Ariyan Javanpeykar
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 2d reviewed Approve Relationship between group $C^\ast$-algebras $C^\ast(G)$ and graph $C^\ast$-algebras $C^\ast(E)$ Jan 26 comment Fiberwise criterion for a stack to be a gerbe @Qfwfq That can certainly happen. Take $S$ to be the spectrum of a field (containing $\mathbb C$) such that there is a smooth conic $C$ over $S$ with no section. Let $G$ be the automorphism group scheme of $C$ over $S$. Then it follows that $BG = BPGL_{2,S}$, with $G$ not isomorphic to $PGL_{2,S}$. If you also want $S$ to be a finite type, then you will have to find a variety $S$ and a smooth conic with no section over $S$. This can be a bit more difficult (because of Tsen's theorem), but it should still be possible if $S$ is of dimension bigger than 1. Jan 26 comment Fiberwise criterion for a stack to be a gerbe @Qfwfq I don't think so. As is explained in Lecture 3 of these notes pub.math.leidenuniv.nl/~holmesdst/Vistoli_torsors.pdf it follows that the group scheme $H$ is an inner form of $G$. But all inner forms are trivial over $\mathbb C$ (i.e., isomorphic to $G$). (I think that should be right, although I didn't really check my last claim. Nonetheless it should follow from the fact that all $G$-torsors are trivial over Spec $\mathbb C$, as they have sections.) What happens in the example over $\mathbb Q$ is that there are non-trivial $PGL_{2,\mathbb Q}$-torsors (non-split smooth conics). Jan 25 comment Fiberwise criterion for a stack to be a gerbe The answer to Q1 should be positive, as checking whether a morphism of algebraic spaces is (formally) unramified can be done on geometric points. Jan 25 comment Fiberwise criterion for a stack to be a gerbe @Qfwfq You are right that $\mu_3$ are $\mathbb Z/3$ are non-isomorphic over $\mathbb Q$. Indeed, the group $\mu_3$ has no non-trivial sections over $\mathbb Q$. On the other hand, it is quite common for two neutral gerbes to be isomorphic even when the relevant group schemes are not. For a geometric example: consider the stack of smooth conics $B(PGL_2)$ over $\mathbb Q$. Let $X$ be a non-split smooth conic over $\mathbb Q$ with automorphism group $G$. Then $B(PGL_2)$ is isomorphic to $BG$, but $G$ is not isomorphic to $PGL_2$. (Here I write $PGL_2$ for what I should write $PGL_{2,\mathbb Q}$. Jan 25 reviewed Approve Dual Borel conjecture in Laver's model Jan 25 comment Curves of higher genus The Fermat curves $x^n+y^n+z^n =0$ are candidates. Their genus grows and their Jacobians have many extra automorphisms. For the Jacobian of a curve to be CM its dimension should match up with the rank of its endomorphism algebra, and that doesn't happen here for large $n$. On the other hand, the Jacobian of the Fermat curve does have CM quotients. This fact is exploited very often; see for instance Section III in jstor.org/stable/2946559?seq=1#page_scan_tab_contents or Lang's book on Complex Multiplication. Jan 5 awarded Popular Question Nov 20 comment Definition of étale (etc) for non-representable morphisms of algebraic stacks? Sorry for that typo. My guess is that the same definition should go for etale and finite. (I guess one should show that if you have $X\to Y \to Z$ such that $X\to Y$ is P and $X\to Z$ is P then $Y\to Z$ is P. If that's the case, then I think this definition makes sense.) Nov 20 comment Definition of étale (etc) for non-representable morphisms of algebraic stacks? I think $X\to Y$ is smooth if for all schemes $U$ and all (representable) smooth $U\to Y$, the (representable) morphism $U\to Y$ is smooth. Similar definition should go for etale. I read this in Toric Stacks II, p. 8 arxiv.org/pdf/1107.1907.pdf Nov 19 awarded Nice Answer Nov 13 awarded Popular Question Nov 8 comment curve over higher dimensional basis with 0-dimensional locus of bad reduction My apologies. I now see the point. Nov 8 comment What's wrong with my understanding of the scheme $\text{Isom}(E_\lambda, E_{\lambda'})$? @BenLim I can't see where your mistake is in your computation at the moment. In any case, there has to be a mistake because $Isom_S(E,E^\prime)\to S$ is a finite unramified morphism of schemes, whenever $S$ is a scheme and $E$ and $E^\prime$ are elliptic curves over $S$. (Note that $Isom_S(E,E^\prime)$ is the sheaf of isomorphisms respecting the zero section.) Are you sure you are computing the scheme-theoretic fibre correctly? Nov 8 comment What's wrong with my understanding of the scheme $\text{Isom}(E_\lambda, E_{\lambda'})$? @BenLim (Here's me trying to explaining degrees.) The map $\{pt\} \to [\{pt\}/G]$ is of degree $\# G$. The composition with the coarse map $[\{pt\}/G] \to \{pt\}$ is the identity. So this means that the "degree" of $BG \to \{pt\}$ is $1/\# G$. You could think of the "point" of $BG$ as a $1/\# G$-th point. There are some other answers on Mathoverflow explaining this. Look up "groupoid cardinality" for instance. Nov 8 comment What's wrong with my understanding of the scheme $\text{Isom}(E_\lambda, E_{\lambda'})$? @BenLim Concerning your confusion: The Legendre map is etale. An elliptic curve in Legendre form has rational $2$-torsion, and any isomorphism of an elliptic curve respecting all of the $2$-torsion points has to be $\pm 1$. A proof of this can be found in Katz-Mazur. (Compare this to the fact that an automorphism of an elliptic curve respecting all of the $n$-torsion points ($n>2$) has to be trivial.) Nov 8 comment What's wrong with my understanding of the scheme $\text{Isom}(E_\lambda, E_{\lambda'})$? @BenLim Concerning degrees of maps of stacks: consider $G$ a finite group. What would you say the degree of $\{pt\} \to BG$ is? And what about $BG\to \{pt\}$? Nov 8 comment curve over higher dimensional basis with 0-dimensional locus of bad reduction ...More precicely, for all schemes $S$ and for all $X$ and $Y$ over $S$ (in $\mathcal M_g(S)$), the morphism $\mathrm{Isom}_S(X,Y)\to S$ is proper. (If $g\geq 2$, this morphism is finite. But this is not the case when $g=1$.) The given isomorphism $X_U \cong X^\prime_U$ induces a section of $Isom_S(X,Y)$ over $U$. Since $S$ is regular (noetherian) this "generic section" extends to a section over $S$. See Gabber-Liu-Lorenzini Prop 6.2 math.u-bordeaux1.fr/~qliu/articles/GLL2-Duke.pdf . Nov 8 comment curve over higher dimensional basis with 0-dimensional locus of bad reduction @TimoKeller In the last part of Moret-Bailly's answer one can also argue as follows. Let $S$ be an integral noetherian regular scheme. Assume $g\geq 1$ and assume that $X_U\cong X_U^\prime$, where $X$ and $X^\prime$ are curves over $S$ in $\mathcal M_g(S)$ and $U\subset S$ is a dense open. The fact that this isomorphism extends to an isomorphism of $X$ and $X^\prime$ over $S$ follows from the separatedness of the stack $\mathcal M_g$ of smooth proper curves of genus $g$... Nov 8 answered What's wrong with my understanding of the scheme $\text{Isom}(E_\lambda, E_{\lambda'})$?