Torelli Over an Algebraically Closed Field

I was originally going to do this post in wordpress but then instead I’m just posting Torelli over k bar; a rough outline based on a proof in Polishchuk’s book.

But I wanted to add some notes here.  At some point I reference this post about chapter 17 in Polishchuk. 

Extensively in the proof base change is used.  This says if there is a diagram

\begin{array}{ccc} A & \xrightarrow{f} & B\\ \downarrow q & \empty & \downarrow p\\ C & \xrightarrow{g} & D \end{array}

where A = B\times_D C and g if flat and p is proper then there is an isomorphism 

Rq_*f^*H \cong g^*Rp_*H

as f,g are flat they don’t need to be derived (Huybrechts pg 85).  As an application consider three schemes A',A,B and a map g \colon A' \to A.  Note A\times B \to B is flat.  Let F \in D^b(A'),\ P \in D^b(A \times B) and consider \Phi_P(Rg_*F) = Rp_{2*}(p_1^*g_*F\otimes P).  

\begin{array}{ccc} A' \times B & \xrightarrow{q_1} & A'\\ \downarrow g\times id & \empty & \downarrow g\\ A \times B & \xrightarrow{p_1} & A \end{array}

This gives R(g\times id)_*q_1^*F \cong p_1^*Rg_*F.  Also q_2 = p_2 \circ (g \times id). So

\Phi_{(g \times id)^*P}(F) = Rq_{2*}(q_1^*F\otimes (g\times id)^*P)

= Rp_{2*}R(g \times id)_*(q_1^*F \otimes (g\times id)^*P)

= Rp_{2*}(R(g\times id)_*q_1^*F \otimes P) (proj. form.)

Rp_{2*}(p_1^*Rg_*F \otimes P) = \Phi_P(Rg_*F)

This is the result for 

A' \to A \leftarrow A \times B \to B

There are similar stories for 

A' \leftarrow A \leftarrow A \times B \to B

A \leftarrow A \times B \to B \to B'

A \leftarrow A \times B \to B \leftarrow B'

 

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Published:
May 21, 2010 / 2:23 am
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