Piotr Achinger
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Registered User
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PhD student, UC Berkeley
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May 8 |
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Shafarevich’s theorem for elliptic curves defined over function field of algebraic curve over algebraically closed field Hmm 2) should be obvious because the automorphism groups are bounded... |
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May 8 |
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Shafarevich’s theorem for elliptic curves defined over function field of algebraic curve over algebraically closed field @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. |
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Apr 14 |
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on flat morphisms There is no reason for $f$ to be flat outside $U$. For example $S=\mathbb{A}^1=U$, $X = \mathbb{A}^1 \sqcup pt$. |
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Apr 10 |
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Reals with integer powers bounded away from integers? Regarding your nickname: it's time. |
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Apr 6 |
answered | Linearly generated embedding? |
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Apr 2 |
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Name for a class of parabolic subgroups Are those the same as minuscule or cominuscule parabolic subgroups? |
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Mar 29 |
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Relatively numerically trivial divisor Related II: maybe we could ask first: "Suppose $f:X\to Y$ is a projective morphism (no assumptions on $Rf_* O_X$), $E$ is an invertible sheaf on $X$ which is trivial on every fiber. Does it follow that $E = f^* F$ for an invertible sheaf $F$ on $Y$?" |
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Mar 29 |
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Relatively numerically trivial divisor 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. "Seesaw theorem" require flatness) and maybe a counterexample to this would give a counterexample to your question? |
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Mar 26 |
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Line bundles on K3 surfaces tags |
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Mar 24 |
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Existence of non-split vector bundles on smooth projective varieties Perfect, thanks!! |
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Mar 24 |
awarded | ● Nice Question |
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Mar 24 |
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Existence of non-split vector bundles on smooth projective varieties @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... |
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Mar 23 |
asked | Existence of non-split vector bundles on smooth projective varieties |
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Mar 19 |
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Smooth function algebra on cartesian product and beyond 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? |
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Mar 17 |
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Decomposition theorem for principally polarized abelian varieties in positive characteristic. Hi Marc. Maybe it would be helpful if you mentioned why the proof known in characteristic zero doesn't work in positive characteristic. |
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Mar 9 |
answered | How many flat connections has a line bundle in algebraic geometry? |
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Mar 4 |
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Can one determine the local structure of a moduli space of bundles just by knowing the Ext-groups? 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 $n>1$ around $E$. I think finding $n$ might be difficult. |
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Mar 2 |
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Coherent Sheaves and Holomorphic Vector Bundles Locally free sheaves don't form an abelian category, and the abelian category "generated" by them is the category of coherent sheaves. |
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Mar 2 |
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Coherent Sheaves and Holomorphic Vector Bundles 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 $>1$, but will still be coherent. |
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Mar 2 |
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Coherent Sheaves and Holomorphic Vector Bundles Dear John, coherent sheaves are locally cokernels of holomorphic maps between two holomorphic vector bundles. That is, $F$ is coherent if and only if locally on $X$, $F$ is the cokernel of a map $O_X^a \to O_X^b$ between two (trivial) vector bundles of finite ranks $a$, $b$. Hope this helps. |
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Feb 24 |
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cohomology of restrictions of vector bundles to deformations In particular this would show that the boundary maps are zero if $H^1(X, O_X)=0$ e.g. $X$ is simply connected. I'm really not sure if my guess is correct, let me know if you can figure it out. |
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Feb 24 |
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cohomology of restrictions of vector bundles to deformations My guess is that 1) $s$ gives you an element $c$ of $H^1(X, O_X)$ (apply $Hom(-, O_X)$ to the extension $0\to O_X \to P\to N^\vee \to 0$ where $P = O_Y/I_X^2$, $N^\vee = I_X/I_X^2$, getting a connecting homomorphism $\delta: H^0(X, N)\to H^1(X, O_X)$ and let $t = \delta(s)$), 2) the boundary maps $H^i(X, E) \to H^{i+1}(X, E)$ are just cup product with $c$. It seems plausible, but I didn't check it. |
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Feb 24 |
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cohomology of restrictions of vector bundles to deformations There was a mistake: the first s.e.s. is not a s.e.s. of O_X-modules |
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Feb 24 |
awarded | ● Organizer |
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Feb 24 |
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cohomology of restrictions of vector bundles to deformations tags |
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Feb 24 |
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cohomology of restrictions of vector bundles to deformations added 877 characters in body |
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Feb 23 |
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cohomology of restrictions of vector bundles to deformations added 753 characters in body; added 10 characters in body; deleted 759 characters in body |
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Feb 23 |
answered | cohomology of restrictions of vector bundles to deformations |
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Feb 23 |
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cohomology of restrictions of vector bundles to deformations How about tensoring the exact sequence $0\to \mathcal{O}_X \to \mathcal{O}_{X'}\to \mathcal{O}_X\to 0$ with $E$ and looking at the long exact sequence? You get a lot of information, e.g. that $H^i(X', E) = 0$ if $H^i(X, E) = 0$. |
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Feb 23 |
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cohomology of restrictions of vector bundles to deformations Dear Eric, what is the generic fiber of $X'$ ($X'$ is not reduced)? |
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Feb 20 |
accepted | Algebraic surface of a line arrangement |
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Feb 19 |
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Reference Request: Vector bundles in rigid analytic geometry I thought this correspondence works the same way for most kinds of ringed spaces. Are there extra difficulties in your setting? |
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Feb 19 |
answered | Algebraic surface of a line arrangement |
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Feb 13 |
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Recovering an abelian category out of its derived category Dear Jacob, you can recover a variety with ample or anti-ample canonical bundle from its derived category alone, you don't need the canonical bundle. |
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Feb 11 |
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Minimal semistable model for K3-surfaces. I think you're right. I never heard of MMP in such context (I thought MMP dealt with birational classification of varieties, not with models over DVRs, but I guess they are related somehow). Sorry for confusion! |
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Feb 11 |
accepted | does there exist a family of objects over the tangent space to the base space of a family of objects? |
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Feb 11 |
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Minimal semistable model for K3-surfaces. I don't think the minimal model program has anything to do with your question. |
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Feb 8 |
accepted | Theta group representation |
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Feb 8 |
answered | Theta group representation |
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Feb 7 |
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does there exist a family of objects over the tangent space to the base space of a family of objects? What I think Dan meant is: if the families you are considering are families of elliptic curves with fixed basis of $n$-torsion, the moduli space will be the modular curve $X(n)$ (or rather its open subset $Y(n)$). After removing finitely many points, it will be a fine moduli space. As computed here en.wikipedia.org/wiki/Modular_curve , the genus of $X(n)$ is nonzero for most $n$. Since $X(n)$ is smooth, for such $n$ there are no non-constant maps $\mathbb{A}^1\to X(n)$, that is, every family of the considered type over $\mathbb{A}^1$ has to be constant. |
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Feb 6 |
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does there exist a family of objects over the tangent space to the base space of a family of objects? deleted 2 characters in body |
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Feb 6 |
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does there exist a family of objects over the tangent space to the base space of a family of objects? added 415 characters in body |
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Feb 6 |
answered | does there exist a family of objects over the tangent space to the base space of a family of objects? |
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Feb 6 |
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Symmetric power of tangent space Where does the second one come from? |
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Feb 6 |
awarded | ● Yearling |
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Jan 31 |
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When does $Aut(X)=Bir(X)$ hold? Milne's notes on abelian varieties, p. 15. |
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Jan 31 |
answered | When does $Aut(X)=Bir(X)$ hold? |
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Jan 30 |
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Formal criterion of flatness @ayanta: Good point! Could you expand that a bit? |
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Jan 30 |
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Formal criterion of flatness Dear Daniel, thank you for the comment! I was not aware of this valuative criterion. What I meant is that a f.g. $M$ is flat over a local ring $(R,m)$ iff $M/m^n$ is flat over $R/m^n$ for all $n$. |
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Jan 30 |
asked | Formal criterion of flatness |
