Polishcuk Ex. 6 pg. 132

Statement:

S is a symmetric nxn matix, L a line bundle on an abelian variety A, set

$\ L^S = \bigotimes_i p_i^*L^{s_{ii}} \bigotimes_{i

where the $\ p_i$ are the various projections $\ A^n \to A$, and similarly $\ p_{ij} \colon A^n \to A\times A$ are projections to the i,j th factors.

An m x n matrix $\ M$ defines a map $\ A^m \to A^n$.  Show $\ M^*L^S$ and $\ L^{M^t S M}$ are algebraically equivalent.  In the case when $\ L \cong [-1]^*L$, show the two line bundles are isomorphic.

Incidentally by May 17th I was already tired of thinking of this problem, and I haven’t even completely solved it, but I think I figured out all the essential ideas…

The results and compatibilities to use (for any line bundle L and $\ a \in A$, $\ \phi_L(a):= t^*_aL \otimes L^{-1}$:

1. Corollary 9.2 pg 112 of Polishchuk: L,M alg. equiv. iff $\ \phi_L = \phi_M$
2. For $\ L \in \mbox{Pic}^0(A)$, $\ (f+g)*L \cong f^*L \otimes g^*L$
3. $\ \phi_{M\otimes L} = \phi_M + \phi_L$, $\ \phi_L(a+b) = \phi_L(a) + \phi_L(b):= \phi_L(a) \otimes \phi_L(b)$
4. $\ (p^*L)^n = p^*(L^n)$, $\ p^*(L \otimes M) = p^*L \otimes p^*M$

Small exercise

To warm up, show $\ L, \mbox{ }, [-1]^*L$ are algebraically equivalent.  The notation $\ (A,L)$ means $\ L$ is a line bundle on $\ A$.  Note that the top row of the diagram ( the vertical maps are $\ A \xrightarrow{-1} A$)

$\ \begin{array}{ccc} (A,L^{-1}) & \xrightarrow{t_a} & (A,L) \\ \downarrow & \mbox{} & \downarrow \\ (A,L^{-1}) & \xrightarrow{t_{-a}} & (A,L) \end{array}$

is what one would use to compute $\ \phi_L(a)$.  Using result 4, pulling back the bottom two line bundles all the way back to the top left corner computes $\ \phi_{[-1]^*L}(a)$. The commutativity of the diagram shows $\ \phi_{[-1]^*L}(a) = [-1]^*\phi_L(-a)$.  Using the various results above:

$\ \phi_{[-1]^*L}(a) = [-1]^*\phi_L(-a) = (\phi_L(-a))^{-1} = \phi_L(a)$

The second equality uses the constant map sending everything to the identity can be written as $\ [1] + [-1]$.  The third equality follows because $\ \phi_L(e) = \phi_L(a + (-a))$ is trivial, so the stated equality holds.

Hence $\ L$ and $\ [-1]^*L$ are algebraically equivalent.

Idea of the Proof

For the first part, all there is to show is that $\ \phi_{B} = \phi_{B'}$, $\ B = M^*L^S$ and $\ B' = L^{M^tSM}$.  Here’s how I’ll proceed

Write $\ B' = \bigotimes_i p_i^*L^{N_{ii}} \bigotimes_{i. Then break apart $\ \phi_{B'}(\vec v)$ as

$\ \phi_{B'}(v') = \sum \phi_{p_i^*L^{N_{ii}}}(\vec v) +\sum_{i

$\ = \sum p_i^*\phi_L(N_{ii} \vec w_i) + \sum p_i^*\phi_L(-N_{ij} \vec u_i) + p_j^*\phi_L(-N_{ij} \vec u_j) + m_{ij}^*\phi_L(N_{ij} \vec u_{ij})$

where the various $\ w_i, u_i, u_{ij}$ are to be determined so that the equalities in fact hold.  Using result 2, $\ m_{ij}^*\phi_L$ can be expressed as $\ p_i^*\phi_L$ and $\ p_j^*\phi_L$.  So this whole big mess can be written as something like

$\ \phi_{B'}(\vec v)$ =  $\ \sum_{i = 1}^m p_i^* \phi_L(\mbox{something depending on } M, \vec v, i)$

A similar decomposition can be done for $\ \phi_B(\vec v)$, and with any luck these two decompositions will be the same line bundle, which would prove the first part.

For the second part, the key is to use an iterated version of the seesaw principle to show $\ B, B'$ are isomorphic on all fibers.

A small case

As it is apparent to actually do this problem the way I’ve outlined would be an enormous computational undertaking which doesn’t sound too appealing.  Instead I’ll do a small example (n = 1, m = 2) and trust that an analogous computation would generalize to actually prove the result.

So $\ L$ be a line bundle on $\ A$, $\ s \in \mathbb{Z}$, $\ M = [ a \mbox{ } b]$, so $\ N:= M^t sM = \left( \begin{smallmatrix} sa^2 & sab \\ sab & sb^2 \end{smallmatrix} \right)$.  Fix $\ \vec v = (v_1, v_2) \in A^2$.

Computing $\ \phi_{M^*L^s}(\vec v)$

From the commutativity of the diagram (vertical maps are $\ A^2 \xrightarrow{M} A$)

$\ \begin{array}{ccc} (A^2,M^*L^{-s}) & \xrightarrow{t_{\vec v}} & (A^2,M^*L^s) \\ \downarrow & \mbox{} & \downarrow \\ (A,L^{-s}) & \xrightarrow{t_{M\vec v}} & (A,L^s) \end{array}$

I conclude

$\ \phi_{M^*L^s}(\vec v) = M^*\phi_L(M\vec v) = M^*\phi_{L^s}(av_1 + bv_2) = M^*\phi_L(sav_1 + sbv_2)$

Now, $\ M = [a]\circ p_1 + [b]\circ p_2$, and using that $\ \phi_L(c) \in \mbox{Pic}^0(A)$ and results (1-4) above:

$\ M^*(\phi_L(sav_1+sbv_2)) = ([a]\circ p_1 + [b] \circ p_2)^*(\phi_L(sav_1+ sbv_2))$

$\ = p_1^*[a]^*(\phi_L(sav_1 + sbv_2)) + p_2^*[b]^*(\phi_L(sav_1 + sbv_2))$

$\ = p_1^*\phi_L(sav_1 + sbv_2)^a + p_2^*\phi_L(sav_1 + sbv_2)^b$

$\ \phi_{M^*L^s}(\vec v) = p_1^*\phi_L(sa^2v_1 + sabv_2) + p_2^*\phi_L(sabv_1 + sb^2v_2)$ (*)

Computing $\ \phi_{L^N}(\vec v)$

Again, using the results (1-4) above,

$\ \phi_{L^N}(\vec v) = \phi_{p_1^*L^{sa^2}}(\vec v) + \phi_{p_2^*L^{sb^2}}(\vec v) + \phi_{\Lambda(L)^{sab}}(\vec v)$

Using the diagram (the vertical maps are $\ A^2 \xrightarrow{p_1} A$)

$\ \begin{array}{ccc} (A^2,p_1^*L^{-sa^2}) & \xrightarrow{t_{\vec v}} & (A^2, p_1^*L^{sa^2}) \\ \downarrow & \mbox{} & \downarrow \\ (A,L^{-sa^2}) & \xrightarrow{t_{v_1}} & (A,L^{sa^2}) \end{array}$

I conclude that

$\ \phi_{p_1^*L^{sa^2}}(\vec v) = p_1^*\phi_{L^{sa^2}}(v_1) = p_1^*\phi_L(sa^2v_1)$

Similarly,

$\ \phi_{p_2^*L^{sb^2}}(\vec v) = p_2^*\phi_L(sb^2v_2)$.

For the last part, recall $\ \Lambda(L) = m^*L\otimes p_1^*L^{-1} \otimes p_2^*L^{-1}$, so I can in turn break this apart

$\ \phi_{\Lambda(L)^{sab}}(\vec v) = \phi_{m^*L^{sab}}(\vec v) + \phi_{p_1^*L^{-sab}}(\vec v) + \phi_{p_2^*L^{-sab}}(\vec v)$

$\ = \phi_{m^*L^{sab}}(\vec v) + p_1^*\phi_L(-sab v_1) + p_2^*\phi_L(-sab v_2)$ (1)

So it remains to compute $\ \phi_{m^*L^{sab}}(\vec v)$, for this I use the diagram (the vertical maps are $\ A^2 \xrightarrow{m} A$)

$\ \begin{array}{ccc} (A^2,m^*L^{-sab}) & \xrightarrow{t_{\vec v}} & (A^2,m^*L^{sab}) \\ \downarrow & \mbox{} & \downarrow \\ (A,L^{-sab}) & \xrightarrow{t_{v_1 + v_2}} & (A,L^{sab}) \end{array}$

from which I conclude

$\ \phi_{m^*L^{sab}}(\vec v) = m^*\phi_{L^{sab}}(v_1 + v_2) = (p_1 + p_2)^*\phi_{L^{sab}}(v_1 + v_2)$

$\ = p_1^*\phi_{L^{sab}}(v_1 + v_2) + p_2^*\phi_{L^{sab}}(v_1 + v_2)$

$\ = p_1^*\phi_L(sab(v_1 + v_2)) + p_2^*\phi_L(sab(v_1 + v_2))$ (2)

Comparing (1) and (2), I conclude

$\ \phi_{\Lambda(L)^{sab}}(\vec v) = p_1^*\phi_L(sabv_2) + p_2^*\phi_L(sabv_1)$

Hence

$\ \phi_{L^{N}}(\vec v) = p_1^*\phi_L(sa^2v_1) + p_1^*\phi_L(sabv_2) + p_2^*\phi_L(sabv_1) + p_2^*\phi_L(sb^2v_2)$

$\ \phi_{L^N}(\vec v)= p_1^*\phi_L(sa^2v_1 + sab v_2) + p_2^*\phi_L(sabv_1 + sb^2v_2)$ (**)

So comparing (*) and (**), we see it works in this case!

Small case when $\ L \cong [-1]^*L$

The trick here is that instead of actually computing what $\ L^N$ and $\ M^*L^s$ are, I should compute $\ L^N|_{A \times w}$ and $\ M^*L^s|_{A \times w}$ for every $\ w \in A$, and show these fibers are all isomorphic.  Then by symmetry, the same will be true when restricting to $\ w \times A$, and from the seesaw principle (see for example pg. 194 of Huybrechts) I can conclude they two line bundles are isomorphic.

The restriction of $\ L^N$ to $\ A \times w$ has three parts coming from the pull backs along $\ p_1, p_2$ and the $\ \Lambda(L)$ term.  I use the diagrams

$\ A \times w \to A \times A \xrightarrow{p_1} (A,L^{sa^2})$

$\ A \times w \to A \times A \xrightarrow{p_2} (A,L^{sb^2})$

To compute the contribution from the projections.  The first line contributes $\ L^{sa^2}$ to the final answer.  The second line contributes the trivial bundle (so essentially nothing).

The $\ \Lambda(L)$ term can be computed by appealing to three more diagrams:

$\ A \times w \to A \times A \xrightarrow{p_1} (A,L^{-sab})$

$\ A \times w \to A \times A \xrightarrow{p_2} (A,L^{-sab})$

$\ A \times w \to A \times A \xrightarrow{m} (A,L^{sab})$

Again, the first two lines effectively contribute $\ L^{-sab}$ to the final answer.  The third line contributes

$\ t_w^*L^{sab} = \phi_{L^{sab}}(w) \otimes L^{sab} = \phi_L(sabw)\otimes L^{sab}$

So putting this all together,

$\ L^N|_{A\times w} = L^{sa^2}\otimes \phi_L(sabw)$

The restriction of $\ M^*L^s$ is simpler to compute; I use the diagram

$\ A \times w \to A\times A \xrightarrow{M} (A,L^s)$

to conclude the restriction is

$\ (t_{bw} \circ [a])^*L^s = [a]^*t^*_{bw}L^s = [a]^*(\phi_L(sbw) \otimes L^s)$

$\ = \phi_L(sbw)^a \otimes [a]^*L^s$

$\ = \phi_L(sabw) \otimes L^{sa^2}$

Where I have used that $\ L \cong [-1]^*L$ implies $\ [a]^*L \cong L^{a^2}$ (see for example Cor 8.5 pg. 103 Polishchuk).  So this more or less completes the problem.