9
$\begingroup$

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?

$\endgroup$

1 Answer 1

10
$\begingroup$

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.

$\endgroup$
4
  • 2
    $\begingroup$ 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. $\endgroup$
    – pgraf
    May 24, 2016 at 11:28
  • 1
    $\begingroup$ By the way, I think the push out is the usual tensor product unless you demand that the topological rings be complete. $\endgroup$
    – Will Sawin
    May 24, 2016 at 14:27
  • $\begingroup$ Forgot to add this, i had in mind the category of complete adic rings. Would this then be the completed tensor product? $\endgroup$ May 24, 2016 at 19:38
  • 1
    $\begingroup$ @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. $\endgroup$
    – Will Sawin
    May 25, 2016 at 2:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.