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.

Advertisements

About this entry