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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 $p^*\pi_* \L$.

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

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?

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?

Edit: I would like to define it as a fiber product without talking about determinantal varieties. So not as it is done in pages 83-84 of [GAC].

My thoughts:

Let $\mathscr{L}$ 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 $Pic^d$.

I think it's a good idea to consider the Grassmanian $ G(\mathscr{L}) := G(r+1, H^0(\mathscr{L})) $. Indeed $G_d^r$ can be seen as the subset of the product $$ Pic^d \times G(\mathscr{L}) $$ consisting of those couples $(L, \mathscr{W})$ such that $\;f^*\mathscr{W} \subset H^0(L)$.

I have the feeling that $G_d^r$ can be defined as a particular fiber product between $Pic^d$ and $G(\mathscr{L})$, but I don't understand how to build the right fiber diagram. Do you think this is indeed possible?

Remark: I believe we have $f^*H^0(\mathscr{L}) \cong H^0(L)$. Do you agree on this?

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 $p^*\pi_* \L$.

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

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?

My thoughts:

Let $\mathscr{L}$ 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 $Pic^d$.

I think it's a good idea to consider the Grassmanian $ G(\mathscr{L}) := G(r+1, H^0(\mathscr{L})) $. Indeed $G_d^r$ can be seen as the subset of the product $$ Pic^d \times G(\mathscr{L}) $$ consisting of those couples $(L, \mathscr{W})$ such that $\;f^*\mathscr{W} \subset H^0(L)$.

I have the feeling that $G_d^r$ can be defined as a particular fiber product between $Pic^d$ and $G(\mathscr{L})$, but I don't understand how to build the right fiber diagram. Do you think this is indeed possible?

Remark: I believe we have $f^*H^0(\mathscr{L}) \cong H^0(L)$. Do you agree on this?

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?

Edit: I would like to define it as a fiber product without talking about determinantal varieties. So not as it is done in pages 83-84 of [GAC].

My thoughts:

Let $\mathscr{L}$ 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 $Pic^d$.

I think it's a good idea to consider the Grassmanian $ G(\mathscr{L}) := G(r+1, H^0(\mathscr{L})) $. Indeed $G_d^r$ can be seen as the subset of the product $$ Pic^d \times G(\mathscr{L}) $$ consisting of those couples $(L, \mathscr{W})$ such that $\;f^*\mathscr{W} \subset H^0(L)$.

I have the feeling that $G_d^r$ can be defined as a particular fiber product between $Pic^d$ and $G(\mathscr{L})$, but I don't understand how to build the right fiber diagram. Do you think this is indeed possible?

Remark: I believe we have $f^*H^0(\mathscr{L}) \cong H^0(L)$. Do you agree on this?