Non hyperelliptic curves of genus 5 form a family of dimension 12 Suppose $C$ is a complete intersection of three quadrics in $\mathbb{P}^4$, how to count naively the dimension of its parameter spaces? One needs $|O_{\mathbb{P^4}}(2)|=14$ parameters to describe one quadric.
Three general quadrics intersect in a smooth curve, thus the dimension for the choice of three quadrics is $14+14+13$ (the third one should not be in the family of the first two). Then modulo the action of $\operatorname{PGL}_4$. So why is the dimension not $14+14+13-24=17$?
 A: The $PGL_5$ action is not the only thing to account for: even after fixing our choice of of projective coordinates, there are many distinct choices of quadrics that give not only the same curve up to isomorphism, but the same embedded curve, meaning $Q_1 \cap Q_2 \cap Q_3$ is the same subvariety.
Formally, the description you're going for is the construction of a flat rational map
$\mathbb{P}^{14} \times \mathbb{P}^{14} \times \mathbb{P}^{14}\ --> \mathcal{M}_5, \qquad (Q_1, Q_2, Q_3) \mapsto \text{isomorphism class of }Q_1 \cap Q_2 \cap Q_2.$
The fiber of this map for a curve $C$ is given not only by the ($PGL_5$) choice of isomorphism $H^0(\mathcal{O}_{\mathbb{P}^{14}}(1)) \to H^0(\omega_C)$, but also by the choice of triple
$$(Q_1, Q_2, Q_3) \in \mathbb{P}H^0(\mathcal{I}_C(2))^{\times3}.$$
Since $H^0(\omega_C(2))$ has 12 sections, $H^0(\mathcal{I}_C(2))$ has at most three sections, hence exactly three. So the fiber above has dimension 6 (as a projective subspace of $(\mathbb{P}^{14})^{\times3}$.)
So the overall count is:
$$14+14+14 - 24 - 6 = 42 - 30 = 12 = \dim \mathcal{M}_5.$$
