I had a number of questionsquestion regarding Gabber's rigidity, I'm not sure whether I am understanding it correctly, so please let me know if I'm making any mistakes.
Let $A$ be a ring (let's assume Noetherian) and $I$ be an ideal, since the pair $(\hat{A},I)$ is a henselian pair ($\hat{A}$ is the completion along $I$), Gabber's rigidity in algebraic $K$-theory for henselian pairs implies that $K_*(A/I, \mathbb{Z}/l\mathbb{Z})\simeq K_*(\hat{A}, \mathbb{Z}/l\mathbb{Z})$. Here algebraic $K$-theory of the completion is taken in the sense that $\hat{A}$ is just a Noetherian ring and we can construct $BGL(\hat{A})^+$ just as we can do for any ring i.e. the $K$-theory of finitely generated projective modules on $\hat{A}$. There is another way to define $K$-theory and that is to take the $K$-theory of formal vector bundles on $\hat{A}$, which means a system of vector bundles on all finite thickenings of $I$ with compatibility conditions under pullbacks. I don't think theseThese two coincide in generalprobably are the same. (Correct me if I am wrong)
Q1) HereThere is my first question: Givenan equivalence between the affine formal scheme $\text{Spf}(\hat{A})$ we can associate to it the scheme $\text{Spec}(\hat{A})$. Does this generalizevector bundles and give a well-defined assignement in the non-affine case? For the formal completion $X_Z$ ($X$ along a closed subscheme $Z$) can we assign a scheme that on affines behaves just like what I described? If so does this functor have name? (I am going to denote this hypothetical functor$\hat{A}$ bundles given by $\mathcal{F}$taking limit and restriction to infinitesimal neighborhoods.)
Q2Q) My original question is about whether there is a global version of this rigidity in the setting of formal completions or not? The following is my attempt on this question. (I am not sure whether I am making any mistakes or not)
Let's assume we have Noetherian schemes $X$ and a closed subscheme $Z$ and we are looking at the formal completion $X_Z$. Well asSince we are dealing with formal schemes it makes sense to consider formal vector bundles and compare its $K$-theory with the $K$-theory of $Z$ with coefficient in $\mathbb{Z}/l\mathbb{Z}$. But this comparison possibly fails even in the affine case according to the observation above. So I think the natural generalization would be instead of considering $X_Z$ asOne question is whether there is a Zariski descent for formal scheme we can consider it as a scheme just as in the affine case (This means if Q1 has a positive answer we consider $\mathcal{F}(X_Z)$). Since completionschemes with respect to an ideal preserves being Noetherian and having a finite Krull dimension and because of Zariski descent of connective algebraic $K$-theory this implies that $K_*(Z,\mathbb{Z}/l\mathbb{Z})\simeq K_*(\mathcal{F}(X_Z),\mathbb{Z}/l\mathbb{Z})$ where $\mathcal{F}(X_Z)$ is the associated scheme to $X_Z$ according to Q1.
Few points about Q1: The relation between $\text{Spf}(\hat{A})$ and $\text{Spec}(\hat{A})$ iscoefficients? if so that the former is the colimit of the schemes $\text{Spec}(A/I^n)$ in the category of ringed spaces andwould imply $\text{Spec}(\hat{A})$ is the colimit of the same schemes in the category of schemes$K_*(X_Z,\mathbb{Z}/l)\simeq K_*(Z, \mathbb{Z}/l) $. ForOr there could be possibly a proof of the latter see heredifferent approach. So question Q1 canThis could also be restated in the following way: Does the colimit of the schemes $X_Z^{(n)}$ ($n$-th thickening of $Z$) exist in the category of schemes? Statedwrong in this way it seems obvious (maybe it is not) because glueing schemes is a colimit and colimits should commute. So we can take the colimit on affine charts and then glue together and this should give the functor $\mathcal{F}$.
Regarding the question when the $K$-theory of formal scheme $X_Z$ should match with its $K$-theory as a scheme i.e. $\mathcal{F}(X_Z)$global case, I think if the pair $(X,Z)$ satisfy the property that every formal vector bundle on $X_Z$ can be extendedwhich I'd like to a neighborhood of $Z$ they should be compatibleknow some examples.