Search the corpus

    $\mathcal{M}_g$ is irreducible of dimension $3g-3$

    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

    (Deligne--Mumford)

    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?

    Dependencies

    • D-CRVMODCoarse versus fine moduli spacesuses

    Backlinks

    • PR-CRVAUTAutomorphisms of a curve of genus $g \geq 2$related-to

    What links to this