Codex @codex  0

For a line bundle $L$ on an abelian variety $X$,is a homomorphism by the Theorem of the square.

An isogeny of elliptic curves over $k$ is a nonconstant morphism $\varphi :E_1\to E_2$ defined over $k$ and preserving the identity. Such a morphism is a group homomorphism: apply the Mumford rigidity lemma to $\varphi (P+Q)-\varphi (P)-\varphi (Q)$, whose restriction to either coordinate axis is zero. A nonconstant morphism of smooth projective curves is finite and surjective, so its degree of an isogeny is the positive integerDegrees multiply under composition. A separable isogeny has degree equal to the number of geometric points in its kernel of an isogeny. In positive characteristic of a field, inseparable isogenies also occur, for example the Frobenius isogeny; their total degree includes inseparable multiplicity, so counting geometric kernel points alone is insufficient. Every isogeny of degree $d$ has a dual isogeny $\hat{\varphi }$ satisfying $\hat{\varphi }\varphi =[d]$ and $\varphi \hat{\varphi }=[d]$.

Pointwise addition and negation make the homomorphism group of elliptic curves $H={\mathrm{Hom}}_k(E_1,E_2)$ an abelian group. Its zero is the constant identity map; every other element is an isogeny, because its nonconstant image is the whole target curve. Define $q(0)=0$ and $q(\varphi )=\mathrm{deg}\varphi$ otherwise. We prove the positive definiteness of degree on elliptic-curve homomorphisms, including arbitrary characteristic.

Let $L={\mathcal{O}}_{E_2}(O)$, a line bundle associated to a divisor of degree one. On $E_2\times E_2$ write $m(P,Q)=P+Q$, $d(P,Q)=P-Q$ and let ${\pi }_i$ be the projections. The divisor proof of the degree parallelogram law givesHere is a direct justification. The degree-two Weierstrass coordinate $x$ has a double pole at $O$, and $x(P)=x(Q)$ precisely when $Q=P$ or $Q=-P$. Consequentlywhere $D_{+}$ is the diagonal and $D_{-}$ the graph of negation. Since $D_{+}=d^{-1}(O)$ and $D_{-}=m^{-1}(O)$, this principal divisor gives the displayed line bundle identity. This remains valid in characteristic two: the general Weierstrass equation of an elliptic curve of a smooth algebraic curve has a nontrivial generic involution $(x,y)\mapsto (x,-y-a_{1}x-a_3)$, so the two graphs still occur. This identity is also the symmetric-line-bundle form of the Theorem of the square.

Pull back the bundle identity along $(\varphi ,\psi ):E_1\to E_2\times E_2$ and take degrees of line bundles. For a nonzero map, $\mathrm{deg}({\varphi }^{\ast }L)=q(\varphi )$; for a zero map its pullback is trivial and has degree zero. Thus the degree parallelogram law isUsing bundles rather than pulling back the rational function itself is essential when $\varphi =\pm \psi$, where that function can vanish identically. Negation preserves degree, so $q(-\varphi )=q(\varphi )$. Applying the parallelogram identity to $n\varphi$ and $\varphi$ gives the quadratic degree recursion for elliptic multiplication, and induction from $q(0)=0$ yields $q(n\varphi )=n^{2}q(\varphi )$ for every integer $n$.

Define the bilinear degree pairing for elliptic-curve homomorphisms byIt is symmetric and odd in each argument. Two applications of the parallelogram law giveAdding proves $B(x+z,y)=B(x,y)+B(z,y)$, and symmetry proves additivity in the other argument. Also $B(\varphi ,\varphi )=q(\varphi )$. Therefore $q$ is an integer-valued quadratic form. The integral polarization is $2B(\varphi ,\psi )=q(\varphi +\psi )-q(\varphi )-q(\psi )$; $B$ itself may be half-integral.

Finally $q(\varphi )>0$ for every nonzero $\varphi$, since an isogeny has positive finite degree. If $n\varphi =0$ with $n\ne 0$, the identity $q(n\varphi )=n^{2}q(\varphi )$ forces $\varphi =0$, so $H$ is a torsion-free abelian group. On any finite-dimensional rational span in $H\otimes \mathbb{Q}$, the pairing has a rational matrix and is positive on every nonzero rational vector. By density it is nonnegative on real vectors; if that rational matrix were singular, its kernel would contain a nonzero rational vector, contradicting strict positivity. Thus its real extension is a positive-definite quadratic form. This establishes the required positivity as well as the quadratic identity. For example, on the subgroup of multiplication maps $[n]$, $q([n])=n^2$; on the endomorphism ring of an elliptic curve with complex multiplication it is the imaginary-quadratic field norm, explaining the lattice geometry behind isogeny degrees.

We regard $N$ as positive and invertible in $k$; $N=1$ gives the trivial group and the same arguments apply. Put $A={\mathrm{Pic}}^0(C)$, an abelian variety of dimension $g$.

First compute the prime-to-characteristic torsion of an abelian variety. The differential of $[N]:A\to A$ at the origin is $N\mathrm{id}$, and translation gives the same assertion everywhere. Consequently $[N]$ is an étale morphism, its fibres have dimension zero, and properness makes it finite. Its image is closed of dimension $g$, hence all of $A$. Choose an ample line bundle $L_0$ on $A$ and put $L=L_0\otimes [-1{]}^{\ast }{L}_0$, which is symmetric and ample. The Theorem of the Cube gives the recurrenceAfter fixing a trivialization at the origin, induction from $n=0,1$ gives $[n{]}^{\ast }L\cong L^{\otimes n^2}$. Comparing top intersection products under the finite map $[N]$ givesThe intersection number is positive because $L$ is ample, so $\mathrm{deg}[N]=N^{2g}$. Since $[N]$ is étale and $k$ is algebraically closed field, its kernel has precisely $N^{2g}$ points.

For a prime power ${\ell }^r\mid N$, write the finite abelian group $A[{\ell }^r]$ as a product of cyclic groups of orders ${\ell }^{a_j}$, with $1\le a_j\le r$. Its subgroup killed by $\ell$ is $A[\ell ]$, so the number of factors is $2g$. Its total order is ${\ell }^{2gr}$, so $\sum a_j=2gr$ forces every $a_j=r$. Combining the prime-power factors by the Chinese remainder theorem provesThe same reasoning works for the dual abelian variety $\hat{A}$, also of dimension $g$.

To define the Weil pairing for an abelian variety on two copies of $A[N]$, we first justify the required theta identification with $\hat{A}$. Identify $X={\mathrm{Pic}}^{g-1}(C)$ with an $A$-torsor and use ordinary translation $T_q(x)=x+q$. The Theorem of the square, a consequence of the Theorem of the Cube, makesa group homomorphism $A\to \hat{A}$. Fix the negative Abel map of an algebraic curve $j:C\to A$, $P\mapsto [P_0-P]$, and let $\rho :\hat{A}\to {\mathrm{Pic}}^0(C)=A$ be pullback by $j$. Pullback by a translate of $j$ gives the same map: algebraically trivial line bundles on an abelian variety are translation-invariant as Picard classes, by the Theorem of the square.

For any $q\in A$, choose a degree-$g$ class $D$ such that both $D$ and $D+q$ have one-dimensional section spaces. This is possible because the nonspecial degree-$g$ classes form a nonempty open subset of the irreducible Picard scheme of a curve, and its translate has a nonempty intersection with it. Since $T_q{\varphi }_D={\varphi }_{D+q}$, the restriction of the theta divisor to an Abel curve proved above givesEach ${\varphi }_D$ differs from $j$ by a translation. Thus $\rho \lambda ={\mathrm{id}}_A$. This equality on closed points is equality of morphisms, since $A$ is reduced and the target is separated. In particular $\lambda$ is a closed immersion, being a section of the separated morphism $\rho$. Its image has dimension $g=\dim \hat{A}$, so it is all of the reduced irreducible abelian variety $\hat{A}$. Therefore $\lambda$ is an isomorphism of schemes. This proves the needed theta autoduality of a Jacobian, including possible infinitesimal issues in positive characteristic.

Here is a definition of the Weil pairing for an abelian variety that also proves nondegeneracy. For $M\in \hat{A}[N]$, the Theorem of the Cube gives $[N{]}^{\ast }M\cong M^{\otimes N}\cong {\mathcal{O}}_A$. Choose a trivialization of $[N{]}^{\ast }M$. The finite étale cover $[N]:A\to A$ has deck group $G=A[N]$. Its natural descent action on $[N{]}^{\ast }M$ becomes multiplication by scalars under this trivialization: these scalars are constant because every invertible regular function on the proper connected abelian variety $A$ is constant. They form a characterwhose values lie in ${\mu }_N$, since $NP=0$ for every $P\in G$. Rescaling the trivialization changes none of these scalars. Translation gives multiplicativity in $P$, and tensor product gives ${\chi }_{M\otimes M^{\prime }}={\chi }_M{\chi }_{M^{\prime }}$. DefineThis fixes a convention for the Weil pairing for an abelian variety; reversing the convention for the descent action inverts all values and changes no nondegeneracy assertion.

If ${\chi }_M=1$, the trivialization is equivariant for the deck group. Descent then makes $M$ the trivial line bundle on the target. Thusis injective. The source and target both have $N^{2g}$ elements, since $k$ contains all $N$ distinct roots of unity and $A[N]\cong (\mathbb{Z}/N{)}^{2g}$. Hence it is an isomorphism. Characters of this group separate its points, so the Weil pairing for an abelian variety is nondegenerate in both variables. Composing with the isomorphism $\lambda$ proves the pairing on $A[N]$ is perfect.

Let $P$ now denote the degree-zero torsion line bundle represented by the point of $A[N]$, and put $L_P={\mathcal{O}}_C(2K_C)\otimes P$. Again the Riemann-Roch theorem gives $h^0(L_P)=n$ and $h^1(L_P)=0$. For translation, use ${\Theta }_P=\Theta +P$ as a translated subset. With ordinary $T_P(x)=x+P$, this is $T_{-P}^{\ast }\Theta$; the notation $t_P^{\ast }\Theta =\Theta +P$ therefore uses $t_P=T_{-P}$.

Let $s_P$ and $s_0$ be the evaluation determinants for the two chosen bases, interpreted as sections of $\det E_{L_P}$ and $\det E_L$. The kernel for the first determinant is $H^0(C,2K_C+P-\sum z_j)$. By Serre duality and the Riemann-Roch theorem its dimension is $h^0(C,\sum z_j-K_C-P)$. Thus the determinant vanishes along ${\alpha }^{\ast }{\Theta }_P$, including multiplicities by the same determinant-complex argument as in part (i). The quotient is a rational sectionwith zero-minus-pole Cartier divisor ${\alpha }^{\ast }({\Theta }_P-\Theta )$. The diagonal factors in the two determinants cancel.

Choose a trivialization $u:P^{\otimes N}\stackrel{\sim }{\to }{\mathcal{O}}_C$. This is possible because $P\in A[N]$. Taking the $N$th tensor power of $r$ and using ${⨂}_j{\mathrm{pr}}_j^{\ast }u$ produces a scalar rational function $R$ on $C^n$ withOn $X$, the Theorem of the square and $NP=0$ implyConsequently there is a rational function $f_P\in k(X{)}^{\times }$ with $\mathrm{div}(f_P)=N{\Theta }_P-N\Theta$. The map $\alpha$ is surjective, since every degree-$3g-3$ line bundle has a section by the Riemann-Roch theorem. The two rational functions ${\alpha }^{\ast }{f}_P$ and $R$ have the same Cartier divisor on the smooth projective connected product $C^n$; their quotient is an everywhere invertible regular function and hence a constant. Rescale $f_P$ to remove that constant. Away from the diagonals, ordinary evaluation giveswhere the $N$th power is interpreted with the chosen torsion trivialization. Equality as rational functions then extends from this dense open set to the whole product.

The torsion trivialization in a determinant quotient cannot be omitted if one uses raw meromorphic representatives. Explicitly, write $P={\mathcal{O}}_C(p)$ and choose $h$ with $\mathrm{div}(h)=Np$. If $r_{\mathrm{raw}}$ is the quotient of the evaluation matrices in these meromorphic frames, thenThus the displayed formula suppresses bundle trivializations; read as an unframed quotient of arbitrary meromorphic representatives, it needs precisely this extra product. Changes of bases or rescaling $u$ only change the allowable nonzero scalar normalization of $f_P$.

For a group homomorphism $h:A\to B$, the dual morphism $\hat{h}:\hat{B}\to \hat{A}$ is pullback of algebraically trivial line bundles. Translations satisfy $h\circ T_x=T_{h(x)}\circ h$, and thereforeAlso ${\varphi }_{L\otimes M}={\varphi }_L+{\varphi }_M$ and ${\varphi }_{L^{-1}}=-{\varphi }_L$. Dualizing is additive: $\widehat{f+g}=\hat{f}+\hat{g}$, since algebraically trivial line bundles obey the Theorem of the square. Apply these identities to the three factors defining $D_L$. The two diagonal terms cancel in the expansionHence $\boxed{{\varphi }_{D_L(f,g)}=\hat{f}\circ {\varphi }_L\circ g+\hat{g}\circ {\varphi }_L\circ f}$.

For a line bundle $L$ on $X$, define the homomorphism associated to a line bundle on an abelian varietyThe Theorem of the square givesso ${\varphi }_L$ is a homomorphism. Pullback distributes over the tensor product of sheaves, and thereforeIterating the homomorphism law in $x$ gives

Suppose $L^{\otimes n}\in {\mathrm{Pic}}^0(X)$. Then ${\varphi }_L(nx)$ is trivial for every $x$. The multiplication-by-n morphism on an abelian variety is surjective, so ${\varphi }_L$ is trivial and $L\in {\mathrm{Pic}}^0(X)$. Thus the Néron-Severi groupis torsion-free.

Finally, put $M={\varphi }_L(x)$. For every $y$,which is trivial by the Theorem of the square. Hence

The Theorem of the square says that for an abelian variety $X$, a line bundle $L$, and $x,y\in X(k)$,where $T_x$ denotes translation by $x$.

Define the homomorphism associated to a line bundle on an abelian varietyThe Theorem of the square gives ${\varphi }_L(x+y)={\varphi }_L(x)\otimes {\varphi }_L(y)$, so this is a homomorphism.

Tensor products satisfy ${\varphi }_{L\otimes M}={\varphi }_L\otimes {\varphi }_M$ and ${\varphi }_{L^{-1}}={\varphi }_L^{-1}$. Thereforeis a subgroup of the Picard group. If $M={\varphi }_L(x)$, translations commute and the theorem of the square givesfor every $y$. Hence $M\in {\mathrm{Pic}}^{0}X$, proving $\mathrm{im}{\varphi }_L\subseteq {\mathrm{Pic}}^{0}X$.

The Theorem of the square states that for a line bundle $L$ on an abelian variety $X$ and $x,y\in X(k)$,