Moduli of curves · Qual Corpus

The coarse moduli space in genus $0$ is a point, while in genus $1$ the $j$-invariant identifies the coarse moduli space with $\AA^1$. For $g\ge2$, the coarse moduli space $M_g$ is an irreducible quasi-projective variety of dimension $3g-3$.

Coarse versus fine moduli spaces (Tag D-CRVMOD)

Let $k$ be an algebraically closed field. A variety $M_g$ over $k$ is a coarse moduli space for curves of genus $g$ when:

the closed points of $M_g$ are in bijection with isomorphism classes of smooth projective curves of genus $g$ over $k$;
every flat family $\mathcal{X} \to T$ whose fibres are such curves induces a morphism $h\colon T \to M_g$ with $h(t)$ the class of the fibre $\mathcal{X}_t$, compatibly with pullback of families along morphisms $T'\to T$; and
for every scheme $N$ with an assignment of morphisms $T\to N$ to families as in (2), compatible with pullback, there is a unique morphism $M_g\to N$ through which that assignment factors.

(Fine moduli space)

$M_g$ is a fine moduli space when it represents the functor sending $T$ to the set of flat families over $T$ up to isomorphism: the classifying maps of (2) give a bijection \[ \operatorname{Hom}(T, M_g) \longleftrightarrow \ts{\text{families over } T}/{\cong} \] natural in $T$. Equivalently $M_g$ carries a universal family $\mathcal{U} \to M_g$ such that every family $\mathcal{X}\to T$ is isomorphic to the pullback of $\mathcal{U}$ along a unique morphism $T\to M_g$.

(Automorphisms and fine moduli)

Let $C$ be a smooth projective curve of genus $g$ over $k$, and let $T$ be a connected scheme over $k$. Families over $T$ that become isomorphic to $C\times T$ étale-locally on $T$ are classified up to isomorphism by $H^1_{\mathrm{et}}(T,\Aut C)$. The classifying map of each such family is the constant map to the point $[C]$. Hence, if some such family is not isomorphic to $C\times T$, then two non-isomorphic families have the same classifying map, and no fine moduli space exists.

(A quadratic twist)

Let $\operatorname{char}k\ne2,3$, and let $E\colon v^2=u^3+au+b$ be an elliptic curve with origin $p_0$ at infinity. Over $T = \AA^1 \smz$ with coordinate $t$, put \[ \mathcal{E}_t : y^2 = x^3 + a t^2 x + b t^3 . \] Over a square root $s$ of $t$, the substitution $x = s^2 u$, $y = s^3 v$ identifies $\mathcal E_t$ with $E$, so every fibre is isomorphic to $E$ and the classifying map is constant. The family is the twist of $E\times T$ by the $\mu_2$-torsor $s\mapsto s^2=t$ over $\GG_m$, acting through $-1\in\Aut(E,p_0)$. That torsor is nontrivial, since $t$ has no square root in $k[t,t^{-1}]$, so $\mathcal E\to T$ is not isomorphic to $E\times T$.

Every genus $g \geq 2$ has hyperelliptic curves, and each carries the hyperelliptic involution $\iota$. For $\operatorname{char}k\ne2$, the twist of $C\times\GG_m$ by the squaring $\mu_2$-torsor acting through $\iota$ is not isomorphic to $C\times\GG_m$. So there is no fine moduli space of curves of genus $g\ge2$, although a general curve of genus $g \geq 3$ has trivial automorphism group.

The moduli stack $\mathcal{M}_g$ (D-ALGSTACK, T-MGSMOOTH) has as objects over $T$ the families over $T$ and as morphisms the isomorphisms of families; the automorphism group of the point $[C]$ of $\mathcal M_g$ is $\Aut C$, and $M_g$ is the coarse moduli space of $\mathcal{M}_g$.

What conditions define a coarse moduli space?

What does a fine moduli space have that a coarse one does not?

What obstructs the existence of a fine moduli space of curves?

Exhibit a nonconstant family all of whose fibres are isomorphic.

A fine moduli space represents the functor of families of smooth curves of genus $g$; a coarse moduli space receives a natural transformation from that functor that is universal among maps to schemes and bijective on geometric points. Let $C$ have an automorphism $\sigma$ of order $2$, with $\operatorname{char}k\neq2$. The quotient of $C\times\mathbb G_m$ by $(x,t)\mapsto(\sigma x,-t)$, mapped to $\mathbb G_m$ by $t\mapsto t^2$, is a family whose fibres are all isomorphic to $C$ but which is not isomorphic to $C\times\mathbb G_m$; a fine moduli space would classify it by a constant map and force it to be trivial.

For genus $1$, elliptic curves gives the coarse moduli space $\AA^1$ with coordinate $j$ and the nontrivial family $ty^2=x^3-x$ with constant fibres.

The dimension count

$\mathcal{M}_g$ is irreducible of dimension $3g-3$ (Tag T-CRVMG)

For $g\ge2$, curve deformations are quadratic differentials: $\dim H^1(T_C)=h^0(\omega_C^{\tensor2})=3g-3$, the dimension of irreducible $\mathcal M_g$.Suggest a slogan

Fix $g \geq 2$ and $k = \kbar$. The coarse moduli space $\mathcal{M}_g$ of smooth projective curves of genus $g$ over $k$ is irreducible and quasiprojective of dimension $3g-3$.

(Small genus)

$\mathcal{M}_0$ is a point, $\mathcal{M}_1 \cong \AA^1$ by the $j$-invariant, $\dim \mathcal{M}_2 = 3$, and $\dim \mathcal{M}_3 = 6$.

The dimension $3g-3$ is reflected infinitesimally: first-order deformations of $C$ are $H^1(C,T_C)$, and by Serre duality this is $\dualof{H^0(C,\omega_C^{\otimes 2})}$. Since $\deg 2K = 4g-4 > 2g-2$ for $g \geq 2$, the divisor $2K$ is nonspecial and Riemann–Roch gives $h^0(\omega^{\otimes 2}) = (4g-4) + 1 - g = 3g-3$ outright. The hypothesis $g \geq 2$ is doing work in that line, which is why $\mathcal{M}_1$ has to be quoted separately: for $g = 1$ the same count returns $0$, while the actual dimension is $1$, because $T_C \cong \OO_C$ makes $h^1(T_C) = 1$ instead.

Two concrete dimension counts recover the formula.

Genus $2$, via the hyperelliptic locus. A hyperelliptic curve of genus $g$ is a double cover of $\PP^1$ branched at $2g+2$ points, and the curve determines that branch set, so the moduli are the configurations of $2g+2$ unordered points of $\PP^1$ modulo $\PGL_2$. Normalizing three of them to $0, 1, \infty$ uses up the $3$-dimensional group and leaves $2g+2-3 = 2g-1$ parameters, so the hyperelliptic locus is irreducible of dimension $2g-1$. Every curve of genus $2$ is hyperelliptic, so at $g=2$ that locus is all of $\mathcal{M}_2$ and $\dim \mathcal{M}_2 = 2(2)-1 = 3$, which agrees with $3g-3 = 3$. This is also the last genus where the two numbers agree: $2g-1 < 3g-3$ exactly when $g > 2$, so for $g \geq 3$ the hyperelliptic curves are a proper closed subvariety and the general curve is not hyperelliptic.

Genus $3$, via plane quartics. A non-hyperelliptic curve of genus $3$ is canonically embedded as a smooth plane quartic, and every smooth plane quartic arises this way. Quartic forms in three variables have $\binom{4+2}{2} = 15$ coefficients, so they are parameterized by $\PP^{14}$, with the smooth ones an open $U$. Two quartics give isomorphic curves exactly when they differ by $\PGL_3$, which has dimension $8$, and the stabilizers are finite because $\Aut C$ is finite for $g \geq 2$; so the orbits are $8$-dimensional and the image of $U$ in $\mathcal{M}_3$ has dimension $14 - 8 = 6 = 3g-3$.

The genus-$3$ curves left out of that count are exactly the hyperelliptic ones, and the count survives them: by the first computation they form a locus of dimension $2(3)-1 = 5$, strictly less than $6$. So they lie in a proper closed subset, the plane quartics are a dense open subset of an irreducible $\mathcal{M}_3$, and the dimension computed on that open set is the dimension of the whole space. The two counts use different coordinate-change groups: $\PGL_2$ acts on the branch points in the hyperelliptic count, while $\PGL_3$ acts on the ambient $\PP^2$ in the quartic count.

What is the dimension of the moduli space of curves of genus $g$?

Why is $\dim \mathcal{M}_2 = 3$?

Why is $\dim \mathcal{M}_3 = 6$?

What is $\mathcal{M}_1$, and why does the formula $3g-3$ not give it?

Deformation theory gives $h^1(C,T_C)=h^0(C,\omega_C^{\otimes2})=3g-3$. In genus $2$, six branch points on $\PP^1$ modulo $\PGL_2$ give $6-3=3$. In genus $3$, smooth plane quartics, an open subset of $\PP^{14}$, modulo $\PGL_3$ give $14-8=6$.

Automorphism groups

Automorphisms of a curve of genus $g \geq 2$ (Tag PR-CRVAUT)

Curves of genus at least $2$ have finite automorphism groups; in characteristic zero, Hurwitz bounds their order by $84(g-1)$.Suggest a slogan

Let $C$ be a smooth projective curve of genus $g \geq 2$ over $k = \kbar$. Then $\Aut C$ is finite, and in characteristic zero \[ \size \Aut C \leq 84(g-1) . \] If $C$ is non-hyperelliptic, every automorphism of $C$ is induced by a linear automorphism of the ambient $\PP^{g-1}$ of the canonical embedding; for $g = 3$ this says $\Aut C$ is the subgroup of $\PGL_3$ preserving the plane quartic. The general curve of genus $g \geq 3$ has $\Aut C = 1$.

Finiteness follows from the vanishing of infinitesimal automorphisms together with a projective realization of the automorphism group. The reason: $T_C \cong \dualof{\omega_C}$ has degree $2-2g < 0$, so $h^0(T_C) = 0$ and there are no infinitesimal automorphisms; $\Aut C$ is an algebraic group of dimension $0$. The argument that makes it a finite group rather than merely $0$-dimensional runs through the canonical embedding: $\omega_C$ is intrinsic, so any automorphism carries $\abs{K}$ to itself and acts on $\PP^{g-1} = \PP(\dualof{H^0(\omega_C)})$. For non-hyperelliptic $C$ the canonical map is an embedding, so $\Aut C$ is exactly the closed subgroup of $\PGL_g$ stabilizing the canonical curve, hence an algebraic group of dimension $0$ in a variety, hence finite. For hyperelliptic $C$ run the same argument with $\abs{3K}$, which is very ample. The genus-$3$ statement is this with $g-1 = 2$: automorphisms are restrictions of automorphisms of $\PP^2$, so a plane quartic has no automorphisms beyond the linear ones visible in its equation.

The bound comes from Riemann–Hurwitz applied to the quotient $C \to C/G$ with $G = \Aut C$ of order $n$. Writing $g'$ for the genus of the quotient and $e_1,\ldots,e_r$ for the ramification indices of the branch points, \[ 2g-2 = n\left( 2g'-2 + \sum_{i=1}^r \left(1 - \frac{1}{e_i}\right) \right) , \] so $n$ is largest when the bracket is smallest, and the smallest positive value the bracket can take is $\frac{1}{42}$, attained at $g' = 0$ with three branch points of orders $(2,3,7)$. That gives $n \leq 42(2g-2) = 84(g-1)$. The bound is sharp: the Klein quartic $x^3 y + y^3 z + z^3 x = 0$ has $g = 3$ and $\Aut = \PSL_2(\FF_7)$ of order $168 = 84 \cdot 2$, and it is a plane quartic, so its automorphisms are linear as the previous paragraph requires. The characteristic hypothesis is not decorative: the count of $1 - 1/e_i$ is the tame formula, and in characteristic $p$ wild ramification breaks the bound.

The two ends of this card are what the moduli statements need. Finiteness of $\Aut C$ is why the $\PGL_3$ orbits in the plane-quartic count are $8$-dimensional, which is what makes $\dim \mathcal{M}_3 = 14-8$ come out right. Nontriviality of $\Aut C$ for some curve is the obstruction to a fine moduli space, and no genus escapes it: the general curve of genus $g \geq 3$ has no automorphisms, but every hyperelliptic curve carries its involution and the hyperelliptic locus is nonempty in every genus. At $g = 2$ there is no general case to appeal to at all, since every genus-$2$ curve is hyperelliptic.

Is the automorphism group of a curve of genus $g \geq 2$ finite, and why?

State the sharp bound on $\size \Aut C$ and the curve that attains it.

Where do the automorphisms of a genus-$3$ curve come from?

Does the general curve of genus $g \geq 3$ have a nontrivial automorphism?

What is Hurwitz's theorem?

For $g\ge2$, $\Aut C$ is finite, so the $\PGL_3$-orbits of smooth plane quartics have dimension $8$. Every hyperelliptic curve has the hyperelliptic involution, so for every $g\ge2$ some curve of genus $g$ has a nontrivial automorphism, and $M_g$ is not a fine moduli space.

Smoothness and properness of the moduli stacks

The moduli stacks of smooth and stable curves (Tag T-MGSMOOTH)

$\mathcal M_g$ is smooth of dimension $3g-3$ but not proper; adding stable curves gives the smooth proper compactification $\overline{\mathcal M}_g$.Suggest a slogan

Let $g \geq 2$.

The moduli stack $\mathcal{M}_g$ of smooth projective curves of genus $g$ is a smooth Deligne–Mumford stack over $\Spec \ZZ$ of relative dimension $3g - 3$, with irreducible geometric fibres.
$\mathcal{M}_g$ is not proper.
It is an open dense substack of the moduli stack $\overline{\mathcal{M}}_g$ of stable curves, which is smooth and proper over $\Spec \ZZ$.

The coarse moduli space $M_g$ is irreducible of dimension $3g-3$ but is singular for $g \geq 4$, at curves with extra automorphisms; smoothness is a property of the stack.

Why is the moduli space of curves $\mathcal{M}_g$ smooth? Is it proper?

Stable maps

Stable maps (Tag D-STABLEMAP)

Let $X$ be a projective variety over $\CC$ and $\beta \in H_2(X, \ZZ)$. A stable map of genus $g$ with $n$ marked points and class $\beta$ is a tuple $(C; p_1, \ldots, p_n; f)$ where $C$ is a connected projective curve of arithmetic genus $g$ with at worst nodes, $p_1, \ldots, p_n$ are distinct smooth points of $C$, and $f \colon C \to X$ is a morphism with $f_*[C] = \beta$, such that the automorphism group of the tuple is finite. Equivalently, every irreducible component of $C$ of genus $0$ contracted by $f$ contains at least three special points (nodes or marked points), and every contracted component of genus $1$ contains at least one.

(Kontsevich)

Stable maps are parametrized by a proper Deligne–Mumford stack $\overline{\mathcal{M}}_{g,n}(X, \beta)$. For $X$ a point it is the moduli stack $\overline{\mathcal{M}}_{g,n}$ of stable marked curves.

The Gromov–Witten invariants of $X$ are integrals over the virtual fundamental class of $\overline{\mathcal{M}}_{g,n}(X, \beta)$ of pullbacks of cohomology classes along the evaluation maps $f \mapsto f(p_i)$. For $X = \PP^2$, $g = 0$ and $\beta = d[\text{line}]$ with $n = 3d - 1$ point classes, the invariant is the number $N_d$ of rational plane curves of degree $d$ through $3d - 1$ general points.

What is a stable map?

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Curves and surfaces