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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 $1-F_q$. This endomorphism is separable: indeed, if $\omega$ is a differential on $E$ then $F_q^*(\omega)=0$, and so $(1-F_q)^* \omega=\omega$ is non-zero. We have thus proved the following proposition:

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 Cauchy--Schwartz 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,

$$ 2 \langle 1, -F_q \rangle = \deg(1-F_q)-\deg(1)-\deg(F_q), $$

and so we have the following theorem

In other words, we can write $\# E(\bF_q)$ as $q+1-a$, 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 $a$:

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 $q^{w/2}$.

The zeta function of a variety $X/\bF_q$ is defined by

$$ Z_X(T) = \exp \left( \sum_{r=1}^{\infty} \# X(\bF_q) \frac{T^r}{r} \right). $$

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:

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:

We will revisit the ordinary/supesingular dichotomy after discussing group schemes. For now, we prove just one more result.

Abelian varieties

A good reference for this section is the first chapter of Mumford's "Abelian varieties" (MR0282985).

Definition and relation to elliptic curves

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 Néron--Severi 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 $H=0$.

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 translation-by-$x$ 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 Appell--Humbert theorem.

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:

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:

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 one-dimensional complex tori are algebraic.

• Most complex tori of higher dimension do not possess even a non-zero 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.

The dual torus

Let $X=V/M$ be a complex torus. Let $\ol{V}^*$ be the vector space of conjugate-linear 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:

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.

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 non-degenerate. 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 $H$ positive-definite (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.