Abelian varieties: 2. Abelian schemes

This posting follows Matt Stevenson's notes for Math 731 (Fall 2017) at the University of Michigan. We also follow Mumford's book.

Abelian schemes. Given a Noetherian scheme $S,$ an abelian scheme $A$ over $S$ is a group scheme $A$ over $S$ such that the structure map $A \rightarrow S$

1. is proper,
2. is smooth, and
3. has geometrically connected fibers.

In particular, when $S = \mathrm{Spec}(k)$ for a field $k,$ an abelian scheme over $k$ (i.e., $\mathrm{Spec}(k)$) is a proper and geometrically connected variety over $k,$ which we call an abelian variety.

Remark 1. Given an abelian scheme over $S$ and a scheme map $T \rightarrow S,$ the base change $A \times_{S} T \rightarrow T$ is proper and smooth, as these properties are preserved under any base change. For any point $x \in T,$ we have $$A \times_{S} T \times_{T} \mathrm{Spec}(\kappa(x)) \simeq A \times_{S} \mathrm{Spec}(\kappa(x)),$$ which is geometrically connected because $A$ is an abelian $S$-scheme. Hence, the base change $A \times_{S} T \rightarrow T$ is an abelian $T$-scheme.

Personal Remark. Property about smooth maps can be reviewed in Chapter 2 of BLR.

Remark 2. An abelian scheme $A$ over $S$ can be seen as a family of abelian varieties. Indeed, if we consider the structure map $A \rightarrow S$ and a point $s \in S,$ the base change $A \times_{S} \mathrm{Spec}(\kappa(s))$ is an abelian variety.

Remark 3. Any abelian variety (over a field) is geometrically integral. More generally, any scheme that is smooth and geometrically connected over a field is geometrically integral. This is because smoothness (or more generally, regularity) ensures that every stalk is a domain. Hence, this comes down to the fact that if there is a point $x$ in the intersection of two irreducible components of a Noetherian scheme $X,$ the stalk $\mathscr{O}_{X,x}$ is not an integral domain (5.3.C, Vakil); this is because a domain cannot have two minimal primes (as $(0)$ is its unique minimal prime).

Analytic example. Let $A$ be an abelian variety over $\mathbb{C}.$ Then by GAGA, we see $A(\mathbb{C})$ is complex manifold that is compact and connected. We shall sketch how to prove that $A(\mathbb{C})$ is an abelian group using analytic technique. For any $x \in A(\mathbb{C}),$ denote by $c_{x} : A(\mathbb{C}) \rightarrow A(\mathbb{C})$ the holomorphic map given by the conjugation: $y \mapsto xyx^{-1}.$ To show that $A(\mathbb{C})$ is an abelian group, we may show that $c_{x} = \mathrm{id}_{A(\mathbb{C})}.$ Consider $$A(\mathbb{C}) \rightarrow \mathrm{Aut}_{\mathrm{LieGrp}}(A(\mathbb{C})) \rightarrow \mathrm{Aut}_{\mathbb{C}}(T_{e}A(\mathbb{C})) \simeq \mathrm{GL}_{g}(\mathbb{C}),$$ given by $x \mapsto c_{x} \mapsto dc_{x},$ where $e \in A(\mathbb{C})$ is the identity with respect to the group structure. One may check that this is a holomorphic map and since $A(\mathbb{C})$ is a compact complex manifold, the map must be constant. This implies that $$(dc_{x})_{e} = (dc_{e})_{e} = \mathrm{id}_{T_{e}A(\mathbb{C})}.$$ Now, the magic happens when we use the exponential map $$\mathrm{exp} : T_{e}A(\mathbb{C}) \rightarrow A(\mathbb{C}),$$ which will allow us to see that $c_{x} = \mathrm{id}_{A(\mathbb{C})},$ showing that $A(\mathbb{C})$ is an abelian group. It follows that $A$ is a commutative group scheme over $\mathbb{C}.$

Exponential map. Given a complex Lie group $G,$ write $\mathfrak{g} := T_{e}G,$ where $e$ is the identity of $G.$ It is known (e.g., p.79 of Hochschild) that for each $v \in \mathfrak{g},$ there is a unique holomorphic group map $\phi_{v} : \mathbb{C} \rightarrow G$ such that $d\phi_{v} : \mathbb{C} \rightarrow \mathfrak{g}$ is given by $1 \mapsto v.$ We then define the exponential map $$\mathrm{exp}_{G} = \mathrm{exp} : \mathfrak{g} \rightarrow G$$ of $G$ by $v \mapsto \phi_{v}(1).$ The map $\mathrm{exp}$ turns out to be holomorphic (e.g., p.80 of Hochschild).

Given any $s \in \mathbb{C},$ we have $$d\phi_{v}(s t) = stv = tsv = d \phi_{sv}(t),$$ so $\phi_{v}(t) = \phi_{tv}(1) = \mathrm{exp}(tv).$ In particular, we have $$\mathrm{exp}(0) = \phi_{0}(1) = e,$$ because the trivial group map $\mathfrak{g} \rightarrow G$ mapping everything to $e$ has the zero derivation. Denoting by $f_{v} : \mathbb{C} \rightarrow \mathfrak{g}$ the map given by $t \mapsto tv$ so that $\phi_{v} = \mathrm{exp} \circ f_{v}.$ We have $$(d\phi_{v})_{0} = (d\mathrm{exp})_{0}(df_{v})_{0}$$ by the Chain Rule, so $$\begin{align*}(d\mathrm{exp})_{0}(v) &= (d\mathrm{exp})_{0}(df_{v})_{0}(1) \\ &= (d\phi_{v})_{0}(1)  \\ &= v,\end{align*}$$ showing that $d\mathrm{exp} = \mathrm{id}_{\mathfrak{g}}.$ By the Implicit Function Theorem, this implies that $\mathrm{exp}$ restricts to a homeomorphism from an open neighborhood of $0 \in \mathfrak{g}$ to an open neighborhood of $e \in G.$ This implies that if $G$ is connected, then $\mathrm{exp}(\mathfrak{g})$ generates the whole $G$ using the fact that in any connected Lie group, a nonempty open subset generates the whole group.

Let $T : G \rightarrow H$ be any map of complex Lie groups and denote by $\mathfrak{g}$ and $\mathfrak{h}$ the tangent spaces of $G$ and $H,$ respectively. We have $$T \circ \mathrm{exp}_{G} = \mathrm{exp}_{H} \circ (dT)_{e},$$ where the first map is given by the composition $\mathfrak{g} \rightarrow G \rightarrow H,$ while the second is $\mathfrak{g} \rightarrow \mathfrak{h} \rightarrow H.$ To check this, fix $v \in \mathfrak{g}$ and let $\phi_{v} : \mathbb{C} \rightarrow G$ the unique holomorphic group map satisfying $(d\phi_{v})_{0}(1) = v$ so that $\mathrm{exp}_{G}(v) = \phi_{v}(1).$ Then $T \circ \phi_{v} : \mathbb{C} \rightarrow G \rightarrow H$ is a holomorphic group map such that $$d(T \circ \phi_{v})_{0} = (dT)_{e} \circ (d\phi_{v})_{0} : \mathbb{C} \rightarrow \mathfrak{g} \rightarrow \mathfrak{h}$$ is given by $1 \mapsto v \mapsto (dT)_{e}(v).$ Hence, we must have $$\mathrm{exp}_{H}((dT)_{e}(v)) = T(\phi_{v}(1)) = T(\mathrm{exp}_{G}(v)),$$ as desired.

We want to consider the above identity when $T = c_{g} : G \rightarrow G$ the conjugation by an element $g \in G.$ If $G$ is compact (complex manifold), then $(dc_{x})_{e} = (dc_{e})_{e} = \mathrm{id}_{\mathfrak{g}},$ so the above identity buys us $$c_{g} \circ \mathrm{exp} = \mathrm{exp}.$$ In other words, for any $v \in \mathfrak{g},$ we see that $g$ commutes with $\mathrm{exp}(v)$ in $G,$ so this shows that $\mathrm{exp}(v)$ is in the center of $G$ if $G$ is compact. Thus, we have shown (modulo some facts about exponential maps) the following:

Theorem. Let $G$ be any complex Lie group. If $G$ is connected and compact, then it is an abelian group.

Personal Remark. This document seems to discuss the same fact. The author of the document says that this fact was an exercise in Chapter 8 of Fulton and Harris.

Torus description of complex abelian varieties. Let $A$ be an abelain variety of dimension $g$ over $\mathbb{C}.$ Writing $T_{e}A(\mathbb{C}) = \mathbb{C}^{g},$ we have noted that $\mathrm{exp} : \mathbb{C}^{g} \rightarrow A(\mathbb{C})$ is a surjective holomorphic group map. We also have noted that the exponential map restricts to a homeomorphism near $0,$ so let $U \ni 0$ be an open neighborhood in $\mathbb{C}^{g}$ such that $\mathrm{exp}$ gives $U \simeq \mathrm{exp}(U) =: V \subset A(\mathbb{C}).$ We have $e \in V$ and no other elements to $0$ in $U$ map to $e.$ Thus, for any nonero $v \in \ker(\mathrm{exp}),$ we have $U \cap (v + U) = \emptyset.$ This implies that $\ker(\mathrm{exp})$ is a discrete subgroup of $\mathbb{C}^{g}$ and that $\ker(\mathrm{exp})$ acts on $\mathbb{C}^{g}$ properly discontinuously. The induced group isomorphism $$\mathbb{C}^{g}/\ker(\mathrm{exp}) \simeq A(\mathbb{C})$$ is a holomorphic map, and since the differential at the identity of this map is an isomorphism, by the Inverse Function Theorem, its inverse is holomorphic at $e \in A(\mathbb{C})$ and hence everywhere holomorphic by applying translation at a point we test the holomorphy. As $\ker(\mathrm{exp})$ is a discrete subgroup of $\mathbb{C}^{g} = \mathbb{R}^{2g},$ it is generated by an $\mathbb{R}$-linearly independent subset of $\mathbb{R}^{2g}$ with size $\leq 2g.$ Since the quotient is compact (as $A(\mathbb{C})$ is compact), this non-strict inequality must be an equality, or in other words, we see that $\ker(\mathrm{exp})$ is a lattice in $\mathbb{C}^{g} = \mathbb{R}^{2g}.$ Thus, we have $$A(\mathbb{C}) \simeq \mathbb{C}^{g}/\ker(\mathrm{exp}) = \mathbb{R}^{2g}/\ker(\mathrm{exp}) \simeq (\mathbb{R}/\mathbb{Z})^{2g} = (S^{1})^{2g}.$$ In particular, denoting $\Lambda := \ker(\mathrm{exp}),$ the $n$-torsion subgroup of $A(\mathbb{C})$ can be computed as as $$A(\mathbb{C})[n] = (1/n)\Lambda / \Lambda \simeq (\mathbb{Z}/(n))^{2g}.$$ Note that when $g = \dim(A) = 1,$ we now know that $A(\mathbb{C})$ is (analytic) topologically a torus $S^{1} \times S^{1}$ and $A(\mathbb{C})[n] \simeq (\mathbb{Z}/(n))^{2}.$ An $1$-dimensional abelian variety is called an elliptic curve.

Remark. One can find a lattice $\Gamma \subset \mathbb{C}^{g}$ such that $\mathbb{C}^{g}/\Gamma$ is not an analytification of a complex variety. There is a classification of Riemann on which lattices give a quotient coming from a complex variety.

Personal Remark. I may find and add the references to the remark above.

Abelian varieties: 1. Group schemes

This posting follows Matt Stevenson's notes for Math 731 (Fall 2017) at the University of Michigan. Various references available in the notes will not be specified here.

Group schemes. Let $S$ be a scheme. A group scheme over $S$ (also called an $S$-group) is a group object in $\textbf{Sch}_{S},$ the category of $S$-schemes. Concretely, this means an $S$-scheme $G$ such that the functor $$h_{G} = \mathrm{Hom}_{S}(-, G) : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}$$ factors through the forgetful functor $\textbf{Grp} \rightarrow \textbf{Set}.$ We say a group scheme $G$ over $S$ is commutative if $G(T) := h_{G}(T)$ is an abelian group for each $S$-scheme $T.$

We now give examples. We will focus on the affine case $S = \mathrm{Spec}(R)$, although it should be possible to use general $S$ by gluing affine pieces.

Example 1. Consider $$\mathbb{G}_{a} = \mathbb{G}_{a, R} := \mathrm{Spec}(R[t]).$$ For any $R$-scheme $T,$ we have $$\mathbb{G}_{a}(T) = \mathrm{Hom}_{\mathrm{Spec}(R)}(T, \mathbb{G}_{a}) \simeq \Gamma(T, \mathscr{O}_{T}),$$ where the isomorphism is taken in $\textbf{Set},$ given by $$\phi \mapsto \phi^{*}(t).$$ This map is functorial in $T,$ varying in $\textbf{Sch}_{R},$ and the right-hand side gives the functor into $\textbf{Grp},$ where the group structure is given by the addition. We see that $\mathbb{G}_{a}$ is a (commutative) group scheme.

Example 2. Consider $$\mathbb{G}_{m} = \mathbb{G}_{m, R} := \mathrm{Spec}(R[t, t^{-1}]).$$ For any $R$-scheme $T,$ we have $$\mathbb{G}_{m}(T) = \mathrm{Hom}_{\mathrm{Spec}(R)}(T, \mathbb{G}_{m}) \simeq \Gamma(T, \mathscr{O}_{T})^{\times},$$ where the isomorphism is taken in $\textbf{Set},$ given by $$\phi \mapsto \phi^{*}(t).$$ This map is functorial in $T,$ varying in $\textbf{Sch}_{R},$ and the right-hand side gives the functor into $\textbf{Grp},$ where the group structure is given by the multiplication. We see that $\mathbb{G}_{m}$ is a (commutative) group scheme.

Example 3. We define $$\mathrm{GL}_{n} = \mathrm{GL}_{n, R} := \mathrm{Spec}(R[t_{ij}]_{1 \leq i,j \leq n}[1/\mathrm{det}]).$$ For any $R$-scheme $T,$ we have $$\mathrm{GL}_{n}(T) = \mathrm{Hom}_{\mathrm{Spec}(R)}(T, \mathrm{Spec}(R[t_{ij}]_{1 \leq i,j \leq n}[1/\mathrm{det}])) \simeq \mathrm{GL}_{n}(\mathscr{O}_{T}(T)),$$ the general linear group of $n \times n$ matrices over the global sections of $T,$ given by $\phi \mapsto [\phi^{*}(t_{ij})]_{1 \leq i,j \leq n}.$ This is functorial in $T,$ and the right hand-side gives the functor into $\textbf{Grp}$ given by the matrix multiplication. This shows that $\mathrm{GL}_{n}$ is a group scheme.

Example 4. We define $$\mu_{n} = \mu_{n, R} := \mathrm{Spec}(R[t]/(t^{n}-1)).$$ For any $R$-scheme $T,$ we have $$\mu_{n}(T) = \mathrm{Hom}_{\mathrm{Spec}(R)}(T, \mathrm{Spec}(R[t]/(t^{n}-1))) \simeq \{f \in \mathscr{O}_{T}(T)^{\times} : f^{n} = 1\},$$ given by $\phi \mapsto \phi^{*}(\bar{t}),$ which is functorial in $T,$ and the right hand-side gives the functor into $\textbf{Grp}$ given by the matrix multiplication. This shows that $\mu_{n}$ is a (commutative) group scheme. The group structures are coming from multiplications of $\mathscr{O}_{T}(T)^{\times}.$

Remark. The behavior of $\mu_{n, R}$ depends on the characteristic of $R.$

First, consider $R = \mathbb{C}.$ Then the equation $t^{n} = 1$ has $n$ distinct solutions in $\mathbb{C},$ so $$\mu_{n} = \mathrm{Spec}\left(\frac{\mathbb{C}[t]}{(t^{n} - 1)}\right) \simeq \bigsqcup_{z \in \mu_{n}(\mathbb{C})}\mathrm{Spec}(\mathbb{C}) \simeq \mu_{n}(\mathbb{C}),$$ which is in particular smooth over $\mathbb{C}.$ It turns out that all group schemes in characteristic $0$ are smooth, by a theorem due to Oort. (We will see this later.)

On the other hand, when $R = \mathbb{F}_{p}$ for a prime $p,$ then $$\mu_{p} = \mathrm{Spec}(\mathbb{F}_{p}[t]/(t^{p} - 1)) = \mathrm{Spec}(\mathbb{F}_{p}[t]/(t - 1)^{p}),$$ so $\mu_{n}$ is a non-reduced scheme. Moreover, we note that $\mu_{p}$ is not smooth over $\mathbb{F}_{p}$ as the derivative of $(t - 1)^{p}$ in $t$ vanishes. In this case, note that $\mu_{p} = \mu_{p, \mathbb{F}_{p}}$ is a point. Note that for any $\mathbb{F}_{p}$-algebra $A$ that is a domain, there is only one $\mathbb{F}_{p}$-algebra map $\mathbb{F}_{p}[t]/(t-1)^{p} \rightarrow A$ because the only $a \in A$ such that $(a - 1)^{p} = 0$ must be $1$ as $A$ is a domain. Hence, we have $\mu_{p, \mathbb{F}_{p}}(A)$ is a singleton. However, when $A$ is not a domain, this may not be the case. For instance, take $A = \mathbb{F}_{p}[\epsilon]/(\epsilon^{2}).$ We have $$\begin{align*} \mu_{p,\mathbb{F}_{p}}\left( \frac{\mathbb{F}_{p}[\epsilon]}{(\epsilon^{2})} \right) &= \mathrm{Hom}_{\mathrm{Spec}(\mathbb{F}_{p})}\left( \mathrm{Spec}\left( \frac{\mathbb{F}_{p}[\epsilon]}{(\epsilon^{2})} \right), \mu_{p,\mathbb{F}_{p}} \right) \\ &\simeq \{1 + b\bar{\epsilon} \in \mathbb{F}_{p}[\epsilon]/(\epsilon^{2}) : b \in \mathbb{F}_{p}\}\end{align*},$$ given by $\phi \mapsto \phi^{*}(\bar{t}).$ The last set has $p$ elements and $p > 1.$

More conventions/definitions. For convenience, we shall call a group scheme over a base scheme $S$ an $S$-group. Again, an $S$-group is by definition a certain functor $\textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Grp},$ and we define an $S$-group map to be a natural transformation between two $S$-groups. With this definition, we see that the $S$-groups form a category. Note that given an $S$-group map $f : G \rightarrow H$ and an $S$-group $T,$ the corresponding map $f_{T} : G(T) \rightarrow H(T)$ is a group homomorphism.