Line bundles on abelian varieties II

This, like the previous and probably the next few posts, is an attempt at understanding some of the ideas in Mumford’s Abelian Varieties.

Let ${A}$ be an abelian variety. Last time, I described a formula which allowed us to express the pull-back ${n_A^* \mathcal{L}}$ of a line bundle ${\mathcal{L} \in \mathrm{Pic}(A)}$ as

$\displaystyle n_A^*\mathcal{L} \simeq \mathcal{L}^{(n^2 + n)/2} \otimes (-1)^* \mathcal{L}^{(n^2 - n)/2}.$

This was a special case of the so-called “theorem of the cube,” which allowed us to express, for three morphisms ${f, g, h: X \rightarrow A}$ , the pull-back

$\displaystyle (f + g + h)^* \mathcal{L}$

in terms of the various pull-backs of partial sums. Namely, we had the formula

$\displaystyle (f + g+ h)^* \mathcal{L} \simeq (f+g)^* \mathcal{L} \otimes (f + h)^* \mathcal{L} \otimes (g + h)^* \mathcal{L} \otimes f^* \mathcal{L}^{-1} \otimes g^* \mathcal{L}^{-1} \otimes h^{*} \mathcal{L}^{-1} .$

In this formula, we take ${f = \mathrm{Id} = 1_A}$ and ${g, h}$ constant maps at ${x,y \in A}$ , respectively. Let ${T_x, T_y}$ , etc. denote the translation maps on ${A}$ . Then ${T_x, T_y, T_{x+y}}$ are the relevant sums, and we have

$\displaystyle T_{x+y}^* \mathcal{L} \simeq T_x^* \mathcal{L} \otimes T_y^* \mathcal{L} \otimes \mathcal{L}^{-1}.$

We get the theorem of the square:

Theorem 1 (Theorem of the square)Let ${A}$ be an abelian variety, ${\mathcal{L} \in \mathrm{Pic}(A)}$ , and ${x, y \in A}$ points. Then

$\displaystyle T_{x+y}^* \mathcal{L} \otimes \mathcal{L} \simeq T_x^* \mathcal{L} \otimes T_y^* \mathcal{L}.$

In the rest of this post, I’ll describe some applications of this result, for instance the fact that abelian varieties are projective.

1. Translation-invariant line bundles

Let ${A}$ be an abelian variety. We will be interested in line bundles on ${A}$ which are translation-invariant, i.e. line bundles ${\mathcal{L}}$ such that ${T_x^* \mathcal{L} \simeq \mathcal{L}}$ for all ${x \in A}$ .

Definition 2 The group of ${\mathcal{L} \in \mathrm{Pic}(A)}$ which are translation-invariant is ${\hat{A}}$ .

In fact, ${\hat{A}}$ turns out to be not just a group, but an abelian variety itself. This is far from obvious, but it turns out to be a corollary of the (non-obvious) fact that line bundles on ${A}$ are themselves parametrized by a scheme, called the Picard scheme. The collection of translation-invariant bundles turns out to form the connected component at zero of the Picard scheme.

What are some examples of translation-invariant line bundles?

Example 1 Let ${\mathcal{L} \in \mathrm{Pic}(A)}$ be any line bundle. Then ${T_x^* \mathcal{L} \otimes \mathcal{L}^{-1}}$ is translation-invariant; this is a consequence of the theorem of the square.

One of the next goals is to show that any element of ${\hat{A}}$ can be obtained in the manner of the previous example.

Example 2 Let ${A}$ have dimension one, so ${A}$ is an elliptic curve. Then the translation-invariant line bundles correspond to divisors of total degree zero. In fact, by the theorem of the square, we find that the line bundle associated to ${(x) - (0)}$ for any ${x \in A}$ is in ${\hat{A}}$ . The other direction will follow because any translation-invariant bundle is algebraically equivalent to zero, as we will see later, and consequently has degree zero.

Let ${m: A \times A \rightarrow A}$ be the multiplication map, and let ${\mathcal{L} \in \mathrm{Pic}(A)}$ . Then the line bundle ${m^* \mathcal{L} \in \mathrm{Pic}(A \times A)}$ has the property that restriction to any fiber ${\left\{x \right\} \times A}$ is precisely ${T_x^* \mathcal{L}}$ . If we think of ${m^* \mathcal{L}}$ as a family of line bundles on ${A}$ , parametrized by ${\mathcal{L}}$ , then to say that ${\mathcal{L} }$ is translation-invariant is a certain trivialty of this family. This is expressed in the following result.

Proposition 3 A line bundle ${\mathcal{L} \in \mathrm{Pic}(A)}$ belongs to ${\hat{A}}$ if and only if

$\displaystyle m^* \mathcal{L} \simeq \mathcal{L} \boxtimes \mathcal{L} \in \mathrm{Pic}(A \times A),$

for ${\mathcal{L}}$ the multiplication map.

In fact, this follows from a “rigidity lemma” on families of line bundles. If we consider the line bundle ${m^* \mathcal{L} \otimes (\mathcal{L} \boxtimes \mathcal{L})^{-1}}$ , then we have seen that it is trivial on each vertical fiber, and similarly on each horizontal fiber. But the rigidity lemma (which is proved in Mumford’s book) states precisely that such a line bundle is trivial. The converse is easy, by restricting to the vertical fibers.

A consequence of this fact is that

$\displaystyle \mathcal{L}^{-1} \simeq (-1)_A^*\mathcal{L} \quad \text{for } \mathcal{L} \in \hat{A},$

because ${(-1)^* \mathcal{L} \otimes \mathcal{L}}$ is obtained by pulling back ${\mathcal{L} \boxtimes \mathcal{L} \simeq m^* \mathcal{L}}$ under the map

$\displaystyle d: A \rightarrow A \times A, \quad x \mapsto (x, -x).$

However, ${d^*(\mathcal{L} \boxtimes \mathcal{L}) \simeq d^* m^* ( \mathcal{L} ) \simeq \mathcal{O}_A}$ because ${m \circ d}$ is constant.

In particular, translation-invariant line bundles are very strongly non-ample: if ${\mathcal{L} \in \hat{A}}$ were ample, then ${(-1)^* \mathcal{L}}$ would be too (since ${(-1)_A}$ is an isomorphism). But, ${\mathcal{L}}$ and ${\mathcal{L}^{-1}}$ cannot both be ample unless ${A}$ is zero-dimensional. In fact, it turns out that ampleness is a polar opposite of translation-invariance: a line bundle ${\mathcal{L} \in \mathrm{Pic}(A)}$ is ample if and only if the set of ${x \in A}$ such that ${T_x^* \mathcal{L} \simeq \mathcal{L}}$ is finite, as it turns out.

2. A little intersection theory

Let ${A}$ be an abelian variety, and let ${D \subset A}$ be a divisor on ${A}$ . We saw in the previous section that

$\displaystyle T_{x+y}^* D + D \sim T_x^* D + T_y^* D,$

where ${\sim}$ denotes linear equivalence, for any ${x, y \in A}$ . In particular, we get

$\displaystyle T_x^* D + T_{-x}^* D \sim 2D.$

This means that if ${D}$ is any effective divisor on ${A}$ , then the linear system ${|2D|}$ is base-point free: given ${y \in A}$ , we can choose ${x}$ such that ${T_x^* D + T_{-x}^* D}$ does not contain ${y}$ . In particular, ${2D}$ defines a map ${A \rightarrow \mathbb{P}^N}$ for some ${N}$ . We’d like to choose ${D}$ to make this map an immersion, or at least finite. We can thus state:

Observation: On an abelian variety, there are a lot of line bundles generated by global sections (and thus a lot of maps to projective space.)

Suppose ${E}$ is any base-point free divisor, defining a map

$\displaystyle \phi_E: A \rightarrow \mathbb{P}^N.$

We’d like to know when ${\phi_E}$ is a finite morphism. By Zariski’s main theorem, finiteness is equivalent to quasi-finiteness. Consequently, ${\phi_E}$ fails to be finite precisely when there exists an irreducible curve ${C \subset A}$ such that ${\phi_E(C) = \ast}$ . This means that every divisor in the linear system ${|E|}$ either contains ${C}$ or fails to intersect ${C}$ . In the language of line bundles, any section of ${\mathcal{O}(E)}$ vanishing on a point of ${C}$ must vanish everywhere on ${C}$ . By changing ${E}$ if necessary, we can assume that ${E \cap C = \emptyset}$ .

Observation: The intersections ${C \cap T_x^* E}$ are either empty or all of ${C}$ , for all ${x \in A}$ . In other words, ${C}$ can intersect no translate of ${E}$ .

In fact, the intersection number ${C. T_x^* E}$ for ${x \in A}$ is constant, and must thus be identically zero. Alternatively, one considers the family of line bundles ${\mathcal{O}(T_x^* E )|_{\widetilde{C}}}$ on the normalization of ${C}$ . When ${x = 0}$ , this line bundle is trivial; it follows that for all ${x}$ , the degree of the line bundle is zero. Since ${E}$ is an effective divisor, this implies that ${C \cap T_x^* E = \emptyset}$ for all ${x}$ such that intersection is not ${C}$ .

This is a pretty useful geometric argument: “wiggling” the divisor ${E}$ by a translation by ${x}$ can’t force it to intersect ${C}$ . Motivated by this, we can deduce something seemingly stronger.

Observation: If ${\phi_E: A \rightarrow \mathbb{P}^N}$ contracts the irreducible curve ${C}$ to a point and ${C \cap E = \emptyset}$ , then ${E}$ is invariant under translation by ${x-y}$ for ${x,y \in C}$ .

In fact, if ${x, y \in C}$ and ${e \in E}$ , then the translation ${T_{x-e}}$ has the property that ${T_{x-e}(E) \cap C \neq \emptyset}$ (because the intersection contains ${x}$ ). In particular, it contains ${y}$ by the previous observation: thus

$\displaystyle y - (x - e) \in E,$

which is what we wanted.

These two observations show that there are strong constraints on a base-point free divisor ${E}$ with the property that the induced map to ${\mathbb{P}^N}$ is not finite: the last observation shows that ${E}$ has to be invariant by a proper curve’s worth of translations. We thus find:

Proposition 4 Let ${E}$ be a divisor on ${A}$ contained in ${A \setminus U}$ for ${U}$ an open affine. If ${E}$ is base-point free, then the map ${\phi_E: A \rightarrow \mathbb{P}^N}$ is finite.

Proof: In fact, we can assume ${0 \in U}$ . In this case, we find that the only translations which leave ${E}$ invariant have to be in ${U}$ (i.e., they have to send ${0}$ into something in ${U}$ ). This means that there can’t be a proper curve’s worth of translations leaving invariant ${E}$ , and by the previous observations, ${\phi_E: A \rightarrow \mathbb{P}^N}$ can’t contract a curve. $\Box$

Finally, putting all this together, we may deduce:

Theorem 5 An abelian variety ${A}$ is projective.

Proof: It suffices to show that ${A}$ admits a finite map ${\phi_E: A \rightarrow \mathbb{P}^N}$ for a suitable divisor ${E}$ , because the divisor ${E}$ is then the pull-back of the ample divisor on ${\mathbb{P}^N}$ by a finite map and is thus itself ample. In order to do this, we need to find a divisor on ${A}$ without base points and which is contained in ${A \setminus U}$ for ${U}$ an open affine. That’s easy: just start by picking an irreducible component of ${A \setminus U}$ , and then take twice that for ${E}$ . The analysis at the beginning of the section shows that ${E}$ is base-point free. $\Box$

In fact, using a refinement of these arguments, one can prove a characterization of ampleness:

Theorem 6 A line bundle ${\mathcal{L} \in \mathrm{Pic}(A)}$ associated to an effective divisor is ample if and only if the set of all ${x \in A}$ such that ${T_x^* \mathcal{L} \simeq \mathcal{L}}$ is finite.

plm Says:

April 16, 2012 at 4:58 am

If you repeat the typo: check the missing L^{-1} in the cube formula.

Do you have other ideas in mind for studying Mumford’s Abelian Varieties?

Akhil Mathew Says:

April 16, 2012 at 8:50 pm
I don’t think it’s a typo: the equation is just cut off. (WordPress margins are a lot shorter than those of a typical LaTeX document.)

I’m not sure — my near goal for the blog is pretty limited (e.g., to prove that $\hat{A}$ is smooth and of the right dimension, and maybe Mordell-Weil a little further down the line). There’s a lot of fascinating stuff in Mumford’s book that I’m hoping to explore over the summer, though.

Climbing Mount Bourbaki