Let $R$ be a complete dvr with fraction field $K$ and residue field $k$, and let $X, Y$ be two smooth projective $R$-schemes with isomorphic generic fibers.



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> Is it true that $[X\_k]=[Y\_k]$ in $K\_0(\text{Var}\_k)$?
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Recall that $K\_0(\text{Var}\_k)$ is the so-called Grothendieck ring of varieties, namely, the free Abelian group on $k$-varieties, modulo the relation $[X]=[Y]+[X\setminus Y]$ for $Y\subset X$ closed.


That's the short version; let me say why I would call this a Cauchy integral formula and say what I know. 


First of all, why is this a Cauchy integral formula? In this analogy, I'm thinking of a morphism $X\to Y$ as a function $Y(T)\to K\_0(\text{Var}\_T)$, sending a point $y$ to $X\_y$. You're supposed to think of $[X\_K]$ as a function $\text{Spec}(K)$ (a puntured disk), and the question asks if we can recover the value at the central point $\text{Spec}(k)$ from this data. The complex-analytic analogue is exactly the Cauchy integral formula.


Let me give a slightly non-trival example. A trivial family of Hirzebruch surfaces $X\_n$ may degenerate to either $X\_n$ or $X\_{n+2}$. But of course $[X\_n]=[X\_{n+2}]=(\mathbb{L}+1)^2$.


Some other examples: if $X\_K, Y\_K$ are isomorphic as polarized varieties and one of them is not ruled, all is good by Matsusaka-Mumford. In particular, if $X, Y$ are canonically polarized, the answer to my question is affirmative.


Likewise suppose $X\_K, Y\_K$ have trivial canonical bundle. Then $X\_k, Y\_k$ have trivial canonical bundle as well (at least if $K$ has characteristic zero) and are birationally equivalent, by spreading out the isomorphism $X\_K\simeq Y\_K$. But birational Calabi-Yau's have the same class in (some mild localization of) $K\_0(\text{Var}\_k)$, by Kontsevich's motivic integration results.


A slightly better version of the question is:



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> Let $X, Y$ be smooth projective $R$-schemes with $[X\_K]=[Y\_K]$. Does this imply $[X\_k]=[Y\_k]$?
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This would give some kind of specialization map $K\_0(\text{Var}\_K)\to K\_0(\text{Var}\_k)$, at least if $\text{char}(K)=0$.


ADDED: The result is also true for curves and surfaces. For curves this is easy; for surfaces, observe that $X\_k, Y\_k$ are birationally equivalent. Thus there is a surface $Z$ obtained by blowing up $X\_k, Y\_k$ at some number of points, so $$[X\_k]+n\mathbb{L}=[Z]=[Y\_k]+m\mathbb{L}.$$ But $n=m$ because $X\_k, Y\_k$ have equal Euler characteristic, so $[X\_k]=[Y\_k]$.



Is there a motivic Cauchy integral formula?
I believe the answer is yes in the residue characteristic zero case. I will show it for a mild localisation of the Grothendieck ring (inverting $[L]$ and $[L]-1$). This may also follow from work of Denef and Loeser on the motivic nearby finer.


The real meat of the argument will be the weak factorisation theorem. 


First, by standard algebraization arguments + removal of singularities, we can reduce to the following problem:


Let $X$ and $Y$ be smooth varieties both mapping to a curve $C$, with the fibrations isomorphic away from a point $x$ in $C$, and both smooth over a neighbourhood of $x$. Show that $[X\_x]=[Y\_x]$.


To solve this, consider the involution of $K^0(Var)[L^{-1}]$ that sends $[X]$ to $[X] /[L]^{d}$ for $[X]$ a smooth projective variety of dimension $d$. We can check that this is well-defined using the blow-up relations of Franziska Bittner/Heinloth: For $Y \to X$ a blow-up with centre $Z$ of codimension $c$ and exceptional divisor $Z$, $$[X] -[L]^c [Z]= [Y]- [L][E]$$ follows from $$[X]-[Z]=[Y]-[E]$$ and $$([L]^c-1)[Z]) = ([L]-1)[Y],$$ both obvious.


Now applying the involution to the known relation $$[X]-[X\_x]=[Y]-[Y\_x],$$ we obtain $$[X]-[L][X\_x]=[Y]-[L][Y\_x].$$ Subtracting, we obtain $$ \left([L]-1\right) X\_x = \left([L]-1\right) [Y\_x]$$ and we win by dividing by $[L]-1$.


I claim the division by $[L]$ and $[L]-1$ in this argument may be removed if desired, to obtain an identity in the unlocalised Grothendieck ring. When working with varieties of dimension $\leq d$, rather than dividing $[X]$ by $[L]^{\dim X}$ we may multiply by $[L]^{d-\dim X}$ to achieve the same effect. This removes the need to adjoin $[L]^{-1}$. Furthermore, rather than working with the involution directly, observe that it only changes a class by a multiple of $[L]-1$ and work with this change divide by $[L]-1$, which is well-defined because we can remove an extra factor of $[L]-1$ from our argument for the relation. This removes the need to adjoin $([L]-1)^{-1}$.


One cool thing about this argument is it tells you something also in the case where $X\_x$ is singular - something like "the integral over $X\_x$ of the motivic nearby fiber of Denef and Loeser depends only on the $X\_\eta$".


