Let me for simplicity start with affine case. If $X=\operatorname{Spec}(A)$ is an affine variety $Z \subset X$ is a closed affine subvariety $Z=\operatorname{Spec}(A/I)$. What conditions are sufficient to say that completion of the (total space of) normal bundle $N\_{Z/X}$ along zero section is isomorphic to the completion of $X$ along $Z$?



Formal completion of the normal bundle
The formal completion of $X$ along $Z$ is always isomorphic to that of the normal bundle $N\_{Z \subset X}$ along its zero section if $X= \mathrm{Spec}(A)$ and $Z = \mathrm{Spec}(A/I)$ are smooth and affine over some base $k$. If the affineness or smoothness assumption is dropped, it is easy to give counterexamples. 




---


It's easier to work in the world of algebra, so I will show that there exists an isomorphism of $k$-algebras 


$$\widehat{\mathrm{Sym}}\_{A/I}(I/I^2) \to \widehat{A}$$ 


compatible with the projection down to $A/I$. Here $\widehat{A}$ is the $I$-adic completion of $A$, and $\widehat{\mathrm{Sym}}\_{A/I}(I/I^2)$ is the completion of the symmetric algebra $\mathrm{Sym}\_{A/I}(I/I^2)$ along the augmentation ideal.


By the infinitesimal criterion for smoothness, one can find a compatible system of $k$-algebra maps $\epsilon\_n:A/I \to A/I^n$ lifting the projection $A/I^n \to A/I$. In the limit, we obtain a $k$-algebra map $\epsilon:A/I \to \widehat{A}$ lifting the projection $\widehat{A} \to A/I$. Using $\epsilon$, we view $\widehat{A}$ and each quotient $A/I^n$ as $A/I$-algebras.


Now the $A/I$-module $I/I^2$ is a projective module as $X$ and $Z$ are smooth. There is an obvious $A/I$-module map $I/I^2 \to A/I^2 \simeq \widehat{A}/I\widehat{A}^2$. By projectivity, one can find an $A/I$-module map $I/I^2 \to \widehat{A}$ lifting the previous one. By the universal property of the symmetric algebra, this extends to an $A/I$-algebra map $\mathrm{Sym}\_{A/I}(I/I^2) \to \widehat{A}$. As both sides are augmented over $A/I$, this naturally extends to an $A/I$-algebra map $\widehat{\mathrm{Sym}}\_{A/I}(I/I^2) \to \widehat{A}$; by filtering both sides suitably, one can check that this is an isomorphism.


