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Let $C$ be a curve of genus $g$ and $L$ a $g^r_d$ on it and assume that we are in the range ${r+2\choose 2}>2d-g+1$. If $C$ and $L$ are chosen to be general then by the maximal rank conjecture (which is a theorem now), there are exactly ${r+2\choose 2}-2d+g-1$ linearly independent quadrics containing the image of $C$ under the embedding given by $L$.

Clearly, one would expect these quadrics to intersect "as transversely as possible". For instance, in the canonical curve case this is true by Petri's theorem, namely, the base locus of the quadrics is the curve itself if it is not trigonal or a plane quintic). I am curious if this is true in general. So, precisely:

Is it true (or is anything known in this direction) that if $(C,L)$ is general as above then the dimension of the base locus of $I_2(C,L)$ is equal to $min\{1,r-\mbox{dim }I_2(C,L)\}$?

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    $\begingroup$ I don't recall off-hand who proved this, but if $2d \geq 2g+2$ , then for any line bundle L of degree d, one has that the base locus of $I_2(C,L)$ is exactly C. Frequently, the non-existence of a 3 pointed secant line is a sufficient (it's certainly necessary!) condition for C to be defined by the quadrics through it. $\endgroup$
    – meh
    Commented Sep 20, 2016 at 19:39
  • $\begingroup$ Thanks for the comment! You probably mean the result by Saint-Donat, Fujita, Green (and Lazarsfeld too I guess), which states that $\varphi_L(C)$ is cut out by quadrics if $d\geq 2g+2$. If you mean something different I would be happy to hear it! Moreover, this is an "absolute" statement in the sense that it is true for every curve and every line bundle. What I am curious about is if one can compromise on that by considering "only" generic curves and line bundles to improve the statement in the way that I described in my question (and possibly for lower degree range). $\endgroup$
    – Irfan Kadikoylu
    Commented Sep 21, 2016 at 8:27
  • $\begingroup$ Ok, actually you probably meant the result of Green, Lazarsfeld, which says that if the curve has no trisecant line and $d\geq 2g+2-2h^1(L)-Cliff(C)$ then it is cut by quadrics. The thing is, I need this result for generic curves of degree $=3g/2$ with $h^1=1$ and for generic curves, the bound in that theorem becomes (strangely enough) $d\geq (3g+1)/2$. It is not my lucky day, I guess! $\endgroup$
    – Irfan Kadikoylu
    Commented Sep 21, 2016 at 9:40
  • $\begingroup$ I was referring to what you call the result of Saint-Donat, Fujita, and Green-Lazarsfeld. It was just meant to be just a comment since you asked 'is anything known' :). I don't have a reference for the Green-Lazarsfeld result. Can you provide one. I am familiar with the analagous statement for normal generation. $\endgroup$
    – meh
    Commented Sep 21, 2016 at 16:19
  • $\begingroup$ Further I think you are out of luck in the sense that the result you mention is sharp. Allow me to do 'approximate math' with fractions. Suppose C is a general curve of Clifford index $\frac{g-1}{2}$ computed by a divisor D of degree $\frac{g+3}{2} $. Pick $\frac{g-3}{2}$ points in D and project from them. The line bundle is of degree $\frac{3g-1}{2} $ and has a trisecant line (the other 3 points of D). If you demand your result be true for every line bundle of a given degree that is the best one can do. $\endgroup$
    – meh
    Commented Sep 21, 2016 at 16:26

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