Let $C$ be a non hyperelliptic complex algebraic curve of genus $g$, then the vector space $I_2(C)$ of quadrics containing the canonical image of $C$ is $\binom{g+1}{2}-h^0(2\omega_C) = (g-2)(g-3)/2$ dimensional. Moreover, if $g > 4$ then $C$ is the intersection of the nulls of all these quadrics (see ACGH VI.4.1 for a proof, I'm not sure how "classical" this is, or who originally proved it).
Question (edited following a comment from David Speyer) what is the least
$d$ so that if $V\subset I_2(C)$ is
any $d$ dimensional vector space, and $X$ is the intersection of the
nulls of the quadrics in $V$, then the
only irreducible component of $X$
which linearly spans $|\omega_C|^*$ is
the canonical image of $C$ ?
I don't even know the generic bound, or indeed what is the bound for hyperelliptic curves (in which case the canonical curve is a rational normal curve).
Best Answer
Petri's theorem (see ACGH III.3) states that if $C\subset P^{g-1}$ is a canonical curve (i.e. the canonical image of a non hyperelliptic curve) of genus $g\ge 4$ then the ideal of $C$ is generated by quadrics iff $C$ is not trigonal or a plane quintic (for trigonal curves or plane quintics the intersection of all quadrics through $C$ is a surface).
If $C$ is hyperelliptic, the dimension of the space of quadrics through the canonical image $\Gamma$ is $(g+1)g/2-2g+1=(g-1)(g-2)/2$. The ideal of $\Gamma$ is also generated by quadrics.
If $g-2$ is not a power of 2, a trivial lower bound is $d\ge g-1$. Indeed by Bezout's theorem (see Fulton, Intersection theory, 8.4) if the intersection of $g-2$ quadrics of $P^{g-1}$ is a curve of degree $d$, then $d$ divides $2^{g-2}$.
The following $2g-5$ quadrics cut the rational normal curve in $P^{g-1}$ set theoretically: $x_0x_2-x_1^2,x_0x_3-x_1x_2, \dots, x_0x_{g-1}-x_{g-2}x_1, x_2^2-x_1x_3, x_3^2-x_2x_4,\dots,x_{g-2}^2-x_{g-3}x_{g-1}$.
I don't know if one can do better.