# Polishchuk Ex. 4,6 pg. 149

### Exercise 4

Prop. $X$ a complete variety, and $f \colon X \to A$ is a map into an abel. var. s.t. $f^* \colon \mbox{Pic}^0(A) \to \mbox{Pic}(X)$ is trivial.  Then $f(X)$ is a point.

proof: Consider the pull back of the normalized poincare bundle

$L = (f, id)^*\mathcal{P}$ on $X \times \widehat{A}$

To calculate $L|_{X \times a}$ I use the diagram

$X \times a \xrightarrow{(f, a)} A \times a \leftarrow \mathcal{P}_a$

I have $\mathcal{P}_a \in \mbox{Pic}^0(A)$ and $L|_{X \times a} = f^*\mathcal{P}_a = \mathcal{O}_X$.  By the see-saw principle this implies $L \cong p_2^*N$ for some line bundle on $\widehat{A}$.  Then $L|_{x \times \widehat{A}} = N$ for all $x$, but using the poincare bundle, I have

$L|_{x \times \widehat{A}} = \mathcal{P}|_{f(x) \times \widehat{A}}$

That is $\mathcal{P}|_{f(x) \times \widehat{A}}$ is constant and since $\mbox{Pic}^0(\widehat{A}) \subset \mbox{Pic}(\widehat{A})$ characterizes line bundles upto isomorphism, it follows that $f(x)$ is constant. QED

Corollary: $f$ is determined by $f^*$ upto translation, i.e upto replacing $f$ with $t_x \circ f$.

proof: Suppose $f,g$ are such that $f^* = g^*$, then for $L \in \mbox{Pic}^0(A)$

$(f - g)^*L = f*L\otimes (-g)^*L$

$f^*L \otimes g^*L^{-1} = \mathcal{O}_X$

So by the proposition, $f - g = pt$, hence $f = t_a \circ g$ for $a = (f - g)(X) \in A$. QED.

### Exercise 6

For this problem, here are two big results that I’ll use heavily but wont prove.  For the first, see Abelian Varieties by Mumford pg. 150.

Thm (Riemman-Roch) Let $L$ be an invertible sheaf on an abelian variety $A$.  If $L = L(D)$, then $\chi (L) = (D^g)/g!$ for $g = \dim A$, and $(D^g)$ is the g-fold self intersection number. [_]

Polishchuk ex 5 pg 149: $\chi(f^*F) = \deg(f) \chi(F)$ where $f$ is an isogeny of abelian varieties and $F$ is a coherent sheaf. [_]

Ex 6: Let $L$ be a line bundle on an abel. var. $A$.  Then for $g = \dim A$, $\chi(L^n) = n^g \chi(L)$ for $n > 0$.  Also the multiplication by n map $[n]\colon A \to A$ has degree $n^{2g}$.

proof: For the first statement apply the Riemann-Roch theorem.  If $L = L(D)$ then $L^n = L(nD)$ and

$\chi(L^n) = ((nD)^g)/g! = n^g(D^g)/g! = n^g \chi(L)$

For the second statement, note that every abelian variety is projective, so $A$ has a very ample line bundle $L$, as $[-1] \colon A \to A$ is an isomorphism, $[-1]^*L$ is also ample and $M = L\otimes [-1]^*L$ is symmetric.  Indeed,

$[-1]^*M = [-1]^*L\otimes ([-1] \circ [-1])^*L = M$

By proposition 8.5 on page 103 of Polishuck, $[n]^*M = M^{n^2}$, so using ex 5,

$\chi([n]^*M) = \chi(M^{n^2}) = (n^2)^g\chi(M)$

$= n^{2g}\chi(M) = \deg([n]) \chi(M)$

Hence $\deg ([n]) = n^{2g}$.QED.