Let $X$ be a variety and $Y \subset X$ a closed subvariety.

Edit: Assume they are both smooth.

Denote $N_{Y / X}$ the normal bundle of $Y$ in $X$. The formal neighbourhood of $Y$ in $X$ is the formal scheme $X_Y = \text{Spec}\left( \lim_{n} \mathcal{O}_X \left/ \mathcal{I}_Y^n \right. \right)$.

The question is the following: how distant is the formal neighbourhood of $Y$ in $X$ from the formal neighbourhood of $Y$ in the total space of $N_{Y / X}$?

The question should be related to the splitting of the tangent sequence $$ 0 \rightarrow T_Y \rightarrow T_X \vert_Y \rightarrow N_{Y / X} \rightarrow 0, $$ and thus to the vanishing of $\text{Ext}^1_Y(N_{Y/X}, T_Y)$, but I can't figure out this relation.


  • $\begingroup$ You are probably assuming that $Y$ and $X$ are smooth, right? $\endgroup$
    – abx
    Apr 1, 2020 at 18:19
  • 2
    $\begingroup$ A good example to keep in mind is where $Y$ is a non-singular fiber of a (non iso-trivial) elliptic fibration $X\to \mathbb{P}^1$. In this case $N_{Y/X}$ is trivial, but the formal neighborhood of $Y$ in $X$ is non-trivial. In this case the Ext group and the extension you write are both non-trivial, but in general I think you need to know about higher order obstructions to fully answer this question. $\endgroup$
    – Jim Bryan
    Apr 1, 2020 at 19:16
  • $\begingroup$ @abx Yes, I am assuming that they are smooth. $\endgroup$ Apr 1, 2020 at 19:48
  • $\begingroup$ @Jim Bryan Thanks, I will have a look a that example. However, at a quick read, it seems it doesn't contradict the fact that the vanishing of that extension group (or of the splitting of the exact sequence), could imply the triviality of the formal neighbourhood. Am I right? $\endgroup$ Apr 1, 2020 at 19:52
  • 2
    $\begingroup$ Note an additional complication in characteristic $p$: take an elliptic fibration as in Jim Bryan's comment and pull it back by the Frobenius on the base. Now Kodaira-Spencer is identically zero and the tangent sequence splits. However, the formal neighborhood remains nontrivial. $\endgroup$ Apr 2, 2020 at 10:06

1 Answer 1


If you look at a family of elliptic curves like $y^2 = x^3 - x-t^n$ as a surface $X$ mapping to a curve $C$ with parameter $t$, and $Y$ the fiber over $0$, then the neighborhood modulo $\mathcal I_Y^n = (t)^n= (t^n)$ is trivial, hence equal to the $n$'th formal neighborhood of $Y$ in the total space of $N_{Y/X}$, but the neighborhood modulo $\mathcal I_Y^{n+1}$ is not. This shows that no finite list of obstruction classes will describe the formal neighborhood.

If the $n$th' order formal neighborhood is isomorphic to the $n$th order formal neighborhood in the normal bundle, then the ring of functions on it is isomorphic to $$\mathcal O_Y \oplus N_{Y/X}^\vee \oplus \operatorname{Sym}^{2} N_{Y/X}^\vee \oplus \dots \oplus \operatorname{Sym}^{n-1} N_{Y/X}^\vee $$ and the $n+1$st order neighborhood is locally the sum of that with $\operatorname{Sym}^n N_{Y/X}^\vee$ and then automorphisms respecting the algebra structure and the filtration but not the direct sum structure are given by $\operatorname{Sym}^n N_{Y/X}^\vee$-values derivations on that sum. A derivation is determined by the Leibnitz rule and its value on the first two factors, so the local automorphisms are given by $$\mathcal T_Y \otimes \operatorname{Sym}^n N_{Y/X}^\vee + N_{Y/X} \otimes \operatorname{Sym}^n N_{Y/X}^\vee$$ and so you obstruction classes in $H^1(Y, \mathcal T_Y \otimes \operatorname{Sym}^n N_{Y/X}^\vee )$ and $H^1(Y, N_{Y/X} \otimes \operatorname{Sym}^n N_{Y/X}^\vee)$, I think.

The first obstruction controls when the formal neighborhood can be viewed as a fibration over $Y$ and the second obstruction controls when you can identify the fibers with the normal bundle.

  • $\begingroup$ Dear Will, thanks for your answer and sorry for the late reply. I am going through the details of what you've written, but I get stuck pretty soon. I have two questions: are you considering $y^2 = x^3 - x -t^n$ inside $\mathbb{C}^3$? Moreover, you say that the $n$th neighbourhood of $Y$ in the total space of its normal bundle is trivial, but I don't see why. If you consider the family in $\mathbb{C}^3$, the normal bundle of $Y$ in $X$ is trivial, and the $n$th neighbourhood of $Y$ in $N_{Y/X}$ should not be trivial. $\endgroup$ Apr 6, 2020 at 14:42
  • $\begingroup$ Finally, in the second line after the direct sum I think there is a typo ("the filtration but not the given by...") that I can't make sense of. Thanks. $\endgroup$ Apr 6, 2020 at 14:43
  • $\begingroup$ @Federico I want to consider it as a family of closed elliptic curves, say in $\mathbb P^2 \times \mathbb A^1$ where the $\mathbb P^2$ has $x,y$ coordinates and the $\mathbb A^1$ has $t$ as a coordinate. As for the typo, I added in the missing words. $\endgroup$
    – Will Sawin
    Apr 6, 2020 at 19:34
  • $\begingroup$ @Will Sawin: are you sure about the second cohomology group? According to Hartshorne "Algebraic Geometry" Exercise 4.10 of Chapter 3 (which deals with $n =1$) first order extensions of $Y$ are in $1$-to-$1$ correspondence with elements of $H^1(Y,N_{Y/X}^{\vee}\otimes T_Y)$. In general to have that the formal neighborhood of $Y$ in $X$ is isomorphic to the formal neighborhood of $Y$ in $N_{Y/X}$ isn't it enough to have that $H^1(Y,\text{Sym}^nN_{Y/X}^{\vee}\otimes T_Y) = 0$ for all $n > 0$? $\endgroup$
    – Arty
    Jan 16, 2022 at 17:31
  • $\begingroup$ @Arty I am not sure that obstructions in the second cohomology group are nontrivial. If you can prove that only the first obstruction is needed, that would be great! But I don't think what happens for $n=1$ is necessarily a great guide to what happens for all $n$. $\endgroup$
    – Will Sawin
    Jan 16, 2022 at 18:13

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