Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 18 (Abelian Varieties)

Let $E=\mathbb{C}/\Lambda$, and let ${\Lambda }^{\prime }$ be the inverse image of the cyclic subgroup $G$ under the quotient map. Then $\Lambda \subset {\Lambda }^{\prime }$ is an inclusion of Euclidean lattices with ${\Lambda }^{\prime }/\Lambda \cong \mathbb{Z}/N\mathbb{Z}$. The Smith normal form supplies a basis $u,v$ of ${\Lambda }^{\prime }$ for whichChoose the orientation of this basis so that $\mathrm{Im}(v/(Nu))>0$. Dividing the coordinate on $\mathbb{C}$ by $Nu$ gives an isomorphism of complex toriHere $u$ maps to $1/N$, and its class generates ${\Lambda }^{\prime }/\Lambda$. Thus the isomorphism carries $G$ to the required cyclic subgroup. Every pair is represented: $\Phi$ is surjective. This argument avoids assuming that an arbitrary integer lift of a primitive vector modulo $N$ is itself primitive.

Suppose ${\tau }_1=(a{\tau }_2+b)/(c{\tau }_2+d)$, with $ad-bc=1$, and put $\alpha =(c{\tau }_2+d{)}^{-1}$. The identitiesshow that multiplication by $\alpha$ takes one period lattice isomorphically to the other: its basis matrix has determinant $ad-bc=1$. If $N\mid c$, the image of $1/N$ is congruent to $a/N$ modulo the target period lattice. Moreover $ad\equiv 1(\mathrm{mod}N)$, so $a$ is a unit modulo $N$. The image therefore generates the same cyclic subgroup.

Conversely, a holomorphic group isomorphism of complex tori lifts to $z\mapsto \alpha z$, with $\alpha {\Lambda }_{{\tau }_2}={\Lambda }_{{\tau }_1}$. Indeed its lift fixing zero is additive: its failure to be additive is a continuous lattice-valued function, hence zero. An additive holomorphic map is complex linear. WriteThe corresponding integral basis matrix has determinant $1$: complex multiplication preserves the real orientation, and both bases $(1,{\tau }_j)$ are positively oriented. Solving gives ${\tau }_1=(a{\tau }_2+b)/(c{\tau }_2+d)$. Preservation of the cyclic subgroup means $\alpha /N\equiv u/N(\mathrm{mod}{\Lambda }_{{\tau }_1})$ for some unit $u$ modulo $N$. Comparing the two real basis coefficients gives $N\mid c$. Thus the matrix lies in the Gamma 0 congruence subgroup, and the classification isThis is the noncompact modular curve classification; no cusp points have been added.

Use the convention that a Hermitian form is complex linear in its first argument. In the integral Hermitian forms on the square complex torus calculation, every Hermitian form on $\mathbb{C}$ has the form $H_t(z,w)=tz\overline{w}$, where $t=H_t(1,1)$ is real. For $\lambda =m+ni$ and $\mu =r+si$ in the period lattice,Consequently integrality holds precisely when $t\in \mathbb{Z}$: necessity follows from $E_t(1,i)=-t$, and sufficiency follows from the displayed formula. Addition of Hermitian forms corresponds to addition of $t$. Positivity is equivalent to $t>0$, since $H_t(z,z)=t|z{|}^2$. The answer isWith the conjugate-linear-first convention the formula is $t\overline{z}w$ and the sign of $E_t$ reverses. The positive integer parameter is unchanged.

For the intended complex unit circle, put $u=\alpha (1)$ and $v=\alpha (i)$, with $|u|=|v|=1$. The complete family of semicharacters of a complex lattice isTo obtain this, the identity at zero gives $\alpha (0)=1$; on each coordinate axis the alternating form vanishes, so $\alpha (m)=u^m$ and $\alpha (ni)=v^n$, also for negative integers. Combining the axes gives the factor $(-1{)}^{mn}$. Conversely, for $\lambda =m+ni$, $\mu =r+si$, the quotient of the proposed values is $(-1{)}^{ms+nr}$, which equals $e^{i\pi (nr-ms)}$. Hence every displayed formula is a semicharacter of a complex lattice and none are missing.

The original PDF actually puts $z\in \mathbb{Z}$ in its definition of $C_1$. Under that literal reading $C_1=\{1,-1\}$, so there are exactly four solutions, given by the same formula with $u,v\in \{1,-1\}$. Under either reading the specified choice is ${\alpha }_1(m+ni)=(-1{)}^{mn}$.

The Appell–Humbert theorem identifies the positive Hermitian form $H_1$ with a holomorphic line bundle whose First Chern class is represented, in our linear-first convention, byIts integral over the square fundamental domain is $1$, so $\mathrm{deg}L=1$. In particular the oriented Chern pairing is $-E_1(1,i)=1$; using $E_1(1,i)$ without adjusting the Hermitian convention would give the wrong sign.

The complex torus has a nowhere-vanishing differential $dz$, so its canonical divisor is trivial and its genus is one. A nonzero section of a line bundle of negative degree would have an effective divisor of negative degree, which is impossible. Thus $H^0(X,L^{-1})=0$. The Riemann-Roch theorem and Serre duality giveTherefore $\boxed{h^0(X,L)=1}$. The choice of semicharacter of a complex lattice changes the degree-zero component of the line bundle, but not this calculation.

A holomorphic group endomorphism of the complex torus lifts to multiplication by a scalar $c\in \mathbb{C}$, as in Question 1. It descends exactly when $c\Lambda \subseteq \Lambda$. Applying this condition to $1$ gives $c\in \mathbb{Z}[i]$. Conversely any Gaussian integer $c$ preserves $\mathbb{Z}[i]$, since the latter is a ring. Addition of endomorphisms is addition of scalars, and composition is multiplication of scalars. Thus, as rings,The wording “endomorphism algebra” here denotes this integral endomorphism ring of an elliptic curve. If it instead meant the rational algebra ${\mathrm{End}}^0(X)=\mathrm{End}(X){\otimes }_{\mathbb{Z}}\mathbb{Q}$, that algebra would be $\mathbb{Q}(i)$.

Pulling back the factors of automorphy in the Appell–Humbert theorem by $z\mapsto cz$, where $c=a+bi$, pulls back both the Hermitian form and the semicharacter of a complex lattice. Thus the explicit answer isWriting $\lambda =m+ni$ makes the second formula completely explicit:The second equality is an equality of signs, obtained by reducing the exponents modulo $2$. The Hermitian form follows from $H_1(cz,cw)=|c{|}^2{H}_1(z,w)$. Its integer parameter equals the norm of the Gaussian integer, agreeing with the degree of this nonzero isogeny of elliptic curves. The formulas also include $c=0$, when the pulled-back line bundle is trivial.

The bundle is on $A$, so the printed, undefined target $\mathrm{Pic}(X)$ must be read as the Picard group $\mathrm{Pic}(A)$. Write this Picard group additively and set $q(h)=[h^{\ast }L]$. The zero morphism pulls back $L$ to a constant one-dimensional vector space tensored with ${\mathcal{O}}_A$, so $q(0)=0$. The Theorem of the Cube, pulled back along $(f,g,h):A\to B^3$, givesThis is the bilinear cross-effect of a line bundle on an abelian variety: set $d(f,g)=q(f+g)-q(f)-q(g)$. Substitution into this identity givesThe definition gives $d(f,g)=d(g,f)$, so additivity holds in both arguments. Additivity includes negative multiples and the zero morphism. Hence $D_L$ is a symmetric bilinear map with values in $\mathrm{Pic}(A)$. The equalities refer to isomorphism classes of line bundles; without chosen rigidifications they are not asserted to be specified canonical isomorphisms of bundles.

Let $m:B\times B\to B$ be addition, and let $p_1,p_2$ be the projections. Consider the line bundleIts restriction to $B\times \{y\}$ has Picard group class $[T_y^{\ast }L\otimes L^{-1}]={\varphi }_L(y)$; the factor $L_y^{-1}$ is a constant line and has trivial class. Its restriction to $\{0\}\times B$ is also trivial. Choose a trivialization of the fiber $L_0$ and use it to normalize along the zero sections. The defining universal property of the normalized Poincaré line bundle then identifiesMore explicitly, both sides have the same restrictions to all fibers of the second projection, and both are normalized on $\{0\}\times B$; the Seesaw theorem makes their quotient trivial. Here ${\varphi }_L:B\to \hat{B}$ is the homomorphism associated to a line bundle on an abelian variety, and $\hat{B}$ is the dual abelian variety. Pull back by $(f,g)$ to obtain the required identityThe choice of fiber trivialization is immaterial to this identity in the Picard group.

For $M\in {\mathrm{Pic}}^0(B)$, translations preserve its line bundle class, so ${\varphi }_M=0$. Equivalently, the kernel of $L\mapsto {\varphi }_L$ is the Identity component of the Picard group. Applying the preceding Poincaré line bundle formula to $M$ yieldssince the normalized Poincaré line bundle is trivial on $B\times \{0\}$. Pullback and tensor products of sheaves also give $D_{L\otimes M}=D_L\otimes D_M$. Thus $D_L$ depends only on $[L]\in \boxed{\mathrm{NS}(B)}$, the Néron-Severi group.

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 $M$ on $C\times T$, put $d(t)=\mathrm{deg}{M}_t$, and choose an ample line bundle $A$ on $C$. Fix $t_0$. For sufficiently large $q$, both $H^1(C,M_{t_0}\otimes A^q)$ and $H^1(C,M_{t_0}^{-1}\otimes A^q)$ vanish, by Serre duality and negativity of the degrees of their dual canonical twists. The semicontinuity theorem for coherent cohomology makes both groups vanish on a neighborhood of $t_0$. The family is flat over $T$: $C\times T\to T$ is flat and the sheaves in question are locally free.

On that neighborhood the Riemann-Roch theorem givesUpper semicontinuity of both zeroth sheaf cohomology dimensions, after shrinking again, gives $d(t)\le d(t_0)$ and $-d(t)\le -d(t_0)$. Hence $d(t)=d(t_0)$. The degree is locally constant, and thus constant on connected $T$. With the usual convention that a variety is irreducible, this proves the assertion as printed. If disconnected varieties are allowed, the necessary qualification is constancy on each connected component: take $T$ to be two points and the bundles ${\mathcal{O}}_{{\mathbb{P}}^1}$ and ${\mathcal{O}}_{{\mathbb{P}}^1}(1)$ to see that a global constant degree need not exist.

Pull $L$ back by the morphism of varietiesThe fiber of this line bundle over $x$ is ${\phi }^{\ast }{T}_x^{\ast }L$. An abelian variety is connected, so the preceding constancy of line bundle degree in a family applies. Consequently $\boxed{\mathrm{deg}({\phi }^{\ast }{T}_x^{\ast }L)=\mathrm{deg}({\phi }^{\ast }L)}$ for every $x\in X$.

The defining section of ${\mathcal{O}}_X(D)$ is nowhere zero on $C$, so its restriction is trivial and has degree zero. Every translated line bundle also has degree zero on $C$, by the preceding part. For any $t\in X$, either $C\subseteq D-t$ or the defining section of $T_t^{\ast }{\mathcal{O}}_X(D)$ restricts to a nonzero section on $C$. In the second case its zero effective divisor has degree zero, so it has no zeros. Thus a translate of $D$ either contains all of $C$ or misses all of $C$.

Take $d\in D$ and $x_1,x_2\in C$. The translate $D-(d-x_1)$ contains $x_1$, hence all of $C$. In particular $d+(x_2-x_1)\in D$. This proves $D+(x_2-x_1)\subseteq D$; exchanging $x_1,x_2$ proves the reverse inclusion. Since the prime Weil divisor $D$ is reduced and translation is an automorphism, equality of these sets gives equality of Weil divisors. Remembering that pullback by $T_t$ translates the divisor by $-t$, the answer is

The line bundle translation stabilizer is the closed subgroupEach difference $x_1-x_2$ lies in $K(L)$, by the equality of Weil divisors just proved. Thus the subgroup they generate lies in $K(L)$, and closedness puts its Zariski topology closure there too. Therefore $\boxed{Y\subseteq K(L)}$. The closure of a subgroup is again a subgroup: multiplication and inversion preserve its closure by continuity. In this setting it is connected as well, since $C-C$ is connected and contains zero, and the increasing finite sums of $C-C$ are connected with a common point. Its reduced closure is an abelian subvariety.

For the curve generation criterion for an abelian variety, first suppose $Y=X$ and an irreducible Weil divisor $D$ avoids $C$. The preceding part gives $K({\mathcal{O}}_X(D))=X$, so ${\varphi }_{{\mathcal{O}}_X(D)}=0$ and ${\mathcal{O}}_X(D)\in {\mathrm{Pic}}^0(X)$. A nonzero effective divisor cannot have this property: if $H$ is an ample line bundle and $g=\dim X$, then the intersection product of Cartier divisors $D\cdot H^{g-1}$ is positive, whereas an algebraically trivial line bundle has zero intersection with every curve. The latter follows from constancy of degree along a connected family defining algebraic equivalence; the former is the positive projective degree of $D$ after replacing $H$ by a very ample power. This contradiction proves that $C$ meets every irreducible Weil divisor.

Conversely suppose $Y\ne X$. Fix $c_0\in C$, so $C\subseteq c_0+Y$. Use the permitted fiber theorem to choose a morphism of varieties $f:X\to Z$ with $f^{-1}(z_0)=Y$. This morphism is nonconstant because $Y$ is proper. Choose an affine neighborhood $U$ of $z_0$. The irreducible image $f(X)$ has positive dimension, so its intersection with $U$ does too; some regular function $h$ on $U$ is nonconstant on this intersection. Then $F=h\circ f$ is a nonconstant rational function on $X$, regular on $f^{-1}(U)$.

The pole divisor avoiding a fiber construction applies: its pole Weil divisor is nonzero. Indeed on the smooth, hence normal, projective variety $X$, a rational function with no codimension-one poles extends to a global regular function; every global regular function on a connected projective variety is constant. Every pole component $D_0$ lies outside $f^{-1}(U)$ and hence avoids $Y$. Therefore $D_0+c_0$ is an irreducible Weil divisor disjoint from $c_0+Y$, and in particular from $C$. This proves the converse using precisely the fiber fact allowed in the question, without requiring a projective target $Z$. The criterion is

Write the projective plane as $\mathbb{P}(V)$ for a three-dimensional complex vector space $V$. A projective line is the zero set of a nonzero linear functional $\ell \in V^{\ast }$. Two such functionals give the same line exactly when they differ by a nonzero scalar. Hence the parameter space of lines is the dual projective space $\mathbb{P}(V^{\ast })$. A choice of basis identifies $V^{\ast }$ with ${\mathbb{C}}^3$, giving $\boxed{({\mathbb{P}}^2{)}^{\vee }\cong {\mathbb{P}}^2}$. The description as $\mathbb{P}(V^{\ast })$ is intrinsic; the identification with the original $\mathbb{P}(V)$ requires a choice.

Represent the smooth plane conic by $v^{\mathsf{T}}Qv=0$, with $Q$ an invertible symmetric $3\times 3$ matrix. The tangent line at $[v]$ has coefficients $\ell =Qv$, because the differential is $2v^{\mathsf{T}}Q$. Thus its image in the dual projective space satisfiesConversely any nonzero $\ell$ satisfying this equation gives $v=Q^{-1}\ell$, a point of the original smooth plane conic whose tangent is $[\ell ]$. This correspondence is the restriction of an invertible projective linear transformation. The dual conic is therefore the smooth conic

To analyze the genus one tangent incidence curve of two conics, choose projective coordinates in which $C_1$ has equation $XZ-Y^2=0$. Its points and tangent lines are parametrized by $[a:b]\in {\mathbb{P}}^1$ asFor $x=[X:Y:Z]\in C_2$, the incidence equation is a nonzero homogeneous quadratic in $a,b$, so the projection $E\to C_2$ is finite of degree two. Its discriminant vanishes exactly when $Y^2-XZ=0$, that is, at $C_1\cap C_2$. There are four such points; Bézout theorem says their intersection multiplicities sum to four, so all four intersections are transverse.

Locally on $C_2$, after choosing an affine tangent-parameter chart and completing the square, the incidence equation is $w^2=u(s)$, where $s$ is a local parameter and $u$ has a simple zero at each intersection. This is smooth, with ramification index two there; away from those points the roots are distinct and the cover is étale. The analogous chart covers a root at infinity. Thus $E$ is smooth everywhere, and the cover has exactly four ramification points. It is connected: the discriminant has odd valuation at each of its four zeros, so it cannot be a square in the function field of $C_2$. The associated quadratic extension is therefore a field, not two separate sheets.

A smooth plane conic over $\mathbb{C}$ is isomorphic to the projective line. Apply the Riemann-Hurwitz formula to this connected degree-two cover:Consequently $\boxed{g(E)=1}$, and $E$ is a smooth projective genus one curve.

The printed construction is not well defined: the next tangent must pass through $x_1$, rather than $x_0$. If $x_0\ne x_1$ and ${\ell }_1\ne {\ell }_0$, a second line through $x_0$ cannot also pass through $x_1$, because the unique line through both points is ${\ell }_0$. Here is a concrete counterexample satisfying all the conic hypotheses. In the affine chart takewith ${\ell }_0:x=1$, $x_0=(1,3\sqrt{3}/2)$ and $x_1=(1,-3\sqrt{3}/2)$. The second tangent through $x_0$ is $-23x+12\sqrt{3}y=31$. It does not contain $x_1$: its left side there is $-77$. The conics meet transversely at the four complex points with $x^2=32/5$, $y^2=-27/5$. Thus this is a defect in the original PDF, not just in its conversion.

For the corrected construction, let ${\iota }_{\ell }$ exchange the two points of $C_2$ on a fixed tangent line, and let ${\iota }_x$ exchange the two tangents to $C_1$ through a fixed point of $C_2$. Both projections from $E$ are degree-two morphisms to a smooth plane conic: for the line projection this follows from intersecting a line with $C_2$, which has no line component. Since $E$ is smooth and the ground field has characteristic zero, their quadratic function-field extensions define regular involutions on the whole curve. At a ramification point “the second” point is the same point, counted with multiplicity. Each switch is an involution of a degree-two map from a genus one curve. Hence the corrected step is the everywhere-defined automorphism $F={\iota }_x\circ {\iota }_{\ell }$.

Choose an origin $O$ on the genus one curve. The Abel-Jacobi map of a genus-one curve identifies $E$ with ${\mathrm{Pic}}^0(E)$ by $p\mapsto {\mathcal{O}}_E(p-O)$. Fibers of each degree-two projection are linearly equivalent Weil divisors, since they are pullbacks of points of ${\mathbb{P}}^1$. Thus their group sums are constant: for suitable $a,b\in E$,This also holds at the ramification points, where $2p=a$ or $2p=b$. Therefore the corrected step is a translation on an elliptic curve. If $F^n(p_0)=p_0$ for one point, then $n(b-a)=0$. It follows that $F^n(p)=p$ for every $p\in E$. The corrected construction has the Poncelet porism: one periodic orbit implies all orbits are periodic, with the same least period. The literal printed construction fails before this conclusion; the proof establishes the intended, explicitly corrected assertion.