Timeline for If a variety $X$ has finite automorphism group, is the same true for its $n$-fold self-products?
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 27, 2018 at 21:29 | history | edited | Wille Liu | CC BY-SA 3.0 |
added 85 characters in body
|
| Feb 27, 2018 at 21:25 | comment | added | Ariyan Javanpeykar | Dear @WilleLiou, the vanishing of $H^0(X,T_X)$ is not equivalent to the finiteness of Aut(X). It is equivalent to the automorphism group scheme being zero-dimensional. However, the automorphism group scheme is not necessarily of finite type, only locally of finite type over $\mathbb{C}$. Thus, it could a priori happen that $Aut(X)$ is finite, and that $Aut(X^n)$ is an infinite countable discrete group. | |
| Feb 27, 2018 at 21:25 | history | edited | Wille Liu | CC BY-SA 3.0 |
added 86 characters in body
|
| Feb 27, 2018 at 21:12 | history | answered | Wille Liu | CC BY-SA 3.0 |