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.



The Poincare Bundle(s) on C \times J
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).


