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?

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Published:
October 12, 2009 / 11:08 pm
Category:
alg. geo., Teleman
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