# The Tate Conjecture over finite fields

In this post we will describe the Tate conjecture for varieties over finite fields, focusing on its various forms, and talking about some applications.

# Motivation

One of the main objects of study in algebraic geometry is the étale cohomology groups of a variety. They are vast importance, for a number of reasons: they are important examples of Galois representations, but they are also just useful in their own right as a means of measuring ‘geometric complexity’ of varieties. That said, studying étale cohomology is hard. Even in the case of curves (for this computation, see the above link), and even if we disregard the Galois action, the computation of $\ell$-adic cohomology is exceedingly difficult. This leads one to wonder if there is a sort sideways means of attacking the understanding of the cohomology groups $H^i(X,\mathbb{Q}_\ell)$.

Now, if we are going to try and push the computation of $H^i(X,\mathbb{Q}_\ell)$ to another field, we better choose one which is better suited to computations. One such possibility is to try and phrase $H^i(X,\mathbb{Q}_\ell)$ in terms of intersection theory. Intersection theory is roughly the study of closed subvarieties of a given variety $X$, and how these subvarieties interact under ‘intersection’ (really a sort of generalized intersection). Not only do we have a fruitful, powerful, and computationally feasible theory of such objects, but they are also geometrically intuitive. In fact, intersection theory is where all of algebraic geometry came from. Questions of how many points of intersection do conics in $\mathbb{P}^2_k$ have (Bezout’s theorem), or how many curves pass through a given point of a surface, were asked far before the invention of many of the modern tools which make up ‘algebraic geometry’.

It would then seem useful if we could somehow try and study these hard, arithmetico-geometric objects $H^i(X,\mathbb{Q}_\ell)$, using this intuitive and rich field of intersection theory. In a very, very rough sense, this is precisely what the Tate conjecture asserts takes place, at least in the case of varieties over finite (or, more generally, finitely generated over their prime subfield) fields. It roughly says that, in some sense, through a dictionary discussed below, every cohomology class $c\in H^i(X,\mathbb{Q}_\ell)$ is just a $\mathbb{Q}_\ell$-linear combination of subvarieties. Moreover, this association will respect the ring-theoretic structure of the $H^i$‘s.

Somewhat astoundingly, the Tate conjecture, and its various related conjectures, have various surprising reinterpretations in terms of number theoretic concepts. These reinterpretations will be far from ‘just cute’, as we will see they intimately relate the Tate conjecture to some of the largest open problems in the number theory of function fields.

Lastly, because I would be remiss if I didn’t mention it, the Tate conjecture is an étale-theoretic analogue of a conjecture which is very, very well known–at least by its name. In fact, proving this analogue will win you a million bucks. I speak, of course, of the Hodge conjecture. This roughly says that we can understand (part of) the singular cohomology (with $\mathbb{Q}$-coefficients!) in terms of intersection theoretic data. The Hodge conjecture is historically the predecessor of the Tate conjecture. That said, if you believe that ‘all Weil cohomology theories’ were created equally, their veracity are inextricably linked, and hinting at some sort of general motivic framework in which all of these questions sit.

# Intersection theoretic preliminaries

So, let us begin by setting up the bridge between intersection theory and $\ell$-adic cohomology that the Tate conjecture will concern. For the whole post, let us a fix a field $k=\mathbb{F}_q$, a variety $X/k$ (smooth, projective, geometrically integral), and let us denote $X_{\bar{k}}$ by $\overline{X}$. We also fix a prime $\ell\ne\mathrm{char}(k)$. We will also use the letter $Y$ to denote an arbitrary variety (smooth, projective, integral) over a field $F$ (so as not to imply that these definitions/notions are particular to the case of varieties over finite fields). Any cohomology theory unadorned by a subscript should be assumed étale.

So, let us begin by recalling the notion of an $r$-cycle. An $r$-cycle $Z\subseteq Y$ will be nothing but a closed integral subscheme of codimension $r$. For example, a $1$-cycle is nothing but a Weil divisor. We will denote the free abelian group on the $r$-cycles by $\mathcal{Z}^r(Y)$. Thinking geometrically (i.e. of the case over the complex numbers), this is nothing but taking a complex manifold $Y$ and looking at the complex submanifolds of a given codimension.

Note we have a natural map $\mathcal{Z}^r(X)\hookrightarrow\mathcal{Z}^r(\overline{X})$ given on $r$-cycles by $W\mapsto \overline{W}:=W_{\overline{k}}$. This map is, in fact, injective. We will call its image the rational $r$-cycles on $\overline{X}$, and still denote it $\mathcal{Z}^r(X)$. Note, moreover, that we have an action of $G_k:=\mathrm{Gal}(\overline{k}/k)$ on $\mathcal{Z}^r(\overline{X})$. Indeed, given $\sigma\in G_k$ we obtain an automorphism of $\overline{X}$ by just acting on $\overline{k}$. Then, the image $\sigma(Z)$, of an $r$-cycle $Z$, under $\sigma$, is another $r$-cycle. Extending linearly does give us an action $G_k\times\mathcal{Z}^r(\overline{X})\to\mathcal{Z}^r(\overline{X})$ which respects addition. Moreover, we see that we can identify $\mathcal{Z}^r(\overline{X})^{G_k}$ with the rational cycles on $X$.

## Rational equivalence and the Chow ring

Now, the object $\mathcal{Z}^r(Y)$ does not, as it stands, tell us that much about the geometry of $Y$. The reason for this is that we have imposed no relation telling us how various cycles interact, and when they can be put together. For example, think about the case $r=1$, so that $\mathcal{Z}^1(Y)$ becomes the familiar group $\mathrm{Div}(Y)$ of divisors on $Y$. This is not the group we usually study, not the one with geometric content. This honor goes to its quotient $\text{Cl}(Y)$ which is $\mathrm{Div}(Y)/\mathrm{PDiv}(Y)$, where $\mathrm{PDiv}(Y)$ is the group of principal divisors on $Y$.

So, what sort of equivalence relation would we like to impose on $\mathcal{Z}^r(Y)$; what subgroup should we mod out by? This is already a somewhat deep question since, there are several important options, some of which are conjecturally the same for most intents and purposes. The tamest of these equivalence relations (i.e. the most restrictive) is the notion of rational equivalence. This is the generalization of linear equivalence of divisors. Namely, we say that $Z,Z'\in\mathcal{Z}^r(Y)$ are rationally equivalent, denoted $Z\sim_\text{rat}Z'$, if there is some $r-1$-cycle $W\subseteq Y$ with $Z,Z'\subseteq W$, and $Z-Z'=\mathrm{div}(f)$ for some $f\in K(W)^\times$. Thus, we will consider two elements of $\mathcal{Z}^r(Y)$ equivalent if they are equivalent as divisors in some subscheme of one higher-dimension.

One needs to be a little careful here though. Indeed, even though $W$ is integral, it may not be regular in codimension $1$, and so the usual definition of $\mathrm{div}(f)$ needn’t work. Instead, for a divisor $D\in\mathrm{Div}(W)$ we associate define

$\displaystyle n_D(f):=\mathrm{length}_{\mathcal{O}_{W,D}}\mathcal{O}_{W,D}/f\mathcal{O}_{W,D}$

and then define

$\displaystyle \mathrm{div}(f):=\sum_{D\in\mathrm{Div}(W)}n_D(f) D$

If $W$ happens to be regular in codimension $1$, this agrees with the usual definitions. In particular, note that if $r=1$, then the only $0$-cycle is $X$ itself, and so rational equivalence on $\mathcal{Z}^1(Y)=\mathrm{Div}(Y)$ is just the usual linear equivalence. To see that the above definitions make sense, and to get a better understanding of the basic properties of rational equivalence, see Fulton’s book Intersection Theory.

We will denote the quotient $\mathcal{Z}^r(Y)/\sim_\mathrm{rat}$ by $\mathrm{CH}^r(Y)$, and call it the $r^\text{th}$ Chow group of $Y$. We will denote by $\mathrm{CH}^\ast(Y)$ the group

$\displaystyle \bigoplus_{r\geqslant 0}\mathrm{CH}^r(Y)$

and call it the Chow ring of $Y$. This is not a misnomer since, in fact, the Chow ring does have the structure of a graded ring with the groups $\mathrm{CH}^m(Y)$ being the graded pieces. This multiplication is a little difficult to define, but in nice cases it’s just intersection: $(Z,Z')\mapsto Z\cap Z'$. Of course, this isn’t going to work in general since $Z^2$ should be in $\mathcal{Z}^{2r}(Y)$, but $Z\cap Z=Z\in\mathcal{Z}^r(Y)$. The actual product on $Y$ (which exists by something called the ‘moving lemma’) should be thought of as the unique way of making such a product exist such that when $Z_1\cap Z_2\in\mathcal{Z}^{i_1+i_2}(Y)$, for $Z_j\in\mathcal{Z}^{i_j}(Y)$, then $Z_1\cdot Z_2=Z_1\cap Z_2$. For more information, see Fulton’s book, or the relevant part of Chapter 5 of Hartshorne which discusses the case of surfaces (although he proves the moving lemma in general!).

Example: One can show that if $Y=\mathbb{P}^n_F$, then $\mathrm{CH}^\ast(Y)$ is the ring $\mathbb{Z}[L]/(L^{n+1})$, with the obvious grading, where $L$ corresponds to a line in $Y$.

Example: If $Y=\mathbb{P}^n_F\times\mathbb{P}^m_F$, then $\mathrm{CH}(Y)=\mathbb{Z}[s,t]/(s^n,t^m)$, such that $s$ and $t$ both have degree $1$.

## Numerical equivalence

As was said at the beginning of this section, the subject matter on equivalence relations on $\mathcal{Z}^r(Y)$ is a complex one. Namely, there are many. We now discuss another, coarser alternative to rational equivalence. For this section, mainly for notational convenience, let’s assume that $F$, the field over which $Y$ is defined, is algebraically closed. We will denote the dimension of $Y$ by $n$.

Let us begin by noting that we have an obvious degree map $\deg:\mathrm{CH}^n(Y)\to \mathbb{Z}$. Indeed, since $Y$ is an integral variety, codimension $n=\dim Y$ closed subvarieties of $Y$ are just closed points, and so $\mathcal{Z}^n(Y)$ is just the free-abelian group on the closed points of $Y$. Of course, two points are not necessarily linearly equivalent. In fact, if $Y$ is a curve, then $p\sim_\text{rat}q$, $p\ne q$ closed points of $Y$, implies that $Y=\mathbb{P}^1_F$. Regardless, the map $\deg$ is defined in the obvious way:

$\displaystyle \deg:\sum_{p\in |Y|}n_p p\mapsto \sum_{p\in |Y|}n_p$

here $|Y|$ denotes the closed points of $Y$.

Note then that for every $0\leqslant r\leqslant n$ we have a pairing:

$\langle -,-\rangle_r:\mathrm{CH}^r(Y)\times\mathrm{CH}^{n-r}(Y)\to\mathbb{Z}$

given by

$\langle Z,Z'\rangle_r=\deg\left(Z\cdot Z'\right)$

where $Z\times Z'\in \mathrm{CH}^n(Y)$ is the product in the Chow ring. This is called the $r^\text{th}$ intersection pairing.

This pairing is unsatisfactorily inexplicit, coming from the inexplicitness of the product on $\mathrm{CH}^\ast(Y)$. That said, if $Y$ happens to be a surface, then we have a very pleasing description of $\langle -,-\rangle_1$. Indeed, $\mathrm{CH}^1(Y)=\mathrm{Cl}(Y)$, and so we’re trying to define a pairing:

$\langle -,-\rangle_1:\mathrm{Cl}(Y)\times\mathrm{Cl}(Y)\to\mathbb{Z}$

which can be described as follows:

$\langle D,D'\rangle_1=\chi(\mathcal{O}_Y)-\chi(\mathcal{O}(-D))-\chi(\mathcal{O}(-D'))+\chi\left(\mathcal{O}(-D-D')\right)$

where $\chi$ denotes the Euler characteristic of a coherent sheaf on $Y$, and $\mathcal{O}(-D)$ is the line bundle associated to the divisor $-D$ (similarly for $\mathcal{O}(-D')$ and $\mathcal{O}(-D-D')$.

Example: Suppose that $Y=\mathbb{P}^2_F$. Then, $\mathrm{Cl}(Y)=\mathbb{Z}$, and the first intersection pairing is just the multiplication map $\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$.

Example: Suppose that $Y=\mathbb{P}^1_F\times\mathbb{P}^1_F$. Then, $\mathrm{Cl}(Y)=\mathbb{Z}^2$, and the first intersection pairing is the map:

$\mathbb{Z}^2\times\mathbb{Z}^2\to\mathbb{Z}^2:((a,b),(c,d))\mapsto ad+bc$

Both of these examples can be computed using the explicit form of the intersection pairing described above, or by the description of their Chow rings given in the last subsection.

Remark: For those that like complex geometry, here is a description of the intersection pairing when $F=\mathbb{C}$. Recall first, using the exponential sequence:

$0\to\underline{\mathbb{Z}}\to\mathcal{O}_{Y^\text{an}}\to\mathcal{O}_{Y^\text{an}}^\times\to 0$

we get a long exact sequence containing the piece

\begin{aligned} 0\to \mathbb{Z} &\to \mathcal{O}_{Y^\text{an}}(Y^\text{an})\to \mathcal{O}_{Y^\text{an}}(Y^\text{an})\to H^1(Y^\text{an},\underline{\mathbb{Z}})\to H^1(Y^\text{an},\mathcal{O}_{Y^\text{an}})\to\\ &H^1(Y^\text{an},\mathcal{O}_{Y^\text{an}}^\times)\to H^2(Y^\text{an},\underline{ \mathbb{Z}})\end{aligned}

We denote the map

$\mathrm{Pic}(Y^\text{an})\cong H^1(Y^\text{an},\mathcal{O}_{Y^\text{an}})\to H^2(Y^\text{an},\underline{\mathbb{Z}})\cong H^2_{\text{sing}}(Y^\text{an},\mathbb{Z})$

by $c^1$, and call it the Chern class map.

But, we have a pairing, the cup product,

$\cup:H^2_\text{sing}(Y^\text{an},\mathbb{Z})\times H^2_\text{sing}(Y^\text{an},\mathbb{Z})\to H^4_\text{sing}(Y^\text{an},\mathbb{Z})\cong\mathbb{Z}$

(since we have a canonical orientation on complex manifolds, the last map is canonical). We then can define the first intersection pairing as follows:

$(D,D')\mapsto c^1\left(\mathcal{O}(D)^{\text{an}}\right)\cup c^1\left(\mathcal{O}(D')^{\text{an}}\right) \in \mathbb{Z}$

So, now we can define the equivalence relation $\sim_\text{num}$. Namely, we say that $Z,Z'\in\mathcal{Z}^r(Y)$ are numerically equivalent, denoted $Z\sim_\text{num}Z'$ if for all $W\in\mathcal{Z}^{n-r}(Y)$ we have that $\langle Z-Z',W\rangle_r=0$. Here we are conflating elements of $\mathcal{Z}^r(Y)$ with their corresponding elements of $\mathrm{CH}^r(Y)$, which is of no consequence since the intersection pairing is insensitive to rational equivalence definitionally. We then define the group $\mathcal{Z}^r(Y)/\sim_\text{num}$ by $\mathrm{Num}^r(Y)$, and call it the $r^\text{th}$ numerical class group of $Y$. In general numerical equivalence is weaker than rational equivalence, and so we have a surjection $\mathrm{CH}^r(Y)\twoheadrightarrow \mathrm{Num}^r(Y)$.

Example: Note that in both of the examples we computed in the last section, we actually have that $\sim_\text{rat}=\sim_\text{num}$. This is not a coincidence, as we shall see below.

## Algebraic equivalence

We define one more equivalence relation on $\mathcal{Z}^r(Y)$. This one, despite its name, is actually very geometric in nature. Similarly, in this section, we assume that $F$ is algebraically closed.

We say that two elements $Z,Z'\in\mathcal{Z}(Y)$ are algebraically equivalent, denoted $Z\sim_\text{alg} Z'$, if there exists a smooth (projective, integral) curve $C/F$ and an element $W\in\mathcal{Z}^{r+1}(Y\times C)$ such that $W\mid_{t_0}=Z$ and $W_{t_1}=Z'$ for some closed points $t_0,t_1\in |Y|$. The definition of the ‘restriction’ of an element of $\mathcal{Z}^{r+1}(Y\times C)$ is more delicate than may want to think, so we leave it as an intuitive definition–just intersection $W$ with the fiber $(Y\times C)_{t_0}\cong Y$. For a more formal definition, see Fulton’s book. The intuition for this definition is clear. Two $r$-cycles should be equivalent if we can put them together in an algebraic family of $r$-cycles; we can algebraically interpolate one cycle to the other.

We shall denote the quotient $\mathcal{Z}^r(Y)/\sim_\text{alg}$ by $\mathrm{NS}^r(Y)$ and call it the $r^\text{th}$ Neron-Severi group of $Y$.

Now, something one can prove which is intuitive, and the source of the term ‘rational’, is that $Z\sim_\text{rat}Z'$ if and only if $Z\sim_\text{alg}Z'$ where we take $C=\mathbb{P}^1_F$. In this way, we see that $\sim_\text{alg}$ is weaker than $\sim_\text{rat}$ and so we get a surjection $\mathrm{CH}^r(Y)\twoheadrightarrow \mathrm{NS}^r(Y)$. And, less obviously, but a good exercise, is that $\sim_\text{num}$ is even weaker than $\sim_\text{alg}$, and so we also obtain a natural surjection $\mathrm{NS}^r(Y)\twoheadrightarrow\mathrm{Num}^r(Y)$.

There are two theorems which rule in the land of the Neron-Severi group:

Theorem(Theorem of the Base): The group $\mathrm{NS}^r(Y)$ is finitely generated.

and

Theorem: The kernel of the map $\mathrm{Pic}(Y)\twoheadrightarrow\mathrm{NS}^1(Y)$ is $\mathrm{Pic}^0_{Y/F}(F)$.

Here $\mathrm{Pic}^0_{Y/F}$ is the connected component of the identity of the group scheme $\mathrm{Pic}^0_{Y/F}$ representing the relative Picard functor $\mathbf{Pic}_{Y/F}$ (see Kleiman’s article The Picard Scheme). In particular, $\mathrm{Pic}^0_{Y/F}$ is an abelian variety, and so $\mathrm{Pic}^0_{Y/F}(F)$ is $\ell$-divisible for any $\ell\ne\mathrm{char}(F)$.

# The cycle class map

Now that we know some basic definitions from intersection theory, we can define the object of central importance to us: the cycle class map. For $X$ (remember our conventions from the last section!) we will define a map:

$c^r:\mathcal{Z}^r\left(\overline{X}\right)\to H^{2r}\left(\overline{X},\mathbb{Q}_\ell(r)\right)$

bridging, as alluded to, intersection theory and $\ell$-adic cohomology.

Before jumping straight into the properties of this map, let us quickly state some intuition as to what happens in the case over $\mathbb{C}$. Namely, let’s assume that we have a complex manifold $M$ of complex dimension $n$. Then, the analogue of the cycle class map will be a map:

$c^r:\mathcal{Z}^r(M)\to H^{2r}_\text{sing}(M,\mathbb{C})=H^{2r}_\text{dR}(M/\mathbb{C})$

where this cohomology on the right means de Rham cohomology of $M$ with respect to smooth complex forms. Also, here $\mathcal{Z}^r(M)$ will denote the free abelian group on $r$-codimensional integral analytic subvarieties of $M$.

So, we want to take an $r$-cycle $Z\in\mathcal{Z}^r(M)$ and get out an element of

$H^{2r}_\text{dR}(M/\mathbb{C})=\left(H^{2n-2r}_\text{dR}(M/\mathbb{C})\right)^\vee$

Well, the map is simple. Namely, let’s define $c^r(Z)$ to the linear form:

$\displaystyle c^r(Z):H^{2n-2r}_\text{dR}(M/\mathbb{C})\to\mathbb{C}:\omega\mapsto \int_Z i_Z^\ast\omega$

here, $i_Z$ is the inclusion $Z\hookrightarrow M$. This is technically a lie unless $Z$ is smooth, but it gives the right idea. Namely, from this, it seems conceivable that we should be able to associate cohomology classes to cycles.

That said, the cycle class map for $\ell$-adic cohomology is not nearly as simple. If one is willing to have no intuition for the map, there is a quick and dirty definition using the Gysin map (see Milne’s Lectures on Étale Cohomology). The best way is to use the intersection (heh heh) of intersection theory and $K$-theory, in particular, to use the notion of the Chern character. That said, one can define this map and it enjoys several important properties.

First, note that both $\mathcal{Z}^r(\overline{X})$ and $H^{2r}\left(\overline{X},\mathbb{Q}_\ell(r)\right)$ naturally have an action of $G_k$. The action on $\mathcal{Z}^r(\overline{X})$ is described above, and the action on $H^{2r}\left(\overline{X},\mathbb{Q}_\ell(r)\right)=H^{2r}\left(\overline{X},\mathbb{Q}_\ell(r)\right)(r)$ is slightly more complicated (it’s briefly discussed here). Thus, one would hope that the map $c^r$ is a map of $G_k$-modules and, in fact, it is.

Second, if denote $\displaystyle \bigoplus_r H^{2r}(X,\mathbb{Q}_\ell(r))$ by $H^{2\ast}(\overline{X})$, the cycle class maps glue together to give us a map $c:\mathrm{CH}^\ast(\overline{X})\to H^{2\ast}(\overline{X})$. Now, $\mathrm{CH}^\ast\left(\overline{X}\right)$ has the structure of a ring, as desired above. One might hope that $H^{2\ast}(\overline{X})$ also has a ring structure, and that $c$ is also a map of rings. This is, in fact, true, the product on $H^{2\ast}\left(\overline{X}\right)$ coming from the cup product:

$H^{2i}(\overline{X},\mathbb{Q}_\ell(i))\times H^{2j}(\overline{X},\mathbb{Q}_\ell(j))\to H^{2i+2j}(\overline{X},\mathbb{Q}_\ell(i)\otimes_{\mathbb{Q}_\ell}\mathbb{Q}_\ell(j))=H^{2(i+j)}\left(\overline{X},\mathbb{Q}_\ell(i+j)\right)$

the cup product being described, for example, in terms of Cech cohomology.

Remark: This sort of compatibility has already been used, implicitly, in the analogue of the cycle class map above. In particular, the complex cycle class map $c^1$ for a complex manifold $M$ is, as the notation suggestions, just the Chern class map. The fact that the intersection pairing, which is, morally, just multiplication in the Chow ring, turned into cup product after taking Chern classes is precisely the manifestation of the fact that the cycle class map preserves products.

Luckily for us, the cycle class map is easy to explicitly describe when $r=1$ (the case that will be of the most interest to us). To understand it, first recall the Kummer sequence

$1\to\mu_{\ell^m}\to\mathbf{G}_m\to\mathbf{G}_m\to 1$

of sheaves of abelian groups on $\overline{X}_\text{et}$. Then, a portion of the long exact sequence gives us a map $H^1(\overline{X},\mathbf{G}_m)\to H^2(\overline{X},\mu_{\ell^m})$. These maps are compatible with varying $m$, and so by passing to the limit we obtain a map $H^1(\overline{X},\mathbf{G}_m)\to H^1(\overline{X},\mathbb{Z}(1))$. But, as is well known, we have the equality $H^1(\overline{X},\mathbf{G}_m)=\mathrm{Pic}(\overline{X})$. Thus, we obtain a map:

$\mathrm{Z}^1(\overline{X})=\mathrm{Div}(\overline{X})\to \mathrm{Pic}\left(\overline{X}\right)\to H^2\left(\overline{X},\mathbb{Z}(1)\right)\to H^2\left(\overline{X},\mathbb{Q}(1)\right)$

and this is the cycle class map $c^1$.

# The Tate conjecture

## Statement

We now finally have enough terminology set to state the Tate conjecture.

Let us begin by defining yet another equivalence relation on $\mathcal{Z}^r(\overline{X})$. Namely, let’s say that $Z\sim_\text{hom} Z'$ if and only if $Z-Z'\in\ker c^r$. We then say that $Z$ and $Z'$ are homologically equivalent. Let us denote the quotient of $\mathcal{Z}^r(\overline{X})/\sim_\text{hom}$ by $A^r\left(\overline{X}\right)$, and let’s denote the image of $\mathrm{Z}^r(X)$, the image of the rational cycles, by $A^r(X)$. Then, we obtain by definition injections:

$A^r(X)\hookrightarrow A^r\left(\overline{X}\right)\hookrightarrow H^{2r}\left(\overline{X},\mathbb{Q}_\ell(r)\right)$

Now, while it is not obvious (we talk more later about why its true for $r=1$, and how you might extend this to work in general) the injection $A^r(X)\hookrightarrow H^{2r}\left(\overline{X},\mathbb{Q}_\ell(r)\right)$ remains an injection after tensoring the left group with $\mathbb{Q}_\ell$. Thus, we obtain an injection

$A^r(X)\otimes_\mathbb{Z}\mathbb{Q}_\ell\hookrightarrow H^{2r}\left(\overline{X},\mathbb{Q}_\ell(r)\right)$

Note though, that since this maps comes from the cycle class map, that this injection is an injection of $G_k$ modules. In particular, since $A^r(X)\otimes_\mathbb{Z}\mathbb{Q}_\ell$ is a trivial $G_k$-module, we see that its image lands in $H^{2r}\left(\overline{X},\mathbb{Q}_\ell(r)\right)^{G_k}$.

Then, finally, the Tate conjecture says the following:

Conjecture(Tate): The map $A^r(X)\otimes_\mathbb{Z}\mathbb{Q}_\ell\to H^{2r}\left(\overline{X},\mathbb{Q}_\ell(r)\right)^{G_k}$ is an isomorphism.

We will denote the statement “The Tate conjecture is true for $X$, $r$, and $\ell$” by $T^r_\ell(X)$.

So, we see that the Tate conjecture fulfills the desire described in the motivation. Namely, we understand the important piece of the cohomology of $\overline{X}$, a complicated object, in terms of intersection theory. More specifically, it says that every cohomology class is not represented by a sum of $r$-cycles, but that it is, in fact, a $\mathbb{Q}_\ell$-linear combination of cycles.

## A related conjecture

Another conjecture which often times gets lumped in with the Tate conjecture (for reasons we’ll soon mention), is the following.

We defined yet another equivalence relation on $\mathcal{Z}^r(\overline{X})$ in the last section, the notion of homological equivalence $\sim_\text{hom}$. One might hope that this equivalence coincides with one of the equivalences on $\mathcal{Z}^r(\overline{X})$ that we’ve already defined. In fact, the conjecture is the following:

Conjecture(Numerical Equivalence=Homological Equivalence): The relations $\sim_\text{num}$ and $\sim_\text{hom}$ are equal on $\mathcal{Z}^r(\overline{X})$. In particular, $A^r\left(\overline{X}\right)=\mathrm{Num}^r(\overline{X})$ and $A^r(X)=\mathrm{Num}^r(X)$.

We shall denote the truth of this conjecture by the symbol $E^r_\ell(\overline{X})$.

This is a highly desirable result, not only because it allows us to not have to remember another equivalence relation on $\mathcal{Z}^r(\overline{X})$. Indeed, if the Tate conjecture is going to be practically useful, besides just bridging the gap between cohomology and intersection theory, we need to be able to, in practice, compute $A^r(X)$. The largest obstruction to this is determining when two elements of $\mathcal{Z}^r(\overline{X})$ are homologically equivalent. Since determining numerical equivalence is, in practice, a much simpler question, knowing conjecture $E^r_\ell(\overline{X})$ is hugely important.

## Highfalutin motivation

Without dwelling too long on such high-minded ideas, let us quickly state a motivation for the Tate conjecture which uses the objects of Grothendieck’s dream: motives (motifs). Let us quickly remind ourselves what the category of motives should be.

Given a smooth variety $Y/F$, there are many cohomology theories of which we can speak of. There is $\ell$-adic cohomology, there is algebraic de Rham cohomology, there is crystalline cohomology, there is Deligne cohomology, Hodge cohomology, and if $F=\mathbb{C}$ we even have the singular cohomology of the analytification. All of these cohomology theories are patently different, not only in the fact that they land in different categories ($\ell$-adic vector spaces, or $\mathbb{C}$-spaces, etc.), but they even have different ‘extra structure’. For example, Hodge cohomology comes with a grading, de Rham cohomology comes with a filtration, etc.

Grothendieck’s dream was to realize some sort of ‘universal cohomology theory’ for smooth varieties over $F$. In particular, some nice ‘linear algebra’ category $\mathsf{Mot}(F)$, and some functor:

$M:\mathsf{SmVar}/F\to\mathsf{Mot}(F)$

over which every cohomology theory factors. Namely, we have a series of functors:

$\begin{matrix} & & &\mathsf{GrVec}_F\\ & &\nearrow &\\ \mathsf{Mot}(F)& & & \vdots\\ & &\searrow &\\ & & &\mathsf{Vec}_{\mathbb{Q}_\ell}\end{matrix}$

such that, for example, the composition

$\mathsf{SmVar}/F\xrightarrow{M}\mathsf{Mot}(F)\to \mathsf{GrVec}_F$

is just the functor $H_\mathrm{Hodge}$ of Hodge cohomology. So it is, in a sense, a ‘universal cohomology theory’.

Note that since we can recover from $M(Y)$ things like the étale cohomology of $Y$, the linear algebra like object $M(Y)$ should, in some way, contain the information contained in $H^i(Y,\mathbb{Q}_\ell)$. So, for example, if $C/F$ is some smooth projective curve, then $M(C)$ should somehow capture the genus of $C$.

As an example of the category of motives, let’s restrict our attention from $\mathsf{SmVar}/F$ to $\mathsf{Curves}/F$ (smooth, projective, geometrically integral curves). Then, we want the category of motives to be something an abelian category. It turns out that a nice candidate for the category of motives be the category $\mathsf{AbVar}/F$ of abelian varieties over $F$. Our functor $M:\mathsf{Curves}/F\to\mathsf{AbVar}/F$ is nothing then but the Jacobian functor: $M(C)=\mathrm{Jac}(C)$. Although, this has some issues, of which I will not discuss.

Another option might be that for $\mathsf{AbVar}/F$ the category of motives might be category of continuous $G_F$-representations, the functor $M$ then being the Tate module functor: $A\mapsto V_\ell(A)$.

It turns out that, in general, there is a well-defined category of pure motives $\mathsf{Mot}(F)$ and a functor $M:\mathsf{SmVar}/F\to\mathsf{Mot}(F)$. It turns out though that there is more than one; we get such category of pure motives for every adequate equivalence relation on $r$-cycles. In particular, we get a category of pure motives with respect to rational equivalence (the so-called Chow motives), or one with respect to algebraic equivalence, or one with respect to numerical equivalence. Thus, in some sense, the theory of motives is the end-all and be-all of the desire to ‘relate intersection theory and cohomology’.

It turns out, in a way which I don’t want to make precise (see, for example, Milne’s article on the Tate conjecture) that conjectures $T^r_\ell(X)$ and $E^r_\ell(\overline{X})$ amount to the statement that $\ell$-adic cohomology forms an equivalence between the category $\mathsf{Mot}(k)\otimes\mathbb{Q}_\ell$ (this just means to ‘enrich’ the morphisms by tensoring all $\mathsf{Hom}$-groups with $\mathbb{Q}_\ell$) and the category of Tate structures which are $\mathbb{Q}_\ell$-spaces together with a fixed endomorphism, called Frobenius, which satisfies some simple properties. More specifically, the equivalence $\mathsf{Mot}(k)\otimes\mathbb{Q}_\ell\to\mathsf{TateSt}_k$ is the association $\mathcal{X}\mapsto (H^i_\ell(\mathcal{X}),\text{Frob}_q)$. Here $H^i_\ell(\mathcal{X})$ is the functor to continuous $\mathbb{Q}_\ell[G_k]$-representations such that $H^i_\ell(M(X))=H^i(\overline{X},\mathbb{Q}_\ell)$.

## Some known cases

### Curves

The only interesting question for curves $C/k$ is whether $T^1_\ell(C)$ and/or $E^1_\ell\left(\overline{C}\right)$ hold. Note though that

$H^2\left(\overline{C},\mathbb{Q}_\ell(1)\right)=H^2\left(\overline{C},\mathbb{Q}_\ell\right)(1)=\left(\mathbb{Q}_\ell(-1)\right)(1)=\mathbb{Q}_\ell$

where, implicitly, we have used the orientation isomorphism and. This is actually an isomorphism of $G_k$-modules, and so we have that

$H^2\left(\overline{C},\mathbb{Q}_\ell(1)\right)^{G_k}=\mathbb{Q}_\ell$

Now, $\mathrm{Div}\left(\overline{C}\right)$ are just the points of $\overline{C}$. The map, $c^1:\mathrm{Div}\left(\overline{C}\right)\to\mathbb{Q}_\ell$ is then, as one can check from our explicit description of $c^1$ above, just the degree map. Clearly then the image of $c^1$ will generate $\mathbb{Q}_\ell$ as a $\mathbb{Q}_\ell$-space.

In fact, the above really shows that $T^{\dim X}_\ell(X)$ is true, since the same argument applies!

### Abelian varieties

Let $A/k$ be an abelian variety. Using the Weil pairing, one can show that $A^1(A)\otimes_\mathbb{Z}\mathbb{Q}_\ell$ is equal to the set of endomorphisms in $\text{End}(A)\otimes_\mathbb{Z}\mathbb{Q}_\ell$ fixed under the Rosatti involution (see Milne’s article in Cornell-Silverman’s Arithmetic Geometry). Moreover, we know that $H^1(\overline{A},\mathbb{Q}_\ell)$ can be identified with $\bigwedge^2 V_\ell(A)^\vee$. Using this, one can show that $T^1_\ell(A)$ if and only if one has the equality $\text{End}(A)\otimes_\mathbb{Z}\mathbb{Q}_\ell=\text{End}_{G_k}\left(V_\ell(A)\right)$. But, this is true! This is just Tate’s isogeny theorem for finite fields. Thus, $T^1_\ell(A)$ holds true for all abelian varieties $A$, and for all $\ell\ne\mathrm{char}(k)$.

### One-dimensional numerical equivalence=homological equivalence

It is known that, in general, $E^1_\ell(X)$ holds for all $\ell$, and for all $X$. The proof of this is possible by the largely explicit description of the cycle class map in the case $r=1$. In fact, for surfaces, we’ll show this result below.

### K3 surfaces

It is known, by deep work of many people (Artin, Swinnerton-Dyer, Ogus, Pera, etc.) that $T^r_\ell(X)$ holds for all $K3$ surfaces, assuming that $\text{char}(k)\ne 2$. The final nail in the coffin was put by Pera in this 2013 paper.

### Products

One can show that $T^1_\ell(X\times_k Y)$ if and only if $T^1_\ell(X)$ and $T^1_\ell(Y)$. The proof of this actually uses the fact that $T^1_\ell(A)$ for abelian varieties $A$. Indeed, the key step is to use the decomposition

$\mathrm{Pic}(X\times_k Y)=\mathrm{Pic}(X)\times\mathrm{Pic}(Y)\times\mathrm{Hom}\left(\mathrm{Alb}(X),\mathrm{Pic}^0_{Y/k}\right)$

where $\mathrm{Alb}(X)$ is the Albanese variety of $X$, and $\mathrm{Pic}^0_{Y/k}$ is the connected component of the (relative) Picard scheme of $Y/k$. Note that both of these latter objects are abelian varieties, and this is where one leverages the knowledge of $T^1_\ell(A)$.

# Tate conjecture for surfaces

I would now like to refocus our attention on the case of surfaces $X/k$. We will be able to rephrase some of the theorems here in a nicer way. But, what is more important to us, why we discuss surfaces specifically, is its relation to number theory that will be discussed later.

## Reexamining the cycle class map

Since surfaces really only have non-trivial even cohomology in $r=2$, let’s focus our attention on $c^1$. Let’s recall how we built this. We started with the Kummer sequence on $\overline{X}$

$1\to \mu_{\ell^m}\to\mathbf{G}_m\to\mathbf{G}_m\to 1$

From there we took one map from the resulting long exact sequence. That said, we were ignoring too much of the picture. Instead of just taking one map, let’s instead include one more term of that sequence. Namely, the long exact sequence gives us the exact sequence

$H^1(\overline{X},\mathbf{G}_m)\xrightarrow{\ell^m}H^1(\overline{X},\mathbf{G}_m)\to H^2\left(\overline{X},\mu_{\ell^m}\right)$

Now, before we just observed that we had a map

$\mathrm{Pic}(\overline{X})=H^1(\overline{X},\mathbf{G}_m)\to H^2\left(\overline{X},\mu_{\ell^m}\right)$

But, inspecting the above, we see that we know much more. Namely, we know what the kernel of this map is: it’s $\ell^m\mathrm{Pic}(\overline{X})$. Thus, we get actually get an injective map

$\mathrm{Pic}(\overline{X})/\ell^m\hookrightarrow H^2\left(\overline{X},\mu_{\ell^m}\right)$

Passing to the limit then gives an injection

$\mathrm{Pic}\left(\overline{X}\right)\otimes_\mathbb{Z}\mathbb{Z}_\ell\hookrightarrow H^2\left(\overline{X},\mathbb{Z}_\ell(1)\right)$

But, note that this is just the map

$A^1(\overline{X})\otimes_\mathbb{Z}\mathbb{Z}_\ell\hookrightarrow H^2\left(\overline{X},\mathbb{Z}_\ell(1)\right)$

and so we see why the injectivity part of the Tate conjecture, as mentioned above, is not difficult, at least for $r=1$. A similar sort of argument, using the fact that we actually build the cycle class map from maps to $H^2\left(\overline{X},\mu_{\ell^m}^{\otimes r}\right)$ for all $m$ (via, say, the notion of Chern characters) gives us the injectivity in general.

That said, the injectivity was not the only thing we gain by looking at this perspective. Namely, let us observe what $\mathrm{Pic}\left(\overline{X}\right)/\ell^m$ is. Namely, let’s recall from the section on the Neron-Severi group that we have a short exact sequence

$0\to \mathrm{Pic}_{\overline{X}/\overline{k}}^0(\overline{k})\to\mathrm{Pic}(\overline{X})\to \mathrm{NS}\left(\overline{X}\right)\to 0$

(since we’ll only be considering $r=1$ for a surface, we use the short-hand $\mathrm{NS}(\overline{X})$ for $\mathrm{NS}^1(\overline{X})$).

But, as we noted above, since $\mathrm{Pic}_{\overline{X}/\overline{k}}^0$ is an abelian variety over $k$, and $\ell\ne\mathrm{char}(k)$, we have that $\mathrm{Pic}^0_{\overline{X}/\overline{k}}(\overline{k})$ is $\ell$-divisible. So, we get, naturally that

$\mathrm{NS}\left(\overline{X}\right)/\ell^m=\mathrm{Pic}\left(\overline{X}\right)/\ell^m$

(more precisely, the inclusion induces an isomorphism). Thus, we see that we now have an injection

$\mathrm{NS}(\overline{X})\otimes_\mathbb{Z}\mathbb{Z}_\ell\hookrightarrow H^2\left(\overline{X},\mathbb{Z}_\ell(1)\right)$

Moreover, it’s clear that the induced map

$\mathrm{NS}(X)\otimes_\mathbb{Z}\mathbb{Q}_\ell\hookrightarrow\mathrm{NS}\left(\overline{X}\right)\otimes_\mathbb{Z}\mathbb{Q}_\ell\hookrightarrow H^2\left(\overline{X},\mathbb{Q}_\ell(1)\right)$

has the same image as the map $A^1(X)\otimes_\mathbb{Z}\mathbb{Q}_\ell$. Thus, we see that the Tate conjecture is equivalent to the statement that the map

$\mathrm{NS}\left(X\right)\otimes_\mathbb{Z}\mathbb{Q}_\ell\hookrightarrow H^2\left(\overline{X},\mathbb{Q}_\ell(1)\right)^{G_k}$

is an isomorphism.

## Replacing Neron-Severi group with numerical class group

What we’d like to do is say that we can naturally replace the object

$\mathrm{NS}(X)_{\mathbb{Q}_\ell}:=\mathrm{NS}(X)\otimes_\mathbb{Z}\mathbb{Q}_\ell$

with, what is ostensibly a proper quotient, namely:

$\mathrm{Num}(X)_{\mathbb{Q}_\ell}:=\mathrm{Num}^1(X)\otimes_\mathbb{Z}\mathbb{Q}_\ell$

Let us begin by writing down a long sequence of quotients that we have, or will soon, establish:

$\mathrm{Pic}(X)\twoheadrightarrow\mathrm{NS}(X)\twoheadrightarrow A^1(X)\twoheadrightarrow\mathrm{Num}(X)$

The first (left-to-right) surjection is clear, and is one we already talked about.

To see that $\mathrm{NS}(X)\twoheadrightarrow A^1(X)$, what we really need to show is that those $Z\in\mathcal{Z}^1(\overline{X})$ with $Z\sim_\text{alg}0$ satisfy $c^1(Z)=0$. But, by construction if $Z\sim_\text{rat}0$ then evidently $c^1(Z)=0$ (think of how we constructed $c^1$!) and then, inside of $\mathrm{Pic}(\overline{X})$ those cycles algebraically equivalent to zero form an $\ell$-divisible group. Thus, they go to zero in $H^2\left(\overline{X},\mu_{\ell^m}\right)$ for all $m$, and thus go to zero in $H^2\left(\overline{X},\mathbb{Q}_\ell(1)\right)$ (note that this is a rehashing of an earlier argument).

To see the surjection $A^1\left(X\right)\twoheadrightarrow\mathrm{Num}(X)$ we need to show that $c^1(Z)=0$ implies that $Z\sim_\text{num}0$. That said, note that if $c^1(Z)=0$ then for any $Z'\in\mathcal{Z}^1(\overline{X})$, we have that

$Z\cdot Z'=c^1(Z)\cup c^1(Z')=0\cup c^1(Z')=0$

which implies the claim.

In fact, we claim that we can extend this sequence of quotients one step further. Namely, we claim that we have a sequence of quotients as follows:

$\mathrm{Pic}(X)\twoheadrightarrow\mathrm{NS}(X)\twoheadrightarrow \mathrm{NS}(X)/\mathrm{NS}(X)[\mathrm{tors}]\twoheadrightarrow A^1(X)\twoheadrightarrow\mathrm{Num}(X)$

To check this it suffices to check that $A^1(\overline{X})$ is torsion free. To see this, we merely note that if $Z\in\mathcal{Z}^1_h(\overline{X})$ and $nZ\sim_\text{hom}0$, then $c^1(nZ)=nc^1(Z)=0$. But, since $n$ is invertible in $H^2\left(\overline{X},\mathbb{Q}_\ell(1)\right)$, this implies that $c^1(Z)=0$ and thus our claim follows.

We can then appeal to a powerful theorem proved by Matsusaka in his 1957 paper The Criteria for Algebraic Equivalence and the Torsion Group. There he proves that the kernel of the map $\mathrm{NS}\left(\overline{X}\right)\to\mathrm{Num}\left(\overline{X}\right)$ is precisely $\mathrm{NS}(\overline{X})[\mathrm{tors}]$. This immediately implies two important facts. First, it shows that $E^1_\ell(X)$ holds for $X$! Namely, just by virtue of sitting in the above sequence of quotients, together with Matsusaka’s theorem, tells us that the map $A^r(\overline{X})\to\mathrm{Num}(\overline{X})$ is an isomorphism, and since this is the canonical quotient, this implies that $\sim_\text{hom}=\sim_\text{num}$. Second, since this tells us that $\mathrm{NS}(X)$ differs from $\mathrm{Num}(X)$ by only torsion elements, and both these groups are finitely generated, we see that the natural quotient maps

$\mathrm{NS}(X)_{\mathbb{Q}_\ell}\twoheadrightarrow A^1(X)\otimes_{\mathbb{Z}}\mathbb{Q}_\ell\twoheadrightarrow\mathrm{Num}(X)_{\mathbb{Q}_\ell}$

are actually isomorphisms.

## Some restatements of Tate’s conjecture

Now with the above analysis, we can fruitfully rephrase Tate’s conjecture in two pleasing ways.

First note that the injection

$\mathrm{NS}(X)_{\mathbb{Q}_\ell}\hookrightarrow H^2\left(\overline{X},\mathbb{Q}_\ell\right)^{G_k}$

gives us an inequality

$\mathrm{rk}\left(\mathrm{NS}(X)\right)\leqslant\dim_{\mathbb{Q}_\ell} H^2\left(\overline{X},\mathbb{Q}_\ell\right)^{G_k}$

(the rank here on the left hand side is well-defined since $\mathrm{NS}(X)$ is a finitely generated group). But, from the discussion in the last subsection, we know that $\text{rk}(\mathrm{NS}(X))=\mathrm{rk}(\mathrm{Num}(X))$. Moreover, we see that equality, in place of this inequality, is actually equivalent to the Tate conjecture.

To continue further, we need to analyze closer the space $H^2\left(\overline{X},\mathbb{Q}_\ell(1)\right)^{G_k}$. Note that since $G_k$ is topologically generated by $\mathrm{Frob}_q$, the geometric Frobenius, and the action on $H^2\left(\overline{X},\mathbb{Q}_\ell(1)\right)$ is continuous,we have that:

$H^2\left(\overline{X},\mathbb{Q}_\ell(1)\right)^{G_k}=H^2\left(\overline{X},\mathbb{Q}_\ell(1)\right)^{\mathrm{Frob}_q=1}$

where the superscript denotes the elements where $\mathrm{Frob}_q$ acts trivially. But, note that

$H^2\left(\overline{X},\mathbb{Q}_\ell(1)\right)=H^2\left(\overline{X},\mathbb{Q}_\ell\right)(1)$

and so

$H^2\left(\overline{X},\mathbb{Q}_\ell(1)\right)^{\mathrm{Frob}_q=1}=H^2\left(X,\mathbb{Q}_\ell\right)^{\mathrm{Frob}_q=q}$

(where, once again, this superscript denotes the set of elements where $\mathrm{Frob}_q$ acts as multiplication by $q$). Note that it was important we used geometric Frobenius here!

To go further, let us recall that since $X/k$ is a smooth variety, we have a well-defined notion of a zeta function: $\zeta(X,s)$. In particular, we have equality $\zeta(X,s)=Z(X,q^{-s})$ where

$\displaystyle Z(X,T)=\exp\left(\sum_n \#(X(\mathbb{F}_{q^n}))\frac{T^n}{n}\right)$

Then, contained in the circle of results known as the Weil conjectures (no longer a conjecture due to Grothendieck, Dwork, Deligne, etc.) is the result that $-\mathrm{ord}_{s=1}\zeta(X,s)$ is precisely the algebraic multiplicity of the eigenvalue $q$ for $\mathrm{Frob}_q$ acting on $H^2\left(\overline{X},\mathbb{Q}_\ell\right)$. In particular, it’s clear then that we now have a longer chain of inequalities:

$\displaystyle \mathrm{rk}\left(\mathrm{NS}(X)\right)=\mathrm{rk}\left(\mathrm{Num}(X)\right)\leqslant\dim_{\mathbb{Q}_\ell}H^2\left(\overline{X},\mathbb{Q}_\ell(1)\right)^{G_k}\leqslant -\mathrm{ord}_{s=1}\zeta(X,s)$

Once again, the first inequality being equality is equivalent to $T^1_\ell(X)$.

In the above chain of inequalities, the second inequality being an equality is equivalent to another conjecture, which some people call $S^1_\ell(X)$. This would follow quite easily from one of the Standard Conjectures stating that $H^2\left(\overline{X},\mathbb{Q}_\ell\right)$ should be a semi-simple $G_k$-module. Thus, one  might be led to make a ‘strengthening of Tate’s conjecture’. Namely, that $\mathrm{rk}(\mathrm{Num})(X)=-\mathrm{ord}_{s=1}\zeta(X,s)$. It’s clear that this is equivalent to $T^1_\ell(X)+S^1_\ell(X)$.

Somewhat surprisingly, this conjecture is no stronger:

Theorem: The veracity of $T^1_\ell(X)$ is equivalent to $\mathrm{rk}(\mathrm{Num})(X)=-\mathrm{ord}_{s=1}\zeta(X,s)$.

Proof: Clearly if the latter condition holds, then $T^1_\ell(X)$ holds, and so it suffices to prove the opposite claim.

We begin by noting that since the intersection pairing and the cup-product pairing are perfect on $\mathrm{NS}(X)_{\mathbb{Q}_\ell}$ and $H^2\left(\overline{X},\mathbb{Q}_\ell\right)$ we get an isomorphism of $\mathbb{Q}_\ell$-spaces:

$\mathrm{NS}(X)_{\mathbb{Q}_\ell}=\mathrm{Hom}_{\mathbb{Z}}\left(\mathrm{NS}(X),\mathbb{Q}_\ell\right)$

and an isomorphism of $\mathbb{Q}_\ell$-spaces

$H^2\left(\overline{X},\mathbb{Q}_\ell(1)\right)\cong \mathrm{Hom}_{\mathbb{Q}_\ell}\left(H^2\left(\overline{X},\mathbb{Q}_\ell(1)\right),\mathbb{Q}_\ell\right)$

Note though that this last isomorphism is not an isomorphism of $G_k$-modules. In fact, one can easily check that under this isomorphism $\mathrm{Hom}_{\mathbb{Q}_\ell}\left(H^2\left(\overline{X},\mathbb{Q}_\ell(1)\right)^{G_k},\mathbb{Q}_\ell\right)$ is taken to $H^2\left(\overline{X},\mathbb{Q}_\ell(1)\right)_{G_k}$–the coinvariants of $H^2\left(\overline{X},\mathbb{Q}_\ell(1)\right)$.

We then have a large diagram:

$\begin{matrix}\mathrm{NS}(X)_{\mathbb{Q}_\ell} & \xrightarrow{\approx} & \mathrm{Hom}_{\mathbb{Z}}\left(\mathrm{NS}(X),\mathbb{Q}_\ell\right)\\ \downarrow & & \uparrow\\ H^2\left(\overline{X},\mathbb{Q}_\ell(1)\right)^{G_k} & \xrightarrow{\alpha} & H^2\left(\overline{X},\mathbb{Q}_\ell(1)\right)_{G_k} \end{matrix}$

Here the top morphism is the isomorphism coming from the perfect pairing as above. The morphism $\alpha$ is the obvious map from the $G_k$-invariants to the $G_k$. The left hand vertical morphism is the cycle class map, and the right hand vertical morphism is the dual of the cycle class map, under the identification of $H^2(\overline{X},\mathbb{Q}_\ell(1))_{G_k}$ with $\mathrm{Hom}_{\mathbb{Q}_\ell}\left(H^2\left(\overline{X},\mathbb{Q}_\ell(1)\right)^{G_k},\mathbb{Q}_\ell\right)$.

One can show (although it takes some checking), that this diagram commutes. So, now, if $T^1_\ell(X)$ holds, then both vertical maps are isomorphisms, and thus $\alpha$ must also be an isomorphism. But, note that $\alpha$ is an isomorphism if and only if $H^2\left(\overline{X},\mathbb{Q}_\ell\right)^{\mathrm{Frob}_q=q}$ is equal to the generalized eigenspace of $q$ for $\mathrm{Frob}_q$ which has dimension precisely $-\mathrm{ord}_{s=1}\zeta(X,s)$. Thus, we see that $T^1_\ell(X)$ implies $S^1_\ell(X)$, and the conclusion follows. $\blacksquare$

Now, one might think this is a nice theorem because it says that the Tate conjecture is equivalent to understanding analytic data of a zeta function (and, in fact, understanding the analytic data algebraically!). And, yes, it is extremely nice for that reason. But, if you look closer, the above theorem implies something else which is also glee-inducing:

Corollary: The veracity of $T^1_\ell(X)$ is independent of $\ell$.

Indeed, the statement $\mathrm{rk}(\mathrm{Num}(X))=-\mathrm{ord}_{s=1}\zeta(X,s)$ makes no reference to $\ell$ anywhere!

Remark: Whenever one has a statement involving $\ell$-adic cohomology, one of the first thing asks is for an ‘independence of $\ell$‘ result, telling us that the choice of $\ell$ is largely unimportant. This isn’t just nice for practical or aesthetic reasons. If one really believes in some sort of ‘philosophy of motives’, then all ‘true geometric results’ should be true about the motives themselves, not just about the individual cohomologies. Then, since each $\ell$-adic cohomology, for varying $\ell$, comes from a different realization of the motive, the theorem being true for the motive implies its true for all $\ell$. This somehow lends credence to the earlier claim that the Tate conjecture is of a motivic nature.

It should also be mentioned that we also have independence of $\ell$ for the Tate conjecture in general. The proof, as above, largely goes through regardless of whether $X$ is a surface or not. The case $r>1$ is slightly more tricky, but has a similar ‘messing around with linear algebra’ type feel.

# The Tate conjecture and the Brauer group

We would now like to discuss a surprising connection between the Tate conjecture for a surface, and the Brauer group of the surface.

## The definition of the Brauer group

Since it’s not entirely standard, let’s begin by recalling how one defines the Brauer group of a smooth variety $X/k$. There are many equivalent ways of defining $\mathrm{Br}(X)$, hopefully one of them resonates with the reader:

\begin{aligned}\mathrm{Br}(X) &= \left\{\text{quasi-coherent }\mathcal{O}_X\text{-algebras, etale locally }U\twoheadrightarrow X\text{ isomorphic to }\mathrm{Mat}_n(\mathcal{O}_U)\right\}/\sim\\ &=\left\{\text{locally free, quasi-coherent }\mathcal{O}_X\text{-algebras whose fibers are central simple algebras}\right\}/\sim\\ &=\varinjlim H^1_\text{et}(X,\mathrm{PGL}_n)/H^1_\text{et}\left(X,\mathrm{GL}_n\right)\\ &= H^1_\text{et}\left(X,\mathrm{PGL}_\infty\right)/H^1_\text{et}\left(X,\mathrm{GL}_\infty\right)\\ &= \left\{\text{Projective bundles on }X\right\}/\left\{\text{Projectivization of vector bundles}\right\}\\ &= H^2_\text{et}(X,\mathbf{G}_m)\end{aligned}

The first two definitions are defining what are sometimes called Azumaya algebras on $X$, and the equivalence $\sim$ is ‘Morita equivalence’, meaning their category of modules on $X$ is isomorphic. For more information on these definitions, and to see their proofs (except the last equality, see below), one can look at the appropriate chapter of Milne’s Étale Cohomology book (NB: it needs to be his book, not his online lecture notes).

Remark: Once one knows what all of these terms means, showing the majority of these equalities is trivial/definitional. The only non-trivial one is the last equality. It’s a conjecture that, in general, if $X$ is, say, a variety, then $\mathrm{Br}(X)$ (defined using the first definition, say) is equal to $H^2_\text{et}(X,\mathbf{G}_m)[\mathrm{tors}]$ (in the above, one can show that $H^2_\text{et}(X,\mathbf{G}_m)$ is torsion, so there is no need to put torsion). This has been proven by Gabber/de Jong in the case that $X$ is quasi-projective. The technique by which de Jong proves this result is by virtue of the theory of twisted sheaves, an indispensable tool in modern arithmetic geometry. One can read about their ubiquity in this article of Lieblich.

It’s nice to note that any geometrically integral separated (possibly non proper!) is quasiprojective, so this is very general in this case. Indeed, using Nagata compactification, we can open embed our surface into a proper surface. Then, by applying resolution of singularities (which holds in characteristic $p$ for surfaces by work of Abhyankar!), we obtain then a proper smooth scheme containing our scheme as a dense open subset. We then use the fact (Theorem 1.2.8 in Badescu’s Algebraic Surfaces) that every smooth proper surface is projective. The result follows.

As an example, one can compute the following:

Example: If $X=\text{Spec}(F)$, for $F$ a field, then $\mathrm{Br}(X)=\mathrm{Br}(F)$ in the usual sense. Thus, all instances of the Brauer groups of fields that one might know are valid examples:

$\mathrm{Br}(F)=0$ if $F$ is one of the following: an algebraically closed field, a finite field, the function field of a curve over an algebraically closed field (Tsen’s theorem), the function field of a curve over a finite field (this is Tsen’s theorem again).

$\mathrm{Br}(\mathbb{R})=H^2_\text{et}(\text{Spec}(\mathbb{R}),\mathbf{G}_m)=H^2(G_{\mathbb{R}},\mathbb{C}^\times)=\mathbb{R}^\times/\mathbb{R}^{>0}=\mathbb{Z}/2\mathbb{Z}$.

-If $F$ is a local field (any characteristic), then $\mathrm{Br}(X)=\mathbb{Q}/\mathbb{Z}$. This is a large portion of local class field theory.

-If $F$ is a global field, then $\mathrm{Br}(F)$ sits in a short exact sequence like

$\displaystyle 0\to\mathrm{Br}(F)\to\bigoplus_v \mathrm{Br}(F_v)\xrightarrow{\mathrm{inv}}\mathbb{Q}/\mathbb{Z}\to 0$

where $v$ ranges over the places of $F$, and $F_v$ is the completion of $F$ at $v$. This is a large portion of global class field theory, and contains (in the exactness at $\mathrm{Br}(X)$) the famous Albert–Brauer–Hasse–Noether theorem which generalizes the Hasse-Minkowski theorem.

Example: If $X$ is regular and integral, with function field $F$, then $\mathrm{Br}(X)\hookrightarrow\mathrm{Br}(F)$. In particular, if $X/F$ is a smooth curve (projective or affine), then $\mathrm{Br}(X)=0$ (by Tsen’s theorem). This also follows, for example, by a direct calculation of $H^2_\text{et}(X,\mathbf{G}_m)$.

## The relationship between the Brauer group and the Tate conjecture

Now that we know the definition, and some basic examples of the Brauer group, we can begin to connect it to the Tate conjecture. A goal, which at first glance, seems formidable.

The key to the bridging of the two notions is, once again, the Kummer sequence. But, now we want to consider the Kummer sequence on $X$. Namely, taking the associated long exact sequence gives us

$H^1\left(X,\mathbf{G}_m\right)\xrightarrow{\ell^m}H^1(X,\mathbf{G}_m)\to H^2(X,\mu_{\ell^m})\to H^2(X,\mathbf{G}_m)\xrightarrow{\ell^m}H^2(X,\mathbf{G}_m)$

Then, by performing some algebra, we obtain the short exact sequence

$0\to\mathrm{NS}(X)/\ell^m\to H^2(X,\mu_{\ell^m})\to \mathrm{Br}(X)[\ell^m]\to 0$

Taking the limit then gives us a short exact sequence

$0\to\mathrm{NS}(X)\otimes_{\mathbb{Z}}\mathbb{Z}_\ell\to H^2\left(\overline{X},\mathbb{Z}_\ell(1)\right)^{G_k}\to T_\ell \mathrm{Br}(X)\to 0$

Here we are implicitly using that $\varprojlim H^2(X,\mu_{\ell^m})=H^2(\overline{X},\mathbb{Z}_\ell(1))$, which is true, I believe, due to Tate; it boils down roughly to the fact that $\mathrm{Br}(k)=0$. Also, here $T_\ell \mathrm{Br}(X)$ is the ‘Tate module’ of $\mathrm{Br}(X)$: the $\mathbb{Z}_\ell$-module $\varprojlim \mathrm{Br}(X)[\ell^m]$.

Finally, tensoring with $\mathbb{Q}_\ell$ gives the sort exact sequence

$0\to\mathrm{NS}(X)_{\mathbb{Q}_\ell}\to H^2(\overline{X},\mathbb{Q}_\ell(1))^{G_k}\to V_\ell \mathrm{Br}(X)\to 0$

here, of course, $V_\ell\mathrm{Br}(X)$ denotes $T_\ell\mathrm{Br}(X)\otimes_{\mathbb{Z}_\ell}\mathbb{Q}_\ell$. Thus, we see that $T^1_\ell(X)$ is equivalent to the condition that $V_\ell\mathrm{Br}(X)=0$.

Now, $\mathrm{Br}(X)_\ell$ is finite, since it is surjected onto by $H^2(X,\mu_\ell)$. Thus, one may decompose $\mathrm{Br}(X)[\ell^\infty]$ as $(\mathbb{Q}_\ell/\mathbb{Z}_\ell)^d\times G$, where $G$ is a finite group. One can then check that $V_\ell \mathrm{Br}(X)=\mathbb{Z}_\ell^d$. Thus, we arrive at the following amazing theorem:

Theorem: Let $X/k$ be a smooth projective surface. Then, the veracity of $T^1_\ell(X)$ is equivalent to the statement that $\mathrm{Br}(X)[\ell^\infty]$ is finite.

Note that since we already have independence of $\ell$ for the above, we see that this implies that $\mathrm{Br}(X)[\ell^\infty]$ is finite for one $\ell\ne p$ if and only if it’s true for all $\ell\ne p$. In fact, working harder, one can even take care of the previously disallowed case $\ell=p$:

Theorem: Let $X/k$ be a smooth projective surface. Then, the veracity of $T^1_\ell(X)$ is equivalent to the statement that $\mathrm{Br}(X)$ is finite.

Wow! I find this just incredible. The Tate conjecture, an incredibly geometric, incredibly famous result, is related to the finiteness of an object of astounding importance in its own right (the Brauer group) but which, on inspection, should have nothing to do with connecting intersection theory and cohomology. In particular, we see that the structure of projective bundles on a surface, namely how close they are to projectivization of vector bundles, informs (is equivalent to!) how well intersection theory approximates $\ell$-adic cohomology on $X$…Wow!

# Relationship to number theory

On top of the amazing story unfolded above, the Tate conjecture goes one step further, inserting itself squarely at the top of (a certain type) of number theorist’s wish list.

## The Birch and Swinnerton-Dyer conjecture for function fields

As the title of this subsection suggests, the number theoretic conjecture related to the Tate conjecture is the Birch and Swinnerton-Dyer conjecture for elliptic curves over function fields. So, we quickly recall what this says.

Let $k$ be as usual, a finite field, and $C/k$ a smooth projective geometrically integral curve. Let $K:=K(C)$ be the function field $C$, and let $E/K$ be an elliptic curve. One of the most important arithmetic objects associated to $E$ is its $L$-function, $L(E,s)$. This is defined as follows:

$\displaystyle L(E,s)=\prod_v L_v(E,q^{-s}),\qquad L_v(E,T)=\begin{cases}\left(1-a_vT^{\deg v}+q_v T^{2\deg v}\right)^{-1} & \mbox{if}\quad v\text{ is good}\\ \left(1-a_v T^{\deg v}\right)^{-1} & \mbox{if}\quad v\text{ is bad}\end{cases}$

where, as usual,

$\displaystyle a_v=\begin{cases}q_v+1-\#(E(K_v)) & \mbox{if}\quad v\text{ is good}\\ 1 & \mbox{if} \quad v\text{ is split multiplicative}\\ -1 & \mbox{if}\quad v\text{ is non-split multiplicative}\\ 0 & \mbox{if}\quad v\text{ is additive}\end{cases}$

$v$ ranges over the places of $K$ (the points of $C$!), $K_v$ is the completion of $K$ at $v$, $\mathfrak{k}_v$ is the residue field of $K_v$ (a local field), and $q_v=\#(\mathfrak{k}_v)$. For the definition of all of these concepts (reduction type), see Silverman’s book on elliptic curves.

Function fields of curves over finite fields are one of the two types of global fields (the other being number fields), and so are of huge interest to number theorists. In particular, elliptic curves have particular arithmetic importance. The $L$-function of an elliptic curve is an analytic way of bundling together this arithmetic information, and is usually (because analysis is so powerful!) the most fruitful way of attacking problems concerning the elliptic curve.

The long-standing conjecture of Birch and Swinnerton-Dyer says that a particular important piece of arithmetic data associated to $E$ is equal to an equally important analytic piece of data of its $L$-function. In particular, just like the case of number fields (which most readers are probably familiar), the group of $K$-rational points $E(K)$ is a finitely generated group. This is a result due to Lang and Neron. It can be proved very similarly to the usual Mordell-Weil theorem (the analogue for number fields). But, one can use the more rich algebro-geometric setting to give a different proof. See this article of Conrad for an exposition in modern language. The conjecture is then the following:

Conjecture(Birch and Swinnerton-Dyer): Let $E/K$ be an elliptic curve. Then,

$\mathrm{rk}E(K)=\mathrm{ord}_{s=1}L(E,s)$

There is a much finer version of the conjecture, also stating what $\mathrm{res}_{s=1}L(E,s)$ is in terms of arithmetic data. I won’t state this here, but I will remark that it is equivalent to the version stated above (somewhat surprisingly).

There are a few equivalent versions of the Birch and Swinnerton-Dyer conjecture over function fields. Some of them involve the Tate-Shafarevich group of $E$. This is usually denoted with the Cyrillic letter Sha, but since Word Press LaTeX cannot support this, I’ll use the notation $\mathrm{Sh}(E/K)$ instead. The definition of this group is somewhat complicated. Essentially, it measures failure of a local-to-global principle of the zeroness of a class $c\in H^1(G_k,E(\overline{K})$. In particular, the elements of $\mathrm{Sh}(E/K)$ are, basically, the elements of $H^1(G_K,E(\overline{K}))$ which become zero in every $H^1(G_{K_v},E(\overline{K_v}))$ but are not themselves zero. This group comes up in the generalized statement of the Birch and Swinnerton-Dyer conjecture as one of the arithmetic terms in the expression of $\mathrm{res}_{s=1}L(E,s)$.

As of now, it is not even known whether $\mathrm{Sh}(E/K)$ is even finite. In fact:

Theorem: Let $E/K$ be an elliptic curve. Then, the following are equivalent:

1. The Birch and Swinnerton-Dyer conjecture holds for $E$.

2. We have inequality $\mathrm{rk}(E(K))\leqslant \mathrm{ord}_{s=1}L(E,s)$.

3. The group $\mathrm{Sh}(E/K)$ is finite.

4. The group $\mathrm{Sh}(E/K)[\ell^\infty]$ is finite, for any $\ell$ (even $\ell=p$!).

## The relationship with the Tate conjecture

We now explain the unexpected, and incredibly gratifying relationship between the Birch and Swinnerton-Dyer conjecture and the Tate conjecture.

To state it though, we need a technical fact:

Theorem: Let $E/K$ be an elliptic curve. Then, there exists a smooth, geometrically integral, minimal surface $\mathcal{E}/k$ and a projective, surjective morphism $\pi:\mathcal{E}\to E$. This $\mathcal{E}$ is unique up to birational equivalence.

Recall that a surface is minimal if it is not the blow-up of another surface at a point or, equivalently (this is Castelnuovo’s criterion), it contains no $(-1)$-curves (curves with self-intersection $-1$).

Proof(Sketch): Lift $E/K$ to a model $M\to U$, over some dense open $U\subseteq C$. Use explicit Weierstrass type form of $M$ to extend this to a, possibly non-smooth, model $\overline{M}\to C$. Use resolution of singularities to obtain a smooth projective model. Show that there are only finitely many $(-1)$-curves to blow-down. $\blacksquare$

The absolutely flooring theorem is then the following:

Theorem: Let $E/K$ be an elliptic curve, and $\mathcal{E}$ the associated surface. Then,

$\mathrm{Br}(\mathcal{E})=\mathrm{Sh}(E/K)$.

In particular, using the equivalent forms of the Tate conjecture and Birch and Swinnerton-Dyer conjecture that we’ve discussed:

Theorem: Let $E/K$ be an elliptic curve with associated surface $\mathcal{E}$. Then, the veracity of the Birch and Swinnerton-Dyer conjecture for $E$ is equivalent to the veracity of $T^1_\ell(\mathcal{E})$.

This theorem is, without a doubt, one of the most beautiful (a word I don’t use lightly!) mathematical facts I know. It, in one-fell swoop, connects almost every subject of modern mathematics: geometry (intersection theory), algebra (cohomology theory), number theory (ranks of elliptic curves), and analysis (orders of meromorphic functions). I don’t know what else needs to be said.

The proof of this theorem though, as one might guess, is a little involved. In fact, one writes down a spectral sequence which contains both of these groups and shows that the situation forces enough of the terms to vanish to force a degeneration giving the desired result. If you want to read it in its entirety, you can find it in Grothendieck’s Le Groupe de Brauer III, section 4.

I’m very late to the party, but in the last example just before §”Numerical equivalence”, I think that one should replace $(s^n, t^m)$ by $(s^{n+1}, t^{m+1})$ — see theorem 2.10 in Eisenbud–Harris.
Thank you for this interesting article. I think that there might be a typo in the last example just before the paragraph “Numerical equivalence” : we should have a quotient by the ideal $(s^{n+1},t^{m+1})$, as mentioned in Theorem 2.10 in “3264 & All That Intersection Theory” by Eisenbud and Harris.