# Are curves with maximal Clifford index Brill-Noether general?

By the Brill-Noether Theorem, a general curve $C$ of genus $g\geq2$ has maximal Clifford index $\lfloor \frac{g-1}{2}\rfloor$. Hence a very naive question is:

(Q1) Is a curve with maximal Clifford index a general curve (in the sense of Brill-Noether theory) ? And if not, what can be said about such a curve (when non-general) ?

Of course a similar setting holds by considering the gonality instead of the Clifford index, but I guess that in this situation the two are completely equivalent, that is:

(Q2) Is the gonality of a curve with maximal Clifford index $c$ always equal to $c+2$ ?

Edit: by curve I understand an irreducible curve.

Q2) as stated is obviously true, because if the curve had a pencil of degree less than or equal to c+1, then the Clifford index would be at most c-1 (the clifford index of this pencil). So you probably meant to state the converse statement, which asks if a curve of maximal gonality has maximal Clifford index.

At least if the genus is odd, the answer to this question is yes. In this case, a curve has maximal gonality if and only if it has maximal Clifford index. This comes out as a corollary of a beautiful paper by Hirschowitz-Ramanan "New evidence for Green's conjecture on syzygies of canonical curves". One knows, by the generic Green's conjecture (proved by Voisin), that a curve has nonmaximal Clifford index iff it has an extra syzygy, in the sense of Green. On the other hand, one can compare the divisor of curves in $\mathcal{M}_g$ with extra syzygies with the Hurwitz divisor of curves with nonmaximal gonality. The two divisors can be shown to be the same using direct computation of their classes, and this gives the result. For more details, see the Hirschowitz-Ramanan paper above.

• Dear mkemeny, I don't understand your first remark. A curve with Clifford index $c$ may have gonality $c+2$ or $c+3$ in general. Commented Dec 12, 2015 at 19:40
• No. If a curve has maximal Clifford index $c=\lfloor \frac{g-1}{2} \rfloor$, then, like I said, the gonality is the maximal value $c+2$ (note the gonality is always at most $\lfloor \frac{g+3}{2} \rfloor$, so $c+3$ is certainly not possible in this case). Assume the gonality were $c+1$ or less. By definition, you would have a special line bundle $A$ of degree $c+1$ or less with 2 sections. This contributes to the clifford index, and $Cliff(A) \leq c-1$, which is a contradiction. The problem is to assume that gonality is maximal, say $k$, and prove that Cliff is $k-2$, and not, say $k-3$. Commented Dec 13, 2015 at 11:17
• The difficulty in proving the nontrivial direction is that there could be line bundles with more than two sections which achieve the Clifford index. Commented Dec 13, 2015 at 11:20
• Ah now I see. Thank you for clarifying this to me. Commented Dec 13, 2015 at 11:24

The answer to (Q1) is definitely no. The first example is in genus 4: a (smooth) complete intersection of a quadric cone and a cubic surface in $\mathbb{P}^3$ has maximal Clifford index (= 1), but is not Brill-Noether general: it has a unique $g^1_3$ whose double is the canonical divisor.

The answer to (Q2) should be yes. I recommend the paper The Clifford dimension of a projective curve, by Eisenbud, Lange, Martens, Schreyer (Compositio Math. 72 (1989), no. 2, 173–204). They conjecture that the Clifford index comes from a $g_r^d$ with $r>1$ only for plane curves and in some very particular cases, where the Clifford index is one less than the maximal one. This would imply a positive answer to (Q2). Of course (Q2) is much weaker, so there may very well be a proof independent of the conjecture -- but I do not see it at the moment.

• Dear abx, thank you for your answer. Could you explain me why having a unique $g^1_3$ whose double is $K_C$ prevents $C$ to be Brill-Noether general? Commented Dec 12, 2015 at 6:28
• Of course this depends on your definition of Brill-Noether general; I took the most commonly used one, which asks that the variety $G^r_d$ of $g^r_d$'s is smooth of the expected dimension. In my example $G^1_3$ is not smooth -- actually not reduced. If you ask only for the dimension to be the expected one, the first counter-example is given by double covering of elliptic curves in genus 6.
– abx
Commented Dec 12, 2015 at 18:55