Timeline for What does the group of automorphisms corresponding to $\mathfrak{g}$
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Aug 4, 2017 at 23:09 | history | edited | YCor |
edited tags
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| Aug 4, 2017 at 19:59 | answer | added | Will Sawin | timeline score: 4 | |
| Aug 4, 2017 at 19:39 | history | edited | Grobber | CC BY-SA 3.0 |
edited text
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| Aug 4, 2017 at 2:47 | comment | added | Grobber | I am not sure why the problem could be reduced to the affine case. | |
| Aug 3, 2017 at 22:34 | comment | added | მამუკა ჯიბლაძე | By definition, $\mathfrak g$ is a finite subgroup of the automorphism group of $X$, so it acts by automorphisms on $X$, and $X$ is in the case considered $\operatorname{Spec}A$ | |
| Aug 3, 2017 at 18:58 | comment | added | Grobber | But why does $\mathfrak{g}$ act on $\operatorname{Spec} A$? | |
| Aug 3, 2017 at 18:51 | comment | added | მამუკა ჯიბლაძე | Evidently $G$ is an isomorphic copy of $\mathfrak g$ acting on $A$. More or less by definition the category of affine $\operatorname{Spec} B$-schemes is opposite to that of $B$-algebras, so naming a subgroup $\mathfrak g$ of the automorphism group of $X=\operatorname{Spec} A$ over $\operatorname{Spec} B$ is the same as naming a(n isomorphic) subgroup $G$ of the automorphism group of the $B$-algebra $A$. | |
| Aug 3, 2017 at 18:14 | comment | added | Grobber | I always forget \mathfrak{} | |
| Aug 3, 2017 at 18:13 | history | edited | Grobber | CC BY-SA 3.0 |
added 15 characters in body
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| Aug 3, 2017 at 18:10 | comment | added | M.G. | The notation is \mathfrak{} (Fraktur "g"). I hope you don't mind me fixing it for you. | |
| S Aug 3, 2017 at 18:09 | history | suggested | M.G. | CC BY-SA 3.0 |
fixed latex for g
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| Aug 3, 2017 at 18:08 | review | Suggested edits | |||
| S Aug 3, 2017 at 18:09 | |||||
| Aug 3, 2017 at 18:05 | history | asked | Grobber | CC BY-SA 3.0 |