow.net/questions/37887/criteria-for-a-map-of-schemes-to-be-an-isomorphism Comment by VA VA 2010-09-06T14:37:02Z 2010-09-06T14:37:02Z See <a href="http://mathoverflow.net/questions/12767" rel="nofollow">mathoverflow.net/questions/12767</a> for a related question. http://mathoverflow.net/questions/37381/integral-positive-definite-quadratic-forms-and-graphs/37590#37590 Comment by VA VA 2010-09-03T13:28:21Z 2010-09-03T13:28:21Z That is very interesting. This sort of implies that there is little hope for a nice classification. On the other hand, graphs for the $A_n$ and $D_n$ lattices do have a clear structure, and in the $E_6$, $E_7$, $E_8$ cases there are only finitely many graphs.