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 $A \rightarrow S$
- is proper,
- is smooth, and
- 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.
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.
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.
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