Lecture 3: Abelian varieties (analytic theory)
This lecture covers two disjoint topics. First, I go over the theory of elliptic curves over finite fields (point counting and the notions of ordinary and supersingular). Then I talk about the abelian varieties over the complex numbers from the analytic point of view.
Elliptic curves over finite fields
A good reference for this section is Chapter V of Silverman’s “The arithmetic of elliptic curves” (MR0817210).
Point counting
Let $E$ be an elliptic curve over the finite field $\bF_q$. Then $E^{(q)}=E$, and so the Frobenius map $F_q$ maps $E$ to itself. A point $x$ of $E(\ol{\bF}_q)$ belongs to $E(\bF_q)$ if and only if it is fixed by $F_q$ (since this is equivalent to it being Galois invariant). Thus $E(\bF_q)$ is the set of $\ol{\bF}_q$points of the kernel of the endomorphism $1F_q$. This endomorphism is separable: indeed, if $\omega$ is a differential on $E$ then $F_q^*(\omega)=0$, and so $(1F_q)^* \omega=\omega$ is nonzero. We have thus proved the following proposition:
Proposition. $\# E(\bF_q) = \deg(1F_q)$.
Recall that we have defined a positive definite bilinear pairing $\langle, \rangle$ on $\End(E)$, and that $\langle f, f \rangle = \deg(f)$. Appealing to the CauchySchwartz inequality, we find $\langle 1, F_q \rangle^2 \le \deg(q) \deg(F_q) = q$, and so $\langle 1, F_q \rangle \le \sqrt{q}$. But, by definition,
$$ 2 \langle 1, F_q \rangle = \deg(1F_q)\deg(1)\deg(F_q), $$and so we have the following theorem
Theorem (Hasse bound). $\# E(\bF_q)q1 \le 2 \sqrt{q}$.
In other words, we can write $\# E(\bF_q)$ as $q+1a$, where $a$ is an error term of size at most $2 \sqrt{q}$. We have $a=\langle 1, F_q \rangle$ by the above. We also have the following interpretation of $a$:
Proposition. We have $a = \tr(F_q \mid T_{\ell}{E})$.
Proof. This is formal: if $A$ is any $2 \times 2$ matrix, then
$$ \tr(A)=1+\det(A)\det(1A). $$Applying this to the matrix of $F_q$ on $T_{\ell}{E}$, the result follows. ◾
A Weil number (with respect to $q$) of weight $w$ is an algebraic number with the property that any complex embedding of it has absolute value $q^{w/2}$.
Theorem (Riemann hypothesis). The eigenvalues of $F_q$ on $T_{\ell}{E}$ are Weil numbers of weight 1.
Proof. The characteristic polynomial of $F_q$ on $T_{\ell}{E}$ is $T^2aT+q$. The eigenvalues are the roots of this polynomial, i.e., $(a \pm \sqrt{a^24q})/2$. The Hasse bound shows that $a^24q \le 0$, and so the absolute value of this algebraic number (or its complex conjugate) is $\sqrt{q}$. This completes the proof. ◾
The zeta function of a variety $X/\bF_q$ is defined by
$$ Z_X(T) = \exp \left( \sum_{r=1}^{\infty} \# X(\bF_q) \frac{T^r}{r} \right). $$Theorem (Rationality of the zeta function). We have
$$ Z_E(T) = \frac{1aT+qT^2}{(1T)(1qT)}. $$Proof. The above results show that
$$ \# E(\bF_{q^r}) = q^r+1\tr(F_{q^r} \mid T_{\ell}{E}). $$Let $\alpha$ and $\beta$ be the eigenvalues of $F_q$ on $T_{\ell}{E}$. Since $F_{q^r}$ is just $F_q^r$, the eigenvalues of $F_{q^r}$ on $T_{\ell}(E)$ are $\alpha^r$ and $\beta^r$. We thus see that
$$ \# E(\bF_{q^r}) = q^r+1  \alpha^r\beta^r. $$We now have
$$ \sum_{r=1}^{\infty} \# E(\bF_{q^r}) \frac{T^r}{r} = \log(1T)  \log(1qT)+\log(1\alpha T)+\log(1\beta T), $$and so
$$ Z_E(T) = \frac{(1\alpha T)(1\beta T)}{(1T)(1qT)}, $$from which the result easily follows. ◾
Corollary. $\# E(\bF_{q^r})$ is determined, for any $r$, from $\# E(\bF_q)$.
Suppose that $f \colon E_1 \to E_2$ is an isogeny. Then $f$ induces a map $T_{\ell}(E_1) \to T_{\ell}(E_2)$ which commutes with Frobenius. Since the kernel of $f$ is finite, the map it induces on Tate modules has finite index image; in particular, it induces an isomorphism after tensoring with $\bQ_{\ell}$. It follows that the eigenvalues of Frobenius on the two Tate modules agree, and so:
Theorem. If $E_1$ and $E_2$ are isogenous then $\# E_1(\bF_q) = \# E_2(\bF_q)$.
In fact, the converse to this theorem is also true, as shown by Tate.
Ordinary and supersingular curves
Let $E$ be an elliptic curve over a field $k$ of characteristic $p$. Then the map $[p] \colon E \to E$ is not separable and has degree $p^2$. It follows that the separable degree of $[p]$ is either $p$ or 1. In the first case, $E$ is called ordinary, and in the second case, supersingular. The following result follows immediately from the definitions, and earlier results:
Proposition. If $E$ is ordinary then $E[p](\overline{k}) \cong \bZ/p\bZ$. If $E$ is supersingular then $E[p](\overline{k}) = 0$.
We will revisit the ordinary/supesingular dichotomy after discussing group schemes. For now, we prove just one more result.
Proposition. If $E$ is supersingular then $j(E) \in \bF_{p^2}$.
Proof. Suppose $E$ is supersingular. Then $[p]$ is completely inseparable, and thus factors as $E \to E^{(p^2)} \to E$, where the first map is the Frobenius $F_{p^2}$ and the second map is an isomorphism (since it has degree 1). Since $j(E^{(p^2)})$ is equal to $F_{p^2}(j(E))$ and $j$ is an isomorphism invariant, we see that $j(E)=F_{p^2}(j(E))$, from which the result follows. ◾
Corollary. Assume $k$ algebraically closed. Then there are only finitely many supersingular elliptic curves over $k$, and they can all be defined over $\bF_{p^2}$.
Proof. An elliptic curve over an algebraically closed field descends to the field of its $j$invariant, which gives the final statement. The finiteness statement follows immediately from this. ◾
Abelian varieties
A good reference for this section is the first chapter of Mumford's "Abelian varieties" (MR0282985).
Definition and relation to elliptic curves
Definition. An abelian variety is a complete connected group variety (over some field).
Example. An elliptic curve is a onedimensional abelian variety.
Proposition. Every onedimensional abelian variety is an elliptic curve.
Proof. Let $A$ be a onedimensional abelian variety. We must show that $A$ has genus 1. Pick a nonzero cotangent vector to $A$ at the identity. The group law on $A$ allows us to translate this vector uniquely to any other point, and so we can find a nowhere vanishing holomorphic 1form on $A$. This provides an isomorphism $\Omega^1_A \cong \mathcal{O}_A$, and so $\rH^0(A, \Omega^1_A)$ is onedimensional. ◾
For the rest of this lecture we work over the complex numbers.
Compact complex Lie groups
Let $A$ be an abelian variety. Then $A(\bC)$ is a connected compact complex Lie group. We begin by investigating such groups. Thus let $X$ be such a group. Define $V$ to be the tangent space to $X$ at the identity (the Lie algebra). Let $g=\dim(X)$. Recall that there is a holomorphic map $\exp \colon V \to X$. We have the following results:

$X$ is commutative. Reason: the map $\mathrm{Ad} \colon X \to \End(V)$ is holomorphic, and therefore constant, since $X$ is compact and $\End(V)$ is a vector space. Since $\mathrm{Ad}$ assumes the value 1, this is the only value it assumes. It follows that $X$ acts trivially on $\End(V)$, and so $V$ is a commutative Lie algebra. The result follows.

$\exp$ is a homomorphism. Reason: this follows from commutativity.

$\exp$ is surjective. Reason: the image of $\exp$ contains an open subset of $X$, since $\exp$ is a local homeomorphism. The image of $\exp$ is also a subgroup of $X$. Thus the image is an open subgroup $U$. The quotient $X/U$ is discrete, since $U$ is open, and connected, since $X$ is, and is therefore a point. Thus $X=U$.

$M=\ker(\exp)$ is a lattice in $V$, and thus isomorphic to $\bZ^{2g}$. Reason: since $\exp$ is a local homeomorphism, $M$ is discrete. Since $X=V/M$ is compact, $M$ is cocompact.

$X$ is a torus, i.e., isomorphic to a product of circles. Reason: clear from $X=V/M$.

The $n$torsion $X[n]$ is isomorphic to $(\bZ/n\bZ)^{2g}$. Reason: $X[n]$ is isomorphic to $\tfrac{1}{n} M/M$ by the exponential map.

$\rH^i(X, \bZ)$ is naturally isomorphic to $\Hom(\bigwedge^i(M), \bZ)$. Reason: a simple application of the Künneth formula shows that if $T$ is any torus then cup product induces an isomorphism $\bigwedge^i(\rH^1(T, \bZ)) \to \rH^i(T, \bZ)$. For our torus $X$, we have $\rH_1(X, \bZ) = M$, and the result follows.
Line bundles on complex tori
Let $X=V/M$, as above. Define $\Pic(X)$ (the Picard group of $X$) to be the set of isomorphism classes of line bundles on $X$. This is a group under tensor product. Define $\Pic^0(X)$ to be the subgroup consisting of those bundles which are topologically trivial, and define $\NS(X)$ (the NéronSeveri group) to be the quotient $\Pic(X)/\Pic^0(X)$. We are now going to describe how to compute these groups in terms of $V$ and $M$.
A Riemann form on $V$ (with respect to $M$) is a Hermitian form $H$ such that $E=\im{H}$ takes integer values when restricted to $M$. (Note: some people include positive definite in their definition of Riemann form; we do not.) Let $\cR$ be the set of Riemann forms, which forms a group under addition. Let $\cP$ be the set of pairs $(H, \alpha)$, where $H \in \cR$ and $\alpha \colon M \to U(1)$ is a function satisfying $\alpha(x+y)=e^{i \pi E(x, y)} \alpha(x) \alpha(y)$. (Here $U(1)$ is the set of complex numbers of absolute value 1.) We give $\cP$ the structure of a group by $(H_1, \alpha_1) (H_2, \alpha_2)=(H_1+H_2, \alpha_1 \alpha_2)$. Let $\cP^0$ be the group of homomorphisms $M \to U(1)$, regarded as the subgroup of $\cP$ with $H=0$.
Theorem (AppellHumbert). We have an isomorphism $\Pic(X) \cong \cP$, which induces isomorphisms $\Pic^0(X) \cong \cP^0$ and $\NS(X) \cong \cR$.
Some remarks on the theorem:

Let $\pi \colon V \to X$ be the quotient map. If $L$ is a line bundle on $X$ then $\pi^*(L)$ is the trivial line bundle on $V$, since all line bundles on $V$ are trivial. Furthermore, $\pi^*(L)$ is $M$equivariant, and $L$ can be recovered as the quotient of $\pi^*(L)$ by $M$. Thus to prove the theorem, it suffices to understand the $M$equivariant structures on the trivial line bundle over $V$.

Let $(H, \alpha) \in \cP$. Define an action of $M$ on $V \times \bC$ by
$$ \lambda \cdot (v, z)=(v+\lambda, \alpha(\lambda) e^{\pi H(v, \lambda)+\pi H(\lambda, \lambda)/2} z ). $$This gives the trivial bundle on $V$ an $M$equivariance. We let $L(H, \alpha)$ be the quotient, a line bundle on $X$. The isomorphism $\cP \to \Pic(X)$ is $(H, \alpha) \mapsto L(H, \alpha)$. The main content of the theorem is to show that the equivariances we just constructed are all of them.

Remark. There is a bijection between Hermitian forms $H$ on $V$ and alternating real forms $E$ satisfying $E(ix, iy)=E(x, y)$. The correspondence takes $H$ to $E=\im{H}$, and $E$ to $H(x,y)=E(ix,y)+iE(x,y)$. Thus a Riemann form $H$ is determined by the associated alternating pairing on $M$.

Let $(H, \alpha) \in \cP$, and let $E=\im{H}$. Then $E$ defines an element of $\Hom(\bigwedge^2(M), \bZ)$. But we have previously identified this group with $\rH^2(X, \bZ)$. In fact, $E$, regarded as an element of $\rH^2$, is the Chern class $c_1(L(H, \alpha))$. We thus see that $L(H, \alpha)$ is topologically trivial if and only if $E=0$, which is the same as $H=0$. This gives the isomorphic $\Pic^0(X) \cong \cP^0$.
Let $x \in X$ and let $t_x \colon X \to X$ be the translationby$x$ map, i.e., $t_x(y)=x+y$. Given a line bundle $L$ on $X$, we get a new line bundle $t_x^*(L)$ on $X$. We thus get an action of $X$ on $\Pic(X)$, with $x$ acting by $t_x^*$. The following proposition describes this action in terms of the AppellHumbert theorem.
Proposition. We have an isomorphism $t_x^* L(H, \alpha) \cong L(H, \alpha \cdot e^{2 \pi i E(x, ) })$.
A few remarks:

First, we note that $\lambda \mapsto e^{2 \pi i E(x, \lambda)}$ makes sense as a function on $M$, since $E$ takes integral values on $M$.

The line bundle $L(H, \alpha)$ is translation invariant (i.e., isomorphic to its pullbacks by $t_x^*$) if and only if $H=0$. Indeed, it is clear that if $H=0$ then $L(H, \alpha)$ is translation invariant. Conversely, if $L(H, \alpha)$ is translation invariant then $e^{2 \pi i E(x, \lambda)}=1$ for all $x \in V$ and all $\lambda \in M$, from which it easily follows that $E=0$, and so $H=0$ as well. We can therefore characterize $\Pic^0(X)$ as the group of translation invariant line bundles on $X$.

Let $L$ be a line bundle on $X$. Then $x \mapsto t_x^*(L) \otimes L^*$ defines a group homomorphism $\phi_L \colon X \to \Pic^0(X)$. Indeed, taking $L=L(H, \alpha)$, we see that $t_x^*(L) \otimes L^*$ is equal to $L(0, e^{2 \pi i E(x, )})$. It follows that, in fact, $\phi_L$ depends only on $c_1(L)$.
Sections of line bundles
A $\theta$function on $V$ with respect to $(H, \alpha) \in \cP$ is a holomorphic function $\theta \colon V \to \bC$ satisfying the functional equation
$$ \theta(v+\lambda) = \alpha(\lambda) e^{\pi H(v, \lambda)+\pi H(\lambda, \lambda)/2}. $$Given a section $s$ of $L(H, \alpha)$ over $X$, we obtain a section $\pi^*(s)$ of $\pi^*(L(H, \alpha))$ over $V$. Identifying $\pi^*(L(H, \alpha))$ with the trivial bundle, $\pi^*(s)$ becomes a function on $V$, and the equivariance condition is exactly the above functional equation. We therefore find:
Proposition. The space $\Gamma(X, L(H, \alpha))$ is canonically identified with the space of $\theta$functions for $(H, \alpha)$.
Suppose that $H$ is degenerate, and let $V_0$ be its kernel (i.e., $x \in V_0$ if $H(x, )=0$). Then $V_0$ is also the kernel of $E$, and since $E$ takes integral values on $M$, it follows that $M_0=V_0 \cap M$ is a lattice in $V_0$. Let $\theta$ be a $\theta$function, and $u$ a large element of $V_0$. Write $u=\lambda+\epsilon$ with $\lambda \in M_0$ and $\epsilon$ in some fundamental domain. Then for any $v \in V$ we have
$$ \vert \theta(v+u) \vert = \vert \theta(v+\epsilon) \vert $$since $H(\lambda, )=0$. It follows that $u \mapsto \theta(v+u)$ is a bounded holomorphic function on $V_0$, and therefore constant. Thus $\theta$ factors through $V/V_0$. In particular, $L(H, \alpha)$ is not ample.
Now suppose that $H(w,w) \lt 0$ for some $w \in V$. Let $t$ be a large complex number and write $tw=\lambda+\epsilon$, similar to the above. Then
$$ \vert \theta(v+tw) \vert = \vert \theta(v+\epsilon) \vert e^{\pi \mathrm{Re}(H(v+\epsilon, \lambda))+\pi H(\lambda, \lambda)/2}. $$The quantity $H(\lambda, \lambda)$ is dominant, and very negative. We thus see that $\vert \theta(v+tw) \vert \to 0$ as $\vert t \vert \to \infty$, which implies $\theta(v+tw)$ is 0 as a function of $t$. Thus $\theta(v)=0$ for all $v$, and so 0 is the only $\theta$function.
We have thus shown that if $H$ is not positive definite then $L(H, \alpha)$ is not ample. The converse holds as well:
Theorem (Lefschetz). The bundle $L(H, \alpha)$ is ample if and only if $H$ is positive definite.
Some remarks:

This theorem shows that $X$ is a projective variety is and only if there exists a positive definite Riemann form on $V$.

In fact, one can show that if $X$ is algebraic then it is necessarily projective, and so $X$ is algebraic if and only if it has a positive definite Riemann form.

One can show that if $H$ is positive definite then $L(H, \alpha)^{\otimes n}$ is very ample for all $n \ge 3$.

Suppose $E$ is the elliptic curve given by $\bC/\langle 1, \tau \rangle$. Then $H(x,y) = \frac{x \overline{y}}{\vert \Im(\tau) \vert}$ is a positive definite Riemann form on $\bC$. This recovers the fact that all onedimensional complex tori are algebraic.

Most complex tori of higher dimension do not possess even a nonzero Riemann form, and so most are not algebraic.
Maps of tori
A map of complex tori $X \to Y$ is a holomorphic group homomorphism. In fact, any holomorphic map taking 0 to 0 is a group homomorphism. W write $\Hom(X, Y)$ for the group of maps. An isogeny is a map of tori which is surjective and has finite kernel. The degree of the isogeny is the cardinality of the kernel.
Example. Multiplicationby$n$, denoted $[n]$, is an isogeny of degree $n^{2g}$.
The dual torus
Let $X=V/M$ be a complex torus. Let $\ol{V}^*$ be the vector space of conjugatelinear functions $V \to \bC$, and let $M^{\vee} \subset \ol{V}^*$ be the set of such functions $f$ for which $\im{f}(M) \subset \bZ$. Then $M^{\vee}$ is a lattice in $\ol{V}^*$, and we define $X^{\vee}=\ol{V}^*/M^{\vee}$. We call $X^{\vee}$ the dual torus of $X$. Note that we have a natural isomorphism $(X^{\vee})^{\vee}=X$.
Formation of the dual torus is clearly a functor: if $f \colon X \to Y$ is a map of tori then there is a natural map $f^{\vee} \colon Y^{\vee} \to X^{\vee}$. If $f$ is an isogeny, then so is $f^{\vee}$, and they have the same degree. Even better:
Proposition. If $f$ is an isogeny then $\ker(f)$ and $\ker(f^{\vee})$ are canonically dual (in the sense of finite abelian groups).
Proof. Write $X=V_1/M_1$ and $Y=V_2/M_2$, and let $g \colon V_1 \to V_2$ be the linear map inducing. Then $\ker(f)=g^{1}(M_2)/M_1$, while $\ker(f^{\vee})=(\overline{g}^*)^{1}(M_1^{\vee})/M_2^{\vee}$. If $x \in \ker(f)$ and $y \in \ker(f^{\vee})$ then $\langle g(x), y \rangle$ is a rational number (since $g(x) \in M_2$ and $y$ is in a lattice containing $M_2^{\vee}$ with finite index), and is welldefined up to integers. We thus have a pairing $\ker(f) \times \ker(f^{\vee}) \to \mathbf{Q}/\bZ$ with $n=\deg(f)$, which puts the two groups in duality. ◾
Applying this in the case where $X=Y$ and $f=[n]$, we see that $X[n]$ and $X^{\vee}[n]$ are in duality. This gives us a canonical pairing $X[n] \times X^{\vee}[n] \to \bZ/n\bZ \cong \mu_n$, which is called the Weil pairing.
Proposition. We have a natural isomorphism of groups $X^{\vee}=\Pic^0(X)$.
Proof. The map $\overline{V}^* \to \cP^0$ which takes $f \in \overline{V}^*$ to the map $\lambda \mapsto e^{2\pi i \im(f(\lambda))}$ is easily seen to be a surjective homomorphism with kernel $M^{\vee}$. It thus descends to an isomorphism $X^{\vee} \to \Pic^0(X)$. ◾
Let $H$ be a Riemann form on $V$. Then $v \mapsto H(V, )$ defines an isomorphism of complex vector spaces $V \to \overline{V}^*$, and carries $M$ into $M^{\vee}$. It thus defines a map $\phi_H \colon X \to X^{\vee}$ of complex tori. This map is an isogeny if and only if $H$ is nondegenerate. Identifying $X^{\vee}$ with $\Pic^0(X)$, $\phi_H$ coincides with $\phi_L$, where $L=L(H, \alpha)$ for any $\alpha$. A polarization of $X$ is a map of the form $\phi_H$ (or $\phi_L$) with $H$ positivedefinite (or $L$ ample). A principal polarization is a polarization of degree 1. We thus see that $X$ admits a polarization if and only if it is algebraic.