# Complement of the Brill Noether locus in $\mathcal{M}_g$

Consider the moduli space $\mathcal{M}_g$ of genus $g$ curves over $\mathbb{C}$. Let $d,r\geq 1$ be integers so that the Brill Noether number $\rho(g,r, d)=g-(r+1)(g-d+r)>0$ . I am mainly interested in the case when $r=1$.

Consider the locus $\mathcal{W}^r_d=\{C\in \mathcal{M}_g: W^r_d(C)\neq\emptyset\}$.

Here $W^r_d(C)=\{L\in Pic^d(C): h^0(L)\geq r+1\}$.

It is known that $\mathcal{W}^r_d$ is an open dense set in $\mathcal{M}_g$. What is the codimension of it's complement in the moduli space?

Thank you.

• If $\rho$ is nonnegative, then the set that you call $\mathcal{W}^r_d$ equals all of $\mathcal{M}_g$. When $\rho$ equals $-1$, $\mathcal{W}^r_d$ is a divisor. This is part of the proof that $\overline{\mathcal{M}}_g$ is of general type for $g$ large (Harris, Mumford, Eisenbud). – Jason Starr Oct 4 '16 at 11:56
• @Jason Starr. Thank you. The purpose of my question is as follows. In ACGH, it says that as a corollary of the Fulton Lazarsfeld connectedness theorem, for a general curve $C$, if $\rho\geq 1$, then $W^r_d(C)$ is irreducible. I thought 'general' here meant Brill Noether general. But all curves in the moduli space are Brill Noether general? – user52991 Oct 4 '16 at 12:12
• No, not all curves are Brill-Noether general. However, you defined $W^r_d(C)$ using an inequality, i.e., $h^0(L) \geq r+1$ rather than $h^0(L)=r+1$. With that definition, whenever $\rho\geq 0$, the set $W^r_d(C)$ is nonempty for all smooth, genus $g$ curves $C$. Saying that $C$ is "Brill-Noether general" usually means that for all $\rho$ with $\rho<0$, $W^r_d(C)$ is empty (some authors would add the condition that the spaces $G^r_d(C)$ are all smooth). – Jason Starr Oct 4 '16 at 12:56
• @Jason Starr. Ah sorry, I don't think I follow your comment. In the corollary to the connectedness theorem, it says for a general curve $C$, if $\rho\geq 1$, then the variety is irreducible. What does general mean in this context? Because $\rho< 1$ is not valid here. – user52991 Oct 4 '16 at 13:02
• The question you asked above is different from the question you asked in your most recent comment. Nonemptiness of $W^r_d(C)$ is different from irreducibility of $W^r_d(C)$. Does that help clear up your confusion? – Jason Starr Oct 4 '16 at 13:08