# Piotr Achinger

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 Name Piotr Achinger Member for 3 years Seen 1 hour ago Website Location Berkeley Age 26
PhD student, UC Berkeley
 May8 comment Shafarevich’s theorem for elliptic curves defined over function field of algebraic curve over algebraically closed fieldHmm 2) should be obvious because the automorphism groups are bounded... May8 comment 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. Apr14 comment on flat morphismsThere is no reason for $f$ to be flat outside $U$. For example $S=\mathbb{A}^1=U$, $X = \mathbb{A}^1 \sqcup pt$. Apr10 comment Reals with integer powers bounded away from integers?Regarding your nickname: it's time. Apr6 answered Linearly generated embedding? Apr2 comment Name for a class of parabolic subgroupsAre those the same as minuscule or cominuscule parabolic subgroups? Mar29 comment Relatively numerically trivial divisorRelated 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 $Y$?" Mar29 comment Relatively numerically trivial divisorRelated: 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? Mar26 revised Line bundles on K3 surfacestags Mar24 comment Existence of non-split vector bundles on smooth projective varietiesPerfect, thanks!! Mar24 awarded ● Nice Question Mar24 comment 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... Mar23 asked Existence of non-split vector bundles on smooth projective varieties Mar19 comment 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? Mar17 comment 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. Mar9 answered How many flat connections has a line bundle in algebraic geometry? Mar4 comment 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. Mar2 comment Coherent Sheaves and Holomorphic Vector BundlesLocally free sheaves don't form an abelian category, and the abelian category "generated" by them is the category of coherent sheaves. Mar2 comment Coherent Sheaves and Holomorphic Vector BundlesThink 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. Mar2 comment Coherent Sheaves and Holomorphic Vector BundlesDear 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. Feb24 comment cohomology of restrictions of vector bundles to deformationsIn 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. Feb24 comment cohomology of restrictions of vector bundles to deformationsMy 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. Feb24 revised cohomology of restrictions of vector bundles to deformationsThere was a mistake: the first s.e.s. is not a s.e.s. of O_X-modules Feb24 awarded ● Organizer Feb24 revised cohomology of restrictions of vector bundles to deformationstags Feb24 revised cohomology of restrictions of vector bundles to deformationsadded 877 characters in body Feb23 revised cohomology of restrictions of vector bundles to deformationsadded 753 characters in body; added 10 characters in body; deleted 759 characters in body Feb23 answered cohomology of restrictions of vector bundles to deformations Feb23 comment cohomology of restrictions of vector bundles to deformationsHow 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 $H^i(X, E) = 0$. Feb23 comment cohomology of restrictions of vector bundles to deformationsDear Eric, what is the generic fiber of $X'$ ($X'$ is not reduced)? Feb20 accepted Algebraic surface of a line arrangement Feb19 comment Reference Request: Vector bundles in rigid analytic geometryI thought this correspondence works the same way for most kinds of ringed spaces. Are there extra difficulties in your setting? Feb19 answered Algebraic surface of a line arrangement Feb13 comment Recovering an abelian category out of its derived categoryDear 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. Feb11 comment 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! Feb11 accepted does there exist a family of objects over the tangent space to the base space of a family of objects? Feb11 comment Minimal semistable model for K3-surfaces.I don't think the minimal model program has anything to do with your question. Feb8 accepted Theta group representation Feb8 answered Theta group representation Feb7 comment 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. Feb6 revised 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 Feb6 revised 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 Feb6 answered does there exist a family of objects over the tangent space to the base space of a family of objects? Feb6 comment Symmetric power of tangent spaceWhere does the second one come from? Feb6 awarded ● Yearling Jan31 comment When does $Aut(X)=Bir(X)$ hold?Milne's notes on abelian varieties, p. 15. Jan31 answered When does $Aut(X)=Bir(X)$ hold? Jan30 comment Formal criterion of flatness@ayanta: Good point! Could you expand that a bit? Jan30 comment Formal criterion of flatnessDear 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 $n$. Jan30 asked Formal criterion of flatness