It is well-known that the moduli space of genus $g\ge 2$ curves $\mathcal{M}_g$ has dimension $3g-3$. This can be computed for example as $\dim_{C} H^1(C,T_C)$. Is is also known that the structure of genus $g$ curve is determined by weight $1$ Hodge structure of $H^1(C,\mathbb{Z})$, i.e. a rank $g$ free abelian group $H^1(C,\mathbb{Z})$ with $$ H^1(C,\mathbb{Z})\otimes \mathbb{C}=H^{1,0}(C)\oplus H^{0,1}(C) $$ where $\overline{H^{1,0}(C)}=H^{0,1}(C)$. Is it possible to compute dimension of moduli space of curves via grassmannian parametrizing such $H^{1,0}(C)$?
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1 Firstly, the Hodge structure on $H^1$ of a curve admits a principal polarization, coming from the cup-product on $H^1(C,\mathbb Z)$, so one should restrict to the family of principally polarized Hodge structures of appropriate type.
But giving such a principally polarized Hodge structure is the same as giving a principally polarized abelian variety of dimension $g$ (as the quotient $H^{0,1}/\text{image of }H^1(\mathbb Z)$); the polarization assures that this complex torus actually admits a projectivce embedding.
In the case of the Hodge structure coming from a curve, this associated abelian variety is precisely the Jacobian of $C$. Thus you are asking about the so-called Torelli map which sends a curve $C$ to its Jacobian.
The Torelli theorem says that $C$ is determined by its Jacobian, so this map is injective. But the moduli of principally polarized ab. varieties is of dimension $g(g+1)/2$, which is typically greater than $3g - 3$ (one has equality only when $g = 2$ or $3$). Thus the Torelli map is not surjective when $g > 3$; most principally polarized ab. varieties of dimension $g$ are not Jacobians of curves. Determining the image of the Torelli map is a tricky question, known as the Schottky problem; it admits various approaches.
- $\begingroup$ Thanks for the detail answer. So it is not possible to compute the moduli dimension by Torelli theorem. $\endgroup$M. K.– M. K.2012-12-05 06:16:29 +00:00Commented Dec 5, 2012 at 6:16