Question

I guess I am not getting something, as it seems elementary the maximum number of quadrics needed is usually g. $g$. I.e. given a canonical curve C $C$ in P^(g-1) $P^{g-1}$ which can be cut out by quadrics, it seems any general choice of g-2 $g-2$ quadrics containing it cuts out a union of curves including C. $C$. Then any general quadric containing C $C$ cuts out C $C$ and a finite set of points on the other curves. Then another general quadric through C $C$ omits those points. Is this nonsense? I see now the question in the title is no longer the same as the edited question.

edit: David, do you really want the property in your question or do you just want to know when every d $d$ dimensional subspace of I2(C) $I_2(C)$ determines the canonical curve somehow? i.e. a Torelli result.

Here is an example suggesting d may be large, a plane sextic, re embedded canonically in P^9 $P^9$ via plane cubics. The image is a del Pezzo surface S $S$ of degree 9, $9$, on which any one quadric cuts out the canonical curve, unless the quadric contains the del Pezzo. But the 55 $55$ dimensional space of quadrics in P^9 $P^9$ cuts out the 28 $28$ diml space of plane sextics, hence a 27 $27$ diml space of quadrics contains the del Pezzo. Since the ideal I2 $I_2$ has dimension 28, $28$, we actually need the whole space I2 $I_2$ to get the curve, or to get any set with the curve as a component.

Is this right? If so, plane curves of other degrees may be problematic as well....The situation seems to improve as the degree goes up. A plane septic seems to lie canonically on an embedded copy of P^2 $P^2$ that is contained in only 75 $75$ independent quadrics among the 78 $78$ containing the curve itself, so d seems to equal at least 76, $76$, maybe 77 $77$ since it seems to need two more quadrics this time. For a plane octic d $d$ seems to be at least 166, $166$, out of a space I2 $I_2$ of dim = 171. $171$. well we're gaining on it, but somewhat slowly.