Timeline for Can you build $\text{Aut}(X)$ using only $\text{QCoh}(X)$?
Current License: CC BY-SA 4.0
6 events
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| Aug 2 at 17:56 | comment | added | Qiaochu Yuan | @Pulcinella: Eilenberg-Watts is true for noncommutative rings so we get that the automorphisms of $\text{Mod}(R)$ for $R$ an arbitrary ring are given by invertible $(R, R)$-bimodules. I have no concrete sense of what these are for $R = U(\mathfrak{g})$, though. If $R$ is a $K$-algebra and you want $K$-linear automorphisms then these correspond to bimodules over $K$ (meaning the left and right actions of $K \subset R$ agree, or equivalently we consider $R \otimes_K R^{op}$-modules). For D-modules we could hope to prove a relative Eilenberg-Watts theorem. | |
| Aug 2 at 9:46 | comment | added | Pulcinella | Amazing, thanks! As for other examples - do you happen to know if this result holds for noncommutative rings, so that we can say what Aut of say $\mathcal{D}_X\text{-Mod}$ or $U_q(\mathfrak{g})\text{-Mod}$ is? | |
| Aug 2 at 9:40 | vote | accept | Pulcinella | ||
| Aug 2 at 1:03 | history | edited | Qiaochu Yuan | CC BY-SA 4.0 |
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| Aug 1 at 22:51 | history | edited | Qiaochu Yuan | CC BY-SA 4.0 |
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| Aug 1 at 22:46 | history | answered | Qiaochu Yuan | CC BY-SA 4.0 |