# Algebro-geometric version of {vector fields} $\longleftrightarrow$ {flows} correspondence?

Main Question: What Is the correpondence between flows and vector fields in algebraic geometry?

Here is a more precise statement could be an answer If it was true (I have no idea it is):

"Proposition": Let $X$ be a nice enough (intentionally ambiguous) scheme over a field $k$ with tangent sheaf $\mathcal{T}_X$. For every point $x \in X$ There is a 1-1 correpsodence between germs of vector fields through $x$ and "germs" of formal flows.

More precisely for every $v \in \mathcal{T}_{X,x}$ there corresponds a (continuous) morphism of complete local rings $\phi_v \in Hom_{cont}(\widehat{\mathcal{O}_{X,x}},\widehat{\mathcal{O}_{X,x} \otimes_k k[[t]]})$ and vice versa.

This "Proposition" was mentioned to me as a brief comment by a very respected mathematician (who's name I won't mention since I'm not completely convinced that the above is a fiathfull interpretation of his original statement). Unfortunately he didn't have time to elaborate on this and couldn't point me to a relevant reference.

Reference request: What are some good sources to turn to considering questions about formal geometry? Specifically the formulation of the classical theory of differential equations (of which this question is a particular case) in formal geometric terms.

Even the "proposition" above isn't really precise since I'm not sure what meaning I should give to the tensor product (some kind of completed tensor product perhaps). Which leads me to the following minor question.

Consider the category whose objects are topological rings $A$ whose topology can be generated by an $I$-adic filtration for some ideal $I \subset A$ and let the morphisms be continuous homomorphisms of rings.

(minor) Question: For arbitrary $A$ and $B$ in this category what is pushout of $A \leftarrow \mathbb{Z} \to B$? should this be the correct interpretation of the tensor product above?

You need a characteristic zero assumption, and you need some additional axioms on the homorphism to make it a bijection.

The map from derivations $D$ to homomorphisms sends a function $y \in \mathcal O_{X,x}$ to $$e^{t D} y = y + (Dy) t + (D^2 y) t^2/2 + (D^3 y) t^3/6 + \dots$$

The inverse map is just going to send a homomorphism $f$ to the derivation that sends $y$ to the coefficient of $t$ in $f(y)$. However we need some axioms to ensure this is actually a derivation and that its exponential is $f$.

You can find the right axioms by thinking of a flow as a (formal) group action by the (formal) group of time translations. The axioms are:

1. Composing $f$ with the projection from $\mathcal O_{X,x}[[t]]$ (which is equivalent to the completed tensor product you wrote down) to $\mathcal O_{X,x}$ should give the identity.

2. Composing $f$ with the map $\mathcal O_{X,x}[[t]] \to \mathcal O_{X,x}[[t_1,t_2]]$ that sends $t$ to $t_1+t_2$ should equal applying $f$ twice, once adding the variable $t_1$ and once adding the variable $t_2$.

These axioms immediately imply that the first coefficient of $f(y)$ is $y$, the second is a derivation of $y$, and the other ones are inductively given by powers of the derivation in this way.

• Since the OP asked for a reference, I suggest "Infinitesimale Transformationsgruppen komplexer Räume" by W. Kaup, Math. Ann. 160 (1965), p. 72-92 (if you read German). A more modern reference would be "Extension theorems for differential forms and Bogomolov-Sommese vanishing on log canonical varieties" by Greb, Kebekus and Kovács, Compos. Math. 146 (2010), p. 193-219. – pgraf May 24 '16 at 11:28
• By the way, I think the push out is the usual tensor product unless you demand that the topological rings be complete. – Will Sawin May 24 '16 at 14:27
• Forgot to add this, i had in mind the category of complete adic rings. Would this then be the completed tensor product? – Saal Hardali May 24 '16 at 19:38
• @SaalHardali Yes, then I believe it would be the completed tensor product. I don't think it's too hard to check explicitly the universal property. – Will Sawin May 25 '16 at 2:20