Abelian Varieties: Some details

Look here, for an overview of this post and basic conventions. The main points of this post happened in the thirds section: Pic Zero.

Connecting points on complete variety.

(Thm of Cube) Let $X,Y$ complete and $Z$ another variety and $L$ a line bundle on $X\times Y \times Z$ . If there exists $x,y,z$ such that the restriction of $L$ to $x \times Y\times Z ,X \times y \times Z, X \times Y \times z$ is trivial, then $L$ is trivial.

A proof of this can be found in many places like Mumford’s book on Abelian Var. An outline of the proof ishere. Essential to this proof is the following claim:

On any complete variety of dimension at least 1 any two points can be connected by an irreducible curve.

Despite being a very reasonable claim, I don’t know of any proofs that don’t involve very powerful results. Without rigor, the idea is this: the result is trivial if the dimension is 1, so assume the variety $X$ is at least 2 dimensional; now use induction to reduce simply finding a proper subvariety containing both points. Use

Chow’s lemma: If $X$ is complete there exists a projective variety $Y$ and a birational surjection $Y \to X$ .

The lemma reduces to the case the variety $X$ is projective. Then blow up at the two points $x,x'$ to get $X'$ . Now use

Bertini’s theorem: If $Y$ is a projective nonsingular variety $\subset \mathbb{P}^n$ then there is a hyperplane $H$ such that all points of $Y \cap H$ are smooth. If $\dim Y \ge 2$ then the intersection is connected hence irreducible hence a variety.

Now back to the blowup $X' \xrightarrow{p} X$ . General theory says $H\cap X',\ p^{-1}(x),\ p^{-1}(x')$ are all divisors of dimension $\dim X -1$ . The assumption $\dim X > 1$ assures that $X' \cap H$ is a proper subvariety intersecting the preimages of $x,x'$ and so $p(X' \cap H)$ finishes the induction step.

Using the theorem of the Cube

Because of the group structure on an Abelian variety $A$ , maps into $A$ can be added. An easy consequence (see corollaries) of the theorem of the cube is that if $f,g,h \colon S \to A$ are morphims and $L \in Pic(A)$ then

$\begin{array}{rl} fgh(L):= & (f+g+h)^*L \\ \mbox{} & \otimes (f+g)^*L^{-1}\otimes(f+h)^*L^{-1}\otimes(g+h)^*L^{-1} \\ \mbox{} & \otimes f^*L\otimes g^*L \otimes h^*L \end{array}$ (*)

is trivial on $S$ .

Easy properties

1) $[n]^*L \cong \bigl(L \otimes [-1]^*L\bigr)^{\frac{n^2 - n}{2}}$

Use induction, write $[n+1] = [n ] + [1] + [-1]$ and use (*).

2) $[n] \colon A \to A$ is surjective.

It suffices to show $\ker [n]$ is finite. Take $L$ ample; as $[-1]$ is an automorphism, $[-1]^*L$ is also ample. Hence by 2) for $n \ge 1$ I have that $[n]^*L$ is ample. But $[n]^*L|_{\ker [n]}$ is the pullback over a point, hence trivial. But an ample line bundle cannot be trivial on a positive dimensional subvariety, otherwise this line bundle would give a closed immersion into projective space with image a point, which is absurd.

Pic Zero

Apply (*) $a,b,[1] \colon A \to A$ where $a,b \colon A \to A$ are the constant maps. Note (a+b)^*L = a^*L = b^*L = O$ because they are all the pull back from a point. The upshot is the following isomorphism known as the theorem of the square:

$t^*_{a+b}L\otimes L^{-1} \cong t^*_{a}L\otimes t^*_bL$

Tensor both sides with $L^{-2}$ to get

$\phi_L(a+b) = \phi_L(a)+\phi_L(b)$

where $\phi_L(x) = t^*_xL \otimes L^{-1}$ . The following properties are discussed here

$\phi_L(a+b) = \phi_L(a)\otimes \phi_L(b) =: \phi_L(a) + \phi_L(b)$
$\phi_{L\otimes M}(a) = \phi_L(a)\otimes \phi_M(a) =: (\phi_L + \phi_M)(a)$
$\phi_{t_a^*L \otimes L^{-1}} = 0$

Define Pic Zero $Pic^0(A) := \{ L| \phi_L \equiv 0\} = \{L | t^*L \cong L \forall x\}$ . Its a natural subgroup of the Picard group. Note it is NOT explicitly equal to the degree zero line bundles. But there is a connection, but more on that in later posts.

Easy Property

For all $L \in Pic(A)$ , $L':= L \otimes [-1]^*L \in Pic^0(A)$ .

pf: use the theorem of the square to show $t^*_xL' \cong L'$ .

Note that 3. above shows for any $L$ the image of $\phi_L$ lands in $Pic^0(A)$ . The goal now it show for a good choice of $L$ (namely $L$ ample) that $\phi_L \colon A \to Pic^0(A)$ is surjective.

(A surjects to Pic Zero) If $L$ is ample then $\phi_L \colon A \to Pic^0(A)$ is surjective.

pf: Proceed by contradiction. Let $M \in Pic^0(A)$ be arbitrary. Consider the following line bundle on $A \times A$

$N:= m^*L\otimes p_1^*L^{-1} \otimes p_2^*(L^{-1}\otimes M)$

the idea is to compute the cohomology of $N$ in two different ways and get two different answers, hence contradiction. The tool used to compute cohomology is the

Leray Spectral sequence: this is a special case of the Grothendiek spectral sequence. I wont explain spectral sequences here because many other people have already done it better than I’m likely to do. A good source is greenberg‘s notes.

Here are the main points. In general if $C \xrightarrow{F} C' \xrightarrow{F'} C''$ are functors between abelian categories with

1) $C$ having enough injectives

2) $F$ sending injectives to $F'$ -acylics, and both left exact (not sure if they need to be left exact, but I wont worry about this now)

then there is a spectral sequence relating the right derived functors:

$E_2^{p,q} = R^pF'(R^qF(B)) \Rightarrow R^{p+q}F'\circ F(B)$

The case of interest is when we have a map $X\xrightarrow{f} Y$ of schemes and $C = Sh(X),\ C' = Sh(Y),\ C'' = Ab$ and the functors are $F = f_*,\ F' = \Gamma_Y$ ; the latter is the global sections functor. Note $\Gamma_Y \circ f_* = \Gamma_X$ . Plugging into the above gives the Leray Spectral Sequence

$E_2^{p,q} = H^p(Y,R^qf_*B) \Rightarrow H^{p+q}(X,B)$

In order to use the Leray Spectral Sequence need to know the cohomology of line bundles in $Pic^0$ .

(Cohom of Pic Zero) If $L \in Pic^o(A)$ is nontrivial then $H^i(A,L) = 0 \ \forall i$ .

pf: use induction starting with $i = 0$ . If $H^0 \ne 0$ then $|L| = \mathbb{P}H^o(L)$ is nonempty so there is an effective divisor $D$ such that $L \cong O(D)$ . $L$ being in Pic Zero means $(f+g)^*L \cong f^*L \otimes g^*L$ ; in particular, $L^{-1} = [-1]^*L = O([-1]^*D)$ . Hence $D + [-1]^*D = 0$ , but both are effective divisor so $D = 0$ which contradicts that $L$ is nontrivial.

For the general case use the maps

$A \xrightarrow{[1],e} A\times A \xrightarrow{m} A$ ;

on cohomology this gives a factorization of the identity:

$H^k(A,L) \leftarrow H^k(A\times A,m^*L) \leftarrow H^k(A,L)$

as $L \in Pic^0$ I can replace $m^*L \cong p^*_1L\otimes p_2^*L$ and use a Kunneth formula

$H^k(A\times A,p_1^*L\otimes p_2^*L) \cong \bigoplus_{i+j = k} H^i(L)\otimes H^j(L)$

So inductively, the left side of the above is zero, hence the identity factors through the identity so everything is zero.

To compute the cohomology of $N$ note its restriction to the fibers:

$N|_{x \times A}=t_x^*L\otimes L^{-1} \otimes M^{-1}$

$= \phi_L(x) \otimes M^{-1}$

$N|_{A \times x} = \phi_L(x)$

As mentioned, the cohomology can be computed two ways:

$H^p(A, R^qp_{1,*}N) \Rightarrow H^{p+q}(A \times A, N)$ (2)

$H^p(A, R^qp_{2,*}N) \Rightarrow H^{p+q}(A \times A, N)$ (3)

Assuming $M \ne \phi_L(x)\ \forall x$ it follows that the restriction of $N$ to all the fibers of $p_1$ are nontrivial elements of $Pic^0$ hence have no cohomology. The the cohomology vanishes on all the fibers implies $H^*(A\times A, N) = 0$ . This uses a basic result in cohomology with base change, but that will have to wait for another post.

Now analyze the cohomology using (3). Now you get nonzero cohomology on the fibers when $t^*xL = L$ . In general define the subgroup $K(L) = \ker \phi_L = \{x| t_x^*L = L\}$ .

(Reason to require ample) If $L$ is ample then $K(L)$ is finite.

pf: Let $K'$ be the connected component of the identity in $K(L)$ . On the one hand $L \otimes [-1]^*L$ is still ample when restricted to $K(L)$ . On the other hand, by definition $L \in Pic^0(K')$ so $m^*L^{-1} \otimes p_1^*L \otimes p_2^*L$ is trivial on $K' \times K'$ . Consider the pull back to $K'$ via

$K' \to K' \times K'$

$k \mapsto (k, - k)$

Its the pullback of the trivial bundle hence trivial, on the other hand its also $L \otimes [-1]^*L$ . But a trivial line bundle can’t be ample on a positive dimensional subvariety, so $K' = pt$ and $K(L)$ is finite. In fact the converse is true but that takes more work.

Look again at (3). The cohomology of the fibers vanish outside of $K(L)$ so $Supp(R_{2,*}N) \subset K(L)$ . The point is its supported on a zero dimensional subscheme so doesn’t have any higher cohomology (the only nonzero terms of (3) are those of the form $H^0(R^{p+q})$ , so

$\oplus_{x \in K(L)}(R^kp_{2,*}N)_x \cong H^k(A \times A, N) = 0$

It follows that $R^kp_{2,*}N = (0)$ , but again a result of cohomology with base change says the cohomology of all the fibers must vanish, but the cohomology of of the fiber $A \times e$ is that of the trivial line bundle, which has nonzero $H^0$ so contradiction.

Next is probably the Poincare Bundle and maybe that Abelian varieties are projective.

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You’re currently reading “Abelian Varieties: Some details,” an entry on Math Meandering

Published:: February 6, 2010 / 1:38 am

Category:: alg. geo., wall scribble

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Math Meandering

Abelian Varieties: Some details

Connecting points on complete variety.

Using the theorem of the Cube

Pic Zero

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