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Let $X$ be an affine variety and $A(X)$ be a ring of regular functions on $X$. Consider a Lie algebra of derivations of $A(X)$ which we denote as $Der(A(X))$. It is known that $Aut(X)$ (group of automorphisms of $X$) can canonically be embedded into $Aut(Der(A(X)))$. (For $\sigma \in Aut(X)$, consider $F_{\sigma}: Der(A(X)) \to Der(A(X))$: $F_{\sigma}(D) = \sigma \circ D \circ \sigma^{-1}$). I consider $Aut(X)$ as a subgroup in $Aut(Der(A(X)))$ under such embedding.

My question is, when $Aut(Der(A(X)))$ and $Aut(X)$) are equal? (for which class of varieties?) (I think it is unknown, but hopefully it is clear for some small class of varieties, for example it is known for affine spaces).

Or, what is known about $Aut(Der(A(X)))$?

Thanks in advance!

share|cite|improve this question
Isn't $Der(X)$ a module, not a ring? – Will Sawin May 29 '13 at 16:30
Or a Lie algebra? – Will Sawin May 29 '13 at 16:30
Sorry! Sure, I meant Lie algebra! (yes, it is never a ring). (By Lie algebra I mean abstract Lie algebra, without any topology). – Andriy Regeta May 30 '13 at 8:22

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