Volumes of p-adic Schemes

I came across this idea a long time ago, but I needed the result that uses it in its proof again, so I was curious about figuring out what in the world is going on. It turns out that you can make “{p}-adic measures” to integrate against on algebraic varieties. This is a pretty cool idea that I never would have guessed possible. I mean, maybe complex varieties or something, but over {p}-adic fields?

Let’s start with a pretty standard setup in {p}-adic geometry. Let {K/\mathbb{Q}_p} be a finite extension and {R} the ring of integers of {K}. Let {\mathbb{F}_q=R_K/\mathfrak{m}} be the residue field. If this scares you, then just take {K=\mathbb{Q}_p} and {R=\mathbb{Z}_p}.

Now let {X\rightarrow Spec(R)} be a smooth scheme of relative dimension {n}. The picture to have in mind here is some smooth {n}-dimensional variety over a finite field {X_0} as the closed fiber and a smooth characteristic {0} version of this variety, {X_\eta}, as the generic fiber. This scheme is just interpolating between the two.

Now suppose we have an {n}-form {\omega\in H^0(X, \Omega_{X/R}^n)}. We want to say what it means to integrate against this form. Let {|\cdot |_p} be the normalized {p}-adic valuation on {K}. We want to consider the {p}-adic topology on the set of {R}-valued points {X(R)}. This can be a little weird if you haven’t done it before. It is a totally disconnected, compact space.

The idea for the definition is the exact naive way of converting the definition from a manifold to this setting. Consider some point {s\in X(R)}. Locally in the {p}-adic topology we can find a “disk” containing {s}. This means there is some open {U} about {s} together with a {p}-adic analytic isomorphism {U\rightarrow V\subset R^n} to some open.

In the usual way, we now have a choice of local coordinates {x=(x_i)}. This means we can write {\omega|_U=fdx_1\wedge\cdots \wedge dx_n} where {f} is a {p}-adic analytic on {V}. Now we just define

\displaystyle \int_U \omega= \int_V |f(x)|_p dx_1 \cdots dx_n.

Now maybe it looks like we’ve converted this to another weird {p}-adic integration problem that we don’t know how to do, but we the right hand side makes sense because {R^n} is a compact topological group so we integrate with respect to the normalized Haar measure. Now we’re done, because modulo standard arguments that everything patches together we can define {\int_X \omega} in terms of these local patches (the reason for being able to patch without bump functions will be clear in a moment, but roughly on overlaps the form will differ by a unit with valuation {1}).

This allows us to define a “volume form” for smooth {p}-adic schemes. We will call an {n}-form a volume form if it is nowhere vanishing (i.e. it trivializes {\Omega^n}). You might be scared that the volume you get by integrating isn’t well-defined. After all, on a real manifold you can just scale a non-vanishing {n}-form to get another one, but the integral will be scaled by that constant.

We’re in luck here, because if {\omega} and {\omega'} are both volume forms, then there is some non-vanishing function such that {\omega=f\omega'}. Since {f} is never {0}, it is invertible, and hence is a unit. This means {|f(x)|_p=1}, so since we can only get other volume forms by scaling by a function with {p}-adic valuation {1} everywhere the volume is a well-defined notion under this definition! (A priori, there could be a bunch of “different” forms, though).

It turns out to actually be a really useful notion as well. If we want to compute the volume of {X/R}, then there is a natural way to do it with our set-up. Consider the reduction mod {\mathfrak{m}} map {\phi: X(R)\rightarrow X(\mathbb{F}_q)}. The fiber over any point is a {p}-adic open set, and they partition {X(R)} into a disjoint union of {|X(\mathbb{F}_q)|} mutually isomorphic sets (recall the reduction map is surjective here by the relevant variant on Hensel’s lemma). Fix one point {x_0\in X(\mathbb{F}_q)}, and define {U:=\phi^{-1}(x_0)}. Then by the above analysis we get

\displaystyle Vol(X)=\int_X \omega=|X(\mathbb{F}_q)|\int_{U}\omega

All we have to do is compute this integral over one open now. By our smoothness hypothesis, we can find a regular system of parameters {x_1, \ldots, x_n\in \mathcal{O}_{X, x_0}}. This is a legitimate choice of coordinates because they define a {p}-adic analytic isomorphism with {\mathfrak{m}^n\subset R^n}.

Now we use the same silly trick as before. Suppose {\omega=fdx_1\wedge \cdots \wedge dx_n}, then since {\omega} is a volume form, {f} can’t vanish and hence {|f(x)|_p=1} on {U}. Thus

\displaystyle \int_{U}\omega=\int_{\mathfrak{m}^n}dx_1\cdots dx_n=\frac{1}{q^n}

This tells us that no matter what {X/R} is, if there is a volume form (which often there isn’t), then the volume

\displaystyle Vol(X)=\frac{|X(\mathbb{F}_q)|}{q^n}

just suitably multiplies the number of {\mathbb{F}_q}-rational points there are by a factor dependent on the size of the residue field and the dimension of {X}. Next time we’ll talk about the one place I know of that this has been a really useful idea.