# Moduli space of algebraic surfaces Vs moduli space of curves

Define the surface $S$ as the complete intersection of four quadrics $Q_i$ with $i=1,2,3,4$ in $\mathbb{P}^6$ (complex six dimensional projective space) i.e. $$S=Q_1 \cap Q_2 \cap Q_3 \cap Q_4$$.
Put $C=H \cap S$ where $H$ is an hyperplane of $\mathbb{P}^6$. So $C$ is a smooth curve on $S$ with genus $g=g(C)$. Set $\eta_{C/S}$ the normal bundle. So you can show that $\eta_{C/S}$ satisfy $\omega_C=(\eta_{C/S})^{\otimes2}$ where $\omega_C$ is the canonical bundle on the curve $C$. So the normal bundle is a theta characteristic on $S$.
My questions:

1) what kind of theta-characteristic? odd or even?

2) suppose that $\mathscr{M}_S$ is the moduli space of the surface $S$ and $\mathscr{L}_C$ is the moduli space of the set $S_g=\{C, \theta_C \}$ where $\theta_C$ is a theta-characteristic on $C$. What are the dimensions of $\mathscr{M}_S$ and $\mathscr{L}_C$? is there a relations between these moduli spaces?

any suggestions are welcome.

• for the question 2). how i can explain both of modular spaces? i think the problem is that i've not a lot of experience with modular spaces but i think that there would be some nice presentations of that sets that is usefull for my problem – dario Jun 9 '15 at 22:22
• if i take the 2-torsion group of the curve $C$, $J(C)= \{ L: deg(L)=0, 2L=0 \}$, there is a bijective map from $J(C)$ to $S(C)$ that is the set of all theta characteristic of the curve $C$. So $J(C)$ is a $\mathbb{Z} / 2\mathbb{Z}$ modulo of rank 2g where g is the genus of $C$. Can i say that the dimension of the moduli space of $C$ is the dimension of this module? – dario Jun 10 '15 at 8:51
• The set $\mathscr{L}_C$ is a subvariey of $\mathscr{M}_C$ that is the moduli space of all curve o genus $g$. So the dimension of $\mathscr{L}_C$ is the dimension of $\mathscr{L}_C$ as subvariety of $\mathscr{M}_C$ – dario Jun 10 '15 at 14:38
• i put this as a comment but i think this is the right way to solve the problem is to compute the dimension of the space $H^1(S,T_S)$ where $T_S$ is the tangent bundle of $S$. So i take the exact sequence $$0 \rightarrow T_S \rightarrow \ T(\mathbb{P}^6)|_S \rightarrow O(2)+O(2)+O(2)+O(2) \rightarrow 0$$. This sequence induces a long exact sequence in coomology. So How can i obtain the dimension of $H^1(S,T_S)$? – dario Jun 13 '15 at 12:54