MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The set up is $C$ is a curve and $J$ is its Jacobian. On the $C \times J$ there is the Poincare bundle $P$ which is the universal family of degree zero line bundles on $C$. For every integer $d$ there is also a line bundle $P(d)$ on $C \times J$ which is a family of line bundles of degree $d$ on $C$.

I've seen a construction of $P$ and given $P$ an example of a $P(d)$ would be $P^L:= q_1^*L\otimes P$ where $q_1 \colon C \times J \to C$ is the projection and $L$ is a line bundle of degree $d$ on $C$. If $L,L'$ both have degree $d$ you can form either $P^L$ or $P^{L'}$. I've seen $P(d)$ defined as the inverse limit of $P^L$ as $L$ ranges over all degree $d$ line bundles. I don't really know how to think of such an inverse limit. Is there a more concrete way to describe $P(d)$? It seems up to isomorphism, $P(d) = P^L$, but this is "very" non canonical which seems bad. For example does $P^L$ have a universal property like $P$ does?

I have a more specific question related to this. This question is coming from Prop. 21.6 of Polishchuk's book on Abelian varieties and the Fourier Mukai tranform, if anyone is curious. If $q_2 \colon C \times J \to J$ is the projection, then apparently $q_{2*}P(g-1) = 0$. Why is this so?

One argument for this (that I don't understand) is the following.

1) Embed $P(g-1) \to F$ with $F$ flat over $J$ and $R^1q_{2*}F = 0$.

2) From $0 \to P(g-1) \to F \to F/P=: E \to 0$ and cohomology we get

$0 \to q_{2*}P(g-1) \to q_{2*}F \to q_{2*}F \to R^1q_{2*}P(g-1) \to 0$

3)The restriction of $R^1q_{2*}P(g-1)$ to $a \in J$ is $H^1(C \times a, P(g-1)|_{C \times a})$ which is $H^1$ of a line bundle of degree $g-1$. This is zero outside the theta divisor and Riemann-Roch says $h^1(L) = h^0(L)$ for degree $g-1$ line bundles, so

$q_{2*}F \to q_{2*}F$

is an isomorphism outside a divisor. And hence

4) $q_{2*}F \to q_{2*}F$ is injective.

Part 4) is the part I don't see.

I don't necessarily want to understand this line of reasoning. But I don't see why $q_{2*}P(g-1) = 0$ since it seems to be supported on the theta divisor.

share|cite|improve this question
up vote 3 down vote accepted

For your first question, if you want a proper universal property it is defined by a variety $J^{(d)}$ and a line bundle $L^{(d)}$ on $C\times J^{(d)}$ which is of degree in the $C$-direction, i.e., of degree $d$ on each fibre $C\times x$. The universality then says that for every $X$ and every line bundle $M$ on $C\times X$ of degree $d$ in the $C$-direction there is a unique morphism $f\colon X\rightarrow J^{(d)}$ such that $(\mathrm{id}\times f)^(L^{(d)})$ and $M$ differ by a line bundle from $X$. Picking any line bundle of degree $d$ on $C$ (which may not exist if the base field is not algebraically closed) allows you to construct such a line bundle on $J$ but different choices will give different line bundles on $J$ and hence an automorphism of $J$. This automorphism is then explicitly given by a translation.

As for your second question $q_{2\ast}P(g-1)$ is a torsion free sheaf on $J$ whose restriction to the complement of the theta divisor is $0$ and must therefore be $0$. What happens is that the base change formula makes $H^0(C\times a,P(g-1)_{|C\times a})$ come from $R^1q_{2\ast}P(g-1)$. An algebraic model is the exact sequence $0\rightarrow k[t]\rightarrow k[t]\rightarrow k\rightarrow0$ where $k[t]\rightarrow k[t]$ is multiplication by $t$ (you should think of the first $k[t]$ as $q_{2\ast}F$, the second as $q_{2\ast}E$ and $k$ as $R^1q_{2\ast}P(g-1)$). When you tensor this with $k=k[t]/(t)$ the map $k[t]\rightarrow k[t]$ grows a kernel, i.e., we have an exact sequence $0\rightarrow k\rightarrow k[t]\rightarrow k[t]\rightarrow k\rightarrow0$. This new kernel is $\mathrm{Tor}^1(k,k)$ which models $H^0(C\times a,P(g-1)_{|C\times a})=\mathrm{Tor}^1(k(a),R^1q_{2\ast}P(g-1))$ (which comes from the base change formula).

share|cite|improve this answer
Just to be clear, you are tensoring over $k[t]$ i.e. $\otimes_{k[t]}k$ so $0 \to k[t] \to k[t] \to k$ becomes $0\to k \cong k \xrightarrow{o} k \cong k \to 0$? Also, is it true that we can't hope to make this example a little better by realizing $k[t] \xrightarrow{\cdot t} k[t] \to k \to 0$ as the pushforward of another s.e.s on another scheme? It seems like if we're dealing with quasicoherent modules on affine shcemes then the pushforward of a nonzero module will be nonzero. – solbap May 11 '10 at 18:08
Yes, to the first question. For the second part I am not suggesting that my example says something about direct images of affine morphisms. It is rather an examples of the local picture on the base (in this case $J$) while the direct image is along a proper map. – Torsten Ekedahl May 11 '10 at 20:54

The point is that $q_{2*}P(g-1)$ is a torsion-free sheaf, because it is the image of a torsion-free sheaf. Its support is concentrated on the Theta divisor; but this means that it is torsion, so it must be 0.

share|cite|improve this answer
Ah of course! I knew it was something simple. Thanks! – solbap May 11 '10 at 5:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.