# Polishchuk Chap. 17 constructions and questions

Polischuk claims if $X$ is smooth and projective then any object in $D^b(X)$ (bounded derived category of coherent sheaves) is isomorphic (in the derived sense) to a finite complex of locally free sheaves.  This is roughly the content of III.6.9 in Hartshorne, this problem requires $X$ to be Noetherian, integral, separated and regular.

Being in this situation means you can take determinants.  For a locally free sheaf its the top exterior power.  For a finite complex of locally free sheaves this is the alternating tensor power of the determinants of each of the pieces.

see this related post regarding the red parts.

### Determinant, section construction

$\pi \colon C \to S$ is a relatively smooth projective curves.  I guess this means $C$ is a curve, and $\pi$ is smooth (of relative dimension zero?; see vakil…),  and assume $F$ is flat over S and  $\chi(C_s, F_s) = 0$ (where does this come into play (think it might show rk $R\pi_*F = 0$)?

Do: $0 \to F \to E_0 \to E_1:= E_0/F \to 0$ such that $E_i$ are flat over $S$, (what’s the reason flatness keeps coming into the picture).  Assuming $R\pi^1_*E_0 = 0$ (apparently can take $E_0 = F(D)$ for $D$ a relatively effective divisor) should imply $R\pi^1_*E_1 = 0$ (how? ; long exact sequence does it $0 = R\pi_*^1E_0 \to R\pi_*^1E_1 \to R\pi_*^2F = 0$; $C$ dim = 1) .

Also that $R\pi_*^1 E_i = 0$ should say $R\pi_*E_i$ are vectors bundles, (cohomology with base change, would want $s \mapsto h^0(C_s, (E_i)_s)$ to be constant, comes from flatness?, or how does this work?)

One important thing I didn’t immediately realize is that $C$ is a curve! So the only potential higher cohomology is $H^1$.

In any case, these claims are meant to justify that $R\pi_*F \cong [\pi_*E_0 \to \pi_*E_1]$ in the bounded derived category.  Does this use some spectral sequence magic, or something like E_i are pi_* acyclic?

In any case, $\det R\pi_*F$ is well defined.  Somehow $\mbox{rk} R\pi_*F = 0$, but $\mbox{rk} R\pi_*F = \mbox{rk} R\pi_*E_0 - \mbox{rk} R\pi_*E_1$, so the pushforwards of $E_i$ have the same rank.  So the map $\pi_*\alpha \colon \pi_*E_0 \to \pi_*E_1 = \pi_*(E_0/F)$ is represented by a sheaf version of a square matrix, hence has a sheaf version of a determinant.  So Polishchuk says $\det \pi_* \alpha$ should be a section of $\det (R\pi_*F)^{-1} = \det(\pi_*E_0)^{-1}\otimes \det \pi_*E_1$, but it seems $\det \pi_* \alpha$ is just a section of $\det \pi_*E_1$.  A section of $\det \pi_*E_0$ should just be an alternating map $\pi_*E_0^{its Rank} \to O_S$, for example taking n vectors in $R^n$ and taking the determinant of the matrix they form.  So can describe sections of each peice individually, but what is section of the product?

Then there is a nontrivial argument that shows this construction doesn’t so much depend on the resolution $E_0 \to E_1$.  But I didn’t really work through this.

### Curve Mapping to an Abelian Variety construction

The setup

1. $a \colon C \to A$ map from curve to abel. var.
2. $F$ cohrent, $S(F):= \Phi_{P_A}(a_*F)$.
3. $S(F)$ lands in the bounded derived category so $\det S(F)$ is defined.

Certainly for a a line bundle $L$ on $C$ there is s.e.s (this is by construction exact when localized at every point)

$0 \to L \to L(p) \to L(p)|_p \to 0$

Some general derived category nonsense says the natural functor from $Coh(X) \to D^b(X)$ is exact meaning s.e.s get sent to distinguished triangles so there is something like an exact triangle

$S(L) \to S(L(p)) \to S(L(p)|_p)$

If you just write everything out (see prop. 17.1 of Polishchuk) then what you want happens:

$\det S(L(p)) = \det S(L) \otimes \det S(L(p)|_p) = \det S(L) \otimes P|_{a(p)\times \hat A}$

The magical thing is that this means $D = \sum_i n_i p_i \mapsto \sum_i n_i a(p_i)$ doesn’t depend on the rational equvivalence class of $D$ so there is a well defined map

$a_* \colon Pic(C) \to A(k)$

Facts

1. $\phi_a := \phi_{\det(S(L)}$ is the composition $a_* \circ a^*$. (note the latter is the map $x \mapsto \mathcal{P}_{a_*(a^*(x))}$.
2. $f \colon A \to B$ is a homo of var then $\phi_{f \circ a} = f \circ \phi_a \circ \hat f$.

### Principal Polarization of Jacobian

Given a curve $C$ with a point $p$ then it has a Jacobian $J = Jac(C)$ and a line bundle $\mathcal{P}_C$ (normalized at p) on $C \times J$.  By the universal property of the Poincare bundle $\mathcal{P}$ on $J \times \hat J$, there is $a \colon C \to \hat J$ such that

$\mathcal{P}_C \cong (a \times id)^*\mathcal{P}$

maps:

1. $a^*_0 \colon Pic^0(\hat J) = J \to J =Pic^0(C)$ is the identity (a tautology).
2. This implies $\phi_a = a_* \circ a^*_0 = a_* \colon J \to \hat J$
3. $i \colon C \to J$ sending $q \mapsto O_C(q - p)$ satisfies $i = i^* \circ a$ (see below)

By universal property of $J(C)$ there morphism $i$ is equivalent to a line bundle $M:= (id \times i)^*P_C$ on $C \times C$.  In fact $M = O_{C\times C}\otimes p_1^*O_C(-p)\otimes p_2^*O_C(-p)$. Tautologically the map $i_* \colon Pic^0(C) \to J(k)$ is the identity.  Putting universal properties together, $M$ is actually the pullback of $\mathcal{P}$ along

$C \times C \xrightarrow{id \times i} C \times J \xrightarrow{a \times id} \hat J \times J$

So $M = (a \times i)^*\mathcal{P}$ this gives statement 3.

The magical thing that happens is $(i^* \circ a)_* \circ Pic^0(C) \to J(k)$ is equal to $i^* \circ a_*$.  But $(i^* \circ a)_* = i_* = id$.  So we get an inverse in one direction, but what about the other direction?