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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.

Is it true that $[X_k]=[Y_k]$ in $K_0(\text{Var}_k)$?

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:

Let $X, Y$ be smooth projective $R$-schemes with $[X_K]=[Y_K]$. Does this imply $[X_k]=[Y_k]$?

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 $[X_k]=[Y_k]$.

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  • $\begingroup$ I suspect this is not true. Begin with $F$ a finite field, $n$ and $d$ positive integers, and $X_F$, $Y_F$ two degree $d$, smooth hypersurfaces in $\mathbb{P}^n_F$ such that $X_F$ has an $F$-point, yet $Y_F$ has no $F$-point. Let $[X_F]$ and $[Y_F]$ be the corresponding $F$-points in the projective space $\mathbb{P}^M_F$ parameterizing degree $d$ hypersurfaces in $\mathbb{P}^n$. Let $\nu_X:P_X\to \mathbb{P}^M_F$, resp. $\nu_Y:P_Y\to \mathbb{P}^M_F$ be the blowing up at $[X_F]$, resp. at $[Y_F]$. (continued) $\endgroup$ – Jason Starr Jun 4 '15 at 0:42
  • $\begingroup$ Let $A_X$, resp. $A_Y$ be the local ring of $P_X$, resp. $P_Y$, at the generic point of the exceptional divisor. Of course $A_X$ is isomorphic to $A_Y$. Moreover, the fraction fields of $A_X$ and $A_Y$ are both $F(\mathbb{P}^M_F)$. The pullbacks over $\text{Spec}(A_X)$, resp. $\text{Spec}(A_Y)$ of the universal family $Z$ in $\mathbb{P}^M_F\times_F \mathbb{P}^n_F$ have isomorphic generic fibers. Thus, let $R$ be the completion of these two DVRs, and let $X_R$ and $Y_R$ be the pullback of the universal family $Z$ in $\mathbb{P}^n_R$. (continued) $\endgroup$ – Jason Starr Jun 4 '15 at 0:45
  • $\begingroup$ The generic fibers of $X_R$ and $Y_R$ are both the base change of the generic fiber of $Z$. Yet the closed fiber $X_k$ has a $k$-point, whereas the closed fiber $Y_k$ has no $k$-point. If I correctly recall the Lang-Nishimura Theorem, I believe existence of rational points should distinguish elements in the Grothendieck ring of varieties. $\endgroup$ – Jason Starr Jun 4 '15 at 0:49
  • $\begingroup$ CORRECTION. I realize now that the generic fibers $X_K$ and $Y_K$ in my example are not $K$-isomorphic. Rather, there is a an automorphism $\sigma$ of $K$, and $X_K$ is $K$-isomorphic to $\sigma^*Y_K$. If $d>n+1$, then degree $d$ hypersurfaces in $\mathbb{P}^n_R$ have ample canonical bundle, so if the generic fibers are $K$-isomorphic, then also the special fibers are $K$-isomorphic. $\endgroup$ – Jason Starr Jun 4 '15 at 3:41
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    $\begingroup$ You might find this paper helpful:arxiv.org/pdf/math/0107134.pdf $\endgroup$ – Wille Liou Jun 4 '15 at 23:51
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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 $X_\eta$".

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  • $\begingroup$ Great! I think this works in the equicharacteristic, algebraically closed residue field case. (Thinking through the "standard algebraization arguments" you allude to, I think you need a field extension.) This is indeed very nice. It would be good to work through exactly how the application of Bittner uses weak factorization etc. here, i.e. what is this argument saying about equidecomposability of the two special fibers? $\endgroup$ – Daniel Litt Oct 15 '16 at 15:53
  • $\begingroup$ @DanielLitt I had an earlier, more complicated version of the argument that made these issues more clear. Basically you can get from $X$ to $Y$ by a series of blow-ups and blow-downs during which $X$ and $Y$ remain smooth and $X_x$ and $Y_x$ are simple normal crossings divisors (this is one version of weak factorization). You can check that the invariant of a normal crossing divisor which is the sum of the Grothendieck classes of the strata, with the codimension $c$ strata weighted by $(1-[L])^c$, is preserved by blowup, so it is the same for $X_x$ and $Y_x$. $\endgroup$ – Will Sawin Oct 15 '16 at 19:54
  • $\begingroup$ Great! This gives a new proof of Kontsevich's theorem that birat'l CYs have the same class in the Grothendieck ring, in the hyperkahler case, I think. $\endgroup$ – Daniel Litt Oct 16 '16 at 17:25
  • $\begingroup$ It might be worth writing out the details of the specialization map you describe and seeing if one can do anything else with it... $\endgroup$ – Daniel Litt Oct 16 '16 at 17:27

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