Timeline for Is the automorphism group of a normal affine scheme a group scheme or an algebraic space?
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| Sep 13, 2021 at 13:55 | comment | added | Jason Starr | There is a cardinality argument that proves that the group functor is not represented by a scheme (unless we take the quotient by all possible nilpotent structures, as indicated by Friedrich Knop below). The usual group functor has $k[\epsilon]/\epsilon^2$-points (at the group identity) that form a countably infinite dimensional $k$-vector space. Yet if the functor is representable, then this equals the $k$-vector space dual of $\mathfrak{m}/\mathfrak{m}^2$. As the dual of an infinite-dimensional $k$-vector space, this should have uncountable dimension. | |
| Sep 13, 2021 at 12:21 | comment | added | Ben McKay | I remember the example of the ``shear'' $(x,y)\mapsto(x,y+p(x))$ where $p(x)$ is any polynomial, an infinite dimensional family of automorphisms of the plane. | |
| Sep 13, 2021 at 12:16 | answer | added | Friedrich Knop | timeline score: 6 | |
| Sep 13, 2021 at 4:36 | comment | added | abx | You might have a look at On some infinite-dimensional groups. II by Shafarevich, Mathematics of the USSR-Izvestiya, 1982, 18:1, 185–194. He defines a notion of "infinite-dimensional algebraic groups", which includes for instance $\operatorname{Aut}(\mathbb{A}^n) $ for $n\geq 2$. | |
| Sep 12, 2021 at 23:04 | comment | added | Jason Starr | The automorphism group functor, defined in the usual way, is almost never representable for affine schemes; it is neither representable as a scheme nor as an algebraic space. If you want a different definition, please specify the definition that you use. | |
| Sep 12, 2021 at 20:00 | history | asked | schemer | CC BY-SA 4.0 |