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Section 1.6 The Adventures of BUNTES (Sachi)

Subsection 1.6.1 In which we are introduced to an important homomorphism, review some concepts and our story begins

Abelian variety \(X\text{,}\) we know this is a complete group variety, our goal is to give an embedding \(X\to \PP^N\) for some \(N\text{.}\) This motivates the study of line bundles.

Last time Ricky proved theorem of cube 1.5.11 and square 1.5.15. For any line bundle \(L\) on \(X\text{,}\) there is a group homomorphism \(\Phi_L\colon X \to \Pic(X)\) via \(x\mapsto T_x^* L\otimes L^{-1}\text{.}\) Be careful \(T_x^*\) is \(-x\text{,}\) convention, who knows why.

Example 1.6.1.

Let \(X =E\) an elliptic curve, \(L = L((0))\text{,}\) \(x\mapsto (x) - (0)\text{,}\) in this case this is in \(\Pic^0(E) \cong E \cong \widehat E\text{,}\)

Translate by \(q\in E\text{.}\) \((x+q) - (q)\) take \(p\) to be the third point on the line with \(x,q\text{,}\) \((x) + (q) + (p) \cong 3(0)\) and \((x+q) + (p) \cong 2 (0)\) subtracting these gives \((x) - (x+q) +(q) \cong (0)\) or \((x) - (0) \cong (x+q) -(q)\text{.}\)

What about the converse of this, what can we say about translation invariant line bundles

\begin{equation*} K(L) = \{x\in X : T_x^* L \cong L\}\text{?} \end{equation*}

Consider \(m^* L \otimes p_2^*L^{-1}\) on \(X\times X\text{,}\) then

\begin{equation*} \{x : \text{this is trivial on }\{x\}\times X\} \end{equation*}

is closed. See-saw 1.7.6 implies restriction is pullback

\begin{equation*} T_x^*L \otimes L^{-1} \end{equation*}

so this is \(K(L)\text{.}\)

Subsection 1.6.2 In which Pooh discovers our main theorem

3. to 4..

Is algebraic geometry.

2. to 1..

Follows as being equal is stronger than being linearly equivalent.

Subsection 1.6.3 In which Owl proves the ampleness of \(L\) implies finiteness of \(K(L)\)

4. to 2. Assume \(L\) ample and \(K(L)\) is infinite. Let \(Y\) be the connected component at 0 of \(K(L)\text{,}\) \(\dim Y \gt 0\text{.}\) Show trivial bundle is ample on \(Y\) implies \(Y\) is affine, But \(Y\) is closed and therefore complete so this is a contradiction. \(L|_Y\) ample \(\lb-1\rb^* L|_Y\) is ample. \(L|_Y\otimes \lb -1\rb^*L|_Y\) is ample, consider

\begin{align*} d\colon \amp Y \to Y\times Y\\ \amp y\mapsto (y,-y) \end{align*}

\(m\circ d = \) constant, \(d^*m^* (L) = \sheaf O_Y\text{,}\) LHS is \(L|_Y \otimes [-1]^* L|_Y\text{.}\)

Subsection 1.6.4 In which Rabbbit sets out on a long journey to prove finiteness of \(H(D)\) implies \(|2D|\) is basepoint free and gives a finite map \(X \to \PP^N\)

Note 1.6.5.

\(|2D|\) is always basepoint free.

Apply the theorem of the square 1.5.15: \(T_{x+y}^*D + D \cong T_x^* D + T_y^*D\text{,}\) let \(y = -x\text{,}\) \(2D \cong T_x^* D + T_{-x}^*D\text{.}\) (\(D\) effective) For any \(y\in X\text{,}\) choose some \(x\) s.t. RHS doesn't contain \(y\text{.}\) \(E = 2D\)

\begin{equation*} \psi_E\colon X \to \PP^N \end{equation*}

can we make this finite? If \(\psi_E\) is not finite then \(\psi(C) = \text{pt}\) for some irreducible curve \(C\) (Zariski's main theorem). For each divisor in \(|E|\) either it contains \(C\) or fails to intersect \(C\) by changing \(E\) if necessary, assume \(E \cap C = \emptyset\text{.}\)

Intersection numbers are constant.

\(\sheaf O(T_x^*E)|_{\widetilde C}\text{,}\) when \(x=0\) this is trivial so \(\deg =0\text{.}\) So \(\deg = 0\) for all line bundles. \(E\) effective implies \(C\cap T_x^* E = \emptyset\) for all \(x\) s.t. \(\cap\) is not in \(C\text{.}\)

If \(e \in E\text{,}\) \(T_{x-e}^*(E) \cap C \ne \emptyset\text{.}\) This is as \(x\) is in it, \(x-(x-e) =e\text{,}\) because it is nonempty it's all of \(C\text{.}\) So \(y\) is in it. So \(y- (x-e) \in E\text{.}\) This is also \(e - (x-y) \in E\text{,}\) so \(E\) is invariant under \(T^*_{x-y}\)

Now assume \(H(E) = \{x\in X: T_x^*E = E \}\) is finite. But if \(\psi_E(C) = \text{pt}\) then \(T_{x-y}^*(E) = E\) for all \(x,y\in C\text{.}\) So \(H\) is not finite, a contradiction. So \(\psi_E\) can't collapse a curve so \(\psi_E\) is finite.

Subsection 1.6.5 In which Piglet discovers a corollary

Let \(X\) be an abelian variety, \(U \subseteq X\) be an open affine set, \(0\in U\text{,}\) \(X\smallsetminus U = D_1 \cup \cdots \cup D_t\) irreducible divisors. Let \(D = \sum D_i\text{,}\) then claim: \(H(D) = \{x\in X: T_x^*D = D \}\) is finite. If \(H\subseteq U\text{,}\) \(U\) affine, then \(H\) closed subvariety of an abelian variety, hence complete, so its finite. If \(x\in H\) then \(-x \in H\text{.}\) Now claim that if \(x\in H\) then \(T_x^*\) preserves \(U\text{,}\) if not let \(u\in U\text{.}\) Suppose \(u-x = d\) for some \(d\in D\) then \(u = d+x\) which is \(d\) translated by \(-x\) so \(d+x \in D\) so \(u\in D\text{.}\) But contradiction, oh no! So \(T_x^*\) preserves \(U\text{,}\) for all \(x\in H\text{,}\) as \(0 \in U\text{,}\) for all \(x\in H\) we have \(0-x \in U\) and \(0+x\in U\) so \(H\subseteq U\text{.}\)

\([n]\colon X \to X\) and \(X[n]\) is the kernel of this. Note that for \(x\in X[n]\)

\begin{equation*} [n]\circ T_x = [n] \end{equation*}

\(y\in X\text{,}\) then \(n(y-x) = ny - nx = ny\) so for all \(L \in \Pic X\)

\begin{equation*} T_x^*([n]^* L ) \cong ([n]^* L) \end{equation*}

which implies

\begin{equation*} K([n]^* L ) \supseteq X[n] \end{equation*}

and we just need to find \(L\) s.t. this is finite. \(X\) projective implies there exists an ample \(L\text{.}\) The theorem of the cube 1.5.11 implies

\begin{equation*} [n]^*L \cong L^{\frac{n^2 + n}{2}} \otimes L^{\frac{n^2 - n}{2}} \end{equation*}

where both terms on the right are ample, hence the left is also.

Subsection 1.6.6 Epilogue: In which we might discuss isogenies

Definition 1.6.10.

\(f\colon X \to Y\) a morphism of varieties, get a field extension \(k(X)/f^*k(Y)\text{,}\) if \(\dim X = \dim Y\) and \(f\) is surjective. Then this is a finite field extension and \(\deg f\) is \(d = \lb k(X) : f^*k(Y)\rb\) and \(d = \#f^{-1}(y)\) for almost all \(y\text{.}\)

Definition 1.6.11.

A homomorphism of abelian varieties \(f\colon X \to Y\) is an isogeny if \(f\) is surjective with finite kernel.

Let \(D\) be an ample symmetric divisor, e.g.

\begin{equation*} D = D' + [-1]^* D' \end{equation*}

know \(\lb n \rb^* D \sim n^2 D\)

\begin{equation*} \deg([n]^*(D\cdot\ldots\cdot D)) = ([n]^*D\cdot \ldots\cdot [n]^*D) = (n^2 D\cdot\ldots\cdot n^2 D)= n^{2g} (D\cdot \ldots\cdot D)\text{.} \end{equation*}