Ariyan Javanpeykar
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 1d answered Families of abelian varieties on the line (or more generally simply connected varieties) 1d comment Families of abelian varieties on the line (or more generally simply connected varieties) @PiotrAchinger The isotriviality of families of ordinary abelian varieties is also proven in Moret-Bailly's 1985 asterisque 129 "Pinceaux de varietes abeliennes" in Thm. 5.1 (see also Thm. 5.2) on page 237 Chapter XI. Note that Moret-Bailly attributes Thm. 5.1 to Raynaud. I guess the paper of Nori-Srinivas came out at the same time. Apr 17 comment Finite group action on quasi-projective varieties You could have a look at Liu's book on Algebraic Geometry, and specifically Chapter 4 in which he discusses flat morphisms, etale morphisms and smooth morphisms. Of course, the same is done in Hartshorne's book. Apr 17 comment Finite group action on quasi-projective varieties The morphism $X\to X/G$ is finite. Therefore, it is generically etale if and only if the extension of function fields $K(X/G) \subset K(X)$ is separable. This is the case if the base field (which you didn't specify) is of characteristic zero, because fields of characteristic zero are perfect. Apr 16 revised Does X(13) have potentially good reduction at 13? Emphasized the role of Neron models for hyperbolic curves Apr 16 comment Reference for Superelliptic Curves Superelliptic curves are also called "cyclic curves". You might have more luck finding references searching for "cyclic curves". Apr 14 comment Uniqueness of smooth compactification upto a smooth morphism The answer is no, I think. Take $X$ to be a smooth projective minimal surface (of positive Kodaira dimension to be safe). Let $D$ be a smooth irreducible curve in $X$, and let $U =X\setminus D$. Then $X$ is a smooth compactification of $U$. Any other smooth compactification of $U$, $X'$ say, will map uniquely to $X$, and this morphism is smooth if and only if it is an isomorphism. Is that convincing? Apr 13 comment Uniqueness of smooth compactification upto a smooth morphism Let $\bar{X}$ be a smooth compactification of $X$ with (non-empty) boundary $D$. Now, blow-up a point in $D$. Write $\bar{X}'\to \bar{X}$ for this blow-up. Then $\bar{X}'$ is another compactification of $X$. This is your $g$, right? This morphism $g$ is not smooth. So what am I missing in your question? Apr 13 awarded Notable Question Apr 7 revised Shafarevich conjecture for abelian varieties deleted 13 characters in body Apr 6 answered Shafarevich conjecture for abelian varieties Apr 3 comment Equivariant Riemann-Roch on DM stacks? You can also consider e-mailing A. Krishna tifr.res.in/People_Finder/compcode.php?param1=39 I think he's working on generalizing work of Edidin; see his abstract here maths-people.anu.edu.au/~alperj/kioloa-abstracts.html#Krishna Apr 3 comment Equivariant Riemann-Roch on DM stacks? Have you had a look at Edidin's paper: arxiv.org/pdf/1205.4742.pdf ? Mar 30 revised Examples of varieties with many automorphisms acting trivially on co-homology added 325 characters in body Mar 30 comment Examples of varieties with many automorphisms acting trivially on co-homology @DavidSpeyer Ok. My reasoning for why $H^2$ is one-dimensional (so that $X$ has Picard rank one) was flawed. So this example doesn't work. But there might be still hope in finding a CY threefold which does the trick. Indeed, there are examples of CY threefolds with Picard number $4$ with infinite automorphism group. They provide possible candidates. But I just realized they don't because of Prop. 2.4 in arxiv.org/pdf/1206.1649v3.pdf Mar 30 comment Examples of varieties with many automorphisms acting trivially on co-homology @DavidSpeyer Thank you for your comment. I thought this would work because the H^{1,1} equals the H^2 and is one-dimensional. So $G$ is mapping to the finite group $GL_1(\mathbb Z)$. Are you saying that $G$ is in fact finite in this example? Mar 30 answered Examples of varieties with many automorphisms acting trivially on co-homology Mar 28 answered Reduction of Abelian Varieties with Complex Multiplication have Complex Multiplication Mar 27 answered Smooth algebraic stacks with precisely two $\mathbb C$-objects Mar 27 awarded Nice Answer