Let $C$ be a smooth curve over complex numbers. Consider the Brill Noether varieties. If $g$ is the genus of $C$. If $r,d$ are positive integers,

$$W^r_d=\{A\in Pic^d(C): h^0(A)\geq r+1\},$$ $$G^r_d=\{(A,V): A\in Pic^d(C), V\in G(r+1,H^0(C,A))\}.$$

We have the projection morphism $p:G^r_d\rightarrow W^r_d$, which is constructed by 'canonical blow up', which doesnt turn out to be blow up in usual sense.

Related question On the construction of the varieties parametrizing special linear series on a curve

Suppose that $W^r_d$ is irreducible, is it true that $G^r_d$ is irreducible as well? If $p$ is the blow up map in the usual sense, this is true. But in this case is this true?

For example if $d\geq r+ g$, then $W^r_d=Pic^d C$ by Riemann Roch. Then is $G^r_d$ irreducible?


1 Answer 1


No in general. For instance, take for $C$ a hyperelliptic cuve of even genus $g=2k$ with $k\geq 6$. Then $G^2_{g+2}$ has a unique component (of dimension $g+2$) which dominates $\mathrm{Pic}^{g+2}(C)$, but $p^{-1}((k+1)g^1_2)\cong \mathbb{G}(3,k+2)$ has dimension $3k-3>g+2$, so it cannot be contained in that component.

  • $\begingroup$ how do you say it has a unique component dominating $Pic^{g+2} C$. Is this uniqueness true for any $G^r_d$ with $d=g+r$? $\endgroup$
    – user52991
    Oct 14, 2016 at 13:47
  • $\begingroup$ Yes. There is a nonempty open subset $U\subset\mathrm{Pic}^{g+r}(C)$ such that $h^0(L)=r+1$ for $L\in U$; the map $p:G^r_{g+r}\rightarrow\mathrm{Pic}^{g+r}(C)$ induces an isomorphism of $p^{-1}(U)$ onto $U$. The closure of $U$ in $G^r_{g+r}$ is the only component dominating $\mathrm{Pic}^{g+r}$. $\endgroup$
    – abx
    Oct 14, 2016 at 15:39
  • $\begingroup$ I was thinking about your example. The fiber is not in the dominating component. The fiber doesn't contain any complete linear series. Is it possible that a component of $G^r_d$ does not contain any complete linear series? $\endgroup$
    – user52991
    Nov 3, 2016 at 15:13

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