Let $A$ be an integral domain which is a finitely generated algebra over an algebraically closed field $k$. Let $\phi :A \to A^{[1]}$ be a $k$-algebra homomorphism and let us write $\phi_t : A \to A[t]$ to emphasize the indeterminate $t$. $\phi$ is called an exponential map if the following two conditions are satisfied:

(i) $\epsilon_0 \phi_t : A \to A$ is the identity map, where $\epsilon_0 : A[t] \to A$ is the evaluation map at $t=0$.

(ii) $\phi_{s} \phi_t=\phi_{s+t}$, where $\phi_s$ is extended to a homomorphism $A[t] \to A[t,s]$ by setting $\phi_s(t)=t$.

Can someone please refer me to some book or lecture note or paper explaining the bijective correspondence between these exponential maps and algebraic actions of $\mathbb G_a=(k,+)$ on $X$ , the affine variety corresponding to $A$ ?

So far I have only been able to locate the bijective correspondence between locally nilpotent derivations on finitely generated domain over algebraically closed field of characteristic zero and the $\mathbb G_a$ actions on affine variety corresponding to the affine $k$-domain, in Gene Freudenberg's book : Algebraic theory Locally nilpotent derivations. But I haven't been able to find explicit detail concerning the correspondence between $\mathbb G_a$-actions and exponential maps (that there is a correspondence is mentioned,for example, in Gupta, N. Invent. math. (2014) 195: 279-288, On the cancellation problem for the affine space $\mathbb A^3$ in characteristic $p$

  • $\begingroup$ I think you are missing main points in Freudenberg's book. He does say explicitly that $\mathbb{G}_a$ actions, locally nilpotent derivations and associated exponential maps are all essentially equivalent (page 32-33, second edition). $\endgroup$ – Mohan May 15 '18 at 19:10
  • $\begingroup$ @Mohan: Yes, but that is only in characteristic zero (section 1.5 of chapter 1 of Freudenberg is all about characteristic zero); exponential maps and LND are not equivalent in positive characteristic as far as I know ... $\endgroup$ – users May 15 '18 at 19:17
  • 3
    $\begingroup$ Isn't this literally the definition of a $\mathbb G_a$ action on a variety, but with all the arrows reversed and products replaced with coproducts by the contravariant equivalence between affine schemes and rings? $\endgroup$ – Will Sawin May 15 '18 at 19:18
  • $\begingroup$ In positive characteristics, these are more complicated and you may find some results for example in Dufresne's article in the arxiv. $\endgroup$ – Mohan May 15 '18 at 20:09
  • 1
    $\begingroup$ As Will Sawin says, this is literally the definition of a $\mathbb{G}_a$ action on the affine scheme $\text{Spec } A$, just interpreted in the opposite category. This means none of your hypotheses are necessary; $k$ can be any commutative ring and $A$ can be any commutative $k$-algebra. The thing that replaces locally nilpotent derivations in positive characteristic can be found here: qchu.wordpress.com/2017/11/26/… $\endgroup$ – Qiaochu Yuan May 15 '18 at 20:28

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.