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Let $X$ be a smooth projective irreducible algebraic curve over field $k$. For $d,r,k,m >0$ the representable Quot scheme $\mathcal {Quot}_X^{r,d}(\mathcal{O}_X(m)^k)$ is given for any test scheme $U \to X$ by

$${\displaystyle {\mathcal {Quot}}_{X}^{r,d}(\mathcal{O}_X(m)^k)(U)=\left\{({\mathcal {F}},q):{\begin{matrix}{\mathcal {F}}\in {\text{Coh}}(X \times U)\\{\text{Supp}}({\mathcal {F}}){\text{ is proper over }}X \times U\\{\mathcal {F}}{\text{ is flat over }}X \times U\\q:(\mathcal{O}_{X\times U}(m))^k\to {\mathcal {F}}{\text{ surjective}} \\ {\text{ rank }}({\mathcal {F}})= r {\text{ and }} {\text{ deg }}({\mathcal {F}})= d \end{matrix}}\right\}/\sim }$$

Now I want to define an subfunctor ${\mathcal {R}}_{X}^{m} \subset \mathcal {Quot}_X^{r,d}(\mathcal{O}_X(m)^k)$ by taking restriction:

$${\displaystyle {\mathcal {R}}_{X}^{m}(U)=\left\{({\mathcal {F}},q):{\begin{matrix}{\mathcal {F}}\in {\mathcal {Quot}}_{X}^{r,d}(\mathcal{O}_X(m)^k)(U) \\ {\mathcal {F}} {\text{ locally free and }} \\ H^1(X \times U, \mathcal {F}(m))=0 \end{matrix}}\right\}/\sim }$$

First obvious problem: is the thing ${\mathcal {R}}_{X}^{m}$ well defined, that is what about the compatibility with pullbacks of the last condition on $H^1$ group? i.e. I want to assure that if $U \to X$ is an arbitrary test scheme and the quotient $\mathcal{F} \in {\mathcal {R}}_{X}^{m}(U)$ then for every $X$-morphism $f:S \to U$ the pullback $\hat{f}^*\mathcal{F}$ is an element in ${\mathcal {R}}_{X}^{m}(S)$, where $\hat{f}: X \times S \to X \times U$.

The crucial obstacle for me is to verify $H^1(X \times S, \hat{f}^*\mathcal {F}(m))=0$. I conjecture that here I can somehow use one of many corollaries of the Semicontinuity Theorem (see e.g. Algebraic Geometry by R. Hartshorne, pp 281), but don't know how it can be applied in this situation. Any idea how to solve it? Can I make a reduction argument to reduce it to a setting where semicontinuity does the job?

Let $X$ be a smooth projective irreducible algebraic curve over field $k$. For $d,r,k,m >0$ the representable Quot scheme $\mathcal {Quot}_X^{r,d}(\mathcal{O}_X(m)^k)$ is given for any test scheme $U \to X$ by

$${\displaystyle {\mathcal {Quot}}_{X}^{r,d}(\mathcal{O}_X(m)^k)(U)=\left\{({\mathcal {F}},q):{\begin{matrix}{\mathcal {F}}\in {\text{Coh}}(X \times U)\\{\text{Supp}}({\mathcal {F}}){\text{ is proper over }}X \times U\\{\mathcal {F}}{\text{ is flat over }}X \times U\\q:(\mathcal{O}_{X\times U}(m))^k\to {\mathcal {F}}{\text{ surjective}} \\ {\text{ rank }}({\mathcal {F}})= r {\text{ and }} {\text{ deg }}({\mathcal {F}})= d \end{matrix}}\right\}/\sim }$$

Now we want to define an subfunctor ${\mathcal {R}}_{X}^{m} \subset \mathcal {Quot}_X^{r,d}(\mathcal{O}_X(m)^k)$ by taking restriction:

$${\displaystyle {\mathcal {R}}_{X}^{m}(U)=\left\{({\mathcal {F}},q):{\begin{matrix}{\mathcal {F}}\in {\mathcal {Quot}}_{X}^{r,d}(\mathcal{O}_X(m)^k)(U) \\ {\mathcal {F}} {\text{ locally free and }} \\ H^1(X \times U, \mathcal {F}(m))=0 \end{matrix}}\right\}/\sim }$$

First obvious problem: is the thing ${\mathcal {R}}_{X}^{m}$ well defined, that is what about the compatibility with pullbacks of the last condition on $H^1$ group? i.e. I want to assure that if $U $ is an arbitrary $k$ scheme and the quotient $\mathcal{F} \in {\mathcal {R}}_{X}^{m}(U)$ then for every $X$-morphism $f:S \to U$ the pullback $\hat{f}^*\mathcal{F}$ is an element in ${\mathcal {R}}_{X}^{m}(S)$, where $\hat{f}: X \times S \to X \times U$.

The crucial obstacle for me is to verify $H^1(X \times S, \hat{f}^*\mathcal {F}(m))=0$. I conjecture that here I can somehow use one of many corollaries of the Semicontinuity Theorem (see e.g. Algebraic Geometry by R. Hartshorne, pp 281), but don't know how it can be applied in this situation. Any idea how to solve it? Can I make a reduction argument to reduce it to a setting where semicontinuity does the job?

Let $X$ be a smooth projective irreducible algebraic curve over field $k$. For $d,r,k,m >0$ the representable Quot scheme $\mathcal {Quot}_X^{r,d}(\mathcal{O}_X(m)^k)$ is given for any test scheme $U \to X$ by

$${\displaystyle {\mathcal {Quot}}_{X}^{r,d}(\mathcal{O}_X(m)^k)(U)=\left\{({\mathcal {F}},q):{\begin{matrix}{\mathcal {F}}\in {\text{Coh}}(X \times U)\\{\text{Supp}}({\mathcal {F}}){\text{ is proper over }}X \times U\\{\mathcal {F}}{\text{ is flat over }}X \times U\\q:(\mathcal{O}_{X\times U}(m))^k\to {\mathcal {F}}{\text{ surjective}} \\ {\text{ rank }}({\mathcal {F}})= r {\text{ and }} {\text{ deg }}({\mathcal {F}})= d \end{matrix}}\right\}/\sim }$$

Now I want to define an subfunctor ${\mathcal {R}}_{X}^{m} \subset \mathcal {Quot}_X^{r,d}(\mathcal{O}_X(m)^k)$ by taking restriction:

$${\displaystyle {\mathcal {R}}_{X}^{m}(U)=\left\{({\mathcal {F}},q):{\begin{matrix}{\mathcal {F}}\in {\mathcal {Quot}}_{X}^{r,d}(\mathcal{O}_X(m)^k)(U) \\ {\mathcal {F}} {\text{ locally free and }} \\ H^1(X \times U, \mathcal {F}(m))=0 \end{matrix}}\right\}/\sim }$$

First obvious problem: is the thing ${\mathcal {R}}_{X}^{m}$ well defined, that is what about the compatibility with pullbacks of the last condition on $H^1$ group? i.e. I want to assure that if $U \to X$ is an arbitrary test scheme and the quotient $\mathcal{F} \in {\mathcal {R}}_{X}^{m}(U)$ then for every $X$-morphism $f:S \to U$ the pullback $\hat{f}^*\mathcal{F}$ is an element in ${\mathcal {R}}_{X}^{m}(S)$, where $\hat{f}: X \times S \to X \times U$.

The crucial obstacle for me is to verify $H^1(X \times S, \hat{f}\mathcal {F}(m))=0$. I conjecture that here I can somehow use one of many corollaries of the Semicontinuity Theorem (see e.g. Algebraic Geometry by R. Hartshorne, pp 281), but don't know how it can be applied in this situation. Any idea how to solve it? Can I make a reduction argument to reduce it to a setting where semicontinuity does the job?

Let $X$ be a smooth projective irreducible algebraic curve over field $k$. For $d,r,k,m >0$ the representable Quot scheme $\mathcal {Quot}_X^{r,d}(\mathcal{O}_X(m)^k)$ is given for any test scheme $U \to X$ by

$${\displaystyle {\mathcal {Quot}}_{X}^{r,d}(\mathcal{O}_X(m)^k)(U)=\left\{({\mathcal {F}},q):{\begin{matrix}{\mathcal {F}}\in {\text{Coh}}(X \times U)\\{\text{Supp}}({\mathcal {F}}){\text{ is proper over }}X \times U\\{\mathcal {F}}{\text{ is flat over }}X \times U\\q:(\mathcal{O}_{X\times U}(m))^k\to {\mathcal {F}}{\text{ surjective}} \\ {\text{ rank }}({\mathcal {F}})= r {\text{ and }} {\text{ deg }}({\mathcal {F}})= d \end{matrix}}\right\}/\sim }$$

Now I want to define an subfunctor ${\mathcal {R}}_{X}^{m} \subset \mathcal {Quot}_X^{r,d}(\mathcal{O}_X(m)^k)$ by taking restriction:

$${\displaystyle {\mathcal {R}}_{X}^{m}(U)=\left\{({\mathcal {F}},q):{\begin{matrix}{\mathcal {F}}\in {\mathcal {Quot}}_{X}^{r,d}(\mathcal{O}_X(m)^k)(U) \\ {\mathcal {F}} {\text{ locally free and }} \\ H^1(X \times U, \mathcal {F}(m))=0 \end{matrix}}\right\}/\sim }$$

First obvious problem: is the thing ${\mathcal {R}}_{X}^{m}$ well defined, that is what about the compatibility with pullbacks of the last condition on $H^1$ group? i.e. I want to assure that if $U \to X$ is an arbitrary test scheme and the quotient $\mathcal{F} \in {\mathcal {R}}_{X}^{m}(U)$ then for every $X$-morphism $f:S \to U$ the pullback $\hat{f}^*\mathcal{F}$ is an element in ${\mathcal {R}}_{X}^{m}(S)$, where $\hat{f}: X \times S \to X \times U$.

The crucial obstacle for me is to verify $H^1(X \times S, \hat{f}^*\mathcal {F}(m))=0$. I conjecture that here I can somehow use one of many corollaries of the Semicontinuity Theorem (see e.g. Algebraic Geometry by R. Hartshorne, pp 281), but don't know how it can be applied in this situation. Any idea how to solve it? Can I make a reduction argument to reduce it to a setting where semicontinuity does the job?

Let $X$ be a smooth projective irreducible algebraic curve over field $k$. For $d,r,k,m >0$ the representable Quot scheme $\mathcal {Quot}_X^{r,d}(\mathcal{O}_X(m)^k)$ is given for any test scheme $U \to X$ by

$${\displaystyle {\mathcal {Quot}}_{X}^{r,d}(\mathcal{O}_X(m)^k)(U)=\left\{({\mathcal {F}},q):{\begin{matrix}{\mathcal {F}}\in {\text{Coh}}(X \times U)\\{\text{Supp}}({\mathcal {F}}){\text{ is proper over }}X \times U\\{\mathcal {F}}{\text{ is flat over }}X \times U\\q:(\mathcal{O}_{X\times U}(m))^k\to {\mathcal {F}}{\text{ surjective}} \\ {\text{ rank }}({\mathcal {F}})= r {\text{ and }} {\text{ deg }}({\mathcal {F}})= d \end{matrix}}\right\}/\sim }$$

Now I want to define an subfunctor ${\mathcal {R}}_{X}^{m} \subset \mathcal {Quot}_X^{r,d}(\mathcal{O}_X(m)^k)$ by taking restriction:

$${\displaystyle {\mathcal {R}}_{X}^{m}(U)=\left\{({\mathcal {F}},q):{\begin{matrix}{\mathcal {F}}\in {\mathcal {Quot}}_{X}^{r,d}(\mathcal{O}_X(m)^k)(U) \\ {\mathcal {F}} {\text{ locally free and }} \\ H^1(X \times U, \mathcal {F}(m))=0 \end{matrix}}\right\}/\sim }$$

First obvious problem: is the thing ${\mathcal {R}}_{X}^{m}$ well defined, that is what about the compatibility with pullbacks of the last condition on $H^1$ group? i.e. I want to assure that if $U \to X$ is an arbitrary test scheme and the quotient $\mathcal{F} \in {\mathcal {R}}_{X}^{m}(U)$ then for every $X$-morphism $f:S \to U$ the pullback $f^*\mathcal{F}$ is an element in ${\mathcal {R}}_{X}^{m}(S)$.

The crucial obstacle for me is to verify $H^1(X \times S, f^*\mathcal {F}(m))=0$. I conjecture that here I can somehow use one of many corollaries of the Semicontinuity Theorem (see e.g. Algebraic Geometry by R. Hartshorne, pp 281), but don't know how it can be applied in this situation. Any idea how to solve it? Can I make a reduction argument to reduce it to a setting where semicontinuity does the job?

Let $X$ be a smooth projective irreducible algebraic curve over field $k$. For $d,r,k,m >0$ the representable Quot scheme $\mathcal {Quot}_X^{r,d}(\mathcal{O}_X(m)^k)$ is given for any test scheme $U \to X$ by

$${\displaystyle {\mathcal {Quot}}_{X}^{r,d}(\mathcal{O}_X(m)^k)(U)=\left\{({\mathcal {F}},q):{\begin{matrix}{\mathcal {F}}\in {\text{Coh}}(X \times U)\\{\text{Supp}}({\mathcal {F}}){\text{ is proper over }}X \times U\\{\mathcal {F}}{\text{ is flat over }}X \times U\\q:(\mathcal{O}_{X\times U}(m))^k\to {\mathcal {F}}{\text{ surjective}} \\ {\text{ rank }}({\mathcal {F}})= r {\text{ and }} {\text{ deg }}({\mathcal {F}})= d \end{matrix}}\right\}/\sim }$$

Now I want to define an subfunctor ${\mathcal {R}}_{X}^{m} \subset \mathcal {Quot}_X^{r,d}(\mathcal{O}_X(m)^k)$ by taking restriction:

$${\displaystyle {\mathcal {R}}_{X}^{m}(U)=\left\{({\mathcal {F}},q):{\begin{matrix}{\mathcal {F}}\in {\mathcal {Quot}}_{X}^{r,d}(\mathcal{O}_X(m)^k)(U) \\ {\mathcal {F}} {\text{ locally free and }} \\ H^1(X \times U, \mathcal {F}(m))=0 \end{matrix}}\right\}/\sim }$$

First obvious problem: is the thing ${\mathcal {R}}_{X}^{m}$ well defined, that is what about the compatibility with pullbacks of the last condition on $H^1$ group? i.e. I want to assure that if $U \to X$ is an arbitrary test scheme and the quotient $\mathcal{F} \in {\mathcal {R}}_{X}^{m}(U)$ then for every $X$-morphism $f:S \to U$ the pullback $\hat{f}^*\mathcal{F}$ is an element in ${\mathcal {R}}_{X}^{m}(S)$, where $\hat{f}: X \times S \to X \times U$.

The crucial obstacle for me is to verify $H^1(X \times S, \hat{f}\mathcal {F}(m))=0$. I conjecture that here I can somehow use one of many corollaries of the Semicontinuity Theorem (see e.g. Algebraic Geometry by R. Hartshorne, pp 281), but don't know how it can be applied in this situation. Any idea how to solve it? Can I make a reduction argument to reduce it to a setting where semicontinuity does the job?