User mp - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T05:26:19Z http://mathoverflow.net/feeds/user/8761 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98050/the-game-of-removing-two-vertices-in-a-graph/98055#98055 Answer by MP for The game of removing two vertices in a graph MP 2012-05-26T18:14:55Z 2012-05-26T18:14:55Z <p>Per Martin's request, here is a more detailed version of my comment. Given a graph $G$, the <a href="http://www.jstor.org/stable/3619902?seq=1" rel="nofollow">independence game</a> is the game in which two players take turn in removing a vertex and all of its neighbors. The game terminates when a player can make no further move, and it is a win for the player who last moved in the normal play and for the other one in the misère play.</p> <p>The game that Martin suggests with starting graph $G$ is the special case of the independence game, played on the line dual of $G$: the line dual of a graph $G$ is the graph whose vertices are the edges of $G$ and whose edges are pairs of non-disjoint edges of $G$.</p> <p>In particular, when the initial graph is a path, its line dual is also a path and the game is the same as what is called Dawson's chess. In this case, the independence game on a path with $n$ vertices with normal play, is a second player win if and only if either $n \in \{ 0, 14, 34\}$ or $n \equiv 4, 8, 20, 24, 28 \pmod{34}$.</p> http://mathoverflow.net/questions/69143/is-every-algebraic-space-the-quotient-of-a-scheme-by-a-finite-group/86669#86669 Answer by MP for Is every algebraic space the quotient of a scheme by a finite group? MP 2012-01-25T23:41:37Z 2012-01-25T23:41:37Z <p>I am not too familiar with algebraic spaces, but maybe this is an example of an algebraic space that cannot be a quotient of a scheme by a finite group. The property of such quotients that I use is that if an irreducible algebraic space $X$ of dimension at least two is the quotient of a scheme $S$ by a finite group $G$, then there is no point on $X$ with the property that every closed curve in $X$ contains that point. The reason is that given any point in $S$, it is certainly possible to find a curve in $S$ missing the $G$-orbit of that point, and thus we obtain a curve in $X$ missing any given point.</p> <p>On the other hand, there are algebraic spaces with a point contained in every closed curve. A classical example is obtained as follows. Let $X'$ denote the blow up of $\mathbb{P}^2$ at 10 points lying on a smooth plane cubic. Suppose that the points on the cubic are chosen so that the 10 divisor classes determined by them and the class of a line in the Picard group of the cubic are independent. It is a fact (proved by Artin, I believe) that a curve with negative definite self-intersection on a smooth surface can be contracted to an algebraic space. In particular, denoting by $C$ the strict transform of the cubic in $X'$, the curve $C$ can be contracted on $X'$ to an algebraic space $X$. It is also a consequence of the construction that the curve $C$ intersects every curve in $X'$, and therefore every curve in $X$ contains the point to which $C$ is contracted.</p> http://mathoverflow.net/questions/86471/can-every-finite-poset-be-realized-as-divisors-of-an-algebraic-curve/86480#86480 Answer by MP for Can every finite poset be realized as divisors of an algebraic curve? MP 2012-01-23T20:17:49Z 2012-01-24T20:21:03Z <p>Choosing a general set of $n$ points on a curve of genus at least $n$, you can assume that the divisor they define has a unique global section. Let $P$ denote the poset generated by all the possible sums with coefficients $0,1$ of these $n$ points. This poset is the same as the poset of subsets of an $n$-element set.</p> <p>Since every finite poset seems to be a subposet of the poset of subsets of a finite set [EDIT: this is true-see the comments below], just embed your poset in a "power set poset" and remove the unwanted divisors, to deduce that what you want is true.</p> http://mathoverflow.net/questions/86047/degree-of-fibers-with-too-large-dimension/86051#86051 Answer by MP for Degree of fibers with too large dimension MP 2012-01-19T00:26:36Z 2012-01-19T10:26:52Z <p>This is not true in general. Take your example and reembed <strong>everything</strong> using the $n$-th Veronese map. A subvariety $F$ of $X$ is reembedded as a variety of degree ${\rm deg}(F) n^{\dim(F)}$. In particular, bigger dimensional fibers will have larger degree than smaller dimensional fibers, provided $n$ is large enough.</p> <p>EDIT: This is what I had in mind. Choose a positive integer $d$. Let $Z=(z_1,\ldots,z_N)$ be a list of all the monomials of degree at most $d$ in $x_1,\ldots,x_n$ such that $(z_1,\ldots,z_M)$ is a list of all the monomials of degree at most $d$ in $x_1,\ldots,x_m$. Using the monomials $Z$ we obtain an embedding $v_d \colon X \to \mathbb{A}^N$, called the $d$-th Veronese embedding. A property of $v_d$ is that for every subvariety $F$ of $\mathbb{A}^n$ the equality $deg(v_d(F)) = deg(F) \, d^{dim(F)}$ holds. Let $\pi_d \colon X \to \mathbb{A}^M$ be the projection on the first $M$ coordinates. The morphism $\pi_d$ is <em>identical</em> to the morphism $\pi$ (whatever that means), but the fibers of $\pi_d$ as subschemes of $\mathbb{A}^N$ are isomorphic to the image under the $d$-th Veronese embedding of the fibers of $\pi$. Thus, for $d$ large enough, the fibers of larger dimension will have degree larger than the fibers of smaller dimension.</p> <p>I hope this clarifies your doubts!</p> http://mathoverflow.net/questions/85758/how-does-one-know-the-following-surface-contains-no-other-lines/85759#85759 Answer by MP for How does one know the following surface contains no other lines? MP 2012-01-15T19:45:07Z 2012-01-15T19:45:07Z <p>Any line on $X$ has at least one point in common with the plane $x_2=0$. Thus, any line on $X$ contains a point whose first two coordinates vanish. In particular, one of the equations of a line in $X$ can be chosen to be of the form $\lambda x_1+\mu x_2=0$ for a non-zero pair $(\lambda,\mu)$. From here, the result is immediate.</p> http://mathoverflow.net/questions/83409/uniformity-of-injectivity-for-maps-associated-to-linear-systems/83412#83412 Answer by MP for Uniformity of injectivity for maps associated to linear systems MP 2011-12-14T11:19:09Z 2011-12-14T11:19:09Z <p>Maybe I am making a mistake, but you could try the following. First, for each $k$ the set $U_k$ of points $x$ for which there is a section of $L^k$ not vanishing at $x$ is open. Moreover, the union of all the $U_k$ is all of $X$, by assumption (case of 0-dimensional $X$ is easy!). Thus, by compactness, there is a $k_0$ such that the line bundle $L^{k_0}$ is base point free. Now use $L^{k_0}$ to give a map to projective space and reduce to the projective case. If you care about the "all greater multiples" business, you can build it in this argument, I believe.</p> http://mathoverflow.net/questions/6108/anticanonical-divisor-of-the-blow-up-of-p2-in-9-points/75622#75622 Answer by MP for Anticanonical divisor of the blow up of P^2 in 9 points MP 2011-09-16T17:21:05Z 2011-09-16T17:21:05Z <p>Just to complement quim's answer, note that if $S$ is defined over the algebraic closure of a finite field, then the anticanonical divisor of the blow up of 9 general points in $\mathbb{P}^2$ is actually semiample!</p> <p>The reason this is not in contradiction with quim's answer lies in the subtlety of the word "general". Here the correct genericity assumption is what is sometimes called 'very general": the points need to lie in the complement of a countable union of closed subsets. Indeed, if you read quim's answer you see that there is a degree zero line bundle that needs to be non-torsion, and this condition translates to not being of order at most $n$ for every positive integer $n$: clearly a countable union of closed conditions. On the other hand, <em>every</em> degree zero line bundle on smooth curve defined over a finite field s torsion!!</p> http://mathoverflow.net/questions/75607/boundedness-of-c-k-on-a-surface-with-k-pseudoeffective/75614#75614 Answer by MP for Boundedness of $C.K$ on a surface with $-K$ pseudoeffective MP 2011-09-16T15:49:59Z 2011-09-16T15:49:59Z <p>Let $-K=N+E$ be the Zariski decomposition of $-K$, so that $N,E$ are $\mathbb{Q}$-divisors with $N$ nef, $E$ effective, such that $N \cdot E = 0$ and the restriction of the intersection pairing to the irreducible components of $E$ is negative definite. It follows that if $C$ is an integral curve such that $C \cdot (-K)&lt;0$, then we must have $C \cdot E &lt; 0$, so that $C$ is a component of $E$. Since $E$ has only finitely many components, we deduce that there are only a finite number of integral curves with negative intersection with $-K$, and the required boundedness follows.</p> http://mathoverflow.net/questions/75142/rotational-symmetry-group-of-qxq/75208#75208 Answer by MP for Rotational symmetry group of QxQ MP 2011-09-12T10:11:47Z 2011-09-12T10:11:47Z <p>Let $G$ denote the group of rotations of the plane fixing the set of rational points. This group has multiple "names": </p> <ul> <li><p>it is $SO(2,\mathbb{Q})$, as Donu already commented; </p></li> <li><p>it is the group $\mathbb{Q}(i)^\times / \mathbb{Q}^\times$, as quid already commented; </p></li> <li><p>it is the group $\mathbb{Z}/4\mathbb{Z} \oplus \bigoplus_{i=1}^\infty \mathbb{Z}$, as André already commented.</p></li> </ul> <p>Here, I want to justify the isomorphisms of $G$ with $\mathbb{Q}(i)^\times / \mathbb{Q}^\times$ and $\mathbb{Z}/4\mathbb{Z} \oplus \bigoplus_{i=1}^\infty \mathbb{Z}$: the argument is a combination of quid's answer and the comments to his answer.</p> <p>The group $G$ is the group of matrices $\begin{pmatrix} x &amp; -y \cr y &amp; x \end{pmatrix}$ with $x,y \in \mathbb{Q}$ satisfying $x^2+y^2=1$. We identify this group with the elements of norm one of the multiplicative group $\mathbb{Q}(i)^\times$ assigning to the above matrix the element $x+iy$. There is a surjective group homomorphism $$q \colon \mathbb{Q}(i)^\times \longrightarrow G$$ mapping $a$ to $q(a) = a \cdot \overline{a}^{-1}$, where $\overline{a}$ is the complex conjugate of $a$. The kernel of $q$ is $\mathbb{Q}^\times$, so that we obtain $G \simeq \mathbb{Q}(i)^\times / \mathbb{Q}^\times$. To conclude we analyze the group $\mathbb{Q}(i)^\times / \mathbb{Q}^\times$.</p> <p>Every element of $\mathbb{Q}(i)^\times$ can be written uniquely as a product of powers of primes of $\mathbb{Z}[i]$ times a unit (of $\mathbb{Z}[i]$). As is well-known, the splitting of the primes of $\mathbb{Z}$ in $\mathbb{Z}[i]$ is of one of three different kinds: </p> <ul> <li><p>primes congruent to 3 mod 4 stay irreducible, </p></li> <li><p>primes congruent to 1 mod 4 split as a product of two distinct primes, </p></li> <li><p>2 splits as $i \cdot (1-i)^2$.</p></li> </ul> <p>Using unique factorisation in $\mathbb{Z}[i]$, we encode elements of $\mathbb{Q}(i)^\times$ by the exponents of their prime factors (the units are irrelevant for our purposes). We obtain that the contributions of the three kinds of primes to $\mathbb{Q}(i)^\times / \mathbb{Q}^\times$ are </p> <ul> <li><p>nothing for the primes dividing a prime congruent to 3 mod 4, </p></li> <li><p>a copy of $\mathbb{Z}$ for the primes dividing a prime congruent to 1 mod 4, </p></li> <li><p>a copy of $\mathbb{Z}/4\mathbb{Z}$ for the prime $(1-i)$.</p></li> </ul> http://mathoverflow.net/questions/73743/countable-union-of-closed-subschemes-over-uncountable-field/73750#73750 Answer by MP for countable union of closed subschemes over uncountable field MP 2011-08-26T09:09:48Z 2011-08-26T09:09:48Z <p>I do not know a reference, but the following short argument seems to work.</p> <p>Assume that the dimension of $X$ is at least 1! Argue by induction on the dimension of $X$. Reduce to the case in which the subschemes are irreducible of codimension one. Shrinking $X$ if necessary, reduce also to the case in which $X$ is quasiprojective. Let $L$ be a pencil of integral divisors on $X$. Since the ground-field is uncountable, the pencil $L$ contains uncountably many elements that are integral; let $D \in L$ be an integral element of $L$ that is different from each of the subschemes you want to avoid. By the inductive hypothesis, $D$ contains a point that is not contained in any of the subschemes and you are done.</p> <p>You can find this stated as a hint in an Exercise V.4.15 (c) in Hartshorne.</p> http://mathoverflow.net/questions/36574/reconstructing-a-graph-given-access-to-its-cut-function/36578#36578 Answer by MP for Reconstructing a graph given access to its cut function MP 2010-08-24T18:17:55Z 2010-08-24T18:17:55Z <p>Asking individual vertices, you figure out the valence of each vertex with $n$ questions. Asking for pairs of vertices, you can then decide if each pair has an edge between them or not: they have an edge if and only if the "degree" on the pair is two less than the sum of the "degrees". Thus you can reconstruct the graph with $n+\binom{n}{2}$ questions.</p> http://mathoverflow.net/questions/36560/a-naive-question-about-homogeneous-polynomials/36563#36563 Answer by MP for a naive question about homogeneous polynomials MP 2010-08-24T15:43:01Z 2010-08-24T15:43:01Z <p>Assume that $p$ is non-zero. If the form $dp/p$ were exact, then locally a primitive would be $log(p)+const$; this is easily seen not to work as soon as you can "loop around" $S$ (e.g. restrict everything to a line intersecting $S$ and see what happens there). Thus the form $dp/p$ is exact if and only if $S$ is empty, and hence if and only if $p$ is constant.</p> http://mathoverflow.net/questions/131830/open-immersion-between-affine-spaces Comment by MP MP 2013-05-25T10:58:56Z 2013-05-25T10:58:56Z Take a look at the the Ax-Grothendieck theorem! http://mathoverflow.net/questions/129706/zeta3-in-terms-of-derivatives-of-zeta-at-1-2-and-pi/129721#129721 Comment by MP MP 2013-05-05T14:29:37Z 2013-05-05T14:29:37Z In the question, the result is checked to $10^4$ digits, which seems more precision that 500! http://mathoverflow.net/questions/129054/enriques-classification-of-algebraic-surfaces Comment by MP MP 2013-04-29T08:04:26Z 2013-04-29T08:04:26Z The second question seems easier. The assertions $K$ nef and $K^2=0$ imply that if a multiple of $K$ moves, then it maps to a curve. Combining this with the inequality on the 12th plurigenus gives you the result. I am not sure how to proceed in the other case: you would need to know something similar to $K^2=0$ to proceed in a similar way. http://mathoverflow.net/questions/128807/noether-lefschetz-over-finite-fields Comment by MP MP 2013-04-26T09:09:34Z 2013-04-26T09:09:34Z (Though let me add that the Picard number over the ground field <i>might</i> be odd, so you could refine the question.) http://mathoverflow.net/questions/128807/noether-lefschetz-over-finite-fields Comment by MP MP 2013-04-26T09:07:57Z 2013-04-26T09:07:57Z K3 surfaces over finite fields have (geometrically) even Picard number: not much hope for Noether-Lefschetz for quartic surfaces! http://mathoverflow.net/questions/128688/gauss-mapping-in-finite-characteristic Comment by MP MP 2013-04-25T10:05:00Z 2013-04-25T10:05:00Z You can start reading here: <a href="http://link.springer.com/article/10.1007%2Fs10711-008-9334-1#page-1" rel="nofollow">link.springer.com/article/&hellip;</a> http://mathoverflow.net/questions/126076/simple-field-extension-and-rational-points Comment by MP MP 2013-03-31T16:52:29Z 2013-03-31T16:52:29Z @wccanard: you have interpreted my comment correctly, and I was wrong! http://mathoverflow.net/questions/126076/simple-field-extension-and-rational-points Comment by MP MP 2013-03-31T11:05:04Z 2013-03-31T11:05:04Z $x^p+sy^p+tz^p \in \mathbb{F}_p(s,t)[x,y,z]$? http://mathoverflow.net/questions/123948/kodaira-dimension-of-symmetric-products-of-curves Comment by MP MP 2013-03-08T07:58:55Z 2013-03-08T07:58:55Z If the Jacobian of the curve is simple, then all its proper subvarieties are of general type; in particular the symmetric product of the curve is of general type, until the Abel-Jacobi map is surjective. By deformation, I would guess that the same is true for all curves, not just the ones that have simple Jacobian. http://mathoverflow.net/questions/121767/resolution-of-singularities-of-this-cubic-surface Comment by MP MP 2013-02-14T10:18:08Z 2013-02-14T10:18:08Z It seems that your surface the Cayley cubic: there are a lot of interpretations of the Cayley cubic! http://mathoverflow.net/questions/120683/rational-groups Comment by MP MP 2013-02-03T17:33:22Z 2013-02-03T17:33:22Z Ok, now I see why my example is not a counterexample! http://mathoverflow.net/questions/120683/rational-groups Comment by MP MP 2013-02-03T17:28:12Z 2013-02-03T17:28:12Z Btw, possibly I am confused: is your $g$ in the normalizer/centralizer formula an $x$? Also, by a $p$-element do you mean an element of order $p$ or of order a power of $p$? http://mathoverflow.net/questions/120683/rational-groups Comment by MP MP 2013-02-03T17:17:02Z 2013-02-03T17:17:02Z Isn't $Z/4$ a counterexample? http://mathoverflow.net/questions/119198/when-is-a-smooth-projective-variety-a-fibration Comment by MP MP 2013-01-18T11:44:56Z 2013-01-18T11:44:56Z <a href="http://mathoverflow.net/questions/35429/which-algebraic-varieties-admit-a-morphism-to-a-curve" rel="nofollow" title="which algebraic varieties admit a morphism to a curve">mathoverflow.net/questions/35429/&hellip;</a> http://mathoverflow.net/questions/118029/lines-on-degree-2n-3-fermat-hypersufaces Comment by MP MP 2013-01-04T09:11:41Z 2013-01-04T09:11:41Z The Fermat threefolds contain several one-parameter families of lines: partition the variables into two sets of size 2 and 3 and set them separately to zero. You obtain $10d$ families of lines in this way. Of I recall correctly, these are &quot;non-reduced&quot; as soon as $d$ is at least 5. You can find out more in papers by Albano-Katz and more recently Candelas and others.