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Let $X$ be a smooth projective curve over $k=\bar{k}$, and $Pic^d$ the $d$-part of the Picard group of $X$ (isomorphism classes of line bundles of degree $d$ on $X$).

I'd like to define a scheme $G_d^r$ parametrizing $g_d^r$'s on our curve $X$, so with set theoretic support given by $$ \operatorname{Supp}(G_d^r) = \{\; (L,W) \in Pic^d\times G(r+1, H^0(L)) \;\}. $$

My question is:

What is the most elegant way to define $G_d^r$ as a scheme?

Further explanations & my idea: $\newcommand{\L}{\mathscr{L}}$ Let $\L$ denote the Poincaré line bundle on $X$ (assuming it does exist), i.e. a line bundle on $X\times Pic^d$ such that for every $L\in Pic^d$ we have $f^*\mathscr{L} = L$, where $f:*\to Pic^d$ is the inclusion of $L$ in $Pic^d$.

In GAC = Geometry of Algebraic Curves by Arbarello & co. they define $G_d^r$ in the following way:

  1. They choose a divisor $\Gamma$ with high degree $m$ on $X\times Pic^d$ and consider the exact sequence of sheaves over $Pic^d$ $$ 0\to \pi_*\L \to \pi_*\L(\Gamma) \overset{\gamma}\to \pi_*\L(\Gamma)/\L \to R^1\pi_*\L\to 0 $$ where the ones in the middle are locally free of ranks $n=d+m-g+1$ and $m$.

  2. They consider the Grassmannian bundle $$G(n-(d+m-g-r),\; \pi_*\L(\Gamma))=G(r+1),\; \pi_*\L(\Gamma)) \overset{p}\longrightarrow Pic^d$$ and pull back via $p$ to get a map $$ \psi: S \hookrightarrow p^* \pi_*\L(\Gamma) \overset{p*(\gamma)}\longrightarrow p^* \pi_*\L(\Gamma)/\L $$ where $S$ fits in the natural short exact sequence $0\to S\to p^* \pi_*\L(\Gamma) \to Q \to 0$.

  3. They define $G_d^r = \ker(\psi)$, so that its points are couples $(L,W)$ where $L\in Pic^d$ is a line bundle and $W$ is an $(r+1)$-dimensional subspace of $H^0(L)$.

In my opinion the fact that we are making a choice in picking $\Gamma$ is annoying. It makes the definition not elegant.

My idea to define $G_d^r$ as something which looks more or less like $$ G_d^r := G(r+1,\; \pi_*\mathscr{L})\,. $$ This should produce something very close to the GAC definition. Indeed $S \to p^* \pi_* \L(\Gamma)$ is an inclusion, so $\ker(\psi) \cong \ker(p^*(\gamma))$. The latter turns out to be simply the pullback via $p$ of the kernel of $\gamma$, which we see from $(1)$ to be equal to $p^*\pi_* \L$.

Moreover, the support of $G_d^r$ are the fibers over closed points $L$ of $Pic^d$, and we find that these are (for $f:L\hookrightarrow Pic^d$) $$ f^*G_d^r \cong G(r+1,\; f^*\pi_*\L) \cong G(r+1,\; \pi_*f^*\L) \cong G(r+1,\; H^0(L)) $$

Do you think my idea makes sense? How can I make it precise?

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    $\begingroup$ All of this is nicely done in "Geometry of Algebraic Curves" (I), by Arbarello et al. $\endgroup$
    – abx
    Commented Dec 12, 2013 at 17:09
  • $\begingroup$ @abx: I know that book, but I'm not satisfied with the way they define it there. I'm sure there's a better, more modern approach to it $\endgroup$
    – Abramo
    Commented Dec 12, 2013 at 17:15
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    $\begingroup$ It seems to me that in many cases of interest the push forward of L will be 0 or perhaps locally free of rank smaller than r+1. The spaces of sections of special linear series don't extend to nearby line bundles, so the push forward of L cannot be of any use. Twisting by a sufficiently ample line bundle kills cohomology so that pushforwards are well behaved; when cohomology jumps uncontrollably the push forward is not well behaved. $\endgroup$
    – Jack Huizenga
    Commented Dec 13, 2013 at 14:06
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    $\begingroup$ In particular, it is certainly not the case that the fiber of the push forward of the Poincare bundle at some line bundle is the global sections of the bundle. $\endgroup$
    – Jack Huizenga
    Commented Dec 13, 2013 at 14:19
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    $\begingroup$ I think Jack hits the key issue. An elementary take on the base divisor D is to produce a sheaf sequence 0-->L-->L(D)-->L(D)|D-->0, such that H^0(L) and H^1(L) are simply kernel and cokernel of the associated restriction map H^0(L(D))-->H^0(L(D)|D). This allows one to model spaces of sections of line bundles nearby L as kernels of matrices of the same dimension. A detailed discussion of the role of base divisors is in Kempf's Abelian Integrals, (or lacking that, some related comments are in Contemp. Math. vol 397, p.215 ff. where I admit we also tried at first to avoid them.) $\endgroup$
    – roy smith
    Commented Dec 13, 2013 at 17:39

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