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.

# More Complicated Brauer Computations

Let’s wrap up some of our Brauer group loose ends today. We can push through the calculation of the Brauer groups of curves over some other fields using the same methods as the last post, but just a little more effort.

First, note that with absolutely no extra effort we can run the same argument as yesterday in the following situation. Suppose ${X}$ is a regular, integral, quasi-compact scheme of dimension ${1}$ with the property that all closed points ${v\in X}$ have perfect residue fields ${k(v)}$. Let ${g: \text{Spec} K \hookrightarrow X}$ be the inclusion of the generic point.

Running the Leray spectral sequence a little further than last time still gives us an inclusion, but we will usually want more information because ${Br(K)}$ may not be ${0}$. The low degree terms (plus the argument from last time) gives us a sequence:

$\displaystyle 0\rightarrow Br'(X)\rightarrow Br(K)\rightarrow \bigoplus_v Hom_{cont}(G_{k(v)}, \mathbb{Q}/\mathbb{Z})\rightarrow H^3(X, \mathbb{G}_m)\rightarrow \cdots$

This allows us to recover a result we already proved. In the special case that ${X=\text{Spec} A}$ where ${A}$ is a Henselian DVR with perfect residue field ${k}$, then the uniformizing parameter defines a splitting to get a split exact sequence

$\displaystyle 0\rightarrow Br(A)\rightarrow Br(K)\rightarrow Hom_{cont}(G_k, \mathbb{Q}/\mathbb{Z})\rightarrow 0$

Thus when ${A}$ is a strict local ring (e.g. ${\mathbb{Z}_p}$) we get an isomorphism ${Br(K)\rightarrow \mathbb{Q}/\mathbb{Z}}$ since ${Br(A)\simeq Br(k)=0}$ (since ${k}$ is ${C_1}$). In fact, going back to Brauer groups of fields, we had a lot of trouble trying to figure anything out about number fields. Now we may have a tool (although without class field theory it isn’t very useful, so we’ll skip this for now).

The last computation we’ll do today is to consider a smooth (projective) curve over a finite field ${C/k}$. Fix a separable closure ${k^s}$ and ${K}$ the function field. First, we could attempt to use Leray on the generic point, since we can use that ${H^3(K, \mathbb{G}_m)=0}$ to get some more information. Unfortunately without something else this isn’t enough to recover ${Br(C)}$ up to isomorphism.

Instead, consider the base change map ${f: C^s=C\otimes_k k^s\rightarrow C}$. We use the Hochschild-Serre spectral sequence ${H^p(G_k, H^q(C^s, \mathbb{G}_m))\Rightarrow H^{p+q}(C, \mathbb{G}_m)}$. The low degree terms give us

$\displaystyle 0\rightarrow Br(k)\rightarrow \ker (Br(C)\rightarrow Br(C^s))\rightarrow H^1(G_k, Pic(C^s))\rightarrow \cdots$

First, ${\ker( Br(C)\rightarrow Br(C^s))=Br(C)}$ by the last post. Next ${H^1(G_k, Pic^0(C^s))=0}$ by Lang’s theorem as stated in Mumford’s Abelian Varieties, so ${H^1(G_k, Pic(C^s))=0}$ as well. That tells us that ${Br(C)\simeq Br(k)=0}$ since ${k}$ is ${C_1}$. So even over finite fields (finite was really used and not just ${C_1}$ for Lang’s theorem) we get that smooth, projective curves have trivial Brauer group.