Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 126 (Abelian Varieties)

Let $X=V/\Gamma$ be a complex torus. A Riemann form on a complex torus is a Hermitian form $H$ on $V$ whose imaginary parttakes integer values on $\Gamma \times \Gamma$. Equivalently, $E$ is an integral alternating form satisfyingfor every nonzero $v$. A polarisation of a complex torus is such a positive Riemann form, or equivalently its integral cohomology class. By the Appell–Humbert theorem, it is the first Chern class of an ample holomorphic line bundle. A polarisation is principal when the homomorphism $\Gamma \to {\Gamma }^{\ast }$ induced by $E$ is an isomorphism.

The rank assumption says that ${\Gamma }^{\prime }=\Gamma \cap V^{\prime }$ is a full lattice in the real vector space underlying $V^{\prime }$. Thus $X^{\prime }=V^{\prime }/{\Gamma }^{\prime }$ is compact, and the inclusion $V^{\prime }\hookrightarrow V$ descends to an injective holomorphic homomorphism $X^{\prime }\hookrightarrow X$. It is therefore a complex subtorus.

Conversely, let $j:Y\hookrightarrow X$ be a subtorus. Its differential at the identity identifies the universal cover of $Y$ with a complex subspace $V^{\prime }\subseteq V$. Lifting $j$ to universal covers shows that the period lattice of $Y$ is precisely $V^{\prime }\cap \Gamma$. Compactness of $Y$ makes this a full lattice of rank $2{\dim }_{\mathbb{C}}{V}^{\prime }$, so every subtorus has the stated form.

Now let $H$ be a polarisation on $X$. Its restriction to $V^{\prime }$ is still positive definite, and its imaginary part remains integral on ${\Gamma }^{\prime }$, so it polarises $X^{\prime }$. Define the Hermitian orthogonal complementBecause $V^{\prime }$ is spanned over $\mathbb{R}$ by lattice vectors and $E=\mathrm{Im}H$ is integral on $\Gamma$, the real equations $E(v,{\gamma }^{\prime })=0$ for ${\gamma }^{\prime }\in {\Gamma }^{\prime }$ make $V^{\prime \prime }$ rational with respect to $\Gamma$. Hence ${\Gamma }^{\prime \prime }=V^{\prime \prime }\cap \Gamma$ is a full lattice in $V^{\prime \prime }$, and $X^{\prime \prime }=V^{\prime \prime }/{\Gamma }^{\prime \prime }$ is a subtorus. Since $V=V^{\prime }\oplus V^{\prime \prime }$, the lattice ${\Gamma }^{\prime }+{\Gamma }^{\prime \prime }$ has finite index in $\Gamma$. Consequently the addition mapis an isogeny of complex tori: it is surjective and has finite kernel. Therefore $X=X^{\prime }+X^{\prime \prime }$ and $X^{\prime }\cap X^{\prime \prime }$ is finite.

Write a lattice vector asA one-dimensional subtorus would give two $\mathbb{Z}$-independent lattice vectors $u,u^{\prime }$ spanning one complex line, so their determinant would vanish. Separating the real and imaginary parts and using the $\mathbb{Q}$-linear independence of $1,\sqrt{2},\sqrt{3}$ givesIf either $b$ or $b^{\prime }$ is nonzero, these relations make $(a,b,c,d)$ and $(a^{\prime },b^{\prime },c^{\prime },d^{\prime })$ rationally proportional, contradicting their lattice independence. Thus $b=b^{\prime }=0$. If either $d$ or $d^{\prime }$ is nonzero, the last two relevant relations again make all four coordinates proportional. Hence $d=d^{\prime }=0$, and both vectors lie in $\mathbb{C}(1,0)$. Conversely, $(1,0)$ and $(i,0)$ lie in $\Gamma$, sois a one-dimensional subtorus. It is the unique one.

If $X$ admitted a polarisation, part (ii) would give a one-dimensional complementary subtorus $X^{\prime \prime }$ with $X=X^{\prime }+X^{\prime \prime }$. Uniqueness would force $X^{\prime \prime }=X^{\prime }$, making $X^{\prime }+X^{\prime \prime }=X^{\prime }\ne X$, a contradiction. Thus $X$ is a nonprojective complex torus and has no polarisation.

A group scheme over $k$ is a $k$-scheme $G$ equipped with multiplication $m:G{\times }_{k}G\to G$, identity $e:\mathrm{Spec}k\to G$, and inversion $j:G\to G$ satisfying the associativity, identity, and inverse diagrams.

ConsiderThe diagonal morphism is ${\Delta }_{G/k}=q^{-1}(e)$. For a finite-type $k$-scheme the rational identity point is closed, so its inverse image is closed. Thus the diagonal is a closed immersion and every such group scheme over a field is a separated scheme.

In characteristic $p>0$, the infinitesimal additive groupis a nonreduced group scheme. Its comultiplication is $t\mapsto t\otimes 1+1\otimes t$, which is well defined because $(t\otimes 1+1\otimes t{)}^p=t^p\otimes 1+1\otimes t^p$.

The Mumford rigidity lemma says that if $A$ is complete, $B$ is connected, and a morphism $F:A\times B\to Z$ maps the fiber over some $b_0\in B$ to a point, then $F$ is constant on every $A$-fiber and factors through $B$.

Put $y=f(e_X)$ and normalizeso $g(e_X)=e_G$. DefineWhen the first coordinate is $e_X$, this is constantly $e_G$. Apply rigidity with the second copy of the complete variety $X$ as the complete factor. It follows that $F(x,z)$ is independent of $z$, and evaluation at $z=e_X$ gives $F(x,z)=g(x)$. Thereforeso $g$ is a homomorphism of group varieties and $f=T_y\circ g$.

Completeness is essential. Take $X=G={\mathbb{G}}_a$. In characteristic different from two, the morphism $f(t)=t^2$ satisfies $f(0)=0$ but is not additive. If it were $T_y\circ g$ with $g$ a group homomorphism, evaluation at zero would give $y=0$ and hence $f=g$, a contradiction. In characteristic two the same argument works with $f(t)=t^3$.

For an abelian variety $X$, consider the commutator morphismIt is the identity whenever either coordinate is the identity. The Mumford rigidity lemma applied successively to the two complete connected factors makes $c$ constant everywhere; its value at $(e,e)$ is $e$. Therefore every pair of points commutes, so the group law is commutative.

Because $p:X_1\times X_2\to X$ is an isomorphism and $X$ is connected, each $X_i$ is connected. They are complete because they are closed in the complete variety $X$. Letbe the two component morphisms.

For $x,x^{\prime }\in X_1$, the morphismfrom $X_1\times X_1$ to $X_2$ is constantly $e$ on either coordinate axis. Rigidity therefore makes it constantly $e$, so $X_1$ is closed under addition. The same argument applies to $X_2$. If $-x=u+v$ is the unique decomposition with $u\in X_1$ and $v\in X_2$, then $e=(x+u)+v$; uniqueness of the decomposition of $e$ gives $v=e$ and $u=-x$. Thus each $X_i$ is also closed under inversion.

The restrictions of the multiplication and inversion morphisms of $X$ now make each $X_i$ a complete connected group variety, hence an abelian variety. Since the group law on $X$ is commutative,Thus $p$ is a homomorphism. It is already an isomorphism of varieties, and its inverse consequently respects the group operations as well, so $p$ is an isomorphism of group schemes.

Let $m_I:X^3\to X$ add the coordinates indexed by a nonempty subset $I$ of the three-element index set. The Theorem of the Cube says that for every line bundle $\mathcal{L}$ on an abelian variety,is trivial, up to the harmless constant line given by the fiber of $\mathcal{L}$ at the identity.

Pull this line bundle back along $(f,g,h):Y\to X^3$. Pullback commutes with tensor products and duals, and $m_I\circ (f,g,h)$ is the corresponding sum of morphisms. The resulting bundle is precisely ${\mathcal{M}}_{f,g,h}$, so it is trivial.

Take $Y=X$, $f={\mathrm{id}}_X$, and let $g,h$ be the constant maps with values $x,y$. All pullbacks along constant maps are trivial line bundles. The formula for ${\mathcal{M}}_{f,g,h}$ then becomesor equivalently

For a line bundle $\mathcal{L}$ on $X$, define the homomorphism associated to a line bundle on an abelian varietyThe translation identity from part (i) givesso ${\varphi }_{\mathcal{L}}$ is a homomorphism.

Suppose $\mathcal{M}\simeq T_y^{\ast }\mathcal{L}\otimes {\mathcal{L}}^{\vee }$ lies in its image. Thenwhich is trivial by the same translation identity. Hence ${\varphi }_{\mathcal{M}}=0$; the image of every ${\varphi }_{\mathcal{L}}$ lies in the Identity component of the Picard group ${\mathrm{Pic}}^{0}X$.

Directly from the definition,AlsoThe bundle in parentheses lies in ${\mathrm{Pic}}^{0}X$ by part (ii), so it is translation invariant. This proves ${\varphi }_{T_x^{\ast }\mathcal{L}}={\varphi }_{\mathcal{L}}$.

Finally, $[n]\circ T_x=T_{nx}\circ [n]$, and thereforeFor $\mathcal{N}\in {\mathrm{Pic}}^{0}X$, the multiplication pullback formula reduces to $[n{]}^{\ast }\mathcal{N}\simeq {\mathcal{N}}^{\otimes n}$. Consequentlyas required.