If $ \operatorname{Spec}(A) $ is a smooth affine scheme over an algebraically closed field $ k $, then is $ \operatorname{Aut}(\operatorname{Spec}(A)) $ a group scheme or an algebraic space?
Please include a reference if you know of one.
$\begingroup$
$\endgroup$
4
-
7$\begingroup$ 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. $\endgroup$– Jason StarrCommented Sep 12, 2021 at 23:04
-
3$\begingroup$ 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 $n\geq 2$. $\endgroup$– abxCommented Sep 13, 2021 at 4:36
-
1$\begingroup$ 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. $\endgroup$– Ben McKayCommented Sep 13, 2021 at 12:21
-
1$\begingroup$ 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. $\endgroup$– Jason StarrCommented Sep 13, 2021 at 13:55
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
The automorphism group functor is always represented by an affine ind-scheme, i.e. a union of a countable chain of closed affine subschemes of finite type. See the paper Furter, Kraft: On the geometry of the automorphism groups of affine varieties, arxiv:1809.0417. It treats only the reduced part of $\mathrm{Aut}(X)$ with $\mathrm{char}k=0$, though.
answered Sep 13, 2021 at 12:16