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 be an abelian variety. Last time, I described a formula which allowed us to express the pull-back
of a line bundle
as
This was a special case of the so-called “theorem of the cube,” which allowed us to express, for three morphisms , the pull-back
in terms of the various pull-backs of partial sums. Namely, we had the formula
In this formula, we take and
constant maps at
, respectively. Let
, etc. denote the translation maps on
. Then
are the relevant sums, and we have
We get the theorem of the square:
Theorem 1 (Theorem of the square)Let
be an abelian variety,
, and
points. Then
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 be an abelian variety. We will be interested in line bundles on
which are translation-invariant, i.e. line bundles
such that
for all
.
Definition 2 The group of
which are translation-invariant is
.
In fact, 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
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
be any line bundle. Then
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 can be obtained in the manner of the previous example.
Example 2 Let
have dimension one, so
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
for any
is in
. 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 be the multiplication map, and let
. Then the line bundle
has the property that restriction to any fiber
is precisely
. If we think of
as a family of line bundles on
, parametrized by
, then to say that
is translation-invariant is a certain trivialty of this family. This is expressed in the following result.
Proposition 3 A line bundle
belongs to
if and only if
for
the multiplication map.
In fact, this follows from a “rigidity lemma” on families of line bundles. If we consider the line bundle , 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
because is obtained by pulling back
under the map
However, because
is constant.
In particular, translation-invariant line bundles are very strongly non-ample: if were ample, then
would be too (since
is an isomorphism). But,
and
cannot both be ample unless
is zero-dimensional. In fact, it turns out that ampleness is a polar opposite of translation-invariance: a line bundle
is ample if and only if the set of
such that
is finite, as it turns out.
2. A little intersection theory
Let be an abelian variety, and let
be a divisor on
. We saw in the previous section that
where denotes linear equivalence, for any
. In particular, we get
This means that if is any effective divisor on
, then the linear system
is base-point free: given
, we can choose
such that
does not contain
. In particular,
defines a map
for some
. We’d like to choose
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 is any base-point free divisor, defining a map
We’d like to know when is a finite morphism. By Zariski’s main theorem, finiteness is equivalent to quasi-finiteness. Consequently,
fails to be finite precisely when there exists an irreducible curve
such that
. This means that every divisor in the linear system
either contains
or fails to intersect
. In the language of line bundles, any section of
vanishing on a point of
must vanish everywhere on
. By changing
if necessary, we can assume that
.
Observation: The intersections are either empty or all of
, for all
. In other words,
can intersect no translate of
.
In fact, the intersection number for
is constant, and must thus be identically zero. Alternatively, one considers the family of line bundles
on the normalization of
. When
, this line bundle is trivial; it follows that for all
, the degree of the line bundle is zero. Since
is an effective divisor, this implies that
for all
such that intersection is not
.
This is a pretty useful geometric argument: “wiggling” the divisor by a translation by
can’t force it to intersect
. Motivated by this, we can deduce something seemingly stronger.
Observation: If contracts the irreducible curve
to a point and
, then
is invariant under translation by
for
.
In fact, if and
, then the translation
has the property that
(because the intersection contains
). In particular, it contains
by the previous observation: thus
which is what we wanted.
These two observations show that there are strong constraints on a base-point free divisor with the property that the induced map to
is not finite: the last observation shows that
has to be invariant by a proper curve’s worth of translations. We thus find:
Proposition 4 Let
be a divisor on
contained in
for
an open affine. If
is base-point free, then the map
is finite.
Proof: In fact, we can assume . In this case, we find that the only translations which leave
invariant have to be in
(i.e., they have to send
into something in
). This means that there can’t be a proper curve’s worth of translations leaving invariant
, and by the previous observations,
can’t contract a curve.
Finally, putting all this together, we may deduce:
Theorem 5 An abelian variety
is projective.
Proof: It suffices to show that admits a finite map
for a suitable divisor
, because the divisor
is then the pull-back of the ample divisor on
by a finite map and is thus itself ample. In order to do this, we need to find a divisor on
without base points and which is contained in
for
an open affine. That’s easy: just start by picking an irreducible component of
, and then take twice that for
. The analysis at the beginning of the section shows that
is base-point free.
In fact, using a refinement of these arguments, one can prove a characterization of ampleness:
Theorem 6 A line bundle
associated to an effective divisor is ample if and only if the set of all
such that
is finite.
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?
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
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.