Timeline for Why are Stab(M) and Aut_R(M) isomorphic as schemes?, and why is Aut_R(M) smooth over F?
Current License: CC BY-SA 3.0
8 events
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Aug 9, 2011 at 14:41 | comment | added | Sean Sather-Wagstaff | Definitely true. But I was only seeing the isomorphisms of the groups, not the schemes. For instance, I was worried about the difference between $\operatorname{Spec}(F)$ and $\operatorname{Spec}(F[x]/(x^2))$. The point is that the defining equations are linear. Thanks again for the help! | |
Aug 9, 2011 at 14:35 | comment | added | user2035 | Giving an isomorphism $X\cong Y$ of schemes is equivalent to giving a compatible collection of isomorphisms $X(S)\to Y(S)$, by Yoneda's lemma. | |
Aug 9, 2011 at 14:34 | comment | added | Sean Sather-Wagstaff | Egads. I think I see it. The point is that the equations defining Stab(M)(S) are also linear: $ABA^{-1}=B$ means that $AB=BA$, and this is a linear system when $B$ is fixed. Ach. | |
Aug 9, 2011 at 14:24 | comment | added | Sean Sather-Wagstaff | This confuses me a bit. I know that the underlying groups for $\operatorname{Aut}_R(M)(S)$ and $\operatorname{Stab}(M)(S)$ are the same. The question is whether their structure sheaves are the same. In particular, I know that the structure sheaf on $\operatorname{Aut}_R(M)$ is reduced, and I need to know whether the same is true for the structure sheaf on $\operatorname{Stab}(M)$. Is this obviously reduced? | |
Aug 9, 2011 at 14:18 | vote | accept | Sean Sather-Wagstaff | ||
Nov 28, 2011 at 3:42 | |||||
Aug 9, 2011 at 9:10 | comment | added | user2035 | As suggested by Peter's answer, you should work out the functor of points for both $\mathrm{Aut}_R(M)$ and $\mathrm{Stab}(M)$, reducing the problem to ordinary groups. For the fibre of $f$, it does not matter whether the target is $G.M$ or $\mathrm{Mod}(R,F^n)$. | |
Aug 9, 2011 at 5:31 | answer | added | Peter McNamara | timeline score: 2 | |
Aug 8, 2011 at 21:06 | history | asked | Sean Sather-Wagstaff | CC BY-SA 3.0 |