User piotr achinger - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T16:07:24Z http://mathoverflow.net/feeds/user/3847 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/126656/linearly-generated-embedding/126683#126683 Answer by Piotr Achinger for Linearly generated embedding? Piotr Achinger 2013-04-06T01:15:57Z 2013-04-06T01:15:57Z <p>How about "linearly/projectively normal and normally generated" or "linearly/projectively normal and $N_0$"?</p> <p>Mumford in "Varieties defined by quadratic equations" defines $X$ to be normally generated if the section ring is generated in degree $1$. Green and Lazarsfeld in "On the projective normality of complete linear series on an algebraic curve" call the same condition $N_0$.</p> http://mathoverflow.net/questions/125401/existence-of-non-split-vector-bundles-on-smooth-projective-varieties Existence of non-split vector bundles on smooth projective varieties Piotr Achinger 2013-03-23T21:19:01Z 2013-03-24T19:42:39Z <blockquote> <p><strong>Question.</strong> Is it known/easy to see that every smooth projective variety $X$ (over an algebraically closed field), except for the point and $\mathbb{P}^1$, has a vector bundle which is not a direct sum of line bundles? </p> </blockquote> <p>I have a result which trivially shows the above fact in positive characteristic, but I don't know whether I should state it as a corollary as it might be well-known or obvious for reasons I don't see. </p> http://mathoverflow.net/questions/123942/how-many-flat-connections-has-a-line-bundle-in-algebraic-geometry/124042#124042 Answer by Piotr Achinger for How many flat connections has a line bundle in algebraic geometry? Piotr Achinger 2013-03-09T03:06:01Z 2013-03-09T03:06:01Z <p>A vector bundle with an algebraic connection has to have vanishing all Chern classes, at least in characteristic zero. I remember that this follows from the vanishing of the "Atiyah class", but I don't know the details. </p> http://mathoverflow.net/questions/122751/cohomology-of-restrictions-of-vector-bundles-to-deformations/122762#122762 Answer by Piotr Achinger for cohomology of restrictions of vector bundles to deformations Piotr Achinger 2013-02-23T23:41:31Z 2013-02-24T00:47:05Z <p>On $X'$, there is an exact sequence <code>$$0 \to \mathcal{O}_X \to \mathcal{O}_{X'} \to \mathcal{O}_X \to 0.$$</code> Note that this sequence does not carry a lot of information about the deformation, for example it often splits (e.g. if $H^1(X, T_X) = 0$) even if $X'$ is a non-trivial deformation.</p> <p>Anyway, we can tensor this sequence with $E$, obtaining a short exact sequence <code>$$0 \to E|_X \to E|_{X'} \to E|_X \to 0.$$</code> Applying cohomology, we get a long exact sequence <code>$$\ldots \to H^i(X, E) \to H^i(X', E) \to H^i(X, E) \to H^{i+1}(X, E) \to \ldots$$</code></p> <p>In particular, if $H^i(X, E) = 0$ then $H^i(X', E) = 0$ as well, and if the initial sequence was split then you get <code>$$H^i(X', E) = H^i(X, E) \otimes_k k[\varepsilon]/(\varepsilon^2) = H^i(X, E)\oplus \varepsilon H^i(X, E).$$</code></p> http://mathoverflow.net/questions/122277/algebraic-surface-of-a-line-arrangement/122280#122280 Answer by Piotr Achinger for Algebraic surface of a line arrangement Piotr Achinger 2013-02-19T08:44:12Z 2013-02-19T08:44:12Z <p>Regarding 1: If $X$ is a variety with function field $K$, $L$ a finite separable extension of $K$, we can construct a normal variety $Y$, "normalization of $X$ in $L$", with function field $L$ and a finite map $Y\to X$. This is done as follows: for any affine open $X_i = Spec(R)$ in $X$, take $Y_i = Spec(S)$ where $S$ is the integral closure of $R$ in $L$, and then glue these $Y_i$ to obtain $Y$. In your situation, $X = \mathbb{P}^2$, $K = \mathbb{C}(z_1/z_0, z_2/z_0)$, $L$ is the field you described and $Y$ is the surface you need.</p> <p>Regarding 2: I cannot say anything explicit, but if $f:Y\to X$ is a finite map and $L$ is an ample line bundle on $X$, then $f^* L$ is an ample line bundle on $Y$. This gives you a projective embedding of $Y$ (up to Veronese). Sorry I can't be more helpful here...</p> http://mathoverflow.net/questions/121137/theta-group-representation/121143#121143 Answer by Piotr Achinger for Theta group representation Piotr Achinger 2013-02-08T02:37:08Z 2013-02-08T02:37:08Z <p>I think the answer is no due to the following counterexample (there might be a mistake somewhere though):</p> <p>Take an elliptic curve $X$ with $L = O_X(3P_0)$ (characteristic $\neq 2, 3$), then $K$ will be the $3$-torsion subgroup and $K_1$ will be generated by a $3$-torsion point $P$. Then in the embedding by $|L|$ of $X$ in $\mathbb{P}^2$, the points $P$, $-P$ and $P_0$ are collinear, that is, $K_1$ is mapped to a lower-dimensional linear subspace.</p> http://mathoverflow.net/questions/120976/does-there-exist-a-family-of-objects-over-the-tangent-space-to-the-base-space-of/120983#120983 Answer by Piotr Achinger for does there exist a family of objects over the tangent space to the base space of a family of objects? Piotr Achinger 2013-02-06T16:04:50Z 2013-02-06T16:19:56Z <p>Imagine that $S$ is an open subset of some fine moduli space which does not contain any rational curves. This does happen although I don't know any examples. Then any family over an affine space will have to be constant, so the answer would be no.</p> <p>On the other hand, of course there is formal family over the completion of $S$ at the given point $p$ (<code>$\hat{S}_p = \lim Spec (\mathcal{O}_{S, p}/\mathfrak{m}_{S, p}^n)$</code>), which looks like the infinitesimal neighborhood of $0$ in the tangent space ($\hat{T}_0 = Spf(k[[t_1, \ldots, t_s]])$ where the $t_i$ form a basis of the dual of $T= T_{p} S$). I think morally this might play the role of what you want.</p> http://mathoverflow.net/questions/120372/when-does-autxbirx-hold/120374#120374 Answer by Piotr Achinger for When does $Aut(X)=Bir(X)$ hold? Piotr Achinger 2013-01-31T01:19:39Z 2013-01-31T01:19:39Z <p>One example: This holds for abelian varieties, because a rational map to an abelian variety is always regular.</p> http://mathoverflow.net/questions/120273/formal-criterion-of-flatness Formal criterion of flatness Piotr Achinger 2013-01-30T00:19:16Z 2013-01-30T00:19:16Z <p>Let $k$ be a field, $S$ and $R$ be local $k$-algebras with residue field $k$ and $\phi:S\to R$ be a local homomorphism. Then $\phi$ induces (obviously) a natural transformation of "functors of points" $\phi^*:h_R \to h_S$ (where $h_A(B) = Hom(A, B)$). Let $Art_k$ be the category of local artinian $k$-algebras with residue field $k$ and let $F$, $G$ be the restrictions of $h_R$, resp. $h_S$ to $Art_k$. </p> <p>There are well-known criteria of (formal) smoothness/etaleness of $\phi$ in terms of the induced transformation $\phi^* : F\to G$. There is also an infinitesimal criterion of flatness, but that is different in spirit.</p> <blockquote> <p><strong>Question.</strong> Is there a criterion on $\phi^*:F\to G$ which ensures that $\phi$ is <em>flat</em>?</p> </blockquote> <p>You can assume that $\phi$ is finite and that $S$ and $R$ are completions of finitely generated $k$-algebras.</p> http://mathoverflow.net/questions/107271/homology-classes-of-subvarieties-of-toric-varieties Homology classes of subvarieties of toric varieties Piotr Achinger 2012-09-15T16:58:52Z 2013-01-17T07:17:06Z <p>Let $X$ be a smooth proper toric variety, $Z\subseteq X$ a smooth subvariety.</p> <blockquote> <p>Is the fundamental class $[Z] \in H_\ast(X) = A_\ast(X)$ nonzero?</p> </blockquote> <p><strong>Background</strong></p> <p>If $X$ is a Kaehler variety, this is of course true, since the intersection of $Z$ with some power of the Kaehler class is positive. But our $X$ is not Kaehler if it is not projective (the Hodge structure is trivial + Kodaira embedding theorem).</p> <p>I don't think that the statement is true for general complex algebraic varieties, for example the "Hironaka twist" (an example of a $3$-dimensional non-projective smooth proper algebraic variety) has two disjoint smooth curves $M_1$, $M_2$ with the property that $M_1 + M_2$ is numerically zero. So the union of these curves should give a zero class in homology. <strong>EDIT.</strong> This wouldn't be possible if $M_1$ and $M_2$ intersected - see Dustin Cartwright's and ulrich's comments below.</p> <p>Chow rings (or homology) of smooth proper toric varieties is very well understood: it is generated by the boundary divisors, which are smooth toric varieties themselves, intersecting transversely. Relations are obvious from the fan description on $X$. In particular, we need to prove that $Z$ has non-zero intersection with some intersection of the boundary divisors. This could allow for an induction-on-dimension argument, if only $Z$ intersected the boundary nicely. We cannot hope for that, but maybe again there is a "perturbation" argument that may ensure this.</p> <p><strong>EDIT (continued).</strong> Dustin Cartwright's comment below shows that homology classes of irreducible curves in smooth proper varieties are nonzero. Therefore I am tempted to ask more than in the original question:</p> <blockquote> <p>Let $X$ be a smooth proper variety, $Z\subseteq X$ a smooth subvariety. Is the fundamental class $[Z] \in H_*(X, \mathbb{Q})$ nonzero?</p> </blockquote> http://mathoverflow.net/questions/119140/what-are-the-general-techniques-for-proving-a-variety-is-not-toric/119141#119141 Answer by Piotr Achinger for What are the general techniques for proving a variety is not toric? Piotr Achinger 2013-01-17T06:49:37Z 2013-01-17T07:06:01Z <p>I'm thinking about the following argument. It's not entirely precise, I will be grateful for comments how to improve it.</p> <p>Suppose such an $X$ is toric. Because the exceptional curves $E_i$ do not move inside $X$, they are fixed by the torus action. Therefore after blowing them down, the torus still acts and their images are fixed points. But there are only 3 fixed points of the torus action on $\mathbb{P}^2$!</p> <p>Different idea (using fans): since $Pic (X) = \mathbb{Z}^{1+s}$, the fan defining $X$ has $3+s$ rays, and $s$ of them correspond to the exceptional curves $E_1, \ldots, E_s$. Let us denote the torus invariant divisors corresponding to the other 3 rays $D_1, D_2, D_3$. Since $\dim X=2$, these $3+s$ rays/divisors are arranged in a circular order, so that only neighbors on this circle intersect. Because the curves $E_i$ and $E_j$ do not intersect for $i\neq j$, no two $E_i$ are neighbors on the circle, which shows that there at most as many $E$'s are $D's$, that is, $s\leq 3$.</p> <p>Note that the first argument works more generally for the blow-up of $\mathbb{P}^n$ in $s > n+1$ points. </p> http://mathoverflow.net/questions/118930/algebraic-definition-of-the-kuranishi-map Algebraic definition of the Kuranishi map Piotr Achinger 2013-01-14T23:38:31Z 2013-01-15T10:56:53Z <p>Let $X$ be a smooth projective algebraic variety over an algebraically closed field $k$. If $k=\mathbb{C}$, we know by work of Kuranishi that the base of the versal deformation of $X$ is the germ at $0$ of the fiber over $0$ of a holomorphic map $K:H^1(X, T_X)\to H^2(X, T_X)$ (defined in the neighborhood of 0), called the <em>Kuranishi map</em>. </p> <p>This means that, if $S_i$ are the power series rings associated to $H^i(X, T_X)$ (i.e., the completions of $Sym^* H^i(X, T_X)^*$), there is a map $k: S_2 \to S_1$ such that $R = S_1 \otimes_{S_2} k$ pro-represents the deformation functor of $X$.</p> <p><strong>Question</strong>. Can one construct the map $k$ using algebraic methods? </p> <p>Probably one should assume that $k$ is of characteristic zero (or replace power series rings by completed divided power algebras...).</p> http://mathoverflow.net/questions/118658/cohomology-of-the-gauss-manin-connection/118671#118671 Answer by Piotr Achinger for cohomology of the Gauss-Manin connection Piotr Achinger 2013-01-11T20:43:13Z 2013-01-11T20:55:53Z <p>The module with integrable connection $(\mathcal{H}^i, \nabla^i_{GM})$ corresponds to the local system <code>$Rf^{an}_* \mathbb{C}_{Y^{an}}$</code> on $S^{an}$, so the cohomology of the de Rham complex of $(\mathcal{H}^i, \nabla^i_{GM})$ will compute cohomology <code>$H^j(S^{an}, R^if^{an}_* \mathbb{C}_{Y^{an}})$</code>. By definition, the stalks of <code>$R^if^{an}_* \mathbb{C}_{Y^{an}}$</code> are (singular) cohomology groups of the fibers of $f^{an}$. There is a Leray spectral sequence <code>$$H^j(S^{an}, R^if^{an}_* \mathbb{C}_{Y^{an}}) \Rightarrow H^{i+j}(Y^{an}, \mathbb{C}_{Y^{an}})$$</code> which these isomorphisms identify with <code>$$H^j(S, (\mathcal{H}^i, \nabla^i_{GM})) \Rightarrow H^{i+j}_{dR}(Y/\mathbb{C}).$$</code> </p> <p>Fix $s\in S$, then $\pi_1(S, s)$ homotopically acts on $Y^{an}_s = (f^{an})^{-1}(s)$, hence you get an action on $H^i(Y^{an}_s, \mathbb{C})$, called the monodromy action. For $j=0$, you get should get <code>$$H^0(S^{an}, \mathcal{H}^{i, an})^{\nabla^{i, an}_{GM}} = H^0(S^{an}, (\mathcal{H}^{i, an}, \nabla^{i, an}_{GM})) = H^0(S^{an}, R^i f^{an}_* \mathbb{C}_{Y^{an}}) = H^i(Y^{an}_s, \mathbb{C})^{\pi_1(S, s)}.$$</code> </p> http://mathoverflow.net/questions/117670/finite-extension-of-projective-space/117677#117677 Answer by Piotr Achinger for Finite extension of projective space Piotr Achinger 2012-12-30T23:23:26Z 2012-12-30T23:23:26Z <p>I assembled the comments above into an answer:</p> <p>Assume that the field of definition is infinite (or allow finite extensions of the base field in the construction). Let $X$ be a projective scheme over a field $k$ and let $n$ be the dimension of $X$. Let $N$ be smallest such that there exists a finite morphism $X\to \mathbb{P}^N$. I claim that $N=n$. Suppose otherwise; let $f:X\to \mathbb{P}^N$ be finite. Since $N>n$, we can find a point $p\in\mathbb{P}^n$ not lying on $X':=f(X)$ (possibly after passing to a finite extension of $k$). Then $p$ defines a projection $\pi: \mathbb{P}^{n}\setminus\{p\} \to \mathbb{P}^{n-1}$ (can assume that $p=(0:\ldots:0:1)$, then $\pi(x_0:\ldots:x_{N-1}:x_N) = (x_0:\ldots:x_{N-1})$). This $\pi$ restricted to $X'$ is quasi-finite (otherwise, $X'$ would contain a line through $p$, hence $p$) and proper, hence finite. The composition $\pi\circ f: X\to \mathbb{P}^{N-1}$ is then finite as well - a contradiction.</p> http://mathoverflow.net/questions/117448/meaning-of-normalization/117450#117450 Answer by Piotr Achinger for meaning of normalization Piotr Achinger 2012-12-28T22:13:44Z 2012-12-28T22:19:57Z <p>Let $X$ be a variety (a separated integral scheme) with function field $K = k(X)$, maybe assumed normal. Let $L$ be a finite separable extension of $K$. From this data, we can construct a variety $Y$ with $k(Y) = L$ together with a finite surjective map $\pi: Y\to X$, called the <em>normalization of $X$ in $L$</em>. </p> <p>If $X$ is affine, equal to $\mathrm{Spec}(A)$, the construction is just $Y = \mathrm{Spec}(B)$ where $B$ is the <em>integral closure</em> of $A$ in $L$. By finiteness of integral closure, the resulting map $Y\to X$ is finite.</p> <p>In the general case, cover $X$ by open affines, their intersections are affine again, so we can glue and obtain $Y$ this way. If you don't like gluing, you could define a sheaf of $\mathcal{O}_X$-algebras $\mathcal{A}$ by $\mathcal{A}(U) =$ integral closure of $\mathcal{O}_X(U)$ (and maybe sheafify) and then define $Y = \mathrm{Spec}_X \mathcal{A}$.</p> <p>In your situation, you already have a part of $Y$ (you have a map $U\to X$ and define $L = k(U)$). This gives a map $Y\to X$ such that $U\to X$ factors through an open immersion $U\to Y$.</p> http://mathoverflow.net/questions/117043/flatness-criterion-on-normal-bases/117055#117055 Answer by Piotr Achinger for flatness criterion on normal bases Piotr Achinger 2012-12-23T00:25:36Z 2012-12-23T00:25:36Z <p>Let $X=Y$ be of characteristic $p$ and let $f:X\to Y$ be the (absolute) Frobenius morphism. Kunz's Regularity Criterion states that $f$ is flat if and only if $X$ is regular. In particular, taking $X$ to be Cohen-Macaulay and normal but not regular/smooth gives another counterexample.</p> http://mathoverflow.net/questions/116847/intuitive-pictures-in-characteristic-p/116850#116850 Answer by Piotr Achinger for Intuitive pictures in characteristic p Piotr Achinger 2012-12-20T10:46:34Z 2012-12-20T10:46:34Z <p>I don't think you can draw something meaningful - I would be surprised if someone made a good drawing of the Frobenius morphism ;).</p> <p>That being said, here is an example (possibly misleading or unrelated to your research) I saw in the slides of Benedict Gross's lectures on the arithmetic of hyperelliptic curves. Take a prime $p$, say $p=57$, and an equation of an hyperelliptic curve $y^2 = x^n + ax^{n-2} + \ldots$ with integer coefficients. Draw a $p\times p$ square and mark the solutions to the above equation mod $p$. The resulting picture exhibits the following:</p> <ul> <li>It is a mixture of chaos and geometry: there is a visible symmetry coming from the hyperelliptic involution $(x, y)\mapsto (x, -y)$.</li> <li>The solutions form a finite set, in particular, it makes combinatorial arguments possible. We can ask how many points are there and whether it gives us some "geometric" information. This is not obvious to someone from other fields, or a non-mathematician.</li> </ul> <p>You can include drawings of the same curve over $\mathbb{R}$ and $\mathbb{C}$. I think the equation $y^2 = \ldots$ and the three pictures together explain pretty well what algebraic geometry is about without going into too much detail.</p> http://mathoverflow.net/questions/115657/what-is-the-algebraic-geometry-version-of-the-spheres/115672#115672 Answer by Piotr Achinger for What is the algebraic geometry version of the spheres? Piotr Achinger 2012-12-07T03:00:32Z 2012-12-07T03:00:32Z <p>Probably not, at least treating the question naively, since algebraic geometry focuses on smooth projective varieties, and these have nonzero cohomology (for basically any cohomology theory) in degrees $0, 2, \ldots 2\cdot \dim$: the class of the hyperplane section in $H^2$ has nonzero top cup product with itself. </p> http://mathoverflow.net/questions/115030/relating-the-toric-rank-of-a-semistable-curve-and-the-first-betti-number-of-its-r/115035#115035 Answer by Piotr Achinger for Relating the toric rank of a semistable curve and the first Betti number of its reduction graph Piotr Achinger 2012-11-30T23:50:14Z 2012-11-30T23:50:14Z <p>I never studied this, but here is I guess the natural approach:</p> <p>Suppost that the reduction is just a cycle of smooth curves $C_1, \ldots, C_k$ with $C_i$ intersecting $C_{(i+1)\mod k}$ in one point $p_i = q_{i+1\mod k}$ with $p_i\neq q_i$. </p> <p>How do we construct a line bundle on this cycle? Such a bundle $L$ gives us by pullback a collection of line bundles $L_i$ on $C_i$, hence a (surjective) map $\tilde J\to A:= \prod Pic^0(C_i)$. What is the kernel? Suppose $L$ is trivial on every $C_i$ and choose a trivialization. This gives us isomorphisms $L_{p_i} \to L_{q_i}$. The composition $L_{p_1} \to L_{q_2} \to L_{p_2} \to \ldots \to L_{q_2} \to L_{p_1}$ is an automorphism of a $1$-dimensional vector space, i.e., an element $t$ of a $1$-dimensional torus $\mathbb{G}_m$. This is all the data we need.</p> <p>In case there are more cycles, we will get such a $t$ for every cycle, hence the dimension of the toric part of $\tilde J$ will be equal to the number of cycles in the graph.</p> http://mathoverflow.net/questions/114790/reference-request-base-point-freeness-of-2-theta/114804#114804 Answer by Piotr Achinger for Reference request: base point freeness of $2\Theta$ Piotr Achinger 2012-11-28T19:05:23Z 2012-11-28T19:05:23Z <p>If $D$ is an ample divisor on an abelian variety, then $2D$ is base point free and $3D$ is very ample. One reference for this is Mumford's book "Abelian Varieties", II 6 and III 17.</p> http://mathoverflow.net/questions/114724/depth-zero-high-dimension/114729#114729 Answer by Piotr Achinger for Depth zero, high dimension Piotr Achinger 2012-11-28T05:04:37Z 2012-11-28T05:04:37Z <p>Let $S = k[x_1, \ldots, x_n, y]/(x_1 y, x_2 y, \ldots, x_n y, y^2)$ and let $R$ be the local ring of $S$ at $0$. Then $\dim R = \dim S = n$, but there are no regular elements, since $y$ annihilates the maximal ideal of $R$ - so $R$ has depth zero.</p> http://mathoverflow.net/questions/114066/singular-points-on-the-hilbert-scheme-of-a-product/114072#114072 Answer by Piotr Achinger for Singular points on the Hilbert scheme of a product Piotr Achinger 2012-11-21T16:08:51Z 2012-11-21T16:08:51Z <p>First, rigidity lemma: let $X\times Y \leftarrow Z\to T$ be a flat family of closed subschemes of $X\times Y$, with the fiber $Z_0$ over $0\in T$ equal to $X\times {y}$. Then the composition $Z\to X\times Y \to Y$ contracts $Z_0$ to a point, therefore it has to contract nearby fibers by the Rigidity Lemma. </p> <p>I omitted a considerable amount of details here, but it should follow that the connected component of the Hilbert scheme corresponding to $P_y$ is isomorphic to $Y$, hence smooth.</p> <p>A good reference for the Rigidity Lemma(s) is Mumford's book on abelian varieties, chapters 2 and 3.</p> http://mathoverflow.net/questions/114034/learning-path-for-the-proof-of-the-weil-conjectures/114070#114070 Answer by Piotr Achinger for Learning path for the proof of the Weil Conjectures Piotr Achinger 2012-11-21T15:53:55Z 2012-11-21T15:53:55Z <p>I'm not an expert, but here is how I would plan my trip:</p> <p>There are obviously two parts: rationality + functional equation + comparison with Betti numbers (which follow from the construction of etale cohomology) and the Riemann hypothesis (which is much deeper).</p> <p>I would therefore follow the following plan:</p> <ol> <li>Learn classical algebraic topology, esp. singular cohomology, Poincare duality, Lefschetz fixed point formula, Leray-Serre spectral sequence.</li> <li>Learn an overview of $\ell$-adic cohomology, without technical details. First, understand how a good cohomology theory like in 1. above will prove the first part of the conjectures. Then understand etale topology and definition of $\ell$-adic cohomology groups and how the Frobenius morphism happens to act on them. </li> <li>Technical machinery underlying $\ell$-adic cohomology. I haven't studied this myself very well, but Milne's book seems to be a standard reference.</li> <li>Read Deligne's Weil I article. It's beautifully written and you don't need much more than 1. and 2. above. The main technical tool is the use of Lefschetz pencils, which is there just to make induction on dimension possible. You can just assume Lefschetz pencils exist, or look to SGA if interested. Note that in Deligne's approach it is crucial to work with constructible sheaves, not just the constant sheaf.</li> <li>Read Deligne's Weil II article. It reproves Weil I and adds much more, but is much longer and more difficult. </li> </ol> <p>Note: 3. and 4. above might be mostly independent. A very good reference is Katz's article "L-functions and monodromy: four lectures on Weil II".</p> http://mathoverflow.net/questions/113858/is-there-an-algebraic-geometry-analogue-of-the-closed-graph-theorem/113860#113860 Answer by Piotr Achinger for Is there an algebraic geometry analogue of the closed graph theorem? Piotr Achinger 2012-11-19T19:29:27Z 2012-11-19T19:29:27Z <p>You might be rediscovering Zariski's Main Theorem, which implies your statement in case $X$ is normal (or just weakly normal) and the projection from the graph $\Gamma$ to $X$ is proper and separable. What you really need is the map $\Gamma\to X$ to be an isomorphism, so the question is equivalent to "when is a bijective map an isomorphism"?</p> <p>You already explained why we need (weak) normality (example with the cuspidal curve) and separability (the Frobenius map). To see why properness is also necessary, look at the map $\mathbb{A}^1\to\mathbb{A}^1$ sending $x$ to $1/x$ for $x\neq 0$ and sending $0\to 0$, whose graph is a union of $(0,0)$ and a hyperbola $xy=1$.</p> http://mathoverflow.net/questions/112329/vector-bundles-on-normal-surfaces/112330#112330 Answer by Piotr Achinger for Vector Bundles on normal surfaces Piotr Achinger 2012-11-14T00:19:45Z 2012-11-14T00:36:23Z <p>Take a singular quadric cone $C$ , then a ruling $L$ of the cone is not a Cartier divisor, hence $\mathcal{O}_C(L)$ is not locally free, but it becomes locally free after removing the vertex of the cone.</p> <p>In any case, the push-forward of a vector bundle from the smooth locus of a normal surface to the surface will be a reflexive coherent sheaf, so you might want to read more about reflexive sheaves and criteria for them being locally free. </p> <p><strong>EDIT.</strong> Regarding principal $G$-bundles, there is a theory of "principal G-sheaves" (Gómez-Sols) and "singular principal G-bundles" (A. Schmitt) and moduli spaces of these have been constructed (at least in the smooth case). So I guess principal $G$-bundles on the smooth locus will extend to this kind of objects on the normal surface.</p> http://mathoverflow.net/questions/111685/on-some-finiteness-properties-for-schemes/111704#111704 Answer by Piotr Achinger for On some finiteness properties for schemes Piotr Achinger 2012-11-07T05:43:05Z 2012-11-08T16:59:05Z <p><strong>A implies B.</strong> True, as you said, because a finitely generated ring is Noetherian, and $X$ is glued from finitely many spectra of such.</p> <p><strong>A implies C.</strong> True (argument as above).</p> <p><strong>B implies A.</strong> False, e.g. $X = \mathrm{Spec }\ \mathbb{Q}$.</p> <p><strong>B implies C.</strong> False (I believe). There are rings $R$ whose spectrum is homeomorphic to the topological space $\{1, 2, \ldots \}$ with open sets $\{n, n+1, \ldots\}$, which is Noetherian but of infinite Krull dimension. I think something like $\mathrm{Spec }\ k[x_1, x_1 x_2, x_1 x_2 x_3, \ldots]$ should work, but I didn't check the details. <strong>EDIT.</strong> This is nonsense - see the comments below and Fred Rohrer's answer.</p> <p><strong>C implies A.</strong> False, e.g. $X= \mathrm{Spec }\ \mathbb{Q}$.</p> <p><strong>C implies B.</strong> False, e.g. $X = \mathrm{Spec}\ k[x, x^{1/2}, x^{1/3}, \ldots]$. </p> http://mathoverflow.net/questions/111614/does-a-section-having-no-zero-locus-divisor-imply-that-it-is-nowhere-vanishing/111615#111615 Answer by Piotr Achinger for Does a section having no zero locus divisor imply that it is nowhere vanishing? Piotr Achinger 2012-11-06T04:46:29Z 2012-11-06T04:46:29Z <p>Yes, a normal domain equals the intersection of its localizations at height one primes inside its fraction field. Hartshorne ("Algebraic Geometry", Prop. II 6.3A) gives the reference: Matsumura "Commutative Algebra", Th. 38, p. 124. We apply this to $s$ and $1/s$.</p> http://mathoverflow.net/questions/111004/good-reduction-for-singular-varieties/111219#111219 Answer by Piotr Achinger for "Good reduction" for singular varieties Piotr Achinger 2012-11-01T21:43:54Z 2012-11-01T21:43:54Z <p>How about the condition: $X/S$ has "good reduction" if the scheme-theoretic singular locus is flat over $S$?</p> http://mathoverflow.net/questions/109649/equality-of-rational-maps/109650#109650 Answer by Piotr Achinger for Equality of rational maps Piotr Achinger 2012-10-14T20:13:54Z 2012-10-14T21:26:19Z <p>The answer all your questions is no. Let $X = \mathbb{P}^1$ and $N=2$. Let $f:X\to \mathbb{P}^2$ be $f(x:y) = (x:y:0)$ and $g:X\to \mathbb{P}^2$ be $g(x:y) = (x^2y:x^3:y^3)$. Then $f$ and $g$ are both bijective, but the image of $g$ is the singular curve $x_0^3 = x_1^2 x_2$.</p> <p><strong>EDIT</strong>. My guess is that when $f$ and $g$ are regular and both images are <em>normal</em> then the answer that $f(X)$ and $g(X)$ are isomorphic could be yes thanks to Zariski's Main Theorem, but I don't see an argument that would show this.</p> <p><strong>EDIT</strong>. Suppose that $f$ and $g$ are both regular. Let $Y_0 = f(X)$ and $Y_1 = g(X)$. Denote the image of $X$ in $Y_0\times Y_1$ under $(f, g)$ by $Z$. Then by your assumption $Z$ projects bijectively onto both $Y_0$ and $Y_1$. If $Y_0$ and $Y_1$ are normal, then by ZMT we have $Y_0 = Z = Y_1$. In any case (even if $f$ and $g$ are rational) we get that $Y_0$ and $Y_1$ are birational.</p> http://mathoverflow.net/questions/109430/projectives-in-the-category-of-coherent-sheaves-on-a-projective-variety/109432#109432 Answer by Piotr Achinger for Projectives in the category of coherent sheaves on a projective variety Piotr Achinger 2012-10-12T05:41:17Z 2012-10-12T06:03:34Z <p>This is never true whenever $X$ has positive dimension. Let $L$ be ample on $X$ and let $E$ be a nonzero coherent sheaf on $X$. Let $P$ be any point of $X$ at which $E$ has a nonzero fiber, so we get a surjection $\mathcal{O}_X \to \mathcal{O}_P$ ($\mathcal{O}_P$ being the skyscraper sheaf at $P$). We can also find a $k>0$ such that the sheaf $\mathrm{Hom}(E, E\otimes L^{-k})= \mathrm{End}(E)\otimes L^{-k}$ has no nonzero global sections. Now tensor the surjection $\mathcal{O}_X\to \mathcal{O}_P$ by $E$ and $E\otimes L^{-k}$, getting surjections $a:E\to E_P$ and $b:E\otimes L^{-k}\to (E\otimes L^{-k})_P = E_P$. We cannot lift $a$ along $b$ because by assumption on $k$ there are no nonzero maps $E\to E\otimes L^{-k}$.</p> http://mathoverflow.net/questions/129993/shafarevichs-theorem-for-elliptic-curves-defined-over-function-field-of-algebrai Comment by Piotr Achinger Piotr Achinger 2013-05-08T01:37:38Z 2013-05-08T01:37:38Z Hmm 2) should be obvious because the automorphism groups are bounded... http://mathoverflow.net/questions/129993/shafarevichs-theorem-for-elliptic-curves-defined-over-function-field-of-algebrai Comment by Piotr Achinger Piotr Achinger 2013-05-08T01:35:14Z 2013-05-08T01:35:14Z @nosr: nice answer! Why post it as comment? There are however two points I didn't understand: 1) why is the degree of $U'\to U$ universally bounded? 2) your proof shows that given $p:E\to U$ we can associate to it a pair $(f:U' \to U, g:U'\to X_1(1728))$ and that there are finitely many such pairs, but I fail to see why there can't be infinitely many $p:E\to U$ to which the same pair $(f, g)$ is assigned - i.e. there could be many ways of descending $g*(\mathcal{E})$ along $f$, where $\mathcal{E}\to X_1(1728)$ is the universal curve. http://mathoverflow.net/questions/129548/group-set-with-1-op-field-with-2-ops-with-3-ops Comment by Piotr Achinger Piotr Achinger 2013-05-03T16:45:12Z 2013-05-03T16:45:12Z What about $A_\infty$-algebras? http://mathoverflow.net/questions/127539/on-flat-morphisms Comment by Piotr Achinger Piotr Achinger 2013-04-14T16:17:39Z 2013-04-14T16:17:39Z There is no reason for $f$ to be flat outside $U$. For example $S=\mathbb{A}^1=U$, $X = \mathbb{A}^1 \sqcup pt$. http://mathoverflow.net/questions/127030/reals-with-integer-powers-bounded-away-from-integers Comment by Piotr Achinger Piotr Achinger 2013-04-10T01:14:00Z 2013-04-10T01:14:00Z Regarding your nickname: it's time. http://mathoverflow.net/questions/126217/name-for-a-class-of-parabolic-subgroups Comment by Piotr Achinger Piotr Achinger 2013-04-02T01:10:02Z 2013-04-02T01:10:02Z Are those the same as minuscule or cominuscule parabolic subgroups? http://mathoverflow.net/questions/125863/relatively-numerically-trivial-divisor Comment by Piotr Achinger Piotr Achinger 2013-03-29T03:18:55Z 2013-03-29T03:18:55Z Related II: maybe we could ask first: &quot;Suppose $f:X\to Y$ is a projective morphism (no assumptions on $Rf_* O_X$), $E$ is an invertible sheaf on $X$ which is <b>trivial</b> on every fiber. Does it follow that $E = f^* F$ for an invertible sheaf $F$ on $Y$?&quot; http://mathoverflow.net/questions/125863/relatively-numerically-trivial-divisor Comment by Piotr Achinger Piotr Achinger 2013-03-29T03:16:19Z 2013-03-29T03:16:19Z Related: is it true that the set of $y$ in $Y$ s.t. $E$ is trivial on $X_y$ is closed in $Y$? Probably not (as you said, the usual results i.e. &quot;Seesaw theorem&quot; require flatness) and maybe a counterexample to this would give a counterexample to your question? http://mathoverflow.net/questions/125401/existence-of-non-split-vector-bundles-on-smooth-projective-varieties/125483#125483 Comment by Piotr Achinger Piotr Achinger 2013-03-24T20:33:46Z 2013-03-24T20:33:46Z Perfect, thanks!! http://mathoverflow.net/questions/125401/existence-of-non-split-vector-bundles-on-smooth-projective-varieties Comment by Piotr Achinger Piotr Achinger 2013-03-24T03:57:48Z 2013-03-24T03:57:48Z @Mahdi: yes. One idea of construction would be to embed $X$ into $\mathbb{P}^N$ and restrict $\Omega^1_{\mathbb{P}^N}$ to $X$ (or pull it back along a finite flat map $f:X\to \mathbb{P}^n$), but I don't know if it can work... http://mathoverflow.net/questions/124999/smooth-function-algebra-on-cartesian-product-and-beyond Comment by Piotr Achinger Piotr Achinger 2013-03-19T18:35:00Z 2013-03-19T18:35:00Z Assume $M$, $N$ compact. Isn't then $C^\infty(M\times N, \mathbb{R})$ the completion of $C^\infty(M, \mathbb{R}) \otimes C^\infty(N, \mathbb{R})$ with respect to the supremum norm? http://mathoverflow.net/questions/124802/decomposition-theorem-for-principally-polarized-abelian-varieties-in-positive-cha Comment by Piotr Achinger Piotr Achinger 2013-03-17T18:32:20Z 2013-03-17T18:32:20Z Hi Marc. Maybe it would be helpful if you mentioned why the proof known in characteristic zero doesn't work in positive characteristic. http://mathoverflow.net/questions/123568/can-one-determine-the-local-structure-of-a-moduli-space-of-bundles-just-by-knowin Comment by Piotr Achinger Piotr Achinger 2013-03-04T21:12:04Z 2013-03-04T21:12:04Z Being smooth and unobstructed are the same. Why is $\dim(M_2) = 0$? With $\dim (M_2) \leq 1$, $M_2$ is either a smooth curve or equal to $Spec(k[x]/(x^n))$ for some $n&gt;1$ around $E$. I think finding $n$ might be difficult. http://mathoverflow.net/questions/123427/coherent-sheaves-and-holomorphic-vector-bundles Comment by Piotr Achinger Piotr Achinger 2013-03-02T20:12:47Z 2013-03-02T20:12:47Z Locally free sheaves don't form an abelian category, and the abelian category &quot;generated&quot; by them is the category of coherent sheaves. http://mathoverflow.net/questions/123427/coherent-sheaves-and-holomorphic-vector-bundles Comment by Piotr Achinger Piotr Achinger 2013-03-02T19:08:27Z 2013-03-02T19:08:27Z Think of a complex submanifold $N\subset M$ and the trivial bundle $O_N$ on $N$, thought of as a sheaf on $M$. This is a coherent sheaf on $M$, but of course does not come from a vector bundle. Another example: the ideal sheaf of $N$ (sections of the trivial bundle $O_M$ on $M$ which vanish along $N$) will not be a vector bundle if $N$ has codimension $&gt;1$, but will still be coherent.